url
stringlengths 64
181
| question_title
stringlengths 15
159
| question_text
stringlengths 47
17.9k
| upvotes
int64 -14
183
| answers
list |
|---|---|---|---|---|
https://chemistry.stackexchange.com/questions/170793/is-there-a-good-synonym-for-a-group-connected-to-a-n-atom
|
Is there a good synonym for "a group connected to a N-atom"?
|
I'm writing a publication in which I compare the effect of alkyl, alkene and similar chains of different length connected to a N-atom of a heterocyclic organic compound. Often I have to refer to these chains, *without the N-atom*.
Right now I'm using something like "the ethyl/propyl/etc. group connected to the N-atom", which feels awkward. Is there a better chemical term I could use? Based on my online research, *amine group* or *amine substituent* appears to be the closest, but it includes the nitrogen itself.
Some examples (not from my work):
[](https://i.stack.imgur.com/RmBOXm.png)
Is there a term that properly describes the encircled parts?
| 3
|
[
[
"\nI think \"N-substituent\" is just fine.\n\n\nA substituent refers to whatever is sticking off something. The prefix \"N-\" doesn't change that; rather, it qualifies what it is sticking off. So the N-substituent doesn't mean \"the nitrogen plus the substituent\", but rather \"the substituent sticking off the nitrogen\".\n\n\n(As a parallel example: in, say, aromatic rings it is common to refer to a C-2 substituent or an *ortho*-substituent or something. This doesn't include the carbon on the ring, it just refers to whatever's sticking off it.)\n\n\n",
"5"
],
[
"\nInspired by the locant *N*- and by biochemical nomenclature for glycolisation, I could see use of the term “*N*-linked moiety”.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/170792/interpretation-of-helmholtz-energy-and-gibbs-energy
|
Interpretation of Helmholtz energy and Gibbs energy
|
I know that this question has many other variations on this site, but I'm trying to see if I understood Helmholtz and Gibbs energy properly or not. The material I'm reading from is ***Physical chemistry*** by Thomas Engel and Philip Reid, third edition, and ***An introduction to Thermal Physics*** by Daniel Schroeder.
In Schroeder's book, Chapter $5$, the author defined Helmholtz energy, $A$, as $A= U-TS$, where,
$U$ is internal energy of the system and
$S$ is the system's final entropy.
The author says that, *It is the total energy needed to create the system minus the heat you can get free from the environment at temperature $T$*. He further states that it is the available or "free" energy.
Then, for a system in constant pressure $(P)$ and temperature $(T)$ environment, he defines Gibbs energy, $G$, as $G= U-TS+PV$, where $PV$ is the atmospheric work term that's in enthalpy,
$H$ $=$$U+PV$.
Also, from Engel and Reid, chapter $6$, we have, for isothermal process,
$dA$ $\le$ $đw\_{total}$ $...$$(1)$,
including expansion and non-expansion work, where the equality is satisfied for reversible process. Equation (1) allows us a way to calculate maximum work that a system can do on the surroundings.
And in a similar manner $dG$ $\le$ $đw\_{non-expansion}$ $....$$(2)$, where the equality is satisfied for reversible process.
It is stated that equation $(2)$ allows one to calculate maximum non-expansion work that can be produced.
---
Now, this is my understanding:
For reversible process, $A$ represents total available internal energy, and a part of $A$, which is $G$, is the available energy to do non-expansion work.For irreversible processes the inequalities (1) and (2) gives the lower bound for expansion and non-expansion work respectively. Is it the correct way of understanding or am I way off?
Also, if $A$ does represent total available internal energy for reversible process then what does it represent for irreversible process? Thanks in advance
| 5
|
[] |
https://chemistry.stackexchange.com/questions/170782/when-naoh-ionizes-in-water-does-the-oh-react-with-the-water-molecules-or-with
|
When NaOH ionizes in water, does the OH– react with the water molecules or with the hydronium ions from the dissociation of water?
|
NaOH in water becomes $\ce{Na+}$ and $\ce{OH-}$ ions. If it's a Bronsted Lowry base then the $\ce{OH-}$ ions will take $\ce{H+}$. But FROM where exactly - from the water molecules or from the hydronium ions (from the dissociation)?
Also do all the $\ce{OH-}$ ions from $\ce{NaOH}$ combine with the $\ce{H3O+}$ to form 2 molecules of water or some of them remain,as there is an excess of $\ce{OH-}$ ?
If $\ce{OH-}$ reacts with the water then the reaction is: $$\ce{OH- + H2O -> H2O + OH-}.$$ So from this reaction we just conclude that $\ce{OH-}$ exist freely in our solution.
| 3
|
[
[
"\nEven pure water contains $\\ce{H2O, OH- and H3O+}$. The two ions can either react with water, or with each other:\n\n\n$$\\ce{OH-(aq) + H2O(l) <=> H2O(l) + OH-(aq)}\\tag{1}$$\n\n\n$$\\ce{H3O+(aq) + H2O(l) <=> H2O(l) + H3O+(aq)}\\tag{2}$$\n\n\n$$\\ce{OH-(aq) + H3O+(aq) <=> H2O(l) + H2O(l)}\\tag{3}$$\n\n\nAll three are fast acid/base reactions. In neutral water, reaction (3) in the forward direction is most unlikely because both reactants are present at very low concentration.\n\n\n\n> \n> If it's a Bronsted Lowry base then the OH− ions will take H+. But FROM where exactly?\n> \n> \n> \n\n\nBoth reaction (1) and (3) will happen. Because of reaction (3), the $\\ce{H3O+}$ concentration will rapidly decrease, making the rate of this reaction very slow. Both reactions (1) and (3) have hydroxide as one of the reactants, but the other reactant is either water (present at high concentration) or hydronium (present at very low concentration in alkaline solution). This makes reaction (1) more likely.\n\n\nAs the OP mentioned, reaction (1) does not use up any water, so that reaction will continue to proceed at a high rate. In fact, some argue that $\\ce{OH-}$ is not the best description, and it should rather be $\\ce{H2O.OH-}$ or $\\ce{H3O2-}$. For the positive ion, $\\ce{H3O+}$ is a compromise, acknowledging that there is no \"naked\" hydrogen ion in water, but not trying to describe higher aggregates.\n\n\n\n> \n> does the OH– react with the water molecules or with the hydronium ions from the dissociation of water?\n> \n> \n> \n\n\nThe short answer is both.\n\n\n",
"8"
]
] |
https://chemistry.stackexchange.com/questions/170778/how-can-i-determine-a-stereogenic-centre
|
How can I determine a stereogenic centre?
|
I have a problem understanding stereogenic centres. A stereogenic centre is specified (in the case of organic chemistry that I am interested to) as a carbon atom which is bounded to four different atoms or groups. However in my teacher's notes in order to make clear the way we find stereogenic centres he gives as an example a cholesterol molecule with its stereogenic centres that are shown in the image:
[](https://i.stack.imgur.com/s2QkM.png)
What I cannot understand is why the 18th carbon atom is said to be stereogenic since it is bounded to four carbon atoms?
Why is the third carbon atom said to be stereogenic since it is bounded to 2nd and 4th carbon atoms which are identical?
I have some confusion regarding the definition of an assymetric carbon atom. Even though it seems to me quite straight forward the way it is implemented seems to not obey to my theoretical expectations.Can someone please justify the way stereogenic centres are specified in the case of cholesterol? Could someone give me some examples of chemistry molecules and the stereogenic centres that we meet there ? Any help is really needed and appreciated!
| 2
|
[
[
"\nAs your teacher's notes says, a carbon atom in a molecule which is bounded to four different atoms or groups can be considered as stereogenic center. An example for a molecule with a carbon atom which is bounded to four different atoms is 1-bromo-1-chloro-1-fluoroethane:\n\n\n[](https://i.stack.imgur.com/5pfZb.png)\n\n\nThe stereogenic center is the $\\ce{C}$1, which is bounded to another $\\ce{C}$, and three halogens, $\\ce{Br, Cl,}$ and $\\ce{F}$, respectively.\n\n\nAn example for a molecule with a carbon atom which is bounded to four different groups is 3-methylhexane:\n\n\n[](https://i.stack.imgur.com/winWm.png)\n\n\nI choose this molecule to clear up your confusion on stereogenic centers in cholesterol. The stereogenic center in 3-methylhexane is $\\ce{C}$3, which attached to 3 different carbon groups, mamely methyl, ethyl, and propyl. The forth group is hydrogen.\n\n\nNow, let's see one of your concerns:\n\n\n\n> \n> Why is the third carbon atom said to be stereogenic since it is bounded to 2nd and 4th carbon atoms which are identical?\n> \n> \n> \n\n\nDo they identical? Not really. What is the next atom to $\\ce{C}$2? It is $\\ce{C}$1 which is $\\mathrm{sp^3}$ hybridized. What is the next atom to $\\ce{C}$4? It is $\\ce{C}$5 which is $\\mathrm{sp^2}$ hybridized. Thus, $\\ce{C}$2 and $\\ce{C}$4 are not really identical such as methyl and ethyl groups are not identical in my second example.\n\n\nAnd, if you really looks carefully there is no steocenter on $\\ce{C}$18 (which is a methyl group). I believe the stereocenter you really talking about is $\\ce{C}$13 (also, if you have problem with $\\ce{C}$13, you should have problem with $\\ce{C}$8 as well). On $\\ce{C}$13, the three other groups are $\\ce{C}$12 ($\\ce{CH2}$ group), $\\ce{C}$14, and $\\ce{C}$17 ($\\ce{C}$18 methyl is the forth). You misunderstood as $\\ce{C}$14 and $\\ce{C}$17 are identicle because both of them are $\\ce{CH}$ groups. It is also true that $\\ce{C}$14 attached to two other groups with $\\ce{CH2}$ ($\\ce{C}$15) and $\\ce{CH}$ ($\\ce{C}$8), as well as $\\ce{C}$17 attached to two similar groups with $\\ce{CH2}$ ($\\ce{C}$16) and $\\ce{CH}$ ($\\ce{C}$20). Thus, to determine whether they are different or not, you need to go one step more. For $\\ce{C}$14, $\\ce{C}$8 attached to $\\ce{CH}$ ($\\ce{C}$9) and $\\ce{CH2}$ ($\\ce{C}$7).\nFor $\\ce{C}$17, $\\ce{C}$20 attached to $\\ce{CH2}$ ($\\ce{C}$22) and $\\ce{CH3}$ ($\\ce{C}$21). That observation exclusively shows $\\ce{C}$14 and $\\ce{C}$17 are not identicle.\n\n\n",
"2"
],
[
"\nConsider the third carbon for instance, As you mentioned, it is connected to three carbon atoms, which are its immediate neighbors. Since the three atoms are carbons, we now go to the next-nearest neighbors, and now we see that the next-nearest neighbors for atom 3 are different. Thus, the local environment for the atom number three has four types of atoms (one O from -OH group, one H, and two distinct carbon atoms). This same reasoning can be extended to atom number 13, where we see that the neighbors of the four carbon atoms connected to atom 13 are in different chemical environment, rendering the atom number 13 to be a chiral center.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170776/how-can-we-make-seacrete-biorock-grow-faster
|
How can we make seacrete / biorock grow faster?
|
Seacrete is an interesting building material (precipitated limestone in seawater) but is quite slow to grow.
<https://naturalbuildingblog.com/seacreteseamentbiorock/>
<https://en.wikipedia.org/wiki/Biorock>
With the purpose of making it grow as fast as possible, for the purpose of building structures, are the variables known (salinity, temperature, exact electric current, any others) that affect its rate of accretion/'growth' to boost the process, i.e., faster in cold or warmer water, salinity, etc.
I.e, have any known studies been made of the optimal conditions for fast seacrete accretion.
| 0
|
[
[
"\nIt is a calcarious deposit that builds up on steel with cathodic protection in seawater. Similar to limestone. It is desirable to reduce current needed for cathodic protection otherwise of no particular value that I have heard of.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170775/why-did-industry-develop-mainly-to-fix-nitrogen-in-ammonia-haber-bosch-instead
|
Why did industry develop mainly to fix nitrogen in ammonia (Haber-Bosch) instead of in nitrate?
|
Since approximately WWI, humanity has been able to fix nitrogen gas from the atmosphere ($\ce{N2}$) into compounds that are much more bio-available, like ammonia and nitrate. The first (hard) step is the Haber-Bosch process $\ce{N2 + 3H2 -> 2NH3}$, which has to be carried out under elevated pressure, temperature and highly anoxic conditions.
Given that our atmosphere is 80% $\ce{N2}$, 20% $\ce{O2}$ and therefore naturally quite oxic, given that nitrate seems to have a much lower enthalpy of fusion(\*), given also that typical fertilizer prescriptions are overweight on $\ce{NO3}$ compared to $\ce{NH3}$ (in ratios like 4/1)... why did humanity not develop an industrial process emulating e.g. fixation in lightning and thus synthesizing useful $\ce{NO\_x}$ for its fertilizer needs as opposed to $\ce{NH3}$?
(\*) In effect, if the Haber-Bosch process is followed by the Ostwald process to make nitric acid, it seems there's going to be a massive loss of exergy (not least because the extraction of $\ce{NH3}$ from the reaction mixture in the Haber-Bosch plant seems to involve a massively wasteful [cooling-down step](https://scitechdaily.com/a-giant-leap-towards-a-greener-future-breakthrough-in-sustainable-ammonia-and-fertilizer-production/)). As an illustration I looked up the standard enthalpy of formation of a mixture (1 mol $\ce{NH3}$ + 1 mol $\ce{KHCO3}$ + 2 mol $\ce{O2}$) vs a stoichiometrically equivalent mixture (1 mol $KNO\_3$ + 2 mol $H\_2O$ + 1 mol $\ce{CO2}$) and the former contains $\pu{448.5 kJ}$ more than the latter mixture!
| 1
|
[
[
"\nI do not address agriculture application, but extreme energetic demand and inefficiency of direct synthesis of nitrogen oxides by the Birkeland–Eyde process, or the Frank–Caro process of thermal capture using $\\ce{CaC2}$. Synthesis of nitric acid via ammonia by H-B process is the most cost and energy efficient process,\n\n\nThere is always long way from labs to industry. Consider how many revolutionary new designs of rechargable cells have been announced during decades.\n\n\nThat quote of the abstract of your linked resource;\n\n\n\n> \n> This analysis shows that the energy consumption for NOX synthesis with plasma technology is almost competitive with the commercial process with its current best value of 2.4 MJ mol N−1, which is required to decrease further to about 0.7 MJ mol N−1 in order to become fully competitive.\n> \n> \n> \n\n\nIndustry has to first see the decisive advantage to change the technology, as aside energy demand there would be new technology challenges and extra investments.\n\n\nMuch more energy than needed for endothermic reaction $\\ce{N2+O2 -> 2NO}$ is wasted, because of creating condition where such reaction can provide sufficient yield by sufficient rate.\n\n\n",
"2"
],
[
"\n**Because the capital investment and energy consumption if the Haber + Ostwald process is still far lower than the alternatives**\n\n\nThe reason why specific industrial processes are chosen are usually economics and practicality. While some chemical reactions look, in principle, to be better alternatives (*surely* direct routes to nitrate are better than making ammonia first and then oxidising it?) they may not be practical or economic.\n\n\nWe also know that some plants can fix nitrogen using enzymatic reactions at normal temperatures and pressures. Every chemist wishes there were a way to industrialise that. It is clearly *possible* but nobody has ever demonstrated it on an industrial scale. And industrial routes to nitrate were known before the Haber process was developed. But those direct processes (using plasma reactions) were *far* more expensive in both capital and energy inputs than the, apparently, indirect, combination of Haber and Ostwald processes. 100 years later this is still true. The [linked paper shows](https://pubs.rsc.org/en/content/articlelanding/2021/ee/d0ee03763j) that the energy cost of plasma processes to nitrate is *still* about five times higher than the the Haber Ostwald process (per mol of fixed nitrogen). Plus, the capital and maintenance costs are higher.\n\n\nIn short, the choice of which processes are used is based on accessible known reactions when the processes is adopted and the overall economics of the competing processes. Plasma processes were and are still far from competitive despite the theoretical advantages of a direct route to nitrate.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170648/why-cant-the-product-of-an-acid-reacting-with-water-react-with-water-itself-and
|
Why can't the product of an acid reacting with water react with water itself and produce OH-?
|
Why can't the product of an acid reacting with water react with water and produce $\ce{OH-}$?
Take the reaction below as an example:
$$\ce{H2CO3(aq) + H2O(l) <=> HCO3-(aq) + H3O+(aq)}$$
$$\ce{HCO3-(aq) + H2O(l) <=> CO3^2-(aq) + H3O+(aq)}$$
Why can't the $\ce{CO3^2-}$ react with water (not $\ce{H3O+}$) and produce $\ce{OH-}$ ions? If it can then why is the solution acidic and not basic?
PS. I am a high school student and would really appreciate a straightforward answer.
| 3
|
[
[
"\nIt definitely does. Let's simplify it to a monoprotic acid (like acetic acid, $\\ce{CH3COOH}$), which I'll just represent as $\\ce{HA}$ for simplicity. In a solution of the acid, you will have *all* of the following species floating around: $\\ce{HA}$, $\\ce{A-}$, $\\ce{H2O}$, $\\ce{H3O+}$, $\\ce{OH-}$. *Every* conceivable reaction between these species will happen, such as:\n\n\n$$\\begin{align}\n\\ce{HA + H2O &<=> A- + H3O+} \\tag{1} \\\\\n\\ce{A- + H2O &<=> HA + OH-} \\tag{2} \\\\\n\\ce{2 H2O &<=> H3O+ + OH-} \\tag{3}\n\\end{align}$$\n\n\nTo obtain a completely accurate value for the pH, and the equilibrium position, it is necessary to take all of these equations into account.\\*\n\n\nBut often we can get away with less work, and only bother with the first. Why? One qualitative way of looking at it is: when we add a weak acid $\\ce{HA}$ to water, we know that *very little of it will dissociate anyway*; this is kind of by definition. So, there isn't much $\\ce{A-}$ in the system anyway; and the second reaction can pretty much be ignored.\n\n\nThat's not to say that there is *no* $\\ce{OH-}$ in the system. There *is* still $\\ce{OH-}$ floating around. For example, in a pH 5 solution (i.e. $[\\ce{H+}] = \\pu{10^{-5} M}$), we have that $[\\ce{OH-}] = \\pu{10^{-9} M}$.\\* It's still there, it's just that there's very little of it.\n\n\nA more formal explanation would involve solving it properly and verifying that the *true* solution is very close to the *approximate* solution obtained by ignoring eqns. (2) and (3).\n\n\nOn the other hand, if $\\ce{HA}$ is a strong acid (like $\\ce{HCl}$), then the reverse reaction $\\ce{Cl- + H2O -> HCl + OH-}$ is completely negligible because $\\ce{Cl-}$ is a pitifully weak base. (Strong acids have very weak conjugate bases.)\n\n\n\n\n---\n\n\n\\* In fact, these equations are not independent; note that (1) + (2) = (3), so there are only two 'pieces of information' to be obtained from these reactions, one corresponding to $K\\_\\mathrm{a}$ and one corresponding to $K\\_\\mathrm{w}$. But that's a story for another day.\n\n\n",
"8"
],
[
"\nI really like this question because it digs at wanting to understand what's going on beneath the equations you've been given to know *why* that equation works.\n\n\nOrthocresl answered it really well, but I might be able to get a little more to what pH and pOH are and give a little more understanding to what's meant by acidic and basic, and what equilibrium is. I feel like that might be as much your question as whether or not H2O can react with CO3 2- (apologies, first time answering a question here, and not sure how to format this correctly).\n\n\ntl;dr version up front:\n\n\nPure water has an equal amount of H+ and OH-, both of them exist in water. We consider this neutral, neither acidic or basic. Anything that moves the concentration to favor more H+ is considered an (Arrhenius) acid, anything that moves the concentration to favor more OH- is considered a base. Even so, there will still be both H+ and OH- in the solution. Individual water molecules will continue to accept protons (H3O+, usually shown as H+) and donate protons (OH-), but there will be more water molecules accepting a proton to form H3O+ than donating them to form OH- at lower pH.\n\n\nnon tl;dr version:\n\n\nVery, very few chemical reactions are truly one way/irreversible unless the products leave the system entirely. Almost all will be occurring in both directions at the same time. But, one will be occurring more/faster. Let's say reaction A occurs twice as fast as the reverse reaction B (A: a + b -> c; B: c -> a + b). Since A occurs twice as fast a B, then there will be more products, c, than reactants, a and b. However, c will still be producing a and b. Eventually you'll get enough c that the reaction will reach a point that it will *look* like the reaction stopped. If reaction A occurs twice as fast as reaction B, this will be when there's twice as much products as reactants.\n\n\nBasically, if a + b produce c twice per minute, and c produces a + b once per minute, you'll eventually get twice as much c as you have a and b, and the reaction will look like it stopped from a macro point of view, but will continue to occur in your solution.\n\n\nTake what I said there from a conceptual view. It's useless for practical purposes. If you want to know the math, this video explains it really well (<https://www.youtube.com/watch?v=2PM1yc_z4Bk&list=PL2ub1_oKCn7qmUZ80MJDPaRTdgJm8ZenX>).\n\n\nI don't see a good way to not discuss different definitions of acids even though I feel it's a little beyond this question. Arrhenius acid is anything that lowers the pH of pure water (base is the opposite). But that's not a good definition. A much better one is the Bronsted definition, anything that donates a proton acts as an acid, and anything that accepts a proton acts as a base (there's a 3rd definition, but it's difficult to understand at first). In this case, your carbonic acid is acting as an acid when it donates a proton to water to form H3O+, and water acts as a base. However, the reverse happens as well. H3O+ will donate a proton to HCO3-. In which case H3O+ is acting as an acid. H2O can also act as a Bronsted acid and donate a proton to form OH-. So yes, CO3 2- can and does act as a base and accept a proton from H2O, and you get OH- floating around.\n\n\nIn *any* aqueous solution, you'll end up with both H3O+ and OH-. This is usually measured with pH. pH is actually the result of this equation pH = -log(10)([H+]). That equation means that pH is the negative base 10 logarithm of the concentration of protons (hydronium ions) in the solution. However, because we know there's an equilibrium between [H+] and [OH-], we can actually measure this in terms of the concentration of OH- ions instead, pOH. pOH is remarkably, the negative logarithm of the concentration of hydroxide ions, pOH = -log(10)([OH-]).\n\n\nSo let's say you have a pH of 5, a slightly acidic solution. You'll have a pOH of 9. The pOH measures in the opposite direction, pOH below 7 is basic, above 7 is acidic. These two measure will equal 14. A pH of 1 will correspond to a pOH of 13. So even at a very high acidity of pH 1, you will still have H2O acting as an acid and donating protons to form OH-. It will just be happening very slowly, and so there won't be many of them.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/170647/uv-absorbance-titration-to-determine-binding-stoichiometry
|
UV absorbance titration to determine binding stoichiometry
|
I previously posted a question but I think I did not explain correctly. I want to construct a Job plot using UV-Vis titration to find binding stoichiometry. I have 20uM of ligand solution and 20uM of DNA solution. Volume of each solution is 800uL. The total concentration of ligand and dsDNA remained unchanged(20uM) during entire titration. Ligand intercalates into dsDNA.
I want to know the protocol to do UV-Vis titration. Like how I should proceed. Do I need to prepare solution of different concentration everytime (for example ligand solution --- 20uM, take UV absorbance,
Now make a complex solution of say 18uM ligand and 2uM dsDNA and so on)
I have no idea what to do.
Thank you.
[](https://i.stack.imgur.com/wyOox.jpg)
| -2
|
[
[
"\nI can advice only in general methodic, I cannot say what you can afford wrt the particular solutions, regarding their stability and interaction.\n\n\nIf the volume of available solutions **is not** the limiting factor, the easiest way is to prepare extra solution mixture for each measurement.\n\n\n* Take $x\\ \\pu{ \\mu L}$ of the ligand solution\n* Take $a - x\\ \\pu{\\mu L}$ of DNA solution\n* It gives $\\frac{x}{a-x}$ ligand/DNA ratio.\n\n\nThen you can measure the series of plot point absorbance vs ligand/DNA ratio.\n\n\nIf the volume of available solutions **is** the limiting factor, you may try the series of incremental additions of the ligand and following measurements.\n\n\nYou would take\n\n\n* $V\\_1\\ \\pu{ \\mu L}$ of solution of the prior ratio $r\\_1$\n* add $V\\_2\\ \\pu{ \\mu L}$ of the ligand solution to get the new ratio:\n\n\n$$r\\_2 = \\frac{r\\_1(V\\_1+V\\_2) + V\\_2}{V\\_1}$$\n\n\nBut there are 2 cons:\n\n\n* Higher cummulative experimental errors with excessive solution manupulations.\n* Possible effects of stability and interaction of incrementally mixed solutions.\n\n\nWhen you have measurement done, you can visially or numerically approximate the poins by 2 lines and to take the intersection as the point of equivalence.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170646/find-percentage-of-carbon-monoxide-reacted-using-ideal-gas-equation
|
Find percentage of carbon monoxide reacted using ideal gas equation
|
**Question**
A palladium or platinum catalyst was used in an automobile to convert carbon monoxide gas to carbon dioxide according to the following reaction:
$$\ce{2CO(g) + O2(g) -> 2CO2(g)}$$
A chemist researching the effectiveness of a new catalyst combines a 2.0 : 1.0 mole ratio mixture of carbon monoxide and oxygen gas (respectively) over the catalyst in a 2.45 L flask at a total pressure of 745 torr and a temperature of 552 K. When the reaction is complete, the pressure in the flask has dropped to 552 torr.
What percentage of the carbon monoxide was converted to carbon dioxide?
(a) 67.7 %
(b) 85.7 %
(c) 77.5 %
(d) 57.8 %
(e) 46.5 %
---
**My attempt**
I calculated the moles of gas in the original system
$$n = \frac{PV}{RT} = \frac{745 \times 2.45}{8.31 \times 552} = 0.398 \, \mathrm{mol}$$
Then I worked out the moles of $\ce{CO}$ and $\ce{O2}$ in the original system using the mole fractions.
$$\ce{2CO + O2 -> 2CO2}$$
moles of $\ce{CO}$ = $(2/3) \times 0.398 = 0.265\, \mathrm{mol}$
moles of $\ce{O2} = (1/3) \times 0.398 = 0.1326\,\mathrm{mol}$
moles of $\ce{CO2} = 0\, \mathrm{mol}$
Then I calculated the number of moles in the new system.
$$n-\frac{PV}{RT} = \frac{552 \times 2.45}{8.31 \times 552} = 0.295\, \mathrm{mol}$$
Then I worked out the moles of $\ce{CO, O2}$ and $\ce{CO2}$ in the new system using the mole fractions.
$$\ce{2CO(g) + O2(g) -> 2CO2(g)}$$
moles of $\ce{CO} = (2/5) \times 0.295 = 0.118\, \mathrm{mol}$
moles of $\ce{O2} = (1/5) \times 0.295 = 0.059\, \mathrm{mol}$
moles of $\ce{CO2} = (2/5) \times 0.295 = 0.118\, \mathrm{mol}$
Therefore percentage of CO reacted
$$\%\_{\ce{CO}} = \frac{0.265 - 0.118}{0.265} \times 100 = 55.5\, \%$$
The correct answer according to the mark scheme is **C) 77.5%**. Is there any other method to get this answer?
| 0
|
[
[
"\nUsing the ideal gas state equation as the starting point:\n\n\n$pV=nRT \\implies \\frac{p\\_1V}{p\\_2V}=\\frac{n\\_1RT}{n\\_2RT} \\\\\n\\implies \\frac{p\\_1}{p\\_2}=\\frac{n\\_1}{n\\_2}$\n\n\nleading to the consequence that the gas pressure is proportional to the total molar amount of present gases.\n\n\nAs there is molar conversion in ratio 2/3, the final/initial pressure ratio 2/3 would mean 100 % conversion.\n\n\nThe rest below is just interpolation between the ratio 1 (0 %) and 2/3 (100 %)\n\n\nIf\n\n\n* $\\alpha$ is the percentual conversion\n* $p\\_\\mathrm{tot,ini}$ is the initial pressure of gases\n* $p\\_\\mathrm{tot,fin}$ is the final pressure of gases\n\n\nthen\n\n\n$$\\frac{p\\_\\mathrm{tot,fin} }{ p\\_\\mathrm{tot,ini}} = 1 - \\frac 13 \\cdot\\frac{\\alpha}{100}$$\n\n\n$$\\alpha = (1 - \\frac{p\\_\\mathrm{tot,fin} }{ p\\_\\mathrm{tot,ini}})\\cdot 300 \\ \\%= (1 - \\frac{\\pu{552 torr} }{ \\pu{745 torr}})\\cdot 300 \\ \\% \\approx 77.7\\ \\% $$\n\n\n",
"3"
],
[
"\nLet:\n\n\nA represent $\\ce{CO}$\n\n\nB represent $\\ce{O2}$\n\n\nC represent $\\ce{CO2}$\n\n\nSo the reaction becomes:\n\n\n$$\\ce{2A + B->2C}$$\n\n\nWe begin by calculating the initial molar fraction of A.\n\n\nSince only A and B are initially present:\n\n\n$$y\\_{Ao}+y\\_{Bo}=1$$\n\n\nSince the initial molar ratio between both species is 2:1, then:\n\n\n$$y\\_{Ao}=2\\;y\\_{Bo}$$\n\n\nUsing both equations to solve for $y\\_{Ao}$, we get:\n\n\n$$y\\_{Ao}=\\frac{2}{3}$$\n\n\nNext, we observe that between the initial and final state, temperature and volume are constant, but there is a change in moles between reactants and products, so:\n\n\n$$T=T\\_o$$\n\n\n$$V=V\\_o$$\n\n\n$$\\Delta n=-1$$\n\n\nIf we divide the ideal gas law in the final state by the initial state, we get:\n\n\n$$\\frac{PV}{P\\_oV\\_o}=\\frac{nRT}{n\\_oRT\\_o}$$\n\n\nCancelling all equal terms:\n\n\n$$\\frac{P}{P\\_o}=\\frac{n}{n\\_o}$$\n\n\nThe molar ratio on the right hand can be expressed in terms of conversion $X$ and expansion coefficient $\\epsilon$ after performing a molar balance of A, B, and C:\n\n\n$$\n\\require{cancel}\n\\begin{align}\nn\\_A=n\\_{Ao}-ax \\\\\nn\\_B=n\\_{Bo}-bx \\\\\nn\\_C=n\\_{Co}+cx \\\\ \n\\hline\nn=n\\_o+\\Delta nx\n\\end{align}\n$$\n\n\nIn this case, $\\Delta n=c-a-b$, and small $x$ represents the non-normalized conversion with respect to the limiting reagent. In order to normalize it, we arbitrarily define A as the limiting reagent, and its conversion as:\n\n\n$$X=\\frac{ax}{n\\_{Ao}}$$\n\n\nSolving for $x$:\n\n\n$$x=\\frac{n\\_{Ao}X}{a}$$\n\n\nSubstituting $x$ into the overall molar balance:\n\n\n$$n=n\\_o+n\\_{Ao}\\frac{\\Delta n}{a}X$$\n\n\nDividing both sides by $n\\_o$:\n\n\n$$\\frac{n}{n\\_o}=1+\\frac{n\\_{Ao}}{n\\_o}\\frac{\\Delta n}{a}X$$\n\n\nOr equivalently:\n\n\n$$\\frac{n}{n\\_o}=1+y\\_{Ao}\\frac{\\Delta n}{a}X$$\n\n\nThen, we define:\n\n\n$$\\epsilon=y\\_{Ao}\\frac{\\Delta n}{a}$$\n\n\nAnd substitute it above to get:\n\n\n$$\\frac{n}{n\\_o}=1+\\epsilon X$$\n\n\nWhich can also be substituted above:\n\n\n$$\\frac{P}{P\\_o}=1+\\epsilon X$$\n\n\nSolving for $X$:\n\n\n$$X=\\frac{\\frac{P}{P\\_o}-1}{\\epsilon}$$\n\n\nCalculating $\\epsilon$:\n\n\n$$\\epsilon=y\\_{Ao}\\frac{\\Delta n}{a}=\\frac{2}{3}\\;\\frac{-1}{2}=-\\frac{1}{3}$$\n\n\nPlugging in all numerical values:\n\n\n$$X=\\frac{\\frac{552}{745}-1}{-\\frac{1}{3}}$$\n\n\nSo we get:\n\n\n$$X=0.777$$\n\n\nOr equivalently 77.7%\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/170643/do-boron-and-silicon-form-metallic-type-bonds-in-alloys
|
Do Boron and Silicon form metallic-type bonds in alloys?
|
For context I am a physics student currently doing a project that involves metallic glasses. I am trying to figure out if you can get Ionic bonds inside an alloy with metalloids. More specifically, I have 2 examples.
* Iron,Nickel,Boron
* Palladium, Copper, Silicon
I have recently found out about Arkel–Ketelaar triangles and assume they answer my question but I'm not entirely sure how to use them in the context of an alloy of three components. I am also not sure if I need to use a particular measure of electronegativity or if any will do.
| 1
|
[] |
https://chemistry.stackexchange.com/questions/170642/equilibrium-constant-g-vs-go
|
Gibbs free energy in standard state vs. equilibrium
|
I have a problem with the definition of the standard Gibbs energy and its connection to the equilibrium constants.
I think, that I've basically understood what the different equation mean but there is one thing, I'm unable to understand:
On the one hand:
One may describe a chemical reaction with $\Delta G=\Delta G^\circ + RT\ln{Q}$. In equilibrium $\Delta G = 0$ and the equation reads $\Delta G^\circ = -RT \ln{K}$.
On the other hand:
The definition of standard state is very clear: pressure = 1 bar and all reactants and products must have activity = 1.
If I consider these two aspects separately, everything seems to be fine. But these two concepts have to be valid at the same time, what leads to $\Delta G^\circ = 0$ (always), since $K=1$ (all activities are per definition = 1).
Therefore, $\Delta G^\circ$ would be always zero. I know that this isn't true, but I don't understand why.
Can anyone explain this to me?
Thanks!
| 2
|
[
[
"\nAs explained in the comments, the standard state conditions lead to $Q=1$ and therefore $$\\Delta G=\\Delta G^\\circ+ RT\\ln{1}=\\Delta G^\\circ$$ On the other hand at equilibrium $Q=K$ and so $$\\Delta G=\\Delta G^\\circ + RT\\ln{K}$$ This of course leads to $\\Delta G^\\circ = -RT\\ln{K}$ since at equilibrium $\\Delta G=0$.\n\n\nSo you might want to think of it as multiple statements:\n\n\n1. For the conversion of reactants to products in their standard states $Q=1$\n2. At equilibrium $\\Delta G=0$\n3. At equilibrium $Q=K$\n4. $\\Delta G^\\circ$ is the free energy change for conversion of reactants to products in their standard states.\n\n\nThe first statement is consistent with the definition of standard states.\nThe second and fourth statements follow from combination of the first and second laws of thermodynamics.\nThe third statement is a definition of $K$.\n\n\n$K$ and $\\Delta G^\\circ$ are very much connected, but $\\Delta G^\\circ$ *does not* describe the change in free energy from reactants to products *at equilibrium*.\n\n\n",
"3"
],
[
"\nWhat you enter into $K$ are *not* the activities of the pure reactants and pure products at standard state (if you did then, yes, $K$ would be 1). Rather, it is their activities *at equilibrium* (raised, of course, to the power of their respective stochiometric coefficients). And, at equilibrium, these activities are generally not equal to one. \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170640/metaphosphoric-acid-can-it-exist-as-monomer
|
Metaphosphoric acid - can it exist as monomer?
|
I read in my textbook that N2O5 can be prepared by reaction of HNO3 and P4O10 . A product HPO3 is also formed.
Then while discussing oxoacids of phosphorus , it said that metaphosphoric acid exists only in polymeric form.
So is there a contradiction ?
| -4
|
[
[
"\n$\\ce{HPO3}$ is an empirical formula for a material that is indeed polymerized. Phosphoric acid is actually [any of several acids containing phosphorus(V)](https://en.wikipedia.org/wiki/Phosphoric_acid); the formula $\\ce{HPO3}$ represents a limiting case for long chains, or cycles containing a finite number (at least three) of $\\ce{HPO3}$ units.\n\n\nWhat is often called phosphoric acid, the monomer $\\ce{H3PO4}$, is technically orthophosphoric acid.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170636/is-there-a-difference-between-a-1g-modes-and-a-g-modes
|
Is there a difference between $A_{1g}$ modes and $A_{g}$ modes?
|
I know that both are totally symmetric and that the '$1$', according to the Mulliken table, refers to symmetry around the $C\_2$ axis but I was reading some papers and found that modes were sometimes referred to as $A\_{1g}$ and sometimes referred to $A\_{g}$, as shown below (difference is between undistorted and distorted unit cell).
[](https://i.stack.imgur.com/xzhW1.png)
Is $A\_{g}$ equivalent to $A\_{1g}$ here because the Raman tensors are identical for both (both are a 3 by 3 maxtrix with a leading diagonal of $a, a, b$) and yet the names are slightly different?
| 2
|
[
[
"\nTo answer your question of A$\\_{1g}$ and A$\\_g$ are the same? No, they are not,\nbecause they correspond to irreducible representations of different symmetry point groups that differ in the allowed symmetry operations. However, they correspond to the same vibrational mode, since the quadratic functions associated with both these modes are same. If you see the Character Table for D$\\_{6h}$ and C$\\_{6h}$, you would see that A$\\_{1g}$ corresponds to the fully symmetric mode of vibration, and there are other modes of A, B and E types. However, once you lower the symmetry to C$\\_{6h}$ symmetry, you loose some elements of symmetry since C$\\_{6h}$ is a subgroup of D$\\_{6h}$. This would be more clear if you see the character tables available online [D$\\_{6h}$](http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=606&option=4) and [C$\\_{6h}$](http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=506&option=4)\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170631/stable-thioesters-in-biological-millieu
|
Stable thioesters in biological millieu?
|
It seems that most thioesters are highly reactive in cells and blood due to the high concentration of biological thiols. Are there any derivatives of thioesters that are stabilized at physiological conditions by certain functional groups? Fatty acid metabolism is mediated by transfer of CoA to the fatty acid, so I imagine this bond must be somewhat stable. Is this stabilized by the nitrogen that sits a couple carbons away from the thioester?
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170629/why-are-proteins-made-up-of-alpha-amino-acids-and-not-beta-amino-acids
|
Why are proteins made up of alpha amino acids and not beta amino acids?
|
Why are proteins made up of alpha amino acids and not beta amino acids? Or gamma amino acids? My idea on this would be that a world with beta or gamma amino acids would be too complicated? Is that true or are there other reasons?
| 0
|
[
[
"\nAdding to Poutnik's useful answer, polyptides/proteins made of alpha-L-aminoacids are capable of participating in a number of [secondary structures](https://proteinstructures.com/structure/secondary-structure/), such as alpha-helix, beta-pleated-sheet, &c. In turn these participate (along with the variety provided by aminoacid sidechains) in the construction of higher-order structural features with biochemical activity.\n\n\nSo one can say at least that things would be different if biochemical evolution had for some reason favoured beta or gamma aminoacids over alphas.\n\n\nAlso, as Poutnik suggested, the betas and gammas might more easily form unwanted cyclic lactams competing with any reaction (if still possible) to form (?pseudo)polypeptide chains. Then, even if such chains could still form in reasonable yield with beta and gamma aminoacids, the added methylene or dimethylene character of each aminoacide residue, (due to CaCb or CaCbCc carbon-chains taking the place of Ca), might restrict the overall range of physical-chemical properties in the pseudopolypeptide, irrespective of its side-chain -- perhaps, for example, it might make all aminoacid contributions more lipophilic than a corresponding alpha-amino-acid residue would do.\n\n\nReally these considerations only add some approach towards exemplifying the difficulties of answer pointed out by poutnik. It may be doubted whether the subject holds enough urgency to persuade anybody to undertake the probably difficult course of synthetic and analytic chemical experimentation that would be needed to test the ideas.\n\n\n",
"1"
],
[
"\nQuestions why? in context of various levels of life design are tricky and rather philosophical.\n\n\nBeta/gamma amino acids would not be optically active, neither (alternative) proteins formed from them. As nucleic acids are optically active due present ribose/deoxyribose, there would be problem with their synthesis.\n\n\nAside of that, considering the length of eventual lactame (cyclical amid) cycle, 3-member cycle of alpha amino acids (as possible unwanted side reaction) is much harder to form than 4/5-member cycle of beta/gamma amino acids.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170624/how-are-the-formulae-of-isochoric-processes-used-in-adiabatic-processes
|
How are the formulae of isochoric processes used in adiabatic processes?
|
We have the formula for an isochoric process:
$c\_V = \frac{\mathrm{d}E}{\mathrm{d}T}$ at constant volume.
$c\_V=$ molar heat capacity at constant volume.
But it's given for adiabatic changes,
$$\delta E=c\_V\cdot \Delta T=W$$
How is this an equation for an adiabatic process when it's derived from equations of an isochoric process?
Is an adiabatic process also isochoric?
Where am I going wrong?
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170621/how-do-you-synthesize-1-fluoro-4-tribromomethylbenzene-from-toluene
|
How do you synthesize 1-fluoro-4-(tribromomethyl)benzene from toluene
|
I got this question from an o-chem II exam which kinda puzzles me a bit[](https://i.stack.imgur.com/xttFs.png)
My proposed path is:
[](https://i.stack.imgur.com/f6Xaf.png)
which doesn't seem to be correct. Is there any other way within the knowledge of o-chem I & II?
| 2
|
[
[
"\nThis route will work.\n\n\np-Nitrotoluene is the major product of toluene nitration and may be isolated by distillation and recrystallisation, procedure [here](https://www.oc-praktikum.de/nop/en/instructions/pdf/1001_en.pdf)\n\n\nThis recent [paper](https://pubs.acs.org/doi/10.1021/acsomega.1c02825) describes low temperature and light catalysed Balz-Schiemann conditions which are rather more user-friendly than some of the older procedures.\n\n\nThis [patent](https://pubchem.ncbi.nlm.nih.gov/patent/EP-0045430-A1) describes a method of brominating the p-F-toluene to the benzotrifluoride\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/170620/if-the-effective-nuclear-charge-of-berylliumbe-is-less-than-sodiumsna-how
|
If the effective nuclear charge of Beryllium(Be) is less than Sodium's(Na), how is the ionization energy of Be higher than Na?
|
According to Slater's rules the effective nuclear charge of Beryllium and Sodium is 1.95 and 2.20 respectively.
That means that the outermost electron of Na feels a stronger attraction from the nucleus than the outermost electron of Be.
But we also know that the atomic radius of Be is smaller than Na and hence Be atom is smaller than Na. That means that the electrons of Be are closer to its nucleus so they must feel a greater attraction than the electrons of Na
The way i think about the above, somewhat contradictory statements, is that the pulling of the outermost electron in Na has to be bigger because there is also a greater shielding effect. Indeed the electron in the 3s orbital of Na feels a greater attraction than the 2s electron of Be but because of the inner electrons pushing it out it can't come closer to the nucleus and so Na ends up being bigger than Be.
Is my way of thinking this right?
My question though has to do with the 1st Ionization energy of Be being greater than Na. Why is that? It's the effective nuclear charge that tells us how much attraction the electrons feel from the nucleus.
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170617/what-is-the-entropy-change-in-isochoric-process
|
What is the entropy change in isochoric process
|
I have studied that entropy increases with increase in temperature and it decreases with increase in pressure but in case of isochoric process both are happening at the same time but still the overall entropy of the system is somehow increased.
I want to know why the increase in entropy with increase in pressure dominates the decrease in entropy with increase in pressure since both are happening at the same time, resulting in an overall increase in entropy of system.
| 1
|
[
[
"\nAt an isobaric process, entropy increases with temperature, as you provide heat to the system.\n\n\nAt an isothermic process, entropy decreases with pressure, as you provide work to the system, which is converted to heat, leaving the system.\n\n\nAt an isochoric process, entropy increases with temperature (which increases the pressure), as you provide heat to the system.\n\n\nWith implied constant heat capacity $C\\_V$:\n\n\n$\\Delta S = \\int\\_{T\\_1}^{T\\_2}{\\mathrm{d}S}=\\int\\_{T\\_1}^{T\\_2}{\\frac{\\delta Q}{T}}=\\int\\_{T\\_1}^{T\\_2}{\\frac{C\\_V\\cdot \\mathrm{d}T}{T}}=C\\_V \\cdot \\ln{\\frac{T\\_2}{T\\_1}}$\n\n\n",
"1"
],
[
"\nFor an arbitrary differential change of a single phase substance, we have $$dS=C\\_p\\frac{dT}{T}-\\left(\\frac{\\partial V}{\\partial T}\\right)\\_PdP$$For an isochoric process, $$dV=\\left(\\frac{\\partial V}{\\partial T}\\right)\\_PdT+\\left(\\frac{\\partial V}{\\partial P}\\right)\\_TdP=0$$or$$dP=-\\frac{\\left(\\frac{\\partial V}{\\partial T}\\right)\\_P}{\\left(\\frac{\\partial V}{\\partial P}\\right)\\_T}dT$$So, for an isochoric process, $$dS=C\\_p\\frac{dT}{T}+\\frac{[\\left(\\frac{\\partial V}{\\partial T}\\right)\\_P]^2}{\\left(\\frac{\\partial V}{\\partial P}\\right)\\_T}dT=\\left[C\\_p-T\\left(\\frac{\\partial V}{\\partial T}\\right)\\_P\\left(\\frac{\\partial P}{\\partial T}\\right)\\_V\\right]\\frac{dT}{T}$$For all materials, the second term in brackets is always less than $C\\_p$, and is equal to $C\\_p-C\\_v$.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170616/do-electroluminescent-quantum-dot-displays-suffer-from-burn-in-effect-like-ole
|
Do electroluminescent quantum dot displays suffer from "burn-in" effect like OLED displays?
|
To my knowledge, compared to oled, elqd displays will be very cheap to mass produce using ink jet printing process. Quantum dots can have very narrow FWHM which allows displays to have wider color gamut of >90% rec2020. The response times of quantum dots are also faster than oled. Most importantly, it is also self-emissive like OLED.
However, one of the biggest concerns I have is if it suffers the same "burn-in" effect like OLED. To my knowledge, the challenge of commercializing ELQD displays is that the lifetime of the blue emitter currently only lasts a couple of hours at 1000 nits. However, I'm not sure if the term "burn-in" is separate from "lifetime." In an OLED display, when a static logo is shown for a really long time without any "burn-in" compensation technologies, the logo will burn in and the respective pixels will permanently shift colors which is very noticeable. I wonder if the same occurs with ELQD displays. In addition, I have also read somewhere that quantum dots are also not truly inorganic materials because the "ligands" are organic.
| 2
|
[] |
https://chemistry.stackexchange.com/questions/170615/what-is-the-highest-volume-change-for-common-miscible-liquid-pairs-and-common-s
|
What is the highest volume change for common miscible liquid pairs, and common solvent/solid pairs? [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Update the question so it focuses on one problem only by [editing this post](/posts/170615/edit).
Closed 8 months ago.
[Improve this question](/posts/170615/edit)
When a solute is added to a solvent, the total volume of the solution is lower than that of the solute+solvent. A good example of this is the addition of salt to water, causing an overall volume decrease. What are the most dramatic example of this for commonly available solvents and solutes, both liquid+solid, and liquid+liquid?
| 1
|
[] |
https://chemistry.stackexchange.com/questions/170613/why-is-the-conductivity-of-nacl-nabr-and-nai-decreasing-with-temperature
|
Why is the conductivity of NaCl, NaBr and NaI decreasing with temperature?
|
I collected the conductivity of NaCl, NaBr, and NaI from 30 celsius to 80 celsius. I used the solutions mixed in water.
In all instances, the conductivity seemed to reduce as the temperature increased. I am unsure if it was an experimental error, or if my results actually make sense. Please state your reasons for thinking its an error or not.
Experiment Details: 50 ml, 0.1 molar of all solutions. I used vernier sensors for detecting the temperature and the conductivity.
For this experiment, I had a 50ml beaker on a hotplate and had the conductivity and temperature vernier sensors dipped in the liquid.
(I made sure the sensors are only touching the solution and not the bottom of the beaker). Then, I just waited for the hotplate to heat up the liquid to 80 Celsius.
From research, the opposite seems to be accurate, but I have conducted many trials giving the same trend of decreasing conductivity.
Source: <https://link.springer.com/article/10.1007/BF02877571>
(This is my first question, please let me know any other details that would be required to answer this question )
| 1
|
[] |
https://chemistry.stackexchange.com/questions/170612/cell-notation-of-all-aqueous-solutions
|
Cell notation of all aqueous solutions
|
If some redox reaction happened with no solid anode or cathode, like all aqueous solutions, something like the unbalanced reaction
$$
\ce {MnO4^- (aq) + C2O4^-^2 (aq) -> Mn^+^2 + CO2 (aq)}
$$ What would an electrochemical cell notation represent in this situation? Is it even **logical** to try to get a cell notation? I'm really confused about that topic. I heard my professor saying something about a *platinum electrode* and I don't think this is rational by any means. Do things like this exists?
| -1
|
[
[
"\nHere is how to explain the use of the platinum electrodes.\n\n\nIf two solutions are prepared separately, one with acidic potassium permanganate, and one with sodium oxalate, they can be joined by a salt bridge. Now if one platinum plate is dipped into either solution, a potential difference can be measured between these two plates, or between these two solutions. Electrons are produced in the oxalate solution according to the half-equation :$$\\ce{C2O4^{2-}-> 2 CO2 + 2 e^-}$$These electrons are collected by the platinum foil, and will go through the electric wire up to the other platinum foil dipped in the permanganate solution. Here they produce the following half-equation $$\\ce{MnO4^- + 8 H^+ + 5 e^- -> Mn^{2+} + 4 H2O}$$ I should be mentioned that in the oxalate solution, the $\\ce{Na^+}$ ions must quit the solution to maintain electric neutrality of the solution, and they do it by crossing the salt bridge joining the two solutions. When arriving in the permanganate solution they compensate the missing charges in solution. It takes some time to explain this movement, due to missing charges in the permanganate solution.\n\n\nFirst, it should be mentioned that, in order to equilibrate the charges in the whole system, $10$ electrons must be produced by oxalate, and $10$ electrons must be consumed in permanganate. The corresponding half-equations become :\n$$\\ce{5 C2O4^{2-} -> 10 CO2 + 10 e-}$$ $$\\ce{2 MnO4^- + 16 H^+ + 10 e- -> 2 Mn^{2+} + 8 H2O}$$\nIf the acid producing $\\ce{H+}$ is $\\ce{H2SO4}$, $16$ $\\ce{H^+}$ ions are consumed in the half-equation. As a consequence, 8 $\\ce{SO4^{2-}}$ ions are left in solution, after the arrival of the electrons through the platinum foil. These $16$ negative charges in excess are compensated by\n\n\n* $2$ $\\ce{K^+}$ remaining after the consumption of the permanganate ions coming from $2$ $\\ce{KMnO4}$,\n* $2$ $\\ce{Mn^{2+}}$ ions produced by the second half-reaction\n* $10$ $\\ce{Na^+}$ ions having crossed the salt bridge,\n\n\nAs a consequence, at the end of the whole process, the composition of the solutions are :\n\n\n* No ions in the former oxalate solution (just some $\\ce{CO2}$), and\n* $\\ce{2 Mn^{2+} + 2 K^+ + 10 Na^+ + 8 SO4^{2-}}$ in the former permanganate solution.\n\n\nThis is also the composition of the final solution one would obtain if the reaction had been done in only one phase, by mixing both solutions, with or without any platinum plates, which are now useless.\n\n\n",
"1"
],
[
"\nNote that an electrode(1) in the wider sense (electrochemical half cell) consist of\n\n\n* an electrode(2) in narrower sense(a conductor)\n* a redox system\n\n\nThe electrode(2) may be\n\n\n* the part of a redox system like $\\ce{Zn(s)|Zn^2+(aq)}$\n* or may stay aside like $\\ce{Pt(s)|H+(aq)/H2(g)}$\n\n\nWithout at least a half cell, a cell notation does not make sense.\n\n\nIf more redox systems exist in parallel in the same place, they have all at assumed equilibrium the same redox potential.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170610/why-is-the-sp3-sp3-reductive-elimination-between-the-r1-and-norbornene-in-the-ca
|
Why is the sp3-sp3 reductive elimination between the R1 and norbornene in the Catellani Cycle ''not favorable''?
|
I have recently started reading a book about C-H activation and I cannot figure out why in the Catellani reaction cycle the R1-norbornene reductive elimination side reaction which can be a major issue isn't favored.The author only briefly mentions sp3-sp3 coupling saying that it isn't favored but it doesn't provide an explanation (Because he probably assumes the reader already knows).
**(Step 6.)**
[](https://i.stack.imgur.com/7Fe9L.png)
The book: C-H Activation (Topics in Current Chemistry, 292)
| 0
|
[
[
"\nAll most all organic reaction mechanisms are given according to the products received after doing variety of kinetic experiments. If you are familiar with reaction progressing energy diagrams, you know all intermediate reactions are reversible to give the thermodynamically most stable product(s). Thus, in the case of choice for $\\ce{C\\_\\mathrm{sp^3}-C\\_\\mathrm{sp^2}}$ bond formation versus $\\ce{C\\_\\mathrm{sp^3}-C\\_\\mathrm{sp^3}}$, the final product(s) rececieved go through more stable $\\ce{C\\_\\mathrm{sp^3}-C\\_\\mathrm{sp^2}}$ bond formation (because reverse reaction require more energy to break).\n\n\n[](https://i.stack.imgur.com/uVfbD.png)\n\n\nThis is the same situation when palladation of norbornene $(\\bf{3})$ ocuurs to give $\\bf{4}$ instead of the presence of the equally reactive $\\ce{R^2-Y}$ in the reaction mixture, which is a terminal alkene.\n\n\nAlso keep in mind that, there is a possibility to receive tricyclic benzo-norbornane product by $\\ce{C\\_\\mathrm{sp^2}-C\\_\\mathrm{sp^3}}$ reductiove elimination from $\\bf{5}$, based on steric hindrence, solvent used, base used, etc. (Ref.1):\n\n\n[](https://i.stack.imgur.com/vy1zc.png)\n\n\n\n\n---\n\n\n**References:**\n\n\n1. Xiaojin Wua and Jianrong (Steve) Zhou, \"An efficient method for the Heck–Catellani reaction of aryl halides,\" *Chem. Commun.* **2013**,*49(94)*, 11035-11037 (DOI: DOI <https://doi.org/10.1039/C3CC46381H>).\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/170608/how-many-carbon-hydrogen-oxygen-nitrogen-and-phosphorus-atoms-are-there-in-t
|
How many carbon, hydrogen, oxygen, nitrogen, and phosphorus atoms are there in the observable universe?
|
If I could somehow reliably count all the carbon, nitrogen, hydrogen, oxygen, and phosphorus atoms in the observable universe, what number would I come up with?
| 3
|
[
[
"\nTwo key numbers can be estimated from known observable things in astronomy (and cosmology where current big bang models explain the processes that created the \"light\" elements and their abundance and observations broadly agree with theory). Spectroscopy allow astronomers to estimate the abundance of many elements in stars and galaxies.\n\n\nThere are also some estimates of how many nuclei there are (based on estimates of the mass of the observable universe.\n\n\nWe can combine the estimates about the number of nuclei with their relative abundance to give some approximate counts for elements.\n\n\nThe number of nuclei [is estimated to be](https://www.universetoday.com/36302/atoms-in-the-universe/) between $\\pu{10^78}$ and $\\pu{10^82}$ (which are both crazy big numbers though the upper estimate is 10,000 times the lower estimate so don't expect precision).\n\n\nBut we also understand the relative abundance, though most estimates count mass proportion (counting nucleons not atoms). Using [the numbers in Wikipedia](https://en.wikipedia.org/wiki/Abundance_of_the_chemical_elements) and adjusting the estimates for the mass of each elemental nucleus gives the following proportions:\n\n\nH 92% \n\nHe 7.5% \n\nO 0.08% \n\nC 0.05% \n\nN 0.01%\n\n\nIf we take the mid estimate of total nuclei as $\\pu{10^80}$ then there are approximately $\\pu{8E76}$ oxygen atoms, $\\pu{5E76}$ carbon atoms and $\\pu{1E76}$ nitrogen atoms in the observable universe (if I've done my maths right).\n\n\n",
"9"
]
] |
https://chemistry.stackexchange.com/questions/170603/predict-who-is-the-acid-and-who-is-the-base-in-an-acid-base-reaction-using-pka
|
Predict who is the acid and who is the base in an acid-base reaction using pKa
|
In my book (Bruice) it is specified how it is possible to compare the pKa of two substances to understand who acts as an acid and who as a base. Here's the example in the book:
$\ce{NH3 + H2O}$.
$\mathrm{p}K\_\mathrm{a}$(NH3) = 36
$\mathrm{p}K\_\mathrm{a}$(H2O) = 15.7
Now it is explained that, being the $\mathrm{p}K\_\mathrm{a}$ of water lower than that of ammonia, water will be the substance that will behave as an acid, so $\ce{NH4+}$ and $\ce{OH-}$ will be the products.
But this reasoning, in my opinion, does not make much sense, as the $\mathrm{p}K\_\mathrm{a}$ of two different reactions are compared: the acid hydrolysis reaction of ammonia, which generates $\ce{NH2-}$ and $\ce{H3O+}$ and whose $\mathrm{p}K\_\mathrm{a}$ is actually 36, and the autoprotolysis reaction of water, which generates $\ce{H3O+}$ and $\ce{OH-}$ and whose $\mathrm{p}K\_\mathrm{a}$ is actually 15.7. So, I wanted to understand if it actually makes sense to compare the $\mathrm{p}K\_\mathrm{a}$ of two different equilibria to predict who the acid is.
Furthermore, also $\ce{NH3}$, and not necessarily $\ce{H2O}$, can act as an acid. Indeed, ammonia can give both acid hydrolysis and basic hydrolysis to which it is possible to associate a $\mathrm{p}K\_\mathrm{a}$ and a $\mathrm{p}K\_\mathrm{b}$.
**Reference** (1) Paula Yurkanis Bruice, Organic Chemistry, 8th ed.; Pearson Education, Inc., 2011, pp. 58.
| 4
|
[
[
"\n\n> \n> But this reasoning, in my opinion, does not make much sense, as the $\\mathrm{p}K\\_\\mathrm{a}$ of two different reactions are compared: the acid hydrolysis reaction of ammonia, which generates $\\ce{NH2-}$ and $\\ce{H3O+}$ and whose $\\mathrm{p}K\\_\\mathrm{a}$ is actually 36, and the autoprotolysis reaction of water, which generates $\\ce{H3O+}$ and $\\ce{OH-}$ and whose $\\mathrm{p}K\\_\\mathrm{a}$ is actually 15.7. So, I wanted to understand if it actually makes sense to compare the $\\mathrm{p}K\\_\\mathrm{a}$ of two different equilibria to predict who the acid is.\n> \n> \n> \n\n\nYou are absolutely right, the argument does not work like that. It is possible to make a different, correct argument based on $\\mathrm{p}K\\_\\mathrm{a}$ values, though.\n\n\nFirst, you write down the two possible reactions:\n\n\n$$\\ce{H2O(l) + NH3(aq) <=> OH-(aq) + NH4+(aq)}\\tag{1}$$\n\n\n$$\\ce{NH3(aq) + H2O(l) <=> NH2-(aq) + H3O+(aq)}\\tag{2}$$\n\n\nI chose the order so that the reactant acting as acid is written first (on the far left) and the conjugate acid produced from the reactant acting as base is written last (on the far right).\n\n\nThen, you compare the strength of the acids on the far left and the far right.\n\n\n(1) $\\ce{compare H2O with NH4+}$\n\n\n(2) $\\ce{compare NH3 with H3O+}$\n\n\nFor reference, here are the $\\mathrm{p}K\\_\\mathrm{a}$ values:\n\n\n$\\mathrm{p}K\\_\\mathrm{a}(\\ce{H2O}) = 15.7$\n\n\n$\\mathrm{p}K\\_\\mathrm{a}(\\ce{NH4+}) = 9.8$\n\n\n$\\mathrm{p}K\\_\\mathrm{a}(\\ce{NH3}) = 36$\n\n\n$\\mathrm{p}K\\_\\mathrm{a}(\\ce{H3O+}) = -1.7$\n\n\nFor (1), $\\ce{NH4+}$ is a stronger acid than $\\ce{H2O}$ (difference of about 6 in $\\mathrm{p}K\\_\\mathrm{a}$), so the equilibrium constant will be smaller than 1. For (2), $\\ce{H3O+}$ is a much stronger acid than $\\ce{NH3}$ (difference of about 38 in $\\mathrm{p}K\\_\\mathrm{a}$), so the equilibrium constant will be much smaller than 1. Neither reaction will be on the side of the products if both reactants are initially present at reasonably high concentrations.\n\n\nExperience confirms this. The species $\\ce{NH2-}$ does not exist in aqueous solution, it would get protonated immediately. Ammonia in water acts as a base, but most of the ammonia stays deprotonated (i.e. ammonia is a weak base). Of course, you can add acid to lower the hydroxide concentration and increase the ratio of ammonium to ammonia to the point where ammonium is the major species.\n\n\n\n> \n> [OP in comments] when we compare the strength of the acids on the far left and the far right, we are still comparing the pKas of two different reactions\n> \n> \n> \n\n\nNo, we are looking at the net reaction of one species donating and one accepting a proton, without showing the proton explicitly. For reaction (1), for example:\n\n\n$$\\ce{H2O(l) <=> H+ + OH-(aq)}$$\n\n\nplus\n\n\n$$\\ce{NH3(aq) + H+ <=> NH4+(aq)}$$\n\n\ngives (after cancelling the protons):\n\n\n$$\\ce{H2O(l) + NH3(aq) <=> OH-(aq) + NH4+(aq)}$$\n\n\nIf you know the equilibrium constants of the \"half reactions\" (or the $\\mathrm{p}K\\_\\mathrm{a}$ values), you can get the equilibrium constant of the acid/base reaction.\n\n\n",
"10"
]
] |
https://chemistry.stackexchange.com/questions/170596/how-exactly-is-heat-supplied-in-an-isothermal-process
|
How exactly is heat supplied in an isothermal process
|
How exactly is heat supplied in an reversible isothermal process. If the temperature of system and surroundings always remains the same, then how come surrounding is supplying heat?
I studied that T of system and surrounding is the same while studying entropy.
| 0
|
[
[
"\nReversibility has two distinct meanings.\n\n\n1. Bidirectionality. You can melt ice to water and freeze it back to ice.\n2. Being in equilibrium along the whole process path. This can be just approximated in real processes by being near equilibrium.\n\n\nThe second meaning requires that reversible process is returned to its initial state by returning the neighbourhood to its initial state.\n\n\nIf the above cannot be done, the process is not reversible(2), even if it is reversible(1).\n\n\n\n\n---\n\n\nThe heat is in your case provided as usually, but infinitely slowly for infinitely long time, with infinitely small temperature difference between the system and its surrounding.\n\n\nA reversible isothermal process is an idealised process lasting infinite (long enough) time, so heat can be provided at zero (small enough to neglect) temperature difference.\n\n\n$$Q \\propto \\Delta T \\cdot \\Delta t$$\n\n\n$$\\lim\\_{\\Delta t \\to \\infty}{\\Delta T} = 0$$\n\n\nE.g., you need 1 hour to transfer heat Q at temperature difference 1 K. \n\nYou need 2 hours for 0.5 K difference. \n\n4 hours for 0.25 K\netc.\nFor time going to infinity and temperature difference going to zero, systems converge to being reversible.\n\n\nWith infinite time, zero temperature difference is needed to provide finite heat.\n\n\nIn reality, scientists choose \"good enough\" approach with small enough differences and long enough time.\n\n\nTruly reversible processes do not exist in reality, as they are assumed to be in equilibrium all the time, having infinite time to achieve anything.\n\n\n",
"1"
],
[
"\nIce fusion at $0°$C in \"hotter\" bath (just hotter than $0°$C) is a good example of a nearly reversible isothermal process. Heat can be introduced in such a system by having a mixture of ice + water in a metallic container, and dipping it into a bigger container full of water at $+1$°C.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170594/weird-needle-like-crystals-formed-in-galvanic-cell-after-drying
|
Weird needle like crystals formed in galvanic cell after drying
|
So I set up a series of galvanic cells for fun; had a brass nut in a glass container and a chunk of zinc in another. I immersed the brass nut and zinc in a solution of saltwater, and connected them with a saltwater-soaked paper towel.
When the side with the brass dried out, a bunch of salt crystals, as well as blue copper carbonate formed. However, a bunch of long, white, needle-like crystals also formed. What could these be?
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170593/why-metal-hydrides-have-low-lattice-enthalpy
|
Why metal hydrides have low lattice enthalpy
|
Lattice enthalpy is inversely proportional to the distance between the ions. And the trend of lattice enthalpies is: $\ce{MF>MCl>MBr>MI}$. But why do metal hydrides have lesser lattice enthalpy than flourides?
[Data](https://www.wiredchemist.com/chemistry/data/lattice-energies)
| 2
|
[
[
"\nAs suggested in the comments, hydride ion size is founded on shifting sands, and even the size of the cation is not really fixed. Everything in ordinary chemistry is at least somewhat squishy.\n\n\nWe should be looking at distances between ion centers and crystal structures. Where the crystal structure is the same we can assess the distance from Lattice constants, which are themselves a little squishy but at least accessible by direct (X-ray diffraction) experiments.\n\n\nLet's look at sodium compounds. Lattice constants in Angstroms (Å) are from the Wikipedia articles on the respective compounds, lattice energies in kJ/mol are quoted from <https://www.wiredchemist.com/chemistry/data/lattice-energies>.\n\n\n$\\ce{NaH}: \\pu{4.98 Å}, \\pu{811 kJ/mol}$\n\n\n$\\ce{NaF}: \\pu{4.62 Å}, \\pu{904 kJ/mol}$\n\n\n$\\ce{NaCl}: \\pu{5.64 Å}, \\pu{769 kJ/mol}$\n\n\nWe would expect that for the same ion charges and crystal structure (the latter true of all the compounds cited above), we would expect smaller lattice constant to correspond to more lattice energy, and this we see. Essentially, the hydride ion holds its valence electrons so loosely that they wander over slightly more volume than the greater number of, but more tightly held, electrons in fluoride ions.\n\n\nThe similar lattice constants for alkali and alkaline earth fluorides versus hydrides facilitates the formation of [solid solutions](https://en.wikipedia.org/wiki/Fluorohydride_salt), which have been considered for hydrogen storage applications.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/170590/reaction-of-btna-dmso-and-phosphate
|
Reaction of BTNA, DMSO, and phosphate?
|
I prepared a solution of n-benzoyl L-tyrosine p-nitroanilide (BTNA) in dimethyl sulfoxide for use in an enzymatic assay. However, mixing it with a solution of phosphate buffer produced a color change for unknown reasons. Here is what I did:
* A 1.4mM solution of BTNA in DMSO was prepared.
* 5mM phosphate buffer was prepared. This was done by dissolving dipotassium phosphate in distilled water and adjusting the pH to 7.59 by addition of hydrochloric acid.
* Into a cuvette was added 0.4mL of the BTNA solution, followed by addition of 0.1mL of the phosphate buffer solution.
When the BTNA solution and phosphate buffer were mixed, the mixture took on a yellow color. I tried mixing the BTNA solution with distilled water without any phosphate added to see if, perhaps, the act of mixing water with the BTNA solution is responsible for the color change. In this case, the mixture did not change color.
My question is thus: Why did mixing 0.4mL of BTNA solution in DMSO with 0.1mL of 7.59 phosphate buffer result in the mixture taking on a yellow color, whereas mixing BTNA solution with water did not result in a color change?
Sidenote: None of the solutions were tested for contaminants.
| 3
|
[] |
https://chemistry.stackexchange.com/questions/170585/is-determination-of-nuclidic-mass-by-nuclear-reaction-currently-state-of-the-art
|
Is determination of nuclidic mass by nuclear reaction currently state of the art for any nuclides?
|
In Section 4.9 of Linus Pauling's General Chemistry Book (Dover 1988 edition, a copy of the WH Freeman and Company 1970 edition) he describes the principle behind the determination of nuclidic mass by nuclear reaction, giving the following beta decay reaction as an example for the determination of the mass of $\ce{{}^{40}\_{19}K\_{21}}$:
$$\ce{{}^{40}\_{19}K\_{21} -> {}^{40}\_{20}Ca\_{20}^+ + e^-}$$
I was wondering if nuclidic masses determined this way remain state of the art (the origin of the currently accepted value of the mass) for any nuclides? More generally, how important is the technique currently?
| 2
|
[] |
https://chemistry.stackexchange.com/questions/170577/electron-orbitals
|
Electron orbitals
|
Can electrons be found anywhere within the space described by a 3D orbital "90% of the time" (as stated in my textbook)? But that would mean they can be found right next to the nucleus or in the space of a lower energy level "90% of their time" (since the spheres and other shapes overlap starting right next to the nucleus).
We also know that electrons cannot jump from one energy level to another without absorbing or releasing energy and that they maintain a specific average distance from the nuceleus as denoted by a 2D depiction of their electron shells in the form of concentric circles (also written in the same textbook).
Even if they only skimmed the surface of the 3D spheres or dumbell shaped orbitals, the latter, e.g. a 2p orbital, still starts close to the nucleus while it should be farther away from the nucleus than a 1s orbital.
PS Thank you for your answers. But please elaborate on them since I did not understand at all. I just finished high school but am extremely curious and confused by the new information we are studying, and I want to understand...
| -1
|
[
[
"\nYes. The electron has a small but not zero possibility to stay quite near the nucleus, and for example at a smaller distance than the traditional radius of the first 1s orbital. The probability of finding an electron is a number that looks a bit like the deviation of a vibrating rope fixed on the nucleus and going to an infinite distance, if all its points are attracted by the nucleus. Hmmm ! Beg the pardon from all real scientists !! It is not like this. It just looks like.\n\n\n",
"0"
],
[
"\n**Classical analogies about simple orbits don't describe electron behaviour well**\n\n\nThinking about electrons as having \"orbits\" is rarely helpful as their behaviour only makes sense in a quantum mechanical (QM) approach. In QM it is meaningless to talk about position and velocity separately and you can't know both precisely at the same time.\n\n\nOrbitals describe the probability of finding electrons in a particular region of space. And, indeed, in some types of orbitals (eg 1s) there is finite probability of the electron overlapping the nucleus (2p orbital have a node there so this isn't true for them): this matters for some other quantum effects involving nuclei and electrons. But the probability described by the shapes of orbitals is not uniform, some areas \"inside\" the area shown in most pictures have a higher probability density than others.\n\n\nSo saying that \"they maintain a specific average distance from the nucleus\" is not really correct: That is a classical analogy that is a poor description of what electrons do. The average distance is a vague way to describe the average of the probability cloud of electron location, but the idea that a distance is \"maintained\" contradicts the quantum picture.\n\n\nAnd the picture involving orbits where being further from the nucleus is \"higher energy\" is far too simple. The energy is a function of the whole probability distributions, not the specific average distance. The electron location clouds of different orbitals **do** overlap.\n\n\nClassical analogies about electron behaviour are not usually helpful. Unfortunately, QM pictures are often not simple until you have reached a certain ability to the complex mathematics behind QM. Until then \"fuzzy clouds of probability\" are the best you can do.\n\n\n",
"0"
],
[
"\n\n> \n> Can electrons be found anywhere within the space described by a 3D orbital \"90% of the time\" (as stated in my textbook)?\n> \n> \n> \n\n\nYes, they can.\n\n\n\n> \n> But that would mean they can be found right next to the nucleus or in the space of a lower energy level \"90% of their time\" (since the spheres and other shapes overlap starting right next to the nucleus).\n> \n> \n> \n\n\nYes, it means that. There is intense overlapping of orbitals.\n\n\n\n\n---\n\n\nParticular points of 3D space around a central force do not belong to any particular energy level, being it around an atom nucleus or a star, because the sum of potential and kinetic energy must be considered.\n\n\nBy other words, the electron energy level does not determine the electron distance to nucleus. It only determines the probability of being at that distance.\n\n\nThe direct analogy is in planetary orbits around the Sun. The Earth is about $\\pu{150000000 km}$ away from the Sun on near circular orbit with orbiting speed near $\\pu{30 km s-1}$. But Earth could have at the same distance higher speed up to $\\pu{42 km s-1}$, being at its perihelion (being the closest to the Sun on an elliptic orbit). Or a much lower speed, being at its aphelion (The most distant point of an elliptic orbit).\n\n\n\n\n---\n\n\nConsider classical gravitational analogy of the probe orbit. The mechanical energy alone(+) of the probe determines the farther distance from the planet the probe can reach. The closest distance(+) is not limited ( but by the radius of the planet ).\n\n\nSimilarly for orbitals. Electrons from all s orbitals can all occur near nucleus, as their zero orbital angular momentum is no limitation. 6s electrons of gold move at relativistic speed there, leading to orbital energy shift and color of gold.\n\n\n(+) - It can be further limited by the conservation of the probe orbital angular momentum.\n\n\n",
"0"
],
[
"\n\n> \n> [OP ...] that would mean they can be found right next to the nucleus\n> \n> \n> \n\n\nThey **are** close to the nucleus at times. The nucleus is tiny compared to the dimensions of a bond (or an atomic radius), so the electron can come pretty close. When it comes too close, it can get captured (electron capture, for certain radioactive nuclei), or gets scattered (by forces that only act at very short range and are usually ignored in chemistry because these are very rare events).\n\n\n\n> \n> [OP ...] or in the space of a lower energy level \"90% of their time\" (since the spheres and other shapes overlap starting right next to the nucleus).\n> \n> \n> \n\n\nThe most powerful way to explain this is by treating electrons as a wave (you might have heard orbitals mentioned together with wave function). Just as we can hear multiple musical notes at the same time in the same place, electrons can occupy the same space. The technical term is superposition of waves. Below are two standing one-dimensional waves, one with a single node and the other with four nodes, superimposed (amplitudes are added).\n\n\n[](https://i.stack.imgur.com/B1Q3Z.gif)\n\n\nEven though the two waves are using the same space, they are independent of each other, and you can separate them mathematically (because they have different wavelength and frequency). Our ears do the same thing with music, taking a single input stream and separating out the different pitches of musical notes and overtones.\n\n\nThe waves describing electrons are more complicated (they are three-dimensional and have an exponential decay away from the nucleus rather than ending at a defined position), but the idea of superposition still holds. Here is a depiction of the 3p wave function (a snapshot, not showing the time-dependency like I did above), showing various nodes (zero probability) and that there is no defined boundary to where electrons might be found (exponential decay means it gets very unlikely for large distances from the nucleus).\n\n\n[](https://i.stack.imgur.com/WIx00.gif)\n\n\nSource: <https://mathematica.stackexchange.com/questions/32378/is-there-something-like-densityplot3d-to-visualize-atomic-orbitals>\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170574/why-is-the-inductive-effect-possible
|
Why is the inductive effect possible?
|
Firstly, I must preface that I am a Biology student. To elaborate my question, I was looking for the reason why carboxylic acids are more acidic than alcohols, despite both compounds containing *—OH* groups. From my understanding it was because of the inductive effect, an example of this effect is *CH3COOH*, shown in the figure below:
[](https://i.stack.imgur.com/hp6u2.png)
The inductive effect in *CH3COOH* occurs because of the electron deficiency in the carbon atom, *C* (which has a partial positive charge, *δ+*). This electron deficiency on the carbon atom is present due to the more electronegative double bonded oxygen atom, *O*. However, the *O* single bonded to the *C* is also more electronegative, as a result, the electron density of the single bonded *O* moves to the *C* in order to 'counteract' the *δ+* on the *C*, leaving the *H* 'less strongly attached' to the *O* (and therefore, the molecule), thus, increasing its acidity.
Now, in an alcohol there is only one oxygen atom, see the figure below:
[](https://i.stack.imgur.com/1RYjN.png)
Because there is no second *O* attached to the *C*, which would cause it to become more positive, the *O* does not become less positive, by sharing electrons to counteract the positive *C*. Thus, the hydrogen atom is 'more strongly attached' to the *O* which is more negative compared to the single bonded oxygen atom in the carboxylic acid.
My **question** is, why does the carbon atom in the carboxylic acid (first figure) counteract the change and increase its pull for the electrons of the single bonded oxygen atom, instead of becoming more positive?
| 1
|
[
[
"\nThat's an interesting reasoning you give here.\nThe point is, the inductive effect is not the major factor explaining the acidity of a compound.\n\n\nElectronegativity will give insights on the position of electrons along the bond, explaining that the hydrogen atom can be detached by a base in both cases (the lengths of carbon-oxygen bonds also need to be considered to understand the interaction between oxygens and carbon).\n\n\nYet, what makes a carboxylic acid more acidic than an alcohol is the stability of the conjugated base : a carboxylate ion, thanks to the $\\ce{C=O}$ bond on the same carbon as the $\\ce{C-O^-}$, is stable by mesomeric resonance (the lone-pair gained by the oxygen atom releasing the hydrogen can 'move', the direct consequence is that both carbon-oxygen bonds will have the same length, between $\\ce{C-O}$ and $\\ce{C=O}$).\nHowever, the alcoholate will not have this resonance, thus will be less stable (see [this link](https://chem.libretexts.org/Courses/Athabasca_University/Chemistry_360%3A_Organic_Chemistry_II/Chapter_20%3A_Carboxylic_Acids_and_Nitriles/20.02_Structure_and_Properties_of_Carboxylic_Acids) for a longer development on that, and images).\n\n\nI don't know if this is what you wanted to know, do not hesitate to tell me otherwise.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/170563/do-formed-galvanic-cells-counteract-electrolysis
|
Do formed galvanic cells counteract electrolysis?
|
I was wondering about the following:
Suppose we carry out an electrolysis with constant applied voltage. During the process, a galvanic cell should be formed by the reactants which are formed during the electrolysis. This galvanic cell should counteract the electrolysis in some way, if it gets discharged.
However is there any way to quantify this matter? I only found the comment on this website:
<https://www.galvanotechnik-for-you.de/lexikon/electrolysis/>
Any help would be appreciated!
| 0
|
[
[
"\nThis is a very good question I don't know why it has close votes. The proper term you should be searching for is called back-emf (emf=electromotive force). It was known long time at least a century ago that if you start electrolysis of water, stop it temporarily, the ammeter will show current moving in the opposite direction for a very short time. This was attributed to the formation of gaseous products at the electrodes, that formed a galvanic cell i.e., adsorbed oxygen and hydrogen want to for form water again. There are many reasons for back-emf and the formation of electrolytic products is one of them.\n\n\nI assume you can read German. One of the finest works is 2-volumes on Electrochemistry History is by Ostwald (*Elektrochemie, ihre Geschichte und Lehre*). It is freely available from the Internet Archive. The English volumes also exist. Check that reference for the development of these ideas. Many electrochemical engineering textbooks will have more details on back-emf.\n\n\n",
"2"
],
[
"\nConsider an electrolytic cell, consisting of identical half cells with copper electrodes and the same electrolyte - e.g. $\\pu{0.1 M}$ $\\ce{CuSO4}$ solution, separated by a salt bridge with the same solution.\n\n\nAny external nonzero DC voltage will start an electrolysis. As $\\ce{Cu^2+(aq)}$ will be depleted at cathode and formed at anode, there will be formed DC voltage of galvanic cell, acting against the external voltage.\n\n\n$$U = \\frac{RT}{2F} \\ln{\\left(\\frac{a(\\ce{Cu^2+aq},\\mathrm{anode})}{ a(\\ce{Cu^2+aq},\\mathrm{cathode})}\\right)}$$\n\n\n\n\n---\n\n\nFor general cases, electrolysis shifts redox systems at each electrode, the one at cathode obtaining more negative equilibrium potential, the one at anode obtaining more positive equilibrium potential.\n\n\nThe consequence is there is progressively needed higher voltage to keep the constant current. There is also higher threshold voltage (equal to the cell equilibrium voltage) as the electrolysis needs voltage higher than this. It was routinely observed e.g. at charging of rechargable cells.\n\n\n\n\n---\n\n\nWhen net electrolysis happens then, at the cathode, the rate of reduction is (much) higher than the rate of oxidation a vice versa at the anode. Without the external voltage and without electrode galvanic connections, both rates are equal with the zero net effect.\n\n\nBy other words, the rate or reduction exponentially grows with decreasing electrode potential, while the rate of oxidation exponentially grows with increasing electrode potential. Both until it hits other limitations, like ion diffusion. At given scenario, an electrode establishes such a potential that leads to particular value of the net current.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170556/why-does-a-symmetric-stretch-mode-not-have-an-imaginary-frequency
|
Why does a symmetric stretch mode not have an imaginary frequency?
|
Currently studying potential energy systems and our professor asked a question at the end of a lecture and I can't wrap my head around it.
He said, "Picture a 2D PES for the simple reaction of H+H2 -> H2 + H. The direction of the symmetric stretch mode would be to the top right corner where both distances between the hydrogens in the transition state are large. This mode does not have an imaginary frequency. Why is that?"
I can't think of a reason that it would be an imaginary (negative) frequency since the TS must already have one for it to be a TS. So my only thought is that the TS already has one and it cannot have two.
Is my thought process too basic? What am I missing here?
| 3
|
[
[
"\nThe path over the transition state is like an inverted parabola so has a negative frequency, i.e. its an upside down harmonic oscillator if you like. But orthogonal to this is a second potential like a Morse potential and this has a normal (positive) frequency, just as is the case of a saddle point. Buy a pack of Pringles if you want to see 3D saddle points :)\n\n\nYou can see this in the energy contour plot calculated for H+H2 below (using the LEPS potentials). The hydrogen molecule is vibrating as it approaches the H atom and the product is similarly vibrating. The diagonal is the symmetric stretch at the saddle point ( circle) and the asym. stretch is along the reaction path, shown as a short line at right angles to the other. The transition state is located at the highest point between the two valleys on approach and retreat and this has the shape of a saddle as you can see from the potential.\n\n\n[](https://i.stack.imgur.com/pR0gG.jpg)\n\n\n(I just noticed that the pic has been clipped on the right shouid read H+H2.)\n\n\n",
"4"
],
[
"\nBelow is a sketch of the atomic arrangements when considering a linear attack.\n\n\n[](https://i.stack.imgur.com/3Rsae.png)\n\n\nThe blue diagonal represents the symmetric stretch. The reaction coordinate (dotted line) takes the lowest possible path from reactant (top left) to product (bottom right). If I stretch the atomic arrangement, I make it worse, and if I scrunch the arrangement, I make it (much) worse. So this is like a vibration in a molecule, with a minimum somewhere in the middle.\n\n\nOn the other hand, starting from the transition state (\"‡\"), the asymmetric stretch makes one bond longer (the two atoms that will no longer be bound) and one bond shorter (the two atoms that will have a bond in dihydrogen). So there is a maximum in the middle, and both directions have lower energy.\n\n\nAnother way of saying it is: The \"middle\" hydrogen \"wants\" to make a single bond, not two or zero. The asymmetric stretch gets the transition state there, the symmetric stretch doesn't.\n\n\nFor an actual potential energy surface, see e.g. [https://people.uleth.ca/~roussel/C4000foundations/slides/07PES.pdf](https://people.uleth.ca/%7Eroussel/C4000foundations/slides/07PES.pdf)\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/170552/is-there-an-international-standard-for-rating-the-danger-of-chemical-elements
|
Is there an international standard for rating the danger of chemical elements?
|
I'm writing a small little tool (something like an interactive periodic table) and I wanted a good "guesstimate" of the danger of certain elements, to visualize across said table.
I could go and guess by manually looking up the physical / chemical / health / other effects of every element, but that would take an obscene amount of time and effort, not to mention it wouldn't be very accurate / precise.
My question is, **is there a standard or "list" recognized internationally as a good scale of the danger of certain elements?**
I know of the NFPA 704 Hazard Classification diamond, and while that would be perfect for my use, it doesn't classify every element, mainly just chemical compounds + a few noteworthy elements.
For example, I could find an NFPA 704 for just about every major compound of arsenic, but not pure arsenic (the element), which would be a pretty obviously dangerous element.
| 0
|
[
[
"\nYou can check the European [ECHA Database](https://echa.europa.eu/en/information-on-chemicals) (or the corresponding original CLP Regulation), which uses the Globally Harmonised System (GHS) managed by the United Nations, if it contains the relevant elements.\n\n\nFor [arsenic](https://echa.europa.eu/en/information-on-chemicals/cl-inventory-database/-/discli/details/85010), you would find that it's\n\n\n* Toxic if swallowed (H301)\n* Toxic if inhaled (H331)\n* Very toxic to aquatic life (H400)\n* Very toxic to aquatic life with long lasting effects (H410)\n\n\n \n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/170542/water-electrolysis-efficiency
|
Water Electrolysis Efficiency
|
During my water electrolysis with NaOH, I measured the voltage and current, and the volume of gases (hydrogen and oxygen) displaced at every 30-second interval. 10g NaOH in 2L water for 8 minutes.
Then, I was able to calculate the rate of reaction of the gases in terms of volume per unit time. Dividing the rate of reaction by power gives the efficiency of the trial (how much oxygen or hydrogen are produced per unit energy). As I vary the voltage from 5V to 10V and then to 15V, there is a decline in efficiency (less gases produced for each joule).
What is the reasoning for this? My hypothesis is that while the rate of production is directly proportional to current, as I increase the voltage, both current and voltage increases which causes the power to increase at a quicker rate. Ex. $P=IV$, then we increase $V$ to 2$V$, $I$ therefore increases to 2$I$ due to $V=IR$. Therefore $4P=2I \cdot 2V$ where the rate of reaction only increases by a factor of 2.
Or is energy lost to heat? But this will not fully account for the large decline as shown in the following picture.
[](https://i.stack.imgur.com/FSHzbm.png)
| 2
|
[
[
"\nOne easier way to measure the amount of products formed (albeit not 100% correct\\*) would be to indeed measure the charge that passed through the circuit as Poutnik mentioned.\n\n\nIn an electrochemical system there are different processes which all lead to losses. Not all of them follow Ohm's law $IR=V$. First there's a minimal potential required by the Nernst equation (approximately 1.23V at STP), the equilibrium potential.\n\n\nHowever there are also kinetic losses, at some small finite potential over the equilibrium potential one will find that the current is still approximately 0. This is because the rate of charge transfer (kinetics at the electrode) at the electrode-electrolyte interface is determined by the applied potential as well. The dependence of this rate on potential follows an approximately exponential course: $i = i\\_0e^{\\eta\\_{kinetic}/b}$ with $\\eta\\_{kinetic}$ called the 'kinetic overpotential' which is the applied potential minus the equilibrium potential (and the potential of the solution $\\eta\\_{kinetic}=U\\_{applied} - U\\_{equilibrium} - U\\_{electrolyte}$).\n\n\nThe next loss is the conduction of the electrolyte, which follows ohmic behavior. So: $\\eta\\_{ohmic} = I\\*R\\_{electrolyte}$, which is the \"Ohmic overpotential\".\n\n\nFinally, we need the hydroxide/water to get to the electrode and O2 and H2 out the solution. This is mass transport and is governed by the Nernst Planck equation and there is not a simple expression for this loss, but only starts being significant when you go to very high currents. At some point you will find that no matter how much potential you apply no increase in current happens, this is due to mass transport losses (as water, H2 and O2 transport is independent of the potential). One gives this also an overpotential: $\\eta\\_{transport}$.\n\n\nThe total current-potential behavior is thus not straight forward, these simple 3 phenomena already are quite complex and I have not described each contribution in enough detail. For more information I would definitely start looking at electrochemistry textbooks. In general the current is some complex function of the potential, and thus the efficiency at different rates (i.e. currents) is different: $I=f(U\\_{applied})$ and $efficiency=\\frac{U\\_{applied}}{U\\_{equilibrium}}$.\n\n\nFinally, all losses end up as heat.\n\n\n\\*(Current going to side reactions or double layer charging of your electrodes)\n\n\n",
"2"
],
[
"\nThe amount of products is determined by passed charge $Q = \\int\\_t{I \\cdot \\mathrm{d}t}$, not by spent energy $E = \\int\\_t{U \\cdot I \\cdot \\mathrm{d}t}$.\n\n\nThe relation of the charge $q$ and amount of substance $n$ is $q = znF$, where F the the Faraday constant $F = e \\cdot N\\_\\mathrm{A} \\approx \\pu{94685 C mol-1}$ and $z$ is the number of electron per particle.\n\n\nMajority of energy even for $\\pu{5 V}$ is wasted. Even worse for $\\pu{10 V}$ and $\\pu{15 V}$. You need not a source of high voltage. You need a low voltage source with low internal resicence, able to provide high current.\n\n\nTheoretical voltage needed to electrolyze water is $U\\_\\mathrm{theor} = \\pu{1.23 V}$. Practical needed voltage lays in range $\\pu{1.5 - 2.0 V}$, due kinetic reasons, depending on used electrodes, solution, geometry of the cell and required production rate.\n\n\nThe power being lost (as heat) $$P\\_\\mathrm{lost} = ( U - U\\_\\mathrm{theor})I \\tag{1}$$\n\n\nand the efficiency $$\\eta = \\frac{U\\_\\mathrm{theor}}{U} \\cdot 100\\ \\% \\tag{2}$$\n\n\nAs first (linearized) approximation:\n\n\n$$\\eta = \\frac{U\\_\\mathrm{theor}}{U\\_\\mathrm{theor} + RI} \\cdot 100\\ \\% \\tag{3}$$\n\n\nwhere $\\mathrm{R}$ is the Ohmic resistance of the electrolytic cell. Note that $\\mathrm{R}$ is not constant, but depends on current, temperature, mixing, geometry, solution and other factors.\n\n\nThere is trade off between the electrolysis efficiency and the production rate. The maximum efficiency is near zero production rate and decreases with the rate.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/170541/how-to-dissolve-beeswax-carnauba-wax-to-spray
|
How to dissolve beeswax/carnauba wax to spray
|
I’m trying to make an edible hydrophobic coating for a tart shell. If anyone knows of a way to lower the viscosity of an edible wax enough to make it sprayable I’d really appreciate it.
I have a paper here that explains that it could be done but it doesn’t state the solvent used
<https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7435775/>
| 0
|
[
[
"\nIn Section 2.2 of the paper cited, the method and solvent are specified:\n\"...soybean wax suspension was prepared through mixture of 1 g of soybean wax and 50 mL ethanol solution followed by heating at 65º C for 3 min.\" The superhydrophobic character of the coating described in the paper may not be necessary for the tart shell (see below for examples). Extra care must be observed when following this recipe, for the flash point of ethanol is only 14º C, but this method has the advantage of easy evaporation and safe consumption of the carrier solvent.\n\n\n[](https://i.stack.imgur.com/hFk8L.jpg)\n\n\nEdible waxes can be emulsified and dispersed in water and are widely used to coat fruit and vegetables. Spraying an aqueous solution onto a tart shell could be adjusted so as not to overwet the shell, but some experimentation will be necessary. Wax emulsions are commercially available from many sources, such as Hexion (Ref 1) and Lubrizol (Ref 2). More commercial sources are listed in Ref 3.\n\n\nOr, with a bit of experimentation, a food-grade wax (e.g., carnauba or polyethylene) and a food-grade emulsifier can be emulsified with intense stirring in hot water (to melt the wax), then cooled to stabilize the formulation.\n\n\nCooking and baking are specialized branches of chemistry, but you can eat your mistakes.\n\n\nRef 1. <https://www.hexion.com/en-US/chemistry/wax-emulsions>\n\n\nRef 2. <https://www.lubrizol.com/-/media/Lubrizol/Coatings/Coatings-Literature/Wax-Additives-Product-Guide---22-122162.pdf>\n\n\nRef 3. <https://www.thomasnet.com/products/fruit-vegetable-coating-wax-93060200-1.html>\n\n\n",
"3"
],
[
"\nI never tried rubbing carnauba wax on a tart and brushing it to a glossy shine, but one other idea came to mind: food grade ethyl acetate.\n\n\nCarnuba wax could be dissolved in ethyl acetate and sprayed on, with a few more minutes in the oven at elevated temperature to evaporate the solvent.\n\n\nEthyl acetate is approved for use in foods as a flavor additive and should be safe in low concentrations.\n\n\nCarnuba wax could also be softened with heat and applied to tarts in a rotating tumbler. This is how it is added to the surface of candies to make them shine.\n\n\nIt should be cautioned that adding excessive amounts of wax may affect the palate feel and texture of the tart. Finding a waterproof wrapping may be better. However, if feeling a bit adventurous, one might consider a chocolate coating as well.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/170538/why-water-volume-went-up-by-almost-the-same-amount-after-adding-salt
|
Why water volume went up by almost the same amount after adding salt?
|
Seems like it's a well-known fact that adding salt to water will not raise the water volume by much (e.g.: [Why there is no change in water level when salt is added?](https://chemistry.stackexchange.com/q/33058) )
I've done this experiment at home and saw unexpected results:
**Setup:**
1. Start with 500 ml of tap water in measuring cup (500 grams)
2. Measure 50 ml of iodized salt in measuring cup (80 grams)
3. Add salt to water, stir until dissolved
**Result**: around 540 ml (give or take) of salt water in measuring cup weighing 580 grams
Weight was measured with kitchen scale, volume measure by eye, but ~40 ml increase in total volume seems way too much.
Followed up by replacing salt with sand, volume went up by about the same amount.
**Video:** <https://www.youtube.com/watch?v=rLTn5jKqipU>
**What am I missing here?**
p.s.: the commenter pointed out that my salt volume is off, I should have looked up the density to go by weight rather than by measuring cup notches. Going by weight, I've added 37 ml or salt, making the result event closer. Volumes were judged by eye in food measuring cups, not exactly the proper lab equipment.
| 1
|
[
[
"\n* You have water volume.\n* Calculate salt volume from its mass and density.\n* Calculate their total volume before dissolution.\n* Calculate solution mass concentration in mass %.\n* From tabulated data or online calculators, obtain its density.\n* Calculate solution volume from its total mass and density.\n* Subtract from it the total volume of separate water and salt.\n* You have the volume difference.\n\n\nHere is the simple formula to calculate the difference:\n\n\n$$\\Delta V = V\\_\\mathrm{solution} - V\\_\\mathrm{solute} - V\\_\\mathrm{solvent} = \\\\\n\\frac{m\\_\\mathrm{solute} + m\\_\\mathrm{solvent}}{\\rho\\_\\mathrm{solution}} - \\frac{m\\_\\mathrm{solute}}{\\rho\\_\\mathrm{solute}} - \\frac{m\\_\\mathrm{solvent}}{\\rho\\_\\mathrm{solvent}}$$\n\n\nFor our particular $\\ce{NaCl}$ case:\n\n\n$$\\Delta V = \\frac{m\\_{\\ce{NaCl}} + m\\_{\\ce{H2O}}}{\\rho\\_\\mathrm{solution}} - \\frac{m\\_{\\ce{NaCl}}}{\\rho\\_{\\ce{NaCl(s)}}} - \\frac{m\\_{\\ce{H2O}}}{\\rho\\_{\\ce{H2O}}}=\n\\\\\n\\frac{m\\_{\\ce{NaCl}} + V\\_{\\ce{H2O}}\\cdot \\rho\\_{\\ce{H2O}} }{\\rho\\_\\mathrm{solution}} - \\frac{m\\_{\\ce{NaCl}}}{\\rho\\_{\\ce{NaCl(s)}}} - V\\_{\\ce{H2O}}$$\n\n\n* $V$ is volume [mL]\n* $\\rho$ is density [g/mL]\n* $m$ is mass [g]\n\n\nVolume changes during dissolution are common, usually contractions. $\\ce{NaCl}$ solution volume decreases, as water molecules as electric dipoles are attracted to ions $\\ce{Na+}$ and $\\ce{Cl-}$. As the result, ions and molecules in solution are packed in average better than salt and water apart.\n\n\nIt can be somewhat illustrated on mechanical macro analogy of 2 jars with marbles and sand. If mixed together, their total loose volume decreases as they better utilized the unused space.\n\n\n*There happened an error in obtaining density for given salt mass fraction, which is smaller than originally provided, therefore the volume of solution is larger.*\n\n\n*The volume contraction $\\pu{10.2 mL}$ is less than 1/3 of the (real) volume of added salt $\\pu{36.9 mL}.$*\n\n\nUsing the link at the solution density (or other density data sources), you can predict volume contraction for any salt/water mass ratio (within salt solubility)\n\n\nHere are calculates values for you particular case:\n\n\n\n\n\n| System | Quantity | Value | Unit |\n| --- | --- | --- | --- |\n| Water | Volume | 500 | mL |\n| - | Density | 0.9982 | g/mL |\n| - | Mass | 499.1 | g |\n| Salt | Volume | 36.9 | mL |\n| - | Density | 2.17 | g/mL |\n| - | Mass | 80 | g |\n| Total | Volume | 536.9 | mL |\n| Solution | Massconcentration | 13.815 | %w/w |\n| - | Volume | 526.7 | mL |\n| - | Density | [1.09947](https://www.handymath.com/cgi-bin/nacltble.cgi?tmptr=20&spcfgrv=&conc=13.815&submit=Calculate) | g/mL |\n| - | Mass | 579.1 | g |\n| - | Volumedifference | -10.2 | mL |\n\n\n ",
"5"
]
] |
https://chemistry.stackexchange.com/questions/170537/chemistry-of-a-proposed-capsaicin-extract
|
Chemistry of a proposed Capsaicin extract [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Update the question so it focuses on one problem only by [editing this post](/posts/170537/edit).
Closed 8 months ago.
[Improve this question](/posts/170537/edit)
Background:
===========
I’d like to make an extract of capsaicin using the following ingredients: Capsicum powder, Fractionated coconut MCT oil, apple cider vinegar, and saline. The idea is for the capsaicin to undergo absorption from the capsicum and into the oil to create an extract. The apple cider vinegar will be used as an emulsifier so I can mix the solution with saline. I chose these specific ingredients because they work well with my particular use case, but let me know if I should use a different solvent and or emulsifier (I'd like it to be food safe, and I'd like to avoid alcohol.)
Proposed process:
=================
1. Combine MCT oil and Capsicum powder and heat to around to around 250° F and let cool
2. Filter out the leftover solids using a coffee filter
3. Combine with apple cider vinegar and saline.
Questions (in order of importance):
===================================
1. Could this combination of ingredients react to create unwanted chemical byproducts?
2. Is capsaicin soluble in Fractionated coconut MCT oil, or will I need to use a different kind of oil?
3. Will this proposed process actually “extract” the capsaicin from the capsicum powder, or is additional processing required?
4. How should I determine the ratio of ingredients?
5. How long should I keep the oil & Capsaicin over heat?
6. I used the term “extract” to describe this, but is it actually a decoction, infusion or something else?
7. Any other tips, suggestions, warnings?
I know just enough about chemistry to know that I should probably ask questions before giving this a go.
| 1
|
[] |
https://chemistry.stackexchange.com/questions/170535/how-did-you-convert-primary-alcohol-using-hcl-with-zncl2
|
How did you convert primary alcohol using HCl (with ZnCl2)?
|
I kinda feel weird for this type of question. When primary OH (alcohol) is converted with HCl (with ZnCl2), how did the free Cl-ion come from? If, it is came from HCl, since when we get rid of the H from HCl?
[](https://i.stack.imgur.com/mLvff.png)
[](https://i.stack.imgur.com/I9Ub5.png)
| 0
|
[
[
"\nThough the [Lucas' reagent](https://en.wikipedia.org/wiki/Lucas%27_reagent) is a solution of *concentrated* $\\ce{HCl}$ and $\\ce{ZnCl2}$, it still is an aqueous solution containing $\\ce{Cl^-}$, because\n\n\n* as a strong acid, practically all $\\ce{HCl}$ in the aqueous phase dissociated \\*\n* $\\ce{ZnCl2}$ is well water soluble\n\n\n\\* One can purchase solutions of HCl in organic solvents which (depending of the solvent) may prohibit the dissociation. An example of such a \"dry solution\" is $\\ce{HCl}$ in diethyl ether ([entry at Sigma](https://www.sigmaaldrich.com/US/en/product/sigald/455180)).\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/170534/how-stable-are-paper-book-matches
|
How stable are paper book matches?
|
I have few hundred paper book match books that I collected about 25 to 45 years ago when many businesses had them custom printed.
They are from around the world although mostly US. They are in an open glass container in a closet. I am asking is there is any risk of them igniting with age
| 3
|
[] |
https://chemistry.stackexchange.com/questions/170533/why-the-change-of-the-position-of-the-salt-bridge-or-metal-piece-leads-to-differ
|
Why the change of the position of the salt bridge or metal piece leads to different voltage value?
|
I am doing an experiment where I use salt bridge to create galvanic cell. The goal of my experiment is to find if there is some correlation between the amount of solvent being used and the voltage being produced which means the accuracy of voltage is very important.
I am using zinc and copper metal pieces, zinc sulfate, and copper(II) sulfate with sodium chloride for salt bridge:
[](https://i.stack.imgur.com/WYe4Y.jpg)
The experiment works, but the problem appears that when I modify the position of either the metal piece or the salt bridge, the voltage value will change, sometimes I will change a lot (e.g. $\pu{0.3 V}$ to $\pu{0.42 V}).$
Does the change of the position of the salt bridge or metal piece leads to different voltage value in galvanic cell?
If not, could it be the reason that my zinc sulfate powder does not fully dissolve in water? So my metal piece could touch the area where there are more Zinc powder which causes it to be more electron transfers happens. It is very hard to get it fully dissolved in distilled water. Any suggestions for improving solubility in water?
I am really looking for way to reduce the uncertainty of my experiment, but the different voltage measured by only changes the salt bridge or metal pieces makes me very concerned.
| 3
|
[] |
https://chemistry.stackexchange.com/questions/170532/colligative-properties-and-binary-phase-diagrams
|
Colligative properties and binary phase diagrams
|
I'm having troubles with colligative properties and how to relate them to binary liquid-vapour and solid-liquid phase diagrams.
First of all, in my book it is indicated that colligative properties are defined in the presence of a non-volatile solute, so much so that it simultaneously determines a lowering of the vapor pressure, an ebullioscopic raising and a cryoscopic lowering of the solution with respect to what happens in the ambit of the pure solvent. But this does not seem entirely correct.
In fact, even a water-ethanol mixture can undergo cryoscopic lowering, i.e. a eutectic is reached, although both constituents are volatile (therefore the solute is volatile, whether it is water or alcohol).
The same argument can be made, for the same volatile mixture, in the context of boiling point:
[](https://i.stack.imgur.com/iw0qJ.jpg)
to the right of the azeotrope, an ebullioscopic rise is observed even though both compounds are volatile.
So, actually, when do these two colligative properties manifest themselves? Is it really necessary for the solute to be volatile?
The other doubt, as already mentioned, concerns about the correlation between state diagrams and the same colligative properties.
[](https://i.stack.imgur.com/BqGG8.jpg)
This state diagram refers to two substances that are miscible in the liquid state, but immiscible in the solid state, and which reach a eutectic, so the graph is based on the cryoscopic lowering. If freezing point lowering is defined only in the presence of non-volatile solutes, how does this graph hold true even in the presence of a volatile solute? (as in the case before, I mention the water+ethanol mixture, which reaches a eutectic)
[](https://i.stack.imgur.com/iyEZK.jpg)
This other graph, on the other hand, is a state diagram which refers to two miscible substances both in the liquid state and in the solid state. In this case a eutectic is not achieved and the graph, at least at first glance, is not based on the freezing point lowering, given that starting from A and progressively adding B, the melting/freezing temperature of the mixture increases.
Why is there this difference?
Thanks for every help, because i'm stuck.
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170526/ambidentate-behavior-of-cyanide
|
Ambidentate Behavior of Cyanide [duplicate]
|
**This question already has an answer here**:
[Adding cyanide vs nitrite for nucleophilic substitution and why does it change when we use silver salts for it](/questions/117754/adding-cyanide-vs-nitrite-for-nucleophilic-substitution-and-why-does-it-change-w)
(1 answer)
Closed 8 months ago.
Why do haloalkanes give alkyl cyanides when treated with KCN but give alkyl isocyanide with AgCN?
| -2
|
[
[
"\nThe $\\ce{KCN}$ substance is made of two ions : $\\ce{K^+}$ and $\\ce{CN^-}$ The active ion is cyanide, and the active atom in cyanide ion is the carbon atom, which is charged $-1$. This negative C atom attacks haloalkanes to produce alkylcyanide.\nOn the contrary, $\\ce{AgCN}$ is not made of two ions. It is not soluble in water. The bond $\\ce{Ag-C}$ is mainly covalent. So the active atom in $\\ce{AgCN}$ is the outer $\\ce{N}$ atom. When interacting with a haloalkane, the $\\ce{N}$ atom has a free doublet which can attack it via a $\\ce{S\\_N2}$ mechanism to produce an isocyanide.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170525/how-does-borane-target-a-specific-double-bond-on-an-indene
|
How does borane target a specific double bond on an indene?
|
[](https://i.stack.imgur.com/TZDrA.png)
According to the text I am using (Organic Chem, Seyan Ege, 3rd ed, problem 12.57) the halide in the electrophile borane (BH3) binds at site G. But I need to make sure I understand why. First, I believe that the double bond F=G is targeted to begin with because it is the double bond furthest from the electron-withdrawing OR group. Next, why does G get the halide and not F? I believe that is because if F gets the hypothetical positive charge, that charge can be distributed, and there is greater resonance than if the positive charge was localized at G. Is that correct? (Please note I am not a student asking you to do my homework, I am an older person working through the text on my own.)
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170521/why-is-sodium-hexafluoroaluminate-formed-instead-of-tetra
|
Why is sodium hexafluoroaluminate formed instead of "tetra"?
|
$$\ce{AlF3 + NaF \to Na3[AlF6]}$$
My question is why don't we get $\ce{Na[AlF4]}$, since when we do the same thing with hydroxides instead of fluorides ($\ce{Al(OH)3}$), we end up with $\ce{Na[Al(OH)4]}$. I know that it is a 3rd period element so it can expand its octet, so boron being in the same group forms $\ce{NaBF4}$. I've also heard that aluminium is larger so it can accommodate more fluorine atoms around it. But why would it want to do that?
| 6
|
[
[
"\nIn fact [sodium tetrafluoroaluminate can be formed[1]](https://doi.org/10.1021/ja01637a003). However, it requires quenching vapor from a temperature of 1000°C, definitely a non-equilibrium process.\n\n\nThe formation of the hexafluoroaluminate under more common conditions correlates with two features of this compound. First, it can form a relatively stable solid lattice, which formerly appeared in nature as the mineral [cryolite](https://en.wikipedia.org/wiki/Cryolite). (Mining of this mineral for use in aluminum smelting has made it rare, and now the flux is made synthetically for industrial use.) The stability of the lattice is evident from cryolite being essentially insoluble in plain water, a rarity for sodium salts.\n\n\nSecond, $\\ce{Al-F}$ bonding is more amenable to forming an octahedral complex than $\\ce{Al-OH}$ or $\\ce{B-F}$. As explained [here](https://chemistry.stackexchange.com/a/121199/17175), when an element with $s$ and $p$ valence orbitals forms an \"extended octet\" species such as $\\ce{SF6}$ or the formally isoelectronic $\\ce{AlF6^{3-}}$, there are antibonding interactions between the ligands which must be minimized. Using a larger central atom (aluminum instead of boron) and ligands that hold the electrons more tightly (fluoride instead of hydroxide) does that minimization.\n\n\n**Reference**\n\n\n1. E. H. Howard (1954). \"Some Physical and Chemical Properties of a New Sodium Aluminum Fluoride\", *J. Am. Chem. Soc.* **76**, 8, 2041–2042.\n<https://doi.org/10.1021/ja01637a003>\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/170513/can-pyruvate-get-reduced-after-glycolysis
|
Can pyruvate get reduced after glycolysis?
|
During cellular respiration, the first step is glycolysis where the glucose is split into 2 pyruvates, 2 NADH, and 2 ATP. The pyruvate either goes through the citric acid cycle or goes through lactic acid fermentation. If you remove the citric acid cycle and fermentation, is there a specific way to reduce pyruvate? I am relatively new to chemistry and I was wondering if pyruvate can either store additional electrons or if it can get reduced to another substance and collect additional electrons. If pyruvate doesn't get reduced in any scenario, can ATP get reduced?
| 2
|
[] |
https://chemistry.stackexchange.com/questions/170512/how-to-determine-the-charge-of-amino-acid-at-certain-ph
|
How to determine the charge of amino acid at certain pH? [duplicate]
|
**This question already has answers here**:
[Calculating charge on amino acid from pKa](/questions/7094/calculating-charge-on-amino-acid-from-pka)
(2 answers)
[How do I calculate the ensemble-average net charge of an amino acid at given pH?](/questions/119920/how-do-i-calculate-the-ensemble-average-net-charge-of-an-amino-acid-at-given-ph)
(1 answer)
Closed 8 months ago.
### Problem
>
> What would charge would you expect on alanine when placed in a solution with a pH of 1.00?
>
>
>
### Answer
>
> +1. Since alanine is nonpolar, we know that the only parts of the amino acid that can be charged are the N-terminus and the C-terminus.
>
>
>
> In an acidic solution, there is an excessive amount of protons available to protonate the amino acid. As a result, the carboxylic acid end and the amine end will both be fully protonated. This will result in an overall charge of +1, due to the nitrogen having three hydrogens attached.
>
>
>
### Question
Let's say I am given a certain pH of 2.00 rather than 1.00 for the "*acidic solution*" and the pH in this example is of that of a non-polar acid (alanine) changes to +1.
Does the pH dictate how many protons are added to amino or carboxyl groups? What determines the number of protons being added to that amino acid on the carboxylic end? I know a carboxylic acid can take up to four protons being added on the molecule.
| -3
|
[
[
"\n**Behaviour of amino acids in water solution**\n\n\nAmino acids exist in the solution in the form of *zwitterions* (molecules that contain an equal number of positively and negatively charged functional groups). Thus, the following equilibrium is established in the alanine solution:\n$$\\ce{H2N-(CH3)CH-COOH <=> ^+H3N-(CH3)CH-COO-}$$\n**Alanine as an ampholyte**\n\n\nSince actually there are two equilibria in the solution, alanine can be considered as an ampholyte:\n$$\\ce{^+H3N-(CH3)CH-COO- + H2O <=> H2N-(CH3)CH-COO- + H3O+}$$\n$$\\ce{^+H3N-(CH3)CH-COO- + H2O <=> ^+H3N-(CH3)CH-COOH + OH-}$$\nAt $\\textrm{pH}\\ 2$ there is an excess of $\\ce{H3O+}$ and a lack of $\\ce{OH-}$ so the first equilibrium shifts to the left while the second equilibrium shifts to the right according to Le Chatelier's principle. Therefore, the main form in which alanine exists in water solution at $\\textrm{pH}\\ 2$ is $\\ce{^+H3N-(CH3)CH-COOH}$. The overall charge of this molecule is $+1$. And vice versa, in an alkali media ($\\textrm{pH}>7$) alanine will have the main form $\\ce{H2N-(CH3)CH-COO-}$ and the net charge equal to $-1$.\n\n\nIt is a possible explanation from the point of view of acid-base theory. Please, check out references for further reading.\n\n\n**References**\n\n\n1. Skoog, Douglas A., West, Donald M., Holler, F. James, Crouch, Stanley R. (2014). *Fundamentals of Analytical Chemistry* (9th ed.). Singapore: Cengage Learning. p. 371-372.\n2. Clayden, J., Greeves, N., & Warren, S. (2012). *Organic Chemistry* (2nd ed.). Oxford University Press. p. 167.\n\n\n",
"6"
],
[
"\nThe change of $\\mathrm{pH}$ from $+1$ to $+2$ does not change much in the main formula of alanine, which is and stays $\\ce{H3\\overset{+}{N}−(CH3)CH−COOH}$, up to $2.34$ which is the first $\\mathrm{p}K\\_\\mathrm{a}.$\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170506/what-is-the-difference-between-ionic-bonding-and-the-general-transmission-of-cha
|
What is the difference between ionic bonding and the general transmission of charge?
|
I understand that in ionic bonding electrons are transferred from atom to atom and these ions are then attracted to each other forming a new compound. How is this different from any other transfer of charge situation? For instance, when silk and glass are rubbed together, electrons are transferred between them. These two items then attract each other but we do not say that a new chemical compound has formed, we still just have silk and glass.
| -2
|
[
[
"\nMaurice's comment gives an idea of magnitude in the glass-wool-cat fur situation. It is quite different in a lightning discharge. Every transfer of electrons, if traced back far enough, involves atoms, molecules, electronic orbitals. It is just that some orbitals are of higher energy and can lose electrons more easily, but those electrons still must have somewhere to go except possibly when in flight in a cathode ray tube. That is why when you walk along a rug in winter and touch a metal post all those electrons stored in your molecules antibonding orbitals readily transfer to empty conduction bands in the metal. Physical forces as well as magnetic and electrical forces can transfer electrons among orbitals. It may be hard to analyze but that is what is happening. Some of that glass and wool is most likely chemically changed. Some of the molecules in the lightning bolt most certainly do get rearranged.\n\n\n",
"1"
],
[
"\nGeneral transmission of change does not specify\n\n\n* the source of the charge\n* the receiver of the charge\n* the amount of transmitted charge\n* the distance of charge transmission\n\n\nOn the other hand ionic bond is *partial* electron transfer between adjacent atoms in contact, being bond by electrostatic force between the opposite charges.\n\n\nNo ionic bond is purely ionic, but is in fact partly ionic, partly covalent. As cations always more or less deform electron density of anions, decreasing the value of the electrostatic dipole.\n\n\nAs the rule of thumb to determine 50% of bond \"ionicity\" is usually considered electronegativity difference 1.7.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170499/is-the-process-of-volume-decrease-in-naoh-solutions-reversible-during-evaporatio
|
Is the process of volume decrease in NaOH solutions reversible during evaporation of water from the solution? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/170499/edit).
Closed 8 months ago.
[Improve this question](/posts/170499/edit)
While preparing w/w 50% NaOH solution I noticed a decrease in solution volume (5-10mL) in a 2L volumetric flask, so I am wondering if there is an increase in water volume during the water evaporation or concentration of NaOH? Is it the same amount and is it measurable?
[QuantumYitian question regarding the decrease of volume during the dissolution of NaOH](https://chemistry.stackexchange.com/questions/167434/why-is-there-a-decrease-in-the-total-volume-when-naoh-dissolves-in-water)
| 0
|
[
[
"\nThere is no water volume in solution. $\\ce{Na+}$ and $\\ce{OH-}$ ions and $\\ce{H2O}$ molecules are packed together in solution in average tighter than is the average packing tightness for $\\ce{NaOH(s)}$ and $\\ce{H2O(l)}$ alone.\n\n\nAt water evaporation, the mass concentration of $\\ce{NaOH}$ obviously increases until solution saturation. After that, extra $\\ce{NaOH}$ crystalizes.\n\n\nIf all water evaporates and then condenses, the initial volumes are obtained, if we neglect experimental troubles of removing all water.\n\n\n\n\n---\n\n\nThe difference of volumes during dissolution (assuming the same initial and final temperature) can be determined from experimental or tabulated values of volume, mass or density:\n\n\n* Measured mass/volume and measured/known density of solute\n* Measured volume/mass and measured/known density of solvent\n* Measured volume or measured/tabulated density of solution at give concentration\n\n\nMany solutions, including $\\ce{NaOH}$ have [tabulated densities](https://www.handymath.com/cgi-bin/naohtble3.cgi?submit=Entry) that can be interpolated and the volume difference calculated.\n\n\nHere is the simple formula to calculate the difference:\n\n\n$$\\Delta V = V\\_\\mathrm{solution} - V\\_\\mathrm{solute} - V\\_\\mathrm{solvent} = \\\\\n\\frac{m\\_\\mathrm{solute} + m\\_\\mathrm{solvent}}{\\rho\\_\\mathrm{solution}} - \\frac{m\\_\\mathrm{solute}}{\\rho\\_\\mathrm{solute}} - \\frac{m\\_\\mathrm{solvent}}{\\rho\\_\\mathrm{solvent}}=\n\\\\\n\\frac{m\\_{\\ce{NaOH}} + m\\_{\\ce{H2O}}}{\\rho\\_\\mathrm{solution}} - \\frac{m\\_{\\ce{NaOH}}}{\\rho\\_{\\ce{NaOH(s)}}} - \\frac{m\\_{\\ce{H2O}}}{\\rho\\_{\\ce{H2O}}}=\n\\\\\n\\frac{m\\_{\\ce{NaOH}} + V\\_{\\ce{H2O}}\\cdot \\rho\\_{\\ce{H2O}} }{\\rho\\_\\mathrm{solution}} - \\frac{m\\_{\\ce{NaOH}}}{\\rho\\_{\\ce{NaOH(s)}}} - V\\_{\\ce{H2O}}$$\n\n\n* $V$ is volume\n* $\\rho$ is density\n* $m$ is mass\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170492/feasibility-comparison-of-symmetric-ether-synthesis-and-hcl-alcohol-sn2-reacti
|
Feasibility comparison of symmetric ether synthesis and HCl + alcohol SN2 reactions
|
According to Solomons & Fryhle 10$^{th}$ edition Ch 11.8,
>
> Because the chloride ion is a weaker nucleophile than bromide or iodide ions, hydrogen chloride doesn't react with primary/secondary alcohols unless ZnCl$\_2$ or a similar lewis acid is added to the mixture as well.
>
>
>
However, we also know that symmetric ethers can be formed by an $S\_N2$ reaction between two alcohols, such as the synthesis of diethyl ether from two molecules of ethanol under [dilute acidic conditions](https://chemistry.stackexchange.com/questions/41574/controlling-the-reaction-of-alcohols-with-acid-to-either-ether-synthesis-or-elim), despite the fact that alcohols are generally bad nucleophiles.
With regards to the [Mayr-Patz equation](https://www.cup.lmu.de/oc/mayr/reaktionsdatenbank2/pages/show_page/7), ethanol has $N = 6.68, s\_N=0.85$ and chloride has $N = 13.00, s\_N = 0.6$ (both in 80% Ethanol, 20% water albeit chloride's $s\_N$ is estimated). Unfortunately, I don't know where to find the E value for a protonated ethanol.
Given this information, I'm now wondering if ethanol is necessarily always a worse nucleophile than chloride (considering it has a higher $s\_N$ value), and if it is worse, how is it possible for symmetric ether synthesis to be feasible with just acid, while $S\_N2$ happens at a negigible rate between ethanol and HCl?
| 2
|
[] |
https://chemistry.stackexchange.com/questions/170491/how-to-change-the-counterion-of-a-nucleotide-salt
|
How to change the counterion of a nucleotide salt?
|
I need to take the sodium nucleotide salts I have available in my lab and modify them to instead contain some more exotic counterions. Can I do this by binding my nucleotides to a strong anion exchange resin, washing, and eluting with a salt of my preferred counter ion? Seems like it should work, but Im a biologist, not a chemist. Any other suggestions would also be appreciated! Thanks very much!
| 1
|
[] |
https://chemistry.stackexchange.com/questions/170488/why-does-the-single-crystal-ferroelectric-have-a-different-hysteresis-curve-comp
|
why does the single crystal ferroelectric have a different hysteresis curve compared to polycrystalline ferroelectric?
|
For the ferroelectric hysteresis curve, is there a reason why single crystal ferroelectrics have a 'hard' ferroelectric hysteresis curve compared to 'soft' ferroelectric hysteresis curve for polycrystalline ferroelectrics?
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170485/confusion-about-equilibrium-constant-in-simultaneous-equilibrium
|
Confusion about equilibrium constant in simultaneous equilibrium
|
Ok, so the equilibrium constant in terms of pressure is the ratio of partial pressures at equilibrium of gaseous products to reactants (all raised to the powers equal to the stoichiometric coefficients). But if I have multiple equilibriums taking place simultaneously, where, let's say, there is one (gaseous) product common to both reactions, then that means the partial pressure of that substance will actually be equal to the sum of partial pressures of the moles of that substance created from both reactions.
For example:
$$\ce{X(s)} \ce{<=>} \ce{Y(g)} + \ce{Z(g)} $$
$$\ce{R(s)} \ce{<=>} \ce{Y(g)} + \ce{S(g)} $$
Here, let's say at equilibrium the partial pressure of Y and Z (the amount of them created from reaction 1) are $P\_1$, and that of Y and S created in reaction 2 is $P\_2$.
1. My teacher said that the partial pressure of Y in the entire reaction mixture would be $P\_1 + P\_2$, which makes sense. But if we write the expression for $K\_1$ and $K\_2$, my teacher says we will take the partial pressure of Y to be $P\_1 + P\_2$. I'm having a hard time understanding why we won't just take the partial pressures of Y created in those specific reactions?
2. Also, is it not true that the equilibrium constant of each of these reactions would change if we put them in a mixture together, since the concentration of $Y$ would change because that is present as a product in both reactions? Or would it just be that the equilibrium will move in the backward direction since more product is added?
3. If the answer to (2) is that **$K$ will not change**, then that means if I *only* had reaction 1 taking place, equilibrium constant for that would also take pressure of Y to be $P\_1 + P\_2$, which seems obviously wrong to me- it should only be $P\_1$.
Overall, stuff is not adding up in my mind for simultaneous equilibriums, if anyone could clear my concepts I would appreciate it!
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170482/why-does-cooking-oil-solidify
|
Why does cooking oil solidify
|
I am in a tropical climate. The vegetable cooking oil I buy remains liquid for about 3 or 4 months and by the time I'm finishing a container, it has solidified.
Why does this happen? Is there any way to avoid this?
---
*It appears to be palm oil. Yes it liquefies again although in my case, it solidifes again. I warm it up every night before cooking supper.*
| 0
|
[
[
"\nAs revealed later, first check, if it is temperature dependent temporary solidification. It is natural and reversible process, the same process as when melted fat solidifies back when cooled down. Prevention is to keep oil warm enough and the cure is warming it up until it melts.\n\n\nEvery fat has its typical melting point ( or range). Solid fats melt at temperature higher than room one, vegetable oils melt at temperature lower than room one. Some fats are in this aspect in a gray zone, melting or freezing according to the current room temperature.\n\n\nNote in the table below **palm oil has exceptionally high freezing point** $\\pu{35 ^{\\circ}C}$. If solidification is an issue, consider using other oil with lower melting point. Oils have tendency to be supercooled (hesitating to start freezing up), so it may need lower than expected temperature to observe solidification, that may be triggered by random events.\n\n\nI also assume some oils may be oil mixtures with lowered freezing point. .\n\n\nSee below the table [digeoris.com: Freezing Point of Vegetable Oil](https://digeoris.com/blog/optimal-temperatures-for-freezing-vegetable-and-animal-fats/)\n\n\n\n\n\n| Oil or fat | Freezing / melting point °C | Transportmin °C | Transportmax °C | Storagemin °C | Storagemax °C |\n| --- | --- | --- | --- | --- | --- |\n| Castor oil | -3 | 20 | 25 | 30 | 35 |\n| Coconut oil | 25 | 27 | 32 | 40 | 45 |\n| Cottonseed oil | 48 | Ambient | Ambient | 20 | 25 |\n| Linseed oil | -24 | Ambient | Ambient | 15 | 20 |\n| Maize (corn) oil | -11 | Ambient | Ambient | 15 | 20 |\n| Olive oil | -6 | Ambient | Ambient | 15 | 20 |\n| Palm oil | 35 | 32 | 40 | 50 | 55 |\n| Canola (rapeseed)oil | -10 | Ambient | Ambient | 15 | 20 |\n| Safflower oil | -17 | Ambient | Ambient | 15 | 20 |\n| Sesame oil | -6 | Ambient | Ambient | 15 | 20 |\n| Sheanut butter | 34-38 | 38 | 41 | 50 | 55 |\n| Soybean oil | -16 | Ambient | Ambient | 20 | 25 |\n| Sunflower oil | -17 | Ambient | Ambient | 15 | 20 |\n\n\n\n\n---\n\n\nIf the above does not apply:\n\n\nIt probably polymerizes due present double bonds, reacting with air oxygen, being a vegetable oil. Famous for this is lineseed oil, that is used in oil paints.\n\n\nI suggest keeping oil stocks in cold place, as the reaction is slower at lower temperature. If applicable, even in fridge, where it may temporarily freeze ( and melt back when warmed up in a room).\n\n\nAnd/or it may be more applicable to buy smaller oil packages or divide their content in several smaller vessels, kept closed, cold if possible, taken sequentially. The remaining vessels would not then be repeatedly exposed to air and if possible, would be staying in cold place.\n\n\n",
"5"
],
[
"\nCooking oils are consist of esters of [fatty acids](https://en.wikipedia.org/wiki/Fatty_acid). Some of them are saturated, others are unsaturated; the later refers to the presence of one, or multiple $\\ce{C=C}$ double bonds. An example for this later is [oleic acid](https://en.wikipedia.org/wiki/Oleic_acid):\n\n\n[](https://i.stack.imgur.com/RpaDB.png)\n\n\n(image credit: English edition of [wikipedia](https://en.wikipedia.org/wiki/Oleic_acid)).\n\n\nOver time, promoted by heat (not only cooking, ambient temperature suffice given enough time), sun light, and traces of metal salts of these fatty acids, exposure to air leads to polymerization. Oxygen (about $\\pu{21 vol\\%}$ in air) is a diradical, which can attack these double bonds. Eventually, individual molecules with $\\ce{C=C}$ double bonds then cross-link with each other on expense of these double bonds. So\n\n\n* the $\\ce{C=C}$ double bonds are consumed\n* individual small molecules bind with each other on positions of former $\\ce{C=C}$ double bonds\n* the new molecules formed are larger\n* compared to the starting material, the newly formed products have a higher melting and boiling temperature.\n\n\nBecause cooking oil is a blend of compounds, oleic acid isn't the only compound undergoing this reaction. Then, the hardened cooking oil equally is a blend of multiple chemicals, too. The principle is not restricted to cooking, though. In painting, this process is triggered by intentional addition of a [siccatives](https://en.wikipedia.org/wiki/Oil_drying_agent), compounds known to accelerate this process.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/170475/diffusion-migration-and-einstein-equation
|
Diffusion, Migration and Einstein Equation
|
In the textbook: Electrochemical Systems by Newman and Alyea, 3rd edition, chapter 11.9: Moderately Dilute Solutions, equation for the mole flux of the component $i$ is given by: $$ N\_i = - \frac {u\_i c\_i} {z\_i F} \nabla \bar\mu\_i\ + c\_i v \tag {1}$$
where $u\_i$ is the ionic mobility, $c\_i$ is the concentration, $\bar \mu\_i$ is the electrochemical potential and $v$ is the velocity of the streaming fluid (assuming a low concentration of the component $i$ as situation becomes more complicated at higher concentrations).
One thing to note is that I wrote ionic mobility $u\_i$ a bit differently compared to the textbook. I defined it as a terminal velocity of the ion in the unitary electric field, so its measuring unit is $[\frac {m^2}{Vs}]$ in my case. In the textbook they defined it also as a terminal velocity in the unitary electric field, but divided by $z\_i F$. This is something to keep in mind to avoid confusion.
Basically, equation (1) states that the mole flux of the component has three contributions: diffusion, migration and convection. Diffusion and migration are accounted for in the first term (electrochemical potential gradient) and convection is accounted for in the second term.
My problem is with the first term $-\frac {u\_i c\_i} {z\_i F} \nabla \mu\_i$ which I know how to derive and I will do this here as I don't understand something about the equation (1).
To start, we define diffusion mole flux from as the 1st Fick's law: $$N\_{i,dif} = -\frac {D\_i c\_i} {RT} \nabla \mu\_i \tag {2}$$
where $D\_i$ is the diffusion coefficient and $\mu\_i$ is the chemical potential of the component $i$
Migration mole flux is given by: $$N\_{i,mig} = -u\_i c\_i \nabla \phi \tag {3}$$
where $\phi$ is the electric potential.
If there is no convection, we can write total mole flux of the component as: $$N\_i = -(\frac {D\_i c\_i} {RT} \nabla \mu\_i + u\_i c\_i \nabla \phi) \tag {4}$$
We recall the definition of electrochemical potential $ \bar \mu\_i$: $$ \bar \mu\_i = \mu\_i + z\_i F \phi \tag {5} $$
If we assume that system is in thermodynamic equilibrium than electrochemical potential gradient and net mole flux vanish: $$ \nabla \bar \mu\_i = 0 \tag {6} $$ $$ N\_i = 0 \tag {7}$$
From equations (5) and (6), we can write: $$\frac {\nabla \mu\_i}{\nabla \phi} = -z\_i F \tag {8}$$
From equations (4) and (7), we can write: $$ \frac {\nabla \mu\_i}{\nabla \phi} = - \frac {u\_i RT}{D\_i} \tag {9}$$
By equating (8) and (9), we get the Einstein equation, which relates diffusion coefficient $D\_i$ and ionic mobility $u\_i$ in the thermodynamic equilibrium: $$D\_i = \frac {u\_i RT}{z\_i F} \tag {10}$$
If we know substitute the equation (10) into equation (4), we can write: $$N\_i = -u\_i c\_i(\frac {1}{z\_i F} \nabla \mu\_i + \nabla \phi) = - \frac {u\_i c\_i} {z\_i F} \nabla \bar\mu\_i\ \tag {11}$$
After the derivation of the first term in the equation (1), I can get to my point. Namely, that the first term in the equation must be equal to zero since it is impossible to write this term in that form (via gradient of the electrochemical potential) without using Einstein equation. We saw earlier that Einstein equation is only valid in the thermodynamic equilibrium and therefore equation (6) applies.
| 5
|
[
[
"\nBefore writing, I understand your derivation, but I do not agree with Eq. (8). The gradient of a scalar field is a vector, so it is not correct to divide a vector quantity like $\\underline{\\nabla}\\mu\\_i$ with $\\underline{\\nabla}\\phi$.\n\n\nIndeed the flux of species $j$, in a $1$-dimensional system, can be written as\n$$ N\\_j = -D\\_j\\dfrac{dC\\_j}{dx} - u\\_j C\\_j \\dfrac{d\\phi}{dx} \\tag{1} $$\n\n\nHere is a somewhat informal derivation of your Eq. (10). Imagine the charged species $ j $ being subjected to two forces, the drag force and the electrical force, but both being balanced. By Newton's 2nd law we have\n\\begin{align}\n F\\_D &= F\\_E \\\\\n 6 \\pi \\mu r\\_j v\\_D &= z\\_j e E \\\\\n \\dfrac{v\\_D}{E} &= \\dfrac{z\\_j e}{6 \\pi \\mu r\\_j} \\\\\n u\\_j &= \\dfrac{k\\_BT}{6 \\pi \\mu r\\_j}\\dfrac{z\\_j e}{k\\_BT} \\hspace{1 cm}\n \\left(D\\_j = \\dfrac{k\\_BT}{6 \\pi \\mu r\\_j} \\hspace{0.5 cm} \\text{(Einstein-Stokes law)}\\right) \\\\\n u\\_j &= D\\_j\\dfrac{z\\_j e N\\_A}{N\\_Ak\\_B T} \\\\\n u\\_j &= D\\_j\\dfrac{z\\_j F}{RT} \\tag{2} \\\\\n\\end{align}\nThis is the Einstein-Smoluchowski equation. Combining Eqs. (1) and (2) we have\n$$ N\\_j = -D\\_j\\dfrac{dC\\_j}{dx} - \\dfrac{F}{RT} z\\_jD\\_jC\\_j \\dfrac{d\\phi}{dx} \\tag{3} $$\nIf you couple Eq. (3) with the convection part, i.e. $ C\\_j\\mathbf{v} $, the result is the famous **Nernst-Planck equation**.\n\n\nEq. (2) was derived by considering that the charged species $ j $ is in an equilibrium between its forces. Although I didn't, you can derive Einstein-Stokes law by doing the same, but considering: (1) the balance between the chemical force $d\\mu\\_j/dx $ and the drag force, and (2) comparing the derived expression with Fick's law.\nThus, **how come this relation, obtained by an equilibrium between forces, is combined with the expression to describe the motion of species in $N\\_j$?**. The easiest answer is the following: **so we can reduce the number of phenomenological and constant parameters, to describe the motion, from two to one**.\n\n\nNernst-Planck equation is a good approximation to the molar flux, if the concentrations are sufficiently low. This means that you can neglect ion-ion interactions in the medium. This is the reason why you are able to change the activity by the concentration in the chemical part of $\\tilde{\\mu}\\_j$. It also assumes that the flux can be written as a linear combination of driving forces, and the phenomenological coefficients, are typically considered as constants. But indeed, coupling it with Eq. (2) goes much further. If you want to be more precise, Eq. (1) should be written\n$$ N\\_j = -D\\_j^\\infty\\dfrac{dC\\_j}{dx} - u\\_j^\\infty C\\_j \\dfrac{d\\phi}{dx} \\tag{4} $$\nSo you need two phenomenological parameters to evaluate the flux. However, sometimes it is hard to obtain both coefficients, $ D\\_j^\\infty $ and $ u\\_j^\\infty $, so we go around with this by applying Eq. (2). I will give you an example of my own. I am modeling lithium-sulfur batteries, and there are publications where $\\ce{Li+}$ (and other sulfur species) are as high as $4-5$ mol/dm$^3$, and other authors and myself still use Eq. (3).\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170471/differentiating-ca2-and-sr2-ion
|
Differentiating Ca2+ and Sr2+ ion
|
$\ce{Ca^{+2}}$ and $\ce{Sr^{+2}}$ both give a bit similar red color in flame test. What is the easiest way (chemical test) to distinguish $\ce{Ca^{+2}}$ and $\ce{Sr^{+2}}$?
| 0
|
[
[
"\nThe video document given by AChem is not really clear. It should be replaced by the following method, used in my university.\n\n\nLet's start from $5$ mL of an initial solution containing possibly any one or all of the three ions $\\ce{Ca^{2+}, Sr^{2+}, Ba^{2+}}$.\n\n\n§$1$. Check if this initial solution contains the ion $\\ce{Ba^{2+}}$ by taking $3$ drops of the initial solution, adding $3$ drops of $0.1$ M $\\ce{K2Cr2O7}$ plus $1$ mL $0.1$ M $\\ce{CH3COONa}$. If a yellow precipitate appears, it means that **Barium** is present in solution : Go to $§2$. \n\nIf the solution remains clear, the solution does not contain $\\ce{Ba^{2+}}$ : skip §$2$, and go to §$3$ with the rest of the initial solution.\n\n\n§$2$. Add the following solutions to the rest of the initial solution : first $3$ mL $0.1$ M $\\ce{K2Cr2O7}$ solution, plus $5$ mL $0.1$ M $\\ce{CH3COONa}$. Filtrate the yellow $\\ce{BaCrO4}$ precipitate. The filtrate must be yellow, and does not contain $\\ce{Ba^{2+}}$ any more.\n\n\n§$3$. Add enough $\\ce{Na2CO3}$ to make the solution basic (pH > $8$). Heat to boiling temperature, cool, filter and wash the precipitate containing $\\ce{CaCO3 + SrCO3}$. Throw away the yellow filtrate.\n\n\n$§4$. Add enough $\\ce{HNO3}$ to dissolve the precipitate. Usually $1$ mL $1$ M $\\ce{HNO3}$ is enough. Drop the liquid in a ceramic capsule. Heat it so as to evaporate it totally, and to eliminate all crystallization waters. Then cool it down and add $1$ mL $100$% pure ethanol. Grind the stuff with a glass stick. Filtrate on a small filter paper. Wash twice with $0.5$ mL pure ethanol ($100$%). The residual insoluble part is made of $\\ce{Sr(NO3)2}$. See how to use the filtrate in $§6$.\n\n\n§$5$. The obtained residue of $\\ce{Sr(NO3)2}$ is washed several times to eliminate traces of Calcium nitrate. Dissolve it in $3$ mL water. Add $1$ mL 1 M $\\ce{H2SO4}$. A white precipitate is produced if the substance contains **Strontium**. Go to §6 with the filtrate.\n\n\n§$6$. The filtrate contains calcium nitrate. It can be proven by first evaporating to dryness. Then the residue is dissolved in $1$ mL water. $1$ drop $1$ M $\\ce{CH3COOH}$ is added to the obtained solution. The $1$ mL $0.1$ M ammonium oxalate is added. A white $\\ce{CaC2O4}$ precipitate shows the presence of **Calcium**.\n\n\n",
"4"
],
[
"\nA good test for distinguishing calcium ions and strontium ions is **potassium ferrocyanide test.**\n\n\n$$\\ce{Ca^2+ + 2K4[Fe(CN)6] -> \\underset{white}{K2Ca[Fe(CN)6]} \\downarrow}$$\n\n\nIn presence of ammonium chloride, test is more sensitive. In this case, potassium is replaced by ammonium ions in the precipitate. However, the downside of this test is barium and magnesium interfere. You can separate them as discussed [in my other answer.](https://chemistry.stackexchange.com/a/167916/17368). You can find more details on this test on a paper [linked in a previous comment of mine](https://chemistry.stackexchange.com/questions/170053/can-i-use-potassium-ferricyanide-instead-of-potassium-ferrocyanide-for-the-ident#comment357032_170053). However, the best test is checking the solubility of the corresponding sulfates (explained in my other answer).\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/170470/endothermic-peak-in-dsc
|
Endothermic peak in DSC
|
I have an endothermic peak, with no mass loss and no change in (organic) crystal structure (monitored by vt-pxrd). What can cause such effect? Also, excluding the options of impurities or instrumental errors.
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170466/why-cant-we-use-concentrated-acids-to-test-for-carbonates
|
Why can't we use concentrated acids to test for carbonates?
|
According to [BBC Bitesize](https://www.bbc.co.uk/bitesize/guides/zxtvw6f/revision/3#:%7E:text=Testing%20for%20carbonate%20ions,the%20gas%20is%20carbon%20dioxide.):
>
> Carbonate ions, $\ce{CO\_3^{2-}}$, are detected using **a dilute acid**. Bubbles are given off when an acid, usually dilute hydrochloric acid, is added to the test compound.
>
>
>
Is there a requirement that we use dilute acids for this purpose? Can't this test be done using relatively concentrated acids with the likely advantage that the reaction happens more vigorously?
| 1
|
[
[
"\nBBC's story is quite an incomplete one. Many anions can form gaseous products when acidified. The reason for using a dilute acid for carbonate test is simply that it works and therefore one does not need a stronger concentration! Even lemon juice will work for carbonates. Since carbonate is a common anion, for elementary purposes, dilute HCl is the acid of choice because it does not form insoluble salts with commonly encountered carbonates such as alkali (Li, Na, K) and alkaline earth metals (Mg, Ca, Ba) etc. Sulfuric acid forms insoluble salt layer with Ca/Ba which can prevent further raction. Nitric acid (dilute or concentrated) is to avoided because it can also generate nitrogen oxide(s) bubbles in contact with certain substances depending on whatever is present along with carbonates.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/170465/find-volume-of-methanol-produced-via-reaction-between-carbon-monoxide-and-hydrog
|
Find volume of methanol produced via reaction between carbon monoxide and hydrogen gases
|
>
> Methanol is manufactured by the reaction of carbon monoxide gas with hydrogen gas. If a $95~\%$ yield is usually obtained, approximately how many liters of hydrogen gas at $\pu{350 °C}$ and $\pu{300 atm}$ are usually required to make $1.0$ liter of methanol measured at $\pu{20.0 °C}?$ The density of methanol at $\pu{20.0 °C}$ is $\pu{0.8 g mL^-1}.$
>
>
> (a) $\pu{0.89 L}$
>
> (b) $\pu{2.5 L}$
>
> (c) $\pu{4.8 L}$
>
> (d) $\pu{5.0 L}$
>
> (e) $\pu{8.1 L}$
>
>
>
>
> The correct answer is (d) $\pu{5.0 L}.$
>
>
>
I got two different answers, both wrong. I have detailed them below.
### Attempt 1
I found the mass of methanol:
$$m(\ce{MeOH}) = \rho V = (\pu{0.8 g mL^-1})(\pu{1000 mL}) = \pu{800 g},\tag{1.1}$$
and calculated its theoretical yield:
$$m\_\mathrm{theor}(\ce{MeOH}) = 0.95\times\pu{800 g} = \pu{842 g}.\tag{1.2}$$
The amount of methanol is
$$n(\ce{MeOH}) = \frac{\pu{842 g}}{\pu{32 g mol^-1}} = \pu{26.3 mol}.\tag{1.3}$$
The amount of hydrogen from balanced equation is $n(\ce{H2}) = \pu{52.6 mol}.$ From here
$$
\begin{align}
V(\ce{H2}) &= \frac{nRT}{p} \\
&= \frac{(\pu{52.6 mol})(\pu{8.31 J mol^-1 K^-1})(\pu{(350 + 273) K})}{\pu{3E7 Pa}} \\
&= \pu{9.08E-3 m^3} \\
&\approx \pu{9.1 L}. \tag{1.4}
\end{align}
$$
### Attempt 2
Assuming the pressure of methanol is $\pu{1 atm} = \pu{1E5 Pa},$
$$
\begin{align}
n &= \frac{pV}{RT} \\
&= \frac{(\pu{1E5 Pa})(\pu{1000 mL})}{(\pu{8.31 J mol^-1 K^-1})(\pu{293 K})} \\
&= \pu{41071 mol}. \tag{2.1}
\end{align}
$$
Theoretical amount of methanol is
$$n\_\mathrm{theor}(\ce{MeOH}) = \pu{41071 mol}/0.95 = \pu{43232 mol}.\tag{2.2}$$
The amount of hydrogen from the balanced equation is $n(\ce{H2}) = \pu{86464 mol}.$ From here
$$
\begin{align}
V(\ce{H2}) &= \frac{nRT}{p} \\
&= \frac{(\pu{86464 mol})(\pu{8.31 J mol^-1 K^-1})(\pu{(350 + 273) K})}{\pu{3E7 Pa}} \\
&= \pu{14.92 m^3}. \tag{2.3}
\end{align}
$$
Attempt 2 is definitely wrong. I don't get what I did wrong in my attempt 1. What am I missing?
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170463/what-processes-generate-entropy-as-heat-flows-across-temperature-gradient
|
What processes generate entropy as heat flows across temperature gradient
|
Suppose, we have a source at high
temperature $T\_\mathrm h$ and sink at temperature lower temperature $T\_\mathrm l$.
If $Q\_\mathrm h$ amount of heat flow from source to sink, then change in entropy of source is $\frac{-Q\_\mathrm h}{T\_\mathrm h}$ and change in entropy of sink is $\frac{+Q\_\mathrm h}{T\_\mathrm l}$. I saw a similar question here:
<https://physics.stackexchange.com/questions/358142/entropy-generation-during-heat-transfer-processes>.
And while I understand that entropy is being generated in the partition, my question is
how does heat conduction across a finite temperature gradient generates entropy
| 2
|
[
[
"\nFor steady state heat conduction in the slab between the source and sink, we have $$\\frac{d}{d x}\\left(k\\frac{d T}{d x}\\right)=0$$where k is the thermal conductivity. If we divide this equation by absolute temperature T, we obtain$$\\frac{1}{T}\\frac{d}{d x}\\left(k\\frac{d T}{d x}\\right)=0$$Next, making use of the product rule for differentiation, we have:$$\\frac{1}{T}\\frac{d}{d x}\\left(k\\frac{d T}{d x}\\right)=\\frac{d}{d x}\\left(\\frac{k}{T}\\frac{dT}{dx}\\right)+\\frac{k}{T^2}\\left(\\frac{dT}{dx}\\right)^2=0$$Assuming that the source is at x = 0 and the sink is at x = L, if we integrate this equation between x = 0 and x + L, we obtain: $$-\\frac{q}{T\\_C}+\\frac{q}{T\\_H}+\\int\\_0^L{\\frac{q^2}{k}dx}=0\\tag{1}$$ where the heat flux q is given by $$q=-k\\frac{dT}{dx}$$If we rearrange Eqn. 1 slightly, we obtain:$$\\frac{q}{T\\_C}=\\frac{q}{T\\_H}+\\int\\_0^L{\\frac{q^2}{k}dx}\\tag{2}$$The quantity $\\frac{q}{T\\_H}$ is the rate at which entropy enters the slab per unit area, and the quantity $\\frac{q}{T\\_C}$ is the rate at which entropy exits the slab per unit area of slab. Eqn. 2 tells us that the rate at which entropy exits the slab is equal to the rate at which entropy exits the slab plus the rate at which entropy is generated within the slab. According to Eqn.2, the rate of entropy generation within the slab per unit volume is $q^/k$. This entropy generation rate is, of course, positive definite.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/170458/lewis-structure-and-resonance-for-arginine-ion
|
Lewis structure and resonance for arginine ion
|
I'm working through a problem set in MIT's open courseware 'Principles of Chemical Science' course [here](https://ocw.mit.edu/courses/5-111sc-principles-of-chemical-science-fall-2014/resources/mit5_111f14_lec10prob/). In question 5, it asks:
>
> The skeletal structures of two amino acids, leucine and arginine, are drawn below. Non-zero formal charges are indicated. Provide the Lewis structure(s), including double bonds and lone pairs, for each of these molecules. If there are equivalent resonance forms (which may include moving the formal charge on N), include them
>
>
>
This arginine structure is given for part (b):
[](https://i.stack.imgur.com/uMkPM.png)
I've been able to produce the resonance structures given in the solution sheet ([here](https://ocw.mit.edu/courses/5-111sc-principles-of-chemical-science-fall-2014/resources/mit5_111f14_lec10soln/)).
[](https://i.stack.imgur.com/IYbet.png)
But I've also produced another resonance structure which I think is valid but it was not included in the solutions. I've put the formal charge on on the non-terminal N atom. My structure seems to satisfy the octet rule for all atoms and has the same number of bonds and electron pairs as the other solutions. Is there something I'm missing here?
[](https://i.stack.imgur.com/f0imt.png)
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170455/why-surface-tension-acts-tangentially-if-there-is-a-net-inward-force
|
Why surface tension acts tangentially if there is a net inward force? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/170455/edit).
Closed 8 months ago.
This post was edited and submitted for review 8 months ago and failed to reopen the post:
>
> Original close reason(s) were not resolved
>
>
>
[Improve this question](/posts/170455/edit)
I understand that surface tension is caused by an imbalance of forces for molecules on the surface in comparison to those that lie within. As a result, this creates a net inward force that seeks to reduce surface area. However, if this force acts inwards, why does surface tension act tangentially? I've seen explanations that state bonds on the surface become stronger when the surface area is minimised. If this is true then why is this the case? It doesn't seem clear to me why a net downward force causes the surface to act as an elastic membrane.
| 1
|
[
[
"\nEach molecule at the surface of a drop of water is pulled towards the others around it (by cohesion, i.e. in this case hydrogen bonds), those at the surface, and those below. There is nothing above it, so the net force is pointing towards the center of the drop.\n\n\nIn the orthogonal surface area, the forces cancel out, except if the surface was previously disturbed/deformed, i.e. the area increased from the planar/spherical ideal.\n\n\nWhen you increase the surface area by deforming the drop, that means you pull out water molecules from the body, against that net downwards force.\n\n\nErgo, surface tension and that perpendicular force are essentially the same thing.\n\n\n",
"-1"
]
] |
https://chemistry.stackexchange.com/questions/170298/white-vs-light-pink-color-of-mgo
|
white vs light pink color of MgO
|
Recently I ordered some high purity MgO, which is light pink in color. However, the old powder I was using was white. MSDS also suggests that it should be white. Can anyone explain the reason of two different colors of MgO of same purity.
Thanks in advance
PS- I suspected it to be due to absorption of moisture earlier, but it is not the case. Even on heating at 800 C, does not change the color.
| 2
|
[] |
https://chemistry.stackexchange.com/questions/170295/experimental-evidence-for-zwitterions
|
Experimental evidence for zwitterions
|
If I have a diprotic acid that is +1 positively charged in its fully protonated state, I can figure out the apparent equilibrium constants by titration with base. The net charge will be neutral after losing one proton, and -1 after losing the second proton.
Examples would be the hydronium ion or the amino acid glycine at a pH < 2.
How could you determine whether the single-protonated state is a zwitterion (glycine in neutral aqueous solution) or a molecule without charged functional groups (glycine in neutral polar aprotic solvent, or water at neutral pH)?
Is it necessary to do an experiment, or is there a simple rule to figure out whether a zwitterion is expected?
| 10
|
[
[
"\nThat is a very difficult problem worth a PhD project in physical chemistry. Intially, I thought one could try capillary electrophoresis at different pHs and if the analyte travels with the electrosmotic flow marker at a certain pH, one can say a zwitterion has existed (or a neutral form) by inference. However, you query is about *distinguishing* the neutral state and the zwitterionic state. One has to think about spectroscopic techniques that rely on dipole moments. Since you are interested in aqueous phase, gas phase microwave spectroscopy of amino-acids is out of question (there are papers on it). Since zwitterion ions can possess large dipole moments as compared to their neutral counterparts, two remaining ones are infrared and photoelectron spectroscopy.\n\n\nZwitterion formation in hydrated amino acid, dipole bound anions: How many water molecules are required? J. Chem. Phys. 119, 10696 (2003); <https://doi.org/10.1063/1.1620501>\n\n\n\n> \n> We utilize the facts that zwitterions possess very large dipole\n> moments, and that excess electrons can bind to strong dipole fields to\n> form dipole bound anions, which in turn display distinctive and\n> recognizible photoelectron spectral signatures. The appearance of\n> dipole-bound photoelectron spectra of hydrated amino acid anions,\n> beginning at a given hydration number, thus signals the onset of\n> greatly enhanced dipole moments there and, **by implication, of\n> zwitterion formation**.\n> \n> \n> \n\n\nIR would be finiky with water, but there are reports that water bound to zwitterions has a different vibrational frequency\n\n\nStability and IR Spectroscopy of Zwitterionic Form of β-Alanine in Water Clusters, J. Phys. Chem. B 2019, 123, 20, 4392–4399, <https://doi.org/10.1021/acs.jpcb.9b00654>\n\n\n\n> \n> We perform an experimental and computational study on the number of\n> water molecules needed for zwitterion formation of β-alanine. Our\n> density functional theory investigation reveals that a minimum of five\n> water molecules are required to form and stabilize the zwitterion. A\n> characteristic connecting water molecule located between the COO– and\n> NH3+ groups is found to enhance the stability. This water molecule is\n> also involved in a characteristic infrared active vibration at ≈1560\n> cm–1, which is slightly shifted with the number of surrounding water\n> molecules and is located in a spectral region outside of water\n> vibrations.\n> \n> \n> \n\n\n",
"8"
]
] |
https://chemistry.stackexchange.com/questions/170288/assigning-configuration-of-chiral-plane-selecting-of-the-pilot-atom
|
Assigning configuration of chiral plane - selecting of the pilot atom
|
I am little confused about when it comes to assigning configuration of planarly chiral cyclophanes - and I think it comes down to selection of pilot atom.
All sources I have came across say that the pilot atom is the first out of the plane atom and it should be selected in order assign configuration of chiral plane. However, these sources present slightly different strategies for selecting a pilot atom.
(a) One says that the pilot atom should be the one which is on the same side as the highest priority substituent which is causing desymmetrization the plane (i.e. the one that is closer).
(b) While the other source says simply that the pilot atom should be selected according to CIP priority rules.
In most cases it doesn't make any differences. Actually, it doesn't make a difference for any of the examples that I have came across. However, when thinking about it a structure came to my mind that brings conflict between these two strategies, it looks like this:
[](https://i.stack.imgur.com/ZdX66.png)
Seems to me that if we followed strategy (a), then the pilot atom would have to be a carbon from the top CH2 group and chiral plane would be assigned with *R*p configuration. However, following strategy (b) means that oxygen atom gets the pilot status and chiral plane has to be assigned with *S*p configuration. Only one approach can (if any of these) can be correct - could you tell me which one is it?
I have checked IUPAC recommendations (<https://iupac.qmul.ac.uk/BlueBook/PDF/P9.pdf>), but I didn't find an answer to my question.
| 3
|
[] |
https://chemistry.stackexchange.com/questions/170285/difference-between-sodium-aluminate-and-sodium-metaaluminate
|
Difference between sodium aluminate and sodium metaaluminate
|
So I was studying Chemistry (Hydrogen - Uses, Properties and Preparation) from 2 different textbooks (based on Grade 9 syllabus, I.C.S.E. Board, India) and I found what seems an astounding anomaly to me.
In the two textbooks, the *same* reaction for liberating Hydrogen gas from a hot concentrated solution of $\ce{NaOH}$ and $\ce{Al}$ is given as follows:-
Textbook 1: $\ce{2Al + 2NaOH + 2H2O -> \underset{\text{Sodium metaaluminate}}{2NaAlO2} + 3H2 ^}$
Textbook 2: $\ce{2Al + 2NaOH + 2H2O -> \underset{\text{Sodium aluminate}}{2NaAlO2} + 3H2 ^}$
If the two reactions are the same, why and how is the name of the same compound ($\ce{NaAlO2}$, *Sodium meta aluminate* / *Sodium aluminate*) different? Secondly, what does 'meta' imply in Sodium metaaluminate?
| 2
|
[
[
"\nWhat appears to be an \"astounding anomaly\" to you is totally normal in chemical nomenclature and biological sciences in general. Remember is chemistry has been studied by millions of people for centuries. One cannot make them follow identical nomenclature rules, although general consesus is developed by IUPAC. Many chemical names have 10 or even more synonyms. If you study botany, a single plant may have plenty of radically different Latin names. Here are some more names (16) identified by SciFinder (Chemical Abstract Service, USA) for the same compound. I list only a few as examples. Now note that that some names may not the be standard ones.\n\n\nIn short, both texts are fine.\n\n\nAluminum sodium dioxide\n\n\nSodium aluminate ($\\ce{Na2Al2O4}$)\n\n\nSodium aluminum dioxide\n\n\nSodium aluminum oxide ($\\ce{NaAlO2}$)\n\n\nSodium metaaluminate ($\\ce{NaAlO2}$)\n\n\n",
"6"
],
[
"\nSeems like just a nomenclature thing to me. Prefixes *meta*, *ortho* and *pyro* are being sometimes used when giving names of inorganic acids to indicate a ratio of hydrogen atoms to the central atom(s).\n\n\nAnd so:\n\n\n* *meta* means an acid with a minimal number of hydrogen atoms in relation to central atom (given its oxidation state), e.g. $\\ce{(HPO3)n}$ is metaphosphoric(V) acid\n* *ortho* is a prefix given to an acid which has emperical formula having an H2O more than its *meta* acid counterpart, e.g. $\\ce{H3PO4}$ is orthophosphoric(V) acid\n* *pyro* denotes an acid formed by condensation of an *ortho* acid, so pyrophosphoric(V) acid is $\\ce{H4P2O7}$\n\n\nIn case of a compound from your question $\\ce{NaAlO2}$ can be treated formally as a salt of \"$\\ce{HAlO2}$\" acid - which would be of *meta* type. However, this a nomenclature thing so this prefix may be skipped.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/170281/buffer-solution-of-nah2po4-and-na2hpo4
|
Buffer solution of NaH2PO4 and Na2HPO4
|
We have this exercise without solutions
>
> From a 0.2 M $\ce{NaH2PO4}$ solution and a 0.2 M $\ce{Na2HPO4}$ solution, a buffer
> solution with pH = 6.8 is to be prepared. The total concentration of
> $\ce{PO\_4^3}$- should be 0.1 mol l-¹. Calculate the volumes of the two solutions needed to prepare one litre of buffer solution.
>
>
>
So my understanding is that the $\ce{Na}$ will dissolve in the water. Also, both $\ce{NaH\_2PO\_4}$ and $\ce{Na\_2HPO\_4}$ are acidic, but $\ce{NaH\_2PO\_4}$ will be more acidic in this case because is has one more $\ce{H}$ than the other. So if a buffer consists of those two substances, then $\ce{NaH\_2PO\_4}$ will be the acid and $\ce{Na\_2HPO\_4}$ will be the base
First, is this correct ?
What I don't really understand is
>
> The total concentration of $\ce{PO\_4^3}$- should be 0.1 mol l-¹.
>
>
>
Does this mean that $c(\ce{NaH\_2PO\_4}) + c(\ce{Na\_2HPO\_4})= 0.1 $ ? In other words, we would need to set up the Henderson Hasselbalch equation
$$pH = pK\_a + \log{\frac{c(\ce{Na\_2HPO\_4})}{c(\ce{NaH\_2PO\_4})}} = pK\_a + \log{\frac{c(\ce{Na\_2HPO\_4})}{c(\ce{Na\_2HPO\_4) - 0.1}}} $$
and then solve for $c(\ce{Na\_2HPO\_4})$, and then we would find $$c(\ce{NaH\_2PO\_4}) = 0.1 - c(\ce{Na\_2HPO\_4}) $$
We have one litre of buffer, so the number of moles is simply
$$n(\ce{Na\_2HPO\_4}) = 1 \cdot c(\ce{Na\_2HPO\_4})$$ respectively
$$n(\ce{NaH\_2PO\_4}) = 1 \cdot c(\ce{NaH\_2PO\_4})$$
and then the volume is found by dividing the number of moles by the original concentration of $0.2 M$
Is this how they expect us to proceed ? We have no solutions so I prefer to ask
Are there some other important things I need to be aware of ? The $\ce{Na}$ seems to dissolve in the water, but what happens to the different hydrogen atoms? Do they simply become $\ce{H+}$ and $\ce{HO-}$ ?
| 1
|
[] |
https://chemistry.stackexchange.com/questions/170279/how-is-there-still-iron-on-earth
|
How is there still iron on earth? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/170279/edit).
Closed 9 months ago.
[Improve this question](/posts/170279/edit)
Iron rusts and the earth is pretty old, so how is it that there is still iron left that has not oxidized(/rusted)?
I tried looking it up, and the amount of iron on earth is mind boggling, but is that it?
Is there simply enough iron that not all of it has oxidized yet?
Maybe it's just that well insulated inside the crust, or does it occur naturally so more is created all the time?
| 2
|
[
[
"\nYou may confuse iron(1) as an element and iron(2) as the metallic form of iron(1). Rusting of iron(2) does not destroy iron(1), but converts it to iron(1) compounds like non-stoichiometric hydrated oxides.\n\n\nIron(1) of Earth is much older than Earth and the Solar system. But most of surface iron(2) is less than 100 years old. All but the one contained in rare meteorites was produced by men from iron(1) ores.\n\n\nIron(1) in [Earth's mantle](https://en.wikipedia.org/wiki/Earth%27s_mantle) and [Earth's crust](https://en.wikipedia.org/wiki/Earth%27s_crust) occurs in oxidized forms in various minerals, ores and rocks. E.g. the primary mineral of Earth upper mantle is [olivine](https://en.wikipedia.org/wiki/Olivine) $\\ce{(Mg, Fe)2SiO4}$\n\n\nThere is also iron(2), together with nickel, in [Earth core](https://en.wikipedia.org/wiki/Internal_structure_of_Earth#Core), at high temperature and extreme pressure, where is no rusting.\n\n\n",
"9"
],
[
"\nYou have it backwards. For about 2000 years people have taken iron oxides (primarily) and converted them into metallic iron and steel. So there is dramatically more metallic iron on the earth surface now than 2000 years ago. about 6 million tons per day of new metallic iron/steel is made on earth each day. Admittedly much of that steel is recycled steel. Some mills only recycle metallic steel. Certainly many tons of oxides, etc, develop each day but I can't guess what that amount is except it is much less than new production of new metallic steel\n\n\n",
"8"
]
] |
https://chemistry.stackexchange.com/questions/170268/when-we-take-baths-how-does-the-water-in-the-bathtub-absorb-only-the-dirtiness
|
When we take baths, how does the water in the bathtub absorb only the dirtiness on our skin?
|
I'm 16. I study chemistry in college along with biology, physics, and maths, and I have a general interest in all of them.
I always found this interesting: sometimes when I left the bathtub, even when I forgot to use soap, the bath would still have a somewhat noticeable, murky tint in the water, indicating that something had dissolved in it, even though I didn't take the time to rub it off manually. When we get in the bath, how does the dirtiness (small amounts of mud, bacteria, pen markings etc.) dissolve in the bath alone without any of our skin dissolving as well so that we can leave clean? Maybe dead skin might dissolve, that would make sense, is there an adaptive mechanism on the skin involved in this sort of process of hygiene? The same thing would happen if I put any other object in the water with something on its surface and the object would come out clean when the thing on the surface dissolved.
How does the water break the attraction? Some sort of cohesive force, between the atoms/molecules of the thing (dirtiness) on the surface of the object and the atoms/molecules of the object (our bodies) and separate them to make it dissolve? And how can it differentiate between the cohesion of the skin and the dirtiness (what determines solubility, if both polar and non polar molecules on our skin will dissolve in the water)?
| 0
|
[
[
"\nWater is called a universal solvent, and perhaps it is rather a miraculous solvent with unusual properties. It is able to \"wet\" a lot a surfaces and thus able to disrupt the adhesive forces. (Cohesion refers to sticking of like molecules only.)\n\n\nHowever, the murkiness in water is not due to dissolution of dead skin or dirt dissolving in water, but rather it is a suspension (hence murky). If you had a microscope, you would rather see particles of dust, mud, soil, oils, dead skin etc if someone has not taken a shower for a while. Just pouring water does not remove these things from objects, mechanical action is also needed to dislodge them from the surface (e.g., dishwashers and washing machines use mechanical forces to clean items with water).\n\n\nSince you appear to like science, next time do a Tyndall light experiment. Take that murky tinted water (as you call it), and fill it in a transparent glass tumbler. Pass light through it, and view that water at ninety degrees (to the light beam). If you see a light beam, it indicates really small particles floating in water. It is not a true solution.\n\n\n",
"4"
],
[
"\nThe outer layer of skin is water-insoluble. According to [this paper](https://rupress.org/jcb/article/179/7/1599/34826/Mice-deficient-in-involucrin-envoplakin-and), there is a structure specifically made in skin cells before they die and become part of the outermost layer of the skin:\n\n\n\n> \n> The outermost, cornified, layers of the epidermis are composed of terminally differentiated keratinocytes known as corneocytes. Corneocytes lack a plasma membrane and instead are encased in a structure known as the cornified envelope (CE). CEs consist of highly insoluble, cross-linked proteins with covalently attached lipids and are essential for the mechanical integrity and water impermeability of the skin.\n> \n> \n> \n\n\nSo not dissolving in water is one function of the skin. It also protects the inside of the body from dehydration by preventing excess evaporation of water on the skin (unless you are hot, when sweating helps to maintain a lower body temperature, also important for survival).\n\n\n\n> \n> [OP] How does the water break the attraction?\n> \n> \n> \n\n\nThat depends on what is on the skin. If you are muddy, the components of the mud are easy to wash off with water because the components are polar, and water is polar. If you are covered in motor oil (which is non-polar), water alone will not be efficient to wash it off, but soapy water will work (with some scrubbing).\n\n\n\n> \n> [OP] When we get in the bath, how does the dirtiness (small amounts of mud, bacteria, pen markings etc.) dissolve in the bath alone without any of our skin dissolving as well so that we can leave clean?\n> \n> \n> \n\n\nWhile the layers of dead skin cells are not water-soluble (or soapy water soluble), we nevertheless continually shed these former cells, and continually replace them. In fact, much of household dust is made of material we shed (e.g. it contains quite a bit of cholesterol, as described [here](https://pubs.acs.org/doi/abs/10.1021/es103894r)).\n\n\n\n> \n> When we take baths, how does the water in the bathtub absorb only the dirtiness on our skin?\n> \n> \n> \n\n\nIt doesn't. When you take long bubble-baths, your skin will also lose oils and other substances (giving you \"dry skin\"). Some people use oils or lotions (\"moisturizer\") to replenish these after a long soak.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170267/why-do-nuclei-move-considerably-slower-than-electrons
|
Why do nuclei move considerably slower than electrons
|
I've been trying to learn quantum chemistry at an introductionary level. While reading I've found out that the Born-Oppenheimer approximation seems to be the reason for the basic and crucial model of molecules for chemists; molecular structure in space as nuclei occupying some positions with chemical bonds between them. Thus, I've tried to learn more about this approximation. However, I've gotten stuck at the reasoning for wanting to try to separate the wavefunction of the molecule into a product of two functions, one depending on the positions of the nuclei and one depending on the positions of the electrons as well as depending parametrically on the positions of the nuclei. All resources state that the reasoning is because nuclei move considerably slower than electrons, but I have not found one that could elaborate why this is the case.
Most books and other resources simply states that "Because the nuclei is over thousand times heavier than an electron, in the timescale of the movements of the electron the nuclei can be considered to be standing still.", or "Because of the great difference in mass, the electrons can respond almost instantaneously to the movements of the nuclei", and then quickly moves on. However, in [this](https://chemistry.stackexchange.com/a/31367) answer on "Born–Oppenheimer adiabaticity" it was said that "nuclei at average are moving considerably slower than electrons" because of the equipartition theorem.
In the book *Introduction to Statistical Physics* by S. R. A. Salinas, the theory of equipartition of energy is formulated by:
>
> Consider a classical system of $n$ degrees of freedom given by the Hamiltonian $$H = H\_0 + \phi p\_j^2$$ where (i) $H\_0$ and $\phi$ are functions that do not depend on the particular coordinate $p\_j$; (ii) the function $\phi$ is always positive; and (iii) the coordinate $p\_j$ is defined from $-\infty$ to $+\infty$.
>
>
>
It then shows how the expectation value of $\phi p\_j^2$ equals $\frac{1}{2}k\_BT$, where $k\_B$ is the Boltzmann constant and $T$ temperature. Thus, we can conclude that for any system (in thermal equilibrium) that has energy contributions that are quadratic in such a variable, the average energy is $\frac{1}{2}k\_BT$ for each of those degrees of freedom. Since kinetic energy is always quadratic in the momentum, it will have an average energy of $\frac{1}{2}k\_BT$ in thermal equilibrium in each degree of freedom if the rest of the Hamiltonian does not depend on the momentum (which is the case for the Hamiltonian of a molecule). This elaborates the reasoning in the linked answer. However, this seems to "only" apply for macroscopic systems of molecules which also have to be in thermal equilibrium. I am not completely satisfied with this, especially considering when this is only about the average kinetic energy. I would assume that the distribution of kinetic energy for the different degrees of freedom in a general system at room temperature is such that a decent portion are thousands of times higher than others. I could not find any answer that rectified this.
One book I read mentioned something different than others; that the nuclei move considerably slower than electrons because of conservation of momentum. However, it gave no elaboration. One text I remember reading somewhere on a website said something along the lines of: "The electrostatic forces on nuclei and electrons from their pairwise interaction are the same, and so the changes in their momenta from these forces must be the same. Therefore, one might assume that the actual momenta of the nuclei and electrons are of similar magnitude." This is the explanation for the reasoning that makes the most sense to me, but I have not seen it anywhere else. Since it came from a source that wasn't necessarily trustworthy it made me unsure, until I saw the statement of conservation of momentum which felt related.
Does the statement of equipartition make sense as the explanation for the reasoning? If yes, why? Does the statement of changes in momenta make sense as the explanation for the reasoning? Or maybe there is something different that is the sensible explanation? Please do help me understand more clearly, I've been at this for weeks and I cannot seem to get any answer from books and other texts.
| 7
|
[] |
https://chemistry.stackexchange.com/questions/170265/why-does-zircon-hate-lead-how-do-these-tiny-crystals-so-effectively-exclude-lea
|
Why does zircon hate lead? How do these tiny crystals so effectively exclude lead atoms during formation enabling accurate uranium-lead (U-Pb) dating?
|
### Background
From the December 29, 2022 NPR article [To peer into Earth's deep time, meet a hardy mineral known as the Time Lord](https://www.npr.org/2022/12/29/1139782011/to-peer-into-earths-deep-time-meet-a-hardy-mineral-known-as-the-time-lord):
>
> "(Zircons) are really the best markers of Earth's time, or the history of the Earth," says [Michael Ackerson](https://naturalhistory.si.edu/staff/michael-ackerson), a geologist with the Smithsonian's National Museum of Natural History.)
>
>
>
From the podcast linked on the same page after 01:00:
>
> He tells me that one of the crystals I'm looking at is 4.32 billion years old. Zircons are the oldest known pieces of Earth that still exist on the surface today; the oldest go back as far as 4.37 billion years.
>
>
> "They are the best markers of Earth's time, (for) the history of the Earth.
>
>
> Ackerson says a zircon crystal originally forms in magma - hot molten rock. Other minerals do too; together they'll make up say, a granite mountain, that slowly weathers away.
>
>
> "Most of the minerals don't survive, they slowly weather away. Quartz, feldspar, they're chemically or physically weathered or eroded, to a point where they're no longer quartz and feldspar. Zircon, and one of the reasons that zircons are so useful, is that zircon is very resilient.
>
>
> The hearty crystals persist, and eventually get incorporated into another rock as it's forming. That means scientists can crush up the Earth's oldest rocks and pick through the debris, to find little grains of zircon, that are even older.
>
>
> To know a zircon's age, they zap it with a laser.
>
>
>
Discussion of the use of lasers, argon plasma and detectors to count uranium and lead atoms.
>
> The important atoms are uranium and lead. A growing crystal of zircon ***loves uranium*** and will take it in, but ***zircon hates lead,*** so if you find lead inside, it pretty surely came from the decay of uranium, which happens at a steady rate, like the ticking of a clock.
>
>
>
### Question:
Why does zircon hate lead? How do these tiny crystals so effectively exclude lead atoms during formation enabling accurate uranium-lead (U-Pb) dating?
I'm interested in a basic understand of the chemistry that allows a growing crystal of zircon in magma to so effectively prevent uranium from being incorporated while allowing so many other metals in.
Besides uranium and obviously zirconium and silicon, Wikipedia's [zircon](https://en.wikipedia.org/wiki/Zircon) gives the formula for zirconium silicate as $\ce{(Zr\_{1–y}, REE\_y)(SiO4)\_{1–x}(OH)\_{4x–y}}$ where $\ce{REE}$ stands for [rare earth element](https://en.wikipedia.org/wiki/Rare-earth_element):
>
> ...a set of 17 nearly-indistinguishable lustrous silvery-white soft heavy metals.
>
>
>
The Wikipedia zircon page also says:
>
> Zircon precipitates from silicate melts and has relatively high concentrations of high field strength incompatible elements. For example, hafnium is almost always present in quantities ranging from 1 to 4%.
>
>
>
---
This seems potentially relevant: [Chapter Six - Computer modeling of Zircon (ZrSiO4)—Coffinite (USiO4) solid solutions and lead incorporation: Geological implications](https://www.sciencedirect.com/science/article/abs/pii/S246851781930005X)
>
> The structure of zircon, ZrSiO4 is modeled using interatomic potentials. The uranium end-member, coffinite (USiO4) and intermediate solid solutions of zircon and coffinite (UxZr1–xSiO4) are then modeled, allowing the prediction of lattice parameters as a function of uranium concentration. Finally, possible structures resulting from the radioactive decay of uranium to lead in coffinite are considered.
>
>
>
| 8
|
[
[
"\nZircon is rather picky because of its rather unusual composition. It has the composition $\\ce{MSiO4}$ where $\\ce{M}$ is predominantly zirconium (which is of course named after the mineral) but generally includes other elements in solid solution. Bevause the $\\ce{SiO4}$ group with the normal oxidation states of silicon and oxygen takes four negative charges, the metal $\\ce{M}$ must match the $+4$ oxidation state of the zirconium, at least on average. The metal must also have relatively large atoms in this oxidation state to substitute comfortably for the zirconium. Only lanthanides, actinides, and a few early-group transition elements in the fifth or later period fit these criteria. [Wikipedia](https://en.wikipedia.org/wiki/Zircon) specifically mentions \"[high field strength incompatible elements\"](https://en.wikipedia.org/wiki/Incompatible_element) including $\\ce{Zr, Nb, Hf, Ta, Th, U}$ and rare earth elements.\n\n\nUranium, which commonly adopts the $+4$ oxidation state, works well, but lead does not. Lead is too difficult to achieve the $+4$ oxidation state because of the relativistic impact that emerges in heavy late-group elements. Wikipedia discusses this relativistic effect:\n\n\n\n> \n> Lead shows two main oxidation states: +4 and +2. The tetravalent state is common for the carbon group. The divalent state is rare for carbon and silicon, minor for germanium, important (but not prevailing) for tin, and is the more important of the two oxidation states for lead.[1] This is attributable to relativistic effects, specifically the inert pair effect, which manifests itself when there is a large difference in electronegativity between lead and oxide, halide, or nitride anions, leading to a significant partial positive charge on lead. The result is a stronger contraction of the lead 6s orbital than is the case for the 6p orbital, making it rather inert in ionic compounds.\n> \n> \n> \n\n\n**Cited Reference**\n\n\n1. Greenwood, N. N.; Earnshaw, A. (1998), *Chemistry of the Elements* (2nd ed.), Butterworth-Heinemann, ISBN 978-0-7506-3365-9, p. 373.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/170264/my-experiemental-density-of-h2-does-not-match-the-published-data-what-causes-th
|
My experiemental density of H2 does not match the published data. What causes this discrpenecy?
|
I placed 1 gram of NaBH4 in a balloon and placed it over a glass round bottle filled with water and acetic acid. While secured, I emptied the balloon into the bottle and made sure to wash the inside of the balloon. The reaction taking place is $\ce{NaBH4 + 2(H2O) -> NaBO2 + 4(H2)}$.
The yield in this case should be 213 grams of $\ce{H2}$. According to Wikipedia, "hydrogen" has a density of 0.08988 g/L. That should leave me with 2.369 L of $\ce{H2}$. But my little experiment showed a balloon with the about 13 cm in diameter, which a volume of sphere that wide would be 1.15 liters of H2. This would mean a yield of .103 grams or 48.5%. But that just doesn't make sense. My NaBH4 has not degraded by half.
If you were to double the given density of $\ce{H2}$ to 0.17976 g/L, the results are much more in line with what I'd expect. The yield of $\ce{H2}$ would be .206 grams or 97%, which would make sense since not everything is ideal and my $\ce{NaBH4}$ is a bit old. Obviously there will be minor errors in measurement, and its not a perfect sphere and I didn't control for temperature, but that cannot account for a 50% reduction in yield. Maybe the balloon is compressing the gas? But I don't think so.
So the only other thing I can think of is that the presented density of 0.08988 g/L for hydrogen is technically only for monoatomic Hydrogen, or $\ce{H}$, which doesn't make sense and no one ever uses. The density of $\ce{H2}$ at STP should be 0.17976 g/L then to match my observations. There must be something I'm missing here entirely that I'm not understanding.
| 3
|
[
[
"\n[The pressure inside your balloon is not approximately 1 atmosphere](https://chemistry.stackexchange.com/a/31958/1499), even at ambient temperature. It is substantially greater, as it has to counteract atmospheric pressure **plus** the elastic force of the balloon latex trying to pull itself together. Therefore, your sample of hydrogen gas is not close to STP conditions, and its molar volume and density should *not* match. Also annoyingly, the pressure inside a balloon is sensitively dependent on how inflated the balloon is, and it varies non-monotonically - it starts at 1 bar, increases up to a maximum, then decreases again. This is not a very good way to measure gas volumes.\n\n\nIf you want a easy and fairly reliable way of measuring gas volumes on the order of a liter, a typical procedure is to fill a container (usually a graduated cylinder for extra precision) with water, invert it, and [measure the water displaced by the evolved gas](https://youtu.be/Bfu7QZqm1b8?t=157s). For volumes on the order of milliliters, you can displace the plunger of a gas-tight syringe.\n\n\n**Edit**: Actually when I wrote this answer, I trusted the calculations were overall correct, but Karsten's comment tipped me off to some errors, so I decided to double-check. So let's calculate the theoretical volume of hydrogen gas produced:\n\n\n$$\\ce{NaBH4 (s) + 4 H2O (l) -> NaB(OH)4 (aq) + 4 H2 (g)}$$\n\n\n(Arguably your proposed sodium metaborate product is actually best described as sodium tetrahydroxyborate in aqueous solution, though that doesn't affect the proportionality between moles of borohydride used and moles of hydrogen gas evolved.)\n\n\n$$\\mathrm{1.00\\ g\\ NaBH\\_4 \\times \\frac{1\\ mol\\ NaBH\\_4}{37.83\\ g\\ NaBH\\_4 }\\times \\frac{4\\ mol\\ H\\_2}{1\\ mol\\ NaBH\\_4}\\times \\frac{22.4\\ L\\ H\\_2(@0\\ °C, 1\\ atm)}{1\\ mol\\ H\\_2}=2.37\\ L\\ H\\_2(@0\\ °C, 1\\ atm)}$$\n\n\n(This volume can change by ~10% depending on the actual temperature of the gas inside your balloon. At 25 °C, you would get 2.60 L instead.)\n\n\nSo even though your calculation is a bit roundabout and contains some mistakes, it does seem the major error is experimental.\n\n\n",
"9"
],
[
"\nFirst, a minor consideration about gases:\nIn an experimental setup like yours, I would not really try to deal with a gas mass and density. The molar volume of any ordinary gas is ~22.4 liter. Easy to remember and one size fits all.\n\n\nSecond, the experiment results:\nArmed with the above consideration about the molar volume, I would ask myself first not \"what's wrong with my hydrogen?\", but instead \"why I am getting too little gas?\" .\n\n\nThis alternative question has two important advantages:\n\n\n* It does not challenge a well-established fact\n* It is way easier to propose answers.\n\n\nThe possible answers may be:\n\n\n1. The purity of the substances used (you state that \"not everything is ideal with my...\" . How can the possible impurities react in similar conditions?\n2. Parasitic reaction paths that can yield less hydrogen (well, I cannot think of one in your particular case)\n3. Did you collect all of the gas? Can some of it be dissolved in the solution? (there is solubility data available online - at least about hydrogen in water, you can get this as an estimate for your final solution)\n4. What are my margins of error in both measurement and calculation? You start with 1.0g +/- 1% or 1.0g +/- 20% ?\n\n\nFor the point 1 question, you can stage a much easier experiment that actually involves measuring the mass of the hydrogen: weight your reacting substances, mix them and wait until the reaction finishes, weight again. Subtract the final mass from the initial mass.\n\n\nYou may, or may not get something like 213mg .\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/170259/length-of-a-1d-box-in-hexa-1-3-5-triene
|
Length of a 1D box in hexa-1,3,5-triene
|
### Problem
From Hayward's *Quantum Mechanics for Chemists* [1, p. 36]
>
> **2.3.** Calculate the wavelength of light that will be absorbed when
> a it electron in hexa-1,3,5-triene, $\ce{CH2=CH—CH=CH—CH=CH2},$ is promoted from the highest occupied level to the lowest unoccupied level. The average $\ce{C—C}$ bond length in hexatriene can be taken to
> be $\pu{144 pm}.$ Compare your answer with the experimentally observed
> wavelength of $\pu{258 nm}.$
>
>
>
### Answer
>
> **2.3.** Wavelength of light is $\pu{352 nm}$ (for box length of $\pu{864 nm})$
>
>
>
### Question
Since $L = 5\times(\pu{144E-12 m}),$
$$E = \frac{7h^2}{8mL^2} = \pu{8.14E-19 J}.\tag{1}$$
After substituting the values I get the answer
$$\lambda = \frac{hc}{E} = \pu{244 nm}.\tag{2}$$
The textbook answer is $\pu{352 nm}.$ What confuses me is that they state that the length of the box is $\pu{864 nm}$ — but there are *five* bonds joining the *six* carbons together. So, would not we multiply the average bond length by $5$ and not $6?$
Is it acceptable to take the length of the box as the average bond length multiplied by the number of atoms?
### Reference
1. Hayward, D. O. *Quantum Mechanics for Chemists*; Tutorial chemistry texts; Royal Society of Chemistry: Cambridge, UK, **2002**. ISBN 978-0-85404-607-2.
| 4
|
[
[
"\nFor this type of question, we not only consider the length of the bonds between the carbon atoms, but we also have to add on the atomic radii at both ends of the structure. I assume the atomic radii of carbon atoms taken in the text is $\\pu{72 nm}$, which means doing $\\pu{144 nm}\\cdot5 + \\pu{72 nm}\\cdot2 = \\pu{864 nm}$.\n\n\nWe can then solve it like you did doing $E = (4^2-3^2)\\frac{h^2}{8mL^2}$ and the applying $\\lambda = \\frac{hc}{E}$. Ultimately, however, the addition of the two carbon radii to the length of the box would explain the textbook result (see page 6 of the link below).\n\n\n**Source:** [https://nanohub.org/courses/OED/01a/asset/5958#:~:text=Using%20the%20simple%20model%20shown,estimated%20to%20be%200.867%20nm](https://nanohub.org/courses/OED/01a/asset/5958#:%7E:text=Using%20the%20simple%20model%20shown,estimated%20to%20be%200.867%20nm).\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170257/why-does-fcc-crystalline-forms-form-hexagonal-crystalline-structures-during-cvd
|
Why does FCC crystalline forms form hexagonal crystalline structures during CVD?
|
During the CVD process, Copper and Nickel which are both Face Centered Cubic crystalline structures are used as substrates for shaping hexagonal atomic structures such as graphene and hexagonal Boron-Nitride. Why is it that these form hexagonal atomic structures when they themselves aren't hexagonal?
| 0
|
[
[
"\n\n> \n> [Ivan Neretin] But they are hexagonal, if you look at them sideways. – \n> \n> \n> \n\n\n\n> \n> [OP] Sorry I dont quite grasp what you mean. Do you have a visual representation?\n> \n> \n> \n\n\nHere is the [visual](http://lampx.tugraz.at/%7Ehadley/ss1/crystalstructure/structures/fcc/fcc_jsmol.php).\n\n\n[](https://i.stack.imgur.com/KJcBk.gif)\n\n\n\n> \n> [Jon Custer] Fcc and hcp structures are similar stacking (fcc you need to look along the 111 body diagonal - a classic issue with the cubic conventional cell).\n> \n> \n> \n\n\nYes, both may be constructed by stacking hexagonal layers, either in an ABABA.. (hexagonal close packing hcp) or an ABCABCA.. (face-centered cubic fcc) fashion.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/170253/i-have-plotted-a-stern-volmer-plot-but-cannot-work-out-the-fluorescence-lifetime
|
I have plotted a Stern-Volmer Plot but cannot work out the fluorescence lifetime to be able to get the rate constant [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/170253/edit).
Closed 9 months ago.
[Improve this question](/posts/170253/edit)
I have plotted a stern Volmer plot and have my straight line equation. I understand that the gradient is kqtf but I need tf to work out kq but I don't have kd to do 1/kd.
How do I work it out?
| -4
|
[
[
"\nIn a Stern-Volmer plot the ratio of fluorescence yield $\\varphi$ is plotted vs the quencher concentration $[Q]$ as $\\varphi\\_0/\\varphi\\_Q =1 +k\\_{SV}[Q] $ where the gradient is the quenching constant which is $k\\_2\\tau$ where $k\\_2$ is the 2nd order rate constant and $\\tau$ the fluorescence decay in the absence of quenching.\n\n\n(If you search this site you will find examples <https://applying-maths-book.com/intro.html> )\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170252/what-happens-if-benzene-reacts-with-cl2-in-presence-of-febr3
|
what happens if benzene reacts with Cl2 in presence of FeBr3?
|
we know that halogenation of benzene adds a halogen to benzene in presence of a strong electrophile. but what if we use chlorine as a halogen but use an electrophile made out of a different halogen, like bromine?
| -1
|
[
[
"\nTypically mixing a halogen electrophile with a catalyst using a different halide ion is not done in Friedel-Crafts halogenations. Here's why in the case described in the question.\n\n\n* The electrophile could be changed through halogen displacement. Here, that would be chlorine reacting with $\\ce{FeBr3}$, independently of the benzene, so the electrophile in subsequent reaction with benzene becomes bromine instead of chlorine. Since the chlorine could alternatively act as an electrophile first, you get a mixture of halobenzene products.\n* When the proton is displaced, it will combine with a halide ion abstracted from the halide molecule/complex anion. But that could be bromide instead of chloride, giving a mixed hydrogen halide product and altering the catalyst (and perhaps its solubility and Lewis acidity).\n\n\nSo you risk mixing up the organic product, the displaced hydrogen halide and the catalyst. Don't go there. Match the halide ion in the catalyst with the halogen electrophile.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/170246/why-dont-serine-and-threonine-have-a-3rd-pka
|
Why don't serine and threonine have a 3rd pKa
|
Serine and threonine both have an extra OH functional group and hence they should be able to dissociate when the pH gets high enough. However multiple amino acid pH charts do not show the 3rd pKa of these 2 amino acids. Why is this so?
| 1
|
[
[
"\n**Serine and threonine have extremely high 3rd pKa values**\n\n\nSerine and threonine have a stable OH group, making it unlikely to dissociate except at extremely high pH values. To put things in perspective, physiological pH (the standard pH of the body) is about 7.4. The C-terminus is generally deprotonated at a pH of 1-3, so it's always deprotonated in physiological conditions. The N-terminus is generally deprotonated at basic conditions of 9-11, so it's unlikely to be deprotonated in the body. However, the pkR (side chain pka) for serine is 13, which makes it physiologically irrelevant. Thus, it is not found on amino acid tables together with other pka values.\n\n\nFor more on pKa values of amino acids, see [here](https://www.labxchange.org/library/pathway/lx-pathway:573b19de-1dda-48de-b82c-7cf4c3524f3c/items/lx-pb:573b19de-1dda-48de-b82c-7cf4c3524f3c:html:6bfe2257).\n\n\n**What makes the pKa of serine/threonine so high?**\n\n\nOther amino acids, glutamic acid and aspartic acid specifically, also have side-chain OH groups. These are stabilized by resonance from the carbonyl carbon. When deprotonated, the negative charge spreads out uniformly over both oxygen atoms. The same holds true for the C-terminus of every amino acid. They are stable in anionic form at low pH because of their strong resonance. However, serine and threonine, which have no resonance, don't have stable conjugate anions at low pH and therefore have high pKa values.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170238/calculate-the-titer-of-a-solution
|
Calculate the titer of a solution
|
I hope this is the right place to ask. We have this exercise without solution
>
> One litre of diluted sodium hydroxide solution (c = 0.15mol/litre) is
> to be prepared in the laboratory. The titer from the finished NaOH
> solution is then determined. For this purpose, 0.62 g potassium
> hydrogen phthalate (M = 204.2 g/mol, $pK\_a = 5.4$) are dissolved in 0.05 litre
> of water and titrated with the prepared sodium hydroxide solution
> until turnover. The consumption is 0.022 litre. Calculate the titer of the
> prepared sodium hydroxide solution.
>
>
>
On Wikipedia, I found "In titration, the titer is the ratio of actual to nominal concentration of a titrant" <https://en.wikipedia.org/wiki/Titer>
So if I understand correctly, the titer is just a fraction of the form $ t = \frac {c\_{real}} {c\_{ideal}} $, so a fraction of the real concentration over the ideal concentration, is this correct ?
In our case, we would simply need to use the titration equation at equivalence point: $$V\_1 \cdot M\_1 = V\_2 \cdot M\_2$$ where $V$ is the volume and $M$ the molarity
So we would solve
$$M\_2 = \frac{V\_1 \cdot M\_1}{V\_2} $$
and then the titer is $$t = \frac{c\_{real}}{c\_{ideal}} = \frac{M\_2}{0.15} $$
because 0.15 mol/litre is given as ideal concentration in the exercise
$V\_1$ and $V\_2$ are given as 0.05 litre and 0.022 litre, and $M\_1$ could be found by dividing 0.62 g by the molar mass (M = 204.2 g/mol), and then divide again by the volume of 0.05 litre
---
I am really unsure because there aren't a lot of results when searching on the Internet, so it seems to me that this isn't a frequently searched topic. Is my understanding above correct ? Why are there relatively few search results for the titer in Chemistry ? Even on this site, I did not find a similar question. Is this topic considered "not important" or am I missing something ?
| 3
|
[
[
"\nYour approach is correct, although there is no information about which solution is considered as ideal, and which is real. It is also a pity you are not able to be more precise than $0.022$ liter as a final result. Two significant figures ($2$ and $2$) are not really reliable. Usually three or four significant figures are easy to obtain with ordinary burettes. The real volume should look like $22.02$ mL and not $22$ mL. Anyway the final titer should be equal to $0.935$.\n\n\n",
"5"
],
[
"\nWikipedia is not the ultimate resource for classical volumetric titrations. This is why textbooks still exist. The definition of titer given in Wikipedia has nothing to do with the classical meaning of titer. There is no ideal or real solution in volumetric titrations. Titer in volumetric analysis means an expression of type:\n\n\n1 mL (Given NaOH conc. X) = Y (weight) of analyte.\n\n\nIn your case, it will be,\n\n\n1 mL (Given NaOH conc. X) = mg of KHP.\n\n\nIt is a quick way of analyzing solutions. Once an analytical chemist determines the titer of a titrant, all he/she need to know is the burette volume, and we directly get the weight of the analyte in one step. I will give you hints:\n\n\nWrite balanced equations of titration reaction. Calculate the actual molarity. This will give you X. From X, calculate the number moles of NaOH in 1 mL of that X NaOH. From the moles determine the moles of your target analyte based on the equation stoichiometry.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/170235/does-propene-have-resonance-structures
|
Does propene have resonance structures? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/170235/edit).
Closed 9 months ago.
The community reviewed whether to reopen this question 9 months ago and left it closed:
>
> Original close reason(s) were not resolved
>
>
>
[Improve this question](/posts/170235/edit)
Propene is an alkene with a double bond between 2 carbon atoms. $\ce{C3H6}$ may have 2 resonance structures due to 2 possible placings of the double bond. E.g. it can be C=C-C or C-C=C (ignoring hydrogen atoms).
Delocalized electrons are a result of resonance structures. Can propene have delocalized electrons?
| 1
|
[
[
"\nPropene actually fulfil conditions for hyperconjugation (an alfa-CH adjacent to sp2 hybrid C atom) It is a delocalization of sigma bond electrons, also called no bond resonance. Mentioned C-H bond align in plane of the pi orbital. In such situation electrons of the sigma bond can be delocalized into the pi orbital. (Propene does not have resonance structures concerning only pi orbitals, as I suspect you think).\n\n\nThese are 3 \"no bond resonance\" structures of propene.\n[](https://i.stack.imgur.com/89KZp.png)\nThis site has diagrams concerning this problem, including mentioned contributing structures:\n[http://www.adichemistry.com/organic/basics/hyperconjugation/hyperconjugation-1.html](https://www.adichemistry.com/organic/basics/hyperconjugation/hyperconjugation-1.html)\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170228/why-are-these-structural-formulas-not-isomers-of-c5h12
|
Why are these structural formulas not isomers of C5H12? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/170228/edit).
Closed 9 months ago.
[Improve this question](/posts/170228/edit)
A Biology textbook question stated that there are three isomers of C5H12, the structural formula of the first two isomers pentane and 2-methylbutane were pictured, the reader was then asked to draw the third isomer of C5H12.
Probably a silly question, nonetheless, why are the following structural formulas I came up with not isomers of C5H12? If someone could elucidate my trouble, I and perhaps others, would be very appreciative.
1:
```
H
|
H — C — H
| H H H
| | | |
H — C — C — C — C — H
| | | |
H H H H
```
2. H atoms removed for simplicity:
```
C
|
C — C — C
|
C
```
3. three lines represent one single bond, so the hydrogen atoms can fit in the structural formula:
```
C — C — C
|
|
|
C — C
```
4:
```
C — C
|
|
|
C — C
|
C
```
5:
```
C — C — C
|
C
|
C
```
6.:
```
C — C — C
| |
C C
```
Edit: I know a similar question has been asked on this website, that bond rotation has to do with my confusion and I think I have just answered my own question 🤦, so I am guessing all structures are pentane, except structure 2 which might be 2-methylbutane. Is this because single bonds allow the atoms they join to freely rotate about the bond axis? My trouble is visualising these structural formulas in 3D.
Finally that leaves us with 2,2-dimethylpropane:
```
C
|
C — C — C
|
C
```
| -1
|
[
[
"\nAccording to IUPAC (in physical organic chemistry as well as sterochemistry), an **isomer** is one of several species (molecular entities) that have the same atomic composition (molecular formula) but different two-dimensional representations in which atoms are shown joined by lines representing single or multiple bonds, **without any indication or implication concerning the spatial direction of bonds.**\n\n\nSo ignoring the spatial direction of bonds, your examples 1, 3, 4, 5, and 6 are all the same; they are all simply **pentane.**\n\n\nOnly your example 2 is a different isomer; it shows **2-methylbutane.**\n\n\nThe third isomer **2,2-dimethylpropane** is missing.\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/170227/spin-preserving-excitations
|
Spin-preserving excitations
|
I'm currently working on quantum chemistry with quantum machine learning and one of the operations considered for the calculation is the excitation (and desexcitation) of an electron from an orbital to the other (which corresponds mathematically to a Givens rotation).
So far, these excitations have been taken as spin-preserving, meaning that : say I have initially two electrons in the ground state orbital, if I excite one of them to the excited state, its spin won't be change.
Apparently this is something common in quantum chemistry and I wanted to know why.
An argument I thought of was that the necessary energy to flip the spin of the electron is way higher than the typical electronic energy scale of the atom/molecule, hence making it impossible to flip.
Thank you for your response!
| 1
|
[] |
https://chemistry.stackexchange.com/questions/170224/does-wet-fabric-filter-smoke-better-than-dry-fabric
|
Does wet fabric filter smoke better than dry fabric?
|
I have heard in various situations people use a wet fabric to breathe better in situations where a lot of fire-based smoke is present (like near wildfires). I wanted to know if there was any truth to this, specifically with regard to wet fabric vs dry fabric.
A quick Google search brought me to [this website from FEMA](https://www.fema.gov/faq/using-wet-towels-chemical-attack) where it says that dry material works better for aerosols and particulates, but it seems the primary focus of the article is chemical attacks, and I can't really find any other sources to back this up; almost ubiquitously across my searches, it seems that everyone is inclined to think that wet fabrics do better.
So my question is that: do wet or dry fabrics work better for filtering smoke? I would greatly appreciate sources when possible, seeing as this seems like an empirically determined fact.
| 8
|
[
[
"\nThe composition of smoke is not well-defined. On the one hand, some smokes could be considered aerosols of solid particles or liquid droplets only (no gaseous material worth considering). These smokes, as well as other aerosols (e.g., sneezes) can be filtered well enough by using fine weave fabrics (e.g., K95 medical masks). There does not seem to be any suggestion that wetting these masks improves their performance, perhaps because the fine weave would be blocked too much, even by a thin film of water or by swelling of the fabric material.\n\n\nOn the other hand, it might seem that a wet mask could trap liquid aerosol droplets (assuming they are aqueous, not oily) more easily, and this might prove advantageous if the mask is less protective or efficient than a K95 mask. Such a less effective mask probably wouldn't close down so much if it became wet.\n\n\nOn the other hand (if you have one), the gases in a fire combustion smoke may include NO2, SO2, HCN, H2S, Cl2 (from PVC), dioxins, methanol, and many others (Ref 1). These might well be trapped better by a wet mask because they won't be \"trapped\" at all by a dry one.\n\n\nSmoke inhalation is the primary cause of death in victims of indoor fires. Particulate inhalation could certainly cause death, but I think the gases would cause incapacitation faster (choking, gasping, collapsing) than suspended particulates.\n\n\nThe graph below shows chemical compositions of volatile organic compounds in smokes from various fuels (Ref 1). The prevalence of oxygenates is obvious; these would likely be better absorbed by a wet mask.\n\n\n[](https://i.stack.imgur.com/v4NeP.jpg)\n\n\nIn a reverse sense, wet masks seem to prevent transmission of large aerosol droplets outward better than dry masks (Ref 2). But for COVID protection, dry masks are preferable (Ref 3). But, wet or dry is better than nothing (Ref 4). And cotton exposed to humidity shows an improvement in filtering, whereas synthetic fabrics do not (Ref 5).\n\n\nIn general, specific testing, against the specific smoke (or its likely composition), of the exact mask or cloth/towel/etc., would be needed to determine what would \"help you breathe better\". In the short run, helping you get out of the smoke would be best; perfect elimination of contaminants with overly restricted airflow isn't the best solution. Perhaps every situation is unique.\n\n\nRef 1. <https://en.wikipedia.org/wiki/Smoke>\n\n\nRef 2. <https://www.sciencedaily.com/releases/2021/11/211122135517.htm>\n\n\nRef 3. <https://www.wusa9.com/article/news/verify/does-a-wet-mask-make-a-difference-or-am-i-safe-wet-mask-efficiancy-vs-dry-masks/65-5d760229-9e3e-4649-b420-969c74f219cb>\n\n\nRef 4. <https://www.popsci.com/science/wet-masks-covid-protection/>\n\n\nRef 5. <https://www.nist.gov/news-events/news/2021/03/study-indicates-humidity-breath-makes-cotton-masks-more-effective-slowing>\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/170220/is-delta-s-of-universe-always-0-when-an-ideal-gas-undergoes-an-isothermal-irreve
|
Is delta S of universe always 0 when an ideal gas undergoes an isothermal irreversible process?
|
Since for an irreversible process,
$$dS\_\mathrm{Surrounding} =-\frac{\text{dq}\_\mathrm{irr,sys} }{T\_{\text {surr }}}\tag{1}$$
where $\text{dq}\_\mathrm{irr,sys}$ is heat exchange of system
and $-dq\_\mathrm{irr}$ is heat absorbed by the surrounding.
Also $d u=d q+d w$.
Now in an irreversible isothermal process involving only ideal gas.
$$
\begin{aligned}
& du=d w+d q \quad . \quad(d v=C v \cdot d T) \\
& -d w=d q \\
& P\_{\text {ext }} \cdot d v=d q
\end{aligned}
$$
Now Substituting in (i) Now $T\_{\text {syst }}=T\_{\text {surr }}$ As the process is Isothermal . Clearly $dS\_{\text {Surr}}=-dS\_{\text {Sys}}$. Does that mean change in entropy of universe in irreversible isothermal process is always $0$?
| -5
|
[
[
"\nFor the irreversible case, we are talking about a situation in which the ideal surroundings reservoir is maintained at the same temperature as the initial temperature T of the system, such that the final equilibrium temperature of the system is also T. This is what we mean by an isothermal irreversible process, even though, within the bulk of the system during the process, the temperature may not be spatially uniform at T.\n\n\nIf, in determining the change in entropy of the system, you devise a reversible path between the same initial and final thermodynamic equilibrium states of the system as for the irreversible process, you will find that the final thermodynamic equilibrium state of the surroundings will not be th[e same as for the irreversible process. So this approach gives you the correct change in entropy for the system in the irreversible process, but it will not give you the correct value for the surroundings. Since the reservoir is ideal, the change in entropy of the surroundings for the irreversible process will be\n$$\\Delta S\\_{surroundings, irreversible}=\\frac{q\\_{surroundings, irreversible}}{T}=-\\frac{q\\_{system, irreversible}}{T}$$\nThe change in entropy of the system for both the irreversible and reversible path will be $$\\Delta S\\_{system,irreversible}=\\frac{q\\_{system,reversible}}{T}$$\nSo, $$\\Delta S\\_{total, irreversible}=\\frac{(a\\_{system,reversible}-q\\_{system,irreversible})}{T}$$\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170219/electrode-potential-and-likeliness-to-reduce-oxidize-electrolysis
|
Electrode potential and likeliness to reduce/oxidize (electrolysis) [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/170219/edit).
Closed 9 months ago.
[Improve this question](/posts/170219/edit)
I'm a little confused about situations where multiple chemical compounds can be reduced/oxidized and the likelihood of this happening to one compound over the others, depending on the electrode potential of that compound. Here are some exercises I'm trying to figure out on this topic and their answers as given to me:
First exercise: electrolysis with inert electrodes and aqueous solution $\ce{CuI\_2}$, the question is: "why do we observe $\ce{I\_2}$ gas release rather than $\ce{O\_2}$".
The answer is: "If $\ce{H\_2O}$ is oxidized at the anode (to form $\ce{O\_2}$) rather than $\ce{I^-}$ (to form $\ce{I\_2}$), we have $E^°\_{cell}$ = $E^°\_{Cu^{++}/Cu}$ - $E^°\_{H\_2O/O\_2} = 0.337 - 1.23 = -0.893V$ which is lower than $E^°\_{cell}$ = $E^°\_{Cu^{++}/Cu}$ - $E^°\_{I\_2/I^-} = 0.337 - 0.5355 = -0.1985V$ and thus, since the potential is higher (in absolute value) for water, no $\ce{O\_2}$ should be released as long as their is $\ce{I^-}$ ions available"
Second exercise: electrolysis with inert electrodes and aqueous solution at $\ce{pH = 0}$, $\ce{T=298,15K}$, $\ce{p=1bar}$ and containing $10^{-2}M$ of $\ce{Ag^+}$ ions and $10^{-2}M$ of $\ce{Cu^{++}}$ ions, the question is: "discuss the possibility of having a copper deposit without impurities from the silver".
The answer is: "[I'm skipping some calculations but it can be provided if required] We get the standard electrode potentials: $E^°\_{Cu^{++}/Cu} = 0.337V$ and $E^°\_{Ag^+/Ag}$ = $0.799V$ and from Nernst equation (so calculation of the actual -non standard- cell potential) we get for $\ce{H\_2O/Ag}$ and $\ce{H\_2O/Cu}$, respectively: $E\_{cell}$ = $- 0.549V$ and $E\_{cell}$ = $- 0.952V$ then we calculate the concentration of $\ce{Ag^+}$ at $- 0.952V$ (using Nernst law again) and we get: $[\ce{Ag^+}] = 1,7\times10^{-9}M$ , which is very low compared to [$\ce{Cu^{2+}}$] = $10^{-2}M$, so almost only $\ce{Cu}$ will deposit on the electrode."
So here is my question on all this: I have read everywhere that "more negative electrode potential means more likely to be oxidised", which actually correspond correctly to my first exercise ($E^°\_{I\_2/I^-} = 0.5355V$ is "more negative" than $E^°\_{H\_2O/O\_2} = 1.23V$), is it correct to apply this reasoning here though? And also that "less negative electrode potential means more likely to be reduced" and there I have a problem, in the second exercise $E^°\_{Cu^{++}/Cu} = 0.337V$ is not "less negative" than $E^°\_{Ag^+/Ag} = 0.799V$ but quite the contrary, still we calculated that almost no $\ce{Ag^+}$ was deposited (so reduced, to my understanding) compared to $\ce{Cu^{++}}$.
Could you help me to understand why we get this ? Simple and general guidelines for this kind of reasonings would really help.
---
**EDIT (for those who would try to understand this topic)**: As Robert DiGiovanni pointed out (see the discussion below his message), the conclusion for the second exercise is incorrect. Since $E^°\_{Ag^+/Ag}$ is more positive than $E^°\_{Cu^{++}/Cu}$, $\ce{Ag}$ will reduce **first**. Furthermore, the result from the Nernst equation actually confirms the fact that copper cannot be deposited pure. Since $\ce{[Ag^+]}$ is found really low from Nernst equation (giving the cell potential at *equilibrium*), it means that most $\ce{Ag^+}$ will be reduced when reaching equilibrium, and thus most of the Ag will be deposited on the electrode.
| 0
|
[
[
"\nThe main theme of the questions is a very good academic exercise but the question writer (whoever wrote the the two questions for students) made them vague and the answers are more convoluted than the question. It is bound to confuse students.\n\n\nFor the first question, $\\ce{CuI2}$ does not exist and it cannot exist in solution. Copper(II) is a strong oxidizing agent that will oxidize iodide to free iodine. Let us correct the question and say, if we had a solution of $\\ce{KI}$ i.e., potassium iodide in water and if we electrolyze it, what products do we expect at the anode?\nThe general theme for addressing such questions would be:\n\n\n0. List all the possible ions and molecules in the solution: We have $\\ce{H2O}$, free $\\ce{K+}$ and $\\ce{I-}$, $\\ce{H+}$ and $\\ce{OH-}$\n1. Since we are interested in oxidation only, we will think about the species which can be oxidized at the *anode*, the only chemically realistic possibilities are that $\\ce{H2O}$ and $\\ce{I-}$. Forget the cathode for the time being.\n2. Now check, which oxidation is *thermodynamically* favored by looking up the electrode potential tables. All modern electrode potential tables are written as *reduction* potentials. Note that all listed electrode potentials are provided under standard conditions, if they have $\\ce{H+}$ or $\\ce{OH-}$ in the equations, then under standard conditions, we mean that they have unit activity or say pH=0 or pH=14 respectively.\n\n\n*Please do not flip signs of electrode potentials. They are sign invariant. This action is frowned upon by modern electrochemists, although some general chemistry textbooks still do that.*\n\n\n$$\\ce{I2 + 2e- ⇌ 2 I- (+0.5355 V)}$$\n\n\nSimilarly,\n\n\n$$\\ce{O2(g) + 4H+(aq) + 4e- ⇌ 2H2O(l) (+1.229 V)}$$\n\n\nIf we interpret the above equations as follows given that we have $\\ce{H2O}$ and $\\ce{I-}$, then at the anode, it will require less energy to oxidize iodide ion to free iodine vs. water to free oxygen. Thus at the anode, thermodynamic oxidation of iodide will be favored instead of water.\n\n\nAs to the second question, the logic in the answer there is also convoluted. It will be far easier to consider cathode reactions only. You have to consider reductions only this time. Follow the steps above list the reduction reactions under non-standard conditions. You may post the second question separately after attempting along similar lines of reasonings.\n\n\n",
"3"
],
[
"\nIodide gives up its electron easier than water, silver accepts electrons before copper.\n\n\nThe best way is to draw the half reactions *in the direction they occur*.\nThen the math is easy. For example:\n\n\n$$\\ce{2Na metal + Cl2 -> 2 NaCl}$$\n\n\nThe oxidation half reaction is:\n$$\\ce{2Na -> 2Na+ + 2e- + 2.71 V}$$\n\n\nYou may find it on a Standard Reduction Potential Chart as:\n\n\n$$\\ce{ Na+ + e- -> Na - 2.71 V}$$\n\n\nbut that is not the direction the reaction is running.\n\n\nSimilarly, the reduction half reaction is:\n\n\n$$\\ce{Cl2 + 2e- -> 2 Cl- + 1.36 V}$$\n\n\n2.71 + 1.36 = + 4.07 V. Half reaction redox potentials are simply added.\n\n\nJust remember to set your redox half reactions up this way and the comparisons will be easy.\n\n\nBackground reading on [Potentiometric Titrations](https://en.m.wikipedia.org/wiki/Potentiometric_titration) may be helpful.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170216/what-is-the-determining-factor-between-two-potential-bond-candidates
|
What is the determining factor between two potential bond candidates? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/170216/edit).
Closed 9 months ago.
[Improve this question](/posts/170216/edit)
Say you have a vacuum (imaginary, of course) with three hydrogen atoms, how do we determine which two will bond and which one will be left out? If the same energy is present (equally) for the entire reaction, is the priority given strictly to the two that are closer to each other?
| -4
|
[
[
"\nIf the collision has too much energy. or if energy released by the bond formation is not dissipated by photon emission, collision, or vibration energy, then the bond is broken again. Similarly, $\\ce{2 H}$ need it too to form $\\ce{H2}$.\n\n\nAs a mechanical, everyday analogy, imagine a marble play with frictionless terrain, balls and a round enough hole. If you push a ball, it would not end in the (empty) hole, but will escape on the other side. It has to lose some energy somehow, not to be able to escape.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170213/is-an-aqueous-solution-of-lithium-superoxide-basic-or-neutral
|
Is an aqueous solution of lithium superoxide basic or neutral? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/170213/edit).
Closed 9 months ago.
[Improve this question](/posts/170213/edit)
Is the aqueous solution of lithium superoxide basic or neutral?
I know that $\ce{Li}$ is a metal, thus its oxide ($\ce{Li2O}$) produces base in reaction with water.
| -5
|
[
[
"\nLithium superoxide is only stable in a dry atmosphere and at temperatures below $\\pu{70 K}$ $(\\pu{-203 °C})$ according to [Wikipedia](https://en.wikipedia.org/wiki/Lithium_superoxide). So it cannot be dissolved in water. And in the presence of water, it gets decomposed into $\\ce{LiOH}$ and $\\ce{O2}.$\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/170209/are-hydrocarbons-compounds-of-hydrogen-and-carbon-or-carbon-and-hydrogen
|
Are hydrocarbons compounds of hydrogen and carbon, or carbon and hydrogen?
|
Is it the same to say "Hydrocarbons are compounds of hydrogen and carbon." as saying "Hydrocarbons are compounds of carbon and hydrogen."
I got a B in my chemistry test just because of writing "hydrogen and carbon" instead of "carbon and hydrogen".
So i need to know if its the same and my teacher made the mistake or if it's me that made a mistake.
| -2
|
[
[
"\nRather, in a way, you have been both right.\n\n\n*While there are subtle differences in context of [pragmatics](https://en.wikipedia.org/wiki/Pragmatics) as below, it should not be the bases of rejecting the answer. It rathers looks like the teacher deliberately marked it as wrong, as it did not literally matched the given answer.*\n\n\n*IMHO, instead of asking, you should rather stand up for yourself and confront your teacher. The fact you had to ask about it is evidence of your failure to be confident in your knowledge.*\n\n\nThe right order *A and B* has two points of view(POV):\n\n\n**POV of logic**: The order *A and B* is equivalent to *B and A*.\n\n\n**POV of pragmatics**: A and B is usually used as \"A is the main descriptor\" while \"B is the modifier\". In similar sense as if A was a noun and B was a descriptive adjective.\n\n\n* Organic compounds are primarily compounds of carbon.\n* Not all carbon compounds are organic ones, but all organic compounds contain carbon.\n* Most of organic compounds contain hydrogen, but not always.\n* Typical exceptions are per-halogenated hydrocarbons, or [mellitic anhydride $\\ce{C12O9}$](https://en.wikipedia.org/wiki/Mellitic_anhydride), containing just carbon and oxygen.\n* It is therefore natural to start with carbon, when specifying a subclass of organic compounds.\n\n\n\n\n---\n\n\nSo, primarily, in context of organic compounds, \"Hydrocarbons are compounds of carbon and .....\"\n\n\nNow, hydrogen comes as the modifier, specifying *which* carbon compounds are hydrocarbons.\n\n\nTherefore, \"Hydrocarbons are compounds of hydrogen and carbon.\" is logically right, pragmatically not exactly wrong, but suboptimal and less usual.\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/170207/when-hydrogen-and-oxygen-reacts-what-defines-if-it-will-turn-into-hydrogen-pero
|
When hydrogen and oxygen reacts, what defines if it will turn into hydrogen peroxide or water?
|
If you have a jar with oxygen and hydrogen inside, I'm interested in knowing what variables are most important to ensure that the mixture will become water (if that was the goal) or hydrogen peroxide, but not both. I understand that at room temperature those two gases will not react to form anything, so the question is more about the details of how the mixture could be manipulated to take a route to become X vs Y.
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170200/what-is-the-highest-temperature-ambient-pressure-type-1-superconductor-as-of-20
|
What is the highest temperature, ambient pressure type-1 superconductor as of 2022?
|
I was curious about the highest known Tc for type 1 superconductors.
[This](http://www.superconductors.org/Type1.htm) list suggests lead near 7K is as good as it gets for type-1 superconductors but I wonder if that is true given the current state of the art.
| 1
|
[] |
https://chemistry.stackexchange.com/questions/170198/synthesis-of-a-substituted-indane
|
Synthesis of a Substituted Indane
|
From the starting reactants given as the sole carbon source:
[](https://i.stack.imgur.com/VFKz1.png)
We need to synthesis the following compound (we can use any inorganic reagents).
[](https://i.stack.imgur.com/o9WtC.png)
Here is what I tried so far, but I don't think it looks "realistic"
[](https://i.stack.imgur.com/Mwhl9.jpg)
[](https://i.stack.imgur.com/lL1Cn.jpg)
| 0
|
[
[
"\nI'll be frank about it....you can't get there from here! In @Waylander's words \" This is ...a poorly thought out question...\". Allow me to pose the question as it should have been asked. How does one synthesize 5-(1-bromo-2-methylpropyl)-2,3-dihydro-1H-indene **1** from benzene, propionic acid, and isobutyric acid? \n \n\n\n\n \n \n\n\n\nInspection of ketone **2** shows that a Friedel-Crafts acylation of indane **3** does not occur at $\\ce{C4}$ but rather at $\\ce{C5}$. The synthesis of bromide **1** from ketone **2** requires reduction to a benzylic alcohol (*e. g.*; $\\ce{NaBH4/C2H5OH}$) followed by treatment with $\\ce{HBr}$. Isobutyric acid and propionic acid are readily converted to their respective acyl chlorides with $\\ce{SOCl2}$. As to the preparation of indane from benzene, [this issue has been addressed previously on ChemSE](https://chemistry.stackexchange.com/questions/91606/propose-a-synthesis-of-indane-from-benzene/91753#91753).\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/170194/what-is-the-meaning-by-species-of-glycine
|
what is the meaning by 'species' of glycine? [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 9 months ago.
[Improve this question](/posts/170194/edit)
The question ask for the species of the (i) glycine hydrochloride (ii) glycine titrated with NaOH. When they mentioned 'species', what did they mean by it?
| -3
|
[
[
"\nThe term species is used to describe a given type of atoms, molecules or ions. As a functional definition, a species is any atom, molecule or ion that could be part of a chemical reaction equation as reactant or product.\n\n\nHere, it refers to the different protonation states of glycine, no matter whether anionic, neutral or cationic.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/170193/can-one-calculate-the-polarizability-of-xe-isotopes-using-gaussian-09-and-the-3
|
Can one calculate the polarizability of Xe isotopes using Gaussian 09 and the 3-21G basis?
|
While trying to understand [this paper,](https://pubs.asahq.org/anesthesiology/article/129/2/271/17992/Nuclear-Spin-Attenuates-the-Anesthetic-Potency-of) I read the following method for calculating the polarizability of different Xe isotopes:
>
> Quantum chemical calculation of the exact polarizability of each
> xenon isotope was performed by using Gaussian 09 software (version
> D.01, Gaussian Inc., USA). The exact polarizability of each xenon
> isotope was optimized by b31yp/3-21G and the density functional theory
> method.
>
>
>
My understanding is that DFT can be used to assess polarizability of larger systems, but I'm a little befuddled by how it's applied here.
I couldn't identify what is meant by "b31yp" but it was fairly easy to read a description of the "3-21G" basis online, and it's apparent that the polarization of Xe atoms won't be included in that basis:
>
> the 3-21G\* basis set has polarization functions on second row atoms only.
>
>
>
So...can someone decipher what is meant here by "b31yp"?
EDIT: I think it must be a typo for the "B3LYP" DFT method, described [here:](https://gaussian.com/dft/)
>
> uses the non-local correlation provided by the LYP expression, and VWN functional III for local correlation (not functional V). Note that since LYP includes both local and non-local terms, the correlation functional used is actually:
>
>
>
>
> C\*ECLYP+(1-C)\*ECVWN
>
>
>
>
> In other words, VWN is used to provide the excess local correlation required, since LYP contains a local term essentially equivalent to VWN.
>
>
>
**Does it seem plausible that the researchers correctly calculated that four different isotopes of Xe have the same polarizabilities to three significant figures?**
| 6
|
[
[
"\nIn addition to the other answers, to do a proper calculation of the polarizability of Xe or any other \"heavy\" element, one needs to perform proper relativistic calculations in a larger basis set - ideally at the basis set limit since it's only a single atom. (The basis set used is entirely too small, as mentioned in other answers.)\n\n\nSo the number in the paper ($3.60 Å^3$) is simply wrong. Atomic polarizabilities, particularly of noble gases have been measured accurately and used for benchmark calculations:\n\n\n* Calculation: [Sakurai, Sahoo, and B. P. Das 2018](https://doi.org/10.1103/PhysRevA.97.062510) - 27.508 a.u. = $4.076 Å^3$\n* Experiment: [Olney, Cann, Cooper, Brion (1997)](https://doi.org/10.1016/S0301-0104(97)00145-6) - 27.03 au = $4.005 Å^3$\n* Experiment: [Hohm and Kerl 1990](http://doi.org/10.1080/00268979000100611) - $4.1217 Å^3$\n\n\n(Incidentally, the [NIST reference](https://cccbdb.nist.gov/pollistx.asp) cites $4.005 Å^3$ as the atoms polarizability of Xe.)\n\n\nMy guess is that one of the authors had used Gaussian, knew it handled isotopes (e.g., for vibrations), saw that it could calculate polarizabilities, and just ran the calculations without thinking too deeply.\n\n\n",
"7"
],
[
"\nAs the other answer already says: it's entirely plausible and not surprising that these calculations came out this way. This maybe answered your question, but I'd like to offer some more context.\n\n\nFor all intents and purposes, B3LYP/3-21G is a terrible choice of computational method. By the way, you have (without me reading this paper) correctly identified this. I'm my personal opinion, using a phone book will produce more consistent results. \n\nThe basis set is way too small for any productive work, and in a computational setup of today's hardware, there's hardly any difference by using a better balanced one. \n\n(If you'd really needed to cut down computations, go with a minimal basis, i.e. STO-3G. 6-31G is already a bad, but popular, standard, and only slightly more demanding than 3-21G. Most other split valence basis sets perform computationally better. With a dual core computer, a gigabyte RAM, for small molecules and on a human time scale, is not noticeable.)\n\n\nGaussian 09 is old and about 6 years past is prime. Revision D.01 isn't even the last one. However, with this level of theory, nothing much has changed, so it likely doesn't matter at all. However, if there were any bugs in the program, they would be in the final numbers.\n\n\nDensity Functional Theory is nothing without calibration. \n\nThe holy grail is still hunted for. And until we know about it with certainty, some functionals will be optimised for one scenario and others for something completely different. That's why there are quite a few benchmarks. Unless there's one that covers your field of study, you (them/they) should do those yourself (themselves/their selves).\n\n\nN.B.: Most other quantum chemical programs use VWN5 for B3LYP.\n\n\nAnyway, answering the titular question:\n\n\n\n> \n> Can one calculate the polarizability of Xe isotopes using Gaussian 09 and the 3-21G basis?\n> \n> \n> \n\n\nYou can calculate a number, but it's not reliable; so: no.\n\n\nAnswering a question you have not asked:\n\n\n\n> \n> Can the exact polarizability of each xenon isotope be optimized with B31YP/3-21G?\n> \n> \n> \n\n\nProbably (not very likely) not. With DFA (density functional approximations) nothing is exact (yet, due to the limitations of the method.)\n\n\n",
"5"
],
[
"\nDifferent isotopes (of Xe but also any other element) only differ in the mass of the nucleus; the charge of the nucleus and the electron shells are exactly the same. In the Born Oppenheimer approximation the electrons have their own wave function which includes point potentials from the nuclei (neglecting electron-nuclei cross terms). Because of that, the mass of the nuclei (or kinetic energy of the nuclei) has no effect on the electrons at all.\n\n\nFrom that perspective, it is not surprising that different isotopes have the same polarizability (or indeed any other properties) when calculated with DFT using the BO approximation. What is maybe surprising is that the authors of the paper would think that is not the case. I haven’t read the paper but this result would absolutely be expected in any formulation which doesn’t involve a wave function for the nuclei.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/170192/determine-enthalpy-and-entropy-from-measurements-of-kd-at-different-temperatures
|
Determine enthalpy and entropy from measurements of Kd at different temperatures?
|
In biochemical experiments, it is very common to measure IC50 (half inhibitory concentration) eg by adding different amounts of an inhibitor to an enzyme+substrate and measuring the reaction rate or amount of products. From this, Kd can be calculated (with some assumptions).
Is possible to calculate dH (change in enthalpy) and dS (change in entropy) from running this experiment several times at different temperatures?
The idea is that dG (Gibbs free energy) can be calculated from Kd (or IC50) as dG = - RT ln Kd. Also dG = dH - TdS, so most of the variation of dG with temperature in the second equation would come from the TdS term which enables dH and dS to be fitted from dG at several temperatures. Can dH be assumed to be relatively flat with temperature? How about dS? (this is over a relatively narrow range where bio experiments can be carried out, let's say 5C to 40C)
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170187/do-pseudo-van-der-waals-gases-exist
|
Do "pseudo Van der Waals" gases exist?
|
In college, when deriving the Langmuir isotherm for gas-solid adsorption, the professor proposed a modified version of the Van der Waals state equation, what he called the "pseudo Van der Waals gas state equation". The derivation implied equaling the chemical potential for a certain compound in the gas and interphase, and when dealing with the volume for the interphase:
$$V\cdot (p+q) = RT$$
Where $V$ is the volume, $p$ the pressure, $R$ the gas constant, $T$ the temperature, and $q$ a correction term to account for the adsorbate cohesion. He argued that the interphase was something similar to a very condensed gas, so that equation of state was valid. Indeed, he obtained correctly the Langmuir isotherm, that later was derived using the kinetic approach.
However, I have been reading a lot of books and articles looking for that "pseudo Van der Waals gas", and I haven't found anything similar. I guess the approximation is just to neglect the excluded volume, but I am not sure whether my professor was right or he took some license by himself.
| 6
|
[] |
https://chemistry.stackexchange.com/questions/170185/dependence-of-solubility-on-ph
|
Dependence of solubility on pH
|
I am studying the effect that pH has on solubility of metal salts. One example I came across is the idea that the solubility of $\ce{PbI2}$ is not strongly affected by the addition of strong acid to solution because $\ce{PbI2}$ contains the conjugate base of a strong acid ($\ce{HI}$). Online references explain that this makes sense, since the presence of additional $\ce{H+}$ ions won't successfully pull the $\ce{I-}$ ions out of solution (since $\ce{HI}$) is a strong acid that wants to dissociate).
However, is this line of reasoning actually correct? For example, what if the strong acid added to solution is $\ce{HI}$ itself? Wouldn't that lead to the presence of additional $\ce{I-}$ ions that increase the reaction quotient above $K\_\text{sp}$ and thereby cause the solubility to decrease (more solid to precipitate out)?
I suppose the general question is: how should I think about the role of pH on solubility when the acid being added to tune the pH contains a compound that is being dissolved from the solid in the first place?
| 3
|
[] |
https://chemistry.stackexchange.com/questions/170184/functional-group-for-glycine-that-had-being-titrated-i-hcl-iinaoh
|
Functional group for glycine that had being titrated (i) HCl (ii)NaOH
|
i am kinda confused with this question. They asked for the functional group of (i) glycine hydrochloride (ii) glycine that titrated with NaOH. I did take a look many times especially for glycine hydrochloride. But the R-group is still the same, that is H, not the NH3+ Cl- (if I'm not mistaken, NH2 is amino group).
Oh, and when they mentioned glycine species from both of these mixtures, what did they mean by it?
Please enlighten me since I am totally lost here.
| -5
|
[
[
"\nGlycine:\n$$\\ce{NH3^{+}-CH2-COO- + OH- -> NH2-CH2-COO- + H2O}$$\n\n\nGlycine hydrochloride:\n$$\\ce{NH3^{+}-CH2-COOH + 2 OH- -> NH2-CH2-COO- + 2 H2O}$$\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170182/what-happens-to-a-solutes-potential-energy-during-diffusion
|
What happens to a solute's potential energy during diffusion?
|
I was thinking, if you have a large amount of water, with an uneven solute concentration, diffusion will lead to an even concentration of solute throughout the solution.
Intuitively,that would mean that our system initially has a large amount of potential energy, and that it decreases to a minimum.
I feel like this raises two questions :
Firstly, from which force does this inital "osmotic" potential energy derive ?
Secondly, under what form is this potential energy realeased once concentration equilibrium is reached ?
What's more, if you consider a cell membrane with a great concentration difference between one side, and another, simple diffusion will lead solute molecules to pass through the membrane. So, in order to pass through the membrane, a solute would have to expend potential energy : the diffusion reaction would have a certain activation energy.
Under what form is this energy then dissipated ? It would first become kinetic enrgy, allowing the solute to pass through the membrane, but what form would it then take ? Simply heat energy ?
Any answer would be greatly appreciated, thanks.
| 0
|
[
[
"\nEnergy *can* be recovered across a concentration gradient, as in [osmotic power production](https://en.wikipedia.org/wiki/Osmotic_power).\n\n\nBut you ask, if I understand correctly what happens if the energy is *not* harvested? The same thing that would happen if you dropped a weight:\n\n\n* If the weight is attached so as to spin a generator, then the potential energy can be converted to electric energy.\n* If the weight is allowed to impact a surface, instead, then the potential energy is converted to heat.\n\n\nIf economics is considered the [dismal science](https://en.wikipedia.org/wiki/The_dismal_science), thermodynamics is equally grim. Consider [Ginsberg's Theorem](https://www.barrypopik.com/index.php/new_york_city/entry/you_cant_win_you_cant_break_even)\n\n\n1. You can’t win.\n2. You can’t break even.\n3. You can’t get out of the game.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170173/how-does-phenol-show-tautomerism
|
How does phenol show tautomerism?
|
I read that atleast one alpha carbon has to be sp3 wrt the functional group (and has to have atleast 1 alpha Hydrogen atom) for it to show tautomerism.
I am able to see that in the keto isomers of phenol.
However looking at phenol (enol isomer), how does one predict it can show tautomerism?
I'm wondering if some method can be used to identify that.
| 0
|
[
[
"\nAll [enols](https://en.wikipedia.org/wiki/Enol) exhibit tautomerism, at least to some extent, but not all ketones can tautomerise.\n\n\nConditions for showing tautomerism (for ketones) are listed below:\n\n\n1. Presence of at least one alpha H at the $\\ce{sp^3}$ hybridized alpha C (i.e. C just next to the carbonyl carbon)\n2. In case of alpha-beta unsaturated ketones/aldehydes at least one alpha H\nshould be present at the gamma carbon (i.e. third to the carbonyl C).\n\n\nAlso keep in mind some basic rules such as [Bredt's Rule](https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Logic_of_Organic_Synthesis_(Rao)/02%3A_Rules_and_Guidelines_Governing_Organic_Synthesis).\n\n\nFor better understanding refer to [this](https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(LibreTexts)/22%3A_Carbonyl_Alpha-Substitution_Reactions/22.01%3A_Keto-Enol_Tautomerism).\n\n\n",
"1"
]
] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.