text:
Do symmetric hydrogen bonds in neutral molecules exist?
As far as I know, there have only few truly symmetric hydrogen bonds been observed. Unquestionable is the existence of it in the bifluoride ion, $\ce{[F-H-F]-}$, see also here. There are a couple of more, like in a water hydroxyl complex $\ce{[HO\bond{~}H2O]- @Cu($110$)}$,[1] or in the hydrogen disuperoxide anion $\ce{[O2-H-O2]-}$.[2]
Long time it was believed that malonaldehyde and acetylacetone[3] form keto-enol tautomers with symmetric hydrogen bonds. The reason for this is the lack of resolution of the experiments.
I recently came across the article about substituted malonaldehydes.[4] According to this, if you chose R1 to be an amine and R2 a nitro group, you could end up with a single well on the potential energy surface.

However, there is one response, that left me a bit puzzled. Henry Rzepa wrote:

Searches of the Cambridge crystal structure database will reveal quite a number of instances of symmetrical hydrogen bonds. These are mostly homonuclear, ie N…H…N or O…H…O. Years ago, we came across a more unusual example, of an (almost) symmetrical N….H….O heteronuclear example (http://dx.doi.org/10.1039/C39890001722 ).

So apparently there exist many. Unfortunately, he doesn't extend us the courtesy of even a single example. If you look on Google Scholar, you do indeed find a few examples of symmetric hydrogen bonds. However, most of them are either "crystallograpically" symmetric, for charged species, or both.
Also Evangelista et al. conclude their paper about malonaldehydes with the following statement, which leads me to believe, that there is no such thing as a symmetrical hydrogen bond in a neutral molecule:

A remaining goal is the identification of a substituted malonaldehyde with no barrier at all, i.e., a C2v equilibrium geometry.

Since then a few more years have passed, but my search did not turn up any more results.
Do symmetric hydrogen bonds in neutral molecules exist?

T. Kumagai, M. Kaizu, H. Okuyama, S. Hatta, T. Aruga, I. Hamada, and Y. Morikawa, Phys. Rev. B 2010, 81, 045402
S.J Knak Jensen , I.G Csizmadia, Chem. Phys. Lett. 2000, 319 (3-4), 220-222.
Daryl L. Howard, Henrik G. Kjaergaard, Jing Huang, and Markus Meuwly, J. Phys. Chem. A 2015, 119 (29), 7980–7990.
Steven Bachrach, 03 Feb 2009, Malonaldehydes: searching for short hydrogen bonds Also: Jacqueline C. Hargis, Francesco A. Evangelista, Justin B. Ingels and Henry F. Schaefer III, J. Am. Chem. Soc. 2008, 130 (51), 17471–17478.

What about diborane? Does that even count as hydrogen bonding?

@DGS This is actually quite an interesting stand. I did not think about that. I really need to wrap my head around that and maybe make a few searches if anyone has ever called it a hydrogen bond.

The big difference there is that normal hydrogen bonds are 4-electron, 3-atoms. In the case of diborane, they are 2-electron, 3-atom bonds. Could almost call them hydride bonds instead of hydrogen bonds.

@DGS Yes that's correct. They are also the prime example for 3c2e-bonding, covalent through and through. For the sake of the question, it is probably better to exclude them.

@SendersReagent IUPAC defined hydrogen bond as "an attractive
interaction between a hydrogen atom
from a molecule or a molecular fragment
X–H in which X is more electronegative
than H..." http://www.currentscience.ac.in/Volumes/110/04/0495.pdf So because boron is less electronegative than hydrogen, diborane does not involve hydrogen bonds.

@DavePhD That's what I figured. On the topic, anyone ever seen anything on a protonated proton sponge salt? Maybe symmetrical?

I think this is a great question but most answers attempt to find examples to prove that there are such cases, rather than exploring what conditions must be satisfied, and why it may be impossible to do so. I suspect the answer might be "no" for theoretical reasons: first, the acceptor/donor heavy atoms have to be placed in the right geometry to be equidistant from the well minimum. That is hard enough, but the many examples show it's (probably) not inherently impossible. Second, if you find an optimal geometry, then "orbitals from the two heavy atoms overlap" at the site of the proton.

I use quotation marks because it's the MOs you need to think about. What I wonder is if exclusion principle might not lead to the impossibility of such sharing, ie the orbitals involved are antisymmetric leading to a node. Out on a limb waving my arms feeding thoughts.

Perhaps you should add modify your question to "have symmetric hydrogen bonds in neutral molecules been reported? If not, are they possible?" Despite being interesting, as it stands the question is flawed because it is too broad: it is impossible to sample the realm of all molecules.

Here’s a recent paper arguing that the hydrogen bond in phthalates is inherently asymmetric (i.e. an equilibrium): https://doi.org/10.1021/jacs.9b08492

It appears that, perhaps, water may have symmetric hydrogen bonds between oxygen atoms, but it would take at least 60 GPa of pressure, to make water to bond like that. I guess water stops being molecular, because of that, though.
Article named very descriptively "Compression of Ice to 210 gigapascals: Infrared Evidence for a Symmetric Hydrogen-Bonded Phase" implies that:

Protonated and deuterated ices ($\ce{H2O}$ and $\ce{D2O}$) compressed to a maximum pressure of 210 gigapascals at 85 to 300 kelvin exhibit a phase transition at 60 gigapascals in $\ce{H2O}$ ice (70 gigapascals in $\ce{D2O}$ ice) on the basis of their infrared reflectance spectra determined with synchrotron radiation. The transition is characterized by soft-mode behavior of the $\nu_3$ $\ce{O-H}$ or $\ce{O-D}$ stretch below the transition, followed by a hardening (positive pressure shift) above it. This behavior is interpreted as the transformation of ice phase VII to a structure with symmetric hydrogen bonds.

On the other hand, article "The Hydrogen Bond in the Solid State" mentions:

First example of an $\ce{O-H-N}$ hydrogen bond with a centered position of
the proton: pentachlorophenol/4-methylpyridine at 100 K characterized
by neutron diffraction studies (...)

Unfortunately:

Even very small chemical changes in the system lead to loss of the
symmetry (...)

so symmetric hydrogen bonds can probably exist also in very low temperatures.
References
Goncharov, A F; Struzhkin, V V; Somayazulu, M S; Hemley, R J; Mao, H K. Science 273.5272
Steiner, Thomas Angew. Chem. Int. Ed. 41 1 71

This is certainly an interesting and unexpected take on the matter. I admit I was not hoping for something like this. Thanks for digging it up.

I think I read sth similar about carboxylic acid dimers (also pressure symmetrization). I think it's quite possible also in normal pressure in certain situations, maybe I'll find more...

@Mithoron I have been reading a bit about ice and I came across this paper (Benoit, M., Marx, D., & Parrinello, M. (1998). Nature, 392(6673), 258.) which uses DFT/PIMD trying to simulate the transition to ice X  (the phase described in the paper you link). They note that this phase actually does not have a single well, but still has a double well with the proton symmetrically delocalized across both wells via zero point energy. They do find a transition to a single well at even higher pressure, but from what I can tell this has not been found experimentally.

This question crossed my mind during a lab today because we were making $\ce{Ni(dmg)2}$, which is shown below.

Well first, this is charged in some sense, but it is net-neutral. Also, the configuration shown above is only one way to draw it, and it seems as if everything should be symmetrical and thus that hydrogen should be equally shared between the two oxygens which I believe ought to be classified as a symmetric hydrogen bond on the basis of the fact there must be 4 electrons present here where that hydrogen could be equally shared. As opposed to something like diborane which was being discussed in the comments and is better classified as a 3-centre 2-electron bond.
Indeed, I dug around and found this paper which is quite old but also very interesting! They measure the infrared spectrum of this compound and a related copper compound and find a peak at about $2322\ \mathrm{cm^{-1}}$ which is an absurd shift in frequency from a normal $\ce{O-H}$ vibration. They verify this by forming the deuterated version and finding a large frequency shift down.
They say themselves, "the infrared data interpreted in terms of the abnormal isotope effect indicate that solid $\ce{Ni(dmg)2}$ and $\ce{Ni(emg)2}$ have symmetrical $\ce{OHO}$ and $\ce{ODO}$ bonds."
So then what is this "abnormal isotope effect"? Well, they also studied $\ce{Cu(dmg)2}$ and measured the $\ce{OH}$ and $\ce{OD}$ frequencies. They measured $2382\ \mathrm{cm^{-1}}$ and $2370\ \mathrm{cm^{-1}}$ respectively. Yes, a shift of $12\ \mathrm{cm^{-1}}$ upon deuteration. This effect had apparently been observed before, and the interpretation is that when the hydrogen is present, the hydrogen bond there is symmetrical, but when the oxygen is deuterated, the deuterium bonds more strongly to the oxygen (for some unknown reason to me) and thus is shared unequally which raises the frequency (despite the lowering due to a larger reduced mass).
So not sure if that's exactly what you're looking for, but I thought it close enough to your description to post here.
Obviously what remains to be seen is whether or not this actually corresponds to a single well on the potential energy surface. I would guess that it does simply because of the symmetry present in the system along with the evidence presented above.

EDIT:
Thanks to Martin for running a calculation to see whether or not there is really a single well, and despite some good experimental evidence, he found the hydrogen bond to be asymmetric.
This paper also indicates the same. So alas we haven't found an answer yet.

References:
Caton Jr, J. E., & Banks, C. V. (1967). Inorganic Chemistry, 6(9), 1670-1675.
Bruce-Smith, I. F., Zakharov, B. A., Stare, J., Boldyreva, E. V., & Pulham, C. R. (2014). The Journal of Physical Chemistry C, 118(42), 24705-24713.

This is very interesting indeed. I will run a calculation on it to check for the single well. It could be a similar case to malonaldehyde/acetylacetone, where it appears to be symmetric at first glance only. For now, have my upvote.

I've done a really quick calculation (DF-BP86/def2SVP) and found that the complex in the equilibrium geometry has only $C_\mathrm{2h}$ symmetry. The bond lengths are $\mathbf{d}(\ce{O-H})=110.2~\mathrm{pm}$ and $\mathbf{d}(\ce{O\bond{~}H})=137.0~\mathrm{pm}$. After that I found a paper, that confirms my result: Ian F. Bruce-Smith et. al, J. Phys. Chem. C 2014, 118  (42), 24705–24713. In conclusion: unfortunately not a single well, not a symmetric hydrogen bond.

Ah bummer. I guess it's never completely safe to trust a paper from 1967 cause they obviously couldn't run the calculations we can. Nonetheless, it's a pretty interesting interaction!

Not trust is maybe too strong. I think the papers published at the time are trustworthy. You just have to factor in development of analysis tools, increased resolution of experiments, etc. Nevertheless, prior to your post I did not think about organometallics to achieve symmetric hydrogen bonds; while not the answer I was hoping for, it provides an interesting new direction. I would like to encourage you to include the reference I found in your last paragraph, just to make it clear. Not all people read the comments. In any case, thank your very much for your effort.

Couldn’t the discrepancy between $\ce{O-H\bond{...}O}$ and $\ce{O-D\bond{...}O}$ be explained by tunneling? Deuterium, being twice as heavy as protium has a reduced probability of tunneling, which could macroscopicly be interpreted as ‘sticking’ to one side more. (CC@Mart)

@Jan So then are you saying that the protium tunnels back and forth between oxygens which macroscopically makes the hydrogen bond appear to be symmetric?

@jheindel Basically, yes.

Protium is lighter. Lighter things have a greater tunneling probability, and more broadly lighter things deviate more from prequantum "classical mechanics".

In Schaefer's paper,[1] which you linked, the authors describe nitromalonamide as being the compound with the lowest calculated barrier to intramolecular proton transfer, or tautomerisation.
Earlier this month (Feb 2019), Perrins and Wu used NMR isotope shifts to show that nitromalonamide possesses a symmetric hydrogen bond.[2] This is "the first case, to be compared with the FHF− anion, of a neutral species with a single symmetric structure in solution and with a centred hydrogen".

The authors showed this by synthesising a mixture of oxygen-18 isotopologues. It's easier to explain this starting with an example of a hydrogen bond that is not symmetrical, such as 4-cyano-2,2,6,6-tetramethylheptane-3,5-dione, which the authors used. Consider the three possible isotopologues:

For the 16O2 isotopologue, because the equilibrium constant $K$ is equal to $1$, both C–3 (indicated by a solid arrow) and C–5 (dashed arrow) have the same chemical shift (they interconvert between a "keto" carbon and an "enol" carbon, but because they spend equal amounts of time in both tautomeric forms, the averaged shift for both carbons is the same).
The same can be said about the 18O2 isotopologue. However, although both carbons here (brown arrows) have the same chemical shift as each other, they do not have the same chemical shift as the carbons in the 16O2 isotopologue (blue arrows). This is because the adjacent 18O leads to a slight isotope shift.
For the 16O18O isotopologue, one might naively think that the purple carbon (C–3) should have the same chemical shift as the brown carbons in the 18O2 isotopologue. This would be true if this equilibrium constant were equal to $1$. However, by virtue of introducing unsymmetrical isotope labelling, the equilibrium constant is no longer $1$; more specifically, it favours the C–18OH tautomer, which has a lower vibrational zero-point energy. Therefore, in this isotopologue, the purple and green carbons differ from each other in chemical shift, and also differ from the shifts in the previous two isotopologues, for a total of four peaks.

On the other hand, in nitromalonamide, which has a symmetrical double bond and therefore does not exist in an equilibrium of two tautomers, we have a different situation:

For the 16O2 isotopologue, both blue carbons are equivalent by symmetry and hence exhibit the same shift.
For the 18O2 isotopologue, both brown carbons are equivalent and hence have the same shift as each other, but differ from the shift in the 16O2 isotopologue because of the aforementioned isotope shift.
In the 16O18O isotopologue, now, because there is no equilibrium to speak of, C–3 (brown arrow) has the same chemical shift as the two carbons in the 18O2 isotopologue.[3] So, we end up with only two peaks.

This is, of course, what is seen in the NMR spectra (of all three isotopologues together):

Notes and references

Hargis, J. C.; Evangelista, F. A.; Ingels, J. B.; Schaefer, H. F. Short Intramolecular Hydrogen Bonds: Derivatives of Malonaldehyde with Symmetrical Substituents. J. Am. Chem. Soc. 2008, 130 (51), 17471–17478 DOI: 10.1021/ja8060672.
Perrin, C. L.; Wu, Y. Symmetry of Hydrogen Bonds in Two Enols in Solution. J. Am. Chem. Soc. ASAP DOI: 10.1021/jacs.8b13785.
Theoretically, there should be a three-bond isotope shift arising from the remote oxygen being of a different isotope. However, I suspect that this effect is too tiny to be measured. Three-bond isotope shifts are known, especially with H/D, but a cursory search of the Internet didn't reveal anything about 16O/18O.

It seems like we should clarify if by a symmetric hydrogen bond, we mean a hydrogen bond where the proton exists in a single symmetric well with the minimum halfway between two groups, or if we mean any symmetrically distributed proton. In reading this paper, it seems to me that they cannot distinguish whether this is a single or double well because if the barrier is as small as some of the theory suggests, the proton could be evenly shared by having a relatively large zero point energy, which I think would result in the same NMR results. I wonder if this makes sense and what you think?

I find this very interesting and enlightening, but I still have doubts. On the NMR (what did they actually measure btw) timescale it appears as symmetrical, but most calculations show a double well with a tiny barrier (small enough that you may pick a DFA and receive varying results). So while this is probably as close as we get so far, it is not really what I was looking for.

@jheindel My interpretation of the paper is as follows: if the potential was a double well, then for the 16O/18O mixed isotopomer the two minima (or the two lowest vibrational states) would have different energies, which would result in different shifts for the 16O-bound carbon and the 18O-bound carbon. This doesn’t matter if the interconversion on the NMR time scale is fast or slow (it is most definitely fast), because the Boltzmann-averaged shifts for those two carbons (which is the relevant quantity for fast equilibria) will still be different (as illustrated by the cyano compound).

On the other hand, if it’s a single well, then there cannot be any such asymmetry, as there is just one minimum. In principle, a perfectly symmetrical double well cannot be differentiated from a single well, which is (I think) your line of questioning (forgive me if I misinterpret). That is true for the 16O/16O and 18O/18O isotopomers, where you only see one peak regardless of double/single well, but the authors rely on the fact that the asymmetry in a double well for the 16O/18O isotopomer is detectable via chemical shifts. @Martin-マーチン (sorry, comment ping limitations; and it was 13C NMR).

I’m not totally sure about the case where the barrier is so shallow that the zero-point energy of the molecule is larger than the barrier. That’s somewhat pathological and I’m not sure if you can measure such a thing experimentally at all, since you can’t actually detect molecules without zero-point energy to probe the shape of the PES underneath that (maybe you can, but certainly not within what I know). Then I suppose the only recourse is to computational methods.

What kind of experimental technique would be able to achieve a resolution that rules out the possibility of any sort of underlying double well existing?

@orthocresol thanks for the explanation from an NMR perspective. I normally think about vibrational problems so this is a bit different for me. It does seem like if there is a double well, this is a case where the ZPE along that proton transfer mode exceeds the height of the barrier by about a factor of 10 based on the numbers from calculations and an estimate of the frequency of this vibration.

There are cases when a double well actually changes to a single well dependent on the level of theory though. Most notably in the case of $\ce{H5O2+}$. Perhaps better calculations are needed in this system to corroborate the existence of a single well. Perhaps I will try.

@jheindel They use a more than solid level of theory in [1]. Next level would probably be full-CI optimisations at the complete basis set limit. Have fun with that, we'll talk again in a year ;P. Also, this is a case where (IIRC) BP86 would predict a single well. I'm not sure whether any experimental technique can resolve single vs. double well, but this is interesting and certainly a mind-twister. And if we indeed observe the symmetric hydrogen bond, why are these elaborate methods failing? (cc ortho, obviously)

I believe the anhydrous Hydrogen Fluoride ring (HF)6 would be an example. I'm not certain, and can't find proof in literature, but it definitely looks the part!

(image was drawn myself)
As we would expect it releases heat when forming from HF, and the reaction 6HF -> (HF)6 has an enthalpy of $\pu{-167kJ/mol}$ according to an experiment by Franck and Meyer as referenced in  Fluorine Chemistry, Volume 5 . This is on the order of what we might expect from 6 Hydrogen bonds. The moiety is stable at STP while in an HF atmosphere with a very neutral Gibbs of formation from HF of $\pu{-0.8kJ/mol}$, according to thermodynamics data taken from JANAF for HF and (HF)6. This said, these tables may have some issues, as we're exploring in another question.
In the gas phase, Hydrogen Fluoride also forms dimers, trimers, tetramers, and chains, but I think the example of the ring is easiest to see as an actual neutral bond, and not something akin a Zwitterion.
I'm still hunting for better references on this, but I did find a bulletin quoting a 1921 paper by Langmuir, of all people, saying:
"In double molecules such as H4O2 (in ice), H2F2, and in compounds such as KHF2, etc., it seems that the hydrogen nuclei instead of forming duplets with electrons in the same atom, form duplets in which the two electrons are in different atoms. The hydrogen nucleus itself thus acts like a bond in such a case. Latimer and Rodebush have made a somewhat similar suggestion in regard to hydrogen nuclei acting as bonds. They consider that the hydrogen nucleus acts on two pairs of electrons: one pair in each of the two atoms... Since the first layer of electrons in all atoms contains only two electrons, it seems probable that the hydrogen in this case also holds only two electrons and that these form the definite stable group which we have termed the duplet."
This sounds very akin to what we're hunting for, but interpreting the spectroscopic data would be unfortunately beyond my skill level. The latest and greatest the 1970s had to offer on the topic are listed in this reference on Fluorine compounds by JANAF with  (HF)6 on page 152 (1202 as listed), where they provide several references for the spectral lines. Thanks to anyone willing to give it a peak, or who may happen to know better on the topic.

This is an interesting approach to the question, and certainly has the potential to contain symmetric hydrogen bonds, as of the nature of the biflouride ion, but that brings us back to being ionic, or having large ionic contributions. I am concerned with the auto-protolysis equilibrium, $\ce{2HF <=> H+ + HF2-}$. I also ran a quick calculation on the system, it appears that a twelve-membered ring is a saddle point. While the idea remains interesting, I'm afraid it appears to not head in the right direction.

Thank you very much for giving it a look! If I'm understanding your point about the Bifluoride ion, it's that the system may be behaving partially as a salt of (HF2)- bound to H+? Electrostatically, I'd be surprised by the stable ring shape as compared to a more compact form. There's also a stable trimer (HF)3, which I think couldn't be rationalized as easily in this way, as there's not an even number of Fluorines to form Bifluorides. Could be a trick of resonance, of course. Happy to go with your instinct regardless, and thanks again for looking more deeply into the idea!

Names of elements and compounds aren't capitalised. As far as your comment goes, F-H-F in pure HF is countered by solvated cation H2F+. Oligomers of HF have "normal" h-bonds, if one would have symmetric one, it would dominate all others.

I'd like to learn more about these gas phase ions, especially their thermodynamics data like Gibbs energies. Do you know of anywhere I can read more?


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Reaction between tubli root and jatropha leaf
What compounds are found in tubli root and jatropha leaf extracts?
Would they react if they were mixed together?

Tubli roots (Derris elliptica) contain rotenone (1), a furocoumarin.

Rotenone that has been used as an insecticide, and in the U.S., it is an EPA-approved fish toxicant. According to an article it SFGate, it has been used in 2014 to completely kill the remaining fish population after a lake became infested with alien species.
Jatropha are known to contain differents toxins, such as derivatives of phorbol, a highly substituted tetracyclic azulene diterpenoid, and toxic proteins. It is worth to be mentioned that phorbol esters are known to promote tumor activity.
You might find further details in the following review, J. Toxicol. Environ. Health B Crit. Rev. 2010 ,13, 476-507, and there are numerous other articles on the toxicology of both compounds.
However, I'm not aware of studies that examined possible reactions between these compounds.

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Why can't this be another structural isomer of ethanol?
Are the two structures depicting the same molecule? Why? Why not?

If ethanol was flat, it would make sense, but it isn't.

They can be stereo isomers instead. See https://upload.wikimedia.org/wikipedia/commons/thumb/1/11/Ethanol-3d-stick-structure.svg/266px-Ethanol-3d-stick-structure.svg.png for 3d structure of ethanol.

If ethanol was static, this would be true, but it isn't.

There could be atropoisomers if rotation was blocked but it would need very low temperature to try to separate what is normally rotamers of ethanol.

Try to imagine it in a 3D fashion. You can look it at molcalc.org. The molecule rotates around the $\ce{C-C}$ bond converting one structure into the other. These rotations are very fast.
Another point is, that the representations you chose are 2D, so the could still represent the same molecule:

I changed the URL for ethanol model, the previous one gives error 404 (probably they change the part in the address that looks like a hash from time to time).

@andselisk thanks. Maybe I should come back to this again and add an animated gif, or even link to some other provider as a backup. The Pittsburgh repository comes to mind, but I guess that was still in its infancy when writing this answer.

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Is gravity an obstacle for homogenous solvation due to diffusion?
If you have a glas with a spoon of salt ($\ce{NaCl}$) in it and you carefully add the water($15^\circ\mathrm{C}$). A lot of the salt will not be dissolved immediatelly, it will take a while for full dissolution.
When the salt dissolves, it is said that the salt is mixed equally in the water. But $\ce{NaCl}$ is split in two ions $\ce{Na+}$ and $\ce{Cl-}$. $\ce{Cl}$ has a mass of $35$ and Na $23$. But due to gravity (or is really curved space in the glass a better word) you should expect that the $\ce{Cl}$ atom will be more at the bottom then the $\ce{Na}$ ion. Is that right or is there someting I'm missing?
So can you say that after an hour the salt is evenly diffused through the water and that there are probably more more $\ce{Na+}$ ions at the top of the glass, so that the top of the salt water will taste more salty due to presence of sodium, which has more power of giving salty tast?

The difference in concentrations caused by gravity is totally negligible for any reasonable purpose, and most certainly can't be detected by taste.

A 50kg ball and a 5gm ball fall at the same rate (acceleration due to gravity is independant of mass). So shouldn't Na+ ion also try to move down with an equal rate?

I presume that it is a rethorical question Surya? But indeed they has the same speed because of gravity and enertia of the equivelence principe but there is still a difference due to upward forces of the second law of Newton or/and Archimedes force which are higher for heavier objects what them make slower falling

Since the salt is dissolving without stirring, a different factor in microgravity (earth orbit) is that the salt wouldn't sink to the bottom. It would float in the solution while dissolving. Since the salt isn't all at the bottom, it should dissolve faster in microgravity.

The normal gravity differential just isn't enough. But if you had NaI and CsCl in a 4 inch diameter pipe which was 100 miles high, then some ionic differentiation would happen due to gravity. So your idea isn't wrong, you just misunderstand the scale needed.

At normal Earth gravity, for all practical purposes homogeneous solvation is not hindered by gravitational force.
However: For some solutes such as sucrose and $\ce{CsCl}$, though, the "artificial large gravity" of centrifugation is capable of establishing concentration gradients in their solutions. One example in common laboratory use is buoyant density centrifugation:

One of the most common methods of fractionating DNA is to exploit differences in density between different types of DNA in concentrated salt solutions. Caesium chloride is the most commonly used. The method involves prolonged high speed centrifugation of purified DNA in concentrated caesium chloride. The centrifugation produces a gradient of salt concentration in the centrifuge tube and the DNA migrates to occupy a position in the gradient which corresponds to its own buoyant density.


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Is there an example of a closed shell molecule with a permanent ground state magnetic dipole?
It seems that molecules of point group C$_{n}$ and C$_{nh}$ can in principle support a permanent magnetic dipole moment (Barron, 2004). However, I cannot think of any closed shell molecules that physically have a magnetic dipole in the ground state.
Is this ever observed, and can you give an example of such a case?
Citation: Barron, Laurence D. Molecular light scattering and optical activity. Cambridge University Press, 2004.

Updated 4/6/2017: It turns out that singlet ($^1\Delta_g$) molecular oxygen is paramagnetic due to orbital angular momentum only, as all electrons are paired. It has a magnetic dipole moment of $1\hbar$ and can be determined experimentally. (This makes sense from a computational perspective, given that the lowest restricted Hartree Fock solution of singlet oxygen is always complex, and only complex wave functions yield non-zero expectation values of orbital angular momentum.)

For closed shell (even number of electrons) molecules the answer is no.
After a little more reading, I found the answer. The reason is that the magnetic dipole is odd under time reversal (Table 4.1, p. 218 in Barron, 2004).
Apparently, (Barron, p. 221) "the expectation value of a time-odd operator [e.g. magnetic dipole] vanishes for states invariant under time reversal, which can always be constructed for an even electron system and hence any nondegenerate state".
I do not believe this applies to transition properties, however, as we can see magnetic dipole allowed transitions.
Citation: Barron, Laurence D. Molecular light scattering and optical activity. Cambridge University Press, 2004.

http://arxiv.org/ftp/arxiv/papers/1208/1208.6097.pdf


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How to assign oxidation number of sulfur in thiocyanate anion?
Recently I was trying to balance a redox chemical equation:
$$\ce{CuSCN + KIO3 + HCl <=> CuSO4 + KCl + HCN + ICl + H2O}$$
The equation is in principle simple to solve; one element ($\ce{I}$) is reduced and two elements ($\ce{Cu}$, $\ce{S}$) are oxidized. The number of exchanged electrons in both cases must be the same. However, my problem is that I don't know how to assign the oxidation number to sulfur in a thiocyanate anion.
Later, I solved the equation using the matrix method and via obtained reaction coefficients ($4$, $7$, $14$, $4$, $7$, $4$, $7$, $5$) and concluded that the oxidation state of sulfur in thiocyanate is zero. This result is very surprising, because the oxidation number zero in heteroatomic molecules is not common.
My questions are:

How to assign the oxidation number in such cases?
How is it possible that oxidation number of an atom bound to a more eletropositive atom is zero?

If you assume that carbon isn't reduced that you'll get such result, but it's wrong assumption.

Thanks for the idea! Only now oxidation states in relation to atom's valency make more sense.
In $SCN^{-} $ I assigned oxidation states as S=-2, C=4, N=-3
and in $CN^{-}$ as C=2, N=-3

In the end the oxidation state of sulfur does not really matter for balancing equations.  Rather, identifying the real oxidizing and reducing agents is.
Clearly the iodine in the iodate ion is the oxidizing agent, dropping from $+5$ to $+1$.  But copper, sulfur, and even carbon are all being oxidized or reduced from the cuprous thiocyanate.   We have multiple parts of the $\ce{CuSCN}$ participating in the electron exchange.
Why not then identify the whole $\ce{CuSCN}$ molecule as the reducing agent, thus covering all the reducing agent atoms in their proper proportions?  With this approach you see that the sum of oxidation states for this reductant goes from $0$ to $+7$ ($+2$ for $\ce{Cu}$ in the copper sulfate, $+6$ for the sulfur in the same compound, and $-1$ for $\ce{CN}$ in the hydrogen cyanide).  Then $\ce{CuSCN}$ gives $7$ electrons, iodate takes $4$ electrons, and thus you get $\ce{4 CuSCN + 7 KIO3}$, just like with the matrix method.

Could you add a bit on the oxidation states of S and N in thiocyanate, anyway? I came here looking for that.

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HSO4- ion equilibrium, which side more acidic
I did an experiment in involving the reaction $\ce{HSO4- + H2O -> H3O+ + SO4^{2-}}$. An acid/base indicator was added to the solution. When more   $\ce{SO4^{2-}}$ was added to the solution, it turned pinkish red indicating that the acidity has increased, pH lowered. more $\ce{SO4^{2-}}$ should mean a shift to the left, producing more $\ce{HSO4-}$ and $\ce{H2O}$. Based on Ka values, I thought the right side of the equation should be the one to show more acidity and lower pH, but this doesn't seem to be the case. Can someone explain this?

You are correct. Adding more $\ce{SO4^{2-}}$ should shift the reaction to the left, and raise the pH. I'd suspect that the $\ce{SO4^{2-}}$ wasn't pure and had some $\ce{HSO4^{-}}$.

The indicator was thymol blue. I changed from a purple blue to a pink when I put the SO4 in.  Am I correct in that the color change was  opposite what should be expected?

Thymol blue is blue at pH > 9. I don't know what you have added to what, but it might have been $\ce{HSO4-}$ to a previously alkaline solution.


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Why can't electrons be transferred using hydrogen ions?

Recognizing how oxidation and reduction occurs is essential for analyzing and understanding biochemical reactions. Which of the following methods would NOT be employed in transferring electrons in biochemical reactions?
a) As hydrogen ions (correct)
b) Through direct combination with oxygen
c) Directly as electrons
d)As hydride ions

My problem is that, it doesn't say why hydrogen ions cannot transfer electrons, also I'm not sure what "hydrogen ions" exactly mean. What if you have a hydrogen ion with 2 or 3 electrons? Why can't that be used to transfer electrons?

While H$^{-}$ exists, H$^{2-}$ would autoionize very very quickly... And, normally I would read 'hydrogen ion' as meaning the positive ion unless it was specifically called out as a negative ion.

H- is hydride, your D option... De/protonation isn't redox reaction.

Firstly, $\ce{H+}$ would be called a hydrogen ion, and $\ce{H-}$ would be called a hydride ion. It would take a lot of energy to get three electrons onto a hydrogen atom, and considering that we are talking about a biological system, this just won't happen.
$\ce{H+}$ would not be employed in the transfer of electrons because it is a lone proton, and has no electrons to give. Whatever electron the hydrogen atom originally had was left behind, and the proton now exists on its own. This, I believe, should explain your homework question.
To answer your question, $\ce{H-}$ ions are common reducing agents in non-biological organic reactions, in the forms of substances like $\ce{NaBH4}$ and $\ce{LiAlH4}$, but these are extremely unstable in aqueous solution.

Instead, cells uses NAD+ to facilitate the transfer of electrons from biomolecules, like those used in cellular respiration, in the form of $\ce{H-}$. During the transfer, one electron from the $\ce{H-}$ is transferred to a nitrogen, and the other electron along with the proton forms a bond to a carbon directly across the ring from the nitrogen, converting NAD+ to NADH, a reaction that is readily reversible.

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Determining where a group will direct an addition
I'm looking at a phenol with a ketone in one of the meta positions being treated with $$ \frac{Cl_2}{AlCl_3} $$
I know the Alcohol is the stronger activator because ketone is a deactivator. I'm wondering how determine the major product of the addition.
I'm thinking the Cl will add to the other meta location of the Alcohol group due to it wanting more space around the ring, but I'm not sure if that's correct, due to not being sure how to determine where something will direct and addition.

No point in adding AlCl3 to phenols.


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Does an irreversible reaction have an equilbrium between reactants and products?
Retrospective analysis 2/13/2017 -- The barium sulfate example is a poor choice. Equilibrium equations should really be defined using activities, and the activity of solid barium sulfate is by definition 1.
A previous question, Is every chemical reaction in equilibrium?, started much discussion. I objected to the answer that Curt F. gave and challenged him to derive the equilibrium reaction for a particular irreversible reaction. He countered with a reply indicating that I should make the specific case a separate question - so here it is. I'm going to change the reaction slightly. Given the following reaction between barium chloride and sodium sulfate, does "the" chemical equilibrium exist?
$$\ce{BaCl2(aq) + Na2SO4(aq) -> BaSO4 v + 2NaCl(aq)}$$
I contend that "the" equilibrum between reactants and products such as $$K_{\text{eq}} = \frac{\ce{[BaSO4][NaCl]^2}}{\ce{[BaCl2][Na2SO4]}}$$ doesn't exist since when the barium sulfate precipitates there could be a microgram or a kilogram as the product. Furthermore adding solid barium sulfate to the product will not shift the reaction to the left.
I'd agree that calling the reaction irreversible and saying that $K_{\text{eq}}$ doesn't exist is a tautology. So given reaction between
$\ce{aA + bB}$ to form products $\ce{cC + dD}$ then the reaction is a reversible reaction if the equilibrium such as $$K_{\text{eq}} = \frac{\ce{[C]^{c}[D]^{d}}}{\ce{[A]^{a}[B]^{b}}}$$ exists and if such an equilibrium doesn't exist then it is an irreversible reaction.
There are obviously "some" equilibrium in the reaction, but not "the" equilibrium between products and reactants.

Water has an autoionization equilibrium and $\ce{H2SO4}$ has two $\mathrm{p}K_\mathrm{a}$'s.
Barium sulfate has a $K_{\text{sp}}$.
Also the barium sulfate precipitate isn't static once formed. It is
dissolving and reprecipitating at the same rate, but the rates depend
on the surface area of the precipitate not the "concentration" (or mass) of the precipitate.

So, if I'm wrong, how do you calculate $K_{\text{eq}}$ for the overall reaction given? Assume you mix $500\mathrm{ml}$ of $0.1$ molar barium chloride with $500\mathrm{ml}$ of $0.1$ molar sodium sulfate. What is $K_{\text{eq}}$?!

I like the question, but I think that it comes down to a purely philosophical question. What do you consider to be the reactants. If you consider $\ce{BaCl2(aq)}$ the reactant, then you need to include, that these are in solution, hence it is $\ce{Ba^2+ + 2Cl- }$. Same for sodium sulfate and sodium chloride. Therefore as reactants only barium ions and sulfate ions are left, and yes, they are in equilibrium.

I understand your point but even if the reaction is assumed to be $$\ce{Ba^{2+}(aq) + SO4^{2-}(aq) -> BaSO4 v}$$ then yes there is a $\ce{K_{sp}}$ but not a $\ce{K_{eq}}$ between reactants and products.

Why not? I would use   $$K^\circ = \exp{−\Delta_\mathrm{r}G^\circ/ R T}$$ and then you can formulate your equilibrium constant in terms of activity $$K=\frac{a(\ce{Ba^2+})a(\ce{SO4^2-})}{a(\ce{BaSO4})}.$$ It obviously boils down to $K_\mathrm{sp}$ with all approximations included, but it is still a reaction which is in equilibrium.

Yes.  For a solid like BaSO4, the standard reference state is pure solid at 1 bar, and the activity to be used in the equilibrium constant expression is  1, irrespective of the amount of solid formed.  Anyway, why would the amount of solid be normalized by the amount of a separate solution phase?

@ ChesterMiller - The notion was that K isn't a "true" equilibrium constant between reactants and products since adding more solid barium sulfate to the products wouldn't shift the equilibrium to the left. // @Curt_F. also has made a detailed analysis of why an equilibrium exists using a thermodynamic argument. http://chemistry.stackexchange.com/a/43262/22102 My thermodynamics is very rusty  and I forgot that the activity of a solid is 1 until Martin posted his comment. So the chemistry works, but sidesteps the amount of barium sulfate ppt produced in the "equilibrium."

Put some water in an empty closed adiabatical box, then it will have a reaction in which a part of the liquid water will be transform as vapor water. The reaction is irreversible and there is an equilibrum betxween the reactant and the product.

@Shadock - the evaporation of water isn't really a chemical reaction of reactants to yield a product.

So explain me what a real reaction is.. :)

@Shadock - Considering thermodynamics you end up with two phases like the barium sulfate. It doesn't matter how much liquid water is left, just that there is some. The activity of liquid water will be 1.

I don't think that it is purely philosophical as suggested by Martin. I agree with @Curt_F that there will be an equilibrium. You could show that this is the case by performing the reaction as suggested then remove the aqueous phase and replace it with pure water. The BaSO$_4$ will come to equilibrium with the aqueous phase. There may not be much but it will easily be measurable.  Just because barium sulphate is rather insoluble does not mean that it is not in equilibrium. If you repeat this experiment eventually all the precipitate will be in the aqueous phase. I'm not planning to try though!

The reaction you are interested in is the precipitation of barium sulfate, a relatively insoluble salt.
The reverse of this reaction is the dissolution of barium sulfate crystals.  The dissolution kinetics of barium sulfate have been studied in a variety of systems over the decades.  Here are a few links:

Kornicker et al. 1991 reported that barium sulfate dissolution into water reached equilibrium in less than 30 minutes at 25 °C and took about 5 minutes at 60 °C.  They also presented a rate law for $\ce{BaSO4}$ dissolution of $r=k A (C_{eq} - C)^2$, where $A$ is the specific area of the barium sulfate crystals, $C_{eq}$ is the equilibrium concentration of barium sulfate, and $C$ the instantaneous concentration of dissolved barium sulfate.  This rate law had been used historically (see refs.) but these authors advocate instead for a first-order rate law.
Dove and Czank 1995 also disagreed with second-order rate law of Kornicker et al. and proposed a first-order reaction for $\ce{BaSO4}$, i.e. $r = k A (C_{eq} - C)$.

I couldn't find any references for barite solubility in sodium chloride solution, but I think it is safe to say that the solubility is finite.  Probably the kinetics of dissolution would not be more than an order of magnitude different than in pure water.
Since this dissolution is the reverse of the precipitation reaction you are interested in, and since the literature is clear that dissolution happens at non-zero rates, then both "reverse" reaction and forward reaction must be happening in your system.  That sounds like the definition of a dynamic equilibrium to me.
Note the appearance of $A$ in the equations.  The surface area of barium sulfate solid involved in the equilibrium determines the kinetics.  Usually we assume that the affect of $A$ is the same for dissolution and precipitation kinetics, so that the effects cancel out.  This is why we say that the activity of a solid phase is 1 and does not vary.  However, the effects of $A$ do not always cancel out.  The solubility of nanoparticles of $\ce{BaSO4}$ is probably higher than the solubility of bulk $\ce{SO4}$ for this reason.
Experimentally, one could probe the dynamics of the equilibrium by mixing $\ce{^131BaSO4(s)}$, i.e. radiolabeled barium sulfate, with a saturated solution of $\ce{^138BaSO4(aq)}$.  This could be done even in a sodium chloride solution.  At regular time intervals, samples of the solution would be taken, which could be filtered to remove any traces of solid particulates, and the levels of the radiotracer that had entered solution could be measured.
I don't know, quantitatively, what the outcome of the experiment would be, especially in sodium chloride solution.  But I am 100% confident that at some time scale, probably within tens of minutes, radioactivity would appear in the liquid solution.  That indicates that there is an equilibrium.
The relevant equilibrium constant in this case is the $K_{sp}$ for barium sulfate, as was alluded to in both your question and in the comments.

To continue with this question but in a more general way after a comment on the particular reaction in question.
The case of $\ce{BaSO4}$ has been chosen carefully to argue a particular point, but it seems to me that there are two equilibria involved, the first is chemical $\ce{Ba^{2+} + SO_4^{2-} <=> BaSO4_{aq}}$ the second involves no chemical reaction but is between equilibrium of $\ce{BaSO4_{aq}}$ and its solid. Because of the huge insolubility the reaction appears to be irreversible as the product is effectively removed from solution which drives the reaction very much towards products. If the reaction were to be studied in another polar solvent, one might try acetonitrile or dichloromethane for example, where insolubility is probably far less, then perhaps the true equilibrium $\ce{Ba^{2+} + SO_4^{2-} <=> BaSO4}$ would be measured.
Technically, all reactions are reversible since thermodynamics does not impose an energy difference or time-scale limit on an equilibrium, and assumes that reactants & products are not artificially separated from one another. One might consider that an irreversible reaction is one in which after some time no appreciable concentration of reactants are left, but this assumes that the forward reaction is rapid enough to occur on what ever time-scale we set ourselves. Thus a better  way is to know the reaction’s free energy and then, as this becomes the minimum  size of the reverse reaction’s activation barrier, subjectively estimate whether the reaction is going to be sufficiently exothermic to make it ‘irreversible’.  As only order of magnitudes of rate constants are required, this can be reasonably be done from a simple Arrhenius type equation. A reaction with a first order rate constant of 1/sec might be considered to be irreversible in some cases, but 1/year in others. Geochemists who study rock formation may perhaps consider these reactions as being very fast.
At the molecular level a reaction occurs because there is always a chance that, by random collisions with the surrounding solvent molecules (or other molecules in a gas or vapour), enough energy is imparted to the reactant molecules to overtop the activation barrier and transform them to products, and similarly for products returning to reactants. As the probability of reacting decreases exponentially with energy (Boltzmann distribution) this probability soon becomes vanishingly small with an increase in energy and hence for practical purposes a  reaction may, in practice, be considered to be  ‘irreversible’  or similarly products ‘un-reactive’ even though there is still technically equilibrium between reactants and products.

Also realised it change from H to Na can't be bothered changing it as it's 11:33pm and I have school tomorrow
BaCl2(aq)+H2SO4(aq)⟶BaSO4(ppt)+2HCl(aq)
Let's just think what (aq) means; it means you have ions floating about in there which have Ks with there solids
(If you start from thinking no ppt just ions dissolved) we have Ba^2+ , H^1+ , Cl^1- , and SO4^2-
Then you consider the Ks of the different salts which are BaCl2 HCl BaSO4 and H2SO4. They will all go to Ks so all the salts would be forming and being dissolved (unless blocked eg things can become supersaturated). BaSO4 has extremely low Ks so most will ppt at the same time BaCl2 will be go to Ks with the ions Ba^2+ and Cl^1- which are still in solution and H2SO4 would also go to Ks, which means BaCl2 is being formed and therefore there is a reverse reaction.
(((note if I had BaSO4 with water which would be in equilibrium (so tiny dissolved/tiny bit of Ba^2+ ions and SO4^2- ions) and I added Cl^-1 ions a negligible amount more BaSO4 will dissolve as BaCl2 would go to equilibrium reducing Ba^2+ ions leading to  negligible amount of BaSO4 dissolving to remain at Ks)))-this also show "
Le Chatelier's principle"  Yea I was trying to post this on original thred but this one everybody seems to be saying this


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Dissimilar metals and thin layers
I have a question about dissimilar metals/galvanic corrosion.
I am designing a printed circuit board (PCB), which should be mounted on bare aluminium. For cost reasons, the aluminium part cannot be treated.
The PCB itself contains copper traces
Until now, I have used HAL plating (Hot Air Leveling) on the copper traces, which is a 1 to 5 um tin plating.
This tin plating should be sufficient to form a barrier between the copper and the aluminium.
The manufacturers are moving to ENIG plating (Electroless Nickel Immersion Gold), which consists of a 3 to 6 um nickel layer on the copper with a 0.1 to 0.2 um gold flash on top.
Gold and Aluminium do not behave too well together, but I heard that since the gold is a very thin layer, one can assume it is not there, and therefore in the galvanic series, only the nickel to aluminium voltage will exist.
It is hard for me to find literature/references to this subject, especially in combination with very thin layers. This may just be because it is not my field of expertise. I myself am an electronics design engineer
Could someone enlighten me about this subject?
PS: I was wondering whether to put this in Chemistry or in Physics. In neither I could find related questions.

you better try this in physics


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why the formula of coca cola can not be find using chemistry?
i mean we have SPECTROSCOPY to stablish what molecules make a compound
and also we have advanced chemistry
so why is SO HARD to find the formula of coca cola ? simply buy a bottle examine it to check what molecules or ingredients have and copy it :D
but why is so hard ?

A rough analogy would be that, unless you are an expert baker, knowing a fancy cake contains flour, sugar, and salt won't get you very far.  I think this is a great question at its core, but probably needs to be narrowed down a bit.

Related question: Why is the recipe of Coca Cola still a secret?

I think this might be a full-on duplicate @NicolauSakerNeto, but I'll see what other people think.  I did not remember we had that question.


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How to calculate pH?
Could you please help me or give me any source to understand the calculation of pH in acid and bases both strong and weak ? And pH in salt solutions ?

https://www.chem.purdue.edu/gchelp/howtosolveit/Equilibrium/Calculating_pHandpOH.htm?

http://chemistry.stackexchange.com/questions/32769/calculating-ph-in-titration-of-monoprotic-weak-acid-with-strong-base

http://chemistry.stackexchange.com/questions/29399/ph-of-a-weak-base-calculation

http://chemistry.stackexchange.com/questions/35266/calculating-ph-weak-acid-strong-base

http://chemistry.stackexchange.com/questions/25203/calculating-vinegar-strength


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Why do the first and second ionizations of a diprotic acid produce the same hydronium ion concentrations in solution?
I was going through Oxtoby Chemistry when I got confused by the following passage in the textbook. It said that "two simultaneous equilibria are involved in the ionization of a diprotic acid such as carbonic acid":

I understand the chemical species for both the first and second ionizations barely dissociate, which means the hydronium ion production is very little.
However, what makes hydronium ion production identical (i.e. important point #1)? Shouldn't the second ionization occur to a lesser extent than the first ionization? Thanks.

Of course it is. That simply means that you put total concentration, as both equilibria happen simultaneously.

Thank you. To confirm my understanding (please correct me if I'm wrong), would this mean [H3O+] would have a coefficient of 1 (i.imgur.com/yQpgxnG.png) vs. 2 (i.imgur.com/MnHoJbL.png) in the overall equilibrium expression?

This is saying both the $\ce{H2CO3}$ and the $\ce{HCO3-}$ are both donating $\ce{H+}$ to the water to form $\ce{H3O+}$. What it means when it says "The $\ce{[H3O+]}$ in the two ionization equilibria are one in the same" is to say that both of these reactions are occurring simultaneously in the same system, and that there is only one measure of the concentration of $\ce{H3O+}$ for the system.
It is not saying that both species are donating $\ce{H+}$ to the same extent. As a matter of fact, the $\ce{H+}$ being donated by $\ce{H2CO3}$ will keep the the $\ce{HCO3- + H2O -> CO3^{2-} +H3O+}$ shifted almost entirely to the left, meaning $\ce{HCO3-}$ will be donating an extremely small fraction of the $\ce{H+}$.

This completely clarifies my understanding (especially with your second paragraph), thank you so much.


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Vapor Pressure - Sealed Vessel
I have read that vapor pressure is only affected by temperature, but all of the examples have been in a sealed vessel. I have also read that pressure does not affect vapor pressure. I am looking for clarification of if and how pressure affects vapor pressure in sealed and unsealed vessels.
Also I would be appreciative if someone could provide an in depth, intuitive explanation of vapor pressure and why it is unaffected by pressure.

I assume you are talking about a 2 phase system of vapor in equilibrium with liquid.  Strictly speaking, the term "vapor pressure" only applies to single-component systems.  If there is a second component present, like air and water in a closed sealed container, then the equilibrium of each species is described by the partial molar free energy (or fugacity) of the species in the vapor as being equal to the partial molar free energy (or fugacity) of the same species in the liquid.  Even in the case of water and air in a closed container at 1 bar (where air is regarded as insoluble), the partial molar free energy of the liquid water will be slightly different from its value in the single component species at the same temperature (Poynting correction).  Even if the gas phase were ideal, such that the partial molar free energy of the water in the gas phase were the same as its value for pure water vapor at the same partial pressure, the effect of the higher pressure on the liquid water would slightly disturb the equilibrium and the partial pressure of the water vapor would differ from its value for a single component system.
If total pressure in the two-component system were high enough (i.e., higher air pressure), the partial molar free energy of the water in the gas phase would also be affected (say by the non-ideal interaction between the water vapor and the air), but not in the same way as the partial molar free energy of the liquid water changes.  So here, even more strongly, the total pressure times the mole fraction of water vapor in the gas (which would typically be defined as the partial pressure of the water vapor for an ideal gas mixture) will not be equal to the equilibrium vapor pressure of water at the temperature of the system.  With all this said, in practice, particularly for air and water, the effect can be expected to be pretty small.

In relation to your point about vapor pressure pertaining to single component systems, if i have just water in a vessel and I increase the pressure in the vessel will the vapor pressure exerted by the water change?

The only way you can increase the pressure is to increase the temperature.  In a single component system at equilibrium, the total pressure is the same as the vapor pressure.

I have a question which says, 'What would happen to the vapor pressure of a liquid if you changed the pressure in the vessel?' what do you think would be the most suitable answer?

If it were single component, you couldn't increase the pressure if the vessel were held isothermal.  If you compressed the contents adiabatically, the temperature would rise and the vapor pressure (aka the total pressure) would rise.  If it were two component and you increased the total pressure isothermally, the vapor pressure would not change (neglecting non-idealities).

If  the container is unsealed, after it reaches equilibrium with a theoretically "infinite" atmosphere, all the vapor would have dispersed and its partial pressure would approximate zero.


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Sodium thiosulfate and Hydrochloric acid
If I was using this method (https://edu.rsc.org/resources/the-effect-of-concentration-on-reaction-rate/743.article) however used pre-prepared concentrations of sodium thiosulfate solution, which intervals would be more suitable, 0.1M, 0.2M, 0.3M, 0.4M or 0.5M, 1M, 1.5M and 2M? Also, what concentration of Hydrochloric acid should be used (I've seen a variety of suggestions elsewhere). This is for a secondary school practical. Thank you.

There's a nice demonstration of the experiment performed by Annis Hapkiewicz.
She uses a 0.15 M $\ce{Na2S2O3}$ stock solution.  She prepares 5 beakers with 50 40, 30, 20 and 10 mL of the stock solution filled up to a total volume of 50 mL with destilled water.
This corresponds to thiosulfate concentrations of 0.15, 0.12, 0.09, 0.06, and 0.03 M.
I couldn't figure out the concentration of the hydrochloric acid used, but the times for the appearance of the sulfur cloud (disapperance of the cross) were given.
The range is between 20 s and 5 min.
I'd probably go with the 2 M HCl for safety reasons and perform the experiment with the thiosulfate concentrations given above. Check the times and adjust the concentrations if necessary.
You don't want it to to fast to be measured by the kids, but you don't want the lowest concentration to take half an hour.


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Which book should I use?
I am going to take PChem I in my biochem dept next sem, which is a watered down version of the PChem course in Chem dept. It basically covers thermodynamics and a little bit of quantum stuff. The syllabus says there is no required text, since course notes will be available once we register for class. But I do want to get ahead of class by reading books. I have several books at hand, please give me suggestions on which one should I use and if you have any good free lectures, feel free to post links as well.I don't feel like a need a Quantum Chem Book, given that how little quantum is actually covered.
Intro to Molecular thermodynamics by Hanson&Green, PChem for Chem and BioSci by Chang,Pchem by Castellan.Molecular Driving Forces by Dill&Bromberg, PChem: A molecular approach by McQuarrie, An introduction to thermal physics by Schroeder, Callen's thermodynamics text and also Fermi's.
Personally, I found Hanson, Castellan, Dill, and Chang are pretty good, haven't read any other

This seems off-topic. Highly opinion-based, there is no "right" answer. You probably won't get much of an answer.

I am new here so sorry If I dont understand the  rule properly. But in fact I dont wnat the "right" answer, because I know there isnn't one. All I want is an answer or  some answers. Like someone could come up and say McQuarrie's book is not good for beginner's because of A B and C. Schroeder's book has good features such as this and that. Answers like this is good enough for me.

@Zeyuan, the problem is that answers like this aren't good enough for this site. See What types of questions should I avoid asking?, specifically a statement on chatty, open-ended questions.

But you're always welcome with such chatty questions in our chat room.

Try out Atkins, Physical Chemistry for the Life Sciences? http://www.amazon.co.uk/Physical-Chemistry-Sciences-Peter-Atkins/dp/0199564280 I can't guarantee you will like it, but it suits your needs. Wild meow is right, come over to chat to discuss this sort of thing.


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Preparation of KHSO4
This is theoretical scheme for preparing $\ce{KHSO4}$:
$$\ce{K2CO3 + 2H2SO4 -> 2KHSO4 + CO2 +H2O}$$
using 1 eq $\ce{K2CO3}$ and 2 eq of $\ce{H2SO4}$ in approx 10 eq of $\ce{H2O}$ one gets $\ce{K2SO4}$ instead. How is it possible that $\ce{H2SO4}$ completely dissociated despite $\ce{HSO4-}$ is very weak acid in water? Does $\ce{K+}$ ion somehow influence the acidity of that anion? You have to use double the amount of sulfuric acid (4 eq) to get desired product.
Also when using 1 eq $\ce{K2CO3}$ and 1 eq of $\ce{H2SO4}$ what product would you expect? Mixture of $\ce{KHCO3}$ and $\ce{KHSO4}$?

The pKa2 of sulfuric acid is 1.99 (Table of pka).  This is definitely less acidic than pKa1, but calling this a 'weak' acid may be a distraction.
The relevant question is: does $\ce{KHCO3}$ readily deprotonate $\ce{KHSO4}$?  The pKa1 of carbonic acid is 6.35 (same source), making it a weaker acid than  $\ce{KHSO4}$.  The difference is more than 4 orders of magnitude, so yes, I expect that $\ce{KHSO4}$ will more strongly favor depronotation than carbonic acid, and hence $\ce{K2SO4}$ is the product I expect.
To get at this quantitatively, consider the sub reaction:
$$\ce{KHCO3 + KHSO4 <=> K2SO4 + H2CO3 }$$
Which leads to the expression:
$$\ce{K_{eq} = \frac{[K2SO4][H2CO3]}{[KHCO3][KHSO4]} }$$
which can be rewritten as
$$\ce{K_{eq} = \frac{[H^+][K2SO4][H2CO3]}{[H^+][KHCO3][KHSO4]} }$$
Which can be rewritten as
$$\ce{K_{eq} = \frac{[H2CO3]}{[H^+][KHCO3]}\cdot\frac{[H^+][K2SO4]}{[KHSO4]}}$$
which can be written as
$$\ce{K_{eq} = \frac{$K$_{a2,\ce{KHSO4}}}{$K$_{a1,\ce{H2CO3}}} }$$
Taking the log of both sides
$$\ce{pK_{eq} = p$K$_{a2,\ce{KHSO4}} - p$K$_{a1,\ce{H2CO3}} = } -4.36$$
Which is $$\ce{K_{eq} \approx 23,000}$$ heavily favoring products, and hence, $\ce{K2SO4}$ is the dominant product


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How to get specific concentration of carbonic acid in solution?
I wanted to know if there is a way of getting specific quantities of carbonic acid? Carbonic acid is water + $\ce{CO2}$ but not all the $\ce{CO2}$ dissolves in water. So if I wanted $5\:\mathrm{g}$ of carbonic acid solution, how much $\ce{CO2}$ should I add to water? Is there a way to know how much has been added?

To get a specific amount of $\ce{H2CO3}$, you should monitor and adjust the $\mathrm{pH}$ of the water to the $\mathrm{pH}$ that would correspond to $5\:\mathrm{g}$ of $\ce{H2CO3}$. For this you would need to know the acid dissociation constant, $K_\mathrm{a}(\ce{H2CO3})=1.738 \times 10^{-4}$, and the following equations:
$$\ce{CO2(g) +H2O(l) ->H2CO3(aq)}$$
$$\ce{H2CO3(aq) +H2O(l) -> HCO3- (aq) +H3O+(aq)}$$
So now we can calculate, using:
$$K_\mathrm{a}=\frac{[\ce{HCO3-}]\cdot [\ce{H3O+}]}{[\ce{H2CO3}]\cdot [\ce{H2O}]}$$
$[\ce{H2O}]$ is omitted, as it remains mostly constant ($55.5084\ \mathrm{M}$). The only time it needs to be considered is when working with extremely concentrated solutions. From this we now get:
$$K_\mathrm{a}=\frac{[\ce{HCO3-}]\cdot [\ce{H3O+}]}{[\ce{H2CO3}]}$$
Disregarding any autoprotolysis, the $[\ce{H3O+}]=0$ before any $\ce{CO2}$ is added and any $\ce{H2CO3}$ is formed. Therefore, we know that all $\ce{H3O+}$ comes from the carbonic acid, and that $[\ce{H3O+}]=[\ce{HCO3-}]$. Therefore:
$$[\ce{H2CO3}]=\frac{[\ce{H3O+}]^2}{1.738 \times 10^{-4}}$$
Here, we can greatly simplify this equation, and introduce a parameter that we can actually measure--$\mathrm{pH}$. We know:
$$\mathrm{pH}=-\log[\ce{H3O+}] \Rightarrow [\ce{H3O+}]=10^{-\mathrm{pH}}$$
Substituting, we get:
$$[\ce{H2CO3}]=\frac{(10^{-\mathrm{pH}})^2}{1.738 \times 10^{-4}}$$
This, however, only gives you the amount of $\ce{H2CO3}$ that is not dissociated into $\ce{H+}$ and $\ce{HCO3-}$. If we want to calculate exactly how much $\ce{H2CO3}$ is in solution, even in the dissociated form, we must add to it the amount that is dissociated, which as you may recall is $[\ce{H3O+}]=10^{-\mathrm{pH}}$:
$$\text{Total}\ \ce{[H2CO3]}=\frac{(10^{-\mathrm{pH}})^2}{1.738 \times 10^{-4}}+10^{-\mathrm{pH}}$$
From this you can calculate the mass of the $\ce{H2CO3}$ in solution by multiplying this value by the volume of solution present.

Sorry but how will I know that since that will only give me concentration of CO2? would I just say 1 mole of H2CO3 gives 1 mol of CO2? So the concentration of H2CO3 will be the concentration of CO2?

or do you mean to say concentration of CO2 = (10^-pH)^2/(4.3 x 10^-7)?

Yes, sorry. They are equivalent, as by the chemical equation: $\ce{CO2(g) +H2O(l) ->H2CO3(aq)}$. I will clarify.

one more question, although H2O is not included in the equilibrium equation because it is a liquid, wouldn't the quantity of water affect how much CO2 has dissolved? If so, then how do I know how much quantity I should take to make 5g of H2CO3?

This pKa value accounts also reaction of CO2 with water. You need here to use pKa = 3,76 - OP's asking specifically about H2Co3 and it's concentration is much lower then CO2

i am not a chemistry genius so can you please explain me what pKa is and where you got 3.76?

can explain me why the assumption that bicarbonate ion concentration and hydronium ion concentration is true


text:
How does charge density affect the rate of a decomposition?
For example, for the thermal decomposition of carbonates of group 2 metals, how does the charge density decreasing down the group affect the rate of this decomposition?

Increase in charge density increases the polarizing capacity of the atom, decreasing the bond strength, thus facilitating easy decomposition

timaeus, you should have posted what you did as a new answer, not as an edit to an existing answer.


text:
Are there no d-electrons in calcium?
For calcium, the electron configuration is $\mathrm{1s^2\:2s^2\:2p^6\:3s^2\:3p^6\:4s^2}$. Does it mean there's no electron in d sub-shell? Also, how would one write the electron configuration for a calcium ion?

What you provided is the ground-state configuration for calcium.
In theory every atom "has" d-electrons, just not in the ground state. When excited (which is usually VERY rare), electrons can bump up to higher energy levels like p,d,f,etc.

Because there is a possibility for electrons to transition to an excited state at all temperatures, other than absolute zero, there are some d-electrons present in $\ce{Ca}$ as well.

An isolated calcium atom has no $d$ electrons, but calcium in at least one compound has been found to incorporate $3d$ orbitals into its bonding. Krieck et al.[1],[2] describe the inverse sandwich complex $\ce{[(thf)3Ca(μ-C6H3}\text{-}\ce{1,3,5}\text{-}\ce{Ph3)Ca(thf)3]}$, a dicalcium complex in which what are ordinarily antibonding orbitals in the organic ligand overlap appropriately symmetric calcium $3d$ orbitals to become bonding between calcium and carbon. Note especially the oxidation state of calcium, $+1$. Calcium(II) would more likely show the classical predominantly ionic bonding in which the metal valence electrons play no role except the $4s$ electrons being largely transferred to the nonmetal atoms.
Reference

Sven Krieck, Helmar Görls, Lian Yu, Markus Reiher, and Matthias Westerhausen (2009). "Stable “Inverse” Sandwich Complex with Unprecedented Organocalcium(I): Crystal Structures of [(thf)2Mg(Br)-C6H2-2,4,6-Ph3] and [(thf)3Ca{μ-C6H3-1,3,5-Ph3}Ca(thf)3]". J. Am. Chem. Soc. 131, 8, 2977–2985.
https://doi.org/10.1021/ja808524y.

Krieck, Sven; Görls, Helmar; Westerhausen, Matthias (2010). "Mechanistic Elucidation of the Formation of the Inverse Ca(I) Sandwich Complex [(thf)3Ca(μ-C6H3-1,3,5-Ph3)Ca(thf)3] and Stability of Aryl-Substituted Phenylcalcium Complexes". Journal of the American Chemical Society. 132 (35): 12492–12501. https://doi.org/10.1021/ja105534w. PMID 20718434.


text:
Creating an exothermic reaction through exposure to oxygen
Is there a way to create a label that when ripped or torn, it's triggered by oxygen and creates an exothermic reaction that heats the object it's surrounding? Say just like a cold cup of coffee or something?

There are products like this already on the market.  A product called Hand Warmers are gloves that make use of exothermic reaction to warm your hands.  And there are cold packs (that make use of endothermic reactions) to put on sports injuries.

A common hand-warmer,  as @Chester Miller mentioned, contains iron filings (with some salt, water and sawdust, which act catalytically) which oxidize ("rust") so rapidly that the packet gets hot.


text:
Brillouin's Theorem: How do other determinants mix with the ground state?
I am a bit confused about Brillouin's Theorem. I get that a singly excited determinant can not mix with the ground state wavefunction because that translates to an off-diagonal matrix element ($\langle \chi_a \chi_b|\mathcal H |\chi_r \chi_b \rangle$) and the Hartree-Fock eigenvalue problem requires off-diagonal elements to equal 0, but how about when a doubly excited determinant mixes with the ground state ($\langle \chi_a \chi_b|\mathcal H |\chi_r \chi_s \rangle$)? Both of them do not have $i=j$ thus both should be off-diagonal entries, yet a doubly excited determinant can mix with the ground state so what am I missing?

Brillouin's theorem is the direct consequence of the so-called Slater–Condon rules which for the case of the HF reference determinant $\Phi_{0}$ and a singly excited determinant $\Phi_{a}^{r}$ show that the Hamiltonian matrix element between these determinants reduces to the Fock matrix element between the corresponding orbitals
$$
  \newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
  \newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle}
  \newcommand{\op}[1]{\hat{#1}}
  \newcommand{\el}{_{\mathrm{e}}}
  \newcommand{\core}{^{\mathrm{core}}}
  \newcommand{\oneeintphys}[3]{\bracket{#1}{#2}{#3}}
  \newcommand{\twoeintphysrantisymm}[2]{\langle{#1}\,||\,{#2}\rangle}
  \bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} }
  =
  \oneeintphys{a}{\op{H}\core}{r} + \sum\limits_{j=1}^{n} \twoeintphysrantisymm{aj}{rj}
  =
  \bracket{ \psi_{a} }{ \op{F} }{ \psi_{r} } \, .
$$
Now taking the HF equations and orthogonality of spin orbitals into account we indeed get zero for $\bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} }$ since
$$
  \bracket{ \psi_{a} }{ \op{F} }{ \psi_{r} }
  =
  \varepsilon_{r} \braket{ \psi_{a} }{ \psi_{r} }
  =
  0 \, .
$$
Now try to use the Slater–Condon rules for the case of the HF reference determinant $\Phi_{0}$ and a double excited determinant $\Phi_{ab}^{rs}$ to convince yourself that $\bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{ab}^{rs} }$ unlike $\bracket{ \Phi_{0} }{ \op{H}\el }{ \Phi_{a}^{r} }$ won't vanish. This is a very well-known result which manifests itself in many different ways. For instance, this is the reason why in order to improve the HF energies in configuration interaction scheme we include doubly excited determinants first (CID) rather than singly excited once (CIS): the later won't have any effect on electronic energy.

Thanks for your answer but I still do not understand. I have used the Slater-Condon rules for the mixture between the HF reference determinant and the doubly excited determinant which yields <-ab||rs-> for the matrix element . But this is also an off-diagonal matrix element so why does it not equate to 0 as in the above case? I also know that the singly excited determinant and the doubly excited determinant can mix resulting in the matrix element <-Xb|f|Xs-> = <-Xb|h|Xs-> + sum_r <-rb || rs-> which also does not equate to 0 but it is an off-diagonal matrix element so it should equal 0 no?

Sorry, but it sounds like you have a mishmash in your head. :| Indeed, you get $\langle ab||rs \rangle$ for the case of reference and double excited determinants, but what do you mean by saying that this is an off-diagonal matrix element? To start from, which matrix you think $\langle ab||rs \rangle$ belongs to?

No need to apologize I think you are right. I think <-ab||rs-> belongs to the HF matrix no?

There is no HF matrix. However, there is a Fock one. It is composed of the elements $F_{ij} = \langle i | \hat{F} | j \rangle$ and is indeed diagonal. Now look at $\langle a b || rs \rangle$. First, it certainly not an element of a Fock matrix: it has totally different form. Secondly, it is not even an element of a square (or, in general, rectangular, or two-dimensional) matrix!

Elements of  square (in general, rectangular) matrices can be label using two indices (say, $i$ and $j$), but for elements of the $\langle a b || rs \rangle$ form you need 4 indices. So, $\langle a b || rs \rangle$ are elements of some 4-dimensional matrix, but not of the Fock matrix. Now think what is the diagonal of a 4-dimensional matrix?

Thanks for the clarification, yes the element <-ab||rs-> is not the correct form because the F operator equates to 0 because 2 spin orbitals are different but the G operator is not 0, even so yes the element <-ab||rs-> is 4-dimensional so it can not be part of the Fock matrix is my understanding correct? However with that in mind if we look at the singly excited determinant mixing with the doubly excited determinant that is an element of the Fock matrix and does have the correct form <-Xb|f|Xs-> so that should be an element of the Fock matrix right?

The diagonal of a 4-D matrix should be a 4-D vector I think

For $\langle S | \hat{H} | D \rangle$, where $S$ and $D$ are singly and doubly excited determinants there are few possibilities. If the determinants differ only by a a single orbital, then you indeed get $\langle \Phi_{a}^{r} | \hat{H} | \Phi_{ab}^{rs} \rangle = \langle b | \hat{F} | s \rangle = 0$, but if they differ by two orbitals $\langle \Phi_{a}^{r} | \hat{H} | \Phi_{bc}^{st} \rangle$ you again don't get the vanishing off-diagonal element of the Fock matrix.

And for the terminological punctuality I must mention that, strictly speaking, matrices are always rectangular (i.e. 2-D) arrays of numbers; 3- and 4-dimensional arrays can be only called cube- and hypercube-matrices. With respect to diagonal of a 4-D array being a 4-D vector, the only think I can say is that I never heard about diagonals of any array with dimension greater than 2. :) I mean, in principle one could possibly define what diagonal means in such cases, but this is certainly 0ff-topic. Whenever we say "diagonal", we mean "diagonal of a matrix", i.e. of a 2-D array.

Ahh okay i got you thanks for the clarification sir

My another mistake in comments above is that $\langle \Phi_{a}^{r} | \hat{H} | \Phi_{bc}^{st} \rangle$ will actually vanish because there are more than two different orbitals in $\Phi_{a}^{r}$ and $\Phi_{bc}^{st}$. The non-zero Hamiltonian matrix element between singly and doubly excited determinant would be, say, $\langle \Phi_{a}^{r} | \hat{H} | \Phi_{ab}^{st} \rangle$ since these determinants differ by two orbitals.


text:
What is a electronic configuration of scandium ion?
The electronic configuration of Scandium is: $\ce{[Ar] 4s^2 3d^1}$
What is the electronic configuration of Scandium ion?
Also about the electronic configuration of Cobalt, which is $\ce{[Ar] 4s^2 3d^7}$. So What is the electronic configuration if Cobalt ion?
Is it $\ce{[Ar] 4s^2 3d^6}$ or $\ce{[Ar] 4s^1 3d^7}$ ? And why?
Thanks for your help.

Strictly speaking, in general you can't determine the ground state electronic configuration of an atom through qualitative arguments, and for ions it gets even harder. We're lucky that the aufbau principle works as well as it does for how simple it is, but it makes many people overconfident in its strength.

Please use latex, and please shorten the configuration to nearest noble gas for brevity.

The electronic configuration of the cobalt is $\ce{[Ar] 3d^1 4s^2}$ the electrons with the higher energy will be snatched and the electronic configuration of Argon is very stable then with scandium you will have easely $\ce{Sc^{3+}}$.
For the Cobalt it is a bit more difficult, like you write the configuration you cannot find a correct answer. For all element first you do first the configuration using the Klechkovsky rule and after you put all orbitals by principal quantum number growing.
So the cobalt configuration is $\ce{[Ar] 3d^7 4s^2}$ then if you snatch two electrons from the $\ce{4s}$ orbital you have a stable configuration for the $\ce{Co(II)}$ ion. You cannot snatch them from the $\ce{3d}$ orbital (even if $\ce{[Ar] 3d^5 4s^2}$ looks stable because the $\ce{3d}$ orbital is half full then the spin is maximal) because its energy is lesser than the energy of the $\ce{4s}$ orbital.
You can find the Cobalt at different oxydation states from $\ce{+I}$ to $\ce{+IV}$ but it depends on what you have in your solution or in your gas if you have a gas.

NB :
Remember that the configuration of the elements are given if gas phase, then for example the more stable configuration of the copper iron is for $\ce{Cu^+}$ and not for $\ce{Cu^2+}$, $\ce{Cu^2+}$ is stable in water, so the answer may depends on the problem you have.

Explaination for the copper :
Stability in aqueous conditions depends on the hydration energy of the ions when they bond to the water molecules (an exothermic process). The $\ce{Cu^{2+}}$ ion has a greater charge density than the $\ce{Cu^+}$ ion and so forms much stronger bonds releasing more energy.
The extra energy needed for the second ionisation of the copper is more than compensated for by the hydration, so much so that the $\ce{Cu^+}$ ion loses an electron to become $\ce{Cu^{2+}}$ which can then release this hydration energy.
I hope it can help you !

"You can find the Cobalt at different oxydation states from +I
to +IV". Now +V too.


text:
Process to separate Lauric Acid from Coconut Oil
I am trying to find a (relatively) simple process to separate lauric acid from other MCTs contained within coconut oil.   I am wanting to obtain a separation of +-5% purity.
One process I am trying is heating the coconut oil until it is completely liquid (~50'C), and then slowly cooling it to ~22'C using a fridge I modified with a digital temperature controller. This process sometimes works in making "solid clouds" of lauric acid in the oil, but not sufficient to separate easily from the remainder of the oil.
Is there something I can try in order to separate lauric acid from the coconut oil, using relatively simple processes?
Any comments welcome.

With respect to economy, waste management and regulations concerning food processing, your current approach, a solvent-free crystallization, is probably the most reasonable approach.
There's a Unilever patent which claims that the crystallization, both stirred and quiescent, is significantly improved by addition of sucrose laureates.
The patent holders attribute this to the observation that the additive leads to the formation of

large and non-porous spherulites, which highly increases separation efficiency


text:
Non-carbon based life forms on hot planets
I am writing a story with people made of elements other than carbon. A couple of planets in this series are really hot, so I need compounds with high melting points.
For example, there is a lone girl who lives on Venus, which is hot enough to melt lead. I decided to use tungsten carbide for her house.
But on another planet, there are humanoids made of silicon who survive at higher temperatures than humans on earth.
Is this scientifically possible?
If you're interested in my book and don't mind a bit of self-promotion, watch for the next few years and keep an eye out for:

The New Worlds Series #1: Into the Flames
The New Worlds Series #2: Call of the Cybans
Milky Fight #1: The Authority Gains Strength.

Google for Refractory metals, Superalloy,Ultra-high-temperature ceramics etc

I don't want to sound mean, but this is borderline self-promotional. We're happy to delve in chemical aspects of a story's background, but I don't see the relevance of your books' names prior to having written them, nor the current title of the question.

Welcome to Chemistry.SE! I like to see writers coming here to get some scientific background for their works. Nevertheless, I took the liberty to (heavily) change your question and make the self-promotion criticized by Nicolau (for good reason) a bit less obvious. If you think that it contradicts your intentions, please feel free to roll back.

If you come here more often, which I hope, because I like the scifi and crime story questions, please consider to register your account and make it permanent. Otherwise, it's just based on cookies and might get lost.

Related http://chemistry.stackexchange.com/questions/18843/what-makes-carbon-special-and-versatile. Not a dupe though I think.

Given that a $\ce{C-C}$ bond dissociation energy is $83\ \mathrm{kcal\ mol^{-1}}$, and $\ce{Si-Si}$ bond dissociation energy is $52\ \mathrm{kcal\ mol^{-1}}$, a silicon based life form is much worse off on a high temperature planet. It's actually for this very reason that there aren't nearly as many silicon based compounds as there are carbon based compounds. Thats not to say they don't exist, however. There are silicon hydrides that in the form of silicon chains, analagous to alkanes, existing up to about 8 silicon long. They are all readily hydrolyzed and are thermally unstable, especially the longer chains.
Silicon is also not quite as versatile as carbon in terms of the types of bonds it can form. Carbon is capable of forming single, double, and triple bonds due to its ability to hybridize its $\mathrm{s}$ and $\mathrm{p}$ orbitals. Silicon is a larger atom than carbon, and as a result it is much more difficult for its $\mathrm{p}$-orbitals to overlap with the $\mathrm{p}$-oribatls of adjacent atoms. Again, this is of course not to say that this doesn't happen, as it does in the case of diselenes and disilynes, but again, these are (very) unstable.
Most often silicon is found combined with oxygen, as in the very stable $\ce{SiO2}$, or as you might know it: glass, sand, or quartz; not exactly ideal for a moving, living creature. Just an idea, but if you're trying to make scientifically feasible life forms on a hot planet, you might just be better off having carbon based life forms in an inert atmosphere. In other words, an atmosphere that doesn't lend to them bursting into flames.
I'm not sure how set you were on making these different races out of other elements, but no element by itself really matches the chemical variability that carbon offers.
EDIT: After doing a little more research, I thought it might be beneficial to include polysiloxanes, or silicone, which consist of $\ce{C-(Si-O)_{n}-Si-C}$ bonds. These actually are actually very interesting because they are extremely elastic, functional over a wide range of temperatures ($-100^\circ \mathrm{C}-300^\circ \mathrm{C}$), resistant to ozone and UV radiation, do not promote bacterial growth (but who knows about extraterrestrial bacteria), have a low chemical reactivity, high electrical resistivity, and most interestingly, have self-healing properties.
I would now change my tune and say that maybe a race of silicon-carbon based life forms isn't too far fetched at all! At least for sci-fi writing purposes.

Polysiloxanes seem pretty heat resistant, fluid and versatile. Maybe mixing silicon, oxygen and carbon could be a way to go? Kind of locks you in a system, but I could see some biological ability. Carbon definitely could do variability better, though, since it wouldn't be locked in that pattern.

@DGS I definitely kept thinking about the possibility of carbon-silicon compounds even after I posted the answer... Maybe I'll make an additional section about them.

Only one can evolve, And you should know which of the two can thrive in the environment suggested.


text:
How is enthalpy change equal to heat supplied at constant volume?
We know that:
$$\begin{align}
H &= U + pV \\
\Delta H &= \Delta U + \Delta (pV)
\end{align}$$
Now,

If $p$ is constant,

$$\begin{align}
\Delta H &= \Delta U + p\Delta V \\
&= q_p + w + p\Delta V \\
&= q_p \qquad\qquad  \text{(since } w = -p\Delta V\text{)}
\end{align}$$

If $V$ is constant,
$$\begin{align}
\Delta H &= \Delta U + V\Delta p \\
&= q_V + w + V\Delta p \\
&= q_V + V\Delta p \qquad \qquad \text{(since } w = -p\Delta V = 0\text{)}
\end{align}$$
We also know that at constant $V$,

$$\Delta U = q_V$$
Now my question is that: my book says that at constant $V$, $q = \Delta H = \Delta U$, which I think is false, since I obtained $\Delta U = q_V$, but not $\Delta H = q_v$.
Please tell me where am I going wrong?

Here is a link to the chapter.
Please refer to Pg 161, right column, first paragraph.
Link to chapter: Thermodynamics

Just a note: from what I've seen, $\LaTeX$ in titles is discouraged.

@DGS Removed latex!

Apparently, it's mostly the "ce" that is a problem. But here are the specifics, if you're interested. Oh, and it looks funky on search engines.

@DGS Thanks ill make a note of it. although at other sites, I can't do without latex in titles ( eg math SE) !

I think in the book they take both pressure and volume to be constant when showing you this result since they refer to $\Delta{H} = \Delta{U} + p\Delta{V}$ (eq 6.8) in their derivation which is enthalpy at constant pressure. I agree it's bit confusing.

The book is indeed wrong

@wuschi Yes exactly. They initially took isobaric condition for $\rm q_p$. Now coming back to $q_v$, they considered conditions to be both isobaric and isochoric, while they should've only considered it to be isochoric only.

The text mentions that for systems of solids/liquids, so I think it's treating both v and p are constant in that one expression due to those conditions. The text goes on further to say that the difference becomes significant when gases are involved (p.161)

No the book is not completely wrong.If you read the next 6 lines then you will find that it wants to point that this approximately true only for solids and liquids, not for gases.
From the First Law,
$$\Delta U = q + w$$
and since work done is $0$ in an isochoric process (constant $V$),
$$\Delta U = q_V$$
Furthermore,
$$\begin{align}
\Delta H &= \Delta U + \Delta (pV) \\
&= q_V + \Delta (pV)
\end{align}$$
Since we are considering solids and liquids, the changes in the volume of these matter upon changing pressure are negligible (as they are incompressible matter) when compared to gases, so the $\Delta (pV)$ term can be considered negligible.
Therefore
$$\boxed{\Delta H \approx \Delta U = q_V}$$
for these incompressible solids and liquids. However this fails when we consider gases as the term $\Delta (pV)$ will no longer be negligible.
Comparison between the $\Delta (pV)$ values for different states of matter, being heated from $20\ ^\circ\mathrm{C}$ to $30\ ^\circ\mathrm{C}$:
$$\begin{array}{c|c}
\text{Substance} & \Delta (pV)\text{ / J} \\
\hline
\text{Air} & 2850 \\
\text{Water} & 0.1 \\
\text{Iron} & 0.0004
\end{array}$$

You are right, its true for solids and liquids, but it wasn't mentioned specifically so I got confused. I think your answer is very relevant to this question. +1


text:
Dow's Process for the manufacture of phenol
When chlorobenzene ($\ce{C6H5Cl}$) reacts with aqueous ($\ce{NaOH}$) solution it first forms sodium phenoxide ($\ce{C6H5ONa}$), and then phenol upon reacting with $\ce{H+}$.
But my question is, why can't $\ce{Na}$ directly replace $\ce{Cl}$ from chlorobenzene by breaking $\ce{Na-O}$ bond instead of breaking $\ce{O-H}$ bond?

You mean the Na from the NaOH?

Can you clarify what you think the mechanism is for this reaction? I don't quite understand your question.

@Simon-Nail-It Yes sir, Na from NaOH

So your question is why can't $Na^{+}$ react directly with chlorobenzene instead of reacting with phenol?

@RokNarobe No sir, i mean to say that why is the long process used when we can just break the C-Cl bond in benzene and then Na+ can react with Cl to form NaCl and phenyl cation can react with OH- to give phenol. Why is Sodium Phenoxide is being formed before giving phenol ?

please  provide a clear idea about the mechanism of reaction in your mind.

The mechanism of this reaction is that $\ce{NaOH}$ being a strong base with some extreme conditions abstracts  H from C adjacent to C containing Cl while Cl also leaves.
This creates benzyne as an intermediate.
Subsequent addition of $\ce{NaOH}$ forms the phenoxide ion.


text:
How to choose which reactants to have in excess?
In a precipitation reaction, how do I know which soluble reactant I should put in excess?
E.G.
$$\ce{Na2SO4 +Pb(NO3)2 -> PbSO4 + 2NaNO3}$$
In my book, $\ce{Na2SO4}$ is added in excess, but can we add $\ce{Pb(NO3)2}$ in excess?

Sometimes the excess is the cheap/unlimited reagent you use to get a high yield on your expensive/limited reagent, and sometimes you use the one that you don't mind/can easily extract from the products.
In this case, if you wanted to collect and record the lead(II) sulphate as a precipitate, using an excess of the lead(II) nitrate would be a waste.  If your intention was to produce a large quantity of sodium nitrate solution, and you don't mind having a load of the lead nitrate contaminating it, then you could use an excess of the lead nitrate.

How will the lead(II)nitrate be a waste?

In that instance it would be wasted because your intention is to convert lead nitrate into lead sulphate.  If you have excess lead nitrate then your % yield would be low and it will not tell you how much lead was present in the solution which may be relevant, even though you produced the maximum amount of sodium nitrate.


text:
Side Products of Dow's process
In Dow's process of manufacture of phenol, $\ce{PhCl}$ is fused with $\ce{NaOH}$ at high temperature and pressure. What are the possible side products? And what is the intermediate formed?
I think the intermediate will be benzyne. After the formation of benzyne, $\ce{OH-}$ nucleophilic attack takes place. the reaction mixture also contains some phenoxide ions. So Diphenyl ether is a possible side product (Formed by nucleophilic attack of $\ce{PhO-}$). But the answer given species two side products: Diphenyl ether and p-phenyl phenol. How can I explain the second product?

$\ce{C6H5ONa}$ might again react with left over $\ce{C6H5Cl + NaOH}$

Before we can consider the reaction, we must consider the conversion of the reactants into the reacting species:

Now, we may consider the reaction occurring between these two species:

The product of this reaction is now easily converted into para-phenyl phenol by an acid workup.

The benzyne should be write between square brakets because it is not an isolated compound.

@Shadock It's intermediate not transition state; otherwise you'd be right.

@Mithoron Non-isolated intermediates are typically written in brackets. Transition states are written in brackets and include the double-dagger notation.

I actually made a lot of technical mistakes. I hope this fixes them. My ChemDraw file got corrupted, so I had to do it all over again from scratch...


text:
Which compound is more reactive towards acetylation, 4-aminophenol or 4-aminocyclohexanol?
Does the delocalisation of the lone pair of electrons in the NH2 group mean that they are less available to attack the carbon making the 4-aminocyclohexanol more reactive? Or does the polarisation of the neighbouring molecule due to the higher electron density of the benzene ring make the 4-aminophenol more reactive?

p-Anisidine and DMAP are especially nucleophilic specifically because of the resonance caused by situations like what you describe. That's also why they're so toxic. Not an answer, but certainly related.

4-aminocyclohexanol should be more reactive towards acetylation because the reactivity depends on the availability of the lone pair on N to be able to attack the C atom during acetylation. Since in 4-aminophenol the lone pair resonates in the benzene ring, its availability to attack will get reduced making it lesser reactive towards acetylation.


text:
Kolbe's electrolysis method for getting alkanes
It involves the electrolysis of "Sodium and Potassium salts of Carboxylic Acids". Why only Sodium and Potassium? Will some other metal like Li, Mg, Zn etc. not work? As long as the element is a better reducing agent than Hydrogen, won't our destination be attained? What is the role of the metal here?

The carboxylates are typically prepared from the corresponding acids using metal hydroxides. Sodium and potassium hydroxides are dirt cheap, think in chloralkali electrolysis.

The metal does not play any role other than to form soluble salts here. The metal cations move to the cathode while carboxylate anions move to the anode where oxidation to a $\ce{R-COO·}$ radical takes place. These radicals decompose to $\ce{CO2}$ and $\ce{R·}$ radicals. Two of latter combine to form a C-C bond.

I see, so this mechanism with work with every other metal whose carboxylic salt is soluble in the solvent? Thanks!


text:
Approximate magnetic susceptibility of these liquid propellants?
This question asks if the comparatively large paramagnetic susceptibility of liquid oxygen (LOX) is considered when designing rocket tank and engine systems. This comment suggests fuels such as RP-1 (kerosene-like) and liquid hydrogen (often used in liquid-propellant rockets) do not have a susceptibility as large.
I'd like to find either the approximate or even ballpark values for the magnetic susceptibility of the following four materials, or at least a source where they might be found:

liquid oxygen (LOX)
liquid hydrogen (LH2)
RP-1 (rocket fuel), and
liquid methane (LNG)

I have tried google searches of various permutations of terms, but so far no luck beyond the one value I was able to cite in that question (LOX susceptibility here).
I'm fairly sure that the susceptibility of the other three are much smaller (100× or more) than that of LOX. For example here is what I can do now:

LOX: use the value reported in the link above; 3.5E-03
LH2: This question links to this table which lists the susceptibility of liquid hydrogen -5.44E-06 cm³/mol. If I use a density of 0.07 g/cm³ and molar mass of 2.0 g/mol (both from here) I get approximately -1.9E-07.
RP-1: decide that for paramagnetic and diamagnetic susceptibility, kerosene and crude oil are very roughly equivalent and use -1E-08 m³/kg from section I of this along with a density of order of magnitude 1000 kg/m³ to get a susceptibility of roughly -1E-05
LNG: - same argument as RP-1, roughly -1E-05

Liquid oxygen will be the only paramagnetic substance in this list (it has two unpaired electrons). The others are all diamagnetic. But oxygen is an oxidiser not a full and the others are all fuels. So is the question whether the oxidisers associated with the fuels are paramagnetic?

@matt_black if you look carefully you'll see I never called oxygen a fuel. I did however call it and all the rest propellants, which is the standard terminology in rocket-speak. There are some great videos in Is the Paramagnetism of Liquid Oxygen Ever Considered in Engine or Tank Design? and it's already helpful that you point out that the others propellants are likely to have much weaker responses to magnetic field than LOX, thanks!

As it has been pointed out, molecular oxygen is paramagnetic. It is exceptional in that. Usually, you require either a transition metal ion, or a lanthanide ion, or a radical, for a molecule to be paramagnetic.
For most other molecules, including as far as I can tell the fuels you mention, you can safely assume that they are diamagnetic. Paramagnetism is relatively tricky and intense, diamagnetism, at least in its magnitude, is weak and comparatively straightforward to estimate.
If one wants an inexpensive, ballpark estimate of diamagnetism, many people use Pascal's constants, additive terms considering the involved atoms, with corrections for certain bonds. Rather than Pascal's original work, maybe check out the widely cited Gordon A. Bain  and John F. Berry, Diamagnetic Corrections and Pascal's Constants, J. Chem. Educ. 2008, 85, 4, 532.


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Role of Cu in Corey-House synthesis
What does $\ce{Cu}$ do? And why do we need it? Can we not do $\mathrm{S_N}2$ without it? And is there any other metal that can help in this? Maybe something like $\ce{Ag}$? I'm guessing this because the special thing about $\ce{Cu}$ is that it is worse reducing agent than $\ce{H}$.

Copper is used to avoid lithium-halogen exchange. Additional information on that reaction can be found in this discussion. Organolithiums will often, instead of acting as a nucleophile, form a new organolithium: $$\ce{RLi + R'X <=> RX + R'Li}$$ Using $\ce{Cu(I)}$ allows the formation of an organocuprate, which acts as a nucleophile, allowing alkylation rather than metallation: $$\ce{2RLi + CuX -> LiR2Cu + LiX}$$ $$\ce{LiR2Cu + R'X -> R'R + RCu + LiX}$$ I'm not sure if silver could be used instead; I'm not seeing any reference for it, but I may just not be searching the right terms. The Wikipedia page on organosilver chemistry isn't terribly extensive, but I'm guessing stoichiometric use of silver, while not terribly expensive as transition metals go, would certainly be more so than analogous use of copper.

Update:
It seems that analogous chemistry cannot be performed with silver or gold. Lithium diorganoargentates ($\ce{LiR2Ag}$) are apparently too unstable. Lithium diorganoaurate ($\ce{LiR2Au}$) complexes are apparently either not good nucleophiles, or the triorganogold(III) ($\ce{R2AuR'}$) intermediate complexes are too stable to perform similar chemistry. This is per this 2005 paper. A fair amount of gold chemistry has come out since then, so I'm not sure how much of this has changed since.

Second update (though I should have answered this to begin with):
Also, I'd like to add that this is not an $\text{S}_\text{N}2$ reaction. The organocuprate undergoes oxidative addition into the $\ce{R'X}$ bond. After that, it loses the $\ce{X^-}$ ligand to form $\ce{LiX}$ and $\ce{R2CuR'}$, then undergoes reductive elimination to form $\ce{R'R + CuR}$.

I see, that helps, but why does RLi undergo that reaction? Isn't R-Li a strongly polarized bond? What kind of bond is it actually? And what reaction is this undergoing?

The Li-X exchange mechanism isn't really known for sure yet. This presentation from the MacMillan group talks about three proposed mechanisms: radical exchange, nucleophilic attack on the halogen, and a four-centered intermediate. I would guess it doesn't undergo normal nucleophilic attack due to hard-soft acid-base chemistry, i.e. poor orbital overlap. I'm not totally sure, though.

I've updated my answer to include a link to another Chemistry.SE answer about lithium-halogen exchange.