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https://chemistry.stackexchange.com/questions/89575/why-is-tautomers-not-possible-in-benzoquinone
|
Why is tautomers not possible in benzoquinone? [closed]
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[](https://i.stack.imgur.com/b7tH7.png)1,4-Benzoquinone, commonly known as para-quinone, is a chemical compound with the formula $\ce{C6H4O2}$
When there are alpha Hydrogen's why can't it show tautomerism?
| 0 |
[
[
"\nYou could sort of devise a tautomer by transferring an $\\alpha$ hydrogen from carbon to oxygen. But if you draw the resulting structure you find a pair of cumulated double bonds. That tends to be unstable, and since the cumulated double-bonded carbons favor a 180° bond angle putting them into a relatively small ring makes them a lot more unstable. Your proposed tautomer is way out there in energy.\n\n\nTautomerization of a ketone works better when you have an $\\alpha$ hydrogen **on a saturated carbon atom**.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/89568/does-acetate-cause-major-e2-on-tertiary-haloalkanes
|
Does Acetate Cause Major E2 On Tertiary Haloalkanes?
|
This question really confuses me. I've narrowed it down to E2 being favored in step one (giving 2-methylpropene), but I don't see how a relatively weak base like acetate could cause E2 to be the major product. The only other explanation I could think of is that E2 is a substantial minority product, so the reaction isn't synthetically useful. Am I on the right track? Thanks!
[](https://i.stack.imgur.com/9qxec.png)
| 1 |
[] |
https://chemistry.stackexchange.com/questions/89567/why-does-buffering-capacity-increase-with-the-pk%e2%82%90-of-the-acid-component
|
Why does buffering capacity increase with the pKₐ of the acid component?
|
For example, the $\mathrm{p}K\_\mathrm{a}$ values of ethanoic, propanoic, and butanoic acid are related in the following way:
$$\text{ethanoic} < \text{butanoic} < \text{propanoic}$$
And their buffering capacities are related in exactly the same way. What causes this?
| 2 |
[
[
"\nBuffer capacity depends on the $K\\_\\mathrm{a}$ of an acid. I will use the explanation that can be found in David Sheehan: Physical biochemistry: principles and applications. John Wiley & Sons, 2013.\n\n\nAny aqueous solution containing both $\\ce{A-}$ and $\\ce{AH}$ is, in principle, capable of resisting change in $\\mathrm{pH}$ according to \n$$\\ce {A^- + H^+ <=>> AH}.$$ \nIf alkali is generated in the solution (that would tend to remove protons) we have \n$$\\ce{AH + OH- <=> A- + H\\_2O}.$$ \n\n\nMost buffers consist of mixtures either of a weak acid and its salt or of a weak base and its salt. The ability to resist change in $\\mathrm{pH}$ is finite, especially if the number of protons involve is especially large. This limit is represented by the *buffering capacity of the buffer* $\\beta$. \nThis is defined as the amount of substance in moles of $\\ce{[H^+]}$ which must be added to a liter of the buffer to decrease the $\\mathrm{pH}$ by one unit. It can be mathematically calculated by \n$$\\beta = \\frac{2.3 K\\_\\mathrm{a}[\\ce{H+}][C]}{(K\\_\\mathrm{a}+[\\ce{H+}])^2},$$ \nwhere $[C]$ is the sum of conentrations of $\\ce{A-}$ amd $\\ce{AH}$. This relationship means that buffering capacity increases with the buffer concentration. Buffers work best at $\\mathrm{pH}$ values around their $\\mathrm{p}K\\_\\mathrm{a}$ most pH are effective one $\\mathrm{pH}$ unit above an one below their $\\mathrm{p}K\\_\\mathrm{a}$. \n\nI hope that the formula previously reported help in clarifying how the buffer capacity is related to $\\mathrm{p}K\\_\\mathrm{a}$ (since they are indeed related).\n\n\n**Example** (based on ZUMDAHL, Steven. World of chemistry. Cengage Learning, 2012.) \n\n\nA chemist needs a solution buffered at $\\mathrm{pH}~4.30$. choosing from ethanoic, propanoic, butanoic acid. We can calculate the ratio $\\ce{[HA]/[A^-]}$ required for each system. Considering that a $\\mathrm{pH}~4.30$ corresponds to $\\ce 5\\times 10^{-5} M$ using the equation \n$$[\\ce{H+}]=K\\_a \\frac{[\\ce{HA}]}{[\\ce{A^-}]}$$ \nwe substitute the required $[\\ce{H+}]$ and $\\ce{K\\_a}$ for each acid to calculate the ration $[\\ce{H+}]/[\\ce{A-}]$ needed in each case is: \n\n\n\\begin{array}{llr}\n\\text{Acid} & \n[\\ce{H+}] = K\\_\\mathrm{a} \\frac{[\\ce{HA}]}{[\\ce{A^-}]} & \n\\frac{[\\ce{HA}]}{[\\ce{A^-}]} \\\\ \\hline\n\\text{Ethanoic} & \n5.0 \\times 10^{-5} = 10^{-4.76} \\frac{[\\ce{HA}]}{[\\ce{A^-}]} & \n2.88 \\\\[1ex]\n\\text{Propanoic} &\n5.0 \\times 10^{-5} = 10^{-4.87} \\frac{[\\ce{HA}]}{[\\ce{A^-}]} &\n3.749\\\\[1ex]\n\\text{Butanoic} &\n5.0 \\times 10^{-5} = 10^{-4.9} \\frac{[\\ce{HA}]}{[\\ce{A^-}]} &\n3.98 \\\\ \\hline\n\\end{array}\n\n\nIn this case the choice would be ethanoic acid since it has the ratio closes to 1. \n\n\nIf we need a buffer near $\\mathrm{pH}~5.5$ the best choice is butanoic acid:\n\n\n\\begin{array}{llr}\n\\text{Acid} & \n[\\ce{H+}] = K\\_\\mathrm{a} \\frac{[\\ce{HA}]}{[\\ce{A^-}]} & \n\\frac{[\\ce{HA}]}{[\\ce{A^-}]} \\\\ \\hline\n\\text{Ethanoic} & \n3.2 \\times 10^{-6} = 10^{-4.76} \\frac{[\\ce{HA}]}{[\\ce{A^-}]} &\n0.18 \\\\[1ex]\n\\text{Propanoic} &\n3.2 \\times 10^{-6} = 10^{-4.87} \\frac{[\\ce{HA}]}{[\\ce{A^-}]} &\n0.23 \\\\[1ex]\n\\text{Butanoic} &\n3.2 \\times 10^{-6} = 10^{-4.9} \\frac{[\\ce{HA}]}{[\\ce{A^-}]} &\n0.25 \\\\ \\hline\n\\end{array}\n\n\n",
"2"
],
[
"\nTaking \"buffering capacity\" to mean the amount of acid or base required to change $\\mathrm{pH}$ by a fixed small amount at $\\mathrm{pH}$ values near one of the $\\mathrm pK$'s of the acid on which the buffer is based buffering capacity is not $\\mathrm pK$ dependent. To see this write Henderson–Hasselbalch for each proton in the form $\\mathrm pr\\_1=\\mathrm pK\\_1-\\mathrm{pH}$, $\\mathrm pr\\_2=\\mathrm pK\\_2-\\mathrm{pH}\\ldots$ where $\\mathrm p(\\cdot)=-\\log(\\cdot)$. Taking the antilog we see $r\\_j = 10^{(\\mathrm{pH}-\\mathrm pK\\_j)}$ Note that $r$ is dependent on the *difference* between $\\mathrm{pH}$ and $\\mathrm pK$. Thus we find that $r$ is the same one $\\mathrm{pH}$ unit either side of $\\mathrm pK$ irrespective of whether $\\mathrm pK\\_\\mathrm a$ be 1 or 7.3. This is your first clue that buffering is $\\mathrm pK$-independent.\n\n\nAssuming you have started with $C$ moles of acid the fraction of those that remain undissociated and uncharged at $\\mathrm{pH}$ is $f\\_0(\\mathrm{pH})=1/(1+r\\_1+r\\_1r\\_2+r\\_1r\\_2r\\_3\\ldots)$, the fraction that has lost one proton and thus carries a single negative charge is $f\\_1=f\\_0r\\_1$, the number that is doubly charged is $f\\_2=f\\_1r\\_2$ and so on so that the total charge on acid anions is $$Q(\\mathrm{pH})=C(-f\\_1-2f\\_2-3f\\_3...)$$ (moles of charges or Eq if you prefer). This is clearly a function of $\\mathrm{pH}$ (shown) as well as each of the $\\mathrm pK$'s (not shown). The number of protons required to shift the buffer's $\\mathrm{pH}$ from $\\mathrm{pH}$ to $\\mathrm{pH}+\\delta$ is $\\Delta Q=Q(\\mathrm{pH}+\\delta)-Q(\\mathrm{pH})$. This is the buffering (alkalinity, acidity) between $pH$ and $pH+\\delta$. The buffering capacity is the slope of the buffering curve. At $pH$ it is:\n$$\\partial Q(\\mathrm{pH})/\\partial\\mathrm{pH}=\\lim\\_{\\delta\\to0} (Q(\\mathrm{pH}+\\delta))-Q(\\mathrm{pH})/\\delta\\approx(Q(\\mathrm{pH}+\\delta)-Q(\\mathrm{pH}))/\\delta$$\nin units of $\\mathrm{mol}\\cdot\\mathrm{pH}^{-1}$. It is the number of moles of protons (1 M strong acid delivers 1 mole of protons per liter) which must be added or absorbed per unit $\\mathrm{pH}$ shift.\n\n\nNow these formulas aren't very complex and if you will take the trouble to put them into a spreadsheet or visualization program and make some plots you will see that the buffering capacity of a buffer near a $\\mathrm pK$ depends only on $C$ as long as the $\\mathrm pK$'s are separated.\n\n\nI have been, for simplicity, ignoring the buffering capacity of water here. When $3\\lt\\mathrm{pH}\\lt10$ water does exhibit appreciable buffering and needs to be considered. The net charge on a liter of water ions is $Q\\_\\mathrm w=10^{-\\mathrm{pH}}-10^{\\mathrm{pH}-\\mathrm pK\\_\\mathrm w}$ and the buffering of water, between $\\mathrm{pH}$ and $\\mathrm{pH}+\\delta$, is $$\\Delta Q\\_\\mathrm w=10^{-(\\mathrm{pH}+\\delta)}-10^{-\\mathrm{pH}}+(10^{\\mathrm{pH}-\\mathrm pK\\_\\mathrm w}-10^{\\mathrm{pH}+\\delta-\\mathrm pK\\_\\mathrm w})$$ $Q\\_\\mathrm w$ is added to $Q$ as given above when differentiating to find buffer capacity. Keep in mind that $Q\\_\\mathrm w$ is per liter and $Q$ per mole.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89566/uv-vis-and-polylactid-acid-pla
|
UV-Vis and Polylactid Acid (PLA) [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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I am trying to use UV-VIS for a 3D printed ~4mm PLA design. However I keep running into a problem because the polymer is not showing detectable under the UV-VIS. Does anyone have any suggestions that would allow for visibility of PLA? Any recommended papers would be of great use as well.
I also thought of possibly placing spider silk inside but this would implement an array of complexity in the product.
Thanks guys
| 1 |
[] |
https://chemistry.stackexchange.com/questions/89564/are-artificial-ingredients-better-than-natural-ones-because-they-are-made-in-lab
|
Are artificial ingredients better than natural ones because they are made in laboratory conditions?
|
If we have something like an artificial vanilla flavouring, is it any worse than a natural one?
Will both the natural and artificial ones have the same chemical formula?
Won't the artificial one be purer since its made in lab conditions?
| 1 |
[
[
"\nThe question is by no means restricted to vanilla. It refers to any natural product that can be reproduced artificially. Ditto for the answer.\n\n\nSure, the artificial product might have the right chemical formula. Also, it can be made as pure as you want. **The problem is that the natural product is *not* pure.** It is a bunch of various compounds, and they all contribute to our perception of the flavour. You identify the compound that makes up 90%, investigate its structure, develop the synthetic route, and finally you hold it in your hands, and it feels just about right, but something is amiss. So you look for the next component, which constitutes 9% and hence is a lot harder to isolate, and after a long and tedious procedure you identify it, and repeat everything, and finally you end up with the product that feels totally right. Well, *almost* so. Is it the remaining 1% that is responsible for the difference? You never know until you try.\n\n\nSo it goes.\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/89562/cell-notation-for-the-lead-acid-battery
|
Cell notation for the lead-acid battery
|
Reactions for the lead acid battery are:
$$
\begin{array}{}
\text{Oxidation}&\ce {Pb(s) + HSO4^-(l) &-> PbSO4(s) + H+(l) + 2e-}\\
\text{Reduction}&\ce{PbO2 + HSO4^-(l) + 3H+(l) + 2e- &-> PbSO4(s) + 2H2O}\\
\text{Total reaction}&\ce{Pb(s) + PbO2(s) +2HSO4^- +2H+ &-> PbSO4 + 2H2O}\\
\end{array}
$$
What will be the cell notation for this battery? My attempt:$$\ce{Pb(s), Pb^2+(s)| HSO4^-(l)| PbO2(s), Pb^2+(s),Pb(s)}$$
The things that I have in mind,
1. In both sides the electrode material is $\ce{Pb(s)}$
2. Separate every element that is in the same phase with a comma
3. Separate every element with different phases with a single bar
4. Separate the two half reactions with a single bar (as I find no salt bridge here)
I am having a tough time learning cell notation. Anyways this was a challenging one for me and I tried my best shot. I am really not sure whether everything is right or not. So please identify my mistakes and show me the right way to do it.
| 2 |
[
[
"\nThere are a couple of things wrong here. First off, your final reaction is **unbalanced**. Once you've fixed the balancing, read the other mistakes:\n\n\n1. The ions do not exist in the **liquid** state! They are solvated/hydrated by the solvent. Since the solvent is water here, we'll say that the ions are in the **aqueous** **(aq)** phase instead.\n2. While there is certainly no salt bridge here, there is still an **electrolyte** - an aqueous solution of sulphuric acid. Hence, you must mention it in your cell notation, between the anode and the cathode.\n3. The $\\ce{PbSO4}$ formed at the anode is in **solid state**. Hence, writing it as $\\ce{Pb^2+(s)}$ is incorrect, as it is *not* dissociated into the ions $\\ce{Pb^2+}$ and $\\ce{SO4^2-}$.\n\n\nWith all these corrections, your final, correct cell representation should be:\n\n\n\n> \n> $$\\small{\\ce{Pb(s), HSO4^-(aq) | PbSO4(s), H+ | H2SO4 ($\\pu{x~M}$) | PbO2(s), HSO4-(aq), H+(aq) | PbSO4(s)}}$$\n> \n> \n> \n\n\n\n\n---\n\n\nSome websites (like [KhanAcademy](https://youtu.be/PQ48N5jaG2w)) and texts (NCERT 12), cite the lead-acid battery reaction as this instead:\n\n\n$$\n\\begin{array}{}\n\\text{Oxidation}&\\ce {Pb(s) + SO4^2-(aq) &-> PbSO4(s) + 2e-}\\\\\n\\text{Reduction}&\\ce{PbO2 + SO4^2-(aq) + 4H+(aq) + 2e- &-> PbSO4(s) + 2H2O(l)}\\\\\n\\text{Total reaction}&\\ce{Pb(s) + PbO2(s) +2SO4^2(aq)- + 4H+(aq) &-> 2PbSO4(s) + 2H2O(l)}\\\\\n\\end{array}\n$$ \n\n\nassuming the bisulphite ion to be further ionized. The cell notation in this case would then be:\n\n\n\n> \n> $$\\small{\\ce{Pb(s), SO4^2-(aq) | PbSO4(s) | H2SO4 ($\\pu{x~M}$) | PbO2(s), SO4^2-(aq), H+(aq) | PbSO4(s)}}$$\n> \n> \n> \n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89557/what-does-po2-of-blood-mean-and-why-do-we-use-it
|
What does pO2 of blood mean and why do we use it?
|
I understand the basic Dalton's law of partial pressures in gases. Also, Henry's law of diffusion, says, the concentration of gas dissolved in a fluid is proportional to the partial pressure above it.
So if we say that the $p(\ce{O2})$ of oxygenated blood is $\pu{100 mmHg}$, where is the free gas existing in equilibrium with dissolved gas? Does it mean that the blood has a concentration of oxygen equal to that when placed in a surrounding of $p(\ce{O2}) = \pu{100 mmHg}$? If yes, why don't we directly report in concentrations instead? Is it easier to measure?
Wikipedia also says that the Henry's law doesn't stand if the gas is reacting. But isn't oxygen reacting with the Haemoglobin?
| 5 |
[
[
"\nThere is a good explanation in [Relating oxygen partial pressure,\nsaturation and content: the\nhaemoglobin–oxygen dissociation\ncurve](https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4666443/pdf/EDU-0014-2015.pdf) *Breathe* 2015; 11: 194–201\n\n\n\n> \n> The partial pressure of oxygen (also known\n> as the oxygen tension) is a concept which often\n> causes confusion. In a mixture of gases, the\n> total pressure is the sum of the contributions\n> of each constituent, with the partial pressure of\n> each individual gas representing the pressure\n> which that gas would exert if it alone occupied\n> the volume. In a liquid (such as blood), the partial\n> pressure of a gas is equivalent to the partial\n> pressure which would prevail in a gas phase in\n> equilibrium with the liquid at the same temperature.\n> With a mixture\n> of gases in either the gas or\n> liquid phase, the rate of diffusion of an individual\n> gas is determined by the relevant gradient of its\n> partial pressure, rather than by its concentration.\n> While in a gas mixture, the partial pressure\n> and concentration of each gas are directly proportional,\n> with oxygen in blood the relationship\n> is more complex because of its chemical combination\n> with haemoglobin. This allows blood\n> to carry an enormously greater concentration\n> (content) of oxygen than, for example, water\n> (or blood plasma). Measurement of $p\\_\\ce{O\\_2}$, therefore,\n> does not give direct information about the\n> amount of oxygen carried by blood. \n> \n> \n> \n\n\nSo blood $p\\_\\ce{O\\_2}$ does not correspond to a particular concentration of oxygen, because the concentration of haemoglobin can vary, and most of the oxygen is bound to the heme iron. \n\n\n$P\\_\\ce{O\\_2}$ is the partial pressure of oxygen in a hypothetical gas phase which would make the blood oxygen and gas phase oxygen be in equilibrium. \n\n\n",
"4"
],
[
"\nI believe medical oximeters read % saturation relative to the partial pressure of oxygen in the atmosphere. Thus a reading of 94 - 95% (normal) implies that the amount of oxygen in the blood is 94 - 95% of that which would be found were the blood in equilibrium with air. I think the reason it's done that way is for convenience. If the oximeter reads 50% the gas passer knows right away to increase the partial pressure of oxygen he is feeding the patient by 45% to get saturation back to around 95%.\n\n\n",
"-4"
]
] |
https://chemistry.stackexchange.com/questions/89550/why-mn-atom-show-maximum-valency-of-4-with-f-atom
|
Why Mn atom show maximum valency of 4 with F atom?
|
Mn ($3d^5$, $4s^2$) can have maximum 7 unpaired electrons in excited state so it should have formed MnF7 molecule but it can form only MnF4 molecule.
Is it because:
$d\_{xy}, d\_{yz}, d\_{zx}$ orbitals are not axially oriented and therefore not suitable for head on sigma overlap but can form pi bonds as in Mn2O7 molecule.
| 2 |
[] |
https://chemistry.stackexchange.com/questions/89549/entropy-is-a-measure-of-unavailable-energy
|
Entropy is a measure of unavailable energy?
|
I came across a statement in my text book which said that entropy is a measure of unavailable energy.
What does this statement signify?
| 3 |
[
[
"\nWhen the first law was first formulated it was assumed that the most efficient utilisation of a chemical reaction to produce work was to use all the heat to produce work. The second law was not fully understood and thus at maximum efficiency $-\\Delta H$ was expected to represent the maximum amount of work. Many experiments failed to confirm this. We now know that it is the free energy which measures the maximum capacity to do work, $\\Delta G=\\Delta H-T\\Delta S$. Clearly $\\Delta G$ and $\\Delta H$ are equal only if there is no entropy change between the beginning and end of an isothermal reaction. If this is not the case the work obtainable in a reversible process can be either greater than or less than the heat of reaction.\n\n\nFrom the first law the external work performed must be equal to the loss in energy of the system, unless some heat is taken from or given to the surroundings. This is exactly the point first clearly seen by Gibbs. In a reversible isothermal reaction $T\\Delta S$ is the heat absorbed from the surroundings and if this is positive the work done will be greater than the heat of reaction.\n\n\nSo 'available/unavailable energy' seems to depend on the particular case. Hopefully some thermo experts will give you a more detailed argument.\n\n\nedit: It occurs to me that perhaps the textbook was referring to the fact that one can 'convert *all* the work into heat but not all the heat into work'. In other words not all the random thermal motions of molecules can be made to do work.\n\n\n",
"4"
],
[
"\nAnother way of seeing this issue is to write the energy conservation law two ways:\n$$dU = \\delta Q - \\delta W= TdS - \\sum\\_{k}Y\\_kdX\\_k $$\nand this is combined with Clausius's inequality\n$$\\delta Q \\le TdS \\\\ \\delta W \\le \\sum\\_{k}Y\\_kdX\\_k$$\nHere $T$ is the temperature at which $\\delta Q$ heat transferred *to* and $\\delta W$ is the work done *by* the system whose internal energy and entropy have thereby changed by $dU$ and $dS$. The $Y\\_k$ and $X\\_k$ are the internal intensive and extensive system parameters.\n\n\nEvidently less work is done by the system for the same amount of change $dU$, $dS$ and $dX\\_k$ when the process is irreversible (strict inequality) than when it is reversible (equality). \n\n\nBecause of the Clausius inequality $0 \\le \\delta \\mathcal{N} = TdS- \\delta Q$ and this, of course, is the same as $\\sum\\_{k}Y\\_kdX\\_k - \\delta \\mathcal{N} = \\delta W$. The quantity $\\delta \\mathcal{N} = T\\delta \\sigma $ with $\\delta \\sigma \\ge 0$ represent the amount of internally generated \"heat\" and entropy, resp., by the irreversible process. Because of the internally produced entropy $\\delta \\sigma $ the total entropy change $dS$ will be greater by this amount than in a reversible process connecting the same states, and the resulting $\\delta \\mathcal {N}$ is then the amount of deliverable work that is \"lost\" by the irreversibility.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89548/how-does-the-mathematical-definition-of-entropy-%ce%94s-qrev-t-give-us-the-degree-o
|
How does the mathematical definition of entropy (ΔS=Qrev/T) give us the degree of disorder? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed 5 years ago.
[Improve this question](/posts/89548/edit)
I think degree of disorder in thermodynamics means the number of possible locations that a molecule can take
How does the mathematical equation signify this?
| -1 |
[
[
"\nDisorder is perhaps not the best term to use in describing what entropy measures. It is perhaps better described as a measure of what we don't know about a system. If the system is likely to be in one of a handful of states then the entropy is low and we have a pretty good idea as to what state, or which of the few states, the system is likely to be. If, OTOH, it can be in any of a multitude of states with roughly equal probability the entropy is high and we have less likelihood of being able to deduce in which one of those states it might be. Consider a stick of dynamite. If we ask, before detonation, \"Where are the carbon atoms in the nitroglycerine?\" we can confidently answer \"They are all in that little cardboard tube\". After the explosion our answer has to be \"They are all over the place - maybe as far away as the next county by now.\" Entropy is related to the dispersion of available states. Distance of a carbon atom from the original location of the dynamite stick is a state variable but clearly not the only one. Nonetheless, it helps me, at least, to understand the concept. Originally the distances of carbon atoms from the center of the stick are in a tight distribution. For purposes of illustration lets assume that this is Gaussian with standard deviation 5 cm. After the explosion it is reasonable to suppose, continuing to assume a Gaussian distribution, that the standard deviation is now 10's of meters. The reason for picking the Gaussian for this ilustration is that the entropy of a Gaussian distribution varies directly as the standard deviation. This falls out of the definition of entropy: $S = -k\\_BE{log p\\_i} =-k\\_B\\sum\\limits\\_{i}^{N}p\\_ilog p\\_i$ where there are $N$ possible states numbered with subscript $i$. The probability the system is in state $i$ is $p\\_i$. $k\\_B$ is the Bolzman constant. $E$ is the expectation operator. If we have, for example, 4 possible states and one of them is much more likely to be occupied than the others then we might have $$ -(.01\\*log(.01) + .01\\*log(.01) + .01\\*log(.01) + .97\\*log(.97) ) = 0.0728314$$ If, on the other hand, each of the 4 was equally likely to be occupied then we'd have $$ - (.25\\*log(.25) + .25\\*log(.25) + .25\\*log(.25) + .25\\*log(.25) ) = 0.60206$$\n\n\nWe can no longer be assured that that one state is the one we would expect the system to be in. We know less about the system. Uncertainty has gone up. Entropy has increased.\n\n\nI'm editing this to add the important fact that if all the possible states are equally likely then $$-\\sum\\limits\\_{i}^{N}p\\_ilog p\\_i = -\\sum\\limits\\_{i}^{N}(1/N)log (1/N) = -log(1/N) = log(N)$$ \n\n\nThia gives the definition often found in the textbooks $S= k\\_BlogN$ usually written as $$S = k\\_Blog\\Omega$$\n\n\nin which thhey use $\\Omega$ instead of $N$ to represent the number of available states. The important thing to realize when confronted with this formula is that the states have to be equally probable. In the Shannon based definition they don't.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89544/dilution-calculations
|
Dilution calculations
|
>
> $\pu{5 mL}$ of $\pu{0.012 M}\,\ce{Pb(NO3)2}$ is mixed with $\pu{5 mL}$ of $\pu{0.030 M}\,\ce{KI}$. $\ce{PbI2}$ forms. Calculate the diluted concentrations of $\ce{Pb^{2+}}$ and $\ce{I-}$ ions.
>
>
>
I couldn't figure out whether it is an equilibrium question, or something else entirely.
I have currently tried using the moles of the limiting reactant (Pb(NO3)2) to obtain the moles of PbI2 formed and using the molar ratio to get concentration of each ion. However I have no idea whether I have done the correct calculation for ion concentration. The values I am getting are 0.003M Pb2+ and 0.012M I-
update: I have looked at the questions that follow and am able to say for certain this is not an equilibrium question.
| 0 |
[
[
"\nI haven't checked your numbers but you're assuming the $\\ce{PbI2}$ to be perfectly insoluble. That's a reasonable assumption but in reality its solubility product is $K\\_{sp}=4.41\\times 10^{-9}$ (Wikipedia), so it does have *some* solubility.\n\n\nTo take this into account it becomes an equilibrium problem. Assume the initial concentrations (just after mixing) to be $\\ce{C(Pb^2+)}$ and $\\ce{C(I^-)}$.\n\n\nAfter precipitation the lead concentration has been reduced by$X$ and the iodide concentration by $2X$.\n\n\nWith the solubility product *after precipation*:\n\n\n$$K\\_{sp}=[\\ce{Pb^2+}]\\times [\\ce{I^-}]^2$$\n\n\nSo we have:\n\n\n$$K\\_{sp}=\\Big(\\ce{C(Pb^2+)}-X\\Big)\\times \\Big(\\ce{C(I^-)}-2X\\Big)^2$$\n\n\nNote that this is a third degree polynomial equation in $X$. Solve for $X$ to find the real equilibrium concentrations $[\\ce{Pb^2+}]$ and $[\\ce{I^-}]$.\n\n\nWe can also use the equation for the case where $K\\_{sp}=0$. Then:\n\n\n$$0=\\Big(\\ce{C(Pb^2+)}-X\\Big)\\times \\Big(\\ce{C(I^-)}-2X\\Big)^2$$\n\n\n**First case:** $\\ce{C(Pb^2+)}-X=0 \\implies \\ce{C(Pb^2+)}=X \\implies [\\ce{Pb^2+}]=0$\n\n\nAlso: $[\\ce{I^-}]=\\ce{C(I^-)}-2\\times\\ce{C(Pb^2+)}$\n\n\nIn the **second case** we assume $\\Big(\\ce{C(I^-)}-2X\\Big)^2=0$ and proceed as above. \n\n\n*Which case applies depends on which reagent is the limiting one.*\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89532/how-strong-is-acetic-acid
|
HOW STRONG IS ACETIC ACID? [closed]
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**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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Will **acetic acid** cause irritation on our skin?
Does it have any specific smell? what is it's **pH** range? Is it too strong to cause irritation in our nasal passage?
| -5 |
[
[
"\nIt totally depends on the concentration of acetic acid you use. We have dilute acetic acid in our kitchens (vinegar) so low concentration (Approx 3-9% v/v) is okay. The smell is acidic but is bearable. The pH of vinegar is around 2.4. On the opposite scale glacial acetic acid (99%), causes severe skin burns and is flammable. It has a very strong acidic smell and will most probably burn your nasal passage or severely irritate it. Hope this helps. \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89528/formatting-a-chemical-equation
|
Formatting a chemical equation [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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I recently came across this chemical equation :
$$\ce{2HNO3 + Na2Co3 -> 2NaNO3 + CO2 + H2O}$$
The equation describes the reaction between Nitric Acid and Sodium Carbonate, the equation states that the resulting compounds $\ce{2NaNO3}$ carbon dioxide and water would be created. However my question is how the formula should be written out. I have always written a chemical equation like this :
$$\ce{(HNO3)2 + Na2Co3 -> (NaNO3)2 + CO2 + H2O}$$
Do both ways of writing the equation state the same thing? If so which one should I use. Which form of writing the equation is considered "more correct", used more commonly or considered "more standardized".
| -5 |
[
[
"\nAccording to the IUPAC guidelines arabic numerals are used:\n\n\n\n> \n> (a) As right subscripts, to indicate the number of individual constituents (atoms or groups of atoms)\n> \n> \n> (c) To indicate the composition of (formal) addition compounds or non-stoichiometric compounds. The numeral is written on the line before the formula of each constituent.\n> \n> \n> \n\n\nSource: IUPAC recommendations found\n[here](http://www.sbcs.qmul.ac.uk/iupac/bibliog/inorg.html)\n\n\nThis essentially means that the first variant is correct.\n\n\n",
"2"
]
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https://chemistry.stackexchange.com/questions/89522/how-long-does-it-take-to-saturate-h2o-with-o3
|
How long does it take to saturate H2O with O3 [closed]
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Let’s assume I have one liter of distilled $\ce{H2O}$ in an stainless steel container – from the bottom side of the container I do bubble ozone trough the $\ce{H2O}$.
Of cause it is not pure O3, so let’s assume that the gas consists of:
7% $\ce{O3}$
83% $\ce{ O2}$
10%. $\ce{ N2 / NO2 / NO}$
The gas flow is about 5 liters per minute and the corona discharge consumes about 500W.
How long will it approximately take, till full saturation of the
$\ce{H2O}$ with $\ce{O3}$ will take place?
| 0 |
[
[
"\nThe solubility of ozone in water as per one website is:\n\n\n[](https://i.stack.imgur.com/h4AgR.jpg)\n\n\nFirst thing is to establish which curve you need. In 1 m$^3$ gas you have 70 L ozone:\n\n\n$\\frac{70\\ L\\ \\ce{O3}}{1\\ m^3}\\frac{1\\ mol}{0.082\\cdot 298\\ L}\\frac{16\\cdot 3\\ g\\ \\ce{O3}}{1\\ mol}= 138\\ g\\ \\ce{O3}/m^3$\n\n\nLet us take the curve of 150 g/m$^3$ which has an ozone solubility of 30 mg/L water. \nThen:\n\n\n$\\frac{30\\cdot 10^{-3}\\ g\\ \\ce{O3}}{1\\ L\\ \\ce{H2O}}\\frac{1\\ mol}{16\\cdot 3\\ g\\ \\ce{O3}}\\frac{0.082\\cdot 298\\ L}{1\\ mol}\\frac{1\\ min}{5\\cdot 0.07\\ L\\ \\ce{O3}}= 0.044\\ min$\n\n\nWhich is about 3 seconds. \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89510/single-crystalline-thin-films-is-detect-by-gixrd
|
Single Crystalline Thin Films is detect by GIXRD? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
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I understand epitaxal thin film, have lattice match with substrate.
But what about those films that doesnt match but have only one XRD peak? Is this proof of single crystale thin film?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/89507/how-are-such-ionic-equilibrium-equations-derived-h3o-concentration
|
How are such ionic equilibrium equations derived? (H3O+ concentration) [closed]
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I missed my classes on ionic equilibrium & I came to know that certain equations were derived for cases such as 'Monobasic acid', 'Two monobasic acids', 'Dibasic acid'etc.
These equations relate H3O+ concentration with the acid's initial concentration, its dissociation constant & water's dissociation constant.
These equations were told to me that, they provide highly accurate answers, instead of the commonly used Henderson–Hasselbalch equation.
These equations are of either cubic or bi-quadratic form, taking the H3O+ concentration as the variable.
Can someone explain this? Unfortunately; I couldn't find any sources on the internet (or perhaps I didn't use the proper keywords?)
If anyone is able to drop by a link to a webpage that explains these derivations and the use of these equations, it would be greatly helpful. Thank-you.
| -2 |
[
[
"\nBefore anything else I must clarify that there is nothing wrong with Henderson–Hasselbalch equation. It is formally exact because it comes from the definition of the $K\\_a$ as long as you ignore ionic strength which I believe is not what you are interested in at least at this stage. If you say there's anything inaccurate about it it is because people are making assumptions when applying the equation.\n\n\nNow for the main matter. If you take a look at [this note](http://www.cameron.edu/~keithv/theory/AC_Acids-Bases.pdf) (not what you are looking for I am afraid, but we'll come to that in a moment), near the end of page 4 we have the quadratic form of the equation\n\n\n$$K\\_a=\\frac{[\\ce{H3O+}]^2}{c(\\ce{HA})-\\ce{H3O+}}$$\n\n\nbased on approximation that $[\\ce{H3O+}]\\approx[\\ce{A-}]$. Now, what you are looking for is the formula without making this assumption. So first examine what's wrong about this assumption.\n\n\nMass balance for $\\ce A$ states,\n$$c(\\ce{HA})=[\\ce{HA}]+[\\ce{A-}],$$\n\n\ncharge balance states,\n$$[\\ce{H3O+}]=[\\ce{A-}]+[\\ce{OH-}],$$\n\n\nyou see that the approximation ignores the second term in RHS,\nand we do know,\n$$[\\ce{H3O+}][\\ce{OH-}]=K\\_{\\rm w},$$\n\n\nNow it is not difficult to see what we need to put into the the definition of $K\\_a$.\n$$K\\_a=\\frac{[\\ce{H3O+}][\\ce{A-}]}{[\\ce{HA}]}=\\frac{[\\ce{H3O+}]\\left([\\ce{H3O+}]-\\frac{K\\_{\\rm w}}{[\\ce{H3O+}]}\\right)}{c(\\ce{HA})-[\\ce{H3O+}]+\\frac{K\\_{\\rm w}}{[\\ce{H3O+}]}}$$\n\n\nThis is the cubic equation you are looking for, it is not that difficult to derive but quite a pain to solve numerically if you don't have a calculator that solves equation for you (there are various numerical techniques but that's beyond the scope of this answer). The reason that it is not explicitly included in many books is that you don't need to go for such complex form under most occasions and, more importantly, these forms do vary a lot if, say I am talking about $[\\ce{H2A}]$ instead of $[\\ce{HA}]$.\n\n\nSo the bottom line: instead of pursuing these complex forms (which may seem interesting, but you just can't remember all of them), **stick to the charge balance equation and mass balance equation and optionally, if you have been taught, the proton balance equation**, and also the **definitions of $K\\_a$'s**. With these in mind you can derive the most accurate form of high-order equations for $[\\ce{H3O+}]$ whenever you want to, and these balance are not that hard to grisp compared to these exotic high-order equations.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89506/angular-momentum-question-on-zares-use-of-the-vector-model-to-estimate-probab
|
Angular momentum - Question on Zare's use of the Vector Model to Estimate Probability
|
My question regards a specific example, although I get the feeling that the answer is simpler than I imagine. In Zare's *Angular Momentum: Understanding Spatial Aspects in Chemistry and Physics pgs 53-54*, He discusses the vector model for coupled and uncoupled representations for the addition of two angular momenta.
I've scanned and included the picture I am going to refer to for clarity:
[](https://i.stack.imgur.com/JqvEJ.png)
The image above is a modified representation of the ***uncoupled representation***, although it appears very similar to the coupled diagram, the difference is that **j1** has been arbitrarily fixed, and **j2** is precessing in the dashed circle, around an axis parallel to z.
Zare expands the dot product of $\mathbf{j\_1}$ and $\mathbf{j\_2}$
(Eq 2.34):
$\mathbf{j\_1 . j\_2} = |\mathbf{j\_1}||\mathbf{j\_2}| cos(\gamma)$
$=|\mathbf{j\_1}||\mathbf{j\_2}| [ cos(\theta\_1) cos(\theta\_2)+ sin(\theta\_1) sin(\theta\_2) cos(\theta\_1) cos(\phi\_1 -\phi\_2)]$
So far, so good, no problem. Then he instructs:
"Let us differentiate Eq (2.34) with respect to time to obtain the rate of change of the length of $\mathbf{j}$. From the law of cosines
$|\mathbf{j}|^2 = |\mathbf{j\_1}|^2 + |\mathbf{j\_2}|^2 - 2 |\mathbf{j\_1}| |\mathbf{j\_2}| cos(\gamma)$
$=|\mathbf{j\_1}|^2 + |\mathbf{j\_2}|^2 - 2 \mathbf{j\_1} . \mathbf{j\_2}$
So that
$\frac d{dt} (\mathbf{j\_1} . \mathbf{j\_2}) = -\frac 12 \frac d{dt} [ |\mathbf{j}|^2 + |\mathbf{j\_1}|^2 - |\mathbf{j\_2}|^2 ]$"
And **THIS** next step is where I get lost:
He ***IMMEDIATELY*** equates that expression with the following:
$$ = -|\mathbf{j}| \frac {d|\mathbf{j}|}{dt} $$
So that there's no confusion, what I do not understand is the statement:
$$\frac d{dt} (\mathbf{j\_1} . \mathbf{j\_2}) = -\frac 12 \frac d{dt} [ |\mathbf{j}|^2 + |\mathbf{j\_1}|^2 - |\mathbf{j\_2}|^2 ] = -|\mathbf{j}| \frac {d|\mathbf{j}|}{dt}$$
---
Now, if I explicitly expand all components of the vector norms: $\mathbf{j\_1}$,$\mathbf{j\_2}$,and $\mathbf{j}$, and try to solve for what's inside the brackets on the LHS, I just regenerate the expression:
$|\mathbf{j}|^2 =|\mathbf{j\_1}|^2 + |\mathbf{j\_2}|^2 - 2 \mathbf{j\_1} . \mathbf{j\_2}$, which is of no use in aiding me as far as I can see.
Furthermore, it is explicitly stated that the only time-dependent component of this system is the dihedral angle $\phi\_1 - \phi\_2$, and in any expansion of the vector $|\mathbf{j}|$ into its components, a time derivative would render the any other independent parts mute.
I am thoroughly confused. I have consulted the following textbooks and not found any sign of this exact derivation:
*Cohen-Tannoudji (volumes I and II), Feynman's lectures on Physics Vol III, Shankar, Atkin's physical chemistry, Engel and Reid's physical chemistry, Gaziorowicz, and Eisberg and Resnick.*
What I suspect is that I'm simply lacking in some fundamental piece of calculus or arithmetic. Whether there's some simple trick of Vector Calculus that I haven't employed to give me the identity placed here? I would love a direct answer, but will also accept a nudge in the right direction. Right now it just feels like I'm leafing through books without a compass.
Best,
and ***THANKS!***
| 7 |
[
[
"\nAs mentioned in the comment by Weijun Zhou, the length of the vectors $\\mathbf{j}\\_1$ and $\\mathbf{j}\\_2$ does not change as $\\mathbf{j}\\_2$ precesses about $\\mathbf{j}\\_1$, only the total vector $\\mathbf{j}$ changes its length. Knowing this just requires application of the sum and chain rule of derivations that is\n\n\n$\n-\\frac{1}{2}\\frac{d}{dt}|\\mathbf{j}|^2=-2|\\mathbf{j}|\\frac{1}{2}\\frac{d}{dt}|\\mathbf{j}|\n$\n\n\nand the time derivatives of $\\mathbf{j}\\_1$ and $\\mathbf{j}\\_2$ are zero.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89503/methods-for-dissolving-scotch-kaptop-etc-tape-for-electron-microscopy
|
Methods for dissolving Scotch/Kaptop/etc. Tape for Electron Microscopy?
|
The famous method for the Nobel prize in studying Graphene is the so-called "scotch tape" method. Here, one takes a weakly-bonded van der Waals-type material and peel a layer off. From that layer, another piece of tape is taped onto the original piece of tape with the layer to split it into two pieces. This is repeated until nanometer thick layers are produced.
At this point, you have a single monolayer on a piece of tape, and need to transfer it onto a grid to look at it with a TEM (transmission electron microscope). However, the monolayer is much too thin to peel off the tape, so the tape must be completely dissolved.
**The question I have is: what are the best options for dissolving tape?**
Obviously there are many, many variables at play here. But here are some relevant constraints:
* Completely dissolves tape, leaving little sticky residue
* Solvent is relatively non-reactive to inorganics, metals
* Safe and inexpensive
For example, I found a question on [Research Gate](https://www.researchgate.net/post/can_polyimide_filmskapton_dissolved) that suggests using an alkaline solution to completely dissolve Kapton. I wonder if there are other methods for scotch tape
Since this method for TEM preparation is so common there must be some great literature resources on this topic, so resource recommendations would be great!
I don't fully understand the mechanics of "stickyness" so a description from the atomic perspective would be useful, especially in understanding what good solvents should do.
| 3 |
[
[
"\nGenerally you use Acetone. The following is taken from \nEnoki, Toshiaki, and Tsuneya Andō. 2018. Physics and Chemistry of Graphene : Graphene to Nanographene p. 94\n\n\nThe Scotch tape method is very common in the fabrication of\ngraphene devices, which are mainly used in research into the\nfundamental properties of graphene. It can produce high-quality\ngraphene with lateral sizes ≤ 0.1 mm; however, controlling the size\nand position of the graphene is almost impossible. The following is\nthe typical procedure of the Scotch tape method:\n\n\n(i) Place a few flakes of graphite (≈ 1 mm) on the adhesive side\nof a plastic sticky tape with tweezers (Fig. 3.1(a)). Scotch tape\n(3M) and Nitto tape are commonly used. Natural graphite,\nKish graphite, or HOPG (highly oriented pyrolytic graphite) is\nusually used for the starting material.\n\n\n(ii) Fold the adhesive side to sandwich the graphite flakes and\npress the tape firmly (Fig. 3.1(b)).\n\n\n(iii) Peel the tape apart slowly, so that the graphite flakes are\ncleaved and attached on both sides of the tape (Fig. 3.1(c)).\n\n\n(iv) Repeat the second and the third steps with slightly shifting\nthe fold line, so that the graphite flakes do not overlap (Fig.\n3.1(d)).\n\n\n(v) Stop repeating the cleavage when graphite flakes spread over\nthe sticky tape (Fig. 3.1(e)).\n\n\n(vi) Prepare a silicon substrate with a silicon dioxide layer on\nthe top surface (Fig. 3.1(f)). Place address markers on the\nsurface in advance using photolithography so that one can\neasily locate the position of a graphene flake in an image.\nThese markers are indispensable for further microfabrication\non graphene. An example of the address markers is shown in\nFig. 3.2\n\n\n(vii) Attach the adhesive side of the tape with the graphite flakes to\na silicon substrate, and gently rub the surface to remove any\nair between the substrate and the tape (Fig. 3.1(g)).\n\n\n(viii) Slowly peel the tape off the substrate (for example, more than\n2 min for a 1 cm substrate) (Fig. 3.1(h)). Not only graphite\nflakes but also some adhesive remains on the substrate. The\nlatter can be removed by submerging the substrate in acetone\nfor a few seconds.\n\n\n[](https://i.stack.imgur.com/IfKnU.png)\n[](https://i.stack.imgur.com/rMjui.png)\n\n\nIt is also reported that in the scotch tape method, not only graphene flakes but also a large amount of adhesive is attached to the substrate. You get rid of it immersing the substrate in Acetone. Other methods are reported in the reference too. \n\n\n**About adhesion**\nThere are lots of aspect about adhesion. About scotch tape the most interesting answer you can find here \"<https://www.scientificamerican.com/article/what-exactly-is-the-physi/>\" is the following\n\n\n\"The simplest answer that I can give to the question is that pressure-sensitive adhesives (which are polymers) are 'tacky' or 'sticky' because they are essentially very high viscosity liquids that also have some elastic characteristics--in technical terms, they are 'viscoelastic.' This property means that they exhibit some of the characteristics of liquids, and so they will 'wet' a surface to which they are pressed. But then, because of their elasticity, they will resist separation when stressed. Thus, 'stickiness' is strictly a physical (viscoelastic) phenomenon, not a chemical one.\"\n\n\n",
"3"
],
[
"\nKapton is a polyimide:\n\n\n[](https://i.stack.imgur.com/Eja8D.png)\n\n\nIt contains an $\\ce{N}$ substituted phtalimide as a substructure. \nThis kind of compound is used in the Gabriel synthesis to make primary amines. \nThe synthesis is very ingenious because no secondary or tertiary amines can be formed, which is normally the problem with direct alkylation. \n\n\nIn the final step the phtalimide is hydrolyzed with hydrazine, a reaction that works very well.\nA 35-50% solution of hydrazine in water is used.\n\n\n[](https://i.stack.imgur.com/JEZjU.png)\n\n\nHydrazine is not the most innocuous of products, although in solution it is not so dangerous, and should be used in a fume cupboard.\n\n\nIf you do not have a minimal Chemistry lab at your disposal, then you can better try to use a potassium hydroxide solution.\nI have never tried it because the hydrolysis with hydrazine gives such good results.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89488/which-solution-is-more-conductive
|
Which solution is more conductive
|
Which of the following solutions is more conductive:
1) 0.1 M sodium hydroxide and 0.1 M hydrochloric acid mixture (effectively NaCl?)
2) 0.1 M sodium hydroxide and 0.1 M acetic acid mixture (effectively NaCH3COO?)
I know the answer is the first one but I can't explain why. I have been taught (high school chemistry) that conductivity of ions is simply relative to [ions]. If that were true, both solutions would be equally conductive. I know that strictly speaking, based on the initial acids and bases, 1) would be more conductive, but I can't understand this because there should be none of the original reactants present anyways. This came up on a test, and I can't justify 1) being correct based on what I have learned.
| -1 |
[
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"\nConductivity of different ions carrying the same amount of charge can vary as per the comment of Ivan. In this question, acetic acid is a weak acid so the final acetic anion will combine with the hydrogen ion from the autodissociation of water molecule to form acetic acid molecules. This is known as the hydrolysis of $\\ce{NaAc}$. Hydrochloric acid, on the other hand, is a strong acid so the hydrolysis won't happen. Hydrolysis is the main reason that $\\ce{NaAc}$ solution is basic while $\\ce{NaCl}$ is neutral. \n\n\n\n> \n> There should be none of the original reactants present anyways.\n> \n> \n> \n\n\nThis is one fallacy in your reasoning. There are some $\\ce{CH3COOH} \\rm(aq)$ formed due to hydrolysis in the final solution since it is weak acid.\n\n\nTo summarize, in the first case the conducting particles are equal amount of $\\ce{Na+}$ and $\\ce{Cl-}$, while in the latter they are $\\ce{Na+}$, $\\ce{Ac-}$ and $\\ce{OH-}$, with the concentration of the first equal the sum of the other two. This is what I believe you need to know. Beyond that without looking up the conductivity table or analyzing the structures of the ions we **cannot deduce** whose conductance is higher.\n\n\nIn paricular, given a strong acid $\\ce{HA}$ and a weak acid $\\ce{HB}$, saying that for the solution of the same concentration, $\\ce{NaA}$ will be more conductive than $\\ce{NaB}$ is **simply wrong** and there **are counterexamples**.\n\n\n",
"2"
]
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https://chemistry.stackexchange.com/questions/89473/how-do-you-arrive-at-the-second-form-of-the-1st-order-integrated-rate-law
|
How do you arrive at the second form of the 1st order integrated rate law? [closed]
|
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As I understand it, the first-order integrated rate law is:
$$\ln[\ce{A}] = \ln[\ce{A}]\_0 - k t$$
However, I'm also told that this can be expressed as a ratio of $[\ce{A}]\_0$ and $[\ce{A}]$, as follows
$$\ln \left(\frac{[\ce{A}]\_0}{[\ce{A}]\,\,}\right) = kt$$
How did they arrive at this expression of the integrated rate law? I don't see a way to arrive at it algebraically, unless I'm greatly missing something.
| -4 |
[
[
"\nTo get to ln([A0]/[A])=kt from ln[A]=−kt+ln[A]0 just requires some understandings of log properties and a few log rules.\n\n\nFirst rearrange ln[A]=−kt+ln[A]0\n\n\nto\n\n\nkt = ln[A]0 - ln[A] \n\n\nFrom there, the log rule\nlog(a/b)=log(a)−log(b) means that you can state this as\n\n\nkt = ln([A]0/[A])\n\n\n",
"1"
]
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https://chemistry.stackexchange.com/questions/89471/difference-between-tendency-of-benzene-and-thiophene-to-undergo-sulfonation
|
Difference between tendency of benzene and thiophene to undergo sulfonation
|
Why does thiophene possess a greater tendency to undergo sulphonation than benzene? I came across this reaction which employs the use of sulphuric acid to separate thiophene from commercially prepared benzene.
| 4 |
[
[
"\nIn the chapter **Aromatic five-membered ring heterocycles with one heteroatom** in *Organic Chemistry* by J. William Suggs (2002) the following explanation can be found (pp. 403-404) which summarizes the thoughts already pointed out in the comments:\n\n\n\n> \n> The resonance stabilization energy of benzene is greater than that of these heteroaromatic compounds.\n> The order of aromaticity is benzene > thiophene > pyrrole > furan.\n> \n> \n> […]\n> \n> \n> All three of these ring systems undergo electrophilic aromatic substitution and are much more reactive than benzene.\n> In part, this reactivity difference arises because the rate-determining step in electrophilic aromatic substitution is the first step, which breaks up the aromatic $\\pi$ system.\n> Since thiophene, pyrrole, and furan have less stabilization to lose than benzene, the intermediate is lower in energy and the overall reaction proceeds more rapidly.\n> \n> \n> […]\n> \n> \n> All three of these heteroaromatic rings undergo electrophilic aromatic substitution, preferentially at C-2. \n> The reactivity order is pyrrole > furan > thiophene because of several factors, including the electronegativity of the heteroatom and the resonance stabilization of the aromatic ring.\n> \n> \n> \n\n\n",
"4"
]
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https://chemistry.stackexchange.com/questions/89470/is-there-a-reliable-way-to-determine-if-a-chemical-system-is-more-suited-to-be-c
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Is there a reliable way to determine if a chemical system is more suited to be calculated using a multireference method instead of DFT?
|
Some time ago I recall hearing a professor say that looking at a DFT result (I don't remember exactly what, a TD-DFT, spin contamination etc...), he could tell that the system was multireferenced.
My question is: do we have anything in a DFT calculation that is indicative of a multireference system?
| 6 |
[
[
"\nYou should have a look at the so called FOD (Fermi Occupied Density) analysis by Grimme et al. That is accessible by ORCA 4 very easy and straight forward by specifying \"FOD\" as a keyword. [1-3]\n\n\n[](https://i.stack.imgur.com/dNFJW.jpg) \n\nFrom ORCA Input Library [1]:\n\n\n\n> \n> The FOD plot above (plotted with Chemcraft) indicates a significant and delocalized FOD. Multiconfigurational methods should be used to study this molecule and any single-reference result (HF, DFT, MP2, CC should be regarded with suspicion)\n> \n> \n> \n\n\n\n\n---\n\n\nVia Multiwfn you can access the \"local total/dynamic/nondynamic electron correlation function\" and \"total/dynamic/nondynamic electron correlation index\" [4,5].\n\n\n[![Local Descriptors of Dynamic and Nondynamic Correlation from [5]](https://i.stack.imgur.com/3FvrT.gif)](https://i.stack.imgur.com/3FvrT.gif) \n\nVisual abstract from [5]. You can see the dynamic and non-dynamic correlation from which you also can see if single reference methods are acceptable or not.\n\n\n\n\n---\n\n\n[1] [ORCA Input Library, FOD analysis](https://sites.google.com/site/orcainputlibrary/orbital-and-density-analysis/fod-analysis) \n\n[2] [S. Grimme, A. Hansen, *Angew. Chem. Int. Ed.* **2015**, *54* (42), 12308-12313](http://dx.doi.org/10.1002/anie.201501887) \n\n[3] [C. A. Bauer, A. Hansen, S. Grimme, *Chem. Eur. J.* **2017**, *23* (25), 6150-6164](http://dx.doi.org/10.1002/chem.201604682) \n\n[4] [E. Ramos-Cordoba, P. Salvadorc, E. Matito, *Phys. Chem. Chem. Phys.* **2016**, *18*, 24015-24023](http://dx.doi.org/10.1039/C6CP03072F) \n\n[5] [E. Ramos-Cordoba, E. Matito, *J. Chem. Theory Comput.* **2017**, *13* (6), 2705–2711](http://dx.doi.org/10.1021/acs.jctc.7b00293)\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89464/is-reactant-concentration-product-concentration-at-equilbrium
|
Is reactant concentration = product concentration at equilbrium?
|
When a reaction attains chemical equilibrium, it means that the forward reaction rate = backward reaction rate. Since rate is defined as the change in concentration per unit time,does it mean that at equilibrium,
1) concentration of reactants = concentration of products?
2) Also, is the time taken for the forward reaction = time taken for the backward reaction?
| -3 |
[
[
"\n1) Not necessarily. In chemical equilibriums Concentrations could be any of following variations after attaining equilibrium. $$\\ce{[reactants] < [products]}$$ $$\\ce{[reactants] = [products]}$$ $$\\ce{[reactants] > [products]}$$\nAlthough reaction rates of forward and backward are equal, something like this could also happen. Therefore concentrations are **not ALWAYS equal** at equilibrium of all reactions, but could be in some reactions.\n[](https://i.stack.imgur.com/Z7E9y.png)\nimage source and more reading:-<http://faculty.chem.queensu.ca/people/faculty/mombourquette/FirstYrChem/equilibrium/index.htm>\n\n\n2) Since the general idea of \"Rate\" means some change (say in this case **reactants consumed or products produced**) divided by time, thus AT EQUILIBRIUM, as you said, $$\\ce{forward reaction rate = backward reaction rate}$$ But I don't see how it can strictly be said like their times are equal. I THINK it depends. Maybe someone can correct me.\n\n\n",
"1"
],
[
"\nI think that it depends on your definition of equilibrium. There is alwasy an equilibrium between reagents and products (if we are considering reversible reactions). It can be thermodinamically shifted in one of the two directions\n\n\nProducts\n\n\n$$\\ce{A + B <=>> C}$$\n\n\nReagents\n$$\\ce{A + B <<=> C}$$\n\n\nSame\n$$\\ce{A + B <=> C}$$\n\n\nit means that in the first case you will have a thermodinamical tendence to one of them or that in the latter case that there are not actually reagents and product but that for the two chemicals (it can be also one form of same chemical) is equal. \n\n\nAs an example C can be in the form of Graphite or Diamond. Only one of them is the thermodimically stable but it will take quite a lot of time to see them chance in to each other.. you can have a look here for a more detailed explaination here <http://www.ch.ic.ac.uk/rzepa/mim/century/html/diamond.htm>\n\n\nIf you reach the equilibrium in a reaction it means that even if the quantity of reagents and products are different the inteconversion (which is still present !) won't alterate their final ratio of concentration of each of them. It is important to notice that there is still an interconversion (that's why reversible equilibrium) . Also you have to keep in mind that changing the condition of your sistem (P,T,conc) can alterate your equilibrium so there is not ONE equilibrium. Just one caveat about concentration. As chemist we generally refer to molarity but it is important to define it before any reasoning ;) Hope this helps\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89460/balancing-other-atoms-in-a-redox-reaction-using-half-reaction-method
|
Balancing other atoms in a redox reaction using half reaction method
|
I have this balanced redox reaction:
$$\ce{2H2O + 8Al + 3NaNO3 + 5NaOH -> 3NH3 + 8NaAlO2}$$
I think its unbalanced form should be (I guess water added to balanced it)
$$\ce{Al + NaNO3 + NaOH -> NH3 + NaAlO2}$$
I wrote the half reactions as
$$
\begin{align}
\ce {6H2O + 8e- + NO3- &-> NH3 + 9OH-}\label{rxn:1}\tag{1}\\
\ce {Al &-> Al^3+ + 3e-}\label{rxn:2}\tag{2}
\end{align}
$$
Then by $3\cdot\eqref{rxn:1} + 8\cdot\eqref{rxn:2}$, final half reaction is:
$$\ce{18H2O + 3NO3- + 8Al -> 3NH3 + 27OH- + 8Al^3+}$$
Now I am really confused how to get above balanced redox reaction by the final half reaction I have obtained. What is the missing idea?
$$\ce{UPDATE:- Anybody please show me how to proceed from here,}$$
$$\ce{18H2O + 3NO3- + 8Al -> 3NH3 + 27OH- + 8Al^3+}$$
$$\ce{to here}$$
$$\ce{2H2O + 8Al + 3NaNO3 + 5NaOH -> 3NH3 + 8NaAlO2}$$
| -1 |
[
[
"\nYour total reaction is:\n\n\n$$\\ce{2H2O + 8Al + 3NaNO3 + 5NaOH⟶3NH3 + 8NaAlO2}$$\n\n\nBut your choice of oxidation half reaction is:\n\n\n$$\\ce{Al -> Al^{3+} + 3e-}$$\n\n\nNotice that the right side of the half reaction doesn't represent the right side of the total reaction (once you've accounted for the sodium spectator ion).\n\n\nYou *must* find the corresponding oxidation half reaction, and if it's not available, then you're out of luck.\n\n\nYour half reaction will have $\\ce{Al}$ on the left, and $\\ce{AlO2-}$ and 3 electrons on the right; with water, hydrogen ions, and hydroxides to balance.\n\n\nIn comments, you also asked for why this is the oxidation product. That is the wrong question to ask. All you're doing is breaking down the total reaction into component half reactions. The simple half reaction ($\\ce{Al|Al^{3+}}$) is totally valid. It's just gives a different total reaction from the one you started with. There's nothing wrong with that reaction. It's just different.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89459/what-is-the-mathematical-relation-for-the-change-in-water-density-at-different-t
|
What is the mathematical relation for the change in water density at different temperatures?
|
I am curious to the relation of the density of water at different temperature and as to also how one would come to an equation that took the temperature of the water and would output it's density? I have graphed density vs temp from tabulated values and it seems to look logarithmic, but wanted to make sure. I am interested in this as I am currently taking analytic chemistry.
| 1 |
[
[
"\nTo do this one must measure density accurately at many temperatures and then fit a curve to the measurements. This has been done by several investigators in several industries. For example in the sugar industry we have\n\n\n$$\\rho =(((((-281.03006e-12\\*t +105.84601e-9)\\*t-46.241757e-6)\\*t-7.9905127e-3)\\*t+16.952577)\\*t +999.83952)/(1+16.887236e-3\\*t)$$\n\n\nwhere $t$ is the temperature in °C and the density $g·cc^-1$. The coefficients were taken from the ICUMSA formula with the sugar concentrations set to 0.\n\n\nAnd \n\n\n$$\\rho = 0.99984+t\\*(6.7715e-05-t\\*(+9.0735e-06-t\\*(1.015e-07-t\\*(+1.3356e-09-t\\*(1.4421e-11-t\\*(+1.0896e-13-t\\*(4.9038e-16-9.7531e-19\\*t)))))))$$\n\n\nreturns the density of water, also in $g·cc^-1$, as a function of centigrade temperature (ITS 1990). It is based on a fit to data from\nBettin, H.; Spieweck,F.: \"Die Dichte des Wassers als Funktion der Temperatur nach Einführung der Internationalen\nTemperaturskala von 1990. PTB-Mitt. 100 (1990) pg 195-196\nThe referenced data set lists densities for each 0.1°C over the range (0,100). The data used in determining this\npolynomial are the subset on integer degree values. Sporadic checking shows that the fit appears to be\naccurate to 1 count in the 6th decimal place. The rms residual wrt the fit points is 3.1E-7 and the peak\nresidual 5.5E-7. The residuals appear noiselike.\n\n\nYou ought to be able to paste those formulas into a visualization program and plot values. I don't see any logarithmic behaviour. I've left the \\* in the formulas so you can do that. Note, when reading them, that 16.887236e−3∗t means (16.887236e−3)∗t.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89449/understanding-the-increase-in-ph-of-a-buffer-solution-upon-incremental-additions
|
Understanding the increase in pH of a buffer solution upon incremental additions of NaOH analytically
|
So let's say we have $200\ \mathrm{mL}$ of $1\ \mathrm M$ $\ce{CH3COOH}$ solution. In this solution we have the equilibrium $\ce{CH3COOH <=> CH3COO- + H+}$. To that we add $100\ \mathrm{mL}$ of $1\ \mathrm M$ $\ce{NaOH}$ solution. Then after reacting we get a buffer solution where $[\ce{CH3COOH}] = [\ce{CH3COO-}] = \frac{1}{3}\ \mathrm{mol\ dm^{-3}}$. In moles, we have $0.1\ \mathrm{mol}$ of both $\ce{CH3COOH}$ and $\ce{CH3COO-}$.
Then suppose we add $x\ \mathrm{mL}$ of $0.1\ \mathrm M$ $\ce{NaOH}$. This is, in effect, reacting $0.1\ \mathrm{mol}$ of $\ce{CH3COOH}$ with $0.0001 x\ \mathrm{mol}$ of $\ce{NaOH}$, the result of which is that now $n(\ce{CH3COOH}) = (0.1 - 0.0001 x)\ \mathrm{mol}$ and $n(\ce{CH3COO-}) = (0.1 + 0.0001 x)\ \mathrm{mol}$.
Could we not then model, by the Henderson-Hasselbalch equation, $\mathrm{pH} = \mathrm pK\_\mathrm a + \log\left(\dfrac{0.1 + 0.0001 x}{0.1 - 0.0001 x}\right) = 4.76 + \log\left(\dfrac{0.1 + 0.0001 x}{0.1 - 0.0001 x}\right)$?
But this would mean that for the pH of the buffer solution to rise by $1$, we would need $818.82\ \mathrm{cm^3}$ of $0.1\ \mathrm M$ $\ce{NaOH}$ solution, which sounds absurd. Moreover, for different acids, like propanoic acid and butanoic acid, the same line of logic could be used to deduce that the volume required to increase the $\mathrm{pH}$ by $1$ would be the same for all, but they have different buffering capacities. So what's wrong with the logic?
| 1 |
[
[
"\nI just answered what I thought was this question but realized that it was essentially the same question asked 5 years ago. So here's that answer modified for the slight difference in the way it is asked here:\n\n\nWe need to know the the pH of the buffer i.e. what does it measure before any NaOH is added. Later we will assume that it is 4.76, the pK of acetic acid.\n$[\\ce{HAc}] = [\\ce{Ac^-}]$ ([·] symbolizes the molar concentration of ·). In order to solve the general problem we need to be able to calculate the fraction of the total Ac that is dissociated and the fraction that isn't.\n\n\nThe fraction that is dissociated comes right out of the Henderson - Hasselbalch equation and is $$f\\_1 = 1/(1 + r\\_1)$$ where $$r\\_1 = 10^{(pH - pK\\_1)}= [\\ce{Ac^-}]/[\\ce{HAc}]$$ as is clear from inspection of the Henderson - Hasselbalch equation. The subscript 1 indicates that $r\\_1$ is the ratio of the number of acid ions that have lost 1 proton to the number that have lost $ 1 - 1 = 0$. In $f\\_1$ the subscript is indicative that $f\\_1$ is the fraction of the total Ac molecules that has become singly charged by loss of a single proton. When dealing with monoprotic acetic acid the subscripts aren't that important as there is only one proton to loose. But if the acid is polyprotic we have unionized, once ionized, twice ionized etc. ions to consider. The fractions of those ions are $f\\_0$, $f\\_1$, $f\\_2$... and we have other ratios as well\n\n\n$$r\\_j = 10^{(pH - pK\\_j)} = [H\\_{n-j-1}Ac^{−j}]/[H\\_{n-j}Ac^{-(j-1}]$$ Here \"Ac\" stands for the acid anion and n is the number of protons it can yield when fully dissociated. For acetic acid. $\\ce{CH\\_3COOH}$, and Ac is $\\ce{CH\\_3COO}$, n = 1 and j only has values of 0 or 1. With phosphoric acid, $\\ce{H\\_3(PO\\_4)}$, Ac is $\\ce{(PO\\_4)}$, n = 3 and j = 0,1,2 or 3.\n\n\nSo suppose now we have a polyprotic acid with values for $r\\_1$, $r\\_2$, $r\\_3$... and that there are x moles of the acid in a solution. Then there would be $xr\\_1$ moles of the singly deprotonated species, $xr\\_1r\\_2$ moles of the doubly deprotonated, $xr\\_1r\\_2r\\_3$ moles of the triply deprotonated and so on.Then the total number of moles of Ac would be the sum of the number of moles of each$$C\\_{Ac} = x + xr\\_1 +xr\\_1r\\_2 +xr\\_1r\\_2r\\_3...$$ The fraction of the total that is undissociated is $$f\\_0 = x/(x + xr\\_1 +xr\\_1r\\_2 +xr\\_1r\\_2r\\_3...) = 1/(1 + r\\_1 +r\\_1r\\_2 + r\\_1r\\_2r\\_3...)$$ The fraction that is singly dissociated is $r\\_1$ times this $$f\\_1 = r\\_1f\\_0$$ and the fraction that is doubly dissociated is $r\\_2$ times that $$f\\_2 = r\\_2f\\_1$$ and, in general $$f\\_j = r\\_jf\\_{(j-1)}$$ \n\n\nAt this point let's remember that $f\\_j$ is a function of the solution pH and all the pK's of the acid in question and that it is the fraction of the anions of that acid that carry charge -j. Thus we can write an expression for the total charge on all species of Ac at a given pH. This is $$Q\\_{Ac} = -C\\_{Ac}(0f\\_0 + 1f\\_1 + 2f\\_2 + ...)$$\n\n\nBefore going on to show you how Q solves buffering problems let me stop to suggest that the simplest way to work with it is to make an Excel (or other) spread sheet. Designate a column for pKs and a cell into which the pH goes. For example, put pH into cell A1 and start the pKs list in A2, Then in B2 put =10^($A$1 - A2). Using $A$1 lets you copy and paste B2 into as many cells as you have pKs. The B column now contains the r corresponding to the pK in the cell to the left of it. Now enter the formula for $f\\_0$ in a cell and make another column with the $f\\_j = r\\_jf\\_{(j-1)}$ formula in it. So how many pK's. I say make the spread sheet for 5 or 6. Why? Well lets go back to $\\ce{CH\\_3COOH}$ for a minute. It doesn't have 1 proton to give, as we have been assuming. It actually has 4. Are the other three ever coming off? Not with any base in my lab but we can model those other protons simply by assigning pKs that are so high (say 50) that the f values for anything other than $f\\_0$ or $f\\_1$ are 0. The point being that if you are going to go to the trouble to make the spreadsheet you might as well make it big enough to handle any acid you may ever encounter as it easily handles anything up to its maximum size using this trick.\n\n\nNow how to use $Q(pH)$. If there are a total of $C\\_{Ac}$ moles of Ac in a solution the negative charge on them at $pH\\_0$ is $C\\_{Ac}Q(pH\\_0)$. At $pH\\_1$ it is $C\\_{Ac}Q(pH\\_1)$. Thus to move the solution from $pH\\_0$ to $pH\\_1$ you must supply or remove charge of $$\\Delta Q\\_{Ac}(pH\\_0\\ce{->}pH\\_1) = C\\_{Ac}Q(pH\\_1) - C\\_{Ac}Q(pH\\_0)$$ by adding or absorbing protons. If $ pH\\_0 > pH\\_1$ then $Q(pH\\_1) > Q(pH\\_0)$ (less negative) and so the difference will be positive indicating that protons (acid) will need to be added to effect this pH shift.\n\n\nThe acid species in the solution are not the only thing that emits or absorbs protons when pH changes. The solvent does too. \n\n\n$$\\Delta Q\\_{W}(pH\\_0\\ce{->}pH\\_1) = 10^{-pH\\_1} - 10^{-pH\\_0} + (10^{(pH\\_0 - pK\\_w)} - 10^{(pH\\_1 - pK\\_w)})$$ represents the number of protons that must be supplied (or absorbed) to change the pH of water from $pH\\_0$ to $pH\\_1$.\n\n\nNow let's use this to solve the original question. We have 300 mL of 4.76 buffer with $C\\_{Ac}$ = 0.2 mol ( 0.1 mol of Ac and 0.1 mol of $Ac^{-1}$. This implies that $f\\_0 = f\\_1 = 0.5$ Also\n$$\\Delta Q\\_{Ac}(pH\\_0\\ce{->}pH\\_1) = 0.02(Q(pH\\_1) - Q(4.76))$$\n$$\\Delta Q\\_{W}(pH\\_0\\ce{->}pH\\_1) = 10^{-pH\\_1} - 10^{-4.76} + (10^{(4.76 - pK\\_w)} - 10^{(pH\\_1 - pK\\_w)})$$\n\n\nAparently we want to know how much NaOH we would need to raise the pH to 5.76. With our formulas in a simple Excel spread sheet and using 5.76 for $pH\\_1$ we quickly find that $\\Delta Q\\_{Ac}(4.76\\ce{->}5.76) = -0.0818 mole. IOW protons must be absorbed. At the same time we find $\\Delta Q\\_{W}(4.76\\ce{->}5.76) = -1.56E-5 mole per liter. We only have 0.3 L so the protons to be removed to adjust the water are insignificant. We apparently need 81.8 mL of 1 M NaOH. Note that your digits are pretty similar so one of us is off by a power of 10. As 0.8 mol OH- to shift an 0.2 mol buffer is a lot, I think it's you. As we both used Henderson Hasselbalch, we had better get the same answer.\n\n\nSo why would you do what I'm suggesting instead of what you are doing (assuming you find the factor of 10). If you do a lot of problems like this it makes cranking them out a snap.\n\n\nNow this post has gone on long enough so I won't have space to tell you about extension of this technique to much more complicated systems. You can use it to find the pH of complicated mixes of weak and strong acids and bases or any materials which have buffering capacity. I ginned it up (this is just using the 'proton condition' to solve a complex system - I didn't invent it) to predict the pH of brewer's mash which is a mixture of weak acid (bicarbonate) and several malts (each of which has its own buffering properties).\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89448/which-easy-to-acquire-clear-liquid-would-be-best-for-testing-whether-or-not-a-tw
|
Which easy-to-acquire clear liquid would be best for testing whether or not a twenty-sided die is balanced?
|
In [another question](https://rpg.stackexchange.com/questions/65206/is-the-saltwater-float-represented-in-this-question-a-good-way-to-test-for-loade) on another stack, I asked if the Saltwater float present in a specific youtube video was capable of determining any differences in mass between the faces of a die in determining the balance of a group of dice, and was met with very good results and a positive affirmation that yes, it was very possible to do so.
Which led me to believe that there was a possibility that denser liquids might have the possibility of testing the balance of the dice even better, with much less mess. This led me to go through and search the densities of many liquids to determine which would be best for the job.
* Water came in at ~1g/mL at room temperature, which was the medium for the test initially.
* Saltwater was at about ~1.028g/mL. Dice *were* able to float in this.
* H2O2 (Hydrogen peroxide) clocks in at ~1.44g/ML, but *I tested this one myself with a commercially available H2O2 and was unable to float dice in it just on its own. It isn't dense enough to float a die on its own.*
* Originally I thought Galium might be a good choice since it has a high density and is very easy to melt, but it isn't transparent.
So the question is this:
**Is there a liquid, that is easy to acquire, that has a higher density than H2O + MgSO4 (Epsom Salt), that it would be easier to test a dies balance in and at what density does a d20 float?**
Generally when referring to dice, the most well known manufacturer is probably *Chessex*. The dice are made out of some kind of plastic polymer of some sort. I don't really know any specifics about their chemical composition, but they could just be injection molded plastic dice.
As for the approximate mass of said dice.. [This](http://www.dice.co.uk/outlines.htm#poly) site states that the dice themselves are ~5mg *(That doesn't sound right. I learned in school that a Paper clip is a gram, Dice weigh much more than a paper clip..)*
| 0 |
[
[
"\n**You don't want a denser liquid.** You want a liquid with *just the right density* so the dice would barely float, almost fully submerged. To this end, you mix the solutions of different concentration and so fine-tune the density, as is beautifully described in the accepted answer to your **rpg.SE** question.\n\n\nA denser liquid is *worse* for your purpose. Ultimately, with a very dense liquid (like mercury) the dice would just sit on one face, much like they do on a tabletop.\n\n\nIn the relatively rare case your dice might be made of different kind of plastic and wouldn't float at all. That's when you genuinely **do** need a denser solution. Chemistry knows a lot of salts; which of these are available to you is another question. Saturated solution of $\\ce{MgSO4}$ has the density of 1.3 or so; $\\ce{CaCl2}$ will get you to ~1.4, and other salts farther yet, but they are also more difficult to come by.\n\n\nAs for the reference books with some density values, you have to read those really, *really* careful to see what they actually refer to, and then read again, just in case. 1.028 looks like the density of *seawater*; it is salty all right, but that's pretty far from saturation. The solution you use is a lot more dense. 1.44 for hydrogen peroxide looks like the density of the *pure* compound, which you surely can't acquire (which in turn is probably good for your survival); what you had is probably the 30% solution which is a lot less dense.\n\n\nAs for the density at which your d20 will float... well, how are we supposed to guess that? For all we know, your die may be carved out of wood; also, chances are it is cast in bronze. The resulting figures may vary pretty widely. **Upd.** So it is plastic, after all. Well, that still leaves a pretty wide room for wiggling. You can't tell the density until you measure it. Chemistry is an experimental science, they say.\n\n\nSo it goes.\n\n\n",
"2"
],
[
"\nNear impossible to say at what density of solution some unknown die will float. I've seen hollow plastic dice and metal dice. \n\n\nI'm also not sure how reliable some float test would be. I'd guess that the various gaming commissions have standard tests for dice. Using such a standard test would be preferable to trying to invent some test of your own. \n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89447/identifying-amphiprotic-species
|
Identifying amphiprotic species
|
I'm a bit confused about amphiprotic species – specifically why some species can be amphiprotic and others can't.
For instance, $\ce{SH-}$ is an amphiprotic species (in aqueous solution).
$$
\begin{align}
&\text{Acid:} &\ce{SH- + H2O &<=> S^2- + H3O+}\\
&\text{Base:} &\ce{SH- + H2O &<=> H2S + OH-}
\end{align}
$$
On the other hand, isn't $\ce{HNO3}$ an amphiprotic species too? I don't understand why $\ce{HNO3}$ isn't amphiprotic whilst $\ce{HS-}$ is; they seem to both be able to have an acid and a base reaction:
$$
\begin{align}
&\text{Acid:} &\ce{HNO3 + H2O &<=> NO3- + H3O+}\\
&\text{Base:} &\ce{HNO3 + H2O &<=> H2NO3+ +OH-}
\end{align}
$$
I hope this makes sense.
| -1 |
[
[
"\nAmphiprotic are species that have both acidic and basic properties. A classic example of an amphiprotic ion is dihydrogen phosphate $\\ce{H2PO4^-}$, which reacts in the presence of a $\\ce{H3O+}$ as\n$$\\ce{H2PO4^- + H\\_3O^+ <=> H\\_3PO4},$$ \nwhere $\\ce{H3PO4}$ is the conjugate acid of the original base. \nIn the presence of an $\\ce {OH-} $ the $\\ce{H2PO4-}$ reacts in the following way \n$$\\ce {H2PO4^- + OH- <=> HPO4^2- + H2O}.$$\n\n\nAminoacids are amphiprotic, as they contain both, a weak acid and a weak base functional group. Also water behaves amphiprotic.\n\n\nThe second reaction you wrote, even if in theory possible does not happen since $\\ce{HNO3}$ is such a strong acid, that in water is totally dissociated and has no tendency to behave as a base.\n\n\nYou can check this by writing down the general equilibria and calculate the equilibrium constant $K$ of each reaction you wrote. The $K\\_\\mathrm{a}$ of nitric acid is $2.4 \\times 10^1$. This means that, since\n$$K\\_\\mathrm{a} \\times K\\_\\mathrm{b} = K\\_\\mathrm{w} = \\pu{1.0\\*10^{-14}}$$\nthe $K\\_\\mathrm{b}$ of nitric acid is at least in the order of $10^{-15},$ which gives you an idea why this reaction won't happen.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89443/why-do-gas-molecules-travel-in-a-straight-line
|
Why do gas molecules travel in a straight line? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
This question does not appear to be about chemistry within the scope defined in the [help center](https://chemistry.stackexchange.com/help/on-topic).
Closed 5 years ago.
[Improve this question](/posts/89443/edit)
As we know, a single particle moves in a straight line until it collides with another particle. The question is how scientist figure out that gas particles move in a straight line?
| -1 |
[
[
"\nNewton's First Law of Motion states that an object at rest stays at rest and an object in motion stays in motion at a constant velocity, unless acted upon by a force. \n\n\nPut more simply, stuff stays still until something else (like a force or another object) makes it move, and moving stuff moves at a constant speed unless something else changes its speed. (I substituted *speed* for *velocity* in this sentence even though they don't really mean the same thing to make my explanation simpler.) \n\n\nFor example, if one were to throw a tennis ball into the air, it would stay moving at the same speed and direction if it weren't affected by the \"forces\" of gravity, air resistance, or wind. (I put *forces* in parentheses because wind isn't actually a force.)\n\n\nNewton's First Law of Motion doesn't just apply to macroscopic objects like tennis balls; it also applies to microscopic objects like particles. (Except perhaps at the quantum level; I'm not an expert in quantum mechanics so I wouldn't know too much about that.) Therefore one assumes that a gas particle will move in a straight line until it collides with another particle, and this assumption will generally work when experimenting with small quantities of gas in labs.\n\n\nOf course, in nature gas particles don't ALWAYS move in straight lines. For example, planets form when gas and dust particles orbiting around a star clump together due to gravity, and those gas particles will be moving in a curved line relative to the star. You could also argue that no gasses on earth are really moving in a straight line since their paths curve around the sun along with everything else on our planet. It really depends on our reference frame, but Newton's First Law of Motion allows us to know a gas particle will move in a straight line until it collides with another particle within our reference frame.\n\n\n",
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https://chemistry.stackexchange.com/questions/89436/will-the-solubility-product-constant-differ-in-this-case
|
Will the solubility product constant differ in this case? [closed]
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**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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I am considering the solubility product of sodium citrate. I need this value, I do not have it. However, I do know that the solubility product of calcium citrate is (7±2)×10^−14.
Is it safe to assume that sodium citrate will have a very similar Ksp?
I am concerned about whether or not the addition of sodium citrate will force borate ions out of the water as sodium tetraborate.
Furthermore, could I hypothesise on the solubility based on the likely strength of the hydrogen bonds? I am able to view the charges of each individual atom within a molecule, and the citrate's oxygen anions are far more charged than those in the borate ions. The sodium citrate's hydrogen bonds should therefore be stronger, and result in a higher solubility?
Thank you.
| -1 |
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"\nNo. The reason you can't find the $K\\_{\\rm sp}$ of sodium citrate is most likely because it is highly soluble. Except for a few exotic cases, sodium salts are very soluble in water.\n\n\nAdding sodium citrate to the solution does reduce the solubility of borax, however this is not going to be a problem unless you are using solution with really high concentration. As a reference, the solubility of sodium citrate and borax are $\\pu{770 g/L}$ at $\\pu{25^\\circ\\!C}$ and $\\pu{51 g/L}$ at $\\pu{20^\\circ\\!C}$, respectively.\n\n\nThe solubility of a solid in water depends on many factors. In this case where the two solids share the same cation, we can say that thermodynamically it depends on the extent that water stabilizes the anion compared to the crystal lattice. Your reasoning that the citrate anion's hydrogen bonding with water is stronger than borate anion's is valid and I think it's fine to say that sodium citrate is more soluble than borax based on this. Just that you need to keep in mind that the crystal energy also plays a role here as well, and need to be examined with care if the oxygens are not \"far more\" charged in one species than in the other.\n\n\n",
"1"
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https://chemistry.stackexchange.com/questions/89430/molecule-where-an-sp3-chalcogen-is-connected-to-an-sp2-atom
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Molecule where an sp3 chalcogen is connected to an sp2 atom
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I am looking for a small molecule where an $\mathrm{sp^3}$ atom from Group 16 (O, S, Se,...) is connected to an $\mathrm{sp^2}$ or resonant atom of another column?
Background: The universal force field defines the above case in its torsion
rules and I would like to find an example/application of such a structure
to understand it better.
Thanks!
| 1 |
[
[
"\nOut of the blue, consider triphenyloxonium cation. It is non-planar thanks to bulkiness of the phenyl groups.\n\n\nAnother possible example is mono-epoxide of allene. Considering oxygen atom as $sp^2$ here is unconvincing since it would involve huge angular strain, much higher than for a structure with $sp^3$ oxygen.\n\n\n",
"0"
]
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https://chemistry.stackexchange.com/questions/89426/why-an-acid-form-h3o-not-h3o
|
Why an acid form h3o+, not h3o [duplicate]
|
**This question already has an answer here**:
[How to know when electrons will be added onto Lewis Structure from surrounding atoms to become ion?](/questions/24696/how-to-know-when-electrons-will-be-added-onto-lewis-structure-from-surrounding-a)
(1 answer)
Closed 5 years ago.
When a lone proton bonds onto an H2O molecule in an acid, why doesn't the proton bond with a pair of the lone electrons in H2O and satisfy its electrons needs by having 3 single bonds? Why is it still charged?
| -2 |
[
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"\nOkay let’s look at the H2O molecule first. The oxygen atom has 6 protons and each of the hydrogen atoms have a proton. Hence the H2O molecule has a total of 8 positively charged protons. Oxygen has 6 valence electrons and each of the hydrogen’s have 1 valence electron, directly involved in covalent bonding with the oxygen atom. Hence the H2O molecule has a total of 8 negatively charged electrons. As the number of protons = number of electrons in the molecule, the molecule has no overall charge. When a lone proton forms a dative covalent bond with a lone pair on the oxygen atom of the H2O molecule, there are now overall 9 protons and still the same 8 valence electrons in the molecule. Hence due to the imbalance between protons and valence electrons in H3O, there is an overall positive charge on the molecule, hence why it exists as H3O+ when a lone proton (no electrons) bonds with H2O.\n\n\n",
"1"
],
[
"\nThere aren't any free protons floating around it water. They are attached to water molecules or acid molecules. If a positively charged proton is transferred to a neutral water molecule (from another water molecule or from an acid molecule) then clearly that molecule becomes positively charged.\n\n\nNow how is the new proton bonded to the water molecule? If you consider it to be ionically bonded then you are assuming the oxygen is hanging onto its electrons (at which it is pretty good) in which case the proton retains its positive charge. But if you assume, as you seem to be suggesting, that the bond is purely covalent, then the bond would be formed by sharing the electrons equally between the proton and the oxygen. In this case you would indeed write $\\ce{H\\_3O^\\oplus}$ i.e. the positive charge becomes a formal charge on the oxygen. \n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89419/calculating-the-density-of-a-saturated-salt-solution
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Calculating the density of a saturated salt solution
|
Is there a way to calculate the density of a saturated salt solution from the solubility limits or does it have to be experimentally determined?
First instinct, is to add the mass of the salt to mass of the water e.g. a solubility of $\pu{80 g}$ of salt in $\pu{100 mL}$ would have a solution density of $\pu{180 g/100 mL = 1.8 g/mL}$. However, it seems the salt should affect the volume of the solution. What additional information is needed or does this quantity just have to be empirically determined?
| 6 |
[
[
"\nOften, solubility is given as mass per volume of solution (not solute). Nevertheless, unless you have the mass percentage (w) too, you are not able to calculate the density of the solution. \n\n\n",
"1"
],
[
"\nThis is something you *can* calculate, provided you know the [partial molar](https://en.wikipedia.org/wiki/Partial_molar_property) volumes of all the substances in your solution. But someone must have measured those at some point, and tabulated it for your conditions (or close enough that you can interpolate). You can probably estimate it closely enough, depending on your needs. \n\n\nThe experiment would be as simple as taking a known volume aliquot of your saturated solution and weighing it. \n\n\n",
"1"
],
[
"\nYour proposed method (adding the masses, but using the volume of the solvent) *can* work under certain circumstances; it is often a good first approximation for the times when it is not a good assumption.\n\n\nFor low solubility salts, you can safely assume that the volume of the solution is the same as the volume of the solvent. \"Low\" here is somewhat open for debate. As a general rule, I feel safer with this assumption for each order of magnitude below 1 molar / molal. So, I consider this method valid to one (or fewer) sigfigs for solutions near 1.0 molar; 0.1 M makes me feel \"2 sig figs safe\", etc. (Again, these are just guidelines, don't get hung up on specific numbers.) With a salt that can only dissolve fewer than 10 grams into 100g of water, this method is usually fine to 2 or 3 sig figs. A salt that dissolves less than 1.0g into 100g of water is now getting into the territory of measurement errors of accurately measuring the volume of the solution (in most high school or even some college labs).\n\n\nSeawater runs about 3% salt to water by mass. Estimating it's density at 1.03 g/mL is pretty valid. In the example in the OP of 80g of salt in 100mL of water (giving a value of 1.8 g/mL) is a good *first order approximation*, but I would take that value with... a grain of salt. \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89321/how-to-dock-aggregated-structures-comprised-of-elementary-protein-units-like-l
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How to dock aggregated structures comprised of "elementary" protein units like LEGO pieces?
|
[There are proteins](http://guava.physics.uiuc.edu/~nigel/courses/598BIO/498BIOonline-essays/hw1/files/HW1-Kanchanawarin.pdf) with intrinsic [symmetries](http://www.rcsb.org/pdb/help/viewers/jmol_symmetry_view.html). For example:
[](https://i.stack.imgur.com/ihbGR.png)
I was wondering how to use transformations such as: rotations, translation, replications to construct possible structures using 2 types of proteins.
Assume I found a binding site of the 2 proteins using Molecular Dynamics simulators.
For example the above one and this one:
[](https://i.stack.imgur.com/J8j3j.png)
* Can I use group theory way to find possible configurations of the 2 proteins?
* How can I use the tensor product of the symmetry groups ($G\_1 \times G\_2$) to find possible structures?
* What graphic/framework would you recommend do perform all the manipulations on the proteins?
* Is it possible to use the [transfer](https://en.wikipedia.org/wiki/Transfer_(group_theory)) to create some algebraic structure that will help to construct a mechanical configurations comprised of proteins?
* Is there any tool from [Baker's Lab](https://www.bakerlab.org) I can use for this task?
* Amy recommendation for a tool I could use to replicate, rotate and shift structure would be appreciated us well.
---
**TL;DR**:
My goal is to use group theory or Monte Carlo simulation driven by group theory insight to create macro molecular structures using multiple (dozens) instances of the two proteins (the LEGO building blocks).
Any idea that relates symmetry, transformations, group theory to create structures is much welcome.
---
**Clarification**:
Given a binding point between 2 proteins, which software can I use to create the dimers into macro-structures like in LEGO namely only by replicating, translating and rotating the Lego pieces?
| 8 |
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[
"\n**Try [DNA origami](https://en.wikipedia.org/wiki/DNA_origami) instead** \n\n\nWhile protein assemblies are well known, rationally using proteins to assemble larger supramolecular structures is still a difficult problem:\n\n\n* Predicting protein folding is an unsolved problem. Sometimes even small modifications to a sequence can change the folding propensity.\n* Predicting [protein-protein interactions](https://en.wikipedia.org/wiki/Protein%E2%80%93protein_interaction) is often difficult, due to challenges in understanding protein dynamics, intermolecular interactions, and accurate predictions of protein electrostatics.\n\n\nThere has been some progress in terms of forcing particular protein-protein assembly through intentional side-chain modification and cross-linking, and reactive end-groups (e.g [protein-based metal-organic frameworks (MOFs)](http://pubs.acs.org/doi/10.1021/acscentsci.5b00315).\n\n\nThat said, use of proteins for supramolecular assemblies is far harder and less prevalent than using DNA assemblies. At the moment, DNA origami is far superior for these purposes.\n\n\n**Edit**: If you're really looking to just copy and displace atoms, I don't think you need any particular software. I'd write a Python (or other script) that reads in the PDB files, and writes out duplicate atoms displaced in the XYZ directions as needed. Group theory can help predict the angles, but you'll still need to know the protein size to work out the geometry.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89303/heterogenous-equilibrium-nernst-equation
|
Heterogenous equilibrium nernst equation
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In the nernst equation, when a gas - ion electrode is used eg when we write the equation for standard reduction potential for a particular electrode : we get a term of activity of h2 gas and h+ ion. We respectively put 1 atm and 1 molar. However how can we use partial pressure and molarity in the same expression for Q? Shouldn't it be one or the other?
| 0 |
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"\nYou can mix activity with fugacity in the expression of equilibrium coefficient or reaction quotient. In fact that is quite common when the reaction involves both solution and gas phase reactants/products.\n\n\nThe definition of the equilibrium constant actually involves dividing the standard pressure $p^\\unicode{x29b5}=1\\,\\rm atm$ or the standard concentration $c^\\unicode{x29b5}=1\\,\\rm mol/L$. In your case, the equilibrium constant will be\n\n\n$$K^\\unicode{x29b5}=\\frac{[a(\\ce{H+})/c^\\unicode{x29b5}]^2}{[f(\\ce{H2})/p^\\unicode{x29b5}]}$$\n\n\nwhich is dimensionless, and is related to the change of Gibbs free energy by $\\Delta G=\\Delta G^\\unicode{x29b5}-RT\\ln K^\\unicode{x29b5}$.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89297/determine-concentration-of-phosphate-after-copperii-oxalate-is-precipitated
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Determine concentration of phosphate after copper(II) oxalate is precipitated
|
Solution with $\ce{Na3PO4}$ with $\pu{0.02 M}$ concentration and $\ce{Na2C2O4}$ with $\pu{0.03 M}$ concentration. I add excess $\ce{Cu(NO3)2}$ very carefully – tiny amounts.
What will be my $\ce{PO4^3-}$ concentration at the moment when precipitation of $\ce{CuC2O4}$ will occur?
Given: $K\_\mathrm{sp}\left(\ce{Cu3(PO4)2}\right) = \pu{1.40e-37}$, $K\_\mathrm{sp}\left(\ce{CuC2O4}\right) = \pu{4.43e-10}$.
I know that the reactions will be:
$$
\begin{align}
\ce{Cu3(PO4)2 &<=> 3Cu^2+ + 2PO4^3-}\\
\ce{CuC2O4 &<=> Cu^2+ + C2O4^2-}
\end{align}
$$
Now, solving some math I find that $\ce{Cu3(PO4)2}$ precipitates first. But then I start asking my self what is happening with the $\ce{Cu}$ amount. Never did lab lessons or thought about the process behind the scene.
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"\nClearly the phosphate will precipitate first. As you add the copper nitrate copper(II) phosphate will precipitate taking equivalent amounts of copper(II) and phosphate. Eventually there isn't enough phosphate left to take out the incremental additions of copper and its concentration begins to rise. Eventually the saturation limit with respect to oxalate is reached. So find that limiting copper concentration (at which oxalate precipitation begins) from $\\ce{[Cu^{+2}] = KspCu2Ox/[Ox]}$. I get 4.92222e-07 M. Then find the saturation level for phosphate at that copper concentration from\n$\\ce{[PO4^{-3}]}$ = $\\sqrt{K\\_{sp\\ce{Cu3(PO4)\\_2}}/\\ce{[Cu]^3}}$\nI get 1.08348e-09 M. A pKsp of nearly 37 suggests that copper phosphate isn't very soluble.\n\n\n",
"-1"
]
] |
https://chemistry.stackexchange.com/questions/89296/is-the-following-procedure-for-antacid-titration-correct
|
Is the following procedure for antacid titration correct? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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I'm trying to check which antacid is more effective by determining the amount of $\ce{HCl}$ they can neutralize.
* Take $\pu{10 mL}$ antacid suspension and add excess $\pu{1 M}$ $\ce{HCl}$.
* Back titrate the excess 1M $\ce{HCl}$ with 1M $\ce{NaOH/Na2CO3}$.
* Do the required calculations.
Also, can I directly titrate the antacid with $\ce{HCl}$? Some solids remain.
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"\nYes, you should titrate to pH 6 (the pH at the entrance to the small intestine, 7.4 (the end of the small intestine) and 4.5. Multiply the mL of HCl used by its normality to get the mEq of protons absorbed by the antacid. Multiply the pH 4.5 result by 84 to get the sodium bicarbonate equivalent (mg of NaHCO3 that neutralizes the same amount of acid as the test sample).\n\n\nThe antacid may contain some insoluble material (such as SiO2) but try to get everything dissolved. Also, when titrating wait a good long time before accepting the pH reading\n\n\n",
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]
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https://chemistry.stackexchange.com/questions/89293/thermal-vibration-of-bonds
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Thermal vibration of bonds [closed]
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**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
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Closed 5 years ago.
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So, the characteristic frequency of thermal motion is around 7E12 Hz at room temperature (300K), but from that information how can we conclude that the bonds are hard; they don't vibrate !!
| -1 |
[
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"\nThe bond energy is related to which kind of molecules are involved in the bond itself (their polarity,...). all the molecular bonds undergo some kind of vibration (they can bend, stretch, rock,...) which peculiar frequency varies according to the type of bond and molecules involved.\n\n\n",
"1"
]
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https://chemistry.stackexchange.com/questions/89292/in-vibrational-spectroscopy-what-affect-do-overtones-have-on-the-spectra
|
In vibrational spectroscopy, what affect do overtones have on the spectra? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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Closed 5 years ago.
[Improve this question](/posts/89292/edit)
I've been in a lecture about vibrational overtones in anharmonic oscillators. How would these affect an IR spectrum?
| 2 |
[
[
"\nSince a spectrum, including its overtones, is essentially a property of a molecule, one cannot really say how the overtones affect a spectrum. So, I will answer this question by first describing what an overtone is, giving an example of an effect that overtones sometimes participate in, and then an example of how overtones can be useful.\n\n\nOvertones are charatceristic vibrational frequencies of a molecule which take place at approximately integer multiples of the $v=0$ vibrational modes. The existence of these overtones demonstrate that molecular vibrations are anharmonic in nature. This follows from the fact that for a purely harmonic vibration, the only non-zero transitions are those for which $\\Delta v=\\pm 1$. That is, harmonic oscillators can only move on state at a time. Thus, the fact that we observe these overtones tell us that the vibrations are not harmonic or anharmonic.\n\n\nNow, in IR and Raman spectroscopy, there is an effect called [Fermi Resonance](https://en.wikipedia.org/wiki/Fermi_resonance). This effect basically allows for the mixing of multiple vibrational states. If I'm not mistaken, this also only happens for anharmonic vibrations because energy needs to be allowed to be shared between the two states which is not possible if the vibrations are pure harmonic, normal modes. I bring this up because it is quite common that one observes a Fermi resonance between an overtone and a higher frequency ground state mode. To quote the Wikipedia page,\n\n\n\n> \n> High resolution IR spectra of most ketones reveal that the \"carbonyl band\" is split into a doublet. The peak separation is usually only a few cm$^{−1}$. This splitting arises from the mixing of νCO and the overtone of HCH bending modes.\n> \n> \n> \n\n\nThus, if you wish, one effect that overtones have on IR spectra is their ability to participate in Fermi resonance, which leaves the appearance of a peak being split.\n\n\nSometimes it can be difficult to know if a peak being observed is in fact an overtone. Generally, overtones have low intensity, but if they participate in a Fermi resonance, they gain intensity and will look like a normal ground state vibration (depending on what mode they come from). \n\n\nOne place I have seen something likely similar to this is in the following paper[1]. They compute the IR spectrum of a water cluster, including the bending overtone, and show that it is likely the experimental spectrum was mis-assigned. In this example, being able to compute a spectrum independent of the bending overtone is quite useful.\n\n\nExperimentally, overtones can be quite useful if there is a large number of peaks around the same frequency as a particular mode you are interested in studying. For instance, it is thought that the water dimer may play a significant role in absorption of radiation in the atmosphere and thus contribute to global warming. It is very hard, however, to determine the abundance of water dimers in the atmosphere because of the large amount of water vapor which covers up its entire infrared spectrum. Thus, ref. [2] used the *fourth* overtone of the $\\ce{O-H}$ stretch to experimentally measure water dimers in the atmosphere.\n\n\nAnother example of the same use of overtones is in ref. [3]. There, the authors are interested in studying a possible intramolecular hydrogen bond, but one cannot just look at the lowest energy vibrations because they are swamped by a regular $\\ce{O-H}$ stretch. Thus, the authors record infrared spectra in the regime where one would expect to find the first, second, third, and fourth overtones. It is confirmed that these are the peaks observed by simply computing the spectra from first principles. Thus, one also learns which geometric configurations cause which overtone peak.\n\n\nIn summary, overtones are not so much something that \"affect\" a spectrum, but are a necessary part of the spectrum because we live in a world where molecules vibrate anharmonically. I have also given examples of when overtones are quite a useful tool.\n\n\nJust to get you thinking a bit, if and when you learn about two-photon or two-dimensional spectroscopies, imagine how overtones can then be used to gain even deeper insight into the molecules of interest.\n\n\n\n\n---\n\n\n[1]: Wang, Y., & Bowman, J. M. (2013). IR spectra of the water hexamer: Theory, with inclusion of the monomer bend overtone, and experiment are in agreement. The journal of physical chemistry letters, 4(7), 1104-1108.\n\n\n[2]: Pfeilsticker, K., Lotter, A., Peters, C., & Bösch, H. (2003). Atmospheric detection of water dimers via near-infrared absorption. Science, 300(5628), 2078-2080.\n\n\n[3]: Howard, D. L., Jørgensen, P., & Kjaergaard, H. G. (2005). Weak Intramolecular Interactions in Ethylene Glycol Identified by Vapor Phase OH− Stretching Overtone Spectroscopy. Journal of the American Chemical Society, 127(48), 17096-17103.\n\n\n",
"3"
]
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https://chemistry.stackexchange.com/questions/89288/separating-sugar-from-soda
|
Separating sugar from soda
|
I want to do a lab with my students in which they will try to prove how much sugar is in a can of coke. I know that distillation is the go-to method for separating out the syrupy sugar mix, but does anyone know of a method that goes further to separate out the pure granulated sugar and not just a syrup?
| 2 |
[
[
"\nThe simplest approach would be to degas a sample and measure its specific gravity with a vintner's hydrometer. It will have scales of specific gravity and Brix (sugar content) on it.The Brix (Bx) reading is the percent of sucrose, w/w, in a solution with the same density/specific gravity as the one you are testing. There are, of course, other things in coke besides sucrose (fructose, coca extract, phosphoric acid) but they should be in small enough concentration that the sugars will swamp everything else and all sugars behave amazingly the same with respect to density of their solutions).\n\n\nBased on the reported sugar content of Coke it contains 108 grams of sugar per liter. Thus its specific gravity should be about 1.041 and the corresponding Bx reading about 10.4. The density of water at 20 °C is 0.998203 so the density corresponding to 1.041 SG is 1.039 and 10.4% of 1039 grams (weight of a liter of 10.4 Bx sucrose solution) is 108 grams.\n\n\nIf you don't have a hydrometer with the Bx (or Plato) scale or if you determine density by weighing a known volume (convert to specific gravity) you will need to be able to determine the sucrose content of a solution of given specific gravity. There is lots of opportunity here to go into the history of wine and beer making, which you male students will love (though I understand that in modern schools discussion of things related to alcohol may not be politically correct) but in any case determining the w/w sucrose content of solutions of given SG is a well studied problem. There are various tables but the easiest way to get w/w sucrose from specific gravity is a polynomial promulgated by the ASBC (American Society of Brewing Chemists). It is\n\n\n°P = ((135.997\\*S -630.272)\\*S + 1111.14)\\*S -616.868\n\n\nI believe that separating out the pure sugars at anything close to 100% recovery would be difficult (barring HPLC or something of that sort). Sugars seem to be assayed by indirect methods such as the gravimetric ones I proposed or measurement of refractive index.\n\n\n",
"3"
],
[
"\nIf the amount of sugar is to be determined precisely in a given solution, the simplest thing is to mix it with a $0.2$ M $\\ce{HCl}$ solution, to heat it up to boiling for ten minutes, and cooling it down to room temperature. This will hydrolyze the saccharose to a mixture glucose + fructose. Then, the glucose is determined by Fehling's solution, producing a $\\ce{Cu2O}$ precipitate, which can be weighed or titrated. The procedure is described in all books of analytical chemistry.\n\n\n",
"0"
],
[
"\nHow about precipitating the sucrose? Acetone and CaCl2 can precipitate up to 98% of the sucrose. There are other precipitation schemes on the web that use complex organic compounds of transition metals, but they seem more useful for analyses than for actual separation.\n\n\nI can't do any better than to copy and paste an abstract (the whole article is available at the reference url):\n\n\n\"The use of methanolic CaCl2 for the non-aqueous extraction of sucrose from dried sugar beet (Beta vul garis L.) cossettes was investigated, with emphasis on the effects of CaCl2 concentration, time of extraction, and operating temperature. Solubility of sucrose from dried cossettes was optimal in range of 10-15% CaCl2 in methanol and increased by a factor of four as the temperature was raised from lC to 60C. Nearly complete extraction (98%) was achieved in batch experiments with four successive extractions. The methanolic CaCl2 extract was found to have both higher purity and greater sucrose stability compared with aqueous diffusion juice. The sucrose could be precipitated completely from the solution by the addition of 5 volumes of acetone at 20C, leaving 9~/o of the CaCl2 in solution. Steady state conditions could not be reached in simulated continuous column experiments, probably because of calcium exchange with other cations in the beet pectin. Pure sucrose was precipitated from these effluent solutions on cooling. The amount of sucrose recovered was pro portional to the decrease in calcium content on pas sage through the column. From this, the molar ratio of calcium to sucrose in the methanolic CaCI2 sucrose complex was calculated to be 2:1.\" Ref 1.\n\n\nRef 1. <https://bsdf-assbt.org/wp-content/uploads/2017/04/JSBRVol30No1and2P57to69SucroseExtractionFromBeetByMethanolicCalciumChloride.pdf>\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89281/why-does-potassium-have-the-electronic-configuration-of-2-8-8-1
|
Why does potassium have the electronic configuration of 2.8.8.1? [duplicate]
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**This question already has answers here**:
[Why does the 3rd electron shell start filling up with scandium?](/questions/8357/why-does-the-3rd-electron-shell-start-filling-up-with-scandium)
(3 answers)
Closed 5 years ago.
We all know that potassium, the 19th element on the periodic table has the electronic configuration of 2.8.8.1. However, why not 2.8.9? In the element scandium, it has the electronic configuration of 2.8.9.2. I would like to know why potassium is not 2.8.9.
| 1 |
[
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"\nYou *could* actually have the 2.8.9 configuration, putting the last electron in a 3d rather than 4s orbital. But the 2.8.8.1 one is lower in energy and thus more stable. How electrons fill subshells in neutral atoms to make the lowest energy configuration can be seen [here](https://en.m.wikipedia.org/wiki/Aufbau_principle).\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89277/formal-charge-and-valence-shell
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Formal charge and valence shell
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Is there a relationship between valence shell and formal charge of the atom......
Let's take an example.
>
> A carbocation for eg $methenium$ has a formal charge of +1,can we tell how many valence electrons it has using this.A carbon atom always forms 4 bonds but here it is only forming 3 bonds then surely there will be change in it's valence shell and the formal charge will be related to it?How about other atoms such as ammonium ion.What will this positive charge signify....decrease in lone pairs or increase.Stuffs like that
>
>
>
I don't know much about formal charge so that's why I am asking such confusing questions.My teacher said to me that carbon always forms 4 bonds and since it is forming 3 bonds in cases like this , a positive charge will come to it.Same for nitrogen but he didn't tell me about the valence shell so that's why i am asking
| 3 |
[
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"\nCarbon doesn't always form 4 bonds. Consider neutral methylene, $\\ce{H2C:}$ in which it forms two leaving two unshared electrons (:). Now suppose we protonate that (with a super acid?). We'd have $\\ce{H2CH+}$ or $\\ce{H3C+}$, a cation just as when we protonate $\\ce{H3N:}$ we get the ammonium cation $\\ce{NH4+}$. In both cases the bond to the proton is made from the unpaired electrons on C or N and not by sharing an electron from C or N with one from hydrogen. \n\n\nIn computing formal charge we assume that all bond electrons are equally shared. The formula for formal charge is V - N - B/2 in which V is the number of valence electrons in the neutral atom (4 for carbon), N is the number of valence electrons not participating in a bond (0 in the case of methenium) and B the number of electrons in bonds (6 in the case of methenium) so that the formal charge is 4 - 0 - 3 = +1.\n\n\nIt's a different way of indicating charge distribution. If we write methenium as $\\ce{H2CH+}$ it implies that the carbon retained the electrons for the most part so that the bond to the proton is largely ionic. In the formal charge scheme we assume that this bond is totally covalent, that the + charge of the proton has transferred to the carbon because the proton got a half share in two electrons thus neutralizing its charge and write it $\\ce{C^\\oplus H3}$. It's a formality.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89266/equilibrium-kp-constant-partial-pressures
|
Equilibrium, Kp constant, Partial Pressures
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A system in equilibrium contains
$$I\_{2\_(g)}\ \text{and} \ I\_{(g)}$$
under pressures $$P\_1=0,21 \,\text{atm. and}\ P\_2=0,23 \,\text{atm.}$$
While its temperature remains constant, we reduce the system's volume to half its original value.
I have to determine $P\_1'$ when the system's equilibrium is reestablished.
I know I can calculate $K\_p$ and I know I somehow have to use $PV =NRT$, but from there on I am lost, I cannot figure out how to come up with enough equations to find all of my unknowns. Can anyone show me an explicit solution to the problem?
What I tried, but got really nowhere:
$$2I\_{(g)} \leftrightarrow I\_{2\_{(g)}}, K\_p=\frac{P\_1}{P\_2^2}=3,97.$$
$$P\_t=P\_1 + P\_2 = 0,44 \,\text{atm.}$$
When the new equilibrium is established :
$$Kp=\frac{P\_1'}{P\_2'^2}=3,97$$
$$\frac{2P\_1 }{P\_1'}=\frac{n\_1}{n\_1'}$$
and same with $P\_2, P\_2', n\_2, n\_2'$, but I don't know where to go from there.
| 0 |
[
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"\nThe method referred to in the comment is not correct and is a common mistake. One may think,\n\n\n$$p\\_1'+p\\_2'=p\\_t\\frac{V\\_t}{V\\_t'}=2p\\_t=0.88 \\rm atm,$$\n\n\nbut this is invalid since $n\\_1+n\\_2\\neq n\\_1'+n\\_2'$, thus $p\\_tV\\_t\\neq p\\_t'V\\_t'$\n\n\nThe correct way is to use the fact that the number of iodine atoms is constant, from which we get,\n$$2n\\_1+n\\_2=2n\\_1'+n\\_2',$$\n$$(2p\\_1+p\\_2)V\\_t=(2p\\_1'+p\\_2')V\\_t',$$\n$$1.3 {\\rm atm}=4p\\_1+2p\\_2=2p\\_1'+p\\_2',$$\n\n\nThen you can go on by combining this with the equation you have already obtained from $K\\_p$ (listed in your question) and solve for $p\\_1'$ and $p\\_2'$.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89264/why-do-we-use-jj1-instead-of-j%c2%b2-in-the-rigid-body-rotor-energy-equation
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Why do we use J(J+1) instead of J² in the rigid body rotor energy equation?
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I looked for the reason, but it seems to be taken for granted that it must be $J(J+1)$ instead of $J$.
I'm making use of the analogy of the particle in a box energy where
$$E=\frac{n^2h^2}{8mL^2}$$
| 10 |
[
[
"\nThis arises from the quantum mechanical properties of the angular momentum operator (*vide infra*). Since the operators for angular momentum and translational energy are different, there is no reason to expect a similarity between the form of the eigenvalues.\n\n\n\n\n---\n\n\n### Spectra of the angular momentum operators\n\n\nFirst we note the classical relationship between the angular momentum $L$ and the rotational kinetic energy\n\n\n$$E\\_\\mathrm{rot} = \\frac{1}{2}I\\omega^2 = \\frac{L^2}{2I} \\tag{1}$$\n\n\nsince $L = I\\omega$, with $I$ being the moment of inertia and $\\omega$ being the angular velocity. By analogy, the quantum mechanical Hamiltonian for rotational motion is\n\n\n$$\\hat{H} = \\frac{\\hat{L}^2}{2I} \\tag{2}$$\n\n\nwhere $\\hat{L}^2$ is the operator for (the square modulus of) orbital angular momentum. Because $L$ is a vector, we have\n\n\n$$\\hat{L}^2 = \\hat{L}\\_x^2 + \\hat{L}\\_y^2 + \\hat{L}\\_z^2 \\tag{3}$$\n\n\nwhere $L\\_x$ is the operator for the projection of orbital angular momentum onto the $x$-axis, and so on. The classical angular momentum is given by $\\vec{L} = \\vec{r} \\times \\vec{p}$; by expanding the cross product we can obtain the equations $L\\_x = yp\\_z - zp\\_y$, $L\\_y = zp\\_x - xp\\_z$, and $L\\_z = xp\\_y - yp\\_x$. Using these relations as well as the classical commutation relationship $[x,p\\_x] = [y,p\\_y] = [z,p\\_z] = \\mathrm{i}\\hbar$, we can derive the commutators of all four operators in equation $(3)$:\n\n\n$$\\begin{align}\n[L\\_x,L\\_y] &= \\mathrm{i}\\hbar L\\_z; & [L\\_y,L\\_z] &= \\mathrm{i}\\hbar L\\_x; & [L\\_z,L\\_x] &= \\mathrm{i}\\hbar L\\_y; \\tag{4} \\\\\n[L^2,L\\_x] &= 0; & [L^2,L\\_y] &= 0; & [L^2,L\\_z] &= 0 \\tag{5}\n\\end{align}$$\n\n\nConventionally, angular momentum eigenstates are chosen to be simultaneous eigenstates of $L^2$ and $L\\_z$: since $[L^2,L\\_z] = 0$, it is guaranteed that there is a complete set of states which are eigenstates of both operators. We denote these states by $|\\lambda,\\mu\\rangle$ where $\\lambda$ and $\\mu$ are related to the eigenvalues of these states:\n\n\n$$\\begin{align}\nL^2|\\lambda,\\mu\\rangle &= \\lambda\\hbar^2 \\tag{6} \\\\\nL\\_z|\\lambda,\\mu\\rangle &= \\mu\\hbar \\tag{7}\n\\end{align}$$\n\n\nIn this paradigm, we then define the raising and lowering operators $L\\_+$ and $L\\_-$ (the meaning will become clear in due time)\n\n\n$$L\\_\\pm \\equiv L\\_x \\pm \\mathrm{i}L\\_y \\tag{8} $$\n\n\nNow consider the effect of $L\\_\\pm$ on any arbitrary state $|l,m\\rangle$ which is simultaneously an eigenstate of $L^2$ and $L\\_z$. Since $L^2$ commutes with both $L\\_x$ and $L\\_y$, we have\n\n\n$$\\begin{align}\nL^2L\\_\\pm|\\lambda,\\mu\\rangle &= L\\_\\pm L^2|\\lambda,\\mu\\rangle \\\\\n&= L\\_\\pm \\lambda\\hbar^2|\\lambda,\\mu\\rangle \\\\\n&= \\lambda\\hbar^2 L\\_\\pm|\\lambda,\\mu\\rangle \\tag{9}\n\\end{align}$$\n\n\nor in other words, $L\\_\\pm|\\lambda,\\mu\\rangle$ is also an eigenstate of $L^2$ with the same eigenvalue $\\lambda\\hbar^2$ as $|\\lambda,\\mu\\rangle$ itself. Also, using the commutators in equation $(4)$,\n\n\n$$\\begin{align}\nL\\_z L\\_\\pm |\\lambda,\\mu\\rangle &= (L\\_zL\\_x \\pm \\mathrm{i}L\\_zL\\_y)|\\lambda,\\mu\\rangle \\\\\n&= [(\\mathrm{i}\\hbar L\\_y + L\\_xL\\_z) \\pm \\mathrm{i}(L\\_yL\\_z - \\mathrm{i}\\hbar L\\_x)]|\\lambda,\\mu\\rangle \\\\\n&= [\\hbar(\\mathrm{i}L\\_y \\pm L\\_x) + (L\\_x \\pm \\mathrm{i}L\\_y)L\\_z]|\\lambda,\\mu\\rangle \\\\\n&= [\\pm\\hbar(L\\_x \\pm \\mathrm{i}L\\_y) + (L\\_x \\pm \\mathrm{i}L\\_y)L\\_z]|\\lambda,\\mu\\rangle \\\\\n&= L\\_\\pm(L\\_z \\pm \\hbar)|\\lambda,\\mu\\rangle \\\\\n&= L\\_\\pm(\\mu \\pm 1)\\hbar|\\lambda,\\mu\\rangle \\\\\n&= (\\mu \\pm 1)\\hbar L\\_\\pm |\\lambda,\\mu\\rangle \\tag{11}\n\\end{align}$$\n\n\nwhich tells us that $L\\_\\pm|\\lambda,\\mu\\rangle$ is also an eigenstate of $L\\_z$, but with a *different* eigenvalue: the new eigenvalue is $(\\mu \\pm 1)\\hbar$. The raising operator (for example) therefore has the effect of conserving the total angular momentum $L^2$, but increasing the $z$-projection of this angular momentum $L\\_z$.\n\n\nHowever, for this to make physical sense, there has to be a limit to how far this can go: or else we will have no maximum value of $L\\_z$, even for a given value of $L^2$. Note that since $L\\_x^2$ and $L\\_y^2$ are both real and non-negative (which follows from the Hermiticity of $L\\_x$ and $L\\_y$), equation $(3)$ sets a maximum limit on $L\\_z$:\n\n\n$$L\\_z^2 \\leq L^2 \\tag{12}$$\n\n\nMathematically, the only way this can be fulfilled is if there is a \"highest\" state $|\\lambda,\\mu\\_\\mathrm{max}\\rangle$ which the raising operator turns into *zero*:\n\n\n$$L\\_+|\\lambda,\\mu\\_\\mathrm{max}\\rangle = 0 \\tag{13}$$\n\n\nNow consider\n\n\n$$\\begin{align}\nL\\_-L\\_+|\\lambda,\\mu\\_\\mathrm{max}\\rangle &= (L\\_x - \\mathrm{i}L\\_y)(L\\_x + \\mathrm{i}L\\_y)|\\lambda,\\mu\\_\\mathrm{max}\\rangle \\\\\n&= [L\\_x^2 + L\\_y^2 + \\mathrm{i}(L\\_xL\\_y - L\\_yL\\_x)]|\\lambda,\\mu\\_\\mathrm{max}\\rangle \\\\\n&= [L^2 - L\\_z^2 + \\mathrm{i}(\\mathrm{i}\\hbar L\\_z)]|\\lambda,\\mu\\_\\mathrm{max}\\rangle \\\\\n&= (L^2 - L\\_z^2 - \\hbar L\\_z)|\\lambda,\\mu\\_\\mathrm{max}\\rangle \\\\\n&= (\\lambda\\hbar^2 - \\mu\\_\\mathrm{max}^2\\hbar^2 - \\mu\\_\\mathrm{max}\\hbar^2)|\\lambda,\\mu\\_\\mathrm{max}\\rangle \\tag{14}\n\\end{align}$$\n\n\nHowever, we already know that $L\\_+|\\lambda,\\mu\\_\\mathrm{max}\\rangle = 0$. When $L\\_-$ acts on zero, it simply returns zero, and so the \"eigenvalue\" in equation $(14)$ must be equal to zero:\n\n\n$$\\begin{align}\n\\lambda\\hbar^2 - \\mu\\_\\mathrm{max}^2\\hbar^2 - \\mu\\_\\mathrm{max}\\hbar^2 &= 0 \\\\\n\\lambda &= \\mu\\_\\mathrm{max}^2 + \\mu\\_\\mathrm{max} \\\\\n&= \\mu\\_\\mathrm{max}(\\mu\\_\\mathrm{max}+1) \\tag{15}\n\\end{align}$$\n\n\nFundamentally, that is where the term comes from. There are several steps now to close some loopholes. I will not go through them in detail because it is far too long, but I will outline what is needed.\n\n\n1. You need to show that for orbital angular momentum the value of $\\mu\\_\\mathrm{max}$ is restricted to non-negative integer values; we define this quantity to be $l$ and we can replace the quantum number $\\lambda$ with $l$ because $\\lambda$ is related to $\\mu\\_\\mathrm{max}$, which is in turn equivalent to $l$. The eigenvalues of the total angular momentum operator $L^2$ are $\\lambda \\hbar^2 = \\mu\\_\\mathrm{max}(\\mu\\_\\mathrm{max}+1)\\hbar^2 \\equiv l(l+1)\\hbar^2$.\n\n\nNote that for *orbital* angular momentum only integer values of $l$ are allowed; half-integer values are not allowed. This condition can only be derived by solving the differential equations corresponding to the eigenvalue equations of the angular momentum operators (see further reading). For example, you need to solve this:\n\n\n$$\\begin{align}\nL\\_z f(x,y,z) &= (xp\\_y + yp\\_x)f(x,y,z) \\\\\n&= \\left(-x\\mathrm{i}\\hbar\\frac{\\partial}{\\partial y} - y\\mathrm{i}\\hbar\\frac{\\partial}{\\partial x}\\right)f(x,y,z) = \\mu\\hbar f(x,y,z) \\tag{16}\n\\end{align}$$\n\n\nHalf-integer values for spin are allowed because the exact form of the spin angular momentum operators are not known. However, they obey the same commutation relations as the orbital angular momentum operators, so the rules governing them are the same except for the constraint on integer values.\n2. You need to show that the allowed values of $\\mu$ are integers between $-l$ and $l$ respectively. We have already shown that $\\mu$ increases in integer steps, and that $\\mu \\leq l$, so what remains is to show that $\\mu \\geq -l$: the proof is exactly analogous to that in equations $(13)$ through $(15)$, except that you need to consider $L\\_+L\\_-|\\lambda, \\mu\\_\\mathrm{min}\\rangle$.\n\n\nConventionally the quantum number $\\mu$ is renamed to $m\\_l$ and the eigenstates are labelled with $|l,m\\_l\\rangle$. So now we have shown that $l$ is the maximal value of $m\\_l$, and that $m\\_l$ ranges from $-l$ to $l$ in integer steps, as we expected\n3. The total angular momentum $J$ is the sum of the orbital and spin angular momenta, i.e. $J = L + S$. For the rigid rotor system where you ignore spin, $J = L$ and hence we can simply replace $L$, $l$, and $m\\_l$ with $J$, $j$, and $m\\_j$: we therefore have $J^2|j,m\\_j\\rangle = j(j+1)\\hbar^2|j,m\\_j\\rangle$.\n4. Finally, the *energies* are then given by\n\n\n$$\\begin{align}\nE &= \\langle j, m\\_j|H|j,m\\_j \\rangle \\\\\n&= \\left< j,m\\_j \\middle| \\frac{J^2}{2I} \\middle| j,m\\_j \\right> \\\\\n&= j(j+1)\\frac{\\hbar^2}{2I} \\tag{17}\n\\end{align}$$\n\n\nwhere $j$ is a non-negative integer (i.e. $j = 0,1,2, \\ldots$).\n\n\n\n\n---\n\n\n### Further reading\n\n\nChapter 4 of [*Molecular Quantum Mechanics*, 5th ed., Atkins](https://global.oup.com/academic/product/molecular-quantum-mechanics-9780199541423?cc=gb&lang=en&).\n\n\nChapter 7 of [*The Physics of Quantum Mechanics*, Binney & Skinner](https://www-thphys.physics.ox.ac.uk/people/JamesBinney/qb.pdf) for an alternative derivation.\n\n\n",
"13"
]
] |
https://chemistry.stackexchange.com/questions/89262/zero-filling-leading-to-increases-in-resolution-nmr
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Zero filling leading to increases in resolution (NMR)
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Zero-filling involves adding data points with zero intensity to the end of an FID. According to this website <https://www2.chemistry.msu.edu/facilities/nmr/handouts/DH%20NMR%20Basics.pdf> zero filling can take a spectrum with low apparent resolution to one that is much more helpful like so[](https://i.stack.imgur.com/ANZj2.jpg)
In this instance, going from bottom to top corresponds with more zero filling. I do not understand how this happens. How does adding data points with zero intensity lead to increases in apparent resolution? I am certain I am missing some crucial point, but I can not pinpoint what that may be. For what its worth, I understand that we are not increasing the resolution of the instrument's measurements, rather we are manipulating the data to view information that would otherwise not be displayed.
| 9 |
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"\nThe key idea behind zero padding or zero filling is that the \"resolution\" or the *step size* of the frequency axis in the Discrete Fourier Transform is dependent on the number of points you have in the time domain. Zero filling is done before doing the Fourier transform. Thus after the FFT the resolution appears to be improved because you have more points in the spectrum and the spacing between each point appears to be decreased.\n\n\nThis is a result of a beautiful theorem (although this is casually mentioned in most textbooks), but never shown as a theorem. I found in Cooley's original paper from 1967 [[1](https://doi.org/10/bmw8z8), p. 80].\n\n\nCooley & Tukey are the key persons who made Discrete Fourier Transform (DFT) possible by computers for lowly mortals like us, otherwise it was an elitist subject among the mathematicians. Here is a snippet of it. It took me several years to find an original paper which showed this theorem. However, math researchers told me this was known in 1754. Full discussion here: [History of Integral Transform Theorem](https://mathoverflow.net/questions/335829/origin-of-the-theorem-related-to-the-integral-transform-pair/335842#335842)\n\n\n### References\n\n\n1. Cooley, J.; Lewis, P.; Welch, P. Application of the Fast Fourier Transform to Computation of Fourier Integrals, Fourier Series, and Convolution Integrals. *IEEE Transactions on Audio and Electroacoustics* **1967**, 15 (2), 79–84. <https://doi.org/10/bmw8z8>.\n\n\n",
"10"
],
[
"\nThis is pretty interesting. This is actually not a property of NMR or even something physical. Rather, this is a property of doing a discrete Fourier Transform. This is just a Fourier transform when the function is composed of points (as all data will be) rather than a continuous function.\n\n\nI will give as intuitive an explanation as I can and point you to [these lecture notes](http://ocw.usu.edu/Electrical_and_Computer_Engineering/Signals_and_Systems/lecture9.pdf) which I found helpful for the more mathematical treatment.\n\n\nBasically, if we take data which measures some amplitude versus time (the raw NMR signal) and Fourier transform to frequency space, then we get peaks which are the frequencies of the motions involved in the physical process. Now, when we do this Fourier transform, we usually have some resolution in mind that we would like to have. This resolution is essentially how far apart we are willing for each of our points to be. If the points are quite far apart, this means we did a very low resolution Fourier transform, and we will get plots looking like the bottom spectrum you have in the question.\n\n\nIntuitively, the faster we take samples, the better resolution we are going to have after we Fourier transform our data. There is [a theorem](https://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem) that says we actually only need to sample twice as fast as the highest frequency signal present in our data, in order to map the discrete data to a continuous signal of finite bandwidth (what we want our spectrum to simulate).\n\n\nNow, what we really want is to choose a resolution, $F\\_0$, and determine the number of data points needed from this resolution. This resolution, in practice, is determined by the quality of the instrument you are using. So, because we are thinking about our own NMR, what this means is that in order to get all of the resolution our machine can offer, we need to take data for a time $T\\_0$, given by,\n$$\nT\\_0=\\frac{1}{F\\_0}\n$$\n\n\nNow, here is where the physical processes enter. We have no control over how quickly nuclear spins flip, so the actual process we are measuring may be quite fast. We already said, however, that we need to have $T\\_0$ amount of signal in the time domain in order to get the full resolution from our NMR after the Fourier transform.\n\n\nThus, the solution is simply to append a bunch of zeros onto the end our signal, and then Fourier transform. We haven't added any data, or gone beyond the resolution of our instrument. All that we've done is performed a discrete Fourier transform properly.\n\n\nAlso, note that you don't actually need to append zeros until you reach a power of two number of data points, but the algorithms used (called Fast Fourier Transforms) work in such a way that they work best on numbers of data points which are a power of two.\n\n\nI hope this is somewhat clear.\n\n\n",
"9"
],
[
"\nAs has already been rightly pointed out, this is a property of the discrete Fourier transform. To supplement jheindel's answer, I will try to give a taster of the mathematics behind this. The conventional NMR books typically don't cover this in sufficient detail, so for a proper treatment it is a good idea to look in engineering books, specifically on signal processing. MIT OCW also has [an excellent course](https://ocw.mit.edu/resources/res-6-007-signals-and-systems-spring-2011/index.htm).\n\n\nWe start our journey at the Fourier *series*, instead of the transform. Consider a continuous-time Fourier series. (Obviously, NMR signals are discrete-time signals, and we will revisit that at the end of this post.) The point here is to express a periodic time-domain function, $f(t)$, as a sum of complex exponentials (or equivalently, sinusoids, although we will stick with the exponentials here). If this function has a period of $T\\_0$, then its Fourier series representation is\n\n\n$$f(t) = \\sum\\_{k = -\\infty}^{\\infty} c\\_k \\exp\\left(\\frac{\\mathrm{i} 2\\pi kt}{T\\_0}\\right) = \\sum\\_{k = -\\infty}^{\\infty} c\\_k \\exp(\\mathrm{i}k\\omega\\_0 t)$$\n\n\nwhere we define the fundamental frequency $\\omega\\_0 = 2\\pi/ T\\_0$. The Fourier coefficients are given by\n\n\n$$c\\_k = \\frac{1}{T\\_0}\\int\\_{T\\_0} f(t) \\exp\\left(-\\frac{\\mathrm{i} 2\\pi kt}{T\\_0}\\right) \\,\\mathrm{d}t.$$\n\n\nWe can plot these coefficients in the frequency domain. Each complex exponential $\\exp(\\mathrm{i}k\\omega\\_0 t)$ is associated with a frequency $k\\omega\\_0$. Since $k$ is restricted to be an integer, we only have discrete values of the frequency, and so the Fourier series can only be graphically represented as a series of points with associated heights $c\\_k$. In other words, we have a discrete function $F[\\omega]$ which is only defined for $\\omega = k\\omega\\_0$.\n\n\n\n\n\n(n.b. this is just for illustration purposes; the coefficients are not meant to be drawn accurately.) Now imagine the effect of padding your periodic time-domain function with extra regions between each period where $f(t) = 0$. That would increase the period $T\\_0$ and decrease the fundamental frequency $\\omega\\_0$. So our discrete function $F[\\omega]$ becomes more and more closely spaced.\n\n\n\n\n\nIn the limit of infinite padding, we have $T\\_0 \\to \\infty$ and $\\omega\\_0 \\to 0$, such that the function $F[\\omega]$ actually becomes a continuous version $F(\\omega)$. The sum becomes an integral and what we have is now a Fourier *transform*.\n\n\n\n\n\nWhat's important to notice here is that if you draw the Fourier transformed function $F(\\omega)$ over the Fourier series coefficients $F[\\omega]$, you will find that it fits perfectly over them. In other words, the Fourier series is a frequency-domain sample of the Fourier transform. This only holds true if your time-domain function doesn't get stretched or otherwise modified, so this framework is not directly capable of dealing with infinitely long time-domain functions, but it suffices since our time-domain FIDs have finite durations.\n\n\n\n\n\nSo, when we pad the periodic function with zero-valued regions, we are not fundamentally changing the frequency-domain information that is present. All we are doing is taking more closely spaced samples of the Fourier transform $F(\\omega)$, and this comes about only because the samples have a spacing of $\\omega\\_0 = 2\\pi/T\\_0$.\n\n\n**This is exactly analogous to zero-filling in NMR.** Imagine that our NMR spectrum is (one period of) the periodic function at the beginning. When you zero-fill an NMR spectrum, you are not actually generating any *new* information about the peaks that are present. If the (continuous) Fourier transform $F(\\omega)$ does not display a small coupling, then no amount of zero-filling your spectrum is going to reveal that small coupling.\n\n\n\n\n---\n\n\nBroadly speaking, the same considerations apply to discrete-time signals. The biggest thing to clear up is that the [discrete Fourier transform](https://en.wikipedia.org/wiki/Discrete_Fourier_transform) (DFT) is **not** analogous to the continuous-time Fourier transform. Instead, the DFT is analogous to the continuous-time Fourier **series**, the sole difference being that it operates on a discrete-time signal. The discrete-time analogue of a continuous-time Fourier transform is instead called the [discrete-time Fourier transform](https://en.wikipedia.org/wiki/Discrete-time_Fourier_transform) (DTFT).\n\n\nThis terminology is partly to blame for the confusion surrounding zero-filling. When you zero-fill a FID and \"Fourier transform\" it, you are not actually changing its *Fourier transform* (in the continuous-time sense); you are just calculating a more closely spaced *Fourier series*.\n\n\nTo sum up and to make the analogy clear:\n\n\n* In continuous time, as you pad a periodic function with zero-valued regions, the Fourier series becomes more closely spaced, tending towards the Fourier transform.\n* In discrete time, as you pad a periodic function with zeroes, the DFT becomes more closely spaced, tending towards the DTFT.\n\n\n",
"7"
],
[
"\nI wonder if the answers so far represent the complete picture. I remember reading that zero-padding up to a factor of 2 recovers information from the imaginary domain into the real-valued power spectrum and therefore can increase the resolution of Fourier-transform spectroscopic data by a factor of up to $\\sqrt{2}$. Unfortunately, I can't find the relevant reference anymore.\n\n\nPelczer and Szalma write in [*Chem. Rev.* **1991,** *91* (7), 1507–1524](https://doi.org/10.1021/cr00007a012): \"Zero filling once prevents loss of half of the information acquired.\" They cite, among others, [Bartholdi, E.; Ernst, R. R. *J. Magn. Reson.* **1973,** *11* (1), 9–19.](https://doi.org/10.1016/0022-2364(73)90076-0) Unfortunately, based on my limited physics background, I can't claim any deeper insights into topic.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89255/how-is-the-water-escaping
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How is the water escaping?
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I have a number of toy balls filled with a liquid-presumably water or something else non-toxic. Over about a year, the balls have gone from firm to indented as shown. The water level in one is visibly low but four others have become indented as if some water were removed without allowing air to take its place. There are no indications of any leaks, balls have all been maintained at room temperature.
I posted this to chemistry because my hypothesis is that there is some sort of diffusion through the ball surface or in to the floating toy contained within. In the case of the ball on the right, the toy inside appears to have a hole so maybe it was originally full of air but the liquid has migrated inside pushing the air out and appearing to lower the water level. Explanation doesn't work for the other balls.
Thoughts?
-Engineer Dad
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"\nThere is a general belief that plastics are impermeable, probably because many liquids are stored in plastic containers. This is, however, not correct. Plastics are permeable to water and also to gases. The permeability is low but is not zero and therefore, when very long periods of time are involved, a perceptible change occurs. If absolute impermeability is required, then metals should be used. \n\n\nThe process involved is diffusion and depends basically on the type of plastic. The ones containing fluorine (Teflon and the like) are the less permeable and the polyesters the most permeable, probably because they are more polar. \n\n\n[](https://i.stack.imgur.com/lkhy3.jpg)\n\n\nWhat has happened to the plastic balls is that some water has permeated out and in some cases some air has diffused in. \n\n\n",
"10"
],
[
"\nAre the balls exactly the same size as when purchased? Perhaps the skin of the ball has expanded by absorption of some of the contents. If this has happened, an indentation could occur even without escape of contents.\n\n\nOn the other hand, I have stored aqueous solutions in PETE bottles (from Diet Coke), and over a couple of years, water has evaporated out and the bottles have collapsed a bit.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89249/how-can-the-number-of-moles-in-the-product-side-be-more-than-reactant
|
How can the number of moles in the product side be more than reactant?
|
>
> A rigid stainless steel chamber contains $\pu{0.32 bar}$ of methane, $\ce{CH4}$, and excess oxygen $\ce{O2}$, at $\pu{200.0 °C}$. A spark is ignited inside the chamber, completely combusting the methane. What is the change in total pressure within the chamber following the reaction? Assume a constant temperature throughout the process.
>
>
>
I wrote out the following chemical equation:
$$\ce{CH4 + 2O2 -> CO2 + 2H2O}$$
I know I have $\pu{32 kPa}$ of methane, so using up all the moles of methane would mean that pressure decreases all the way to $0$.
The solution writes the following: it multiplies the 1 mol of $\ce{CH4}$ by $1$ to get the number of moles of $\ce{CO2}$ and by $2$ to get the moles of $\ce{H2O}$ produced.
Why can we do that? I would assume that by multiplying in such a way, you're implying that all of the $\ce{CH4}$ is being used up to create $1$ mol of $\ce{CO2}$. This means that you have no more moles to spare for creating $\ce{H2O}$.
I know my conceptual understanding is wrong, but this is what I thought while doing the question and the solution doesn't intuitively make sense to me.
I was wondering if someone could clarify my misunderstanding for me.
| 2 |
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"\nA mole is a way of counting *things*, in the case of chemistry those things are molecules. \n\n\nBut the number of things is a chemical reaction are not necessarily preserved: some reactions create more molecules (electrolysis of water (H2O), for example, starts with two molecules of water but delivers one molecule of oxygen gas (O2) and two molecules of hydrogen gas (H2)). \n\n\nSo moles are not *conserved* if the structure of the molecules changes because of the reaction. The point of writing equations is to match conservation of matter (atoms don't disappear) to the structures of the molecules involved in the reaction. Sometimes reactions just rearrange the atoms and give the same number of moles of product, sometimes they create more molecules, sometimes they create fewer molecules.\n\n\nIn an ideal gas there is a neat relationship between the number of moles and the volume or pressure (at constant pressure the volume is proportional to the number of moles of gas; at constant volume the pressure is proportional to the number of moles of gas). So your question is asking you to compare the number of moles *resulting* from the reaction to the number *going into* the reaction. since they are the same, the pressure won't change.\n\n\n",
"4"
],
[
"\nAre you supposed to consider that these are not ideal gasses? I ask this because if you are allowed to treat them as ideal each mole of methane reacts with 2 moles of oxygen to create 1 mole of CO2 and 2 of water. At 200°C and 0.32 bar the water isn't going to condense so 3 molecules react to form 3 molecules and as PV = nRT for an ideal gas there would be no pressure change.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89246/how-is-the-avagadros-constant-derived
|
How is the avagadros constant derived? [closed]
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**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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İ heard that :
Charge of a mole of electrons/charge of single electron =avagadros constant
How is this derived ?
| 0 |
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[
"\nThe key is to determine the charge of a *mole* of electrons vs the charge of *one* electron.\n\n\nWe get the charge of one electron from the [Milliken oil drop experiment](https://en.wikipedia.org/wiki/Oil_drop_experiment).\n\n\nFor a mole of electrons you can start uf you firzt get the molar mass of water, call this $18.016$ grams per mole. Now electrolyze water with an inert electrolyte such as sodium sulfate. Work out the charge transfer required to decompose one mole = $18.016$ grams. This is two moles of electrons based on the stoichiometry of the electrochemical reaction. Half of this divided by the single electron charge is Avogadro's number.\n\n\n",
"0"
],
[
"\nI believe it is the number of carbon atoms in 12 grams.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89244/vsepr-theory-chemical-bond-and-quantum-mechanics
|
VSEPR theory, chemical bond and quantum mechanics
|
VSEPR theory correctly predicts the shapes of many symmetry-broken molecules such as $\ce{H2O}$ and $\ce{NH3}$. Take $\ce{NH3}$ for example. In VSEPR theory, the nitrogen atom is (approximately) at the center of a tetrahedron, the three $\ce{N-H}$ bonds point to three of the four vertices of the tetrahedron, and the lone pair of nitrogen points to the $4$th vertex. But quantum mechanically speaking, the electrons should all be delocalized in the entire $\ce{NH3}$ molecule. How do I unify the two pictures to understand the concept of chemical bonds and VSEPR theory in quantum mechanics? Does VSEPR correspond to some kind of trial wave function (e.g. antisymmetrized geminal power (AGP))?
Note: when I say how to understand chemical bonds in quantum mechanics, I mean in the chemical bond description of molecules, electron pairs are localized at the bonds while quantum mechanics again says everything can be delocalized. So it's the same discussion as VSEPR v.s. QM. If there are only two atoms and one bond, the quantum mechanical meaning of the chemical bond is clear.
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"\n**VSEPR is a simple and generalised approach based** (mainly) **on empirical observation.** It is a great theory in predicting, in a first order approximation, the geometrical shape of molecules. That is a lot harder with other methods. It is of course based on a physical foundation, not only empiricism. You might want to explore electron-domains and the Pauli principle in this context.\n\n\nThere have been a lot of post-rationalisations that cause more harm than good, because they clearly go beyond the limitations of the theory. One of the most miss-taught ones is the involvement of d-orbitals in hybridisation schemes, but that comes from the lack of understanding of some instructors, and its confusion with valence bond theory. \n\nOne of the most prominent examples where it (basically) fails is outlined in my answer here: [Are the lone pairs in water equivalent?](https://chemistry.stackexchange.com/q/50906/4945) At this point it should be noted, that the general topology of the electron density is quite well reproduced.\n\n\nWith that in mind, a reconciliation of this empirical approach with quantum theory to understand bonding is dangerous, if not futile. It should be used for providing a reasonable guess for a molecular structure and its principle explanation. Anything beyond that might lead to wrong conclusions. It cannot, in any way, be used to generate a guess for a wave function, because it is not based on it.\n\n\nYou will have to use another method for that. Often confused with VSEPR is the valence bond theory (VBT). It cannot be stressed enough, that the two are completely independent (even at its crudest level). There has been a lot of development in that field, and what is often taught as VBT can only be classified as a very crude approximation. It is in principle an exact theory, but carried out at the theoretical limit is not as easily understood as that what is taught commonly. (Just have a look at resonance, and its misconceptions as are outlined here: [What is resonance, and are resonance structures real?](https://chemistry.stackexchange.com/q/51632/4945)) \n\nAnother way to describe bonding is molecular orbital theory (MOT), which already comes with the necessary properties of electron delocalisation. Unfortunately, this theory is more difficult to get started in, and does not provide an easy picture to follow. The good thing about this is, that it doesn't get much more complicated at its theoretical limit. \n\nThe two approaches are at their respective theoretical limit (VBT decribes electron correlation via resonance structures, MOT need multiple determinants for this) equivalent.\n\n\nUnderstanding bonding is not easy and it is by far not without controversy. Even at approximate VBT or MOT there can be many misconceptions, and incorrect deductions, conclusions, rationalisations. For everything in the *\"grey\"* areas, there are opinions, interpretations, and opinions about interpretations. \n\nIn any and every case one should always be aware of the limitations of the model used. One should also be critical about the found results. One should always expect everything to be a lot more complicated than expected (point in case: [CO2](https://chemistry.stackexchange.com/q/30797/4945)). \n\nIf you keep all of that in mind: **VSEPR is an awesome model system.**\n\n\n",
"8"
]
] |
https://chemistry.stackexchange.com/questions/89241/negative-gradient-graphs
|
Negative gradient graphs
|
So basically I have a graph of Absorbance on y-axis against Time on x-axis and it is a negative curve. So this will give me a negative slope gradient. My question is how to reason the negative gradient. Do I write down the rate of concentration as a negative value and just reason it by saying that by being negative it means that the concentration is decreasing?
| 2 |
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"\nYes. Beer's law tells us that the absorption is the product of path length, extinction and concentration. Thus concentration is absorption divided by the product of path and extinction. The rate of change of concentration (first derivative with respect to time of concentration is the rate of change of absorption divided by the product of path and extinction. If the rate of change of absorption is negative the rate of change of concentration is negative. Concentration is decreasing.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89240/bond-strength-of-pi-acceptors-and-metals-with-filled-dpi-orbitals
|
Bond strength of pi acceptors and metals with filled d(pi) orbitals
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I was reading in The Organometallic Chemistry of the Transition Metals (Robert Crabtree) that lewis acids like BF3 often accept electrons from d(pi) orbitals. Is this generally true for all pi acceptor ligands? That is, do metals with filled d(pi) orbitals form thermodynamically stable bonds with pi acceptor ligands? I would imagine that d(pi) electrons would only be available for donation if said metal was in a low oxidation state. Correct me if I'm wrong.
If this post merits two different question threads, I'd be happy to separate them.
| 4 |
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[
"\n**TL;DR** Yes, you need low oxidation states unless there is a significant $\\ce{M\\bond{->}L}$ σ contribution to the bonding.\n\n\n\n\n---\n\n\nThere are a few kinds of π acceptor ligands to distinguish between. These classifications are quite arbitrary and the line is not always clear, but their behaviour is sufficiently different and it is hopefully an instructive endeavour.\n\n\nFirstly you have ligands which are mostly σ donors, but are also capable of acting as π acceptors, albeit weak ones. Two common examples are the cyanide anion and alkyl phosphines. Because the $\\ce{M-L}$ bond is mostly donation from ligand to metal, you don't necessarily need particularly low oxidation states for these complexes to form: so $\\ce{[Fe(CN)6]^3-}$ is OK despite having a relatively high oxidation state of iron, $\\ce{Fe(III)}$.\n\n\nNext up you have ligands which are poor σ donors and pretty good π acceptors. Carbon monoxide falls into this category: it is not really all that good a σ donor and backbonding is often required for a stable $\\ce{M-L}$ bond to be formed (for some exceptions, google \"non-classical carbonyl complexes\"). Effective backbonding, in turn, requires high-energy d electrons on the metal. So, this usually precludes higher oxidation states. Many metal carbonyls have the metal in a negative oxidation state or zero, and $\\ce{[Fe(CO)6]^3+}$ most certainly doesn't exist.\n\n\nLastly you have weird ligands such as $\\ce{BF3}$ which are both σ and π acceptors. In the case of $\\ce{BF3}$, it is a σ acceptor via the empty p orbital on boron, and a π acceptor via the $\\ce{B-F}$ σ\\* orbitals. These are pretty obscure - I have never heard of them before - and are called Z-ligands ([Wikipedia](https://en.wikipedia.org/wiki/Z-Ligand)). Obviously any bond that forms must involve significant contribution of electron density from the metal. It turns out that as of 2011 there were no examples of unsupported $\\ce{M-BF3}$ bonding,1 but there are quite a few examples of similar molecules $\\ce{EX3}$ where $\\ce{E}$ is a heavier Group 13 element.\n\n\n[](https://i.stack.imgur.com/gw8r4.png)\n\n\nIn all these cases the metal fragment is very electron-rich and is again in a low oxidation state (zero or negative). I am sure the electron density on the metal required for these to form are even stricter than for carbonyls; otherwise we would undoubtedly have heard about $\\ce{[M\\_pZ\\_q]}$ complexes, since there are already lots of $\\ce{[M\\_p(CO)\\_q]}$ carbonyls known.\n\n\n\n\n---\n\n\n**Reference**\n\n\n1. Amgoune, A.; Bourissou, D. σ-acceptor, Z-type ligands for transition metals. *Chem. Commun.* **2011,** *47* (3), 859–871. [DOI: 10.1039/C0CC04109B](https://doi.org/10.1039/C0CC04109B).\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89238/charge-on-an-atom
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Charge on an atom [closed]
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[](https://i.stack.imgur.com/iAUik.jpg)
I am really confused at this part.The resonance as shown above causes the generation of a positive charge on nitrogen atom but the fact is that it's octet is still complete even after that resonance....what is the significance of this positive charge on nitrogen?What will it do to the nitrogen atom?
| 1 |
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"\nIt doesn't result in the generation of a positive charge on the nitrogen. It results in the generation of a formal charge on the nitrogen. The formal charge is defined as\n\n\nFC = V - N - B/2\n\n\nwith V being the number of valence electrons in the neutral, ground state atom, N being the number of valence electrons not involved in covalent bonds and B/2 being half the number of shared electrons in covalent bonds. Formal charge is the charge on the \"ion\" that is formed when it is assumed that all electrons \"lost\" are shared equally with the \"ion\" that captures them and it, in turn, lends an equal share of its bonding electrons.\n\n\nAt the left side nitrogen has 5 valence electrons. The pair is unshared and the other 3 participate in bonding. Were the bonds ionic and the electrons completely lost nitrogen would have a charge of +3. But we assume that the three that are lost are equally shared with the carbons that also contribute shares of their valence electrons so the formal charge on the nitrogen \"ion\" is 5 - 2 - 6/2 = 0.\n\n\nTurning to the carbon which is marked in the drawing with a formal charge I get\n4 - 1 - 3 = 0 so unless I'm not seeing something that's an error.\n\n\nNow when the unshared pair moves over to form the second bond nitrogen loses two electrons but gets back a half share in each (and still has the half share in the pair from the original single bond) so FC = 5 - 0 - 4 = 1 \nFor the carbon in this case: FC = 4 - 0 - 4 = 0\n\n\n",
"3"
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https://chemistry.stackexchange.com/questions/89222/physical-significance-of-magnitude-of-%ce%94g-change-in-gibbs-free-energy-of-a-reac
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Physical significance of magnitude of ΔG (change in Gibbs free energy) of a reaction [closed]
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**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
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In thermodynamics, we know that if $\Delta G$ of a reaction is negative, the reaction is spontaneous.
Suppose we are given with two spontaneous reactions ($\Delta G\lt0$) with corresponding magnitudes of $\Delta G$. Can we extract any other information from these magnitudes?
| 0 |
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"\nIt is only a rule of thumb that a negative value for $\\Delta G^0$ of a reaction means that a reaction is spontaneous. If you start out with pure reactants (and no products present), all reactions are spontaneous, irrespective of the $\\Delta G^0$. A negative value for $\\Delta G^0$ only signifies that the equilibrium constant is greater than 1. \n\n\n",
"1"
],
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"\nThe change in the Gibbs free energy for a process at a constant temp. and pressure is equal to how much non-expansion work it can do.\n\n\n",
"-1"
]
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https://chemistry.stackexchange.com/questions/89214/draw-a-diagram-to-show-the-shape-of-chloromethane-and-explain-why-it-has-this-sh
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Draw a diagram to show the shape of Chloromethane and explain why it has this shape. [closed]
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[](https://i.stack.imgur.com/tKQAD.jpg)
I know it has this shape, but I don't know how to explain it. Any help is appreciated, thanks.
| 2 |
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"\nCarbon has in his outer shell 4 electrons and needs 4 more electrons to get the full shell like the noble gas **Ne**.\n\n\n[](https://i.stack.imgur.com/PvtZbm.jpg)\n\n\nChlorine has in his outer shell 7 electrons and needs one electron to get the full shell like the noble gas **Ar**. \n\n\nI give you an answer because the question about the tetrahedral structure of Chloromethane contradicts the perceptions of physics about a s- orbital with 2 electrons and 3 p-orbitals with 2 electrons in each p-orbital. To avoid the contradiction between the claimed s- and p-orbitals and the empirical founded tetrahedral structure the [sp-hybridizations](https://en.wikipedia.org/wiki/Orbital_hybridisation) were postulated. The carbon in Chloromethane has four sp3-hybridizations with all atoms around.\n\n\nIf you have a look around you will see that the hybridizations are the norm. Let the s- and p-orbitals be a toy for physicists.\n\n\nWhat you have to check is, always to get pairs of electrons. That electrons like to be pairwise was expressed by the [Pauli exclusion principle](https://en.wikipedia.org/wiki/Pauli_exclusion_principle). It is a principle and not an explanation why this happens this way in atoms. But if you need an imagination of how the electrons form pairs *you* could think *for yourself* of the electrons as tiny bar magnets (they indeed have a [magnetic dipole moment](https://en.wikipedia.org/wiki/Electron#Fundamental_properties)) and two of them form a magnetic pair. This pairs around the carbon will be in some equilibration to each other. Since chlorine has a lot of other electron pairs it pushes the 3 hydrogen atoms. So you get this shape:\n\n\n[](https://i.stack.imgur.com/WVJGxm.jpg)\n\n\nAll images from Wikipedia\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89211/if-ethyl-bromide-is-treated-with-alcoholic-silver-nitrate-which-product-will-be
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How exactly does the reaction between alkyl halides and silver nitrite occur?
|
$$\ce{R-X + AgNO\_2 -> RNO\_2}$$
Now depending on the $\ce{R}$ and solvent the reaction can occur via $\mathrm{S\_N1}$ or $\mathrm{S\_N2}$.
When the reaction proceeds via $\mathrm{S\_N1}$ the product formed is $\ce{R-O-N=O}$ whereas for $\mathrm{S\_N2}$ it is $\ce{R-NO\_2}$.
Why is this so? According to me as the $\ce{Ag-O}$ bond is covalent the lone pairs on nitrogen must attack in both the cases forming the product $\ce{R-NO\_2}$.
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"\n\n> \n> *The concept of hard and soft acids and bases (HSAB) proved to be\n> useful for rationalizing stability constants of metal complexes. Its application\n> to organic reactions, particularly ambident reactivity, has led\n> to exotic blossoms. By attempting to rationalize all the observed\n> regioselectivities by favorable soft–soft and hard–hard as well as\n> unfavorable hard–soft interactions, older treatments of ambident\n> reactivity, which correctly differentiated between thermodynamic and\n> kinetic control as well as between different coordination states of ionic\n> substrates, have been replaced. By ignoring conflicting experimental\n> results and even referring to untraceable experimental data, the HSAB\n> treatment of ambident reactivity has gained undeserved popularity. In\n> this Review we demonstrate that the HSAB as well as the related\n> Klopman–Salem model do not even correctly predict the behavior of\n> the prototypes of ambident nucleophiles and, therefore, are rather\n> misleading instead of useful guides. An alternative treatment of\n> ambident reactivity based on Marcus theory will be presented.*\n> \n> \n> — H. Mayr’s, M. Breugst’s and A. Ofial’s introductory remark in their review pulished in *Angew. Chem. Int. Ed.* **2011**, *50*, 6470. DOI: [10.1002/anie.201007100](http://dx.doi.org/10.1002/anie.201007100).\n> \n> \n> \n\n\nWithin the review, in the section on the nitrite ion, the authors give two competing effects that control product distribution.\n\n\nFor electrophiles with an electrophilicity parameter $E < -3$ (the *bis*-(4-methoxyphenyl)carbenium ion has $E = 0$, lower values are more stabilised carbocations such as *bis*-(4-dimethylaminophenyl)carbenium) the reactions proceed at equilibrium under thermodynamic control and the product yields are equivalent to their relative stabilities. Since nitro compounds are more stable than nitrite esters, they predominate.\n\n\n$$\\ce{R-ONO <=> R+ + NO2- <=>> R-NO2}$$\n\n\nFor electrophiles with a parameter $E > 0$, the reactions are essentially diffusion controlled, i.e. the free electrophile will capture the first nucleophile it can find. This includes tertiary alkyl carbocations ($E \\approx 7{-}8$). The reactions **do not** go through classical transition states, and any means of describing them *via* frontier orbitals must fail. **That explicitly includes describing them according to $\\mathrm{S\\_N1}$ or $\\mathrm{S\\_N2}$.** In these reactions, kinetic control determines that the $\\ce{O}$-attack is slightly more likely with $k\\_\\ce{O} / k\\_\\ce{N} \\approx 3.4$. (This corresponds to computational models that show a slightly smaller activation barrier for the oxygen attack.)\n\n\nReality is more complex than these simple models imply especially for the diffusion controlled reaction. While the attack of oxygen is favoured kinetically, the attack of nitrogen results in less required reordering of the molecule (the two $\\ce{N-O}$ bonds would still be identical while they become $\\ce{O-N=O}$ in the nitrite ester) which reduces the expected favourism for oxygen. Effects such as solvent and counterion slightly affect the product distribution, and the highest ratio of nitromethane : methyl nitrite can be achieved using $\\ce{Bu4N+ NO2-}$ and $\\ce{MeI}$ in $\\ce{CDCl3}$ (a product ratio of $70\\,:\\,30$ in favour of nitromethane). Note that $\\ce{AgNO2}$ and $\\ce{MeI}$ in DMSO yield more methyl nitrate ($54\\,:\\,46$) although the same electrophile is used. The authors conclude the nitrite section with the statement:\n\n\n\n> \n> It should be noted, however, that the selectivities\n> shown in [the table presented in the review] are also not related to the hardness of the electrophiles.\n> \n> \n> \n\n\n\n\n---\n\n\n**Note**\n\n\nThroughout this answer, electrophilicity parameters according to Mayr’s equation are used:\n\n\n$$\\lg k\\_{20~\\mathrm{^\\circ C}} = s(E + N)$$\n\n\n$E$ is a nucleophile-independent electrophilicity parameter while $s$ and $N$ are electrophile-independent nucleophilicity parameters.\n\n\n\n\n---\n\n\n**Bibliography**\n\n\nA paper related solely to studying the nitrite anion: A. A. Tishkov, U. Schmidhammer, S. Roth, E. Riedle, H. Mayr, *Angew. Chem. Int. Ed.* **2005**, *44*, 4623. DOI: [10.1002/anie.200501274](http://dx.doi.org/10.1002/anie.200501274).\n\n\nThe review mentioned in the opening, concerning all sorts of ambient nucleophiles often explained *via* the HSAB concept: H. Mayr, M. Breugst, A. Ofial, *Angew. Chem. Int. Ed.* **2011**, *50*, 6470. DOI: [10.1002/anie.201007100](http://dx.doi.org/10.1002/anie.201007100).\n\n\n",
"6"
],
[
"\nPearson's [HSAB theory](https://en.wikipedia.org/wiki/HSAB_theory) provides a simple explanation. When the reaction proceeds via $S\\_N1$, a hard acid (electrophile) is firstly generated (the carbocation) which interacts with the harder part of the nitrite anion an that's the oxygen-hard nucleophile. The $S\\_N2$ proceeds smoothly for the pair soft nucleophile–soft electophile. \n\n\nIn the given example the soft electrophilic carbon reacts with the soft center of the nitrite anion, and that's the nitrogen atom. In the former case, the silver coordinated to the oxygen definitely weakens its nucleophilicity. The formed carbocation has to be pretty hard and not so much stabilized by the solvent for the O-attack to be operative.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89207/reaction-characteristics-of-alkaline-battery
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Reaction characteristics of alkaline battery [closed]
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**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
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I'm currently struggling with understanding why electrochemical reactions take place the way they do.
For example, in an alkaline-manganese battery manganese-dioxide is getting reduced by reacting with water whereby the electron is actually absorbed by the hydroxide molecule (?):
$$\ce{MnO2 + H2O + e- -> MnO(OH) + OH-}$$
Essentially, my question is why does manganese give up the bond with the two oxygens and instead bonds with one oxygen which itself is bonded with an hydroxide *and* why does the other hydroxide absorb the electron instead of the other molecules.
Would manganese-dioxide and water also react without the transferred electron?
| -1 |
[
[
"\nChemistry can be difficult because it has lots of pieces of different importance, some of which are not depicted in equations. In this case, the important pieces are the ones that undergo the greatest change.\nThe hydrogens don't change much at all - they are always bonded to oxygen. The oxygens have their outer shells filled, and they sit in a crystal lattice (MnO2) or a liquid (H2O); there's a change, but it's moderate (in this case, anyway). \n\n\nBut the manganese changes from +4 to +3! Big change! It took a lot of energy to rip 4 electrons off the Mn, so there is a driving force to get at least one back. The MnO2 would become MnO2-; this is a nice place for a proton. The proton comes from a water molecule, leaving OH-. \n\n\nMnO2 + H2O don't react together. MnO2 is stable unless it can oxidize something and water is quite inert. \n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89203/is-pdcl2pme32-diamagnetic-or-paramagnetic
|
Is [PdCl2(PMe3)2] diamagnetic or paramagnetic?
|
Please tell me where I am wrong. The complex will have $\ce{Pd^{2+}}$ ion, which has a $\mathrm{d^8}$ configuration. So, it will have $2$ unpaired electrons. Unpaired electrons will mean that it is paramagnetic.
My reference book has this line:
>
> $\ce{[PdCl2(PMe3)2]}$ is a diamagnetic complex of Pd(II)
>
>
>
Shouldn't it say "paramagnetic"?
| 2 |
[
[
"\nYou have not considered the d orbital splitting which has occurred due to the presence of the bonded ligands. Yes, the $\\ce{Pd^2+}$ ion adopts the $\\mathrm{4d^8}$ electronic configuration with two unpaired electrons, as shown below. However, when it forms the square planar complex, the d orbitals split in energy levels and the electrons now occupy the new energy levels differently, still abiding by Hund's rule and the Aufbau principle. These d orbitals no longer possess any unpaired electrons and thus, the complex is not paramagnetic, but diamagnetic. \n\n\n[](https://i.stack.imgur.com/IJ3Is.png)\n\n\nImage source: [Wikimedia Commons](https://commons.wikimedia.org/wiki/File:Electron-config-d8-s2.png)\n\n\n[](https://i.stack.imgur.com/YjJzG.png)\n\n\nImage source:[Wikimedia Commons](https://upload.wikimedia.org/wikipedia/commons/a/a6/Chem507f09sqvstet2.png)\n\n\n",
"4"
],
[
"\nthis is a special case of jahn teller distortion....in pd(II), weak ligands can remove the degeneracy of eg orbitals & as the energy difference between two eg orbitals are high, the ekectrons pair up in the dz2 orbitals. Thus form a diamagnetic complex [](https://i.stack.imgur.com/6NqTh.jpg)\n\n\n",
"-4"
]
] |
https://chemistry.stackexchange.com/questions/89201/concentration-of-solute-and-solvent
|
Concentration of Solute And Solvent [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed 5 years ago.
[Improve this question](/posts/89201/edit)
Say I create a solution consisting of one liter of motor oil dissolved in three liters of acetone, and I leave the solution uncovered and allow it to evaporate.
As the solution evaporates, will the percentages of the solute and solvent remain the same?
| -1 |
[
[
"\nThe mixture, originally 4L of 25% motor oil, loses acetone by evaporation until you have 1L of 100% motor oil. The % of motor oil increases from 25% to 100% as the acetone evaporates off.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89196/gelatination-of-solution-in-saponification
|
Gelatination of solution in saponification
|
When I performed a saponification experiment in school, I added some ethanol to the solution (glycerol ester fatty acid + sodium hydroxide) to change the equilibrium of the reaction. (Molecules of salt of fatty acid made some chunks so they don't react efficiently.) But accidentally I spilled a lot of ethanol and the solution became hard, like a hair gel. And after I added some water to it, it became liquid again.
What I want to know is why the solution became a gel. I know that ethanol is used for an efficient reaction because ethanol can be a common solvent of grease and water and it can help sodium hydroxide and glycerol ester meet, but I don't know why they became harder when ethanol was overused.
Of course, the solution was in the water bath.
| 0 |
[
[
"\nSodium stearate is not very soluble in water (above about 2%, it becomes viscous but not a firm gel). Ethanol can reduce the viscosity at first, but sodium stearate is not soluble in ethanol either, so an excess of ethanol causes a precipitation of solid that firms up the aqueous gel. More water dissolves that precipitate and restores the fluidity.\nThe solubility and rheology of fatty acid soap solutions depends on temperature, concentration, solvents and the structure and purity of the fatty acid group. Stearate is very different from oleate, and both are different from laurate.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89191/what-does-forces-mean-in-the-output-of-gaussian-calculations
|
What does 'forces' mean in the output of Gaussian calculations?
|
In the output of a Gaussian calculation of frequencies, I can see a table listing all the 'forces' of all the atoms. It looks like this:
----------------------------------------------------------------------------------------------------------------------------------------
Center Number Atomic Number Forces (Hartrees/Bohr) X Y Z
---
```
1 6 -0.001094502 -0.000768130 -0.006919164
2 6 0.000528733 -0.000637636 0.005949765
3 6 0.000424174 -0.000366750 -0.002904382
4 6 -0.002619381 -0.000001885 0.005496979
```
...
What are these 'forces'? Are they related to those vibrations? But during vibrations, the force on each atom by other atoms is always changing, so why is only one constant 'force' value assigned to each atom?
| 4 |
[
[
"\nThe forces you have listed are the derivatives of the energy with respect to a displacement of a particular atom in each Cartesian direction. These forces are useful for finding stationary points on the potential energy surface, which occur when the force on each atom is zero. Computationally, it's not possible or worthwhile to try to get all the forces to be exactly zero, so stationary points are determined by a combination of [criteria](http://www.cup.uni-muenchen.de/ch/compchem/geom/basic.html) involving the forces and the displacement between optimization steps.\n\n\nThe vibrations in the system are related to the second derivatives of the energy with respect to the atomic coordinates, which are contained in the Hessian matrix. \n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89186/relationship-between-yield-and-applied-potentials
|
Relationship between Yield and Applied Potentials
|
This question revolves around this question from Atkins' *Chemical Principles* 5th ed., Chapter 13, Question 13.116:
>
> Consider the electroplating of a metal $+1$ cation from a solution of unknown concentration according to the half-reaction $\ce{M+ (aq) + e- -> M (s)}$, with a standard potential $E^\circ$. When the half-cell is connected to an appropriate oxidation half-cell and current is passed through it, the $\ce{M+}$ cation begins plating out at $E\_1$. To what value ($E\_2$) must the applied potential be adjusted, relative to $E\_1$, if $\pu{99.99 \%}$ of the metal is to be removed from the solution?
>
>
>
How is this problem solvable? I know that at applied potential $E\_1$ the cell potential is $0$ ($\Delta G = 0$), so that means at $E\_2$, the equilibrium constant must demand a yield of $\pu{99.99 \%}$. However, doesn't this yield depend on what the oxidation half-cell is?
For example, the oxidation reaction could be one from cerium(III) to cerium(IV) or from chloride ions to chlorine gas, both of which have different effects on the equilibrium constant. I've gotten as far as $-nFE = RT\ln{K}$, where I do believe $E = E\_2 - E\_1$ and $E$ is positive to ensure negative Gibbs free energy change.
This is an even number question so there is no given answer. I would really appreciate a solution or at least explanations clearing up any misconceptions that I may have / clues to how to solve this problem. This is not a homework problem.
| 1 |
[] |
https://chemistry.stackexchange.com/questions/89184/d6h-symmetry-clusters-orbital-classification-confusion
|
D6h symmetry clusters- orbital classification confusion!
|
I was am simulating high symmetry clusters D6h in several programs such as molcas and gaussian. I understand that the highest symmetry used in these programs is D2h and therefore there would be mixing of the different symmetry representations upon correlating from D6h to D2h. However upon simulating Mn@Si12 cluster, I'm seeing that the silicon atoms in the silicon cage have for all representations: s, px, py and pz orbitals despite there being restrictions according to D2h e.g b3u representation has only px functions and b2u only having py functions. Could anyone please shed some light on this phenomenon?
| 7 |
[
[
"\nThe symmetry operations are usually defined with respect to the origin of the coordinate system. For single atoms one naturally chooses $(0,0,0)$, which coincides with the symmetry center.\n\n\nFor example the $\\sigma\\_{xy}$ mirror plane in matrix representation (with $(0,0,0)$ being the symmetry center) would be:\n\n\n$\\sigma\\_{xy}=\\pmatrix{1&0&0\\\\0&1&0\\\\0&0&-1}$\n\n\nwhich means: invert every $z$ component. If applied to some object centered on $(0,0,0)$ (e.g. the AO of an atom on this position), the result would still be centered on $(0,0,0)$.\n\n\nIn case of molecules we can have at most one atom sitting on the symmetry center. Sometimes none, e.g. in $\\ce{C2}$ where the symmetry center is in the middle between both atoms. If you apply the above matrix here (assuming the molecule axis is aligned with the $z$ axis), then it would map the AO from one to the other atom. Therefore it is not symmetric (with respect to the above defined symmetry operation) anymore.\n\n\nThus you cannot relate the IRREP of an atomic orbital in the single atom, with its IRREP in the molecule. The symmetry center changed! For example take an $s$ atomic orbital of any of your $\\ce{Si}$ atoms. The position of the atom does not coincide with the center of the molecule. Therefore any symmetry operation (except for $E$) would map the $s$ orbital to some other $\\ce{Si}$. The AO is no longer characterized by an *irreducible* representation.\n\n\nWhat the code does instead, is first generating Symmetry Adapted Linear Combinations (SALC) of the atomic orbitals. Those are then again characterized by IRREPs (of the molecule) and used for the calculation. Now each obtained MO is indeed only constructed from SALC of the same IRREP. In other words: the Hamiltonian, and therefore the MO coefficient matrix as well, become block diagonal. However, in the output the MO coefficients are transformed back from the SALC basis set to the AO basis set, which is not block diagonal and may contain contributions from any AO for each IRREP of the MOs (an exception would be the AOs of an atom located in the center of the molecule).\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89182/concentration-of-electrons-in-silicon-sample-with-a-phosphorus-dopant
|
Concentration of electrons in silicon sample with a phosphorus dopant
|
I'm studying electrical and computer engineering and all I remember from chemistry is from high school. We have a $\ce{Si}$ sample and we want to replace one in every million atoms with a phosphorus atom.
I want to determine the concentration of conduction electrons after this process and I'm given the molar mass $\pu{0.028 kg/mol}$ and the density $\pu{2300 kg/m^3}$.
It might be very simple, but what can I get from these two? I checked molar mass since I haven't heard about it ever, but I'm getting more confused.
| -1 |
[
[
"\nUnless there is something I'm missing entirely, the answer is quite straightforward. Lets assume that each phosphorus atom contributes one free electron. Then the concentration of electrons is equal to the concentration of $\\ce{P}$ atoms doped to the silicon crystal:\n\n\n$$c\\_\\mathrm{e} = c(\\ce{P})\\label{eq:1}\\tag{1}$$\n\n\nConcentration of phosphorus $c(\\ce{P})$ can be found from its amount $n(\\ce{P})$ and volume $V$:\n\n\n$$c(\\ce{P}) = \\frac{n(\\ce{P})}{V}\\label{eq:2}\\tag{2}$$\n\n\nOn the other hand, volume can also be determined from the mass $m$, density $\\rho$ and molar mass $M$ of silicon:\n\n\n$$V = \\frac{m(\\ce{Si})}{\\rho(\\ce{Si})} = \\frac{n(\\ce{Si})\\cdot M(\\ce{Si})}{\\rho(\\ce{Si})}\\label{eq:3}\\tag{3}$$\n\n\nNow substituting unknown parameters in \\eqref{eq:1} and \\eqref{eq:2} with \\eqref{eq:3}, and knowing that $n(\\ce{P}) : n(\\ce{Si}) = 1 : 10^6$, concentration of electrons can be found using the values you provided:\n\n\n$$c\\_\\mathrm{e} = \\frac{n(\\ce{P})}{n(\\ce{Si})} \\cdot \\frac{\\rho(\\ce{Si})}{M(\\ce{Si})} = 10^{-6}\\cdot\\frac{\\pu{2300 kg m-3}}{\\pu{0.028 kg mol-1}} = \\pu{8.21e-2 mol m-3}\\tag{4}$$\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89174/what-is-equivalents-per-mole
|
What is equivalents per mole?
|
>
> Which has maximum number of equivalents per mole of the oxidant?
> $$
> \begin{align}
> \ce{MnO4- + H2O2 &-> Mn^2+ + O2}\tag{i}\\
> \ce{Cr2O7^2- + Fe^2+ &-> 2 Cr^3+ + Fe^3+}\tag{ii}
> \end{align}
> $$
>
>
>
The answer is the second option. I wanna know why
If there's another term different from equivalent mass like equivalents, then I tried doing it using the formula = No.of moles of oxidant x n factor (no of electrons gained)
| 0 |
[
[
"\nThe question can be understood as:\nWhich one of the following oxidants has a greater n-factor as there's only 1 mole each of permanganate ion and dichromate ion.\n\n\nThe n-factor of permanganate ion is (7-2)=5\n\n\nwhile \n\n\nThe n-factor of dichromate ion is (6\\*2-3\\*2)=6\n\n\nSo, the answer is the 2nd reaction.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89167/why-cant-amylase-digest-glycogen
|
Why can't amylase digest glycogen?
|
Amylase is an enzyme that breaks down starch in the form of amylopectin and amylose. Both amylose and amylopectin are formed by alpha glucose joined together by (1-4) and (1-6) glycosidic bonds. Glycogen is no exception, just that it has more branching.
However, why is it that a Google search shows that it is hydrolyzed by glycogen phosphorylase rather than amylase? Also, how can amylase digest both (1,6) and (1,4) glycosidic bonds?
| 3 |
[
[
"\nAmylase can’t digest glycogen because of its inability to attack the branching (1→6) linkages. \n\n\nPerhaps, another very important reason is controlling the rate of glycogen metabolism through **glycogen phosphorylase**. Just like any other biological system, regulation of metabolic substrates and/products is crucial to maintaining the balance (homeostasis) so to prevent excess glucose production from glycogen metabolism or to little according the needs of the organism. \n\n\n\n\n---\n\n\nTl;dr\n\n\nThere are actually three forms of amylases:\n\n\n* α-amylase (an endoglycosidase, which can hydrolyze a glycosidic linkage anywhere along the chain to produce glucose and maltose).\n* β-amylase (an exoglycosidase that cleaves from the nonreducing end of the polymer).\n* gamma amylase\n\n\nNow starch consists of two main components amylose and amylopectin; \n\n\n* Amylose is a linear polymer of several thousand glucose residues linked by (1→4)\nbonds:\n\n\n[](https://i.stack.imgur.com/FisXD.png)\n\n\n* Amylopectin consists mainly of (1→4)-linked glucose residues but is a branched molecule with (1→6) branch points every 24 to 30 glucose residues on average.\nThe primary structure of **glycogen resembles that of amylopectin**(not amylose), but glycogen is more highly branched, with branch points occurring every 8 to 14 glucose residues:\n\n\n[](https://i.stack.imgur.com/2KOFE.png)\n\n\n**Digestion**\n\n\nThe digestion of starch, the main carbohydrate source in the human diet, begins in the mouth. Saliva contains α-amylase, which randomly hydrolyzes all the (1→4) glycosidic bonds of starch **except its outermost bonds and those next to branches**. \nOn the other hand glycogen being a highly branched molecule, it is evident that amylase won’t be a good enzyme to digest it due to physical structure constraints.\n Substrate specificity of the enzymes ensure that the appropriate substrate which is highly optimized for the active site will be broken down. \n\n\nAlthough glycogen posseses parts similar to starch's amlyopectin its structure is not optimized to be broken down by amylases. *Note: even these amylases only partially break down starch to different oligosaccharides* \n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89165/why-does-silver-plating-require-undercoats-of-copper-and-nickel
|
Why does silver plating require undercoats of copper and nickel?
|
I came across a question that was about silver plating (of steel), but I couldn't help but wonder why the sequence was illustrated as this:
step 1 A coating of copper is applied to the object.
step 2 A coating of nickel is applied to the object.
step 3 The coating of silver is applied to the object.
Why are these initial coats of copper and nickel required, and why this seemingly random sequence (since nickel is in between but has the highest reactivity)?
| 4 |
[
[
"\nI have read that the nickel plating helps removing oxide layers when working with stainless steel which is beneficial for the silver plating process.\n\n\nSadly I do not know why the copper is used but I suspect the context could explain the copper.\n\n\n",
"1"
],
[
"\nI never saw an explanation, only that ( historically) that is the way that works. I expect it has something related to getting a good bond . Although the nickle makes the substrate ( copper) \"silver\" color so the silver layer does not need to be as thick. Both the copper and nickle also provide corrosion protection to the steel. Chromium plating is done the same way.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89163/why-the-dmm-multimeter-register-negative-resistance-when-measure-a-drop-of-hyd
|
Why the DMM (multimeter) register negative resistance when measure a drop of hydrochloric acid [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed 5 years ago.
[Improve this question](/posts/89163/edit)
I noticed that when I put the multimeter electrodes (in the ohmmeter mode) in a drop of hydrochloric acid with mild concentration (15%) the multimeter displays negative values. Can someone explain this behavior?
| 0 |
[
[
"\nThe multimeter probe tips are dissolving and setting up a galvanic cell that augments the current flow thru the ohmmeter. \nThe probe tips may be of the same metal, but the ohmmeter voltage will favor the dissolution of one over the other, and the cell thus formed will assist the applied voltage.\nQuickly switching to voltmeter (or milliammeter) mode will show a cell voltage (current).\nUsing the ohmmeter to measure resistance of a non-corrosive solution (e.g., NaHCO3) will show some (positive) resistance.\nInert electrodes (Pt, Au) will show a positive resistance.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89158/is-there-an-example-where-interpreting-reaction-coordinate-as-time-leads-to-an-i
|
Is there an example where interpreting reaction coordinate as time leads to an incorrect conclusion?
|
I have heard that we should not interpret the reaction coordinate in an energy diagram as time, but are there any situations in which we would arrive at an incorrect conclusion if we did so?
When I say interpreting reaction coordinate as time, I really mean as $\frac{\text{time so far}}{\text{total time for reaction}}$ (I would have written this instead of "time", but I'm not sure of a word or phrase for this).
| 6 |
[
[
"\nTransition states are very short lived species, generally lasting less than a picosecond. At the single molecule level, crossing a transition state is a very fast event: it goes from reactants to products in less than a picosecond.\n\n\nSome reactions are very slow because the probability of crossing the transition state is very low, but never because they spend a long time at the transition state. Thus, interpreting reaction coordinate as time is incorrect.\n\n\nIn a slow reaction, a small number of molecules will react early, but the majority will spend a long time in the reactants state, waiting to cross the transition state. If one could take a static picture of an ensemble of molecules, some would be in the reactants state, some would be in the products state, and it would be almost impossible to see any molecule at the transition state.\n\n\n",
"3"
],
[
"\nOne of the oldest reactions to be understood in modern terms qualifies. Heat mercury in the presence of oxygen, the mercury is first oxidized only to be reduced again at higher temperature as the equilibrium constant shifts. The reaction coordinate, defined as the amount of mercury oxidized, doubles back as time advances in this nonisothetmal process.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89154/can-we-use-liquid-hydrogen-to-power-a-conventional-engine
|
Can we use liquid hydrogen to power a conventional engine? [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
---
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Closed 5 years ago.
[Improve this question](/posts/89154/edit)
Hydrogen has the best fuel efficiency, so if we were to use it in our engines wouldn't their efficiency increase as well.
The engines would require changes and all but would it be plausible to have that?
| 1 |
[
[
"\n**The efficiency that matters is the efficiency of the *vehicle* not the efficiency of the *engine***\n\n\nThe problem of designing more efficient transport is bigger than the problem of designing better engines. Conventional combustion engines can be adapted to burn a wide variety of fuels (alcohols, natural gas and hydrogen are not too big a challenge). And there are alternative ways to move vehicles than combustion engines (fuel cells, for example, which run on hydrogen). So it looks like we can solve the engine problem.\n\n\nBut that isn't enough. To have a viable vehicle we have to solve other problems as well. How to store and deliver the fuel to the engine, for example. And how to distribute the fuel across the country so vehicles can be refilled. The second problem is probably easier to solve if the first problem–a viable vehicle–has a solution. Unfortunately, it isn't that easy.\n\n\nTo have a hydrogen powered car we would need a way to store enough hydrogen in the vehicle to allow it to drive several hundred miles. \n\n\nThis is easy to do with liquid hydrocarbon fuels in conventional vehicles (petrol/gasoline and diesel have very high energy densities and can be stored in a simple unpressurised tank). The weight of the fuel alone is larger then the weight of the tank that contains it and the volume is small as the energy density is very high. \n\n\nIt isn't so easy with hydrogen. While there are several existing ways to store hydrogen and some proposed solutions that are not yet operational, all of them involve big issues with volume and weight. Compressed hydrogen will involve very heavy thick-walled tanks and even then the energy density of the hydrogen gas alone will be lower than a tank full of petrol. The weight and volume of the tanks required to store as much energy as a tank of petrol will very large. Liquid hydrogen has a different set of compromises but isn't much better as it requires cryogenic system to keep the liquid hydrogen cool. There is considerable interest in clever chemical ways to store hydrogen (clathrates, metal organic frameworks and various molecular systems than can reversibly store hydrogen are all being researched).\n\n\nTo get an idea of the scale of the challenge it is worth looking at the mass and volumetric characteristics required to compete with conventional fuel. Diesel (slightly more efficient than petrol) carries about 12kWh/kg and about 10kWh/L. The current *targets* for research on hydrogen storage for the [Fuel cell Technology Office](https://energy.gov/eere/fuelcells/hydrogen-storage) are 1.5kWh/kg and 1kWh/L. This is about 10 times worse than diesel even ignoring the fact that fuel-cell power vehicles might have other advantages partially compensating the overall vehicle when compared to combustion engines. And the best technology solutions are much better than liquid hydrogen storage.\n\n\nSo you can use liquid hydrogen (or other ways of storing hydrogen) to power an conventional engine or vehicle (and such vehicles exist). But the cost and design compromises to make this work for a whole vehicle given that you need 10 time the volume and weight to store the equivalent energy, make such solutions not competitive for mass market vehicles.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89153/is-it-possible-to-have-anions-of-metals
|
Is it possible to have anions of metals? [duplicate]
|
**This question already has an answer here**:
[Can metals have a net negative charge](/questions/27407/can-metals-have-a-net-negative-charge)
(1 answer)
Closed 5 years ago.
I've been wondering if it is possible to have negatively charged metal ions. Most ions tend to have electron configuratons of noble gases. So then is it possible to obtain metal ions from metal atoms by changing their electron configuration to that of the **next** noble gas in the periodic table? (e.g. something like Be(6-))
| 2 |
[
[
"\nYes. And, ironically, the most clear-cut examples involve [*alkali* metals](https://en.wikipedia.org/wiki/Alkalide). In these compounds you have the cation, often an alkali metal itself, stabilized in a complex so that electrons can be transferred to the recipient alkali metal (Na or heavier, based on today's known compounds), and the latter is a full-fledged alkali metal anion.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89150/bromonium-ion-or-mesomeric-effect-intermediate-during-addition-to-alkenes
|
Bromonium ion or Mesomeric effect ( intermediate during addition to alkenes)
|
It's mentioned in *Peter Sykes , A Guidebook To Mechanism In Organic Chemistry , 6th edition, 13th impression by Pearson, Pg. No. 185* , that on addition of $\ce{HBr}$ to bromoethene the only product obtained is 1,1-dibromoethane. It's due to the positve mesomeric effect of bromine to stabilize the carbocation in this case, however the rate is remarkably slower as compared to addition to ethene.
[1](https://i.stack.imgur.com/NXHT4.jpg)
Another reaction corresponding to addition of $\ce{Br2}$ to ethene proceeds with the formation of a bromonium ion intermediate.
[2](https://i.stack.imgur.com/3KIeT.jpg)
My question is, why can't the first reaction just proceed with a bromonium ion intermediate, like *(28)* above can form a bromonium ion and give 1,2-dibromoethane as a product (there may be a possibility that the carbocation may be formed away from bromine to counteract the strong negative inductive effect and in that process it may interact with the lone pairs of bromine to form bromonium ion)? And if bromonium ion is unstable (that is if it can't be formed in 1st reaction) why does it form at all in reaction 2? Why can't the Bromonium ion in the 2nd reaction just undergo a 1,2- hydride shift and convert into the "more stable" ion like 29b. It should also lead to formation of geminal dibromide .It's a circular argument.
It's clearly mentioned that bromonium ion and carbocation are canonical forms of each other , which means you expect them to convert into bromonium ion whenever there's a positively charged carbon near a bromine atom.
[3](https://i.stack.imgur.com/hvWWnm.jpg)
Here's a similar example of phenonium ion.
[4]: <https://i.stack.imgur.com/0j5Ww.jpg> .
The carbocation doesn't stay the way it is in structure *4*. It rather forms *3*. Notice the bond between phenyl ring and the carbon atom on the right. It's not a completely solid bond.
| 9 |
[
[
"\nYou are not going to see the formation of a primary carbocation—the activation energy is too high. Even though it is true the following are resonance structures of one another:\n\n\n[](https://i.stack.imgur.com/wt6sJ.png)\n\n\nthe geometry in which the primary carbocation is formed does not allow for the simultaneous formation of the bromonium ion:\n\n\n[](https://i.stack.imgur.com/PjKhD.png)\n\n\nThe bromonium would only be able to form after a rotation of the $\\ce{C-C}$ bond, so we need to consider the primary carbocation sans-bromonium as the real product of the protonation of the alkene. This is a very high energy product, so it will be slow to form at low temperatures and is out-competed by the other reaction. Since positive charge on carbon-1 can immediately be delocalized to the bromine, this product of protonation is more stable (meaning it forms faster).\n\n\n[](https://i.stack.imgur.com/lE8WA.png)\n\n\n[](https://i.stack.imgur.com/r7TcK.png)\n\n\n",
"4"
],
[
"\nFirst of all, structure (9)-bromonium ion intermediate and structure (10)-the open intermediate aren't canonical forms of each other simply because \"In most resonance, **$\\sigma$ bonds are not involved**(except in hyper-conjugation but what we have here is not hyper-conjugation), and only the $\\pi$ or unshared electrons are put in different ways and canonical forms don't really exist(its our imagination) but reaction intermediates like bromonium ion and structure (10)-type intermediates have been isolated under special conditions.\" \nHad structure(9) and (10) been canonical forms of the same molecule, we would've never been able to isolate structure(9) and (10). **The positive charge in structure(10) might be stabilized by an attraction for Br but that does not involve a full bond.**(see the image below)\n\n\n[](https://i.stack.imgur.com/BipLs.png)\n\n\nThe possibility that \"the carbocation may be formed away from bromine to counteract the strong negative inductive effect and in that process it may interact with the lone pairs of bromine to form bromonium ion\" sounds unreasonable to me because the carbocation will always form on the bromine side as the **+M effect of bromine that stabilizes (29a) & (29b) is way more than its destabilizing -I effect**.\n\n\nIn no way is bromonium ion intermediate less stable than structure (10). \n\n**Ab initio molecular orbital studies show that bromonium ion intermediate is more stable than the intermediate in reaction 1(structure-(10)).**\n\n\nIn the addition of HBr to bromoethene, it follows the convention: **\"the electrophile attacks the substrate before the nucleophile does\"**. Electrophile need not actually be a positive ion but can be the positive end of a dipole or an induced dipole, with the negative part breaking off either during the first step or shortly after. So, the formation of the carbocation after the attack of the proton is inevitable. \n\n\nBut due to the fact that in reaction 1, the electrophile is hydrogen ion, the reaction takes place through the mechanism involving structure (10) and not the one in reaction 2 involving the bromonium ion intermediate(9).\n\n\nI hope you realize that (2) never forms (4). Your statement saying that (4) forms (3) interconvert among themselves is not correct rather (3) and (4) are separate species formed through different set of mechanisms. None of the structure is a canonical form of the other.\n\n\n[](https://i.stack.imgur.com/TK3jp.png)\n\n\nThe above reaction clearly shows that (12) and (9a) are not really canonical forms of each other. I really don't get the sense in which that statement on pg. 180 has been written but it definitely doesn't specifically mean that (9) and (10) are canonical forms of each other.\n\n\nDon't get confused by the reaction that I mentioned above, Peter Sykes says *This neighbouring group participation by bromine does not ofcourse prove that addition to alkenes proceeds via cyclic bromonium ions*, in the past,the reaction was just used to show that such species(cyclic bromonium ions) are not just mere assumptions. That's all!\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89147/which-is-discharged
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Which is discharged? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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Why Na+ ion is discharged at cathode rather than H+ ion when Mercury is used as cathode in electrolysis of aqueous NaCl solution ?
| 2 |
[
[
"\nMercury has a high overvoltage for hydrogen evolution; that means a barrier exists so that molecular hydrogen is not formed easily. Evolving hydrogen requires two protons to be neutralized, get together to form a molecule and clump up with a few more hydrogen molecules to form a gas bubble. If a sodium cation is neutralized right next to mercury liquid, it can quickly amalgamate/dissolve into the mercury, where its activity is much reduced. Sodium amalgam reacts with water, but much slower than metallic sodium reacts with water.\nSimilarly, ammonia amalgam can be made by electrolysis, and the mercury will foam up - it's not very stable.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89143/mixing-supplements-will-the-distributions-remain
|
Mixing Supplements - Will the Distributions Remain?
|
I am combining various vitamins and minerals into a food supplement for my dog and I currently measure each powder individually for each batch of food each week. The supplements are in powder form but have varying grain sizes. Examples include copper, zinc, dicalcium phosphate, kelp powder, table salt, Vitamin E, Vitamin B-12.
If I were to make a giant batch of the supplements above using the proper proportions can I safely measure out of that large batch the appropriate volume each week or will the various consistencies / densities of the powders unevenly distribute making the weekly scoops proportionally unbalanced?
Is there some piece of equipment that could be used to mix the powders guaranteeing even distribution for batch usage?
Thank you in advance for any advice.
| 3 |
[
[
"\nIt's hard to maintain homogeneity for the mix of powders. The bigger the batch, the less homogeneous it's going to be. Ideally you want to disperse solid ingredients; there are several techniques to do that. Two of the most popular ones are:\n\n\n* dissolving all the components in a non-reactive solvent and they dry out or precipitate the powder;\n* using a [ball mill](https://en.wikipedia.org/wiki/Ball_mill).\n\n\nWith supplements, however, it's somewhat trickier, as they may react with each other over time, with the solvent, or be contaminated in grinder. Also they are often granulated for a good reason: this reduces the contact with atmosphere and prolongs shelf life.\n\n\nIf you decide to avoid the dispersion part, then due to [Brazil nut effect](https://en.wikipedia.org/wiki/Granular_convection) the largest powder particles will remain on top of the batch, the smallest – on the bottom. This is especially noticeable when the batch has been shaken, e.g. after being transported in a car trunk, or just being kicked a few times lying on the kitchen's floor.\n\n\nAll in all, I wouldn't recommend doing this. The probability of suffering from an imbalanced diet is too high. There are some multivitamin and mineral complexes designed for humans, maybe look for something similar for your pet.\n\n\n",
"8"
]
] |
https://chemistry.stackexchange.com/questions/89142/dropping-borosilicate-glass-mug
|
Dropping Borosilicate glass mug
|
I recently got a glass tea mug. It's has double wall construction with a vacuum insulation (advertised as such). I noticed on the label, it specifically mentioned it's made of borosilicate glass, and to "discard if dropped".
I understand the purpose of making it out of borosilicate glass (thermal expansion when adding hot tea). But what would be the reason for the directions to link the material to needing to discard the mug if dropped (and not cracked or broken)?
Does the fact that it's made of borosilicate mean it's structural or thermal properties might be compromised if it's dropped?
| 3 |
[
[
"\nPyrex glass labware can be dropped, scratched, even used if cracked (I've started filling a beaker, then noticing that there was a crack - but it still seemed OK).\nBorosilicate glass for commercial (non-lab) use is often tempered, so that a scratch is a potential hazard - the item might fail unexpectedly. The issue is not so much the borosilicate but the tempering.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89137/what-type-of-polyester-is-most-commonly-used-for-clothing
|
What type of polyester is most commonly used for clothing?
|
Many polymers have the name of the monomer in their names, e.g. polyethylene, polyvinyl chloride, polypropylene, polystyrene etc. But the ester in polyester refers to a group of chemicals (having an hydroxyl group replaced by an alkoxy group1) rather than a single chemical compound.
[This article](https://en.wikipedia.org/wiki/Polyester) from Wikipedia states that polyethylene terephthalate (PET or PETE) is the most common type of polyester. PET is the plastic from which almost all plastic soda and beverage bottles are made. This leads me to wonder if the polyester from which clothes are commonly made are PET, and if not, what types of other polyesters (e.g. polyglycolide, polycaprolactone, polyethylene adipate, polybutylene succinate, etc.) clothes may be made from.
I'm aware that most plastics can be stretched into long thin filaments/strands and wound into threads in a process similar to how natural-fiber threads are made. My question is specifically whether or not PET is commonly used as the polyester in clothes, and if not, what is? I've done some Google searches but can't find any references; maybe I just haven't gotten the search string right. Thanks.
---
1. <https://en.wikipedia.org/wiki/Ester>
| 4 |
[
[
"\nBased on several sources (shown below), it appears that polyethylene terephthalate (PET) is in fact used in the production of fabrics, including textiles. PET is marketed under names such as Dacron, Terelene, or Lavsan by various companies depending on your location. As to if it is the only textile polyester, I was unable to be completely certain. However, it appears to be a major textile polyester at the very least, as 39 million megagrams were produced in 2008 for fabrics alone.\n\n\n**Though it was not part of your question, I thought that I should include the following, as I found it interesting:**\n\n\nAs it turns out, not only is PET used as a fabric, but it also makes a very nice fabric that is highly resistant to wrinkling. This is due to the presence of an aromatic ring within the monomers, which makes it quite strong/stiff; especially in this fibre form. These wrinkle-resistant fabrics are appropriately known as \"permanent press fabrics.\"\n\n\nTo find out more information, I have included some sources below:\n\n\n* <https://en.wikipedia.org/wiki/Polyester>\n* <https://en.wikipedia.org/wiki/Polyethylene_terephthalate>\n* <https://www.britannica.com/science/polyethylene-terephthalate>\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89132/how-to-balance-equation-of-alcohol-oxidation-with-potassium-permanganate
|
How to balance equation of alcohol oxidation with potassium permanganate?
|
I need to balance the equation for the oxidation of primary alcohol. The media is neutral to start out with, but the product will be *basic*. I assigned oxidation numbers and added stoichiometric coefficients accordingly:
$$\ce{3 R-\overset{-1}{C}H2OH(aq) + 4 \overset{+7}{Mn}O4^-(aq) -> 3 R-\overset{+3}{C}OO^-(aq) + 4 \overset{+4}{Mn}O2(s)}$$
The next step is balancing the charge(?): currently, net charge on the left is −4, and −3 on the right. From what I know, a reaction like the above will be balanced in a neutral media by adding $\ce{H+}$ or $\ce{OH-}$ on the *right* side of the equation. To balance that out, you will instead add $\ce{H2O}$ on the *left* side.
We, therefore, need to add $\ce{OH-}$ on the *left* side of the equation:
$$\ce{3 R-\overset{-1}{C}H2OH(aq) + 4 \overset{+7}{Mn}O4^-(aq) -> 3 R-\overset{+3}{C}OO^-(aq) + 4 \overset{+4}{Mn}O2(s) + OH^-(aq)}$$
Now that it is balanced, we add $\ce{H2O}$ to equal… oh wait, that's not possible. We are missing $\ce{H}$ on the right side, and we can't add more $\ce{OH-}$ since that would make the charges(?) not be balanced.
I need an advice on this with an explanation.
| 0 |
[
[
"\nThe equation proposed by deLange is interesting, but it is not an answer, because it requires $\\ce{H+}$ ions to proceed, and Oliver, the question's author, states that the reaction should not occur in an acidic solution.\n\n\nSo the $\\ce{H+}$ ions have to disappear from the deLange's equation, which I repeat here.\n$$\\ce{4H+ + 4 MnO4^- + 3 RCH2OH -> 4 MnO2 + 5 H2O + 3 RCOOH}$$ The only way of getting rid of these $\\ce{H+}$ ions is to add $\\ce{4 OH-}$ ions on both sides of deLange's final equation. This will transform the $\\ce{4 H+}$ into $\\ce{4 H2O}$ on the left-hand-side. On the right-hand side, this will transform $\\ce{3 RCH2COOH}$ into $\\ce{3 RCH2COO- + 3 H2O}$ and the fourth $\\ce{OH-}$ ion will remain in the solution, which becomes basic. The obtained equation will be$$\\ce{4 H2O + 4 MnO4^- + 3 RCH2OH -> 4 MnO2 + 5 H2O + 3 RCOO- + 3 H2O + OH-}$$which can be shortened according to $$\\ce{4 MnO4^- + 3 RCH2OH -> 4 MnO2 + 3 RCOO- + 4 H2O + OH-}$$If the initial solution was neutral, it becomes basic at the end.\n\n\n",
"1"
],
[
"\nThe half reaction for the reduction of permanganate is\n$\\ce{3e- + 4H+ + MnO4^- -> MnO2 + 2H2O}$\n\n\nThe half reaction for the oxidation of an alcohol to an acid is\n$\\ce{RCH2OH + H2O -> RCOOH + 4e- + 4H+}$\n\n\nAdd $4$ times the reduction half reaction to $3$ times the oxidation\n\n\n$\\ce{16H+ + 4MnO4- + 3RCH2OH + 3H2O -> 4MnO2 + 8H2O + 3RCOOH + 12H+}$\n\n\nCancelling things that appear on both sides\n\n\n$\\ce{4H+ + 4MnO4- + 3RCH2OH -> 4MnO2 + 5H2O + 3RCOOH}$\n\n\nand Bob's your uncle. Writing it this way also shows that adding acid pushes it to the right, IOW this reaction should be carried out in acid solution.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89119/why-in-this-reaction-acetic-acid-is-strong-acid-and-nh3-is-strong-base-please-e
|
Why In This Reaction Acetic Acid is strong acid and NH3 is strong base ?please explain in details and thanks for answer [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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the acetic acid and amonia reaction why the reaction is towards completion and why nh3 and ch3cooh are strong
| -2 |
[
[
"\n\"Weak\" and \"strong\" are relative terms. In the example shown in the question, ammonium ion is stronger than the acid on the other side of the equation, which is just water. So it looks like a strong acid. But if you rendered the following reaction instead things appear different:\n\n\n$\\ce{H\\_3O^+}+\\ce{NH\\_3}\\leftrightharpoons\\ce{H\\_2O}+\\ce{NH\\_4^+}$\n\n\nNow the ammonium ion is compared with a hydronium ion, which is a much stronger acid than either plain water or ammonium ion. Suddenly ammonium ion looks like only a weak acid.\n\n\nWe really should be using **comparative** terms. In the reaction between ammonia and plain water **ammonium ion is a stronger acid than water.** But in the reaction between ammonia and a mineral acid, **ammonium ion is a weaker acid than the hydronium ion.**\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89117/coordination-isomerism-variation
|
Coordination isomerism variation
|
The school book and [Wikipedia](https://en.wikipedia.org/w/index.php?title=Coordination_isomerism&oldid=765885835) state that coordination isomers are:
>
> In a coordination isomer the total ratio of ligand to metal remains the same, but the ligands attached to a specific metal ion change.
>
>
> For example, a solution containing $\ce{[Co(NH3)6]^3+}$ and $\ce{[Cr(CN)6]^3−}$ is a coordination isomer with a solution containing $\ce{[Cr(NH3)6]^3+}$ and $\ce{[Co(CN)6]^3−}$.
>
>
>
So in this definition, consider $\ce{[Co(NH3)5(CN)][Cr(CN)5(NH3)]}$. Is it coordination isomer of $\ce{[Co(NH3)6][Cr(CN)6]}$ or not?
My teacher says yes, but I'm not quite sure according to Wikipedia.
| 3 |
[
[
"\nYes, it is.\nIn both the cases, the number of ligands remains same, which is according to the definition in Wikipedia.\n\n\nFor any type of structural isomerism, the premise is that the primary valence and secondary valence (as defined by Werner) of central metal/ ion remains same for all the isomers.\n\n\nIn this case, the primary valence is 3 and secondary valence is 6, which remains same for all isomers.\n\n\nOther possible coordination isomers are:\n\n\n$\\ce{[Co(NH3)4(CN)2]+[Cr(NH3)2(CN)4]-}$\n\n\n$\\ce{[Cr(NH3)4(CN)2]+[Co(NH3)2(CN)4]-}$\n\n\n$\\ce{[Cr(NH3)5(CN)]^2+[Co(NH3)(CN)5]^2-}$\n\n\n(cation is always written first followed by anion)\n\n\nI don't think $\\ce{[Co(NH3)3(CN)3][Cr(NH3)3(CN)3]}$ can exist because, now both parts become neutral.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89116/which-orbital-lobes-are-occupied
|
Which orbital lobes are occupied?
|
In the p-orbitals, there are 3 dumbbell-shaped lobes, each of which can contain 2 electrons. In the diagrams of the d- and f-orbitals, there are two or more lobes (or orbital shapes) or each value of m. In all of these cases, which lobes/shapes do the electrons occupy for any given element?
| 2 |
[
[
"\nThe question implies a misunderstanding about orbitals and their representation as dumbbell-shaped lobes. These plots show the relative orientation of orbitals to each other by showing regions that represent some threshold probability (say, 90%) of finding an electron contained therein. The underlying probability distribution has the same symmetry as the orbital itself because it defines the orbital. Any (spatial) orbital can hold up to two electrons$^1$, regardless of its shape. It thus makes little sense to ask in which lobe (or, further, which half of the dumbbell, since there is a nodal plane in between the halves) an electron is.\n\n\nMoreover, any single atom is spherically symmetric in the absence of external fields, meaning that the shapes of the orbitals are not actually important to understand electron distribution in a free atom. The shapes are useful tools to understand certain bonding concepts such as crystal/ligand field theory nevertheless.\n\n\n$^1$ a spin orbital holds one.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/89115/what-is-the-mass-of-nh42co3-need-to-add-to-250-g-water-in-order-to-lower-the-f
|
What is the mass of (NH4)2CO3 need to add to 250 g water in order to lower the freezing point approximately 5.5 °C? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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Closed 5 years ago.
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What is the mass of $\ce{(NH4)2CO3}$ need to add to $250\ \mathrm g$ water in order to lower the freezing point approximately $5.5\ \mathrm{^\circ C}$?
$K\_\mathrm f=1.86\ \mathrm{^\circ C\ kg/mol}$
What would be the boiling point?
$K\_\mathrm b=0.51\ \mathrm{^\circ C\ kg/mol}$
What is the osmotic pressure (the solution in the first question) in $14.5\ \mathrm{^\circ C}$?
| -3 |
[
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"\nThe van 't Hoff factor $i$ of $\\ce{(NH4)2CO3}$ is $3$ in water. So $\\Delta T\\_\\mathrm f = ik\\_\\mathrm fm$ can be used.\n\n\n$$\\Delta T\\_f = 3 \\cdot 1.86 \\cdot \\frac{w}{0.25 \\cdot M\\_{\\ce{(NH4)2CO3}}} = \\frac{3\\cdot 1.86 \\cdot w}{0.25\\cdot 96} = 5.5$$\n\n\nSolving gives $w = 23.66\\ \\mathrm g$.\n\n\nSimilarly you can use $\\Delta T\\_\\mathrm b = ik\\_\\mathrm bm$ to get:\n\n\n$$\\Delta T\\_b = 3 \\cdot 0.51 \\cdot \\frac{w}{0.25 \\cdot M\\_{\\ce{(NH4)2CO3}}} = \\frac{\\Delta T\\_f \\cdot 0.51}{1.86} = 1.51$$\n\n\nFor osmotic pressure use $\\Pi = C R T$ where $C$ is molarity and is equal to approximately $$C = \\frac{w}{M\\_{\\ce{(NH4)2CO3}}}\\cdot \\frac{1}{V}$$\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89112/conductivity-variation-on-caustic-naoh-dilution-system
|
Conductivity variation on Caustic (NaOH) Dilution System
|
I am reading the drawing for an automatic caustic dilution system where 50% NaOH is diluted down to 20%.
I was expecting to see a density sensor on the outlet to provide feedback to the dilution pumps for ratio-ing. i.e. To keep the density at 20% and prevent it from moving to say 19% or 21% by adjusting the amount of water.
However, the vendor seems to have used a conductity sensor on the outlet instead. I am skeptical whether the conductivity of NaOH will vary a lot between 19% to 21%. Isn't it already fully ionized?
Thoughts? Will this work?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/89108/alkane-free-radical-halogenation
|
Alkane free radical halogenation
|
I do not understand how we obtain the product in option one.
According to Free radical stability order tertiary carbon radical is more stable than secondary carbon radical which is more stable than primary carbon radical. So hydrogen selectivity should also follow the same order. Then why is it that we get option(1) instead of option (2).
I’m only in high school so please try to keep it simple.
| 1 |
[
[
"\nThis question is based on relative rates of reaction of different $\\ce{H}$ with halogens. In the free radical mechanism of chlorination of alkane, relative rate of reaction of $\\ce{3^\\circ : 2^\\circ : 1^\\circ}$ $\\ce{H}$ is in order $5:3.8: 1$.\n\n\nUsing this, we get an idea about product amount. To get a rough estimate of relative amounts of different products, we multiply *number of identical hydrogen and its rate of reaction with chlorine*.\n\n\nSo there is only $1$ $3^\\circ$ $\\ce{H}$ and its rel. rate is $5$. So relative amount is $5$. For the $2^\\circ$ $\\ce{H}$, there are $2$ of them and their relative rate is $3.8$. So its relative amount is $2 \\cdot 3.8 = 7.6$. Similar calculation for rest of the hydrogens will show you why given answer is correct.\n\n\nRefer: [Wikipedia](https://en.wikipedia.org/wiki/Free-radical_halogenation#Control_of_halogenation)\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89102/what-is-the-correct-iupac-name-of-this-compund
|
What is the correct IUPAC name of this compund?
|
[](https://i.stack.imgur.com/X4okY.png)
I thought C=C is the "functional group" here and Br is a substituent so the numbering should start from the top left C, making the name 4-Bromo-3-methylbut-2-ene, but my book says 1-Bromo-2-methylbut-2-ene. Which is correct?
| 1 |
[
[
"\nAccording to **P-14.4** in *[Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book)](https://doi.org/10.1039/9781849733069)*:\n\n\n\n> \n> **P-14.4** Numbering\n> \n> \n> When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:\n> \n> \n> [...]\n> \n> \n> (e) saturation/unsaturation: [...]\n> \n> \n> (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;\n> \n> \n> (g) lowest locants for the substituent cited first as a prefix in the name;\n> \n> \n> \n\n\nIn this compound, unsaturation, in the form of the double bond, has the priority for low locants. Since both numbering schemes (shown below) assign the locant '2' to the double bond, criterion (e) is not sufficient for the purposes of deciding between the two options.\n\n\n[](https://i.stack.imgur.com/q76WV.png)\n\n\nHowever, criterion (f) does allow for a choice to be made. The detachable prefixes in this name are 'bromo' and 'methyl'. Since the locant set (1,2) is lower than the locant set (3,4) (note that both sets are ordered in increasing numerical order as stipulated), the preferred IUPAC name (PIN) will use the locant set (1,2).\n\n\nNote, however, that your double bond also contains configurational information. According to the Cahn–Ingold–Prelog rules (see **P-92** of the Blue Book), the double bond in your molecule as drawn is (*E*)-configured. Stereodescriptors such as '*E*' are cited together with the locant describing the position of the stereogenic unit (see **P-91.3**).\n\n\nHence, the PIN of the molecule as depicted in your question is **(*2E*)-1-bromo-2-methylbut-2-ene**.\n\n\n[](https://i.stack.imgur.com/f36HF.png)\n\n\nNote that contrary to existing answers, there is no inherent preference between \"bromo\" and \"methyl\" for low locants, **unless** the application of criterion (f) does not lead to a difference. In that case, criterion (g) must be invoked. Since 'bromo' is earlier than 'methyl' in alphanumerical order, it will be the prefix cited first in the name, and will therefore have priority for a lower locant. For example:\n\n\n[](https://i.stack.imgur.com/zchc8.png)\n\n\n",
"9"
],
[
"\nYou are correct that double bond gets preference over halogen. But after that, we seek to give small locant to bromo and methyl. Since in second name, both substituents get smaller locants, it is the correct name. \n\n\nThat is correct name is *1-Bromo-2-methylbut-2-ene*\n\n\nHad the question been this instead:\n\n\n[](https://i.stack.imgur.com/LIRQx.png)\n\n\nThen the name would have been *1-Bromo-3-methylbut-2-ene* because in this case as orthocresol comments:\n\n\n\n> \n> The name *1-bromo-3-methylbut-2-ene* is preferred over *4-bromo-2-methylbut-2-ene* because the locant set (1,3) is lower than the locant set (2,4). Alphabetical order only comes into play when the locant sets are the same.\n> \n> \n> \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89100/changing-temperature-of-a-galvanic-voltaic-cell-should-i-wait-for-the-electrode
|
Changing Temperature of a Galvanic/Voltaic Cell: Should I wait for the electrodes to also change in temperature?
|
I'm conducting an experiment where I change the temperature of the sulfuric acid in a lead acid storage cell. My one question in controlling the variables is that should I let the electrodes also change in temperature, or should I keep them at room temperature and measure the cell potential? Does it even matter at all?
What I'm guessing is that a change in temperature affects the resistance of a metal, which would then affect the ability of the electrodes to transfer the electrons involved; therefore, the electrodes should be kept at room temperature.
If I should do one or the other, could someone explain to me why?
| 0 |
[
[
"\nIn general the higher the temperature of the electrolyte, the more power you can draw from the cell.\n\n\nThe reason for this is, because the ions of the electrolyte can move more freely when the temperature is higher and chemical reactions do in general happen quicker with increased temperature. \n\n\nOn the other hand side the hotter a metal gets, the bigger its resistance will become. This is because electrons can’t move as freely when the temperature is increased.\n\n\nHowever this effect usually is of importance when temperatures are quite high, like in an light bulb.\n\n\nIf you just heat the electrodes for 30-40°C, I doubt you would be able to even measure the difference in resistance (even though it does exist).\n\n\nSo to sum it up, the output power of a cell in relation to the temperature will mainly be altered by the temperature of the electrolyte. \n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89098/separation-of-solvent-from-residue-of-material
|
Separation of solvent from residue of material. [closed]
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I have a residue of 2kl in which 1.8kl is solvent (EDC) and rest material is residue. First i was thinking to go with normal distillation to recover our solvent. But i heared somewhere that there is some kind of RESIN by which if it got passes solvent will be pass only and residue will collect on another side of that resin. if there is any concept like this please help me.
or there is any other to do?
| 2 |
[
[
"\nDichlorethane has a boiling point of about 84°C - If I would have to separate EDC from a product / residue, I would use a rotovap with water bath configuration. This distillation is a very gentle way to separate solvents from product / residue.\n\n\nSet the temperature of the water bath slightly higher than the boiling point of the EDC. \n\n\nIf you want to get fancy to make your solvent super pure again after the rotovap distillation, you could perform a fractional distillation on it - however in most cases I think this might not be necessary.\n\n\n",
"0"
],
[
"\nDistillation is the way to go. Especially if you have to do it on industrial scale (if we have understood well, you have 2000 L to process). \nNormally it yields recovered solvent with low level of impurities. The operation is easy to perform and there are many installations available to do it. \n\n\nResins are not an option unless the only impurities in the solvent are acidic or basic, and preferably strong ones. Then you can use ion exchange resins. \nOtherwise the situation is just on the reverse side of what you want to do. \nSince your impurities are soluble in an apolar solvent, they are apolar too. \nThere are resins for adsorbing these kinds of compounds, but then they must be in a polar solvent. Usually water is the option because even in alcohols the impurities do not get adsorbed. At least not completely. \n\n\nIf your impurities are not very apolar, they might be adsorbed on silica, but it is not likely that all of them will do so and probably none will do it completely. \n\n\nFurther, there is the economic aspect. Resins and silica are much more expensive than energy, so distillation is always preferred unless unfeasible. Moreover, for using adsorbents, the most efficient is to have them in a column with specific dimensions and a well designed bottom plate. Since the adsorption is an equilibrium, working in a column reduces the residual impurities much more. Unfortunately, these kind of installation are much less common than distillation units. \n\n\nIf you do not have such an installation yourself, it is quite easy to externalize the operation to some contract manufacturing unit. \n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89092/electrophilic-addition-to-alkenes
|
Electrophilic addition to alkenes
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Consider the addition of HBr (in absence of any radical initiators or environment that promotes radical mechanistic pathway), to 3-bromocyclopentene.
[](https://i.stack.imgur.com/ritMft.jpg)
The first step is the attack of $\ce{H^+}$
On the exposed π electron cloud. But my query is whether it forms a classical carbocation like this one to oppose the negative inductive effects of the halogen atom:
[](https://i.stack.imgur.com/O0G07t.jpg)
Or does it proceed with the formation of bromonium ion like this one to partially complete its octet using lone pairs on Br.
[](https://i.stack.imgur.com/U8dj5t.jpg)
These two would lead to different products but I'm interested in the major one.
The first one would lead to 1,3-dibromocyclohexane, while the second one would lead to 1,2-dibromocyclohexane.
| 2 |
[
[
"\nSince it is electrophilic addition, it can go via 2 ways. Either by forming classical carbocation or cyclic carbocation. Since the electrophile will be H+ it will go with classical carbocation. Therefore product will be 1,3 dibromo cyclopentane. Also taken in account the mesomeric effect of bromine atom\n\n\n",
"-1"
]
] |
https://chemistry.stackexchange.com/questions/89087/why-the-ph-of-xanthan-gum-in-aqueous-solution-is-7-neutral-when-it-has-glucuro
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Why the pH of xanthan gum in aqueous solution is 7 (neutral) when it has glucuronic and pyruvic acids in its chemical structure?
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Since [xanthan gum](https://en.wikipedia.org/wiki/Xanthan_gum) has both glucuronic and pyruvic acids in its chemical structure, it should release $\ce{H+}$ ions upon solvation in an aqueous solution and make it acidic. However, the meausured $\ce{pH}$ is $7$.
| 4 |
[
[
"\nPyruvic and glucuronic acids are both pretty strong acids (pK's of respectively 2.5 and 3.28) but that does not mean that the polymer of xanthan gum will be. Those protons are still there but they may be more tightly bound in the polymer than in the free acids. To find out try titrating a solution with strong base. Eventually you should start prying some of those protons off and you should see an inflection point. OTOH as there are so many carboxylic protons in the polymer the inflection points would probably smear and not be visible but you should see some buffering above some pH if my hypothesis is correct.\n\n\nI should make clear that this is purely speculation on my part but it sounds like an interesting experiment.\n\n\n",
"1"
],
[
"\nXanthan gum disperses difficultly in water; the dispersion process can be speeded up by raising the pH to about 10-11. This implies that acidic groups on the polymer are not much ionized, but addition of alkali puts a lot of negative charge on the polymer, helping it to unfold.\n\n\nMeasuring pH of gums, goos, slimes and muds is difficult, especially when the concentration of protons or hydroxide is low. The more xanthan gum you put into water, the goopier it gets; 1% is quite goopy. A measurement of 7 could be +/- 0.25 at least. \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89083/how-isotopes-atomic-mass-was-defined
|
How isotopes atomic mass was defined?
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So we say that 1 Amu (u) = 1/12 of the mass of a Carbon 12 atom.
Im not sure about this, but I believe that first we measured the mass of 1 carbon 12 atom (This includes the mass defects that occur during the formation of nucleus) and then defined that 1 atomic mass unit has a value of 1/12 of the carbon 12 atom mass... Therefore one Carbon 12 atom has **exactly** 12 atomic mass units.
Until here everything is fine.
Now my question is... To define all the other elements **isotopes** atomic mass did we first measured **each** isotope atom mass and then said " If 1/12 of the Carbon 12 atom is equal to X grams and this X grams are equal to 1 atomic unit the then my isotope atom which has Y grams must have Y atomic mass units " did we work it out this way or did we use calculation to define other isotopes mass based on carbon 12 atomic mass ?
Basically what I want to know is, do we need to have the value of an atoms mass in grams or kilograms before we know how many Amu´s it has, or can we calculate it ?
| 0 |
[
[
"\nHistorically, the relative masses of the isotopes were measured relative to O=16. \n\n\nThere is a good historical explanation in [A New Mass-Spectrograph and the Whole Number Rule](http://www.jstor.org/stable/94803) (1927). \n\n\nAt this time (1927), though it was understood that many elements had more than one isotope, only one isotope of oxygen was known. \n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89081/potassium-citrate-in-water
|
Potassium citrate in water
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I take potassium as a supplement in the form of potassium citrate. I dissolve a small amount in tap water and I drink it. Usually, I need to stir a lot, and sometimes I warm the water a little to make it easier to dissolve the potassium.
Recently, I bought some sparkling water (San Pellegrino, but I don't think that makes a difference) and when I put the potassium in it, it reacted making it bubble a little and it dissolved immediately.
Why is that? Did the potassium react with the CO2 or something else in the sparkling water?
| 0 |
[
[
"\nBubbles need nucleation points to form. \nThe growth of bubbles is limited by the surface tension of the liquid. The surface tension acts stronger on smaller bubbles and less strong on bigger bubbles.\nAdding a solid salt to sparkling water means adding a lot of new nucleation points, which subsequently will lead to bubbling.\nThe small bubbles will combine to bigger bubbles, these bigger bubbles are not so much effected by surface tension and so even more gas is produced.\nIn other words, adding the salt will trigger a kind of avalanche effect, which weakens either if the dissolved CO2 gets towards used up or the salt gets dissolved completely.\n\n\nIt is not an reaction of dissolved potassium ions with the CO2, nor is it a reaction of dissolved citrate ions with the CO2.\nWhat happens here is a physical reaction and not an chemical reaction.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89080/how-to-remove-nitrose-gasses-from-ozone-gas-mixture
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How to remove nitrose gasses from ozone gas mixture?
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I produce ozone gas with an ozone generator based on corona discharge principle.
These kind of ozone generators have the disadvantage that if $\ce{N2}$ is present, a bit of $\ce{NO}$ and $\ce{NO2}$ will be produced too.
To produce oxygen I use a oxygen concentrator, with an output of about $90\%$ $\ce{O2}$ and about $10\%$ $\ce{N2}$.
For that reason I have a little quantity of $\ce{N2}$ entering the ozone generator, which subsequently lead to unwanted side reactions like $$\ce{N2 -> NO -> NO2}.$$
The $\ce{NO2}$, even in that small quantity as it is, does influence the following processes in an negative way.
The easiest way to circumnavigate this issue would be to take pure oxygen – however for cost reasons I cant use LOX, so I have to stay with the concentrator.
I am searching for an reagent or an catalyst, which is not effected by the ozone but does absorb or disintegrate the $\ce{NO}$, and especially the $\ce{NO2}$.
To stripe $\ce{NO2}$ from a normal gas is not so difficult, but the ozone does make trouble.
I thought about dry $\ce{NaOH}$, soda lime, or dry $\ce{KOH}$, but I have the feeling that these chemicals will react with $\ce{O3}$.
Any suggestions?
| 5 |
[
[
"\nThere is no need to use caustic since $\\ce{NO2}$ reacts with water itself. As per [1] it is even better absorbed in water than in $\\ce{NaOH}$. \n\n\nOzone is soluble in water depending on the concentration of $\\ce{O3}$ in the gas phase. It is much more soluble than oxygen so you will loose some, but it will have much lower concentration of $\\ce{NO2}$. Of course, the washed ozone will be moist. If you need it dry, you can use a column of molecular sieves or some other desiccant. \n\n\n[1] *Absorption of Nitrogen Dioxide by Aqueous Solutions*,\nF. S. Chambers Jr., T. K. Sherwood,\n*Ind. Eng. Chem.*, 1937, **29** (12), pp 1415–1422\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/89077/sodium-chlorite-vs-chlorine-dioxide
|
Sodium Chlorite vs. Chlorine Dioxide
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I ordered Sodium Chlorite in the mail and received Chlorine Dioxide. Can they be used interchangeably, or are they different? My background in chemistry is limited so laymen's terms would be appreciated. I appreciate any responses or feedback.
| 1 |
[
[
"\nSodium Chlorite is a white powder, which dissolves in water.\n\n\nChlorine Dioxide is a gas. This gas also dissolves in water to some degree and gives a yellow color.\n\n\nSodium Chlorite is used to make Chlorine Dioxide in the laboratory.\n\n\nI strongly doubt, that someone has sent you Chlorine Dioxide, as it is to dangerous to store and to send.\n\n\n",
"7"
],
[
"\n**Warning**: Both sodium chlorite and chlorine dioxide are toxic. There are numerous reports of nausea, diarrhea, severe vomiting, and life-threatening low blood pressure caused by dehydration as a result of chlorite poisoning. Sodium chlorite is a strong oxidizing agent and upon slight contamination would lead to explosion.\n\n\n\n\n---\n\n\nChlorine dioxide is the product formed from the action of chlorine on sodium chlorite (laboratory process of making chlorine dioxide). This reaction is relatively safe. The reverse reaction of converting chlorine dioxide to sodium chlorite is dangerous and I highly recommend against it. This reaction is a highly oxidizing one and may ignite or explode if something goes wrong.\n\n\nChlorine dioxide can be revert back to sodium chlorite by reacting it with sodium peroxide.\n\n\n$$\\ce{Na2O2 + 2ClO2 → O2 + 2NaClO2}$$\n\n\nBut [sodium peroxide can enhance combustion](https://www.cdc.gov/niosh/ipcsneng/neng1606.html) if contact with any combustible substance. Also, it causes redness of skin and pain. **Don't try this.**\n\n\nOther preparation include reacting with sodium carbonate or sodium hydroxide but it tends to form a mixture of sodium chlorite and sodium chlorate.\n\n\nSo my opinion would be to return the chlorine dioxide. Don't try to store it or convert it to sodium chlorite as the reactions are dangerous.\n\n\n",
"2"
],
[
"\nsimply marketing.\n\n\nOP didn't receive Chlorine Dioxide gas, likely didn't receive Chlorine Dioxide solution of any concentration, and either received technical grade sodium chlorite flakes 'marketed' as chlorine dioxide which was misleading, or a solution of sodium chlorite labeled ACD/SCD \"chlorine dioxide\" which while industry practice, confused OP.\n\n\n\"Acidified Chlorine Dioxide\" or \"Stabilized Chlorine Dioxide\" are both industry terms for a Sodium Chlorite solution. See Closys mouthwash ingredients as an example:\n\n\n\n> \n> Purified water, stabilized chlorine dioxide, trisodium phosphate, etc.\n> \n> \n> \n\n\nI would assume OP received a solution. If you received the powder though, my answer is the same. The seller is marketing the end product you would use the chlorite salt to produce. Nevertheless you should be able to ask for the MSDS/SDS[^0] and inquire as to the concentration. Sodium Chlorite (NaClO2) is generally sold at 80% technical grade.\n\n\nsome of the newer Closys competitors utilize a 2 bottle system where you mix two solutions, usually one consisting of sodium chlorite and another of phosphoric acid yielding their proprietary concentration of Chlorine Dioxide with leftover TSP[^1] (same as in Closys). The ClO2 is usually in the range of 20-50ppm. There are many such products like this now.\n\n\nAs for topics brought up in other answers/comments. Chlorite is an oxidant, so is Chlorine Dioxide. Chlorine Dioxide is also a \"free radical.\" Other common oxidants include Sunlight, Ozone (O3) Hydrogen peroxide (H202), and Nitric Oxide (NO), the latter also being a free radical.\n\n\nChlorine Dioxide is a more selective oxidant than Hydrogen Peroxide which one could argue makes it safer, and at appropriate concentration solutions (0.3%/3000ppm) is nearly as easily stored and used.[^2] (air-sealed container in the fridge)\n\n\nIf the Chlorine Dioxide gas is red you are in danger. Red would be highly concentrated. the color ranges from invisible to yellow, green, brown, red. The appropriate storage concentration of 3000ppm in solution is a neon yellow which could indeed be shipped with proper precautions taken.\n\n\nall you might want to know about [chlorite](https://pubchem.ncbi.nlm.nih.gov/compound/Sodium-chlorite) and [chlorine dioxide](https://pubchem.ncbi.nlm.nih.gov/compound/chlorine-dioxide). Both are used regularly and safely in water treatment, food industry, medicine\n\n\nI was going to make this a comment, but I don't have the rep for that. So here's an \"answer\" instead.\n\n\n[^0]: \"(M)SDS: (Material) Safety Data Sheet - chemical information on the product. when you ask for this the provider may also give information on the process used to derive the product and what concentration or grade it is.\"\n\n\n[^1]: TSP: Trisodium Phosphate - a rinsing agent which used to be in dishwashing detergent (still is in the commercial stuff) and is commonly sold at hardware stores for painters who use it to wash/rinse walls before painting.\n\n\n[^2]: Chlorine dioxide and chlorite solutions are used for wound healing (just like H2O2 used to be commonly used) in many countries both in medical and veterenary practice\n\n\nhm, I see this markdown doesn't do footnotes.\n\n\n\n> \n> ¯\\_(ツ)\\_/¯\n> \n> \n> \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/89073/classical-least-squares-in-uv-vis-absorbance-experiment
|
Classical Least Squares in UV-Vis Absorbance Experiment
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How do I find the minimum number of standards needed to calibrate a spectrophotometer using CLS? We never talked about CLS in class so I'm not sure of my approach. Would 1 standard solution per analyte be enough? We'll be plotting absorbance vs. concentration then finding the residuals from the best fit line.
| 2 |
[
[
"\nThat depends on the nature of the analyte and matrix. Ideally (Beer's Law) A should be linearly proportional to concentration but that may not be the case. As an example of this in the water industry barium chloride is often added to samples to determine how much sulfate ion they contain. Spectrophotometers are used to determine how much barium sulfate is in suspension. The relationship between A and sulfate concentration is not linear in this case.\n\n\nWhat you should do is make up a series of standards spanning the range of concentrations you are interested in, measure the absorptions and plot A vs. c. Now start fitting curves, starting with linear and progressing to higher order until doing so does not make the residuals look any noisier. That gives you the order of the response curve you need and the number of standards required is one more than the order.\n\n\nIdeally, the matrix wont interfere and Beer's law applies. In that case when c = 0, A = 0 and you only need one standard.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/89070/ice-crystal-matrix-measurements-of-h2o-molecules
|
Ice Crystal Matrix measurements of H2O molecules?
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I want a mathematically generated [3D STL copy of this image](http://www.snowcrystals.com/science/Animation2.gif), however it's represent algorithmically.
If I do all the H2O's as individual pieces with female connections prepared for chopped bike spokes, are all the H2O's the same and what angles do they have?
Is there a written array of positions and degrees of rotation of the water molecules of a regular ice matrix?
| 4 |
[
[
"\nI decided to summarize all my comments in a form of answer and add some illustrations. Note that I have no experience in 3D printing, so I mainly focus on crystallographic part.\n\n\nSince you asked for\"*a mathematically generated*\" 3D structure, you ought to know that water crystallizes in $P\\mathrm{6\\_3cm}$ (# 185) space group having a hexagonal symmetry. Detailed information about symmetry generators and matrix transformations can be found [online](http://img.chem.ucl.ac.uk/sgp/large/185az1.htm) or in International Tables for Crystallography [1, pp. 582-583]:\n\n\n\n> \n> [](https://i.stack.imgur.com/qqQtK.png)\n> \n> \n> \n\n\nThere is *already enough information* to create a 3D model. If you want a more convenient way, you can get a [CIF file for ice (COD-1011023)](http://www.crystallography.net/cod/1011023.html) (which already embeds all the information above), and load it with [Mercury](https://www.ccdc.cam.ac.uk/support-and-resources/Downloads/) (free, available on Windows, Linux, MacOS). From here on it's just a screencast of what's to click to get the desired molecular pattern like [the one on your animated GIF](http://www.snowcrystals.com/science/Animation2.gif):\n\n\n1. Load the structure (drug-n-drop CIF file, or via `Ctrl`+`O`). What you see now is an [asymmetric unit](http://reference.iucr.org/dictionary/Asymmetric_unit), in simple terms – a seed which an entire crystal structure can be grown from: \n\n[](https://i.stack.imgur.com/gExb6.png)\n2. Let's expand the structure beyond those 6 atoms you've stuck with. Go to `Calculate` > `Packing/Slicing...`. Tick `Pack` option, and click `+ 0.5` boxes next to $a$ and $b$ axis: \n\n[](https://i.stack.imgur.com/GFMfk.png)\n3. Rotate the grown structure approximately like it's shown below: \n\n[](https://i.stack.imgur.com/Mexei.png)\n4. Note that you have 4 water molecules on each side that you don't need. Those can be deleted by right-clicking on them and selecting `Delete this molecule`: \n\n[](https://i.stack.imgur.com/jBnCX.png)\n5. At this point you should have exactly the same 3D representation of the crystal structure: \n\n[](https://i.stack.imgur.com/SNYoi.gif)\n6. Now you can print a 3D model directly from Mercury (`File` > `Print in 3D...`). The one with supporting framework looks like this: \n\n[](https://i.stack.imgur.com/KojkF.png) \n\n\n6.1. Alternatively, you can save the structure as XYZ (`Ctrl`+`S`, choose `Xmol files`) and do some post-processing work in Blender if needed.\n\n\n### Reference\n\n\n1. International Tables for Crystallography: Space-group symmetry, 1st ed.; Hahn, T., Ed.; Fuess, H., Hahn, T., Wondratschek, H., Müller, U., Shmueli, U., Prince, E., Authier, A., Kopský, V., Litvin, D. B., Rossmann, M. G., et al., Series Eds.; International Union of Crystallography: Chester, England, **2006**; Vol. A.\n\n\n",
"3"
],
[
"\nYou can produce STL (and VRML) files directly from Mercury - see <https://www.ccdc.cam.ac.uk/Community/blog/post-56/>\n\n\nI'm not sure how best you'd go about including female sockets for bike spoke segments, but be aware that for some 3D printing types you're charged for the total volume of material used, so you might be better off trying to create a model with multiple molecules in a smaller area with less empty space.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/89069/galvanic-voltaic-cell-does-cell-voltage-change-when-a-resistor-is-connected
|
Galvanic/Voltaic Cell: Does Cell Voltage Change When A Resistor Is Connected?
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I tried making a lead acid battery. After charging for about 10 minutes, I connect the voltmeter in parallel to the cell and see that there is around a 1.2V (which is quite lower than what I predicted). However, when connecting to a lightbulb, the voltage seems to be smaller. Is this supposed to happen? Shouldn't the voltage be the same with the cell providing an appropriate current to match the resistance of the bulb?
| -1 |
[
[
"\nYes, this happens because there is an internal resistance within the battery. Look up how batteries are modeled within circuits. \n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89067/order-of-acidity-of-water-an-alcohol-an-amine-and-acetylene
|
Order of acidity of water, an alcohol, an amine, and acetylene
|
>
> What is the order of acidity of $\ce{H2O}$, $\ce{ROH}$, $\ce{RNH2}$, $\ce{C2H2}$?
>
>
>
I began by forming the conjugate base of each. The conjugate bases are as follows $\ce{-OH}$, $\ce{-RO}$, $\ce{-RNH}$, $\ce{-C2H}$.
Between stability of $\ce{-OH}$ and $\ce{-ROH}$, $\ce{-ROH}$ is more unstable. Also, in $\ce{-RNH}$, the negative charge is on nitrogen atom making it more unstable than the other two groups.
In $\ce{-C2H}$ the pi elctron cloud will face repulsion of with the negative charge on the carbon atom.
So by this logic, decreasing order of acidity should be:
$$\ce{H2O > ROH > RNH2 > C2H2}$$
But the $\mathrm{p}K\_\mathrm{a}$ values suggest the very reverse order. What is the fault in my reasoning?
| -3 |
[
[
"\nAccording to\n\n\n$$\n\\begin{array}{lc}\n\\hline\n\\text{Compound} & \\mathrm{p}K\\_\\mathrm{a} \\\\\n\\hline\n\\text{Water} & 15.7 \\\\\n\\text{Methanol} & 16.0 \\\\\n\\text{Methylamine} & 36.0 \\\\\n\\text{Ethyne} & 25 \\\\\n\\hline\n\\end{array}\n$$\n\n\nthe order of acidity is\n\n\n$$\\text{water} > \\text{methanol} > \\text{ethyne} > \\text{methylamine}$$\n\n\nYou logic is absolutely correct, but only there is an exception in case of electronegativities of $\\mathrm{sp}$ carbon and $\\mathrm{sp^3}$ nirogen.\n\n\nThe $\\mathrm{sp}$ carbon is more electronegative than $\\mathrm{sp^3}$ nitrogen and hence ethyne is more acidic here. The $\\mathrm{p}K\\_\\mathrm{a}$ values also tell you the same story.\n\n\n",
"1"
],
[
"\npKa values are -logKa.\n\n\nKa is the dissociation constant of the acid\n\n\nSmaller the Ka value, weaker is the acid.\n\n\nTherefore, pka values will go on decreasing from strong acid to weak acid.\n\n\n",
"-3"
]
] |
https://chemistry.stackexchange.com/questions/89063/convert-the-lowest-ir-wavelength-to-energy-in-j-frequency-in-hz-and-wavenumber
|
Convert the lowest IR wavelength to energy in J, frequency in Hz, and wavenumbers in cm-1
|
Look up the range of wavelengths in nm in the IR region of the electromagnetic spectrum. Convert the lowest IR wavelength to energy in J, frequency in Hz, and wavenumbers in cm-1.
I think I’m going at this the wrong way but if anyone can explain or help with steps that would be really helpful.
$$
E = \frac{(6.676 \times 10^{34})(\pu{3.0 \times 10^8 m/s})}{(\pu{700 \times 10^{-9} nm})} = \pu{0.28 \times 10^{-19} J}
$$
$$
V = \frac{\pu{0.28 \times 10^{-19} J}}{6.626 \times 10^{-34}}= 4.28 \times 10^{13}
$$
| -2 |
[
[
"\nWell, you took $\\lambda=700nm$ which is correct. \n\n\n\n> \n> if anyone can explain or help with step\n> \n> \n> \n\n\nThere isn't really any explanation involved here. The question is very straightforward. Recall these three formulae from your chemistry lectures:\n\n\n$$\\text{Energy}=hv=\\frac{hc}{\\lambda}$$\nand$$\\text{wave number}=1/\\lambda$$\n\n\nand $$\\text{frequency}=c/\\lambda$$\n\n\nSubstitute the necessary values of the constants in them, and now you are good to go!\n\n\n**PS:** $1nm=10^{-9}m\\neq10^{-9}nm$! Be careful with how you inter convert the units.\n\n\nHope it helps!\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/89061/how-is-the-number-of-successful-collisions-in-a-time-step-distributed
|
How is the number of successful collisions in a time-step distributed?
|
In a spatial homogeneous reaction system with two different molecules, how is the number of collisions which will lead to a reaction in a fixed time-step distributed?
My guess: Poisson distribution, as collisions are Bernoulli-experiments, usually have a low probability for success and the number of trials are huge.
| 3 |
[] |
https://chemistry.stackexchange.com/questions/89057/calculation-of-total-pressure-given-ratio-of-pressure-equilibrium-constants
|
Calculation of total pressure given ratio of pressure equilibrium constants
|
>
> The equilibrium constants $K\_{\mathrm{p},1}$ and $K\_{\mathrm{p},2}$ for the reactions $\ce{X(g) <=> 2 Y(g)}$ and $\ce{Z(g) <=> P(g) + Q(g)}$, respectively, are in the ratio of $1:9$. If the degree of dissociation of $\ce{X}$ and $\ce{Z}$ are equal then the ratio of total pressure at these equilibria are?
>
>
>
**My Approach:** Let the degree of dissociation be $\alpha$
So $K\_\mathrm{c,1}$ for the first reaction is
$K\_\mathrm{c,1} = \frac{(2\alpha)^2}{1+\alpha}$
Similarly, $Kc\_2$ for the second reaction is $K\_\mathrm{c,2} = \frac{(\alpha)(\alpha)}{1+\alpha}$
I don't know how to proceed from here and relate these to the ratio of $K\_{\mathrm{p},1}$ and $K\_{\mathrm{p},2}$. I need help here.
| 0 |
[
[
"\nThe equilibrium expressions for these reactions are:\n\n\n$$ K\\_{\\text{p},1} = \\frac{p\\_{\\text{Y}}^2}{p\\_{\\text{X}}}$$\n$$ K\\_{\\text{p},2} = \\frac{p\\_{\\text{P}}p\\_{\\text{Q}}}{p\\_{\\text{Z}}}$$\n\n\nSince $p\\_\\text{A} = x\\_\\text{A}p\\_\\text{T}$ for component $\\text{A}$, we can rewrite them as:\n\n\n$$ K\\_{\\text{p},1} = \\frac{(x\\_\\text{Y}p\\_{\\text{T},1})^2} {x\\_\\text{X}p\\_{\\text{T},1}} = \\frac{x\\_{\\text{Y}}^2p\\_{\\text{T},1}}{x\\_\\text{X}}$$\n$$ K\\_{\\text{p},2} = \\frac{(x\\_\\text{P}p\\_{\\text{T},2})(x\\_\\text{Q}p\\_{\\text{T},2})}{x\\_\\text{Z}p\\_{\\text{T},2}} = \\frac{x\\_{\\text{P}}x\\_{\\text{Q}}p\\_{\\text{T},2}}{x\\_\\text{Z}}$$\n\n\nTaking the ratios:\n\n\n$$ \\frac{ K\\_{\\text{p},1} }{K\\_{\\text{p},2}} \n = \\frac{x\\_{\\text{Y}}^2x\\_\\text{Z}}{x\\_\\text{X}x\\_\\text{P}x\\_\\text{Q}}\\cdot\\frac{p\\_{\\text{T},1}}{p\\_{\\text{T},2}}\\label{eqn}\\tag{1}$$\n\n\nWe're asked the ratios of the total pressures, and we're given the ratios of the equilibrium constants, so we just need to find the molar fractions ratio in that equation. We know that the degree of dissociation, let's call it $\\alpha$, is the same for both reactions.\n\n\nAssume the first equilibrium starts with $m$ moles of $\\text{X}$:\n\n\n$$\\begin{array}{cccc}\n &\\ce{X(g) &<=> &2 Y(g)}\\\\ \\hline\n\\text{I} &m & &0\\\\\n\\text{C} &-\\alpha m & &+2 \\alpha m\\\\\n\\text{E} &m - \\alpha m & & 2 \\alpha m\n\\end{array}\n$$\n\n\nThen, the molar fractions are:\n\n\n$$x\\_\\text{X} = \\frac{n\\_\\text{X}}{n\\_\\text{T}} = \\frac{m - \\alpha m}{m - \\alpha m + 2 \\alpha m} = \\frac{1 - \\alpha}{1 + \\alpha}$$\n$$x\\_\\text{Y} = \\frac{n\\_\\text{Y}}{n\\_\\text{T}} = \\frac{2\\alpha m}{m - \\alpha m + 2 \\alpha m} = \\frac{2\\alpha}{1 + \\alpha}$$\n\n\nSimilarly, for the second equilibrium, assuming we start with $r$ moles of $\\text{Z}$:\n\n\n$$\\begin{array}{cccccc}\n &\\ce{Z(g) &<=> & P(g) &+ &Q(g)}\\\\ \\hline\n\\text{I} &r & &0 & & 0\\\\\n\\text{C} &-\\alpha r & &+\\alpha r & &+\\alpha r\\\\\n\\text{E} &r - \\alpha r & &\\alpha r & & \\alpha r\n\\end{array}\n$$\n\n\nAnd we calculate the mole fractions of each species:\n\n\n$$x\\_\\text{Z} = \\frac{n\\_\\text{Z}}{n\\_\\text{T}} = \\frac{r - \\alpha r}{r - \\alpha r + \\alpha r + \\alpha r} = \\frac{1 - \\alpha}{1 + \\alpha}$$\n$$x\\_\\text{P} = \\frac{n\\_\\text{P}}{n\\_\\text{T}} = \\frac{\\alpha r}{r - \\alpha r + \\alpha r + \\alpha r} = \\frac{\\alpha}{1 + \\alpha}$$\n$$x\\_\\text{Q} = \\frac{n\\_\\text{Q}}{n\\_\\text{T}} = \\frac{\\alpha r}{r - \\alpha r + \\alpha r + \\alpha r} = \\frac{\\alpha}{1 + \\alpha}$$\n\n\nPlugging the found molar fractions into $\\ref{eqn}$, we have:\n\n\n$$\\frac{ K\\_{\\text{p},1} }{K\\_{\\text{p},2}} \n = \\frac{\\left(\\frac{2\\alpha}{1 + \\alpha}\\right)^2 \\left(\\frac{1 - \\alpha}{1 + \\alpha}\\right)}{\\left(\\frac{1 - \\alpha}{1 + \\alpha}\\right)\\left(\\frac{\\alpha}{1 + \\alpha}\\right)\\left(\\frac{\\alpha}{1 + \\alpha}\\right)}\\cdot\\frac{p\\_{\\text{T},1}}{p\\_{\\text{T},2}}$$\n\n\nWhich simplifies to:\n$$\\frac{ K\\_{\\text{p},1} }{K\\_{\\text{p},2}} \n = 4\\cdot\\frac{p\\_{\\text{T},1}}{p\\_{\\text{T},2}}$$\n\n\nSince $\\displaystyle \\frac{ K\\_{\\text{p},1} }{K\\_{\\text{p},2}} = \\frac{1}{9}$, we get:\n\n\n$$\\frac{1}{9} \n = 4\\cdot\\frac{p\\_{\\text{T},1}}{p\\_{\\text{T},2}}$$\n\n\nAnd finally:\n\n\n$$\\frac{p\\_{\\text{T},1}}{p\\_{\\text{T},2}} = \\frac{1}{36}$$\n\n\nSo the total pressures are on a ratio of $1:36$.\n\n\n",
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https://chemistry.stackexchange.com/questions/89054/decarboxylation-of-alpha-amino-acids-to-amines
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Decarboxylation of alpha-amino acids to amines [closed]
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**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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Closed 5 years ago.
[Improve this question](/posts/89054/edit)
In this reaction, what are the ideal circumstances?
[](https://i.stack.imgur.com/Iz0Vb.png)
1. Which base would be a great choice?
2. How much time does it take for tryptophan to fully decarboxylate?
3. What is the ideal temperature for this type of reaction?
| 0 |
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"\nJust heating with base is ineffective.\nLaval and Golding [[1](https://doi.org/10.1055/s-2003-37512)] describe a general procedure for the decarboxylation of α-amino acids using *N*-bromosuccinimide and Ni(0) in aqueous solution. The yields are pretty good.\n\n\n### Reference\n\n\n1. Laval, G.; Golding, B. T. One-Pot Sequence for the Decarboxylation of α-Amino Acids. *Synlett* **2003**, No. 4, 0542–0546. DOI: [10.1055/s-2003-37512](https://doi.org/10.1055/s-2003-37512).\n\n\n",
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https://chemistry.stackexchange.com/questions/89049/why-does-ph-change-with-temperature
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Why does pH change with temperature? [duplicate]
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**This question already has an answer here**:
[Effect of Temperature on pH of Water](/questions/39608/effect-of-temperature-on-ph-of-water)
(1 answer)
Closed 5 years ago.
Why does pH change with temperature? I recently read up on some chemistry notes, and found out that the higher the temperature of distilled water, the lower the pH. Why? Does this apply to other fluids too? Does this also mean dipping a litmus paper into 2 beakers of water of different temperature would yield different results?
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"\nWater always contains a certain amount of $\\ce{OH-}$ and $\\ce{H3O+}$ ions. The pH is the negative decadic logarithm of the $\\ce{H3O+}$ concentration (pH = - log(c($\\ce{H3O+}$)). As chemical equilibria might change with temperature so does the concentration of $\\ce{H3O+}$ and therefore the pH.\n\n\nAlthough I never tried, I suppose this could affect measurements using pH paper. On the other hand the pH changes might very well be to small to be detectable using this method.\n\n\nHow this affects other fluids / equilibria strongly depends on how each equilibrium is affected by temperature. At higher temperatures equilibria are shifted in the endothermic direction and at lower temperatures they are shifted in the exothermic direction. You will need to look up the details for each equilibrium I suppose. \n\n\n",
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"\n\n> \n> ...the higher the temperature of distilled water, the lower the pH. Why? \n> \n> \n> \n\n\n$$\\ce{2H2O(l)<=> H3O+(aq) + OH-(aq)}$$ ΔH = +ve\n\n\nThe autoionization of water is endothermic therefore the increase in temperature will cause the equilibrium to shift to the endothermic direction, to oppose the increase in temperature.\n\n\nThis increases the amount of H3O+ ions in the system.\nsince [pH=-log [H3O]](https://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_pHandpOH.htm) this decreases the pH.\n\n\n\n> \n> Does this also mean dipping a litmus paper into 2 beakers of water of different temperature would yield different results?\n> \n> \n> \n\n\nYes it does, if you have very good litmus paper and a sharp eye!, if one beaker was 273K (pH 7.47) and another at 373K (pH 6.14) the difference in pH would be [1.33](https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Acids_and_Bases/Acids_and_Bases_in_Aqueous_Solutions/The_pH_Scale/Temperature_Dependence_of_the_pH_of_pure_Water).\n\n\nFor this very reason, a worker would have to take temperature into account [when doing pH measurements.](https://www.westlab.com/blog/2017/11/15/how-does-temperature-affect-ph)\n\n\n\n> \n> A common oversight is when you take a sample from a process tank and make the pH measurement in the laboratory. At that time, you are probably not measuring pH at the same temperature as the temperature in the process tank, this means that you will not have the correct pH value for the sample. **Thus a pH value without a temperature value is meaningless**. This can be simply overcome by testing pH on site and at the source of the sample.\n> \n> \n> \n\n\n",
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