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https://chemistry.stackexchange.com/questions/170171/if-you-give-an-electron-in-a-subshell-more-energy-does-it-simply-jump-to-the
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If you give an electron in a subshell more energy, does it simply "jump" to the next energy level?
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Suppose you have an electron in the $\ce{2s}$ subshell of an atom. If energy is given to it, does it simply jump to the next energy level (into the $\ce{3s}$ subshell), or does it move into $\ce{2p}$?
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[
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"\nDepends on the amount of energy you're supplying.\n\n\nFor example, if you supply light of different wavelengths, the amount of energy each holds is different. Your electron will jump to a subshell corresponding to that energy difference.\n\n\n",
"-1"
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https://chemistry.stackexchange.com/questions/170170/which-water-is-easier-to-filter-rain-water-or-tap-water
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Which water is easier to filter? Rain water or tap water [closed]
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>
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>
>
>
[Improve this question](/posts/170170/edit)
I understand the water greatly varies depending on the region and filtration differs as well. Is it better to filter rain water or tap water from contamination like hormones, pesticides, and cloud seeding, which are hard to filter? Is there a better place/SE to ask this?
| -5
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[
[
"\nFirst thing: All surface water and high ground water started out as evaporated ocean water that condensed on a nanoparticle, scrubbed the air as the droplet formed and fell, landed, extracted whatever it flowed on, eventually collected in streams, ponds, lakes, rivers temporarily, picking up ions, biomass, possibly suspended solids and exchanging gases on its way back to the Oceans. This IGNORES industrial pollution that can be pervasive [Remember how everything cleared up in the start of the pandemic].\n\n\nEach phase has its own purification problems. First step is an analysis. The least complicated is rainwater collected and settled; then new surface water, Municipal sources [tap water] usually have some pretreatment and usually supply an analysis; then high-level ground water such as springs, then shallow wells, deeper wells; Finally sinks such as the Dead Sea, the Great Salt Lake, the ocean, and sewage [space station].\n\n\nA rain water story: I taught high school chemistry for 2 years at a school 2 blocks from Kodak Park, Eastman Kodak's industrial complex. While in class a sudden thunderstorm caused a deluge and everyone said, \"Let's check it for acid rain\". A girl sitting next to the window held out a bucket to collect water; after several seconds she screamed and pulled her arm in! There were black tendrils hanging all over it. We quickly wiped it off and got her to the girl's room to wash her arm and to the nurse to get checked out, no irritation. We never checked for acid rain or had the substance analysed.\n\n\n",
"0"
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"\nYou provided incomplete detail (region you live on, what the quality of your tap water is and it origins from, and if cloud seeding is really relevant on your area, or if there are other industrial sources that emit particles and aerosol nearby) to get more than a generic answer.\n\n\nIn Germany for instance the answer would be different as in the USA or in agricultural regions with heavy pestcide use elseshere.\n\n\nUsing ultra-filtration membrane combined with activated carbon filters often is good enough, and proven to improve typical tap water. If you have a problem with heavy metals a little more effort is needed.\n\n\nQ: rain or tap water easier to filter?\nA: it depends, see above\n\n\nQ: which contaminants are more complicated to get rid of by filtering?\nA: nanoparticles, nitrates, heavy metals,...\n\n\n",
"-1"
]
] |
https://chemistry.stackexchange.com/questions/170167/can-nh4cl-react-with-hno3-to-give-nh4no3
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Can NH4Cl react with HNO3 to give NH4NO3? [closed]
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I assume that the reaction would be
$\ce{HNO3(aq) + NH4Cl(aq) -> HCl(aq) + NH4NO3(aq)}$
But will this reaction really happen?
| -4
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[
[
"\nBe aware that all 4 compounds, dissolved in water, are dissociated to respective ions. E.g. $\\ce{NH4Cl(aq)}$ is just a shortcut for $\\ce{NH4+(aq) + Cl-(aq)}$.\n\n\nSo it is not combination of $\\ce{HNO3}$, $\\ce{NH4Cl}$, $\\ce{HCl}$ and $\\ce{NH4NO3}$, but of (hydrated) ions $\\ce{H+}$, $\\ce{NO3-}$, $\\ce{NH4+}$ and $\\ce{Cl-}$.\n\n\nAmmonium nitrate cannot be formed even by crystallization of the less soluble salt:\n\n\n$$\\ce{NH4Cl(s,aq) + HNO3(aq) -> NH4NO3(s) + HCl(aq)},$$\n\n\nbecause it is more soluble than ammonium chloride.\n\n\nOTOH, if the mixture has minimum of water, there can be reaction like:\n\n\n$$\\ce{NH4Cl(s,aq) + HNO3(l,aq) <=> NH4NO3(s,aq) + HCl(aq,g)}$$\n\n\nBut, in such high concentrations, there would be significant oxidation of chlorides, forming nitrogen oxides, chlorine and nitrosyl chloride.\n\n\n",
"4"
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"\nDissolving those *two* compounds in water would lead to *four* species of ions floating around (without discussing solvation and oxidation).\n\n\nTo get the solid compounds back again would require *removing* the water, e.g., with a vacuum pump, to get the anions and cations to link (crystallize) again. Then the question arises, *which* of the resultant compounds are least soluble, and would precipitate first?\n\n\nIn this way, brines are separated into (relatively) purer substances through [fractional crystallization](https://en.wikipedia.org/wiki/Fractional_crystallization_(chemistry)). For example, [solar evaporation ponds](https://www.compassminerals.com/who-we-are/locations/ogden-utah/) are used commercially to separate $\\ce{NaCl}$, $\\ce{K2SO4}$, and $\\ce{MgCl2}$ from the mix of ions in Utah's Great Salt Lake brine.\n\n\n",
"2"
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[
"\nRedox reactions aside, the reaction would tend to go the other way, precipitating ammonium chloride.\n\n\nBelow, we note the solubilities of a few ammonium salts in water at 20°C, translated into moles of ammonium ions per kilogtlram of water. Mass solubility from which the molar solubilities are computed are given in the linked sources.\n\n\n$\\ce{NH4Cl}: 6.8$ (<https://en.wikipedia.org/wiki/Ammonium_chloride>) (interpolated between 0 and 25°C)\n\n\n$\\ce{NH4HSO4}: 10.0$ (<https://www.scbt.com/p/ammonium-bisulfate-7803-63-6>) (it is assumed that in the acid solutions described below sulfate ions would be converted to bisulfate)\n\n\n$\\ce{NH4NO3}: 18.7$ (<https://en.wikipedia.org/wiki/Ammonium_nitrate>)\n\n\nThus given sufficient salt concentrations, the addition of $\\ce{HCl}$ would precipitatethe less soluble ammonium chloride from solutions of the other salts listed:\n\n\n$\\ce{NH4HSO4 + HCl -> NH4Cl(s) + H2SO4}$\n\n\n$\\ce{NH4NO3 + HCl -> NH4Cl(s) + HNO3}$\n\n\nAs described by Maurice in a comment, given the high solute concentrations required plus the formation of nitric acid, the reaction with ammonium nitrate would likely be accompanied by oxidation of the ammonium and chloride ions. With sulfuric acid the oxidizing action is not as strong, so the (bi)sulfate reaction would be cleaner.\n\n\nThe reduced solubility of ammonium chloride may be attributed to its relatively efficiently packed [lattice structure](https://en.wikipedia.org/wiki/Ammonium_chloride). This salt has a caesium chloride type simple cubic structure, with ammonium ions oriented so that their protic hydrogen atoms are facing alternating chloride ions. With this arrangement both ionic bonding and hydrogen bonding serve to stabilize the solid. The bisulfate and nitrate, with bulkier and nonspherical anions, do not offer as good a fit in the solid lattice and thus the solid is more easily attacked by the water solvent.\n\n\n",
"2"
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https://chemistry.stackexchange.com/questions/170163/should-i-give-an-endothermic-reaction-the-exact-energy-it-needs-to-occur-or-can
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Should I give an endothermic reaction the exact energy it needs to occur or can I give it in parts?
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I know this might be a silly question, but should I sustain the exact energy or more for the enthalpy of endothermic reaction to occur or can I give it in parts?
Let's say a certain reaction requires $100\ \mathrm{kJ}$ to occur, do I have to give it $100\ \mathrm{kJ}$ or more to occur or can I give it $50\ \mathrm{kJ}$, two times? Assuming that no energy is lost. If it is possible, what about real-life, where energy is in fact lost?
| 1
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[
[
"\n**No reaction has to be given the exact activation energy to proceed**\n\n\nThere is a deep misunderstanding here about what activation energy means. It isn't a property of the *bulk* reaction that is happening, but more like a property of the individual molecules undergoing reactions.\n\n\nThe point is that all the molecules in a bulk compound or mixture do *not* have the *same* energy. Simplistically consider a reaction where two different molecules need to collide with more than a certain energy to cause a reaction (the net result of which might absorb more energy (\"endothermic\") or release more energy (\"exothermic\"). Many of the individual molecules will bang together without enough energy and nothing will happen. Some will have enough energy and the reaction will proceed. But the individual molecules will always have a *wide range* of individual energies (As Geoff says, in a Boltzmann distribution). What matters for whether the reaction happens is not how much total energy there is, but what *proportion* of the individual molecules have more than the activation energy threshold.\n\n\nThat proportion is what varies with the energy you add to the system. It isn't about getting the added energy *exactly* right: it is about getting enough to achieve some proportion of the molecules with the right level.\n\n\nIn an endothermic reaction this is still true. But if the final products absorb energy, the amount of energy remaining in the system will fall and fewer molecules will have enough energy to leap the activation energy barrier, so the reaction rate may fall (perhaps low enough to stop the reaction). In which case you can add incrementally more external energy to compensate for the internal loss and keep the reaction rate up.\n\n\nBut, at no point can you add the exact \"right amount\" because energy is always distributed among all the molecules and the issue is what *proportion* have enough energy for the reaction to happen.\n\n\nSo, in the bulk, it doesn't matter how the energy is added (partwise or all at once) as long as there is enough for some proportion of the molecules to react.\n\n\n",
"3"
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"\nIn *principle*, you could add 50% energy to the motion that goes over the reaction barrier, which moves you up some of the vibrational energy levels .. and then another 50%. But in reality, that's going to be hard because in most molecules, there are dissipation mechanisms -- so it would be very hard to \"keep\" the energy in the motion (vibration) needed to overcome the reaction barrier.\n\n\nSometimes it's useful to use metaphors when talking about kinetics.\n\n\nLet's consider that you need to jump over a barrier to get from a starting point to another place:\n\n\n[](https://i.stack.imgur.com/dN3Not.png)\n\n\nIf the figure attempts to jump 50% of the height, it's not that they can jump another 50% to get over the wall.\n\n\nIn detailed chemical kinetics theory, molecules will have a distribution of different energies (the Boltzmann distribution) but either they have the energy to go \"up and over\" or they do not and stay on one side of the reaction barrier.\n\n\nNow this metaphor isn't perfect because some chemical processes (e.g., heating water) can use repeated additions of energy. Also, few chemical reactions are one dimensional and some reactions occur via catalysts or quantum tunneling, so they're not really \"jumping over\" a barrier.\n\n\nYou might wonder, what happens to the first 50 kJ/mol. In the metaphor, the figure \"falls back\" because it does not get over the wall. There are a few mechanisms in which the molecule may dissipate energy, for example going into translational or rotational kinetics, or other low-energy vibrational modes.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170162/what-is-the-maximum-number-of-silver%e2%85%a0-ions-in-one-litre-of-a-0-003-m-sodium-su
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What is the maximum number of silver(Ⅰ) ions in one litre of a 0.003 M sodium sulfide solution?
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>
> What is the maximum number of silver(Ⅰ) ions that can be present dissolved in one litre of a $\pu{0.003 M}$ $\ce{Na2S}$ solution?
>
>
>
According to my book, silver(I) reacts with sulfide producing $\ce{Ag2S}$ with the solubility product constant
$$K\_\mathrm{sp}(\ce{Ag2S}) = [\ce{Ag+}]^2\,[\ce{S^2-}] = \pu{8E-51}.$$
Now here is what I don't understand. I have seen several questions where they ask to solve for the maximum solubility of a substance in water, in this case $\ce{Ag2S}$. The usual thing they do is set up an equation with $x$, like $K\_\mathrm{sp} = (2x)^2\,x.$
But in our case, we are given an initial concentration $c\_0(\ce{Na2S}) = \pu{0.003 M}$, from which we know that the concentration $c\_0(\ce{S}) = \pu{0.003 M}$. We also know the $K\_\mathrm{sp}$ value, and they ask specifically for the *silver(I) ions*, not silver(I) sulfide.
So I am not sure if I need to proceed like above, i.e. set up an equation with $x$ and solve for it, because it seems to me that would give the maximum solubility of $\ce{Ag2S}$, not the maximum number of silver ions.
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"\nProduct solubility problems with non-zero initial concentrations can be generalized for a dissociation reaction of the form:\n\n\n$$\\ce{A\\_aB\\_b(s)<=>aA^{b+}(aq) + bB^{a-}(aq)}$$\n\n\nWith the condition: ($a≠b$) or ($a=b=1)$, and $K\\_{sp}>Q\\_{sp}$\n\n\nAnd a general equilibrium expression:\n\n\n$$K\\_{sp}=[A^{b+}]^a\\;[B^{a-}]^b$$\n\n\nSince initial concentrations are non-zero:\n\n\n$$[A^{b+}]=[A^{b+}]\\_o + ax$$\n\n\n$$[B^{a-}]=[B^{a-}]\\_o + bx$$\n\n\nThe resulting expression can be solved for x:\n\n\n$$K\\_{sp}=\\left([A^{b+}]\\_o + ax\\right)^a\\;\\left([B^{a-}]\\_o + bx\\right)^b$$\n\n\nIn our particular case, the dissociation reaction is:\n\n\n$$\\ce{Ag2S(aq)<=>2Ag+(aq) + S^2-(aq)}$$\n\n\nSo we can define $a$ and $b$ from the stoichiometric coefficients:\n\n\n$$a=2$$\n\n\n$$b=1$$\n\n\nInitially, no $\\ce{Ag+}$ is present, but $\\ce{S-}$ is, so in terms of initial concentrations we have:\n\n\n$$[A^{b+}]\\_o=[\\ce{Ag+}]\\_o=0$$\n\n\n$$[B^{a-}]\\_o=[\\ce{S^{2-}}]\\_o=\\pu{0.003M}$$\n\n\nSo the resulting equilibrium expression would be:\n\n\n$$K\\_{sp}=\\left(2x\\right)^2\\;\\left(0.003+x\\right)=8\\cdot10^{-51}$$\n\n\nSolving for $x$:\n\n\n$$x=\\pu{8.2\\cdot10^{-25}M}$$\n\n\nCalculating the resulting concentration of $\\ce{Ag+}$ at equilibrium:\n\n\n$$[\\ce{Ag+}]=2x=\\pu{1.64\\cdot10^{-24}M}$$\n\n\nFinally, the number of silver ions can be calculated using Avogadro's constant and the volume of the solution given:\n\n\n$$N\\_{\\ce{Ag+}}=[\\ce{Ag+}]\\;V\\;L=(\\pu{1.64\\cdot10^{-24}mol/L})(\\pu{1L})(\\pu{6.022\\cdot10^{23}ions/mol})≈\\pu{1 ion}$$\n\n\n",
"1"
]
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https://chemistry.stackexchange.com/questions/170158/chemistry-thermodynamics-and-sign-convention
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Chemistry Thermodynamics and Sign Convention [closed]
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**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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Can anyone pls explain me the sign conventions that we use in chemistry thermodynamics for heat and work...also kindly explain how to identify what will happen to heat in positive or negative work and vice-verse...
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[
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"\nAccording to the international standard ISO 80000 *Quantities and units* – Part 5: *Thermodynamics*,\n\n\n\n> \n> for a closed thermodynamic system\n> \n> \n> $$\\Delta U=Q+W$$\n> \n> \n> where $Q$ is amount of heat transferred to the system and $W$ is work done on the system provided that no chemical reactions occur\n> \n> \n> \n\n\nThe same sign convention is used in the IUPAC Green Book *Quantities, Units, and Symbols in Physical Chemistry.*\n\n\n\n> \n> The given equation in integrated form is $\\Delta U=Q+W$. $Q\\gt0$ and $W\\gt0$ indicate an increase in the energy of the system.\n> \n> \n> \n\n\n",
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"\nThere exist two ways for defining the sign of the work, depending who was defining it.\n\n\nThe definition proposed by the chemists implies that all energies that are added to a system are positive. $\\Delta U = Q + W$\n\n\nThe approach of physics teachers is different. It is a mechanical approach. It implies that the system is like a steam machine which is heated to produce useful work. The heat is positive when entering the system. But the work is positive when it produces some useful effect (with this heat) outside of the machine. For physicists, the internal energy $U$ is this fraction of the heat that remains inside the system and does not produce useful work. For physicists, $\\Delta U = Q - W$. So as the work produced by the system is positive, it means that if some work is entering the system, this work must be negative.\n\n\nEdit : More details for AtharvZope. Imagine a syringe half filled with a volume $V\\_o$ of a gas (air or water vapor). If the syringe is heated by introducing a heat quantity $Q$ under constant atmospheric pressure $p$, its gas volume increases to $V\\_1 > V\\_o$. So $\\Delta V$ is equal to $\\Delta V = V\\_{fin} - V\\_{in} = V\\_1 - V\\_o > 0$. This sign is the same for both definitions of work. As a consequence, in such a phenomena, $p\\Delta V $ is positive all over the world, for chemists or for physicists. But, if $p\\Delta V $ is always positive, the work $W =|p\\Delta V|$ has not the same sign everywhere. $p\\Delta V$ is always positive, but $W$ is negative for chemists as the corresponding energy is lost by the system ($W\\_{chem} = - p\\Delta V$), and the same work is positive for physicists, because the syringe has produced a work which could be used for moving a car or producing electricity, etc. ($W\\_{phys} = + p\\Delta V$)\n\n\nWhatever the choice of the sign of $W$, the change in internal energy $\\Delta U$ is the same for physics and for chemistry : $\\Delta U = Q + W\\_{chem} = Q - W\\_{phys} = Q - p\\Delta V$\n\n\n",
"0"
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https://chemistry.stackexchange.com/questions/170156/regarding-colour-visibility-due-to-f-centre
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Regarding colour visibility due to F-centre
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Below shown image is a cut out from the NCERT book.
[](https://i.stack.imgur.com/zFWZi.png)
Referring to the highlighted sentences:
I know that an $\ce{e^-}$ absorbs and emits light of a particular wavelength when bound in an atom and corresponding to that wavelength a particular colour is visible.
But in the case of F-centres as NCERT explains.....
that an $\ce{e^-}$ in the F-centre when excites(i.e. absorbs a photon of a particular wavelength) and then de-excites, emits that photon and due to that ionic crystals possess some color.
But here comes the problem, how can an free $\ce{e^-}$ (present in the F-centre) excite?
What's the phenomenon here which is going on?
I am not able to understand that how will an $\ce{e^-}$ present in the anionic vacancies excite?
Overall how F-centres impart color?
[Link](https://ncert.nic.in/textbook.php?lech1=1-9) to the book.
| 0
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"\nA \"sea\" of free electrons looks silvery, i.e., reflective, as in solid and liquid metals, such as aluminum or gallium, and even in [ionized dissolved metals](https://pubmed.ncbi.nlm.nih.gov/26886153/), such as [sodium in $\\ce{NH3}$](https://www.youtube.com/watch?v=Gi19MYsB56w) (although the $\\ce{Na+ in NH3}$ looks blue, at first, as electrons are interspersed by $\\ce{NH3}$ molecules, as it becomes more saturated, it appears metallic, i.e., specularly reflective). The key thing to understand is that [delocalized electrons can reflect EMR](https://physics.stackexchange.com/questions/424962/why-do-metals-have-high-optical-reflectivity).\n\n\nHowever, in the F-center, the electrons are *not* free, but are bound by the surrounding ions. These point defects not only stay in place, but can even be used for long-term data storage, as in [nitrogen vacancies in diamond](https://theconversation.com/turning-diamonds-defects-into-long-term-3-d-data-storage-67685).\n\n\n",
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"\nYou can explain the phenomenon using both crystal field theory and band theory.\n\n\nCrystal field theory: The electron occupies the atomic site of a vacant Cl-; it will occupy the position that minimizes the repulsion of the surrounding Cl- ions. When you excite it, the electron will move and different higher energy levels will be available, dependent on the symmetry of the anion's environment and the strength of the crystal field.\n\n\nBand theory: In NaCl there's wide gap between valence and the conduction band, so no visible radiation is absorbed. Vacancies in the Cl-substructure (filled by electrons) forms new levels in the band-gap; hence lower-energy excitations become allowed and some radiation in the visible spectrum can be absorbed.\n\n\nsee K. Nassau: The origins of color in minerals. American Mineralogist, Volume 63, pages 219-229, 1978\n\n\n",
"1"
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https://chemistry.stackexchange.com/questions/170152/what-does-chemically-happen-to-rubber-when-it-soaks-in-mineral-spirits-type-1
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What does chemically happen to rubber when it soaks in mineral spirits type 1? [closed]
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Mineral spirits(mineral turpentine)type 1 is a mixture of aliphatic and alicyclic hydrocarbons with a maximum content of 25% aromatics
and less than 0.1% benzene and all kind of impurities.
I have noticed that if i soak piece of rubber(2cm^3 cube)in mineral spirits type 1,the rubber swells.I assume that a specific type of hydrocarbons diffused into the rubber and cause it to swell.
Which type of hydrocarbons are diffused into the rubber and responsible for the swelling?(the aromatic ones?)and is there a way to remove the mineral spirit(residues)from the swollen rubber or the hydrocarbons in the mineral spirits are actually bonding to the rubber molecules,so it can't be removed?
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"\nMostly aromatics, some alicyclic.\n\n\nProbably the only way is evaporation, at temperature being trade off between patience and thermal damage.\n\n\nBe aware that of the rubber object may not revert exactly to original shape or size. Some residues may remain and arrangement of macromolecules may change.\n\n\n",
"1"
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https://chemistry.stackexchange.com/questions/170151/carbocation-rearrangement-in-dehydration-of-an-alcohol
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Carbocation rearrangement in dehydration of an alcohol [closed]
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[](https://i.stack.imgur.com/DyEK2.jpg)
In this problem, I know that initially a positive charge is formed at the carbon bearing the hydroxyl group , yielding a secondary carbocation.
But, I can't go any further with that. What kind of ring expansion could this be ?
How would we end up with 2 fused rings of different sizes and why?
Is there any alkyl shift or hydride shift preceding the ring expanison?
I don't know how to predict the direction in which the expansion proceeds after the first step.
Can someone pls help me out with this?
Thanks in Advance.
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"\nAnswer should be 1.\n\n\nRefer to the picture below, this may help you:\n\n\n[](https://i.stack.imgur.com/eSkTu.jpg)\n\n\n",
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https://chemistry.stackexchange.com/questions/170146/in-combustion-method-for-analysing-molecular-formula-of-gaseous-hydrocarbon-is
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In combustion method for analysing molecular formula of gaseous hydrocarbon, is pressure of the gaseous mixture always constant?
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I understand the we have to look at the volumes, but the fact that volume of a gas is directly proportional to the number of moles of the gas holds only at constant pressure. Also we are applying Gay-Lussacs law, so shouldnt the pressure be constant at all times?
I don't understand why the gaseous mixture is taken to be at constant pressure, the diagram of the eudiometer tube is not able to convince me that the pressure of the mixture is constant.
If the volume of the gas is changing, the height of the water column inside the tube is also changing. Since atmospheric pressure is constant then on changing height of water column shouldn't the pressure of the gaseous mixture also change?
Please explain how the pressure of the mixture is constant. [](https://i.stack.imgur.com/VXotI.jpg)
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"\nMathematically constant is not the same as constant in context of measurement.\n\n\nFor the latter,\n\n\n* below the error threshold it is considered constant\n* above the threshold are applied corrections.\n\n\n\n\n---\n\n\nThe volume measurement can be adjusted in such a way outer and inner water levels are aligned anth therefore volumes are measured at constant pressure(safe the trend of atmospheric pressure).\n\n\nOr, corrections can be made for hydrostatic pressure.\n\n\nOr, deviations can be neglected, depending on scenario demand on result accuracy.\n\n\n\n\n---\n\n\nThe topic has in fact nothing to do specifically with chemistry. It is the general topic of mathematical versus scientific precision.\n\n\nA measured quantity is within a given scenario considered constant if:\n\n\n* Deviations are below measurement resolution threshold\n* Deviations are observed but negligible compared to other error sources.\n* Deviations are not negligible wrt other error sources, but can be neglected in context of required accuracy of the result vulue.\n\n\n",
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],
[
"\nIs pressure in the [eudiometer tube](https://en.m.wikipedia.org/wiki/Eudiometer) *always* constant? Probably not if combustion is involved.\n\n\nHowever, once allowed to cool, pressures will be constant. Volumes change. This is done so *volume* (as residual oxygen) can be measured and compared with the original volume of oxygen.\n\n\nHydrocarbon analysis is based on ratio of carbon (atomic weight 12) and hydrogen (atomic weight 1).\n\n\nMore moles of oxygen are consumed for a given mass of hydrocarbon if the *ratio* of hydrogen content is higher, as in CH4 (4:1) as compared with C8H18 (2.25:1).\n\n\nAlkali is used to strip CO2 from the combustion mixture, leaving only residual oxygen.\n\n\nAtmospheric pressure is used to keep pressure in the reactor constant (just hope the barometer does not change much while running the experiment). Volume changes based on moles Oxygen consumed by combustion.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/170141/what-is-the-chemical-equation-for-enthalpy-of-solution-of-hydrated-salts
|
What is the chemical equation for enthalpy of solution of hydrated salts? [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/170141/edit).
Closed 9 months ago.
[Improve this question](/posts/170141/edit)
If I take the anhydrous salt $\ce {CuSO\_4}$, the equation for its dissolution in water would be:
$$ \ce {CuSO\_4}\_{(s)} + \ce {aq}.\rightarrow \ce {CuSO\_4}\_{(aq)} $$
Now for the hydrated salt $\ce {{CuSO\_4}\cdot {5H\_2O}}\_{(s)}$, is it this:
$$ \ce {{CuSO\_4}\cdot {5H\_2O}}\_{(s)} + \ce {aq}.\rightarrow \ce {CuSO\_4}\_{(aq)} $$
or is the output the "aqueous version" of the hydrated salt ($\ce {{CuSO\_4}\cdot {5H\_2O}}\_{(aq)}$), if that's even a thing? I think it's the first one, but I don't understand why the salt would become anhydrous when dissolved.
| -2
|
[
[
"\n$\\ce{CuSO4(aq)}$ is just a shortcut for $\\ce{Cu^2+(aq) + SO4^2-(aq)}$.\n\n\n$$\\ce{CuSO4(s) ->[H2O]Cu^2+(aq) + SO4^2-(aq)}$$\n\n\n$$\\ce{CuSO4 . 5 H2O(s) ->[H2O]Cu^2+(aq) + SO4^2-(aq)}$$\n\n\nAs the symbol (aq) means implicitly involved, indefinite amount of ion-hydrating water, the reactions do not need to be enumerated wrt the water molecule count.\n\n\nCrystal water abandons ions and becomes ordinary water, possibly but not necessarily taking first dibs in ion hydration.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170133/how-to-write-formula-for-potassium-hydrogenpyrophosphite
|
How to write formula for potassium hydrogenpyrophosphite?
|
I know that pyrophosphite by itself is $\ce {({P\_2}{O\_5})^{4-}}$. My question is about the hydrogen part- how to know how many hydrogens to add? The answer key for this question says the compound is $\ce {KH\_3P\_2O\_5}$ implying that I should add 3 hydrogens. But why not add 1 hydrogen and have $\ce {K\_3}$ instead?
| 0
|
[
[
"\nAs phosphorous acid is biprotic:\n\n\n$$\\ce{HO-PH(=O)-OH},$$\n\n\nthe respective pyrophosphorous acid is biprotic too:\n\n\n$$\\ce{HO-PH(=O)-O-PH(=O)-OH}$$\n\n\nand pyrophosphite is:\n\n\n$$\\ce{^{-}O-PH(=O)-O-PH(=O)-O-}$$\n\n\nTherefore, potassium hydrogen pyrophosphite would be:\n\n\n$$\\ce{KHP2H2O5}$$\n\n\nrespectively\n\n\n$$\\ce{KH3P2O5}$$\n\n\nBy other words, from 4 hydrogens of pyrophosphorous acid, only 2, bound to O, are acidic. The other 2, bound directly to P, are not.\n\n\nSimilarly, hypophosphorous acid is monoprotic, as only 1 of 3 hydrogen atoms is acidic:\n\n\n$$\\ce{HO-PH2=O}$$\n\n\n\n\n---\n\n\nPyrophosphite is not analogous to pyrosulphite, nor phosphite to sulfite. There is no $\\ce{PO3^3-}$ nor $\\ce{P2O5^4-}$.\n\n\nPyrophosphorous acid is in this context similar to acetic acid. Not all their hydrogen atoms are acidic, so even if fully neutralized, some remain. Sodium acetate is $\\ce{CH3COONa}$, not $\\ce{CNa3COONa}$.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/170130/how-to-calculate-absolute-charge-of-a-protein
|
How to calculate absolute charge of a protein [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
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Closed 9 months ago.
[Improve this question](/posts/170130/edit)
My understanding is that hydrogen bonds formed by proteins require the NH2 / COOH to be neutral. Hence to find out when the hydrogen bonds are the strongest will depend on the absolute charge of the protein and not the net charge. Hence how do you calculate this absolute charge or is there another way to determine when the strongest hydrogen bonds will be formed when changing the pH condition?
Note: For absolute charge I do not need the actual charge in coulombs but more of the relative charge the NH2 / COOH will bear when changing the pH conditions
| 1
|
[] |
https://chemistry.stackexchange.com/questions/170124/what-is-the-decomposition-of-potassium-nitrite
|
What is the decomposition of potassium nitrite?
|
I can find plenty of information on the decomposition of potassium nitrate into potassium nitrite and free oxygen, but apparently the resulting potassium nitrite likes to further decompose and produce even more free oxygen. I cannot however find the reaction for this, and its products. Do any of you know?
| -1
|
[
[
"\n$\\ce{4 KNO2 -> 2 K2O + 2 N2 + 3 O2}$\n\n\nTemperature is too high for $\\ce{K2O2}$ or $\\ce{KO2}$\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/170122/is-there-a-difference-between-hartree-fock-method-and-lcao
|
Is there a difference between Hartree-Fock method and LCAO?
|
I have to dive in some quantum chemistry for a quantum machine learning project and I came across the so-called Hartree-Fock method.
In one of the reference I used, they considered electrons as independent so that the many-electron wave function (**is it of the full molecule or only for one atomic nucleus ?**) can be described as a Slater determinant.
So far no problem but then I came across the linear combination of atomic orbitals method and I wanted to know how it relates to Hartree-Fock. Because looking at some quantum chemistry textbook, it appears that for $n$ AO we can produce $n$ MO, say in the case of $\ce{H2O}$ it would *a priori* be 3 (for the oxygen) and 2 (for the $\ce{H2}$ fragment). But the Slater determinant accounts for all doesn't it ?
Thank you for your answer :)
| 2
|
[
[
"\nObviously, this is a topic which much can be written on. I'll try to keep it short and simple. The LCAO concept merely says that MOs are formed from AOs through linear combination, i.e.\n\n\n$$\\psi\\_n = \\sum\\_i c\\_{ni}\\phi\\_i$$\n\n\nwhere $\\phi\\_i$'s are AOs, $\\psi\\_n$'s are MOs, and $c\\_{ni}$'s are the coefficients of the $i$-th AO in the $n$-th MO.\n\n\nIt offers no way of determining the coefficients $c\\_{ni}$. Sometimes it can be determined through symmetry (let's say $\\ce{H2}$, the bonding MO obviously has equal contributions from both hydrogen 1s orbitals). And sometimes you can get qualitative results (in $\\ce{HF}$, the bonding orbital is mostly fluorine 2p).\n\n\nBut in general, simply knowing the *form* of the MOs isn't enough, and you need some kind of *quantitative* method to determine the coefficients — which is where methods such as Hartree–Fock come in. Using the iterative method prescribed in Hartree–Fock theory you can actually obtain values for $c\\_{ni}$ (and from there, orbital energies, etc.)\n\n\nThe concept of a Slater determinant is completely separate. It is just a convenient way of representing the *total* electronic wavefunction as a product of one-electron wavefunctions (the $\\psi\\_n$'s above are one-electron wavefunctions), but it still has nothing to do with finding the coefficients $c\\_{ni}$. The total electronic wavefunction refers to all electrons in the molecule, by the way—it would not be correct to say that an electron belongs to one specific nucleus.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/170119/why-wavelength-of-rubidium-flame-testred-violet-is-more-than-lithiumcrimson-r
|
Why wavelength of Rubidium flame test(red violet) is more than Lithium(crimson red)
|
My teacher show this table from my book
[](https://i.stack.imgur.com/t89Zu.jpg)
Here, you can see Red violet wavelength is more than Lithium, which is against order of visible light which increase according to acronym 'VIBGYOR'
According to it violet has lesser wavelength than red(or it's shades)
**what is the basic reason?**
| 0
|
[] |
https://chemistry.stackexchange.com/questions/170118/why-is-silver-cyanide-used-for-electroplating
|
Why is silver cyanide used for electroplating
|
(Partially answered by [Safe silver electroplating solutions](https://chemistry.stackexchange.com/questions/55423/safe-silver-electroplating-solutions))
Why is silver cyanide used as an electrolyte in silver plating? ([source](https://en.wikipedia.org/wiki/Silver_cyanide)) There are other (cheaper, less toxic) silver salts, such as silver nitrate. The linked question says that using the nitrate results "makes a fairly rough film that can be somewhat easily removed with abrasion". How does the anion affect the adhesion of the film of silver to the cathode? Why does cyanide result in a strong stable layer of silver, but nitrate (or otherwise) does not?
| 9
|
[
[
"\nNote the **free** cyanide is highly toxic. Cyanide bound to metal ions is less toxic, often much less or even less toxic than table salt. [Potassium\\_ferrocyanide](https://en.wikipedia.org/wiki/Potassium_ferrocyanide) $\\ce{K4[Fe(CN)6]}$ has LD50 $\\pu{6400 mg/kg}$ (oral, rat), what would be about $\\pu{500 g}$ for $\\ce{75 kg}$ person, assuming the same LD50. $\\ce{NaCl}$ has LD50 $\\pu{3000 mg/kg}$ (oral,rat).\n\n\n[Silver cyanide](https://en.wikipedia.org/wiki/Silver_cyanide) AgCN is hardly used, being insoluble. It would be soluble potassium silver cyanide K[Ag(CN)2](commercially available), with silver strongly bound to cyanide.\n\n\nSilver (or metal in general) ions too easily available for deposition on cathodes, as are silver ions from silver nitrate, lead to diffusion controlled metal deposition. This prefers exposed metals surfaces which are amplified. This is taken care, aside of solution thickeners, by using metal complexes like $\\ce{[Ag(CN)2]-}$ that have to be broken. Additionally, cathodes do not attract anions that migrate to cathodes only to address concentration gradient or to maintain electro-neutrality of solutions.\n\n\n\n\n---\n\n\nFor curiosity, metallic silver dissolves in potassium cyanide solution, evolving hydrogen. Gold needs to be pushed by oxygen to be dissolved in cyanide.\n\n\n$$\\ce{2 Ag(s) + 4 CN-(aq) + 2 H2O(l) \\\\\n-> 2 [Ag(CN)2]-(aq) + 2 OH-(aq) + H2(g)}$$\n\n\n$$\\ce{4 Au(s) + 8 CN-(aq) + 2 H2O(l) + O2(g) \\\\\n-> 4 [Au(CN)2]-(aq) + 4 OH-(aq)}$$\n\n\n",
"15"
],
[
"\nSilver cyanide is used for silver electroplating because of the concentration of the free silver ions in solution which is so low that it is not far from zero. Due to the high values of its complex equilibrium formation constant, the complex $\\ce{[Ag(CN)2]^-}$ is practically not dissociated in $\\ce{Ag^+}$ ions. If such a complex solution is electrolyzed, only the rare free silver ions are discharged at the cathode. So at the beginning of the electrolysis, these free $\\ce{Ag^+}$ ions produce a discontinuous deposit of metallic atoms on the cathode.\n\n\nLet's consider the mechanism of electrolysis in the microscopic level. As the first metallic atoms produced on the cathode by the beginning of the electrolysis look like small points or bumps, these tiny bumps will attract electric field lines. As a consequence, the next positive ions will get discharged most probably on this bump, making it bigger, if the solution is concentrated enough. More and more positive ions will be attracted and discharged on this point. The metallic deposit will be localized around specific points. The surface of the cathode becomes rough and course.\n\n\nBut if the free ion concentration is extremely low, as with $\\ce{Ag^+}$ ions in $\\ce{[Ag(CN)2]^-}$ solutions, the first $\\ce{Ag}$ atom is too far from the next $\\ce{Ag+}$ ion. This metallic ion is not attracted by the $\\ce{Ag}$ \"bump\". It is attracted by the whole cathode. It touches the cathode and get discharged equally on its surface, independently from the position of the previously deposited Ag atom. The Ag deposit makes a smooth layer which looks like a mirror.\n\n\nTo summarize, the formation of a mirror can only occur in a solution of a silver complex where the free (non-complexed) ion concentration is extremely weak, like in $\\ce{[Ag(CN)2]^-}$ solutions. Solutions of other silver compounds, like silver nitrate, will only produce rough silver deposits by electrolysis. No mirror !\n\n\n",
"12"
],
[
"\nThe problem with electroplating the more noble metals such as copper, silver and gold is that these metals will react with the substrate [cathode] with a competitive immersion plating reaction. This corrodes the surface, causes an uneven, less adherent plate and contaminates the solution. There are two common methods, usually used in conjunction, to combat this: The first is to use a thin electroless coating [or strike plating] of copper or even nickel that will not conversion plate under the conditions. This is also the method to plate a non-conductive surface. [Tollens reagent is an example of an electroless plating method]. The second method is to reduce the chemical activity of the metal ion with an appropriate complexing ion. The complexing agents cause ion transport problems and require good mixing and appropriate anodes, pure silver is best. There are proprietary baths to do this some of which use cyanide as the complexing agent. Electroplating has a high art content and you will have to contact the manufacturers for the actual proprietary mixes and recommended procedures.\n\n\nThe classical method to measure quantitative current flow is the silver coulometer, a cell placed in series with a silver anode, a silver nitrate solution and a platinum cathode. No danger of immersion plating and I understand the plating adhered well. Electrolytic deposition of copper on a platinum cathode can give a quantitative removal of copper [beautifully even and adherent] with appropriate control of voltage and pH.\n\n\n",
"2"
],
[
"\nSilver cyanide is commonly used for electroplating because it is a highly conductive and efficient electroplating solution. When used in electroplating, silver cyanide ions are deposited onto a substrate by passing an electric current through the solution. The silver ions are attracted to the cathode (negative electrode) and are reduced to metallic silver, which is then deposited onto the substrate.\n\n\nThere are several reasons why silver cyanide is a good choice for electroplating:\n\n\nHigh conductivity: Silver cyanide is a highly conductive solution, which makes it an efficient choice for electroplating. This is because the higher the conductivity of the solution, the faster the electroplating process can be completed.\n\n\nHigh silver content: Silver cyanide contains a high concentration of silver ions, which allows for a thin and uniform layer of silver to be deposited onto the substrate.\n\n\nEasy to use: Silver cyanide is a stable and easy-to-use electroplating solution, making it suitable for use in a variety of electroplating applications.\n\n\nGood corrosion resistance: Silver has excellent corrosion resistance, making it a good choice for electroplating applications where corrosion resistance is important.\n\n\nOverall, silver cyanide is a popular choice for electroplating due to its high conductivity, high silver content, ease of use, and good corrosion resistance.\n\n\n",
"-2"
]
] |
https://chemistry.stackexchange.com/questions/170111/why-does-the-ozone-layer-deplete-faster-in-winter-months-over-the-poles
|
Why does the ozone layer deplete faster in winter months over the poles?
|
Why is the depletion of the ozone layer greater in the winter months over polar regions?
Is it because there is more release of $\ce{NO\_x}$ due to the increased use of fossil fuels in heating which depletes the $\ce{O3}$ layer?
| 4
|
[
[
"\nAs mentioned in the comments, \"polar vortex\" is one of the prime reason ozone layer depletes faster in winter months over the poles. This is beautifully explained here [1]:\n\n\n\n> \n> Under normal conditions, the amount of stratospheric ozone depends on the amount of sunlight reaching a certain geographic area of the atmosphere. The seasonal variation is low in low latitudes because sunlight is fairly constant year-round. In high latitudes, sunlight goes way down in the winter months. Ozone typically \"builds up\" to higher values over the poles during the winter and early spring in each hemisphere. Because this season is offset by 6 months in the Northern and Southern hemispheres, the effect is seen at the North and South poles roughly 6 months apart.\n> \n> \n> [...] The air masses above the poles become isolated from the rest of the atmosphere during their winter and early spring seasons due to a phenomenon known as the \"polar vortex\". In simplest terms, this vortex is a spinning, funnel shaped region of the atmosphere that forms in late fall and early winter over a pole, allowing chemical reactions in the enclosed air mass to be enhanced due to the lack of mixing with other, lower latitude, air masses. The effect of the pollutants we have added to the atmosphere are thus enhanced in these isolated regions of the atmosphere. The Antarctic vortex over the South Pole is more effective at isolating this region of the atmosphere during the austral winter than is the corresponding arctic vortex. A second feature of the polar stratosphere that is unique and probably aids the polar ozone depletion is polar stratospheric clouds. These very high altitude clouds are composed of ice crystals, sometimes greatly enriched in nitrogen oxide species (\"$\\ce{NO\\_x}$\") that can enhance the ozone degradation reactions discussed above. These ice particles can react with various forms of Chlorine in the atmosphere and accumulate the molecule $\\ce{ClONO2}$, which is a source of ozone depleting Cl radicals. Once spring time comes, this $\\ce{ClONO2}$ decomposes and allows ozone degradation reactions can occur.\n> \n> \n> \n\n\nOne of the other reason is polar stratospheric clouds (PSCs).\n\n\n\n> \n> The very low winter temperatures in the Antarctic stratosphere cause polar stratospheric clouds (PSCs) to form. Special reactions that occur on PSCs, combined with the relative isolation of polar stratospheric air, allow chlorine and bromine reactions to produce the ozone hole in Antarctic springtime.\n> \n> \n> \n\n\n**References**:\n\n\n1. <https://www.soest.hawaii.edu/GG/ASK/ozonehole.html>\n2. <https://csl.noaa.gov/assessments/ozone/2010/twentyquestions/Q10.pdf>\n3. <https://uk-air.defra.gov.uk/research/ozone-uv/moreinfo?view=antarctica-hole-explained>\n4. <http://cimss.ssec.wisc.edu/wxwise/ozone/OZONE5.html>\n5. <https://scied.ucar.edu/learning-zone/atmosphere/ozone-layer>\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/170107/why-should-we-not-add-the-h-ion-conc-coming-from-water-before-calculating-ph-of
|
Why should we not add the H+ ion conc coming from water before calculating pH of acidic/basic solution?
|
For eg:
If we are calculating pH of the soln of a strong acid, we shall do this:
pH= - log [H+]
Where H+ is the concentration coming from that acid.
Why shall we not add to this concentration, the H+ ions coming from water itself (By the autoionization of water) before calculating the pH?
Does water not ionize in presence of the acid/base?
Thank you.
| 0
|
[
[
"\nPure water is very weakly dissociated, meaning that the amount of H+ and OH- is very small. If we look at the autoprotolysis equilibrium of water\n\n\n$$\\ce{H2O <=> H+ + OH-}$$\n\n\nand its autoprotolysis constant at 25 °C\n\n\n$$\\ce{K\\_w = [H+][OH-] = 1.01\\times10^{-14}}$$\n\n\nAccording to the equation above we know that the H+ and OH- are the same, so we can easily calculate that the concentration of H+ is around $\\ce{1.005\\times10^{-7}}$ mol/L\n\n\nNow let's assume you add a small amount of a strong acid, such as HCl, so that the concentration of this acid solution is 0.001 mol/L. If we assume a complete dissociation of HCl according to the following equation\n\n\n$$\\ce{HCl -> H+ + Cl-}$$\n\n\nwe know that the concentrations of H+ and Cl- ions are the same, and equal to 0.001 or $\\ce{1.0\\times10^{-3}}$ mol/L. Thus, several orders of magnitude higher than the amount of H+ ions coming from water.\n\n\nSo when calculating the pH we can use only the concentration of H+ from HCl dissociation, where $\\ce{pH = -log\\_{10}(1.0\\times10^{-3}) = 3}$. Or you can sum up the concentration of H+ coming from HCl and $\\ce{H2O}$, where $\\ce{pH = -log\\_{10}(1.0\\times10^{-3} + 1.005\\times10^{-7}}) \\approx 3$ or more specifically 2.99996. But of course, this amount of decimal numbers makes no sense.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170101/can-ice-tables-give-two-answers
|
Can ICE tables give two answers?
|
I was recently given the question:
>
> The equilibrium constant, K, for the following reaction:
>
> N₂(g) + O₂(g) ↔ 2NO(g)
>
> is 4.00 x 10⁻² at a very high temperature. The reaction is at equilibrium at this temperature with [N2] = [O2] = 0.100 M and [NO] = 0.0200 M in a 2.00 liter flask. If 0.120 mol of NO is suddenly added to the reaction mixture what will be the concentrations of all species when equilibrium is re-established?
>
>
>
I think filling out the ICE table would look like this:
| | N₂ | O₂ | 2NO |
| --- | --- | --- | --- |
| I | .100 | .100 | .080 |
| C | +x | +x | -2x |
| E | .100+x | .100+x | .080-2x |
Which would make the the equilibrium formula:
$$
4.00\cdot 10^{-2}=\frac{(.080-2x)^{2}}{(.100+x)(.100+x)}
$$
However, when solving for $x$ I get $x=\frac{8.2\pm\sqrt{7.84}}{198}$ or $.0\overline{5}$ and $.0\overline{27}$.
nvm I get it now but as this is my first time using stackexchange and this syntax took me too long I feel like posting it because the TeX looks nice to look at :)
noo why does the table only work in preview mode :(
| 1
|
[
[
"\nScenarios leading to a quadratic equation have two mathematical solutions, real or complex. But, one of them often does not have interpretation in scientific context, as it's value is out of the acceptable value range.\n\n\n\n\n---\n\n\nTypically, it provides, directly or indirectly, a negative value of a quantity that cannot be negative. A trivial example:\n\n\nWhat is the length of the shorter side of a rectangular with area $\\pu{30 cm2}$, if the other side is longer by $\\pu{1 cm}$?\n\n\n$$x(x+1)=30$$\n$$x^2 + x - 30 = 0$$\n$$x=\\frac{-1 \\pm \\sqrt{1+120}}{2}=\\frac{-1 \\pm 11}{2}$$\n\n\nA side of the length $\\pu{-6 cm}$ does not obviously make physical sense.\nBut $\\pu{5 cm}$ does.\n\n\n\n\n---\n\n\nIn your particular case, the greater result value would lead to negative $\\ce{NO}$ concentration, so it is not a valid result. There is the constraint $x \\le \\pu{0.04 mol L-1}$ for $[\\ce{NO}] \\ge \\pu{0 mol L-1}$.\n\n\n",
"7"
],
[
"\n$$4.00\\cdot 10^{-2}=\\frac{(0.080-2x)^{2}}{(0.100+x)(0.100+x)}=\\frac{(0.080-2x)^2}{(0.100+x)^2}=\\left(\\frac{0.080-2x}{0.100+x}\\right)^2=0.040$$\n\n\nTaking the square root on both sides:\n\n\n$$\\frac{0.080-2x}{0.100+x}=0.2$$\n\n\nSolving for x:\n\n\n$$x=0.0273$$\n\n\nNotice we did not consider the negative root above, since an equilibrium constant (or its square root) having a negative value is nonsense.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/170099/standard-suffixes-for-compounds
|
Standard suffixes for compounds?
|
Long ago, I learned that suffixes like -ide, -ate, -ose, etc. had specific meanings.
Now, I'm seeing all these drug ads on TV with generic names that all end in -ab or -ib. Do these have a standard meaning, or is it a matter of conforming to the crowd?
| 3
|
[
[
"\nAnything ending in -ib is a kinase inhibitor e.g. [Dasatinib](https://en.wikipedia.org/wiki/Dasatinib)\n\n\nAnything ending in -mab is a monoclonal antibody e.g [Rituximab](https://en.wikipedia.org/wiki/Rituximab)\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/170094/resonance-structures-of-benzene
|
Resonance structures of benzene [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed 9 months ago.
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Does benzene have more than two resonance structures? I draw a third resonance structure keeping 1 bond fixed and rotating other two bonds I get opposite charges on two para position. Is this structure wrong?
| -5
|
[
[
"\nThere are many resonance structures which contribute to the overall reality of benzene.\n\n\nSome of these contain charges, the higher the energy of a resonance form the less it contributes to the overall reality. Consider the \"normal\" resonance forms of benzene which most people think of.\n\n\n[](https://i.stack.imgur.com/s2ioA.png)\n\n\nNext we have some with two charges in them, I have not drawn all the possible resonance forms with two charges. Sadly if I draw all of the resonance forms possible I have to stay up very late and I will miss out on my beauty sleep (maybe at midnight I will turn into a pumpkin).\n\n\n[](https://i.stack.imgur.com/vRwge.png)\n\n\nHere are the two highest energy resonance forms which I can think of, these will make a minimal contribution to the overall reality. But it is good to be aware that they exist.\n\n\n[](https://i.stack.imgur.com/Uoycu.png)\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/170093/equality-of-partial-derivatives-schwartz-s-theorem-and-total-differential
|
Equality of partial derivatives (Schwartz’s theorem) and total differential
|
I'm reviewing some Physical Chemistry concepts and the two books I'm using show the total differential and right away they present the Schwartz's theorem. I don't get the relationship. In the screenshot below you can see that the author mentions that the mixed partial second derivatives can be obtained from the total differential. What is the relationship between equations like 1.7 and 1.8, in general?
I got the screenshot from this excellent book: <https://peverati.github.io/pchem1/SystemVariables.html#thermodynamic-variables>
Atkins does the same think on Mathematical Background 2, pg 91, ninth edition.
[](https://i.stack.imgur.com/HJGhC.png)
| 1
|
[] |
https://chemistry.stackexchange.com/questions/170090/aqueous-solution-of-fecl3-is-yellow-whereas-other-metals-having-the-same-d5-conf
|
Aqueous solution of FeCl3 is yellow whereas other metals having the same d5 configuration are pale violet coloured. Explain the origin of the colour [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
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The aqueous solution of the other metals with the same d5 configuration give pale violet colour.
Pale colour could be caused by the d-d transitions.
i got to know that Ferric chloride forms a self ionizing complex in aqueous medium as [Fe(H2O)4Cl2]^+[FeCl4]2-
Now i dont know what type of transitions are taking place in this compound which is giving out the yellow colour.
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[
[
"\nIf, in fact, ferric chloride firms $\\ce{FeCl4^-}$, then this ion could indeed be responsible for the different color of the chloride in solution.\n\n\nTextbooks often describe high-spin $d^5$ complexes as only pale colored because the $d-d$ transition is spin-forbidden (in a nonrelatvistic theory), and what little transition (and thus coloration in white light) cones from (relativistic) spin-orbit coupling. Less attention is paid to the fact that with the most common metal centers that give this configuration, we typically have octahedral complexes, whose center of symmetry or near-symmetry (with nonidrntucal ligands) also renders the $d-d$ transitions forbidden due to lack of an inherent dipole component in the transition (Laporte forbidden). Imperfect symmetry with different ligands, or molecular vibrations, are necessary to get around this barrier.\n$\\ce{FeCl4^-}$, however, has tetrahedral geometry instead, so no center of symmetry. The spin limitation is still there, but without the Laporte forbidden character we see more color from the spin-orbit coupling. The more intense color of the tetrahedral complex then overwhelms anything the proposed octahedral solution complex may offer.\n\n\n",
"3"
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https://chemistry.stackexchange.com/questions/169915/hydrogen-bond-in-water-vapor
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Hydrogen bond in water vapor [closed]
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As the temperature increases and liquid water changes to gas, are ALL the hydrogen bonds broken or they are just weaker?
| -2
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[
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"\nUnder commonly encountered conditions in industrial applications, essentially all hydrogen bonds are broken. Only at high pressures is hydrogen bonding likely to occur.\n\n\nWhat happens is something of a Catch-22. If hydrogen bonding in water is significant, then it tends to form cross-linked structures because each molecule can both donate and accept two bonds, so the hydrogen bonding tends to form cross-linked structures that condense into a liquid. So to generate steam at near-atmospheric pressure you need enough thermal energy (temperature) to essentially break all the hydrogen bonds at once. Only at high pressure, where the gas becomes denser and more like a liquid, can hydrogen bonding become evident in the gas phase.\n\n\nWe may compare this with hydrogen fluoride, where chainlike structures form instead of cross-linked ones because each molecule donates and accepts only one hydrogen bond. The chainlike structures can be supported in the gas phase, so hydrogen fluoride gas can show hydrogen-bonded oligomers at relatively low pressures. See [here](https://chemistry.stackexchange.com/a/163076/17175).\n\n\n",
"4"
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https://chemistry.stackexchange.com/questions/169901/in-gas-law-equations-if-volume-and-temperature-are-directly-proportional-why-c
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In gas law equations, if volume and temperature are directly proportional, why can't we write the equation as temperature divided by volume? [closed]
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In gas law equations, volume, temperature, pressure and amount of substance may vary. In the simpler equations, like Charles' and Amonton's, temperature and pressure/volume are directly proportional. Why are the equations specifically with the temperature in the denominator, like this one with the reciprocal being untrue ( I'm asking a conceptual, not mathematical question)?
$$\frac{V\_1}{T\_1} = \frac{V\_2}{T\_2}$$
| 0
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[
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"\nInteresting! This bothered me long time ago in elementary school. Anyway, later I learned elsewhere that the mathematical convention is to write the dependent variable first in ratio and proportion with a proportionality symbol. Such relations appeared all the time in chemistry and physics. It is very sad that modern high-school algebra has eliminated entire chapter on ratio and proportion properties (esp. see *invertendo* property\\*).\n\n\nSo, $$ V \\propto T $$ which means\n\n\n$V$ is a function of $T$, not the other way round. Temperature dictates what the volume of gas will be.\n\n\n$$ V =c T $$\n\n\nwhere $c$ is a constant of proportionality. Following the graphing convention, this means that one will plot $T$ on the x-axis and volume $V$ on the y-axis. This is nothing but a linear equation with zero intercept when $T$ is in Kelvins.\n\n\nWe can also write,\n\n\n$$\\frac{ V\\_1}{T\\_1} =c $$\n\n\nand\n$$\\frac{ V\\_2}{T\\_2} =c $$\n\n\nSince both fractions are equal to $c$,\n\n\n$$\\frac{ V\\_1}{T\\_1} =\\frac{ V\\_2}{T\\_2} $$\n\n\n\\****Invertendo*** property of ratios tells us that a/b: c/d is the same as b/a:d/c.\n\n\n$$\\frac{ T\\_1}{V\\_1} =\\frac{ T\\_2}{V\\_2} $$\n\n\nis also correct. It is not wrong algebraically. Here you don't care which one is the independent variable.\n\n\nAs another unrelated example is Ohm's law:\n\n\nOne can say, $$ I \\propto Voltage $$ which means current depends on the applied voltage.\n\n\nSo, $$ I = c Voltage $$\n\n\n$c$ turns out to be the inverse of resistance $R$.\n\n\n",
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"\nThe questions in the title and the body ask different things.\n\n\nWe cound write $\\frac{T}{V}$ instead of $\\frac{V}{T}$, but there is, perhaps unwritten, convention using the cause as the denominator.\n\n\nThe equation $\\frac{V\\_1}{T\\_1}=\\frac{V\\_2}{T\\_2}(=\\frac{nR}{p})$ is obviously valid only for ideal gases ($pV=nRT$) at isobaric conditions.\n\n\n",
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https://chemistry.stackexchange.com/questions/169899/battery-using-zinc-copper-and-vinegar
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battery using zinc, copper, and vinegar
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I filed one side of a penny to reveal the zinc; then I bent the coin and dropped it into a cup of white vinegar. Shortly afterward, many tiny bubbles were forming on the zinc side. Later on, I noticed a black residue on the zinc side. What might be the reaction that occurred? *[I believe that the gas was either H2 or CO2 gas. I also believe that the black residue was either CuO or carbon]* **{Zn + Cu + H ion + CH3COO ion --> ???}**
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[
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"\nCopper surface was probably a little bit oxidized into $\\ce{CuO}$. Dipped into vinegar, this oxide is quickly dissolved by the acetic acid, producing $\\ce{Cu^{2+}}$ ions in solution $$\\ce{CuO + 2 CH3COOH -> Cu^{2+} + 2 CH3COO^-}$$Then, zinc may react with these copper ions, producing copper metal which looks black if the dimensions of the grains are small enough.\n$$\\ce{Cu^{2+} + Zn ⟶ Cu + Zn^{2+}}$$\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/169898/role-of-spin-of-electron-in-splitting-of-orbitals
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Role of spin of electron in splitting of orbitals
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What will happen if there are electrons with opposite spins present in the orbitals of ligand and metal atom during the splitting of orbitals in crystal field splitting?
Magnetic fields of two electrons with opposite spins should cancel out each other, thus its net magnetic field would be zero, there would be no repulsion between orbitals of the ligand and metal atom, and splitting of the orbitals $\mathrm{e\_g}$ and $\mathrm{t\_{2g}}$ would not happen.
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[
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"\nThe *repulsion* between two electrons due to being in the same spin-state is only a part of the net repulsion between the electrons. Coulombic repulsion is present between pairs of electrons with same and opposite spins. The Hamiltonian includes the Exchange integral, the Coulombic integral, etc. In certain cases, it is even favorable for electrons to occupy parallel spins (see magnetism). The only condition in which Pauli exclusion principle applies is when the electrons are present in the same orbital, having all other quantum numbers identical.\n\n\n",
"1"
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https://chemistry.stackexchange.com/questions/169894/can-it-be-said-that-an-electron-transfer-between-two-neutral-atoms-comes-from-th
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Can it be said that an electron transfer between two neutral atoms comes from the neutral atomic electric field? [closed]
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Let us consider the example of Na and Cl. Both atoms are called "neutral" simply because they have the same total number of protons as electrons, yet they have a very weak electric field which is hardly measurable. The electron transfer between two neutral atoms is ultimately driven by the electrostatic force.
Arguments like: This is because atoms and molecules are always trying to achieve the most stable, lowest energy state that they can don't really adress my problem, "lower energy" is not a fundamental force but the outcome of a process *caused by electrostatic force*.
So, in the end, can we say that this very weak electric field of a neutral atom is the main cause that it can attract electrons of other neutral atoms? Simply because the electrons are not classically like point charges in the same place as the protons?
Is there a more detailed book that deals with this in particular? Because I think this is fundamentally important.
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[
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"\nYou don't need to consider quantum mechanics to arrive at a rudimentary understanding of how two neutral objects containing charges can attract, but it is important nonetheless to understand that quantum mechanical properties are very different from classical ones, as explained for instance in answers in [this](https://physics.stackexchange.com/questions/290209/is-field-a-more-fundamental-quantity-or-forcein-classical-mechancis) post. In what follows I start with the classical picture.\n\n\nTwo neutral objects can attract each other because the cancellation of attractive and repulsive contributions from the positive and negative charges is not exact, particularly at close distances, and varies with distance. At long distances the total field of one object becomes very similar to that of a neutral point particle, and is therefore very small and nearly invariant with distance between the objects. Therefore both field and force are negligible. At close distances those contributions don't cancel, in particular because the strength of the field falls with distance. This is why two (real) dipoles can attract (ideal dipoles can only induce mutual alignment). To look into this in more detail you can consult the EM textbook by Griffiths.\n\n\nWhat Fritz London et al accomplished was to place these basic ideas from EM on a more solid QM footing by introducing concepts such as the atomic polarizability. This was in part necessary because in the absence of point particles you need a different way of computing the field generated by the electrons.\n\n\n",
"1"
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https://chemistry.stackexchange.com/questions/169888/strange-intermediate-product-when-benzaldehyde-undergoes-cannizzaro-reaction
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Strange intermediate product when benzaldehyde undergoes Cannizzaro reaction
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In the school laboratory, a friend tried the famous Cannizzaro reaction for benzaldehyde. Some $\ce{NaOH}$ was added to benzaldehyde, and the solution was stirred for 30 minutes. What came as a surprise was that, after this time interval, a big, bright yellow, gooey-looking blob appeared in the flask.
As can be [seen](https://drive.google.com/file/d/1oJOuG74YdOe662LMkXuOvVfzQ7zSGP9A/view?usp=sharing), the blob does not dissolve upon shaking. Whatever reaction occurred here was probably reversible, though, as the blob burst after the solution was transferred into a separation funnel, and the experiment went on to give normal products and reasonable yields, without any obvious anomaly.
Does anyone have any idea about what this strange intermediate product might be, or what might have caused it?
[](https://i.stack.imgur.com/d6DJJm.jpg)
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[
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"\nRobert DiGiovanni's answer comes close. Benzyl alcohol can indeed precipitate during the intermediate stage, but [it is colorless](https://en.wikipedia.org/wiki/Benzyl_alcohol). Because benzyl alcohol has some weak acidity, it can be partially deprotonated by a base as strong as sodium hydroxide, giving $\\ce{C6H5-CH2-O^-Na^+}$. This indeed can impart a yellow color according to [Sigma-Aldrich](https://www.sigmaaldrich.com/US/en/specification-sheet/aldrich/379484), which sells this salt as a solution in the alcohol. When the sodium hydroxide is neutralized upon workup, the benzyl alcohol fully resumes its neutral form $\\ce{C6H5-CH2-OH}$, as if nothing untoward had happened.\n\n\n",
"5"
],
[
"\nBenzyl alcohol has a lower solubility (4 g/100 ml) than sodium benzoate\n(63 g/100 ml) or benzaldehyde (7 gram/100 ml) in water.\n\n\nThe presence of NaOH contributed to the \"salting out\" of the more nonpolar benzyl alcohol, plus some other impurities, forming your \"blob\".\n\n\nBenzaldedyde and benzyl alcohol are also (surprisingly) slightly denser than water.\n\n\n",
"4"
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https://chemistry.stackexchange.com/questions/169881/how-strong-molarity-does-naoh-solution-need-to-be-to-saponify-fats
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How strong (molarity) does NaOH solution need to be to saponify fats?
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I have some large cast iron hob grilles of an awkward shape, and want to clean the burnt-on fat from them using NaOH solution. I have two options
* Use a small amount of a strong solution, with a brush or turkey baster to keep them wet, while they sit on a plastic cement-mixing board with a raised edge, or
* Use a large amount of a weak solution, in a plastic dustbin or suitable trough, to immerse them either fully, or partially and turn them over
The former would require a lot of hands-on time, and lots of opportunity for splashes, all the worse for being of a strong solution. The latter I could just leave to sit for 24 hours, as long as I knew the solution was strong enough to work, or at least work fast enough. I don't really want to buy, or dispose of, large amounts of NaOH.
I've search around but only found the SV, Saponification Value, for various fats and oils. This will at least give me a lower bound for how much I need, if I can estimate the amount that's there.
I have the gut feeling that there is a threshold pH below which saponification won't occur. However, the pH of even 0.01M NaOH is around 12. So my gut says concentration won't matter, and that any practical concentration of NaOH will work, albeit just affecting the speed of reaction.
I have found sources that suggest that 2.5M (10% by weight) will work in 10 minutes. I would expect the speed to vary simply as 1/concentration.
Are my suppositions on threshold and speed correct? I don't really want to spend a day on tens of litres of a weak solution that doesn't work.
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[
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"\n2.5 M might be a bit strong. 1.0 M is 40 g NaOH per liter of water, or around 4% solution.\n\n\nStart with 1 % solution and leave it soak. As with [oven cleaners](https://www.rd.com/article/best-oven-cleaners), a little heat should help move things along, like around 150 F. No need to boil it.\n\n\nI would check it every 2-3 hours. It might be done by then. A little dish soap, rubber gloves, and an abrasive pad should finish the job.\n\n\nIf you need to neutralize, some phosphoric acid from the local hardware store may be of use. Please be careful and wear eye protection. Caustic solution is very corrosive to skin as well.\n\n\n",
"2"
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https://chemistry.stackexchange.com/questions/169877/in-which-orbital-is-the-positive-charge-of-central-nitrogen-present-in-diazometh
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In which orbital is the positive charge of central nitrogen present in diazomethane? [closed]
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[](https://i.stack.imgur.com/rdxjG.png)
Where is the positive charge of central nitrogen located?
It cannot be in the pure p orbital because in that case there will only s-orbital left for hybridization which is not possible. If it's present in the sp hybrid lobe which is also not possile because both sp hybrid orbital are being overlapped and formed sigma bonds with neighbouring atoms.
So in which orbital is this positive charge present?
More detail:
In ground state, valence shell electronic configuration of nitrogen was: 2s2 2p3. After internal excitation, it became 2s1 2p4. Two unpaired electrons made π bonds with neighours and the rest one 2s1 and one 2p2 orbitals hybridized to form two sp hybrid orbitals. Now one electron is lost from one of the sp hybrid orbital and so now both sp orbitals have one unpaired electron each. Now both of these hybrid orbitals made sigma bonds with neighbours. After overlapping with neighbours the sp hybrid orbitals cannot contain positive charge.
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[
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"\n\"Positive charge being present in an orbital\" is weird formulation. Orbitals have zero charge at the best if empty.\n\n\nBut as all 4 orbital pairs contributing to 4 bonds have 2 electrons, it cannot be even said a missing electron in one of them caused the positive charge.\nIt is rather that the middle N provided both electrons for one of its bonds and the other N forms another free electron pair.\n\n\nBy other words, the middle N uses 2 its electrons for the double bond with C and 3 its electrons for the double bond with the other N. The other N formally forms from its 5 electrons the 2 free electron pairs. The remaining unpaired electron and 3 electrons from the middle N form the double N=N bond. This all causes the charge shift.\n\n\n",
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https://chemistry.stackexchange.com/questions/169862/does-aliquot-matter-for-final-concentration
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Does aliquot matter for final concentration?
|
I am wondering if my calculation for getting the final concentration is right:
40 mL of solvent A was taken to extract a sample xy. From that 30 mL were taken and evaporated to dryness. Afterwards, 1 mL of a solvent was added for reconstitution. 100 uL were taken and were diluted to 1000 uL, analysis was performed with external calibration.
How to calculate the final concentration?
My thoughts were the following:
Take the obtained concentration from the calibration curve and multiply by the dilution factor 10 since 100 uL was diluted to 1000 uL. Now, I am struggling a bit. Is this already the final concentration or should I also consider the factor from taking 30 mL out of 40 mL?
Thank you very much for your ideas on this question!
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[
[
"\n$$c\\_\\mathrm{r} = \\frac{V\\_\\mathrm{d} }{ V\\_\\mathrm{i}} \\cdot c\\_\\ce{s}$$\n\n\n$$c\\_\\mathrm{e} = \\frac{V\\_\\mathrm{r} }{ V\\_\\mathrm{a}} \\cdot c\\_\\mathrm{r}$$\n\n\n* $c\\_\\mathrm{e}$ is analyte [molar concentration](https://en.wikipedia.org/wiki/Molar_concentration) in extract\n* $c\\_\\mathrm{r}$ is analyte molar concentration in reconstituted extract\n* $c\\_\\mathrm{s}$ is analyte molar concentration in the final sample, \n\nbeing compared to the calibration curve.\n* $V\\_\\mathrm{a}$ is volume of the extract aliquote\n* $V\\_\\mathrm{r}$ is volume of the reconstituted extract\n* $V\\_\\mathrm{i}$ is volume of the extract injected for final dilution\n* $V\\_\\mathrm{d}$ is volume of the final diluted solvent\n\n\n$$n\\_\\mathrm{e} = V\\_\\mathrm{e}\\cdot c\\_\\mathrm{e} = V\\_\\mathrm{e} \\left(\\frac{V\\_\\mathrm{r} }{ V\\_\\mathrm{a}} \\cdot \\frac{V\\_\\mathrm{d} }{ V\\_\\mathrm{i}} \\cdot c\\_\\ce{s}\\right)$$\n\n\n* $n\\_\\mathrm{e}$ is the analyte molar amount in the whole extract\n* $V\\_\\mathrm{e}$ is the total extract volume\n\n\nAll can be rearranged for mass concentrations and masses.\n\n\n$$m\\_\\mathrm{e} = V\\_\\mathrm{e}\\cdot \\rho\\_\\mathrm{e} = V\\_\\mathrm{e} \\left(\\frac{V\\_\\mathrm{r} }{ V\\_\\mathrm{a}} \\cdot \\frac{V\\_\\mathrm{d} }{ V\\_\\mathrm{i}} \\cdot \\rho\\_\\ce{s}\\right)$$\n\n\n* $m\\_\\mathrm{e}$ is the analyte mass in extract\n* $\\rho\\_\\mathrm{e}$ is the analyte [mass concentration](https://en.wikipedia.org/wiki/Mass_concentration_(chemistry)) in extract\n* $\\rho\\_\\mathrm{s}$ is the analyte mass concentration \n\nin the final sample, being compared to the calibration curve.\n\n\n",
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"\nIs it also necessary to consider a third equation like:\n\n\n$$c\\_{fe} = \\frac{c\\_{e}\\* V\\_{e}}{V\\_{fe}} \n$$\n\n\nwith $ c\\_{fe}$ as finale analyte molar concentration in extract\n\n\nwith $ V\\_{fe}$ as total original volume of the extract\n\n\n",
"0"
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[
"\n30 ml of extraction solution was dried and reconstituted to 1 ml, then diluted 100 ul to 1000ul so the *concentration* is now:\n\n\nExtraction solution × 30 ml/1 ml × 100ul/1000 ul = Concentrated 3x\n\n\n\n```\n ----> Sample run in analytical instrument ---->\n\n```\n\nWhat ever result you get from your analysis as grams analyte/ml must be *divided* by 3 to get the *concentration* of analyte in the extraction solution.\n\n\nNow to find out *total grams* of analyte extracted in your 40 ml of extraction solution:\n\n\n\n> \n> grams analyte/ ml × 40 ml = grams analyte extracted\n> \n> \n> \n\n\nNow, in order to determine the *original* concentration of whatever solution you extracted *from* (assuming 100% extraction efficiency$^1$), you take grams analyte and divide by volume of original solution *before* the 40 ml extraction solution was mixed with it (also assuming a 2 phase extraction with a separatory funnel).\n\n\nSo you need to know your original volume of solution.\n\n\n$^1$ good luck with that, HPLC may be a better choice\n\n\n",
"0"
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[
"\nIt's good to keep in mind the distinction between ***amount*** and ***concentration***.\n\n\nTake the step where you go from 40 mL to 30 mL. Since you're not diluting, you're not changing the concentration of the solution, but you are changing the amount. The way you figure out the reduced amount of compound you have in the new sample is by looking at the ratio of volumes.\n\n\nBut take the 30 mL to 1 mL step. You (hopefully) haven't changed the amount of substance you have when you evaporated and reconstituted things, but you have changed the concentration. To figure out the new concentration, this is where you pull out the old `c1 * v1 = c2 * v2` -- this isn't a magic invocation, it's just a representation of the fact that the amount of substance hasn't changed. (The amount of substance in a solution being the concentration times the volume - you do the calculation before and after and set them equal to each other, as the amount of substance hasn't changed.)\n\n\nYour whole process can be thought of as a series of changes in the amount and concentration:\n\n\n* 40 mL -> 30 mL -- Constant concentration; change of amount\n* 30 mL -> 1 mL -- Constant amount; change of concentration\n* 1 mL -> 100 uL -- Constant concentration; change of amount\n* 100 uL -> 1000 uL -- Constant amount; change of concentration\n\n\nSo what you can do is take the value (the assay concentration) for the final solution and then work backwards through the steps, converting to concentration or total amount as necessary.\n\n\nIf you think about it stepwise, making sure you understand whether each step is a constant concentration step or a constant amount step, you can apply the relevant conversion factors backwards for each step until you reach the value you want. The benefit here is that this process is robust to (and forces you to think about) whether you want the amount of substance in the original sample or the concentration in the original sample (or if, for some reason, you need the amount in your 30 mL sample) - just add in an extra conversion at the appropriate location. It also generalizes to more complex sampling procedures.\n\n\n(There's certainly potentials for shortcuts and rules of thumbs here to speed up calculations -- but if you're confused or uncertain, it's best to go back to first principles and work things out stepwise.)\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/169858/halogenation-of-organic-acid-derivatives
|
Halogenation of organic acid derivatives
|
Chloropicrin can be produced through the reaction of sodium hypochlorite with nitromethane. On the surface, this seems to me to proceed through a mechanism analogous to the haloform reaction, with one notable difference being that, in the case of the generation of chloropicrin, the C-N bond is not cleaved.
On a related note, I found a paper that claims that acetate esters are also capable of undergoing the haloform reaction.
Because of these two facts, I am wondering if a process analogous to the creation of chloropicrin can occur with other organic acid derivatives. As an example, could methyl methanesulfonate react with bleach to give methyl trichloromethanesulfonate? If yes, are there any other organic acid derivatives for which this process can occur?
**Source:**
Nagata, D., & Nishiwaki, N. (2021). Are acetic acid derivatives really negative to the iodoform test? SN Applied Sciences, 3(9). doi:10.1007/s42452-021-04777-0
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https://chemistry.stackexchange.com/questions/169855/reaction-of-aluminium-chloride-with-alkali-metal
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Reaction of aluminium chloride with alkali metal [closed]
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I wonder if we can get aluminium metal by adding alkali metal (ex. Li,Na) to aluminium chloride solution?
or
maybe what will happen if we mix aluminium chloride hexahydrate with sodium metal?
| -2
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[
[
"\nAdding some sodium metal to any aqueous solution will only produce the reaction $\\ce{2 Na + 2 H2O -> 2 NaOH + H2}$, whatever the nature of the solute (containing Al ions or not). The same reaction happens with aluminium chloride hexahydrate. To get an aluminium production with metallic sodium, the aluminium chloride has to be anhydrous. This was discovered by Davy in the years 1800 - 1810.\n\n\n",
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https://chemistry.stackexchange.com/questions/169850/how-can-atoms-get-very-close-together-even-without-direct-covalent-bonds-between
|
How can atoms get very close together even without direct covalent bonds between them? [closed]
|
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Generally speaking, because of the Pauli exclusion principle, it’s difficult for the electron clouds of two atoms to overlap unless they are covalently bound together. The degree of overlap can be estimated with the van der Waals(vdw) radius. When the distance between two atoms is smaller than the sum of their vdw radii, the repulsion between them increases steeply.
However, atoms within the same molecules can be much closer than the vdw radii implied. For example, the half distance between the hydrogen atoms in CH4 is 0.89 angstrom, which is 74% of the vdw radius of hydrogen atoms. The half distances between the fluorine atoms and chlorine atoms in CF4 and CCl4 are 73% and 82% of their vdw radii, respectively (data sheet: C-H: 1.09, C-F: 1.32, C-Cl: 1.77, vdw H: 1.2, vdw F: 1.47, vdw Cl: 1.75). So how can these atoms get so close together despite the lack of covalent bonds between them? My guess is that vdw radius measures the distance between atoms of separate molecules. Because neutral molecules are bound together by the very weak dispersion force while in CX4 the four terminal atoms are tightly bound to the central carbon atom, they can overcome much larger repulsion between them.
My second question is whether there is a lower limit on how close atoms can get together. For example, C60F60 is much larger than CF4, which means the angles between adjacent C-F bonds is much smaller in C60F60. If the lengths of C-C bonds and C-F bonds are maintained, the distance between fluorine atoms may be unacceptably small, which may slightly stretch the C-C and C-F bonds to accommodate the fluorine atoms. Can we predict the stability of perhalogensted compounds (such as the instability of polytetrachloroethylene) by the atomic distances?
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https://chemistry.stackexchange.com/questions/169849/what-form-of-energy-is-produced-by-2h2-o2-2h2o-reaction
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What form of energy is produced by 2H2 + O2 -> 2H2O reaction? [closed]
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**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
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I'm curious about are the details of exothermic nature of 2H2 + O2 -> 2H2O reaction. The explanations I've seen just use a term "energy release" without specifying what type of energy it is. Assume that we have a perfect mixture of H2 and O2 in a closed vessel and produce a spark, which will initiate the reaction. What is the mechanism, which increases the velocity (temperature) of the molecules participating in the reaction?
| -3
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[
[
"\nYou're asking two separate questions.\n\n\n1. What form of energy is produced when combining hydrogen and oxygen?\n2. How does that become *thermal* energy?\n\n\nThe answer to 1. is that energy can be released in different forms, such as electricity in a hydrogen-oxygen [fuel cell](https://en.wikipedia.org/wiki/Fuel_cell), or [light](https://www.researchgate.net/figure/Excitation-emission-spectrum-of-a-hydrogen-oxygen-flame-The-excitation-wavelength-is_fig8_265821077). An efficient fuel cell produces little heat.\n\n\nQuestion 2. is best considered through [thermodynamics](https://www.britannica.com/science/thermodynamics/Entropy-and-heat-death). Theoretically, it *might* be possible to predict the position and velocity of a few unrestrained molecules of $\\ce{H2}$ and $\\ce{O2}$ combining to form $\\ce{H2O}$ (ignoring quantum uncertainty). However, *en masse*, this is effectively *random*. Ah... and what is the [random motion of particles](https://www.coursehero.com/study-guides/boundless-physics/introduction-6/) called?\n\n\n",
"1"
],
[
"\n**The form of energy depends on the reaction conditions but the molecular mechanisms are complex**\n\n\nWhen water is formed from hydrogen reacting with oxygen, a large amount of energy must be released (that's thermodynamics). *How* it gets released depends on the conditions of the reaction.\n\n\nWhen oxygen and hydrogen are reacted in a controlled way on a catalytic surface inside a fuel cell, much of the energy can be extracted as electricity. But the detailed mechanism involves complex reactions on a specific catalytic surface.\n\n\nWhen hydrogen and oxygen gas is ignited, the reaction is fairly uncontrolled and the energy emerges as both light and heat. How this happens is not easy to describe in a simple, single reaction, not least because in a gas there are many molecular collisions happening all the time and it isn't just about two molecules colliding and yielding the product. Plus, the reaction will usually have multiple steps.\n\n\nFor example, a hydrogen molecule might collide with an oxygen molecule to yield an H• radical and an HOO• radical (I'm making up possibilities for illustrative purposes rather than trying to give a realistic description). Both species might have more kinetic energy than the original molecules so will move faster, but they are also, both, highly reactive so further collisions will often yield more reactions, some of which will release more energy. In some cases, the reactive species generated will decay spontaneously, emitting light. In many other cases they will bang into other molecules distributing the excess energy as kinetic energy throughout the remaining gas molecules. This will continue until only stable molecules are left.\n\n\nThe point of this is that excess energy from the formation of new molecules can be released as light or can be distributed very rapidly as kinetic energy to other molecules in the gas (and more kinetic energy in the gas is *heat*). At a molecular level, the excess energy from a specific chemical reaction can appear as rotational energy or vibrational energy in bonds or as kinetic energy from faster movement of the molecules. But the *specific* amounts of energy will distribute very rapidly across all the types of energy because molecular collisions tend to redistribute energy very, very rapidly to an equilibrium across all the types on energy in all the gas molecules.\n\n\nUltimately, at least in uncontrolled reactions, this means we see the excess as heat.\n\n\n",
"1"
],
[
"\nThermal energy is released from the reaction $\\ce{2H2 + O2 -> 2H2O}$. Before the reaction takes place the system of $\\ce{H2}$ and $\\ce{O2}$ molecules are in a metastable state:\n[](https://i.stack.imgur.com/fRh1J.png)\n\n\nGiven enough energy > activation energy the system gains enough energy to overcome the potential barrier and falls into a state of lower energy than the initial state.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/169840/does-the-equilibrium-constant-of-an-ideal-liquid-mixture-reaction-depend-on-pres
|
Does the equilibrium constant of an ideal liquid mixture reaction depend on pressure?
|
I want to calculate the equilibrium constant K[T] for a the following reaction:
$$\ce{CH3OH(l) + C4H8(l) -> C5H12O(l)}$$
It is assumed that the mixture is ideal and the reaction takes place at 14 bar pressure.
K[T] is defined as the product of the thermodynamic activities to the power of the stoichiometric coefficient:
$$\prod\_j {a\_j^{\nu\_j}} = \prod\_j {\left( \frac{\hat f\_j[T, p, \mathbf{x}] }{ f\_j^{\mathrm{ref}}[T, p^{\mathrm{ref}}, \mathbf{x}^{\mathrm{ref}}]} \right)}^{\nu\_j}$$
The fugacity is given by Raoult's law,
$$f\_i = x\_i \cdot p\_i^{\*}$$
where x is the molar fraction and p\* the vapor pressure, which are given.
How can I determine the reference fugacity, and where in the calculations do I use the 14 bar pressure at which the reaction takes place?
| 0
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[
[
"\nThe equilibrium constant for a chemical reaction is only a function of temperature. However, if you want to calculate the equilibrium constant $K$ via an equilibrium calculation, e.g. using the equation you posted, then we need to find the pressure dependence on the fugacity of the species $i$ as a liquid.\n\n\n\n\n---\n\n\nWe write the relationship between the fugacity of a pure species $i$ with the Gibbs energy, for two different situations distinguished by a zero superscript\n\\begin{align}\n g\\_i &= \\Gamma(T) + RT\\ln(f\\_i) \\tag{1} \\\\\n g\\_i^0 &= \\Gamma(T) + RT\\ln(f\\_i^0) \\tag{2}\n\\end{align}\nSubtracting Eq. (2) from Eq. (1)\n$$ g\\_i - g\\_i^0 = RT\\ln\\bigg(\\frac{f\\_i}{f\\_i^0}\\bigg) \\tag{3} $$\nEquation (3) links two different states:\n\n\n1. The fugacity of a pure species at pressure $p$ and temperature $T$.\n2. The fugacity of a pure species at pressure $p^0$ and temperature $T$. The reference state for a pure liquid, in the context of this equation, is generally chosen as that where the substance at temperature $T$ is a **saturated liquid**, and thus $p^0 = p^\\mathrm{sat}$.\n\n\nNow we use the fundamental thermodynamic relation for pure substances\n$$ \\mathrm{d}g = v\\mathrm{d}p - s\\mathrm{d}T \\tag{4} $$\n\n\nThe evolution $(p^\\mathrm{sat},T)\\to(p,T)$ at constant temperature, for a species labeled with $i$, is carried out by integrating Eq. (4)\n$$ g\\_i - g\\_i^\\mathrm{sat} = \\int\\_{p\\_i^\\mathrm{sat}}^p\n v\\_i\\mathrm{d}p \\tag{5} $$\n\n\nThe liquid molar volume will change between these two pressures. For a liquid, we make the assumption that if the pressure range is not that large, it will remain constant and equal to the *liquid saturated volume* at the saturation pressure $ v\\_i = v\\_i^\\mathrm{sat} $. Hence, Eq. (5) turns into\n$$ g\\_i - g\\_i^\\mathrm{sat} = v\\_i^\\mathrm{sat}(p - p^\\mathrm{sat}) \\tag{6} $$\nCombining Eqs. (3) and (6)\n\\begin{equation}\n RT\\ln\\bigg(\\frac{f\\_i}{f\\_i^\\mathrm{sat}}\\bigg) = v\\_i^\\mathrm{sat}(p - p^\\mathrm{sat})\n \\to\n \\frac{f\\_i}{f\\_i^\\mathrm{sat}} = \\exp\\bigg[\\frac{v\\_i^\\mathrm{sat}\n (p - p\\_i^\\mathrm{sat})}{RT}\\bigg] \\tag{7}\n\\end{equation}\nThe right-hand side of Eq. (7) is the famous **Poynting factor**.\n\n\n\n\n---\n\n\nWe return to the equilibrium constant. As stated, the mixture is ideal. We deal with liquids, so the fugacity coefficient of species $ i $ in a mixture is $\\hat{f}\\_i = x\\_i\\gamma\\_if\\_i= x\\_i f\\_i $, because $ \\gamma\\_i = 1 $. Summarizing the reaction in the fashion $\\ce{A + B -> C}$\n\\begin{equation}\n K = \\dfrac{\\dfrac{\\hat{f}\\_C}{f\\_C^\\mathrm{sat}}}\n {\\left(\\dfrac{\\hat{f}\\_B}{f\\_B^\\mathrm{sat}}\\right)\n \\left(\\dfrac{\\hat{f}\\_A}{f\\_A^\\mathrm{sat}}\\right)} =\n \\dfrac{\\dfrac{x\\_Cf\\_C}{f\\_C^\\mathrm{sat}}}\n {\\left(\\dfrac{x\\_Bf\\_B}{f\\_B^\\mathrm{sat}}\\right)\n \\left(\\dfrac{x\\_Af\\_A}{f\\_A^\\mathrm{sat}}\\right)} \\tag{8}\n\\end{equation}\nNow we use Eq. (7) for the three ratios in Eq. (8)\n\\begin{equation}\n \\boxed{K =\n \\dfrac{x\\_C \\exp\\bigg[\\dfrac{v\\_C^\\mathrm{sat}\n (\\color{blue}{p} - p\\_C^\\mathrm{sat})}{RT}\\bigg]}\n {x\\_A \\exp\\bigg[\\dfrac{v\\_A^{sat}(\\color{blue}{p} - p\\_A^{sat})}\n {RT}\\bigg] \\cdot\n x\\_B \\exp\\bigg[\\dfrac{v\\_B^{sat}(\\color{blue}{p} - p\\_B^{sat})}\n {RT}\\bigg]}} \\tag{9}\n\\end{equation}\n\n\nIn blue is emphasized the pressure dependence in the fugacity.\n\n\nEq. (9) shows that additional information is needed when we deal with liquid-phase reactions:\n\n\n* The saturation pressure $p\\_i^\\mathrm{sat}$ for every species.\n* The saturated liquid molar volumes $v\\_i^\\mathrm{sat}$ for every species, evaluated at $p\\_i^\\mathrm{sat}$.\n\n\n",
"0"
]
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https://chemistry.stackexchange.com/questions/169834/difficulty-in-e-z-nomenclature-and-counting-number-of-geometrical-isomers
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Difficulty in E-Z nomenclature and counting number of geometrical isomers
|
[](https://i.stack.imgur.com/iyxqk.jpg)
[](https://i.stack.imgur.com/hZIU9m.jpg)
I have found above four molecules as the geometrical isomers just by drawing them and checking if they are superimposable. I'm not sure that these are the only ones.
I have tried using E and Z but couldn't go ahead with it because configuration around each double bond depends on configurations of all other double bonds and it gets circular.
How do I assign E and Z configurations to each of the four double bonds in this case to distinguish between all possible geometrical isomers?
| 2
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[
[
"\nRishi Shekher: You are correct that there are only four stereoisomers (**A**, **B**, **C**, **D**) of this tetraethylidenecyclobutane. Loong has given you a lead as to how to apply CIP rules to the assignment of the configurations of the double bonds. This method can appear confusing and it is certainly not intuitive. The configurations were generated with ChemDraw 21. I will show you how the CIP algorithm works.\n\n\n \n\n\n\n**Stereoisomer A**: The configuration of each double bond must be determined independently. They are labeled in red in each of the digraphs **1-4**. The digraph is constructed by following a path around the ring, CW or CCW, from the non-duplicate carbon (black dot) to the duplicate carbons (red dot), which are designated as being attached to three atoms of atomic number zero. \n \n\nFocusing on digraph **1** (*vide infra*) and $\\ce{C1}$ (black dot), the double bond immediately to its left ($\\ce{C4}$) is assigned the temporary *Z*-configuration because the path \"around the ring\" to the right is longer, i.e., more carbons than the path leading to the left. This method is used to temporarily assign the five positions. The left hand chain has three *Z*'s while the right hand chain has all *E*'s. One proceeds out each chain from the black dot making a one-to-one comparison until a Z>E is achieved ([CIP Rule 3](http://ursula.chem.yale.edu/%7Echem220/chem220js/STUDYAIDS/isomers/CIP%20rules%20NEW.html)). For $\\ce{C1}$ this is accomplished at $\\ce{C4}$ and $\\ce{C2}$ where Z>E, respectively. Determinant double bonds are shown in blue. \n \n\n\n\n \n \n\n\n\n**Stereoisomer B:** All positions in this isomer are equivalent. The double bonds are all of the *E*-configuration with *Z>E*.\n\n\n \n \n\n\n\n**Stereoisomers C and D:** Stereoisomer **C** has a plane of symmetry. $\\ce{C1}$ is equivalent to $\\ce{C4}$ and $\\ce{C2}$ is equivalent to $\\ce{C3}$. Stereoisomer **D** has four equivalent double bonds owing to two planes of symmetry. Given that both stereoisomers **C** and **D** have the *Z*-configuration, one would would be hard pressed to know which one to draw if asked to do so. This situation is unfortunate.\n\n\n \n \n\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/169830/why-dont-we-use-sodium-bicarbonate-to-neutralise-base-spills
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Why don't we use sodium bicarbonate to neutralise base spills?
|
According to this document on lab safety (<http://faculty.washington.edu/korshin/Class-486/AEESP-safety-notes.pdf>), it says to not use $\ce{NaHCO3}$ to neutralise specifically base spills, stating "Do not use acetic acid or
sodium bicarbonate to clean a base spill. The sodium bicarbonate will not neutralize the spill, and acetic
acid could react strongly with the base."
Could someone explain why sodium bicarbonate won't neutralise a base spill? I have always learnt that it is amphiprotic and therefore can be used to neutralise BOTH acid AND base spills.
| 4
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[
[
"\nSuppose we have a strong basic aqueous solution $B$ that we attempt to neutralize with $\\ce{NaHCO3}$ working as an acid $A$.\n\n\nThe source of excess $\\ce{OH-}$ ions comes from the base, and since it's strong, we can consider:\n\n\n$$n\\_{Bo}=n\\_{\\ce{OH-}}\\implies C\\_{Bo}{V\\_B}=C\\_{\\ce{OH-}}V\\_B$$\n\n\nThe source of excess $\\ce{H+}$ ions is bicarbonate, but since it's a weak acid, we have to consider the equilibrium:\n\n\n$$\\ce{HCO3-(aq)<=>H+(aq) + CO3^{2-}(aq)}$$\n\n\nThe equilibrium expression in terms of moles after mixing both solutions is:\n\n\n$$K\\_n=K\\_a\\;(V\\_A+V\\_B)=\\frac{x^2}{C\\_{Ao}V\\_A-x}$$\n\n\nFor complete neutralization to take place, both excess $\\ce{H+}$ and excess $\\ce{OH-}$ ions need to completely react with each other, so:\n\n\n$$x=C\\_{Bo}V\\_B$$\n\n\nSubstituting above:\n\n\n$$K\\_a\\;(V\\_A+V\\_B)=\\frac{(C\\_{Bo}V\\_B)^2}{C\\_{Ao}V\\_A-C\\_{Bo}V\\_B}$$\n\n\nFor simplicity, let's consider equal molar concentrations of both solutions and a volume of $\\pu{0.1L}$ for our basic solution that needs neutralizing:\n\n\n$$C\\_{Ao}=C\\_{Bo}=\\pu{1mol/L}$$\n\n\n$$V\\_B=\\pu{0.1L}$$\n\n\nThe acid dissociation constant of bicarbonate at 25°C is approximately:\n\n\n$$K\\_a=5.012\\times10^{-11}$$\n\n\nSubstituting all values:\n\n\n$$5.012\\times10^{-11}\\;(V\\_A+0.1)=\\frac{0.1^2}{V\\_A-0.1}$$\n\n\nSolving for $V\\_A$:\n\n\n$$V\\_A=\\pu{14125.2L}$$\n\n\nIn other words, bicarbonate is considerably weaker as an acid (about 400 times) than it is as a base, so attempting to neutralize a strong basic solution with it would require an enormous amount.\n\n\n",
"5"
],
[
"\nI am going to argue that the currently-accepted answer is incorrect.\n\n\nAs the current accepted answer correctly pointed out, the bicarbonate ion dissociates into a proton and a carbonate ion, and that an equilibrium between the bicarbonate ion and the carbonate and hydrogen ion is established.\n\n\nHowever, Sam202's calculation assumes that there is no driving force that would shift the equilibrium towards one side or the other. In the case of an acid/base reaction, this is simply not true. Following from [Le Chatelier's principle](https://en.wikipedia.org/wiki/Le_Chatelier%27s_principle), the equilibrium can be shifted to favor the products by removing products as they are formed. In the case of sodium bicarbonate reacting with sodium hydroxide, hydrogen ions formed by the dissociation of bicarbonate are removed by reaction with hydroxide ions to form water, causing the equilibrium to shift to favor greater dissociation of bicarbonate, driving the reaction to completion.\n\n\nAs a sidenote, acid/base titrations assume that the reaction between the acid and base is stoichiometric. If Sam202's answer were correct, titrating a weak acid with a strong base or vice-versa would be impractical.\n\n\nAs to the question at hand, the reaction of sodium bicarbonate with sodium hydroxide would produce water and sodium carbonate as pointed out by commenter Poutnik. While sodium carbonate is a weaker base compared to sodium hydroxide (pKa of sodium carbonate is 10.33 compared to 15.7 for sodium hydroxide, according to Wikipedia) it is still basic (For reference, pKa of ammonia is 9.25, pKa of sodium bicarbonate is 6.3). Thus, the reaction of sodium bicarbonate with sodium hydroxide would still leave a base spill, as Poutnik correctly deduced.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/169829/why-are-certain-hot-polycarbonate-films-stickier-than-others-molecular-weight
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Why are certain hot polycarbonate films stickier than others? Molecular weight? Hydrolysis?
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In van der waals dry transfer, we pick up flakes of materials such as graphene using polymer films. The material is exfoliated onto an SiO2 covered wafer and picked off using the polymer film. It's obviously pretty bad if the film sticks to the SiO2 itself, otherwise you wind up tearing and distorting your film trying to get it back off.
It's often necessary to try and pick up the flakes at elevated temperatures -- around 100-120 degrees C. It's at this point that some films get extremely tenacious and will not peel off the SiO2. The films that are the best seem to be the ones made on days with high humidity. Occasionally, water steam treatment seems to help. It would therefore seem to me that the PC is being hydrolyzed and broken into lower molecular weight pieces. Just a guess -- but if so, why would this make the film less sticky? If not, I would love to know why!
| 3
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https://chemistry.stackexchange.com/questions/169827/where-are-the-n-and-c-termini-located-in-hemoglobin
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Where are the N and C-termini located in hemoglobin?
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I am trying to locate the N and C terminus of hemoglobin and cannot tell if the C is at Val-1 and the N is Arg-141.
| 0
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[
[
"\nThe N-terminus has the lowest residue number, and the C-terminus has the highest residue number. When showing a sequence, it is given from N-terminus to C-terminus when writing left to right. The specific sequence depends on whether you are looking at the alpha or beta chain, whether you are looking at fetal or adult hemoglobin, and which organism you are looking at.\n\n\n",
"2"
]
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https://chemistry.stackexchange.com/questions/169822/how-do-we-know-the-orbitals-of-multielectron-atoms-look-like-the-orbitals-of-hyd
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How do we know the orbitals of multielectron atoms look like the orbitals of hydrogen?
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How do we know that the solutions of the time-independent Schrodinger equation for multielectron atoms are the same with the solutions for the hydrogen atom if we add a effective potential due to the shielding of the charge of the nucleus from inner shell electrons?
| 1
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[
[
"\nThe electronic Hamiltonian for a hydrogen atom can be written as\n\n\n$$\\hat{H}\\_e = -\\frac{Ze^2}{r\\_{Ne}} - \\frac{\\hbar^2}{2m\\_e} \\nabla^2$$\n\n\nThe relative nuclear charge for an H-atom is $Z=1$ (the atomic number). When approximating the single electron Hamiltonian using an effective nuclear potential and otherwise ignoring electron-electron repulsion, which is missing in the above Hamiltonian, $Z$ is set to $Z\\_{eff}$, the difference of the nuclear charge and the screening constant.\n\n\nThe effect of an increase in $Z$ is to *contract* the size of an atom. The Hamiltonian could be written as\n\n\n$$\\hat{H}\\_e = \\frac{e^2}{r^\\*\\_{Ne}} - \\frac{\\hbar^2}{2m^\\*\\_e} \\nabla^{\\*2}$$\n\n\nwhere $r^\\*\\_{Ne} = r\\_{Ne}/Z$ is an effective distance from the nucleus, $m^\\*\\_e = m\\_e Z^2$ is an effective electron mass and $\\nabla^\\* = \\sum\\_i \\vec{e}\\_i \\frac{\\partial}{\\partial x\\_i^\\*}$. The change in the dimensions of $r$ therefore does not alter the mathematical form of the eigenfunctions of the Hamiltonian above, it only rescales the mass and distance dimensions. The solutions will be the single electron hydrogenic wavefunctions but with the effective radius $r^\\*\\_{Ne} = r\\_{Ne}/Z$ in place of $r$ and the momentum of the electrons altered by the change in effective mass. An effective Bohr radius can be computed under these circumstances as\n(setting $\\mu=Z^2 m\\_e$) $$a\\_\\circ ^\\* = \\frac{4 \\pi \\epsilon \\_\\circ \\hbar^2}{Z^2 m\\_e e^2} = \\frac{a\\_\\circ}{Z^2}$$\nillustrating the contraction in the extent of the wavefunction due to the increase in $Z$.\n\n\nNote that this is an oversimplification of the true situation, but the approximation suggests for instance that Slater-type orbitals might be useful as a first step to describing the properties of atoms. For molecules it is likewise inaccurate but can be helpful to grasp how behavior scales with nuclear charge.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/169820/qm9-atomization-energies
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QM9 atomization energies
|
I am doing experiments with the QM9 dataset from
* <http://quantum-machine.org/datasets/>
and began with some simple checks on results from the literature. Consistent with what the authors of the paper
* C.R. Collins, G.J. Gordon, O.A. Von Lilienfeld and D.J. Yaron, Constant size descriptors for accurate machine learning models of molecular properties. J. Chem. Phys. 148 (2018), 241718.
report, I find 3993 small QM9 molecules with 7 or fewer heavy atoms, which serves as their training set.
But I cannot reproduce the Null row of their Table II, probably because I don't understand well enough what they actually do there. In Section V.A they say:
>
> A “null” model is used to provide a baseline measure of the difficulty of the prediction task. The null model always predicts the mean value of the training data.
>
>
>
They don't specify precisely which quantity they predict, but on p.6 they report values in kcal/mol, and in Table II they talk about atomization energies, so I suppose the target is one of the energies provided by QM9. But there energies are in Hartree and hence must be converted to kcal/mol.
Using the atomization energy (QM9 property 13), I get 319.48 Ha for the mean atomization energy on the training set. With this mean as constant null predictor I obtain an MAE of 30.84 Ha = 19,353 kcal/mol on the training set, and of 31,319, 48,632, and 58,000 kcal/mol, on the first 20,000, 50,000, and 133,000 QM9 molecules, respectively. (The data used are in <https://arnold-neumaier.at/nullQM9.txt>. Calculating the mean and the MAEs for the mean is a standard triviality.) But these numbers are over 100 times larger than the numbers stated in the Null row of their Table II. (The table is also on p.10 of the arXiv preprint <https://arxiv.org/abs/1701.06649> - Section V.A is there called 5.1.)
Without conversion, the numbers would be far too small. The other properties from QM9 don't fare better.
Perhaps I misunderstood what they meant by the Null model, or there is an undocumented preprocessing step for getting the target energies for prediction?
Could anyone please resolve this discrepancy?
| 1
|
[
[
"\nFirst off, let me stress that it's important to have a null model or baseline model for comparison in machine learning. In other words, does the ML model presented in the paper do better than a simple alternative.\n\n\nAtomization energies are calculated from the electronic structure calculation of the molecule and the relevant atoms. In other words, how much energy does it take to dissociate the molecule into the constituent atoms (e.g., 6 carbon and 6 hydrogen atoms for benzene). **Consequently, atomization energies depend on the number and type of atoms in the molecule.** The atomization energy for benzene is higher than that of methane, because you're breaking more bonds.\n\n\nMy reading of this paper, was that the null model consisted of the average energies for each atom, e.g. $E\\_{C\\_6H\\_6} = 6E\\_C + 6E\\_H$ and $E\\_{C\\_5NH\\_5} = 5E\\_C + E\\_N + 5E\\_H$ with the various $E\\_C$ for each element fit from least squares. (Certainly that's what I'd consider as a null model for this, because standard atomization energies use the isolated atoms, so least-squares fit would at least adjust for atoms in a molecule.)\n\n\nYour comment suggests that least-squares null model performs better than the reported null model in the paper.\n\n\nIn that case, I don't know, but if/when I see David Yaron next, I'll ask.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/169819/when-refining-copper-through-electrolysis-do-the-copper-ions-from-the-electroly
|
When refining copper through electrolysis, do the copper ions from the electrolyte copper sulfate get reduced at the anode?
|
Copper sulfate is the electrolyte that is brought up most commonly when we talk about the refining of copper. Do any of the copper ions from the copper sulfate solution reduce at the anode like the copper ions that were oxidized from the cathode?
| 1
|
[
[
"\nLike commenters pointed out the direction of the process is the other way round.\n\n\nApart from that the answer is yes, copper ions from the sulfate will be reduced at the cathode but at the same time copper ions from the anode replace the now \"missing\" ions from the sulfate.\nSo at the beginning of the process only ions from the original copper sulfate are reduced and soon after the amount of ions originating from the anode increases continuously until they all the ions in the solution have been replaced.\n\n\n",
"1"
],
[
"\n\n> \n> \"Cathode is where the cations go\" to be reduced.\n> \n> \n> \n\n\nIn order to keep charge balance oxidation takes place at the anode.\n\n\nWith aqueous CuSO4 solution, if copper is not present in the anode, one can very conveniently use the reaction:\n$$\\ce{2 H2O -> 4 H+ + 4 e- + O2 gas - 1.23 V}$$\n\n\nWith an impure copper anode, copper and any metal with a less negative oxidation potential, such as iron, will go into solution preferentially:\n$$\\ce{Cu -> Cu++ + 2 e- - 0.34 V}$$\n$$\\ce{Fe -> Fe+++ + 3 e- + 0.04 V}$$\n$$\\ce{Al -> Al+++ + 3 e- + 1.66 V}$$\n\n\nThat goes with reduction at the cathode:\n\n\n$$\\ce{Cu++ + 2 e- -> Cu metal + 0.34 V}$$\n\n\nMetal drops out, O2 gas goes away (or other metals stay in solution), H+ replaces Cu ++, SO4 does not react.\n\n\nNote that, at the cathode, copper will \"go first\" because it has a higher positive reduction potential than H+:\n\n\n$$\\ce{2H+ aqueous + 2 e- -> H2 gas 0.00 V}$$\n\n\nas compared with + 0.34 V for Cu ++.\n\n\nNote: reactions are written in the order they proceed at anode and cathode. More positive V values signify order of reaction.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/169815/effect-of-coal-and-natural-gas-burning-on-particulate-matter-pollution
|
Effect of coal and natural gas burning on particulate matter pollution
|
I sometimes hear people talking about how we should replace coal burning plants with natural gas ones, to alleviate the case of particulate matter pollution. What exactly is the difference between coal fuel and natural gas that makes the latter seem "cleaner"?
| 3
|
[
[
"\n* At the same energy outcome, natural gas produces less carbon dioxide than coal. In a way, natural gas is half way between coal and hydrogen.\n* Coal produces smelly smoke, solid particles, sulfur dioxide and minor or trace heavy metal pollutants.\n* It is less known to common people, but power plants burning coal are more significant source of radioactive pollution than nuclear plants. This pollution is very diluted, but rather significant in absolute amount. Coal ash, used in past as a filler for some construction materials, has lead in some cases to significantly increased content of radium-226 in building walls. This radium is a product of long term decay of natural uranium. It further decays while producing radioactive gaseous radon-222, which is dangerous in long term inhalation because of lung cancer. As it stays in lungs as polonium-218 and its decay products.\n\n\nSee e.g. [Uranium produced from coal ash](https://atomicinsights.com/uranium-produced-from-coal-ash/)\n\n\n\n> \n> ... the uranium concentration in the ash pile is about 150-180 parts per million, about 1/4th of the concentration often thought of as commercially viable for ISL[In Situ Leaching] mining. However, coal ash piles have some physical characteristics that might help overcome that disadvantage since they may be easier to drill and it might be easier to protect the local groundwater from contamination. ...\n> \n> \n> \n\n\nSee [Radon in building materials](https://www.suro.cz/en/prirodnioz/building-materials) by Czech government agency for radiation protection.\n\n\n",
"9"
],
[
"\nDealing specifically with particulates would depend on the coal and burning conditions. You could look up a specific coal you are interested in; However generally it will contain a few % of each sulfur, iron, silica, alkali earths, and smaller amounts of things like nickel. The burning conditions make a great difference whether the contaminant become fly ash or slag and clinker in the furnace. As a boy in Chicago in the 50s, our apartment was heated with coal as were the great majority. That meant someone had to dig about a cubic foot of ash and clinker from under the furnace firebox each month. Something missed with gas furnaces. I am tying to say there is hardly a comparison of relative particulates produced by gas versus coal.\n\n\n",
"5"
],
[
"\nWhile the details depend strongly on the specific burning conditions, gas is almost always cleaner than coal and will produce far fewer particulates.\n\n\nThere are two main reasons for this. One is related to the level of non-carbon contaminants in the two fuels. Gas contains far fewer contaminants and most of them are gases many of which are not sources of pollution of any kind (gas is mostly methane but some gas has CO2, some has small levels of higher hydrocarbons like ethane). There is usually very little sulfur (if the original source has a lot it is usually stripped before distribution unless you count the very low levels of sulfur odorants added for safety reasons). Coal, even the best anthracite, can be thought of as mostly carbon but often contains several percent of a variety of contaminants. Some will be non-volatile minerals; some will be sulfur or nitrogen compounds and a variety of other non-carbon materials. These often burn to create particulate-forming compounds (sulfates often react to create particulates in the atmosphere).\n\n\nBut the second reason is the way even the pure carbon in the fuels burns. No burning reaction is \"perfect\" and side reactions leave stuff other than carbon dioxide (and water). Since burning the polymeric solid carbon in coal involves breaking up large chains or networks of carbon atoms, one of the common side reactions leaves incompletely burnt up lumps of carbon or partially reacted carbon. These form very small dust particles (in other words particulate pollution) and there is no perfect way to scrub them from the emissions. The combustion of methane is far cleaner as there are no long molecules of carbon that need to be broken up (though a very poorly controlled burning process *could* encourage some to form). So the expected level of imperfect reactions leading to particulate pollution will usually be far, far smaller when gas is the fuel.\n\n\nIn short, coal is a much dirtier fuel, even if all you are worried about is particulates. Add to this that, for a given amount of energy output, coal will produce a lot more CO2, and gas is always a better fuel for the environment.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/169814/i-tested-the-effect-of-temperature-on-the-ph-of-carbonated-water-but-the-results
|
I tested the effect of temperature on the pH of carbonated water but the results seem to go opposite to what was expected. Could anyone explain?
|
We carried out an experiment to investigate the effect of temperature on the acid carbonate equilibrium in carbonated water. We tested 5 different temperatures (room temp, 2 above and 2 below) by heating carbonated water in a closed test tube. After leaving the test tube for 10 minutes in the water bath, we opened it and quickly recorded the temperature and pH of the solution using probes.
We got the results that as temperature increases the pH decreased (so the solution got more acidic) and vice versa, however published results show that the pH should have increased for an increase in temperature. I understand this is because the increase in temperature decreases the solubility of carbon dioxide, decreasing the amount of carbonic acid and hydronium ions in solution and therefore increasing the pH.
The entire system should have shifted as per the equation:$2CO\_{2(g)}+3H\_2O\_{(l)}\rightleftharpoons CO\_{2(aq)}+2H\_3O^{+}\_{(aq)}+CO^{2-} \_{3(aq)}$
I originally thought that the opposite trend in my data was due to the endothermic nature of the ionisation of carbonic acid making the system overall endothermic, but published data indicates that the system is overall exothermic.
It would be great if anyone had any ideas about how to explain the opposite shift in equilibrium that our results show.
| 0
|
[
[
"\nThe experimental design is unfortunately not robust. Certainly, the pH of pure water decreases with temperature (slowly) but the biggest source of error is most like the pH probe. If you are using a pH-meter that does not have a built in temperature compensation, the pH readings are not reliable at any temperature except the one where it was calibrated. Secondly, pH meters operate in a limited temperature range and the pH calibration buffers are not meant for different temperature.\n\n\nSecondly, pH probes have a response time, i.e., it take a certain amount of time to produce a stable reading, usually a minute or less. Since you mentioned \"quickly recorded\", this is another source of error.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/169810/difficult-in-grasping-the-extent-limit-of-conformational-space-of-a-compound-un
|
Difficult in grasping "the extent/limit of conformational space of a compound until it change its stereochemistry"
|
I have heard that a conformational landscape encompasses all conformers that a compound has for *a specific stereoisomer*. I think it makes sense *verbally*, because if all conformers of a stereoisomer include all conformers for all stereoisomers and one (mistakenly) designates those all stereoisomer conformations to be conformers for a specific stereoisomer, it will require an interconversion (or change in permutational position) of one or multiple bonds to be broken to make a different stereoisomer.
While it seems to make *verbal* sense to me, my visual thinking is really against this. Searching on Google, "stereoisomer conformation" and "conformer" are always treated as two different realms having no connection in between.
I feel there should be an image (or someone has done a research article about this) showing a conformational landscape with a functional group bonded to a particular atom, showing differences in spatial 3D position. It suffices to show that "this coordinate position is still the same stereoisomer" but that if the functional group is extended, whether freely translated or rotated in drawing to somewhere, it will make a different stereoisomer and *it is convincing enough that its stereochemistry interconversion will have to break some bonds first.*
Any example would be fine, but I prefer the case of cis-trans conversion of a cyclohexane in its chair conformation with few functional groups, because my visualization is that *its functional group stereochemistry interconversion still does not need bond breaking (but why it would have a different stereoisomer though?).*
Thus, here is the general question. **What is the extent or limit of conformational space of a molecule (or specifically, a particular stereoisomer) that is spatially unique to it, but if further extended to all random positional permutations will change its stereochemistry so that it will have to break some bonds first to let that interconversion happen?**
| -1
|
[
[
"\nYou have the right idea that conformational space can be partitioned by configurational isomers. The clearest cases are where you have to break bonds to end up with another configurational isomer, but that is actually not always the case. Atropisomerism also gives rise to different configurational isomers, even though they are just separated by a bond rotation with a very high barrier.\n\n\nYour example with cyclohexane is actually a clear-cut case, since you really need to break bonds to switch from *cis* to *trans*. I visualized all possible chair conformations and the partitioning into configurational subspaces. There are [more conformers](https://chemistry.stackexchange.com/questions/32970/conformations-of-cyclohexane) that I left out for simplification.\n\n\n[](https://i.stack.imgur.com/aE4AA.png)\n\n\nIf you still don't see why you need bond breaking, I recommend playing around with a molecular model kit.\n\n\nIn summary, the partitioning of conformers into distinct configurational isomers happens when they are interconverted by a process with a high barrier (*i.e.* bond breaking, hindered rotations, polytopal rearrangements, *etc.*).\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/169809/hydroboration-oxidation-reaction-without-thf
|
Hydroboration–oxidation reaction without THF
|
The first step of [hydroboration–oxidation](https://en.wikipedia.org/wiki/Hydroboration%E2%80%93oxidation_reaction) usually involves $\ce{BH3}$ accompanied by THF. Is it possible for hydroboration to occur without the presence of THF? Borane is incredibly reactive. Wouldn't it just react by itself without being stabilized by THF?
| 1
|
[
[
"\nBorane dimerises to [diborane](https://en.wikipedia.org/wiki/Diborane) which does all the reactions of borane, including hydroborating alkenes. Diborane gas is commercially available if you wish to use it rather than stabilised borane, and have the equipment and skills to handle it. A number of other stabilisers are used for borane and are commercially available e.g. dimethyl sulfide, ammonia, triethylamine, pyridine.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/169805/are-mixtures-of-finely-ground-solid-fertilizer-compounds-stable
|
Are mixtures of finely ground solid fertilizer compounds stable? [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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If I had a combination of the following as finely ground powders, would it be relatively stable, or would they react to form other compounds?
Sodium Nitrate
Ammonium Nitrate
Calcium Nitrate
Potassium Nitrate
Magnesium Sulfate
Potassium Chloride
Ammonia Sulfate
Copper Sulfate
Zinc Sulfate
Ferrous Sulfate
Manganese Sulfate
Sodium Molybdate
Monoammonium Phosphate
Boric Acid
Sodium Hydroxide
(Some of course being hydrates)
| -2
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[
[
"\nAll depends on amounts, concentrations and ratios. It is a big difference between:\n\n\n* being solid/just being dissolved together\n* being together in diluted solution\n\n\nEven in cooking recipes, conditions of mixing ingredients do matter.\n\n\nReactions and precipitations often have concentration thresholds. Higher concentrations lead to more intense reactions or precipitations, undergoing in higher degree, or even to qualitatively new reactions.\n\n\n\n\n---\n\n\nAside of chemical stability, there is danger it would absorb air humidity and all the fine powder could form a single big block.\n\n\n\n\n---\n\n\nSodium hydroxide releases ammonia gas from all ammonium salts when they are all solid or being dissolved.\n\n\nWhen being dissolved:\n\n\n* Sodium hydroxide precipitates hydroxides of transition metals like copper, iron, zinc, manganese.\n* Sulfates precipitate calcium as calcium sulfate.\n* Phosphate can precipitate calcium and will precipitate copper, zinc, iron, manganese.\n* Molybdate may precipitate calcium, copper, zinc, iron, manganese.\n* Boric acid may precipitate calcium, copper, zinc, iron, manganese.\n\n\nGenerally, it is not good to mix together\n\n\n* alkaline compounds (or ones preferring alkaline conditions)\n\t+ sodium hydroxide, sodium molybdate\n* acidic compounds (or ones preferring acidic conditions)\n\t+ copper sulfate, zinc sulfate, ferrous sulfate, manganese sulfate, monoammonium phosphate, ammonia sulfate, ammonium nitrate\n\n\nIt may be advantageous to have\n\n\n* major nutritients ($\\ce{Ca}$,$\\ce{N}$, $\\ce{P}$, $\\ce{K}$) in solid state, and rather course than fine.\n* minor ones in a stock solution(s)\n\n\n\n\n---\n\n\nIf you insist on mixing them, then I would suggest to keep sodium hydroxide alone in its original state and keep the rest in separate compatible mixtures, in the sense of above.\n\n\n",
"3"
],
[
"\nLookup Galveston explosion or the Halifax explosion on the web. The city of Texas City or Halifax was destroyed, and I am not sure if the chemical was \"finely ground\". There are oxidizing and reducing agents in the mix, release of ammonia gas or precipitation of metal ions are your least concerns. Knowing the names of chemicals is not sufficient. You must know the properties and you have quite a list.\n\n\nIf you must mix these do so as a dilute solution paying particular attention to the order of mixing to prevent unwanted intermediate reactions and avoid cross contamination. Again learn the properties.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/169803/why-is-there-a-discrepancy-between-predicted-and-nominal-voltages-of-alkaline-ba
|
Why is there a discrepancy between predicted and nominal voltages of alkaline batteries?
|
According to [Wikipedia](https://en.wikipedia.org/wiki/Alkaline_battery), the sum of standard reduction potentials for the half reactions for alkaline battery yield $\pu{1.43 V}:$
$$\begin{align}
\ce{Zn(s) + 2 HO–(aq) &-> ZnO(s) + H2O(l) + 2 e–} &\quad E^\circ &= \pu{+1.28 V} \\
\ce{2 MnO2(s) + H2O(l) + 2 e– &-> Mn2O3(s) + 2 HO–(aq)} &\quad E^\circ &= \pu{+0.15 V} \\
\hline
\ce{Zn(s) + 2 MnO2(s) &-> ZnO(s) + Mn2O3(s)} &\quad E^\circ &= \pu{+1.43 V}
\end{align}$$
However, the reported voltage of Duracell or other alkaline batteries is $\pu{1.5 V},$ and the actual zero-load voltage of a new alkaline battery is measured up to $\pu{1.65 V}.$ Why?
| 0
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[] |
https://chemistry.stackexchange.com/questions/169799/why-is-a-protonated-carboxylic-acid-less-acidic-than-a-protonated-ester
|
Why is a protonated carboxylic acid less acidic than a protonated ester?
|
To my understanding, both have analogous resonance structures except that the ester has on -OR and the carboxylic acid has an -OH. Why is a positive charge on C-O-R more stable than on C-O-H?
| 2
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[] |
https://chemistry.stackexchange.com/questions/169798/why-does-benzaldehyde-turn-yellow
|
Why does benzaldehyde turn yellow?
|
We used benzaldehyde for a demonstration of the Cannizzaro reaction in school, but the liquid we used appeared visibly yellow, as opposed to the *colourless liquid* described by [Wikipedia](https://en.wikipedia.org/wiki/Benzaldehyde).
A few reasons I could think of include:
* Reaction with oxygen to form quinone-type compounds
* Polymerisation induced by exposure to light
* Impurities
I haven't found much discussion about this, nor do I have any candidate compounds for if my guesses above turn out to be right.
| 6
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[
[
"\n**Tl;DR:** All three factors are responsible for coloring of benzaldehyde (and pretty much any organics)\n\n\n\n\n---\n\n\nFor the first two factors, I would mention the paper1 linked by @Andrew. Benzaldehyde is prone to atmospheric oxidation; the reaction is accelerated in presence of light to form primarily benzoic acid mixed other aromatics and polymerized aromatics causing darkening.\n\n\nFor the impurity aspect, let me quote the following from a paper2 discussing about benzaldehyde purification process:\n\n\n\n> \n> [...] It is clear that traces of impurities, namely metals, influence markedly\n> the benzaldehyde activity. The active impurities present in pure benzaldehyde or\n> in that of anal, grade become visible after distillation leaving the distillation residue brown to yellow coloured. The criterion for the good quality of benzaldehyde is the colorless distillation residue.\n> \n> \n> \n\n\nMy previous discussions:\n\n\n1. [Phenoquinone vs benzoquinone](https://chemistry.stackexchange.com/questions/135877/phenoquinone-vs-benzoquinone/135878#135878)\n2. [Air- and light-sensitivity of phenol](https://chemistry.stackexchange.com/questions/78556/air-and-light-sensitivity-of-phenol/78563#78563)\n\n\n**References**\n\n\n1. THE ABSORPTION OF OXYGEN BY BENZALDEHYDE, H. J. Almquist and G. E. K. Branch\nJournal of the American Chemical Society **1932** 54 (6), 2293-2302\nDOI: [10.1021/ja01345a018](https://pubs.acs.org/doi/pdf/10.1021/ja01345a018)\n2. Benzaldehyde oxidation test, a model reaction with radical\nmechanism. II. The purification of benzaldehyde and stabilization of its activity by\nJ. GAŠPERÍK, Chem. zvesti 29 (6) 808-810 **1975** ([PDF](https://www.chemicalpapers.com/file_access.php?file=296a808.pdf))\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/169792/how-to-calculate-the-buffer-capacity-for-polyprotic-acids-specifically-h2po4
|
How to calculate the buffer capacity for polyprotic acids? specifically H2PO4 --- HPO42- + H30+
|
as the title says. Is b= amount of OH-/ H3O divided by volume of buffer x change in pH wrong to use with polyprotic acids? Can I calculate the buffering capacity of a sodium phosphate buffer made up of NaH2PO4 and NaH2PO4 using this formula?
| 0
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[] |
https://chemistry.stackexchange.com/questions/169789/how-to-differentiate-coenzyme-and-prosthetic-group
|
How to differentiate Coenzyme and Prosthetic group [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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[Improve this question](/posts/169789/edit)
Coenzyme is a non protein molecules which bind loosely with enzymes, its function is help catalytic reaction proper functioning.
Prosthetic group covalently and tightly bind with protein.As my understanding, I knew that the Heme group in hemoglobin is a good example explaining prosthetic group.
Here is my confusion:
1.Does Coenzyme and Prosthetic group both exist in Enzyme simultaneously?
2.Is there any example about Enzyme with prosthetic group?
3.Does Enzyme without Prothetic group affect catalytic reaction proceeding?
| -3
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[
[
"\nOne example is the enzyme methionine synthase. It has vitamin B12 as a prosthetic group and S-adenosyl methionine as a cofactor. Without vitamin B12 present, you would expect no catalysis (in fact, the protein might not even fold properly). For some reading with interactive 3D figures, see <https://proteopedia.org/wiki/index.php/Methionine_synthase>.\n\n\nAccording to IUPAC, these are the definitions of prosthetic group and cofactor:\n\n\n\n> \n> Prosthetic group: A tightly bound, specific nonpolypeptide unit in a protein determining and involved in its biological activity. See also cofactor.\n> \n> \n> \n\n\n\n> \n> Cofactor: An organic molecule or ion (usually a metal ion) that is required by an enzyme for its activity. It may be attached either loosely (coenzyme) or tightly (prosthetic group).\n> \n> \n> \n\n\nSource: [IUPAC Recommendations 1997](https://web.archive.org/web/20121128190330/http://www.chem.qmul.ac.uk/iupac/bioinorg/CD.html#34)\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/169788/if-liquids-absorb-heat-from-their-surroundings-to-become-gas-why-does-gas-feel
|
If liquids absorb heat from their surroundings to become gas, why does gas feel hotter than liquid?
|
I know what you're thinking, "OP, are you for real!? Gas is obviously hotter because it absorbed heat from the environment!"
But hear me out. Yes it is hotter in absolute terms because it absorbed heat to become that way and if you placed one thermometer in the liquid and one in the gas then you'll get a higher temperature reading from the gas.
However, when you read the question again it says "absorbs heat from its environment" thus heat is removed from the environment, thus the environment is less hot than before, and if that "environment" was the palm of our hand (and it was heavier than air so it didn't fly off) then that gas should feel cooler than the liquid because it absorbed heat from our skin thus the nerve endings in our skin had temperature removed from them and now give us the sensation of being cooled down. (assume that the boiling point of the substance is our body temperature).
| 0
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[] |
https://chemistry.stackexchange.com/questions/169786/nucleophilic-substitution-in-ethers-and-alcohols
|
Nucleophilic substitution in ethers and Alcohols
|
In Alcohols(secondary,tertiary,benzylic,allyic) on reaction with HX, X=(Cl,Br,I) SN1 takes place as H2O leaves and forms a carbonation.But in ethers on reaction with HX, X=(Cl,Br,I) (except tertiary,benzylic) SN2 takes place. Is ROH not as good leaving group as H2O.what is the reason for SN1 in Alcohols and SN2 in ethers??
| -2
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[
[
"\nFrom what I understand, the generalization you have constructed is not based on any gross rule that - \"ROH not as good leaving group as H2O\". I think what you mean is that the protonated alkoxy group is a poorer leaving group than the OH2+ group, in which case I would still not agree to a generalization unless I know the identity of the R group. There are many ethers that show SN1 substitution reactions as well, especially under acidic conditions. A good example is the highly strained oxirane ring, which opens via an SN1 reaction in low pH.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/169784/are-solid-fuels-with-calorific-values-10-000-kcal-kg-42-mj-kg-physically-po
|
Are solid fuels with calorific values >= 10,000 kcal/kg (42 MJ/kg) physically possible?
|
I have seen laboratory documentation from a reputable lab in southeast Asia that a company has achieved a solid fuel product made from landfill material with a high heating calorific value ($CV$) of approximately 10,000 kcal/kg (42 MJ/kg). The technology is referred to as Refuse Derived Fuel (RDF). From what I understand, such calorific values are only even possible with fluid fuels such as liquefied natural gas (etc.) which can get up to 12,000 kcal/kg.
Doing some calculations around the fixed carbon composition of landfill as similar to a question [I've asked previously](https://chemistry.stackexchange.com/questions/168959/how-can-combusted-methane-from-landfill-samples-be-quantified), I found that the fixed carbon ($FC$) content of the feedstock composition must equal approximately 130% to have the CV mentioned above. This number is obviously impossible but I'm very surprised that it was reported by this lab. Similar products usually get a CV of around 2500 - 5000 kcal/kg if they're lucky. There are several formulas to calculate CVs, the one I've used is below and notably only a function of the fixed carbon content to calculate the calorific value [kcal/kg], so perhaps my equation is too simplified?
$$CV = 0.196FC + 14.119$$
The company of course masks the idea saying it's their intellectual property (etc.) but I'm just curious of if this is even possible or should the laboratory's reputation be called into question?
| 3
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[
[
"\n**It is clearly possible for *fuels* to have higher calorific values per kg, even solid ones**\n\n\nCalorific values of fuels are well known and available from many sources. For example, [this source](https://www.forestresearch.gov.uk/tools-and-resources/fthr/biomass-energy-resources/reference-biomass/facts-figures/typical-calorific-values-of-fuels/) list the values of fuel oil as 42.5 MJ/kg (same value if you prefer GJ/tonne) and LPG as 46.3 MJ/kg so there doesn't appear to be a simple obvious limit for fuels. Obviously fuel oil is a liquid and LPG a mildly compressed gas under normal conditions.\n\n\nBut one of the reasons for their higher calorific values is that compounds with larger proportions of hydrogen tend to have higher values than ones with more carbon (which when mostly pure has about 33 MJ/kg).\n\n\nSo solid hydrocarbon compounds should be better. And they are with paraffin wax [coming in at about](https://en.wikipedia.org/wiki/Paraffin_wax) the 42 MJ/kg level.\n\n\nSo there is no good reason to expect even slightly higher values to be impossible. we already have examples fuels with that calorific capacity.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/169782/what-to-do-when-there-is-white-ppt-formation-even-after-adding-dil-sulphuric-ac
|
What to do when there is white ppt formation even after adding dil. sulphuric acid in the mixture of aluminium sulphate and water?
|
I heated the solution containing aluminium sulphate and water with 3 ml dil. sulphuric acid. But I didn't get a clear solution because the solution got milky in color due to the hydrolysis of aluminium sulphate to aluminium hydroxide.
I heated the solution to get a clear solution but couldn't succeed.
I tried to filter the solution but in the filtrate, I got the same milky filtrate.
What to do to get a clear solution?
| 0
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[] |
https://chemistry.stackexchange.com/questions/169781/derivation-of-the-hartree-fock-equation
|
Derivation of the Hartree-Fock equation
|
I have to do some tutoring about basic quantum chemistry and the main purpose is to give a deep insight into Hartree-Fock theory. My plan is to start from expressing the many-electron wavefunction as a Slater determinant and why this is even necessary, etc. From this starting point, one intermediate objective is to give an elaborate and crystal clear derivation of the Hartree-Fock equation and I used my old lecture notes and some textbooks to work out a detailed - and most importantly - comprehensive approach.
But I have to consider some constraints. **Firstly**, the majority of the students are likely to struggle with more sophisticated aspects like the formalism of second quantization (like used in the textbook by Hegaker, Jorgensen and Olsen). Things like functional variation or Lagrange mulitpliers shouldn't be a problem because they were covered in prequisited math lecture series. **Secondly** I realised that every textbook I have used for preparation is leaving out some so called "trivial" parts of the derivation that I need for my own understanding.
So far I have worked out the following key steps to arrive at the general form of the Hartree-Fock equation:
1. Establishing the first approach to the many-electron wave function (the Hartree product)
2. Considering the indistinguishibility of electrons and thus the antisymmetry of the wavefunction with the regard of the exchange of two particles
3. Setting up the Slater determinant to come up for the antisymmetry
4. Deriving the expression for the energy of a Slater determinant (keeping in mind the mean-field approximation)
5. Minimization of the energy with the constraint that the spinorbitals remain orthonormal
I am struggling to formulate the last part as detailed and clear as possible. The best approach I found so far is in *Introduction to Computational Chemistry* by Jensen, but even in this textbook, there are some mathematical "leaps" I don't get.
So the **essence of my question** is: Do you know any resource (textbook or lecture notes, etc.) in which this last step (minimization with the constraint of orthonormality yielding the general form of the Hartree-Fock equation) is explained very detailed and clear ?
Any help is heavily appreciated.
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https://chemistry.stackexchange.com/questions/169780/why-does-glass-etching-cream-contain-sulfuric-acid
|
Why does glass etching cream contain sulfuric acid?
|
The [MSDS](https://www.glassetchingsecrets.com/wp-content/uploads/Armour-etch-SDS-2015.pdf) for various glass etching creams all show that they're primarily composed of 4 substances (the composition percentages vary):
* Ammonium Bifluoride - 20 - 40%
* Sodium Bifluoride - 5 - 10%
* Sulphuric Acid - 1 - 5%
* Barium Sulfate - 1 - 5%
The purpose of the fluoride ingredients is obvious. They decompose into hydrofluoric acid (presumably when reacting with water in the air) and do the actual etching. But why is there Sulphuric Acid? It doesn't react with glass at all.
My speculation would be that perhaps it's a strong desiccant and it serves to slow the reaction?
| 2
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https://chemistry.stackexchange.com/questions/169776/is-this-sulphur
|
Is this sulphur? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
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Closed 9 months ago.
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I was electroplating with a copper sulfate electrolyte, and now a yellow substance has settled at the bottom of the container, is it sulphur from the copper sulfate or something else from the screw I electroplated?
If it is sulphur, how can I safely remove it?
Thanks for any help you can give me.
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[
[
"\nThe yellow precipitate is probably cuprous oxide $\\ce{Cu2O}$ which is insoluble, and which is produced in the electrolysis of $\\ce{CuSO4}$if the solution is not acidic enough. Check by adding some acidic solution, like concentrated HCl. If the yellow precipitate is sulfur it will not be attacked by the acid. If it is $\\ce{Cu2O}$, il will soon be dissolved and produce a colorless solution, due to : $$\\ce{Cu2O + 4HCl ⟶ H2CuCl2 + H2O}$$\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/169775/difference-between-pre-equilibrium-and-steady-state-approximation
|
Difference between pre-equilibrium and steady-state approximation?
|
Both are reaction mechanisms in which there are several steps and at least one intermediate present. However, I am confused about the differences of both methods, especially when using them to determine rate laws. Also, in which types of reactions can we apply each of the methods?
| 1
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https://chemistry.stackexchange.com/questions/169773/making-a-reagent-preparation-simpler-that-requires-ph-adjustment
|
Making a reagent preparation simpler that requires pH adjustment
|
I have a basic question on a reagent (solution) preparation. Here are the steps that I take to prepare the reagent:
1. Weigh the chemical (solid form, ammonium bicarbonate) and dissolve with DW to make 100 mM solution. The weight (or volume needed to make 100 mM) can vary depending on the # of the samples to be analyzed
2. Calibrate the pH meter and adjust the pH of the above solution to a certain pH
I would like to make my life easier by not using the pH meter (gets tedious having to wait for the pH calibration and whatnot). I do know that theoretically, if I make the weight of the chemical the same each time I prepare the solution, I can add the same amount of base to set the pH. But is it correct that the volume of base added changes depending on the amount of chemical weighed?
I don't remember my chemistry very well, but from my memory, pH is concentration dependent so I was wondering if adding the same volume of base to a solution of desired concentration (but may have different volumes as I may weigh the chemical differently) is reproducible.
So basically, I want to simply weigh the chemical, dissolve with required volume of DW to make 100 mM concentration, and then add the same volume of base without having to use my pH meter, which will save a lot of time in the morning.
| 1
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https://chemistry.stackexchange.com/questions/169767/show-that-if-h-is-hermitian-then-%ce%a8h%ce%a6-h%ce%a8%ce%a6
|
Show that if H is hermitian, then <Ψ|H|Φ> = <HΨ|Φ>
|
Since H is hermitian, it is equal to its complex conjugate (shown below), a property which I used to do the proof:
[](https://i.stack.imgur.com/409uF.png)
I had to prove that <Ψ|H|Φ> = <HΨ|Φ>, which I did as below:
[](https://i.stack.imgur.com/mGOgM.png)
My question is whether my proof is valid - are the two things equal because the pink bits equal each other?
| 1
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[
[
"\nAs the operator $H$ is linear, when taking it out of the bra symbol it must be replaced by its adjoint and placed on the right, i.e. $\\langle H\\psi|=\\langle \\psi|H^\\dagger$ and so $\\langle H^\\dagger\\psi|=\\langle \\psi|(H^\\dagger)^\\dagger=\\langle\\psi| H$. In you case the operator is hermitian so $H=H^\\dagger$\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/169762/can-the-overall-order-of-a-reaction-ever-be-strictly-less-than-1
|
Can the overall order of a reaction ever be strictly less than $1$?
|
In A-level chemistry we are taught that:
>
> To every reaction: $$\sum\_{i=1}^n c\_iA\_i\to\sum\_{j=1}^m d\_jB\_j$$There is a rate equation: $$r=\kappa[A\_1]^{\alpha\_1}[A\_2]^{\alpha\_2}\cdots[A\_n]^{\alpha\_n}$$With the $(\alpha\_i)$ called the partial orders and $\sum\_{i=1}^n\alpha\_i$ the overall order, $\kappa$ some constant.
>
>
>
Since the case of multiple reactants is confusing (e.g. I have not been able to get a clear answer of how 'rate' is defined in general) I'll stick to a reaction with only one reactant, $A\to bB+cC+dD+\cdots$.
We can then explicitly solve the differential equations, for $[A]\_t$ treated as a function in time $t\ge0$, with $[A]\_0>0$: $$\frac{\mathrm{d}[A]\_t}{\mathrm{d}t}=\kappa\cdot[A]\_t^\alpha\implies\kappa t+C=\begin{cases}\frac{1}{1-\alpha}[A]\_t^{1-\alpha}&\alpha\neq0\\\ln[A]\_t&\alpha=0\end{cases}$$After some shuffling, you get: $$[A]\_t=\begin{cases}\{(1-\alpha)(\kappa\cdot t+C)\}^{\frac{1}{1-\alpha}}&\alpha\neq1\\Ce^{\kappa t}&\alpha=1\end{cases}$$Where $C$ is some constant in both cases. If $\alpha=1$, we get $C=[A]\_t$: otherwise, $C=\frac{1}{1-\alpha}[A]\_0^{1-\alpha}$.
I played with this a little, plotting concentration-time reaction curves on Desmos, and I quickly noticed a significant difference between the cases $\alpha<1$ and $\alpha\ge1$: if the order $\alpha$ is less than $1$, the model predicts the reaction terminates in finite time, at $t=-C/\kappa$. Else, the reaction never terminates!
Intuitively, I don't think a chemical reaction can ever *fully* terminate, so I'm inclined to believe the following conjecture:
>
> In a reaction with only one reactant, the order of reaction must be greater than or equal to $1$.
>
>
>
More generally, I suppose:
>
> In any reaction for which the rate equation holds, the overall order must be greater than or equal to $1$.
>
>
>
Does that make sense? I imagine I'm quite wrong, since I'm doing this on the back of relatively little chemical knowledge. In particular, it's hard to picture what solving the differential equations looks like for multiple reactants.
| 1
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https://chemistry.stackexchange.com/questions/169760/why-concurrent-reading-instead-of-mean-in-titration
|
Why concurrent reading instead of mean in titration? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
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While doing titration we take concurrent reading like.
Reading..........Concurrent reading
1. 3.6
2. 3.4=====> 3.4
3. 3.4
Why don't we take the mean value as we do in physics?
| -1
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[
[
"\nWe usually refer to the *[median](https://mathworld.wolfram.com/StatisticalMedian.html)* rather than the mean. In the case shown, when you have two readings that are identical and one that's different, the median is the reading with two values (here, 3.4) rather than averaging all three (here, that would be 3.5 to two significant figures).\n\n\nThe trade-off goes like this: when we are sure that all the data are normally distributed in one population, the mean represents the best estimate of the true average value. But in many types of experiments, including titration, we have to guard against being influenced by a sample where we went wrong with the procedure or there was some contamination in the sample we couldn't see beforehand. The median is less sensitive to such abnormality in the experiment, and the resulting non-normality in the statistics, than the mean. This is in exchange for being a less optimal estimate if we could be sure everything were perfect/normal.\n\n\n",
"4"
],
[
"\nFirst! A buret reading should be to the estimated 0.01 mL, the first Insignificant figure not to the gradations of 0.1 mL. One titration is one data point. If you want to check your overall precision the procedure is to run complete replicate samples on the one sample or [even on simultaneous replicate samples to check on the sampling technique]. Since only 2 or 3 replicates are usually run, when readings are to the first insignificant figure a simple average is sufficient and the range is a good approximation to the error. In your case the buret should have been read to 3.61 3.43 3.46 [making up last digit] average is ~3.50 error is ~0.15; ~5% relative error quite unacceptable in titrations. To apply correct statistics more data points are needed. Data such as this simply means that your technique needs serious improvement. Once you have the technique down pat such data could be meaningful and there is something going on with your sample. This is why I strongly advise that standard solutions be standardized by the user. First to ensure the source is competent and second to ensure one's technique and equipment are up to par on a known reaction.\n\n\nI worked as an applications chemist for a instrument manufacturer and was dismayed on finding out how many of the labs I visited were working with uncalibrated or improperly calibrated instruments.\n\n\nConclusion: run 2-3 complete replicates. Read the buret to the first estimated digit. If the data has no variation or if the variation is in the significant digit or relative error is greater than 1%, something is wrong.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/169752/what-does-acidic-mean-in-an-article-on-hydrogen-boride-nanosheets
|
What does "acidic" mean in an article on hydrogen boride nanosheets?
|
From Rojas et al. [[1](https://doi.org/10.1038/s43246-021-00184-5)] (emphasis mine):
>
> On the other hand, we found that proton exchange with HB occurs in water with the estimated $\mathrm{p}K\_\mathrm{a}$ of $3.5\pm 0.2,$ even after reactive sites for the hydrolysis reaction are terminated. Thus, for the practical application, one should keep in mind that HB sheets have a chemical stability against hydrolysis reaction, but become *acidic* in the presence of water
>
>
>
Does “*acidic*” here refer to the solution? Or acidic sheets that are gaining protons?
### Reference
1. Rojas, K. I. M.; Cuong, N. T.; Nishino, H.; Ishibiki, R.; Ito, S.; Miyauchi, M.; Fujimoto, Y.; Tominaka, S.; Okada, S.; Hosono, H.; Arboleda, N. B.; Kondo, T.; Morikawa, Y.; Hamada, I. Chemical Stability of Hydrogen Boride Nanosheets in Water. *Commun. Mater.* **2021**, 2 (1), 1–8. DOI: [10.1038/s43246-021-00184-5](https://doi.org/10.1038/s43246-021-00184-5). (Open Access)
| 4
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https://chemistry.stackexchange.com/questions/169750/why-do-vessels-get-a-better-shine-when-tea-leaves-are-soaked-overnight-in-it-bef
|
Why do vessels get a better shine when tea-leaves are soaked overnight in it before washing?
|
The other day I dumped consumed tea leaves into a brass utensil because I needed the tea-pan urgently. The vessel with the tea leaves didn't get around for cleaning until the next day. Ergo, the tea leaves soaked overnight in dirty water. When finally the vessel was cleaned (as in hand-washed with utensil cleaner), the quality of shine was exceptional.
Piqued, I repeated the tea-soaking in another vessel — and was rewarded with the same high gleam. This vessel was made of aluminium.
Why do vessels get a better shine when tea-leaves are soaked overnight in it before washing?
| 4
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[] |
https://chemistry.stackexchange.com/questions/169742/if-i-have-a-mixture-of-potassium-hydroxide-and-sodium-hypochlorite-what-will-be
|
If I have a mixture of potassium hydroxide and sodium hypochlorite, what will be the result of adding citric acid or ascorbic acid? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed 9 months ago.
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I have a cleaning chemical that is $\pu{10\%}$ potassium hydroxide ($\ce{KOH}$}) and $\pu{5\%}$ sodium hypochlorite ($\ce{NaClO}$). What is the result if I add citric acid ($\ce{C6H8O7}$) or ascorbic acid ($\ce{C6H8O6}$)? My goal is to neutralize the high pH of the potassium hydroxide without creating dangerous byproducts. I do not have a background in chemistry. How do I determine the result of the mass balance equation?
$\ce{KOH + NaClO + C6H8O7 -> ?}$
$\ce{KOH + NaClO + C6H8O6 -> ?}$
| -3
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[
[
"\nThe reason hypochlorites are kept alkaline is to prevent evolution of chlorine gas, $\\ce{Cl2}$.\n\n\n* Any attempt to make the cleaners less alkaline will release poisonous $\\ce{Cl2}$, *unless*\n* A reactant combines with the chlorine, preventing its evolution... thus making the cleaner useless as a bleach and disinfectant, *or*\n* The worst outcome: the reactants eliminate the chlorine, but produce even *more toxic and volatile products*. For example, the reaction of hypochlorites and household (aqueous) ammonia produces hazardous chloramine, or even hydrazine. See [Science ABC](https://www.scienceabc.com/pure-sciences/what-happens-when-you-mix-bleach-and-ammonia.html) for more detail on that reaction.\n\n\n",
"4"
]
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https://chemistry.stackexchange.com/questions/169741/which-is-a-better-poison-to-the-habers-process-ceco-or-ceh2s
|
Which is a better poison to the Haber's process, $\ce{CO}$ or $\ce{H2S}$? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
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I am studying about catalysts and their promoters and poisons and came across two of the possible poisons for Fe used during Haber's process and wanted to compare their effect on the rates of reaction. Which of them would lower the rate of reaction of Haber's process more, $\ce{CO}$ or $\ce{H2S}$?
| 2
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https://chemistry.stackexchange.com/questions/169734/why-is-hcl-a-stronger-acid-than-h2s-on-the-basis-of-electronegativity-but-hf-is
|
Why is HCl a stronger acid than H2S on the basis of electronegativity, but HF is a weaker acid than HCl? [duplicate]
|
**This question already has answers here**:
[Why can the acidic strength between H2S and HCl be explained on the basis of electronegativity but H2S and H2O can't?](/questions/107516/why-can-the-acidic-strength-between-h2s-and-hcl-be-explained-on-the-basis-of-ele)
(3 answers)
Closed 9 months ago.
We know that $\ce{HCl}$ is a stronger acid than $\ce{H2S}$. This can be explained due to that fact that chlorine is more electronegative than sulfur, which means the hydrogen ion has a much easier time leaving when bonded to chlorine than sulfur. However, when looking at $\ce{HCl}$ and $\ce{HF}$, $\ce{HCl}$ is actually a stronger acid, even though $\ce{HF}$ is more electronegative and polar. Is there is flaw in my reasoning?
| -2
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[
[
"\nThere are several factors involved in acidity in water. The most important is probably the X-H bond strength. This effectively explains the increasing acidity from HF to HI The bonds become weaker. It also explains the increasing acidity from H2O to H2S, H2Se and H2Te. A second factor is solvation of the anions. There probably is a sweet spot here for the surrounding water molecules to disperse the charge and minimize crowding. This might also involve the shape of the ion. Another consideration is the ability to hydrogen bond. The hydrogen bond will give a reprotonation reaction more opportunity and should lower the acidity of HF, H2O and possibly a bit for HCl.\n\n\nAll the factors should be considered in making comparisons especially when comparing between groups. Also the extent of chemical reactions is governed by the Second Law energetics\n\n\n",
"1"
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https://chemistry.stackexchange.com/questions/169733/point-group-of-harmonic-oscillator
|
Point group of harmonic oscillator
|
The book *Molecular Quantum Mechanics* by Atkins and Friedman [1] says the point group of a harmonic oscillator is $C\_\mathrm{s},$ composed by the identity operator $E$ and a reflection $\sigma\_\mathrm{h}$.
But I don't understand why the rotation of $\pu{{\pi} rad}$ about the $y$ axis is not a symmetry operation in this case. It looks like this operation also leaves the hamiltonian invariant, since $V(x) = V(-x)$ for a harmonic oscillator. So, why can't the point group in this case be $C\_2,$ composed by $E$ and a rotation $C\_2?$
### Reference
1. Atkins, P. W.; Friedman, R. *Molecular Quantum Mechanics*, 5th ed.; Oxford University Press: Oxford ; New York, **2011**. ISBN 978-0-19-954142-3.
| 7
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[
[
"\nMaybe, go one step further and consider what the difference between $C\\_\\mathrm{s}$ and $C\\_2$ really is?\n\n\nHere are the [character tables](https://chemistry.meta.stackexchange.com/questions/3435/group-theory-tables) for both point groups:\n\n\n$$\\begin{array}{c|cc|cc}\n\\hline\nC\\_\\mathrm{s} (= C\\_\\mathrm{h}) & E & \\sigma\\_\\mathrm{h} & & \\\\ \\hline\n\\mathrm{A'} & 1 & 1 & x,y, R\\_z & x^2, y^2, z^2, xy \\\\\n\\mathrm{A''} & 1 & -1 & z, R\\_x, R\\_y & xz, yz \\\\\n\\hline\n\\end{array}$$\n\n\n$$\\begin{array}{c|cc|cc}\\hline\nC\\_2 & E & C\\_2 & & \\\\ \\hline\n\\mathrm{A} & 1 & 1 & z, R\\_z & x^2, y^2, z^2, xy \\\\\n\\mathrm{B} & 1 & -1 & x, y, R\\_x, R\\_y & xz, yz \\\\ \\hline\n\\end{array}$$\n\n\nThe terms to the right describe rotations and translations in 3D molecules, and the harmonic oscillator isn't a 3D molecule, so these are irrelevant. Once we strip that away, we're left with the conclusion that *except for the choice of labels*, $C\\_\\mathrm{s}$ and $C\\_2$ are really equivalent.\nIn group theory terminology they are [isomorphic](https://en.wikipedia.org/wiki/Group_isomorphism).\n\n\n* There are two symmetry operations, the identity operation $(E)$ and another one $(C\\_2$ or $\\sigma\\_\\mathrm{h})$\n* There is one irreducible representation which is totally symmetric $(\\mathrm{A'}$ or $\\mathrm{A})$\n* There is one irreducible representation which is asymmetric with respect to the non-identity operation $(\\mathrm{A''}$ or $\\mathrm{B})$\n\n\nSure, the labels are not the same, but we can only draw a clear distinction between the labels in 3D anyway. In the case of the harmonic oscillator there is only one non-trivial symmetry element and that doesn't change whether we call it $\\sigma\\_\\mathrm{h}$ or $C\\_2$. (This echoes the point in [Hans Wurst's comment](https://chemistry.stackexchange.com/questions/169733/point-group-of-harmonic-oscillator#comment355785_169733).)\n\n\nSo, it's entirely up to you whether you want to call the point group $C\\_\\mathrm{s}$ or $C\\_2$. The maths works out the same in the end — as it *has* to.\n\n\n",
"4"
],
[
"\nAs @Hans Wurst already mentioned, the authors are referring to 1D simple harmonic oscillator and the symmetry of the potential and wavefunction.\n\n\nAs a chemist the first think that comes to our mind when we think of a 1D harmonic oscillator is a diatomic molecule $(C\\_{\\infty\\mathrm v}$ or $D\\_{\\infty\\mathrm h}$. But it can also describe an atom adsorbed to a (*flat*) surface $(C\\_\\mathrm{s}$ symmetry?).\n\n\nHowever, I think the authors are referring to the symmetry of the harmonic oscillators potential and wavefunctions irrespective of the system to which they are applied. In this sense, you only can define two elements of symmetry, the identity and the symmetry around the origin, a mirror plane, thus you could say the oscillator belongs to the $C\\_\\mathrm{s}$ symmetry.\n\n\nThis is what I understand from the book [1, p. 158]:\n\n\n\n> \n> ### 5.17 Symmetry and degeneracy\n> \n> \n> We have already mentioned (in Section 2.13) that the presence of degeneracy is a consequence of the symmetry of a system. We are now in a position to discuss this relation.\n> To do so, we note that the hamiltonian of a system must be invariant\n> under every operation of the relevant point group:\n> \n> \n> $$(RH) = H \\tag{5.40}$$\n> \n> \n> A qualitative interpretation of eqn 5.40 is that the hamiltonian is the operator for the energy, and energy does not change under a symmetry operation.\n> An example is the hamiltonian for the harmonic oscillator: the kinetic energy operator is proportional to $\\mathrm d^2/\\mathrm dx^2$ and the potential energy operator is proportional to $x^2$.\n> Both terms are invariant under the replacement of $x$ by $−x,$ and so the hamiltonian spans the totally symmetric irreducible representation of the point group $C\\_\\mathrm{s}$.\n> Because $H$ is invariant under a similarity transformation of the group (that is, any symmetry operation leaves it unchanged), we can write\n> \n> \n> $$RHR^{−1} = H$$\n> \n> \n> Multiplication from the right by $R$ gives $RH = HR$, so we can conclude that symmetry operations must commute with the hamiltonian.\n> \n> \n> \n\n\n### Reference\n\n\n1. Atkins, P. W.; Friedman, R. *Molecular Quantum Mechanics*, 5th ed.; Oxford University Press: Oxford ; New York, **2011**. ISBN 978-0-19-954142-3.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/169729/solvay-process-and-its-relationship-with-sodium-chloride-and-calcium-carbonate
|
Solvay process and its relationship with sodium chloride and calcium carbonate
|
I’m studying some chemical engineering process and my teacher made a question in class that nobody knew how to answer it and then she asked to made a search after class. However, I don’t even found nothing related.
$$\ce{2 NaCl + CaCO3 -> Na2CO3 + CaCl2 (do not occur spontaneously)}$$
Do you know why $\ce{NaCl + CaCO3}$ reaction do not occur in a spontaneous form in a process? I know that in Solvay Process we need to produce $\ce{CO2}$ from $\ce{CaCO3}$ from heating, but I really don’t know why we couldn’t do this reaction right away.
| 0
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[
[
"\n\n> \n> Calcium Carbonate $\\ce{CaCO3}$ is almost insoluble\n> \n> \n> \n\n\nHere a little digging is required as $\\ce{NaCl + CaCO3 -> Na2CO3 + CaCl2}$ of the\n[Solvay Process](https://www.chemeurope.com/en/encyclopedia/Solvay_process.html) leaves out a few steps.\n\n\nThe primary driver of the process is that sodium bicarbonate is less soluble in $\\ce{NaCl}$ brine than $\\ce{NH4Cl}$, and the starting materials are very inexpensive salt brine and limestone. The product is widely used \"soda ash\" $\\ce{Na2CO3}$.\n\n\nThe more detailed steps are as follows:\n\n\n$$\\ce{CaCO3 ->[\\Delta] CaO (s) + CO2(g)}$$\n\n\n$$\\ce{NaCl(brine) + NH3 + CO2 -> NaHCO3 \\downarrow + NH4Cl (aq)}$$\n\n\nAmmonia is recovered using highly basic $\\ce{CaO}$ as:\n\n\n$$\\ce{2NH4Cl + CaO -> 2 NH3(g) + CaCl2 + H2O}$$\n\n\nBicarbonate is calcined to soda ash as:\n\n\n$$\\ce{2 NaHCO3 ->[\\Delta] Na2CO3 (s) + CO2 + H2O}$$\n\n\nAmmonia and $\\ce{CO2}$ are thus recovered and recycled. By product $\\ce{CaCl2}$ is road salt.\n\n\nProduction and isolation of desired products by *phase change* is highlighted in these reactions: $\\ce{limestone -> CaO + CO2 (reactant)}$, $\\ce{NaHCO3}$ by precipitation, recovery of ammonia gas, and conversion of bicarbonate to solid soda ash by *removing* $\\ce{CO2}$ and $\\ce{H2O}$ as gasses.\n\n\n",
"1"
],
[
"\nThe proposed reaction $\\ce{2NaCl + CaCO3 -> Na2CO3 + CaCl2}$ does not occur, and cannot occur by simple mixing in solution. On the contrary the reaction will be reversed, because $\\ce{CaCO3}$ is the only insoluble product in the reaction. Let me explain the reason why.\n\n\nIf you mix two solutions, one of $\\ce{Na2CO3}$ and one of $\\ce{CaCl2}$, both solutions contain ions, namely $\\ce{Na^+, CO3^{2-}, Ca^{2+}, Cl^-}$, and nothing can prevent the following reaction to occur :$$\\ce{Ca^{2+} + CO3^{2-} -> CaCO3(s)}$$ So the substances formed on the right-hand-side of the proposed equation cannot occur simultaneously in an aqueous solution. If they were present simultaneously in a solution, their ions would interact. $\\ce{CaCO3}$ will precipitate and the proposed reaction will occur in the reverse direction $$\\ce{Na2CO3 + CaCl2 -> 2 NaCl + CaCO3(s)}$$\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/169725/can-we-use-potassium-hydroxide-to-decarboxylate-potassium-benzoate-to-benzene
|
Can we use potassium hydroxide to decarboxylate potassium benzoate to benzene?
|
I know the famous reaction:
[](https://i.stack.imgur.com/hN58b.png)
I was thinking of extension of this reaction to other salts of Benzoic Acids such as:
$$\ce{Potassium Benzoate + KOH ->[CaO,Δ] Benzene + K2CO3}$$
Is this reaction really carried out in Lab or there is not a favorable outcome/different product/unfeasible reaction such as in case of Kolbe's Reaction we obtain different products(basically isomers) for Na containing compound and for K containing compound.
| 0
|
[
[
"\nThe reaction is famously known as soda-lime decarboxylation because of its use of [sodalime](https://en.wikipedia.org/wiki/Soda_lime) which is mixture of $\\ce{CaO}$ (75%) and $\\ce{NaOH}$ (3%). It also contains trace amount of $\\ce{KOH}$ (0.1%). You can however alter the composition where $\\ce{KOH}$ is in majority while $\\ce{NaOH}$ is in trace and the reaction will work in similar manner since the chemistry is not altered.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/169724/why-does-copperii-sulfate-react-with-potassium-iodide-in-aqueous-solution
|
Why does copper(II) sulfate react with potassium iodide in aqueous solution?
|
According to the list of standard reduction potentials [1, p. 5-79] $E^\circ(\ce{Cu^2+/Cu^+}) = \pu{+0.153 V},$ while $E^\circ(\ce{I2/I^-}) = \pu{+0.5355 V}.$ Doesn't it mean that iodine has more tendency to get electron and form $\ce{I-}$? If yes, then why does the reaction between $\ce{CuSO4}$ and $\ce{KI}$ occur?
### Reference
1. Haynes, W. M.; Lide, D. R.; Bruno, T. J. *CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data*, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, **2016**. ISBN 978-1-4987-5429-3.
| 2
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[
[
"\nThis is literally a complex reaction! If I remember correctly when these are mixed there is an immediate color change that I attributed to a fast complex formation. Someone must check this out. I think I remember this.\n\n\n$\\ce{Cu^{2+} + I- <=> CuI+}$ ; \n\n$\\ce{CuI+ + e- <=> CuI}$\n\n\nCombine these two reactions to give:\n\n\n$\\ce{Cu^{2+} +I- + e- <=> CuI } ; ~~~~ \\pu{ E^\\circ = +0.877 V}$ [Langes Handbook]\n\n\nWe have our oxidant!\n\n\n$\\ce{I- <=> 1/2 I2 + e-} ; ~~~~ \\pu{ E^\\circ = -0.535 V}$\n\n\nCombine these two reactions to give:\n\n\n$\\ce{Cu^{2+} +2 I- <=> CuI + 1/2 I2 } ; ~~~~ \\pu{ E^\\circ = = +0.342 V}$\n\n\nThis predicts a mixed product. I can't find a way to convert the $\\ce{CuI}$ to $\\ce{Cu}$ and $\\ce{I2}$.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/169720/calculating-acid-volumes-needed
|
Calculating acid volumes needed
|
I collect water quality samples used for irrigation. The samples are sent to a lab to measure $\ce{Ca^{2+}, Mg^{2+}, CO3^{2-}, HCO3^-}$, pH, and total dissolved solids.
I need to use acid to prevent scaling on my water emitters, so I want to calculate, based on the results of each sample, how much HCl I need to add to lower the pH to 6 and to 4.
I understand that the target pH gives the amount of H+ but how do I add to that the buffer capacity of alkalinity?
Thanks
| 0
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https://chemistry.stackexchange.com/questions/169718/does-addition-of-solute-affect-the-rate-of-condensation
|
Does addition of solute affect the rate of condensation?
|
I know that addition of a non-volatile solute in a pure solvent effects the rate of evaporation as the number of molecules leaving the liquid bulk per unit time decreases. However when the rate of evaporation decreases immediately , rate of condensation should automatically increase , this is because, if we consider a liquid 'X' in equilibrium as:
$$\ce{X(l) <=> X(g)}$$
Now when some amount of solute is added in the pure $\ce{X}$ , this dynamic equilibrium is obviously disturbed. $\mathrm{(Rate)\_{evap}} \downarrow$ so to attain equilibrium $\mathrm{(Rate)\_{cond}} \uparrow$ as reaction shifts backward.
also when equilibrium is attained the number of molecules entering a liquid has decreased , so at equilibrium rate of condensation has decreased. So, overall it has been affected.
But when it browsed various sites , all agreed that it is unchanged by solute.
So, where's the problem? Isn't my logic correct?
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[
[
"\nWhen a solvent is at isobaric conditions in equilibrium with its gaseous phase, the rate of evaporation and condensation are equal.\n\n\nIf we add a solute to the solvent, the activity, chemical potential, saturated vapor pressure and rate of evaporation decreases. As the rate of condensation remains the same, the **net** condensation increases from its equilibrium zero value.\n\n\nIf the system is isolated, then while there is ongoing condensation, temperature raises, saturated vapor pressure increases (due both increasing temperature and diluting by the condensate), while pressure of vapor decreases, until there is reached the new equilibrium with higher temperature but lower gas(vapor) pressure. This presure is now equal to saturated vapor pressure for the final solution and temperature.\n\n\nIf the system is also isothermic, then the heat from ongoing condensation is dissipated. The saturated vapor pressure and the gas (vapor) pressure at the new equilibrium are lower, compared to pure solvent, according to the Raoult law and eventual its deviations.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/169715/how-are-fundamental-equations-valid-for-both-reversible-and-irreversible-process
|
How are fundamental equations valid for both reversible and irreversible processes?
|
We know for an irreversible process, $\mathrm dS\gt\mathrm dq/T$.
And if the process is done at constant pressure we can take the equation as $\mathrm dH-T\,\mathrm dS\lt0$.
And we defined Gibbs energy, $G=H-TS$. At constant temperature and pressure $\mathrm dG\le0$.
But the fundamental equation of Gibbs energy $\mathrm dG$, in terms of temperature and pressure is given by $\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$.
And as per our original conditions, i.e. at constant pressure and temperature for an irreversible process the value $\mathrm dG$ should be less than zero.
I cannot understand for the same condition the two equations give different answers.
| 0
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[
[
"\nSo, let's look how much is the equation fundamental, using as the starting point the definition of the Gibbs free energy:\n$\\require{cancel}$\n\\begin{align}\nG &= H - TS\\\\\nG &= U + pV - TS\\\\\n\\mathrm{d}G &= (\\delta Q + \\delta W) + (p\\mathrm{d}V +V\\mathrm{d}p) - (T\\mathrm{d}S + S\\mathrm{d}T)\\\\\n\\mathrm{d}G &= (\\delta Q -\\cancel{p\\mathrm{d}V} + (\\delta W\\_\\mathrm{nonV})) + (\\cancel{p\\mathrm{d}V} +V\\mathrm{d}p) - (T\\mathrm{d}S + S\\mathrm{d}T)\\\\\n\\mathrm{d}G &= \\cancel{\\delta Q} + (\\delta W\\_\\mathrm{nonV}) + V\\mathrm{d}p - \\cancel{T\\mathrm{d}S} - S\\mathrm{d}T\n\\end{align}\n\n\nFor reversible processes is $\\delta Q=T\\mathrm{d}S$, therefore\n$$\\mathrm{d}G= V\\mathrm{d}p - S\\mathrm{d}T \\ (+ \\delta W\\_\\mathrm{nonV}) $$\n\n\nFor irreversible processes is $\\delta Q \\lt T\\mathrm{d}S$, therefore\n\n\n$$\\mathrm{d}G < V\\mathrm{d}p - S\\mathrm{d}T \\ (+ \\delta W\\_\\mathrm{nonV})$$\n\n\nand for $T$,$p$ constant\n\n\n$$\\mathrm{d}G < 0 \\ (+ \\delta W\\_\\mathrm{nonV})$$.\n\n\n",
"1"
],
[
"\nThe starting equation should read $dS>dq/T\\_B$, not dS>dq/T, where $T\\_B$ is the temperature at the boundary between the system and surroundings (usually an a reservoir temperature) through which the heat flow dq occurs, and T is the average temperature of the system (which may not be spatially uniform). The open literature often fails to make this important distinction. With this correction, all the OP issues are resolved.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/169714/change-in-chemical-properties-on-polymerization
|
Change in chemical properties on polymerization [closed]
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**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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Closed 9 months ago.
This post was edited and submitted for review 9 months ago and failed to reopen the post:
>
> Original close reason(s) were not resolved
>
>
>
[Improve this question](/posts/169714/edit)
What I want to know is how should I expect polyesters, polyamides or other polymers derived from a monomer containing a certain functional group to behave and react chemically vs the monomer, as well as the effect of their structure in their fibre form on their chemical behaviour. Differences in reaction kinetics, and general trends in chemical behaviour to expect and be aware of. I.e In what ways would they be similar and differ chemically from the properties of the monomer from which they were derived
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https://chemistry.stackexchange.com/questions/169708/why-is-fe-p-in-naoh-part-of-hieltjes-lijklema-1980-p-fractionation
|
Why is Fe-P in NaOH part of Hieltjes & Lijklema (1980) P fractionation?
|
**Background:**
In order to separate the different phosphorus (P) species in lake or marine sediments, one can use sequential P fractionation method. Its main idea is to use extractants with weak or strong affinity to P to extract the sediment sample step by step. By doing this, different operationally defined P fractions are extracted sequentially.
There are many methods available, and my question is about the method proposed by Hieltjes & Lijklema (1980). Its procedure is shown in the first screenshot below.
**Question:**
I'm wondering why Fe-P is in the "NaOH extraction" part, which is "17 hours with 50 ml 0.1N NaOH".
In the Table 2 (see second screenshot below) of [the original paper](https://acsess.onlinelibrary.wiley.com/doi/abs/10.2134/jeq1980.00472425000900030015x) (Hieltjes & Lijklema, 1980), we can see that 98.9% of Fe-P is extracted from the 0.1 N NaOH solution. The authors mentioned that this synthetic Fe&P compounds is "*... a suspension of freshly precipitated ironhydroxy-phosphate, aged by standing or moderate heating.*" However, they didn't explain why this synthetic Fe hydroxides can be dissolved in the NaOH solution.
I understand that Al oxides can be dissolved by NaOH, so Al-P is in the "NaOH extraction" part. But Fe oxides can not be dissolved by NaOH, for example [in this posted question](https://chemistry.stackexchange.com/questions/13901/can-i-use-sodium-hydroxide-naoh-to-remove-oxides-from-metals). Then why Fe-P is in this extration?
[](https://i.stack.imgur.com/UQDUP.png)
[](https://i.stack.imgur.com/NCUYO.png)
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[] |
https://chemistry.stackexchange.com/questions/169702/daltons-law-clarification
|
Dalton's law clarification
|
I'm looking at *Elements of Physical Chemistry* by Atkins and de Paula. In section 1A.3, they state Dalton's law as
>
> The pressure exerted by a mixture of perfect gases is the sum of the pressures that each gas would exert if it were alone in the container at the same temperature:
>
>
> $$ p = p\_A + p\_B + ... \tag{1}$$
>
>
> In this expression, $p\_J$ is the pressure that the gas $J$ would exert if it were alone in the container at the same temperature. Dalton's law is strictly valid only for mixtures of perfect gases [...].
>
>
>
They then go on to define the partial pressure as
$$ p\_J = x\_J p \tag{2}$$
where $x\_J$ is the mole fraction of $J$ and $p$ is the total pressure of the mixture.
So my question is this: Eq. 1 holds for *all* gases given the definition in Eq. 2. So when they say Dalton's law only holds for perfect gases, do they mean that because $p\_J$ in Dalton's law is not the same as Eq. 2, Dalton's law doesn't hold for all gases? Using the same notation for the two different meanings for $p\_J$ seems to be widespread and potentially quite confusing.
| 5
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[
[
"\nAfter discussion with other commenters and answers, here is a summary of the issues with the text quoted in the question.\n\n\nFirst, there are two non-equivalent definitions of the partial pressure $p\\_i$ being used.\n\n\n### Definition 1\n\n\nThe first is given only in the text immediately preceding and following Eq. 1 in the quoted text in the question. In mathematical form, this definition is\n\n\n$$ p\\_i = \\frac{n\\_i RT}{V} \\: .$$\n\n\nUsing this definition for the partial pressure, we can show that\n\n\n\\begin{align}\np\\_\\mathrm{tot}^{(id)} &= \\frac{n\\_\\mathrm{tot}RT}{V} \\\\\n&= \\frac{\\sum\\_i{n\\_i}RT}{V} \\\\\n&= \\sum\\_i p\\_i \\: . \\tag{A1}\n\\end{align}\n\n\nThis is Dalton's law, which is only true for ideal (perfect in Atkins and de Paula's parlance) gases.\n\n\n### Definition 2\n\n\nThe second definition of $p\\_i$ is based on the actual pressure of the real gas and the mole fraction of species $i$, and it is given by Eq. 2 in the question\n\n\n$$p\\_i = x\\_i p \\: .$$\n\n\nFrom this definition, and using the fact that $\\sum\\_i x\\_i = 1$, we can express the total pressure $p$ as\n\n\n\\begin{align}\np &= p\\sum\\_i x\\_i \\\\\n&= \\sum\\_i x\\_i p \\\\\n&= \\sum\\_i p\\_i \\: . \\tag{A2}\n\\end{align}\n\n\nThis expression holds for all gases, ideal or not.\n\n\n### Summary\n\n\nNote that the final lines of Eqs. A1 and A2 look similar, but A1 is the ideal gas pressure for $n\\_\\mathrm{tot}$ molecules, whereas A2 is based on the true pressure of the system and does not depend on any assumptions about the form of the equation of state. The confusion that led to my question is the fact that $p\\_i$ was used with these two different meanings in very close proximity and no special attention paid to differentiating them.\n\n\n",
"2"
],
[
"\nThere is no inconsistency as far the pasted text goes. Before studying *any* derivation or law, we should know their assumptions. *Once* we accept those conditions and the assumptions, the derived equations become valid. With the following assumptions, Equation (1) and (2) are perfectly consistent as long as the following hold.\n\n\nSo what are assumptions for Dalton's law?\n(0) The gases in the mixture will not react, even if they are ideal gases.\n(1) There is no interaction among gas molecules\n(2) Molecular volume is negligible. This does not hold true at higher pressures.\n(3) Temperature is not too high or too low (say, we are not at the Sun's surface or Pluto)\n\n\nIf we start talking about real gases for equation (1) and ideal gases for equation (2), it is comparing apples and oranges.\n\n\nBTW there are better textbooks on Physical Chemistry and it is a good habit to read the same topic from multiple books! It sometimes gives interesting insights.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/169698/salt-analysis-cation
|
Salt analysis (cation) [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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Closed 9 months ago.
[Improve this question](/posts/169698/edit)
I'm reading about salt analysis and I've got some questions:
[](https://i.stack.imgur.com/4rIyT.jpg)
[](https://i.stack.imgur.com/HqgAl.jpg)
Image source: [pdf](https://ncert.nic.in/pdf/publication/sciencelaboratorymanuals/classXII/chemistry/lelm107.pdf)
Why are some sulphides (group II) precipitated in presence of HCl while others (group IV) in presence of ammonia solution? Can NaOH be used in place of ammonia solution? What's the role of NH4Cl in testing for group III? Will Na2CO3 work for group V?
(The groups here are not related to groups of periodic table.)
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[
[
"\nLet me answer your three questions.\n\n\n1. Some sulphides are soluble in acidic solutions (group IV). Some are insoluble in the same sort of acidic solution (group II). It is a question of solubility product. It can be taken as an experimental result, a fact.\n2. If no $\\ce{NH4Cl}$ were present in the solution tested for Group III, the precipitates $\\ce{Al(OH)3}$ and $\\ce{Fe(OH)3}$ may remain in a colloidal state, and will cross the filter paper. This is a serious drawback. In the presence of enough ions, like $\\ce{NH4^+}$ and $\\ce{Cl-}$, the colloid is transformed into a precipitate.\n3. $\\ce{NaOH}$ can replace ammonia solution, but it will prevent the precipitation of $\\ce{Al(OH)3}$ in group III and $\\ce{Zn(OH)2}$ in group IV. These precipitates will redissolve in $\\ce{NaOH}$ to form a solution of aluminate $\\ce{Al(OH)4^-}$ and zincate $\\ce{[Zn(OH)4]^{2-}}$. So the method would not detect $\\ce{Al}$ or $\\ce{Zn}$.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/169695/what-will-be-the-major-product-in-the-nitration-of-2-methoxyethylbenzene
|
What will be the major product in the nitration of (2-methoxyethyl)benzene?
|
In this question, I am not quite sure what will be the major product in the reaction and on reacting (2-methoxyethyl)benzene with $\ce{NO2-O-NO2}$ ($\ce{N2O5}$). I know $\ce{NO2}$ will act as electrophile and gets attached to ortho or para. In this will it be ortho position instead of para due to hydrogen bonding with $\ce{CH2CH2OCH3}$?
[](https://i.stack.imgur.com/RM5On.jpg)
| -1
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[
[
"\nThis is a case of the nitration of alkylbenzene by dinitrogen pentoxide (DNP), which can be considered as an anhydride of nitric acid. The substrate in the question is (2-methoxyethyl)benzene, which is closely related to toluene except for the extra bulkiness of the substitution. Thus, one can except nitration mechanism would be closely related to that of toluene.\n\n\nWhat's lacking in this question is the conditions for the nitration. The alkyl group is basically a *o,p*-director. However, DNP-based nitration reactions are dramatically influenced by the medium (solvent) used (Ref.1 & 2). That means, the *ortho* to *para* products ratio is different from condition to condition (e.g., solvent). For example, it was 1:2 if the reaction performed at $\\pu{-10 ^\\circ C}$ as Poutnik pointed out in his comment above (in this case, I assumed the reaction is in toluene). When the nitration of toluene by $\\ce{N2O5}$ is done in an organic solvent other than toluene such as liquefied 1,1,1,2-tetrafluoroethane (TFE), the product ratio varies (Ref.2):\n\n\n[](https://i.stack.imgur.com/20Flg.png)\n\n\nIf the reaction is conducted in the presence of a solid Zeolite with small pores and with DNP as the nitrating agent and dichloromethane as the solvent, the reaction products have a very high proportion of *para*-nitro substituted isomer (Ref.3).\n\n\n**References:**\n\n\n1. R. W. Millar, M. E. Colclough, P. Golding, P. J. Honey, N. C. Paul, A. J. Sanderson, and M. J. Stewart, \"New synthesis routes for energetic materials using dinitrogen pentoxide,\" *Philos. Trans. R. Soc., A* **1992**, *339(1654)*, 305-319 (DOI: <https://doi.org/10.1098/rsta.1992.0037>).\n2. Alexandr K. Kharchenko, Ruslan V. Fauziev, Mikhail N. Zharkov, Ilya V. Kuchurov, and Sergei G. Zlotin, \"Nitration of aromatics with dinitrogen pentoxide in a liquefied 1,1,1,2-tetrafluoroethane medium,\" *RSC Adv.* **2021**, *11(42)*, 25841-25847 (DOI: <https://doi.org/10.1039/D1RA04536A>).\n3. Reddy Damavarapu, Keerthi Jayasuriya, and Thomas J. Kwok, \"Regioselective nitration of aromatic compounds by dinitrogen pentoxide and the reaction products thereof,\" *US Patent* **1999**, 5,977,418 ([PDF](https://patentimages.storage.googleapis.com/cc/d8/43/13ff8abe6639e5/US5977418.pdf)).\n\n\n",
"3"
],
[
"\nNitrogen pentoxide is a good source of $\\ce{NO2^+}$ (in fact [the solid form directly contains $\\ce{NO2^+}$](https://en.wikipedia.org/wiki/Dinitrogen_pentoxide)). So, as you imply, we would expect a nitration reaction especially with an activating, ortho-para directing substituent such as the alkoxy group shown.\n\n\nThe ortho product would be favored by hydrogen bonding, but here there is no good mechanisms for that. Hydrogen bonding with hydrogen attached to carbon may actually be possible, but it is weak, and the nitrogen group will not provide more strongly protection hydrogen. Therefore the more sterically favored para product would be major.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/169694/why-is-polyester-hydrophobic-as-well-as-insoluble-in-various-organic-solvents
|
Why is polyester hydrophobic as well as insoluble in various organic solvents?
|
Polyester is known to be hydrophobic as shown by it's low moisture regain.
However, esters are at least somewhat hydrophilic and soluble in water. This decreases with the hydrocarbon chain length increase, but in the case of polyesters, this is accompained with a proportionate increase in the number of ester groups.
My attempts to explain this were as follows
1. The reason nonpolar molecules are not soluble in water is the H-bonded structure of water is disrupted - there's a loss in entropy; water molecules have to orient themselves in the form of a cage around the non polar regions.
In the case of a polymer chain of PET, the loss in entropy is greater than if we tried dissolving the individual monomer esters that constitute it in water. Therefore one is somewhat soluble, the other is not.
2. Since like dissolves like, and a polar aprotic solvent would not have the H-bonded structure of water that would be disrupted (and in fact since they are both polar aprotic, one would presumably become well incorporated into whatever bonding structure the other has), I would think a polar aprotic solvent would dissolve polyester.
Turns out that this isn't the case. In fact, "polyester fibres are generally resistant to organic solvents" (A. Bendak and El-Marfasi; [pdf link](https://www.google.com/url?sa=t&source=web&rct=j&url=https://jag.journalagent.com/ias/pdfs/IAS_4_4_275_284.pdf&ved=2ahUKEwiAir2T5d37AhWEeN4KHSPpAhcQFnoECAkQBg&usg=AOvVaw3sg51EkbnIfegmqv8ih5YN))
Perhaps, I thought, this might be because of the relatively crystalline structure of polyester that doesn't allow solvent molecules to penetrate it below its glass transition temperature.
However, polyester fibres are soluble in phenol and m-cresol. They [here](https://www.researchgate.net/figure/Relative-solubilities-of-Polyesters-in-different-solvents-at-30-o-C_tbl2_286892808) also appear to be soluble in THF and DMF, but not the remaining organic solvents.
Why wouldn't the same issues that were occurring with water occur with these polar protic solvents (phenol and m-cresol), and why are polyesters soluble in THF&DMF but not other polar aprotic solvents.
Also, what is a good general framework to be thinking about the solubilities of substances in various solvents?
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https://chemistry.stackexchange.com/questions/169692/what-is-the-mechanism-of-salkowski-reaction-of-%d1%81holesterols
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What is the mechanism of Salkowski reaction of сholesterols?
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I recently came across the Salkowski reaction of cholesterols. Cholesterol in chloroform is treated with concentrated sulfuric acid. A positive test exhibits two distinct layers, the upper chloroform layer gets a red to violet colour, while the lower sulfuric acid layer exhibits a greenish glow.
1. The chemistry behind the reaction is dehydration using $\ce{H2SO4},$ forming cholestadiene.
[](https://i.stack.imgur.com/jyI4s.png)
2. In the second step, the cholestadiene undergoes dimerization.
[](https://i.stack.imgur.com/HOZXc.png)
3. In the last step, the cholestadiene undergoes sulfonation at the 7,7' position. (I believe I drew the correct structure)
[](https://i.stack.imgur.com/ojs82.png)
However, I couldn't find a *valid mechanism* for the second and the third steps (dimerization and sulfonation). Can someone explain the mechanism?
| 5
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[
[
"\nThe mechanism of dimerization is now clear, Thanks to Mathew Mahindratne.\nRef.[Yoshihisa, *et.al*](https://doi.org/10.1016/0039-128X(78)90110-1)\n\n\nThere are two different isomers of Cholestadiene present, namely 3,5 Cholestadiene **[ I ]**, and 2,4 cholestdiene **[ IV ]**. The plausible mechanism of their formation is shown below. They can undergo dimerization to yield [ II ] & [III].\n[](https://i.stack.imgur.com/oxbdF.png)\n\n\n[](https://i.stack.imgur.com/0XK40.png)\n\n\nDimer [ II ] was formed from two pathways, which were the conversion of the dimer [ III ] and the dimerization of [I] (3,5-cholestadiene).\n\n\nThe dimer [ III ] was formed from [ IV ] (2,4-cholestadiene).\n\n\nThe mechanism of their formation is shown below.\n[](https://i.stack.imgur.com/RIKSx.png)\n\n\nThe formation of the dimer [ II ] and/or the dimer [ III ] varies with the kind of acids employed.\nUsing a **Bronsted acid** medium results in the formation of **[ II ]**.\nFor eg: In presence of Trichloroacetic acid and hydrochloric acid (10:1 ratio), dimer [ II ] was obtained.\n\n\nUsing a **Lewis acid** results in the formation of **[ III ]**. For eg: *Zinc chloride* and *Acetyl chloride* (Tschugaeff reaction), \n\n*Antimony trichloride* and *Acetyl chloride*, \n\n*Ferric chloride* and *Sulfuric acid* (Zak-Henly reaction), \n\n*Ferric chloride*, *Perchloric acid*, and *Phosphoric acid* results in the formation of [ III ].\n\n\nIf the solution contains both **Bronsted and Lewis acid**s, then a **mixture** **of products** is formed.\n\n\nEg: Both [ II ] and [ III ] are formed from the reactions with\n\n\n1. Sulfuric acid and Acetic anhydride (Liebermann-Burchard reaction)\n2. sulfuric acid (Salkowski reaction)\n3. Trichloroacetic acid-SbCl3\n\n\nSince sulfuric acid, having both Bronsted and Lewis acid character is used in our reaction, a mixture of products will be formed.\nPs: All Bronsted acids are Lewis acids, but not vice versa.\n\n\nHowever, the third step (Sulfonation) needs further explanation.\n\n\n",
"4"
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https://chemistry.stackexchange.com/questions/169690/carbon-monoxide
|
Carbon Monoxide [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
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Closed 9 months ago.
[Improve this question](/posts/169690/edit)
In the compound $\ce{CO}$ I know that the carbon has an oxidation number of $+2$ and the oxygen has an oxidation number of $-2$. However up until now I have thought of oxidation number and charge being the same, and it is apparent that the $\ce{CO}$ molecule is most definitely covalent and has no ionic character hence the molecule is neutral and each atom should not have an actual "charge" or we would call the molecule ionic. So are oxidation number and charge different or do $\ce{C}$ and $\ce{O}$ have charges in carbon monoxide.
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"\nOxidation State and formal charge are most definitely not the same thing; in fact they make opposite assumptions to exist.\n\n\nThe molecule CO is, as you correctly identified, covalent, and you’re likely to only see oxidation states used when it comes to ionic molecules, so focusing on covalent molecules:\n\n\nFormal charge totally neglects the existence of electro negativity in covalent bonds. It makes the assumption that it any covalent bond, shared electrons belong equally split between the two atoms that share them.\n\n\nHence the formal charge of an atom can be defined as :\n\n\nNumber of valence electrons in uncharged atom - number of electrons in lone pairs in the atom in the molecule - (1/2 \\* the number of electrons shared in covalent bonds by the atom in the molecule).\n\n\nOxidation states, however, make the assumption that electronegativity is an all-powerful effect, such that in a covalent bond with an electro negativity difference, shared electrons belong entirely to the more electronegative element.\n\n\nHence the oxidation state of an atom in a covalent bond can be defined as:\n\n\nNumber of valence electrons in uncharged atom of element - number of electrons in atom in the molecule (assigning electrons to more electronegative atoms)\n\n\nThe ONLY real similarity between the two concepts is that in any molecule/ion, the sum of formal charges is equal to the charge of the molecule, and so is the sum of the formal charges. But the individual charges on atoms will often differ from the individual oxidation states on atoms.\n\n\nConsidering the molecule CO, where there is a triple bond between oxygen and carbon , and each atom has one lone pair:\n\n\nOxygen is more electronegative than carbon and oxygen normally has 6 VE and carbon has 4VE. So for oxidation states, all 6 electrons in the triple bond as considered to belong to the oxygen; for formal charge they are split 3 each. Hence:\n\n\nFormal charges:\n\n\nCarbon: 4 - 2 - (0.5\\*6) = -1\n\n\nOxygen: 6 - 2 - (0.5\\*6) = 1+\n\n\nOxidation states:\n\n\nCarbon: 4 - 2 = 2+\n\n\nOxygen: 6 - 2 - 6 = -2\n\n\nNote that although these are different, both pairs of values sum to 0 as your molecule is neutral.\n\n\nFinally, you’re right that the molecule is neutral and does not have an overall charge, although it does obviously have a dipole towards the oxygen nonetheless.\n\n\nBut the truth is that these formal charges and oxidation states are both just numerical but artificial concepts. They’re mainly taught as they can be extremely useful in other areas of chemistry, such as organic synthesis in the case of formal charge or knowing when acidic conditions are needed in the case of oxidation states.\n\n\n",
"1"
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] |
https://chemistry.stackexchange.com/questions/169687/in-the-synthesis-of-azo-compounds-given-the-strongly-basic-reaction-conditions
|
In the synthesis of azo compounds, given the strongly basic reaction conditions, why is the azo product favoured?
|
I was reading about the chemical synthesis of aromatic azo compounds, wherein, first a diazo cation is formed from an aromatic primary amine:
$$\ce{Ar–NH2 + NaNO2 + 2HCl -> Ar–N=N+Cl– + 2H2O + NaCl}$$
This cation is subsequently reacted with a phenol under strongly alkaline conditions. EAS takes place with the diazo cation as the electrophile and we get the azo compound at ortho and para positions.
What I was wondering is, since $\ce{OH-}$ is strongly nucleophilic as well (in fact, I would guess more so than ortho and para positions of the phenol anion since the electron density is being distributed via resonance), how is a favourable yield of the product we want obtained in reactions in like this? Also, why would the the ortho and para product be preferred over the cation reaction with the $\ce{O-}$ of the phenol anion?
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https://chemistry.stackexchange.com/questions/169684/how-are-polyatomic-ions-formed
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How are polyatomic ions formed? [closed]
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**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
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I know that there should be a covalent bond between polyatomic ions, such as $\ce{SO4^2-}$. But what causes it to suddenly gain electrons and become an ion?
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[
[
"\n\n> \n> But what causes it to suddenly gain electrons and become an ion?\n> \n> \n> \n\n\nIn principle, you can get a polyatomic ion by adding atoms to a monoatomic ion, or by changing the number of electrons in a (neutral) molecule, or by splitting a (neutral) molecule into positive and negative ions, or by reactions of other ions.\n\n\n\n> \n> [...] such as $\\ce{SO4^2-}$\n> \n> \n> \n\n\nI am not aware of the existence of $\\ce{SO4}$, so just adding two electrons to a molecule probably doesn't happen in this case. Here are some formal scenarios to get sulfate, some of them more conceptual than realistic:\n$$\\ce{HSO4-(aq) <=> H+(aq) + SO4^2-(aq)}$$\n$$\\ce{S^2-(aq) + 2O2(g) -> SO4^2-(aq)}$$\n$$\\ce{Na2SO4(s) -> 2Na+(aq) + SO4^2-(aq)}$$\n$$\\ce{S2O8^2−(aq) +2I-(aq) -> 2SO4^2−(aq) + I2(aq)}$$\n$$\\ce{SO3(g) + 2OH-(aq) -> SO4^2- + H2O(l)}$$\n\n\nSome of these reactions are redox (electron transfer), some are acid/base (proton transfer), some are dissociation or hydration, and some are combinations of these.\n\n\nHere is a reference for the second scenario (which looked unlikely to me so I looked it up), with a lot of different sulfur-containing ions as intermediates: <https://doi.org/10.5935/0103-5053.20160197>\n\n\n",
"1"
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https://chemistry.stackexchange.com/questions/169680/how-do-common-histological-stains-work
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How do common histological stains work?
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In histology there are commonly used acidic (= negatively charged) dyes, like eosin, erythrosine, orange G, and basic (= positively charged) dyes, like haematoxylin, toluidine and methylene blue, azure II. Acidic dyes stain positively charged structures, like some cytoplasmic proteins, and basic dyes stain negatively charged molecules, like RNA and DNA. Structures stained with acidic dyes usually appear pink, with basic - blue.
That's what commonly offered in histology course. I could not find any information on what type of bonding these dyes use (is there a covalent bonding or non-covalent interaction?). Nor do I know why cell structures stain that way. Can you explain these properties of the dyes?
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https://chemistry.stackexchange.com/questions/169677/how-can-one-obtain-the-same-spin-and-opposite-spin-terms-of-the-mp2-energy-by-in
|
How can one obtain the same-spin and opposite-spin terms of the MP2 energy by integrating out the spin?
|
Just as specified on the psi4 website here: <https://psicode.org/psi4manual/master/dfmp2.html>, the Moller-Plesset Perturbation Theory states that the second order correction to the ground state energy, $E\_0^{(2)}$, is given by
$$E\_0^{(2)} = \sum\_{i<j}\sum\_{a<b} \frac{\left|\,[ia\|jb]\,\right|^2}{\epsilon\_i - \epsilon\_a + \epsilon\_j - \epsilon\_b} = \sum\_{i<j}\sum\_{a<b} \frac{\left( [ia|jb] - [ib|ja]\right)\left( [ai|bj] - [bi|aj]\right)}{\epsilon\_i - \epsilon\_a + \epsilon\_j - \epsilon\_b}\, ,$$
where $[ia\|jb]$ is the antisymmetrized two-electron integral over spin-orbitals and $\epsilon$ are the different orbital energies. As far as I know, one can integrate out the spin variable, $\sigma$, of said integrals in the following way:
$$\int\int d\omega\_1 d\omega\_2 [ia|jb] = \int\int d\omega\_1 d\omega\_2 \sigma\_i^\*(\omega\_1)\sigma\_a(\omega\_1)\sigma\_j^\*(\omega\_2)\sigma\_b(\omega\_2) (ia|jb) = \delta\_{\sigma\_i,\sigma\_a}\delta\_{\sigma\_j,\sigma\_b}(ia|jb)\,,$$
thus obtaining chemist’s integrals over spatial orbitals, $(ia|jb)$. Following this path, $E\_0^{(2)}$ can be divided into a same-spin, $E\_{0,SS}^{(2)}$, and an opposite-spin, $E\_{0,OS}^{(2)}$, terms by integrating the spin out such that
$$E\_{\rm 0,\,SS}^{(2)} = \sum\_{ij}\sum\_{ab}\frac{(ia\mid jb)[(ia\mid jb) - (ib\mid ja)]}{\epsilon\_i - \epsilon\_a + \epsilon\_j - \epsilon\_b}\, ,\qquad
E\_{\rm 0,\,OS}^{(2)} = \sum\_{ij}\sum\_{ab}\frac{(ia\mid jb)(ia\mid jb)}{\epsilon\_i - \epsilon\_a + \epsilon\_j - \epsilon\_b}\,.$$
By doing this, I was able to get the expresion for $E\_{0,OS}^{(2)}$ without much trouble. However, when trying to perform the integration for same spins, I haven't been able to get rid of one of the terms and get the expresion I want (which has three terms only). This happens because all four orbitals have the same spin function, so every term should mantain after the integration. I haven´t been able to find ways to get these expressions without just using algebra like many demonstrations have.
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https://chemistry.stackexchange.com/questions/169672/silicon-crystal-orientation
|
Silicon crystal orientation
|
I'm confused about how [110] direction is determined for (100), (110) or (111) wafers. I found a book chapter which just confused me even more. From the image below, I understand how [110] is determined on the (110) wafer but not the other two. I'm also having a hard time understanding what different planes would look like on the (111) and (110) wafer. I would appreciate a resource for silicon wafers specifically (not necessarily crystallography). Here are things I'm not understanding.
How is the [110] direction determined and why is it different for each of the three wafers below? For the (110) wafer [110] is between x and y axis which makes sense. But for (100) it seems to go from z to y and for (111) it's between x and z. From my understanding [110] intercepts x and y axis at 1 and doesn't intercept z axis but that doesn't seem to be the case for (100) and (111) wafers.
Does the definition of x,y and z axis change when we're talking about (100), (111) or (110) wafers? I know that whichever wafer it is, that direction will be pointing out from the wafer surface but when I draw planes with respect to [110] and [111] (for (111) wafer), I don't know where the (100) plane should be. (110) plane would be perpendicular to the [110] direction and would be normal to the surface and (111) would be 35° from it. I can't find a drawing for it which makes me think we don't need to show the (100) plane? But if not then what if I need to align my mask to (100) plane on (111) wafer?
How would I define higher miller index planes on the 3 wafers? For example I need to align my mask to (122) or (411) plane, how would I start with it? I understand that this might become clear once I learn about the primary planes on the wafers.
[](https://i.stack.imgur.com/lhWUp.png)
| 3
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[
[
"\n\n> \n> I would appreciate a resource for silicon wafers specifically (not necessarily crystallography).\n> \n> \n> \n\n\n[This video](https://youtu.be/vXk6Uhq74nU) is fun to watch (the difference between a [111] and a [100] wafer is striking) and it points at further resources.\n\n\nThis [interactive Jmol site](http://lampx.tugraz.at/%7Ehadley/memm/materials/silicon/silicon.php) lets you select a plane while also showing the unit cell orientation. For the image below (which is an interactive 3D model on the live site), first select to show 5x5x5 unit cells. Then, enter (1,1,1) for the plane, increase the thickness with the slider and click \"draw atoms in the plane\". Once you have a model of your wafer, change to another plane and click \"show HKL plane\". If you look closely, you can see the unit cell in the center in thin black lines (you have to turn it on manually, by right clicking and selecting Style->Unicell or Axes), allowing you to verify the orientation of the two planes.\n\n\n[](https://i.stack.imgur.com/KYITo.png)\n\n\n\n> \n> How is the [110] direction determined and why is it different for each of the three wafers below?\n> \n> \n> \n\n\nSilicon has cubic symmetry, so the three directions [100], [010], and [001] are equivalent. What is confusing is that the images in the question show three different planes intersecting with the wafers: (0,1,1), (1,1,0), (1,0,1) from left to right, but they are labeling all of them <1,1,0> because they are indistinguishable once you take away the axis notation.\n\n\n[](https://i.stack.imgur.com/m1ZeP.png)\nSource: <https://journals.aps.org/authors/crystallographic-notation-h1>\n\n\n\n> \n> Does the definition of x,y and z axis change when we're talking about (100), (111) or (110) wafers?\n> \n> \n> \n\n\nNo, in all three pictures, the x axis goes along [100], the y-axis along [010], and the z-axis along [001].\n\n\n\n> \n> For example I need to align my mask to (122) or (411) plane, how would I start with it?\n> \n> \n> \n\n\nLet $\\vec{a}$ be a vector perpendicular to the surface of the waver, and $\\vec{b}$ be a vector perpendicular to the surface used in the cutoff, and $\\vec{c}$ be the vector perpendicular to the desired alignment plane. The cross product of $\\vec{a}$ and $\\vec{b}$ is parallel to the cutoff edge. The cross product of $\\vec{a}$ and $\\vec{c}$ is parallel to intersection of the wafer surface and the desired alignment plane. To get the angle between the two, you use the usual dot product method.\n\n\nFor a (111) wafer (rightmost diagram), for example, the cutoff line is parallel to (1,1,1) x (1,0,1) = (1,0,-1). The intersection between the wafer surface and the (122) plane is a line parallel to (1,1,1) x (1,2,2) = (0,-1,1). The angle between these two is 120 degrees. Because of the symmetry of the wafer, other solutions are 0 or -120 degrees. (You would get those by dialing in the (212) or the (221) plane.)\n\n\n### Why does it matter?\n\n\nFrom uhoh's comments, slightly edited:\n\n\n\n> \n> In one of my previous lives I had to align contact photomasks to silicon wafers in order to open holes in photoresist, used to etch through a thin silicon nitride layer (10 to 100 nm) which was then used as a mask for directional wet etching of the silicon. KOH can have a 100:1 or larger wet etch anisotropy for certain crystal directions, and you can get pyramidal or even very vertical (wrt wafer surface) sidewalls from anisotropic wet etches if you rotate your mask to get specific orientations of the pattern edges with respect to crystal directions. [...] [Anisotropic wet etching](https://en.wikipedia.org/wiki/Etching_(microfabrication)#Anisotropic_wet_etching_(Orientation_dependent_etching)) (Orientation dependent etching)\n> \n> \n> \n\n\n",
"7"
]
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https://chemistry.stackexchange.com/questions/169666/correct-iupac-name-of-3-chloro-1-phenylprop-1-ene
|
Correct IUPAC name of 3 - chloro-1-phenylprop-1 -ene
|
[](https://i.stack.imgur.com/4sPUh.png)
In this molecule, what should I take the principal group as - the benzene or the propene chain?
I feel that we should take Benzene as the prinicipal ring and the name should be 3-chloropropen-1-yl benzene. However, I am also confused about 3 - chloro-1-phenylprop-1 -ene being a correct IUPAC name .
What takes priority in this case , the unsaturation of the chain or the size of the ring?
What is the relevant IUPAC rule here?
Can someone please help me out on this?
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[
[
"\nThe compound that is given in the question doesn’t have a principal characteristic group, which would be expressed as a suffix and that would determine the senior parent structure. Therefore, the ring has seniority over the chain when selecting the preferred IUPAC name. In general nomenclature and depending on the context, however, the chain may be favoured to recognize the unsaturated acyclic structure.\n\n\nThe corresponding subsection in the current version of *[Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)](http://dx.doi.org/10.1039/9781849733069-fp001)* reads as follows:\n\n\n\n> \n> **P-44.1.2.2** Systems composed of rings and chains (exclusive of linear phanes)\n> \n> \n> Two methods are recognized to name systems composed of rings and chains\n> (exclusive of linear phanes).\n> \n> \n> (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.\n> \n> \n> (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.\n> \n> \n> (…) For selection of a preferred IUPAC name, see P-52.2.8.\n> \n> \n> \n\n\n\n> \n> **P-52.2.8** Selection between a ring and a chain as parent hydride\n> \n> \n> Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).\n> \n> \n> \n\n\nMethod (1): [(1*E*)-3-chloroprop-1-en-1-yl]benzene (preferred IUPAC name; ring preferred to chain)\n\n\nMethod (2): (1*E*)-3-chloro-1-phenylprop-1-ene (unsaturated acyclic structure may be recognized)\n\n\nTherefore, the preferred IUPAC name for the compound that is given in the question is **[(1*E*)-3-chloroprop-1-en-1-yl]benzene.**\n\n\n![[(1E)-3-chloroprop-1-en-1-yl]benzene](https://i.stack.imgur.com/4iFo6.png)\nNote that parentheses are used around substituent prefixes to separate locants referring to different structural elements.\n\n\n",
"3"
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https://chemistry.stackexchange.com/questions/169659/why-are-the-vapor-pressure-of-toluene-and-benzene-different
|
Why are the vapor pressure of toluene and benzene different?
|
We learnt to apply Raoult's law for the ideal solutions. When the questions are asked, they tell that the intermolecular forces of toluene and benzene are equal. But in the same question, they provide two vapor (saturated) pressures for toluene and benzene.
**Question**: If the intermolecular forces of toluene and benzene are equal, how do they have different vapor pressure?
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[
[
"\nMost likely your \"they\" are exam-paper setters in some public exams. The fact is that the moment you deal with a different molecule, the intermolecular interactions cannot be the same. Benzene and toluene are different molecules just like two different persons on Earth, they have different molecular weights and not surprisingly different molecular interactions. Benzene boils at 81 °C and toluene boils at 110 °C. In general, a bigger/heavier organic molecule has a lower vapor pressure and a higher boiling point. Intermolecular forces is an umbrella term, so there is no single number, hence this can never be equal for two different molecules.\n\n\nIn the comments you asked about molar mass. In general heavier molecules have a higher boiling point but one cannot make a prediction. Compare benzene with a molar mass of 78 g/mol, and another metal such as platinum with a molar mass of 195 g/mol. The former boils below 100 °C and platinum boils at several thousand degrees. Molar mass alone cannot let you predict a number related to melting or boiling. Many software do predict the boiling and melting point of compounds, more often that is quite wrong.\n\n\n",
"3"
],
[
"\nPerhaps, by saying benzene and toluene have \"equal intermolecular forces\", maybe they meant they both have London dispersion forces, NOT that they have the same degree of LDF. Then it would be up to you to explain why they have different vapor pressures by saying the toluene would have more sites for LDF than benzene.\n\n\n",
"3"
],
[
"\nBenzene and toluene do not have exactly the same intermolecular forces. Both have Van der Waals - London forces from temporary induced dipoles and toluene has also a slight dipole-dipole interaction from its very small dipole moment[toluene is small while benzene is vanishingly small effectively zero]. The difference in intermolecular attractions is not the force but the energy of attraction or the work required to separate the molecules. This energy is expressed in the heat of fusion and evaporation. If the forces are about equal, then the area in contact is determined by the molecular size and shape and these affect the energy of attraction. The differences in molecular sizes between toluene and benzene are apparent. Therefore very similar intermolecular forces result in different energies of attraction.\n\n\nFinally, the boiling point or temperature of equilibrium depends on the heats and entropies of the transitions: at equilibrium delta H = T Delta S; T = Delta H/Delta S\n\n\n",
"1"
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https://chemistry.stackexchange.com/questions/169657/lewis-structure-of-xef2-that-comes-closer-to-reality
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Lewis structure of XeF2 that comes closer to reality?
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Meanwhile, the hypervalency of Xe in XeF2 is explained by the 3c4e bond, where the p orbitals of the 3 atoms overlap to form a delocalized system of 3 MO's in which 4 electrons (two from the Xe p orbital and one each from the p orbitals of the F atoms). This leads to a bond order of 1/2 per Xe-F group.
Obviously, the lewis representation that can be found everywhere on the Internet cannot be correct:
[](https://i.stack.imgur.com/EI4kp.png)
That would indicate a bond order of 1 per F-Xe Group, not 1/2.
As far as I know, resonance drawings are used to indicate delocalized bonds, so they also represent bond orders of 1/2. Shouldn't XeF2 then be represented as a lewis diagram as follows:
[](https://i.stack.imgur.com/V7WM0.jpg)
And I also wonder what orbitals are those 3 lone pairs at Xe? s,p,p? Or 3 sp2 hybride orbitals (which would make more sense for the geometry but less sense because hybridisation just makes sense for bonds)?
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"\nJust after XeF2 was discovered, D. F. Smith published a short communication [Xenon Difluoride](https://aip.scitation.org/doi/pdf/10.1063/1.1733476) *J. Chem. Phys.* 38, 270 (1963) mentioning the electronic structure of your second drawing.\n\n\nSmith conducted infrared spectroscopy of XeF2 and wrote:\n\n\n[](https://i.stack.imgur.com/BeZLR.png)\n\n\nThen C. A. Coulson published [The nature of the bonding in xenon fluorides and related molecules](https://pubs.rsc.org/en/content/articlelanding/1964/JR/JR9640001442) *J. Chem. Soc.*, 1964, 1442-1454, which considered the validity of four models (plus some sub-models) of XeF2 bonding.\n\n\nCoulson adds more resonance structures in addition to the two mentioned by Smith, including a purely ionic F- Xe2+ F- resonance structure.\n\n\nFor a more recent article on the topic, see [The essential role of charge-shift bonding in hypervalent prototype XeF2](http://jupiter.chem.uoa.gr/thanost/papers/papers7/NatChem_5(2013)417.pdf) *Nature Chemistry* 5, 417–422 (2013). ([official link](https://www.nature.com/articles/nchem.1619))\n\n\nOverall, ionic contributions do need to be considered.\n\n\n",
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https://chemistry.stackexchange.com/questions/169653/do-nahco3-and-cyanoacrylate-react
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Do NaHCO3 and cyanoacrylate react? [closed]
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**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
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I just saw [this video](https://youtu.be/ImLAmfM_AgA) about using baking soda + cyanoacrylate(CA) glue as a substrate+resin.
* Neat, but do CA and $\ce{NaHCO3}$ react in any way?
* If so, what does the equation look like and what is the resulting structure?
* If not at room temperature, do they react at higher temperatures because CA curing is exothermic.
* Is the CA + $\ce{NaHCO3}$ stable once it dries, is there any residue that would be toxic or irritating?
(not sure what to tag this with, feel free to add tags)
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"\n\"Baking soda\"(\\*) speeds up the cyanoacrylate curing, as it's basic anion initiates anionic polymeration. Similarly as benzoyl peroxide initiates radical polymeration of styrene to polystyrene.\n\n\n$$\\ce{HO-CO-O- ->[CH2=CR1R2] \\\\\nHO-CO-O-CH2-CR1R2- ->[CH2=CR1R2] \\\\\nHO-CO-O-CH2-CR1R2-CH2-CR1R2- ... etc}$$\n\n\nWhile curing, there is expected stronger evaporation of monomeric cyanoacrylate, compared to standard curing, due faster reaction, which may be irritating. When cured, the safety concerns are the same as for ordinary cyanoacrylate curing and usage, which has medicinal applications too.\n\n\nNote that advices about medical safety, acute and especially long term one, is explicitly off-topic on this site, which would comment just chemical aspects of safe manipulation and precautions.\n\n\nThe higher temperature is issue for the structural stability, as the remaining encapsulated baking soda may decompose, forming gaseous carbon dioxide.\n\n\n\n\n---\n\n\n(\\*) - \"Baking soda\" = sodium bicarbonate, $\\ce{NaHCO3}$. Not to be confused with sodium carbonate, $\\ce{Na2CO3}$ resp. $\\ce{Na2CO3 . 10 H2O}$ aka 'washing soda\", which is caustic, while baking soda is not.\n\n\n",
"4"
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https://chemistry.stackexchange.com/questions/169640/are-there-any-specific-requirements-for-ipso-substitution-in-aromatic-compounds
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Are there any specific requirements for ipso substitution in aromatic compounds?
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In one of the questions in my textbook, (bromine + water) is able to replace the -SO3H group attached to phenol via ipso substitution, but in another question, (bromine + Fe) cannot replace the same group when attached to toluene.
Why does this happen? Is it because the water causes greater activation of the ring in the first case by phenoxide ion formation? Or is it because -OH is a better activating group in general?
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[
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"\nIn each case you have an electron donating (Me/OH) and withdrawing (SO3H) group playing a tug-of-war in terms of activating or deactivating the ring towards EAS. A methyl is **significantly** less EAS activating than hydroxyl, which can formally donate electron density (anion) into the ring through resonance. Methyl's are just slightly activating by induction so toluene rarely undergoes EAS efficiently (personal experience) without forcing conditions, and this is without a deactivating group making it worse.\n\n\nThe introduction of iron would generate Fe(iii) bromide, as the other commenters have said. This species is usually great for helping out EAS brominations on the electrophile's side (increases polarization of remaining Br2). HOWEVER, even this great electrophile setup cannot overcome how poor the \"nucleophile\" is, being a weak tolule to start off with, further deactivated by the sulfonic acid.\n\n\n",
"1"
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https://chemistry.stackexchange.com/questions/169627/what-is-the-difference-between-smell-odor-and-vapor-of-a-substance
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What is the difference between "smell/odor" and "vapor" of a substance? [closed]
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**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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A solid chemical substance with a very high vaporization temperature can still produce potent smells at room temperature. What, then, produces smell, and could smelling some potent solid substance pose any danger, even if it hasn't vaporized at all?
Thanks!
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[
[
"\n\n> \n> What is the difference between \"smell/odor\" and \"vapor\" of a substance?\n> \n> \n> \n\n\nIt is assumed that the vapor of a given compound/element is the gas phase of the same pure compound/element. By condensing the vapor, you can obtain the same stuff in liquid or solid form.\n\n\nSmell on the other hand is a human/animal perception. Although it is thought that plants can also smell. It is a biochemical interaction of that component with our smell receptors inside our nose. A very interesting example is that of a enantiomeric molecules. These are two different compounds with *identical vapor pressure* at a given temperature yet they *smell very differently*. One enantiomer of limonene smells like lemons and the other smells like oranges! Same elemental composition, same boiling point but a different orientation in space.\n\n\nYour next point is how something which has very high pressure can still have a smell. For example, a very high boiling point metal like osmium (the name means smell) has a very bad odor, but this is not due to osmium metal itself but rather due to its very thin volatile oxide coating which forms on the metal.\n\n\nAlternatively, there is another possibility which has been linked by Karsten in the comments. Metals can react with sweat components and body oils and make volatile smelly compounds. This is why we have a *metallic odor*, again this has nothing to with the metal vapor itself, but some organic compounds which have originated by the reaction of the metal with our fluids present on our skin in very very small amounts. Human nose is an excellent sensitive detector and it is used in analytical chemistry in something called *Olfactory Chromatography*! Perfume makers and food flavor chemists love this technique.\n\n\n",
"1"
],
[
"\nA vapor is a gas made by evaporation of a liquid or of a solid (by sublimation). It may have an odor or not. Most substances in a gaseous state do have an odor. Water exists in a gaseous state, but it does not have any odor.\n\n\n",
"0"
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https://chemistry.stackexchange.com/questions/169623/reactivity-of-central-carbon-of-ketone-vs-alchohol
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Reactivity of central carbon of Ketone vs Alchohol [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
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I am trying to measure the reactivity of the carbon (of C=O and C-OH bonds) as described below. Lets assume the groups $R,R'$ are identical in both molecules.
[](https://i.stack.imgur.com/qnLNh.jpg)
Now I am trying here to to figure out the reactivity of the central carbon towards nucleophilic attack. Now from steric point of view, the ketonic carbon is less hindered, hence more reactive.
Now on charge separation (polarity) point of view, and this is where I am not clear, in case of alcohol, the O-H bond is polar, but to make the central carbon electron deficient the oxygen of O-H group has to pull electrons from $\sigma$ bond which is harder than $\pi$ bond of ketone.
Hence I think central carbon of ketone (all else identical) is more susceptible to nucleophilic attack.
Can someone kindly comment on my reasoning?
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[
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"\nYou are throwing snippet facts about and refusing to look at a totality. Sure ketone is less hindered [draw more appropriate structures to compare] but do not compare it only to an alcohol but also to an alkene. Next you are concerned about the polarity of the OH making the C-O bond more polar. This is confounding separate things and makes conclusions more difficult. In both cases the Carbon-Oxygen bond is polar with the carbon more positive. Your reasoning makes it at best a draw although thinking about the pi bonds has merit. You must think more about the possible mechanisms of carbonyls and monosubstituted alkanes, the intermediates that might be involved, and the possible transition states. Instead of alcohols consider sulfonate esters\n\n\n",
"2"
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https://chemistry.stackexchange.com/questions/169622/molecular-orbital-diagram-of-no
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Molecular Orbital Diagram of NO
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When we draw the molecular orbital diagram for molecules
For those that have less than or equal to 14 electrons we use the order "σ1s σ∗1s σ2s σ∗2s π2px,π2py σ2pz π∗2px,π∗2py σ∗2pz"
and for molecules that have electrons from 15 to 20 have the order "σ1s σ∗1s σ2s σ∗2s σ2pz π2px,π2py π∗2px,π∗2py σ∗2pz".
Now when we consider the case of NO or nitrogen monoxide the number of total electrons is 7+8=15 so by the above stated logic we should use the second order but from all the sources I have gone through all of them have used the first order as shown below
[](https://i.stack.imgur.com/3PWJg.png)
Why is this?
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