text:
Why does ethene not disintegrate upon excitation?
The average c=c double bond has somewhere around 614 kJ/mol of energy, if we convert it to eV, then the bond strength is roughly 6.4 eV.
Additionally, Ethene absorbs light at 171 nm.
However, if we calculate the energy of 171 nm it is around 7.3 eV, hence ethene absorbs energy that is higher than its bonding energy, which should lead to a complete disintegration of the double bond. So why does it not happen?
I'm fully aware that this does not happen, otherwise most bonds would break at around 300-400nm light excitation, however, I have not seen the reason why discussed.

Who said all that energy must go into this one bond?

Well, no one, however, the opposite is also true, I have not seen a discussion about such dissipation. It should also translate in Stokes shift change, however, in publications, I have never seen someone state or imply that the stokes shift or a part of the Stokes shift is caused by such mechanisms.

On uv excitation an excited state will be produced so that reaction will not occur from the ground state. The dissociation of the excited state could should be less than that of the ground state but if, for example as a guess, it is found at 6 eV above the ground state it could be stable with a dissociation energy of say 2 eV and decay by fluorescence or isomerisation (by curve crossing/conical intersection)  or other non-radiative pathways.

Yes, such mechanistic pathway is probably expected and is somewhat supported by the Vavilov's rule, however the ''probably'' is my main issue. Does anyone perhaps have any sources on mechanistic studies etc. on this?


text:
What is total enthalpy?
I've seen that it's defined as $h_0 = h + \frac{v^2}{2}$, where $h_0$ is the total enthalpy per unit mass. I can see that the $\frac{v^2}{2}$ term probably has something to do with kinetic energy, but I am still really confused. Doesn't $h$ already account for flow work? And also, how can both sides of the equation have the same units if $h_0$ is "per unit mass"?

In chemistry, we don't usually throw things around, but if we will, they'll surely have an additional energy of $mv^2\over2$.

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text:
CO-vibration in metal carbonyl
I'm struggling to understand the trend of $\ce{CO}$-vibration in isoelectronic metal carbonyls.
$$
\begin{array}{lr|lr}
\hline
\text{Compound} & ν/\pu{cm-1} & \text{Compound} & ν/\pu{cm-1} \\
\hline
\ce{Ni(CO)4} & 2060 & \ce{[Mn(CO)6]+} & 2090 \\
\ce{[Co(CO)4]-} & 1890 & \ce{Cr(CO)6} & 2000 \\
\ce{[Fe(CO)4]^2-} & 1790 & \ce{[V(CO)6]-} & 1860 \\
  & & \ce{[Ti(CO)6]^2-} & 1750 \\
\hline
\end{array}
$$
Since the complexes are isoelectronic, the reason behind the trend should lie in the atomic charge $(\ce{Ni} > \ce{Co} > \ce{Fe}).$ Higher atomic charge means more attraction between metal cation and the ligand electron pair → higher electron density → more π-backbonding → weaker $\ce{C#O}$ bond → lower wavenumber.
But it is not the case here. I'm not sure why. I think it has something to do with the energy level of HOMO.

I think the mistake you're making is this conclusion: "Higher atomic charge. . .  -> higher electron density -> more pi-backbonding". With higher (ie more positive or less negative) atomic charge, there is less electron density donation towards the ligand, because the electron density is pulled to the metal center more strongly. Thus, $\pi$ back-bonding is weaker and C-O bond is stronger.
The more positively charged metal centers do not pull enough electron density from CO in the sigma interaction to be able to donate more back through pi interaction, which seems to be the way you're looking at it.

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text:
Which compounds react positively in the aniline dye test?
I read that to test for the presence of phenol groups in an organic compound the aniline dye test can be performed. If the solution has phenol present it gives a red or orange precipitate to confirm the presence, however as a side note it was mentioned that this test is applicable for the detection of small quantities of phenols, which have a free para position. Why is it so?

Umm, why don't you just read up about synthesis of azo compounds? That should make things clear.

First, the aniline is converted to a diazonium halide salt in the presence of acid and sodium nitrite, per this reference.
Then, the diazonium salt reacts with the phenol to give an azo compound, per this reference (see the section entitled Coupling reactions of diazonium ions
Note that the diazonium ion is shown reacting with the para position of the phenol. That is because the -OH group in phenol is an ortho,para-director (reference), meaning that it enhances the nucleophilicity of the ortho and para positions on the phenol ring. However, the para position is significantly more nucleophilic because of the relative lack of steric hindrance in comparison to the ortho positions, which are hindered by the proximity of the -OH group.
So, having a free para position on the phenol means that the phenol is especially reactive towards electrophiles, including the diazonium salt in this example, meaning that it will react faster and in higher yields - increasing the sensitivity of the aniline dye test.


text:
Searching for a solvent with a low carbon dioxide solubility
I would like to use a liquid solvent for some experimentation with a carbon dioxide solubility as low as possible under usual pressure and temperature conditions (say, $\pu{1 atm}$ and $\pu{293 K}$).
Until then, I was unable to find much information on the topic:

A study of $\ce{CO2}$ solubility in water and seawater by Weiss [1], indicating that $\ce{CO2}$ is less soluble in saline water than in pure water.
Publications concerning several organic solvent (acetone, cyclohexane, ethanol, methanol, etc.), a sum-up in IUPAC SDS Vol 50 [2]. To the naked eye, it seems that all of them can dissolve much more $\ce{CO2}$ than water, being poor candidates to answer my needs.
Publications concerning olive, linseed or other origin oils (see Simon and Gutknecht [3] for instance). However such fats seems to also have a higher solubility coefficient than water.

I do not know if such liquid does exist, but it would be great. Up to now, my best candidate is salty water.
References

Weiss, R. F. Carbon Dioxide in Water and Seawater: The Solubility of a Non-Ideal Gas. Marine Chemistry 1974, 2 (3), 203–215. https://doi.org/10/cb6n2j.
Carbon Dioxide in Non-Aqueous Solvents at Pressures Less than 200 kPa; Fogg, P. G., Clever, H. L., Eds.; Solubility data series; Pergamon Press: Oxford, 1992. ISBN 978-0-08-040495-0. PDF
Simon, S. A.; Gutknecht, J. Solubility of Carbon Dioxide in Lipid Bilayer Membranes and Organic Solvents. Biochimica et Biophysica Acta (BBA) - Biomembranes 1980, 596 (3), 352–358. https://doi.org/10/bj647m.

Update
I understand the "like dissolves the like"[1] rule of thumb so that a highly polar solvent would be the worst for a non-polar molecule such as $\ce{CO2}$. Water as a solvent would then be the best choice (possibly salted to further reduce its $\ce{CO2}$ solubility) for my needs.
But as Rolf pointed out, the dissociation of $\ce{CO2}$ into carbonate ions also happens and might increase its solubility. For such reasons an aprotic solvent could be useful.
However, when comparing between water (molar fraction 0.00061) which is a strongly polar (dielectric constant: 80 and dipole moment 1.85$~$D) but protic solvent and dimethyl sulfoxide (molar fraction 0.0091[2]) which is also polar (dielectric constant: 47 and dipole moment 3.96$~$D) but aprotic solvent, dimethyl sulfoxide as a higher solubility for carbon dioxide than water.
What can be the reasons for such unexpected outcome? Are my reasonings and assumptions correct or am I flawed in my thinkings?
References

Polar Protic? Polar Aprotic? Nonpolar? All About Solvents. - Master Organic Chemistry. https://www.masterorganicchemistry.com/2012/04/27/polar-protic-polar-aprotic-nonpolar-all-about-solvents/
Li Hua & Chen Wanren, 2005, Solubility of dilute SO2 and CO2 in dimethyl sulfoxide, Physics and Chemistry of Liquids, 43:3, 289-298, DOI: 10.1080/00319100500083583

You're going the wrong way; check out highly polar solvents.

Solubility of CO2 in water (total dissolved inorganic carbon, or DIC) also controlled by stepwise formation of carbonic acid (i.e. H2O + CO2(aq) = H2CO3 = H+ + HCO3– = ...) bicarbonate, and carbonate ion.

Liquid mercury is a well-established gas seal, it doesn't solve any gases. Though it isn't exactly an organic solvent =). Gallium would work too.

To find a solvent with low solubility for non polar gases like CO2 you have to compare the polarity of the solvents. A very polar solvent like water is the best choice.
There's a related question here: Dissolving of non polar gases in water (liquid)
At the end of this answer I have added two links. The first one gives you the solubility of CO2 in water. According to this source, at room temperature and around atmospheric pressure of Co2, the molar fraction in water is 0.00061.
The second link compares many other solvents. At room temperature and 1atm of CO2 the molar fraction is 0.012 in hexane that is the lowest for straight chain alkenes(Fig.1). Overall, among alkanes and cyclic alkanes, cyclohexane is the only one that at room temperature has a CO2 solubility below 0.01(fig.2).
Among aromatic hydrocarbons, you can reach a molar fraction as low as 0.006 if you use 1-Methylnaphthalene(fig.3) and this value is interestingly similar to what you can reach using methanol(you can find a graph comparing alcohols there as well(pag.129). Apparently the CO2 solubility with aromatic compounds is comparable to alcohols(polar).
With all this in mind, water is the best option and if you add salts the solubility of CO2 decreases further as you mentioned.

Solubility of Carbon Dioxide in Water:
https://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/solubility%20of%20carbon%20dioxide%20in%20water.pdf
CARBON DIOXIDE IN NON-AQUEOUS SOLVENTS:
https://srdata.nist.gov/solubility/IUPAC/SDS-50/SDS-50.pdf


text:
Doubts Regarding Ceric Ammonium Nitrate Test for Alcohols
I read about Ceric Ammonium Nitrate Test for Alcohols and a red colour is produced if the solution has an alcohol functional group however as a side note it was mentioned that this test is only useful for compounds having less than 10 Carbon atoms. Why is it so?

I have a strong feeling you could have put those two questions together - while you could think that reasons would be different, that's virtually impossible - such difference in length of carbon chain simply doesn't affect the reactivity of OH group, so solubility is the thing here.

@Mithoron ok I've been told in the past to make separate questions so to avoid that I broke it into 2

Ah, it may be problematic to decide, but these should be  at least linked: https://chemistry.stackexchange.com/questions/119157/why-is-the-lucas-test-not-recommended-to-differentiate-higher-alcohols

The question does not mention in what the ammonium cerium nitrate is dissolved in, but I guess it is water. If so, one expects that sufficiently large compounds that one wishes to test are not sufficiently soluble in water, such as decanol. Obviously, this is not entirely correct because disaccharides such as lactose do are soluble in water.
I expect a sufficient solubility to be important for the test such that it finishes in a timely manner and does not become inconclusive due to prolonged reactions with air, water etc.

Is there any reference you can provide?


text:
Why is the Lucas test not recommended to differentiate higher alcohols?
I read about the Lucas test for alcohols in which a cloudy precipitate is produced with a tertiary or secondary alcohol. However, on a side note it was mentioned that this test is useful only for compounds having less than six carbon atoms. Why is this so?

This "degre" nomenclature for alcohols is terribly outdated. Profs teaching that today are really old and have learned it from teachers which were already oldfashioned then.

The problem is that the alcohols themselves might be insoluble. Ethanol is soluble, yes, but as the alkyl chain length increases, the hydrophobic nature due to the chain starts dominating over whatever hydrophilicity the $\ce{-OH}$ group had been providing for lower homologues.
From source [1]:

.. any alcohol that is insoluble in the reagent might appear to be giving a positive 3° test.

The paper goes on to properly quantify several things:

Based on the results of this study, using 1 drop of alcohol added to 10 drops of reagent, it can be said that, at the very least, all saturated acyclic monofunctional alcohols having six or fewer carbons are soluble in the Lucas reagent.

It is clear that the relative amounts of the sample and the reagent become important in such borderline cases (~6 carbon long chains). If too much alcohol is added, it may cause an illusion of positive lucas test.
From source [1]: (emphasis mine)

That statement may not be true when 3 or 4 drops are used
  per 10 drops of reagent. In fact, sometimes when 10 drops of
  reagent are added to 3 or 4 drops of some of the alcohols, an
  interesting event occurs after a delay of a couple of minutes
  or so. Apparently, some undissolved alcohol that has been
  clinging to the bottom of the test tube suddenly breaks loose
  and, by streaming to the surface of the very dense Lucas
  reagent, gives a momentary illusion of a positive test. The
  real cloudiness, alkyl chloride formation, appears later—all
  the more reason to use a high reagent/alcohol ratio.

References:
[1] "A study of the Lucas test"
R. A. Kjonaas and B. A. Riedford
J. Chem. Ed. 1991 68 (8), 704
DOI: 10.1021/ed068p704

To get a cloudy "precipitate", you need to dissolve the alcohol in the acid solution. Alcohols get less soluble in water as their chain length increases. By the time you pass 6 carbons, the alcohol itself is more or less insoluble. If it reacts along the interface, the chloride product would just mix into the alcohol layer, with no visible change.

Please share the reference you've used

I had no specific reference in mind, just general knowledge. The other answer is more complete than anything I had in mind.

no problem still I appreciate your efforts


text:
How should I interpret and present my total protein amount?
I have a problem. I have purified a protein with method A and then B. After method A I have a protein concentration of 20 mg/ml and after method B a concentration of 18 mg/ml. Method B is dialysis and my sample was diluted, so the volume went from 5 mL (A) to 6 mL (B).
When I present my total amount of protein at these stages I have a higher amount of protein after step B (108 mg) than after step A (100 mg), which is ridiculous, just because of the dilution.
How should I think around this and present this data in my purification table? I know that there can't be more protein after the second step. Should I just calculate the amount of protein after step B, with the volume of step A? Since I know I haven't lost or gained any proteins during the dialysis, I have only gained more buffer which gives me a lower protein concentration.
Please, any advice is valuable to me! Thanks

What is method A? And obviously you've got a wrong concentration, because you did an error in your measurements or in your calculations.  How shoud we know, we've seen neither.

How are you determining protein concentrations? Repeat that measurement if possible on your final protein sample, and get an appropriate uncertainty estimate for this measurement.

This is a great introduction to the concept of lack of precision (ie error) in measurements. It is unlikely that your volumes are exactly 5 mL and 6 mL or that your protein concentrations are exactly 20 mg/mL and 18 mg/mL. The errors in these measurements combine to give different calculated protein amounts even though the protein amounts are presumably the same (unless you've lost some protein in handling or your protein mass is low enough that some could pass through the dialysis membrane you are using).
For example, if the actual values were A: 5.1 mL and 20.4 mg/mL and B: 5.9 mL and 17.8 mg/mL, you'd get the same value of 104 mg total protein for both.
So to answer your question about reporting, it's important to have a sense of how precise your measurements are. A common way to do this is to repeat the measurement multiple times and calculate a standard error. That way, you can report the value as something like 100 +/-10 mg, and you would expect that the range for the two samples would usually overlap. And next time, you might want to measure the volume and concentrations more precisely. Volume certainly should be easy to determine to at least the nearest 0.1 mL.


text:
How to avoid cannizaro reaction during aldol reaction?
I am doing some Aldol reaction in lab,problem is **basically one of my Aldehyde is converting largely in cannizzaro products.**Please suggest me which base I can use so that I will not get cannizzaro products or get them in very less amount.

And what is the reaction, exactly?

I suspect that the rogue aldehyde in consideration has highly substituted alpha carbon/s..
Refer https://www.organic-chemistry.org/namedreactions/aldol-addition.shtm
for the proper mechanism for aldol addition.
Note that the enolate of the other carbonyl reactant is a nucleophile which adds to the electrophilic carbonyl atom. Naturally, steric hinderance plays a big part in this nucleophilic addition to the aldehyde. Thus, a highly substituted aldehyde shows less tendency towards aldol addition. (Trivia: this same steric interaction is also responsible for the fact that ketones are less reactive towards aldol addition that aldehydes)
Excpetional case: Formaldehye
Formaldehyde is the ideal aldehyde for aldol addition (least steric hinderance). But it also has a high tendency to get oxidised to formic acid (Cannizarro product).
Imo, you are carrying out the reaction in a highly basic medium, which will naturally favour cannizzaro in highly substituted aldehydes and formaldehyde. Mildy basic conditions might help.


text:
Calculating ionic strength and pH of a buffer solution
Im trying to calculate ion strength and pH of solution.
Components of the solution:
47 ul of 1000 pmol DNA in Elution Buffer (10 mM Tris-HCl pH 8.5) was bound to 3 ul of 10 µM YoYo1 (flourescent dye) in DMSO and was mixed with 50 ul of Tris-EDTA buffer (100 mM Tris-HCl, 0.01 M EDTA pH 8.0).
About calculation of ion strength, Im confused what Ions are in the solution too.
Tris-HCl => Tris-H + Cl- ?
EDTA => Mg2+ + Ca2+??
DMSO => ??
Furthermore, how can I calculate it?
I think the pH range is about 8.0~8.5, because of two buffers.
Is there any method for calculating more exact pH value?
Thank you so much for your answers!

Why do you suppose TrisHCl EDTA buffer contains Ca/Mg ?


text:
Preparation of 3-ethylpentan-3-ol
Is it possible to use CI3TMS (a carbon with three iodo groups and a trimethylsilyl protecting group) and three equivalents of an ethyl grignard reagent to prepare 3-ethylpentan-3-ol? By deprotecting the product after the grignard reaction using tributylammonium fluoride, is it possible to produce the desired product? I have studied a more lengthy method of arriving at the product; however, I cannot find any reference to my proposed mechanism, which seems shorter and more efficient. Is there some sort of flaw in my mechanism, or would it theoretically work?
CI3TMS --- 1) BrMgEt, diethyl ether 2) TBAF --->  3-ethylpentan-3-ol

I think you're missing an oxygen, when an alcohol is protected by TMSCl, it becomes a TMS ether, thus the reagent you need is I3COTMS.

Did you mean 3-ethylpentan-3-ol? Much easier with 2 equivalents of ethyl magnesium bromide and methyl or ethyl propionate if a different method is appropriate.

I'm pretty sure he did


text:
Why is tert-butoxide often used in elimination reactions when it is not necessary?
I understand that tert-butoxide is used to favor the Hofmann product in elimination reactions due to its steric bulk (and thus inability to abstract a proton in a more substituted position); however, I have often seen proposed organic synthesis involving the use of tert-butoxide when it is seemingly irrelevant. For example, 1-bromobutane will undergo an elimination reaction with or without a sterically hindered base (such as tert-butoxide); however, anytime I see reactions involving elimination reaction with primary substrates, such as the bromo group in 1-bromobutane, tert-butoxide is used rather than ethoxide. Is tert-butoxide simply used to disfavor the formation of substitution products. Is its steric bulk enough to not allow it to act as a good nucleophile in substitution reactions?

Several reasons:

t-BuOK is especially known as a strong base, and a poor nucleophile. Its large, bulky structure causes it to perform exceptionally poorly in substitution, literally eliminating any side reactions when the desired product is the elimination product.
It is easily available, like Raphaël insists.
It's basic strength depends on the medium. It is very strong in DMSO, where the solvent complexates well with $\ce{K+}$ ions. However it is relatively weaker in benzene. This flexibility gives it inevitable popularity.
Many condensation reactions (Stobe's, Darzen's) are seen to have greater yield when t-BuOK is used.

Generally, it's just because there's a big bottle of $\ce{t-BuOK}$ lying around, so why bother using something else?
In the same range of ideas, I've seen publications where they used $\ce{n-BuLi}$ to deprotonate an alcohol; it's completely overkill, but was probably just more convenient in their case than $\ce{NaH}$ or something else.


text:
What produces this NMR spectrum?
I'm having difficulty deciphering this 13C and 1H-NMR spectra. I've tried figuring it out, and have gotten close and am pretty sure it contains a methoxy group attached to a benzyl group, but hit a rut in regard to the other structure. Any help would be super appreciated!

What would really help is a mass ion. Is this a spectrum from a compound you can reanalyse? If so, run a DEPT so you can pinpoint the inverted peaks as CH2. Possibly also run COSY /HMQC correlation experiments.

@Beerhunter the formula is on the top in red and everything looks really nicely formatted, so I think this might be a homework problem of some sort

@szentsas nice spot. Thanks. I was being dopey

@Beerhunter, yes the formula is in the top right. It's a practice problem for an exam. It's difficult for me to do these problems when the formula gets big. I could do others, but not this one.

Did they provide any more information?

@szentsas, no they did not, just the 1H and 13C NMR data and the actual chemical formula, as you noted in the top left. I tried simulating potential chemical structures to see if I could get the right 1H and 13C outputs, but I couldn't get it to coalesce

One more hint I was given... apparently this molecule is a product of a reaction that includes 1,3,5-pentanetriol. Maybe 1,3,5-pentanetriol that has hydroxy's methylated and connected to a benzene with a methoxy group?

Also the benzene ring has to be para-substituted because of the equivalence of the carbons and the positions and also their coupling pattern

@szentas, definitely. That's what explains the 1H peaks around 7, and the double C peaks


text:
How to calculate the relationship between the pressures of zinc, carbon monoxide, and carbon dioxide based on mole balance?

Zinc oxide is reduced at a constant temperature in a closed reactor using $\ce{ZnO(s)}$ and $\ce{C(s)}$ as the only starting materials. The following reactions are assumed to be at thermodynamic equilibrium.
  \begin{align}
\ce{ZnO(s) + C(s) &-> Zn(g) + CO(g)}\\
\ce{2CO(g)   &-> CO2(g) + C(s)}
\end{align}
Assume ideal gas behaviour. Based on mole balance, the relationship applicable to the system at equilibrium is:
  \begin{align}
\tag{A} P_{\ce{Zn}} &= P_{\ce{CO}} + 2P_{\ce{CO2}}\\
\tag{B} P_{\ce{Zn}} &= 2P_{\ce{CO}} + P_{\ce{CO2}}\\
\tag{C} P_{\ce{Zn}} &= P_{\ce{CO}} + P_{\ce{CO2}}\\
\tag{D} P_{\ce{Zn}} &= 0.5P_{\ce{CO}} + 2P_{\ce{CO2}}
\end{align}

In this problem seeing that none of the given options consisted of $\ce{C(s)}$, so in order to get rid of that I just added the two equations and arrived at the equation $$\ce{ZnO(s) + CO(g) = Zn(g) + CO2(g)}.$$
Now just sort of comparing with the options I find that $\ce{CO(g)}$ and $\ce{CO2(g)}$ are on the opposite sides. It sort of gives me the intuition that may be the pressure term should have had a minus. Anyway I know that I have gone horribly wrong somewhere with regard to my concepts. Please tell me how to solve the problem henceforth.

First of all, let me explain why your approach is a bit wrong...
Addition of steps will only give you the net reaction, which is what you got. But in the process, the mechanism becomes blurred. For example, the net reaction which you have expressed, doesn't tell us the full story. One completely overlooks the fact the CO is also recombining to give CO2, and that is exactly where the fallacy arises.

My solution to the problem. Hope it helps.

Tell me something why are you equating number of moles of Zn(g) with CO(g) in the last step?

From the first equation.. it is obvious that equal number of moles of Zn and CO are formed... The amount of CO consumed is inferred from the second equation... Now just add the results to get the residual moles of CO in the mix.

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors.


text:
What is the equation for solving the concentration of the unknown from a Single-point Standard Addition Calibration for Potentiometry?
I know how to use the standard addition equation for a single-point analysis from an equation I got from the Analytical Chemistry book "Quantitative Chemical Analysis" 9ed by D. Harris.
However, solving for a potentiometry problem, the equation doesn't seem to be effective anymore. I keep on getting negative value for the initial concentration of the unknown sample.
The problem goes like this:
The Na+ concentration of a solution was determined by measurement with a sodium ion-selective electrode. The electrode system developed a potential of -0.2462 V when immersed in 10.0 mL of the solution of unknown concentration. After addition of 1.00 mL of 2.00 x 10-2 M NaCl, the potential changed to -0.1994 V. Calculate the Na+ concentration of the original solution.
Is there a specific equation I can use so I can correctly solve for the concentration of the unknwon solution?

The little problem with the standard addition in potentiometry is that the response to the addition is not linear, but logarithmic.
If we consider the simplified Nernst equation in context of parameters $A$ and $B,$ and if we consider $c$ and $c_0$ as unknown concentration and  concentration increment, respectively, we can follow this derivation:
$$\begin{align}
E_1 &= A + B \log c \label{eqn:1}\tag{1}\\
E_2 &= A + B \log (c + c_0) \label{eqn:2}\tag{2}\\
\end{align}$$
Subtracting \eqref{eqn:1} from \eqref{eqn:2} yields:
$$\begin{align}
E_2 - E_1 &= B \log\left(1 + \frac{c_0}{c}\right) \tag{3}\\
10^{(E_2 - E_1)/B} &= 1 + \frac{c_0}{c} \tag{4}\\
c &= \frac {c_0}{10^{(E_2 - E_1)/B} - 1}, \tag{5}
\end{align}$$
where $B = \displaystyle\frac{RT}{nF}\ln 10.$


text:
Increasing concentration of acetylcholine via SPE
I am trying to increase the final yield of acetylcholine (ACh) in my biofluid samples. I need to get the yield high enough for high confidence annotation of ACh (currently all I can really say is that I've got something that matches the ACh expected mz).
My first samples were prepared using C18 solid phase extraction using a water liquid phase and methanol to elute. Based on my understanding of C18 SPE, I assume I've lost a lot of ACh at this step, because I assume the positively-charged ACh would have been washed away during the wash step.
Now that I have a polar analyte to target, I need to prepare more samples with a higher ACh concentration. I should be able to increase the concentration by using an SPE protocol more amiable for ACh extraction, but I can't find a protocol like this in the literature. How does one design a targeted SPE protocol? Can I assume any HILIC procedure that targets charged molecules will be appropriate here, or is there more to consider?
My problem is made more complicated because I need the final, concentrated version of my extract to be compatible with my downstream bioassay (i.e. no carry over from acid, no strong solvent that can't be removed)

Maybe you can clarify what you mean by this: "Now that I have a polar analyte to target, I need to prepare more samples with a higher ACh concentration." Are you quantifying ACh in your samples? Maybe you should be using a cation exchange column?

Thanks for pointing that out. What I need to do is identify this mass trace as acetylcholine unambiguously, and for that I will use either FTICR-MS/MS or LC-MS. We've tried MS/MS and found that the concentration wasn't high enough, which is why I am trying to target it directly. Ultimately, I would like to quantify the amount of acetylcholine in my samples.

I found the following supelco catalog informative : http://www.supelco.com.tw/C-2%20%20150-163%20New.pdf I can't claim to be an expert but you may want to try something like the DSC-WCX columns.


text:
Acidic cleavage of ether with a 2° alkyl group
I know that an ether with a 3° alkyl group and a 1° alkyl group is cleaved by $\mathrm{S_N1}$ mechanism to give a tertiary alkyl halide and a 1° alcohol.
And an ether with a two 1° alkyl groups is cleaved by $\mathrm{S_N2}$ mechanism to give a 1° alkyl halide and a 1° alcohol.
But what happens if a ether with a 2° alkyl group and a 1° alkyl group is cleaved with a halogen acid?

what happens if a ether with a 2° alkyl group and a 1° alkyl group is cleaved with a halogen acid?

To this question I have drawn a scheme below. The given ether is $\ce{2^0}$ group to the left and methyl to right. Acid cleavage with a halogenic acid involves an initial protonation of ether. In an $\ce{S_N^2}$ reaction of this ether, nucleophile $\ce{I^-}$ attacks via path 1 and path2.

In path 1, $\ce{I^-}$ approach is sterically hindered by phenyl  and Methyl groups. This leads to a high energy transition state.
In path 2, $\ce{I^-}$ approach is from right of protonated ether where steric hindrance is comparatively less( only $\ce{H}$). A more stable transition state favors $\ce{S_N^2}$ reaction .

In each of these cases, the ether will generally be cleaved to produce 2 alkyl halides. Acidic clevage can occur via Sn2 or Sn1 depending on the nature of the substrate. A bulky group, such as tert-butyl, will undergo an Sn1 mechanism in which the protonated ether group leaves before the nucleophile attacks. The same process is generally consistant with secondary substrates. I have drawn a mechanism involving both a primary and tertiary substrate to demonstrate this point; however, it is important to note that in the presence of most concentrated halogen acids and heat, ethers will almost always undergo complete acidic clevage in which 2 alkyl halides are generated. I have also included another mechanism involving a secondary and primary substrate. The primary substrate will undergo an Sn2 reaction due to the lack of steric hindrance; however, it is a little less clear with the secondary substrate. Will it undergo Sn1 since Sn2 would be more sterically hindered or will it undergo  Sn2 since the carbocation intermediate in Sn1 is secondary and thus less stable? Since we are using an aqueous halogen acid in this case, it will most likely undergo Sn1 since it is in the presence of a polar protic solvent, which will thus stabilize the carbocation intermediate in the process of forming.


text:
What are the equations giving the molar flow of different molecules linked by a reaction?
Let consider a steady state CSTR in which occurs the following reaction:
$$\ce{aA + bB \leftrightarrow cC}$$

Notations:
$F_i$ is the molar flow of $i$, $r$ is the rate of reaction, $V$ is the volume, $X$ is the conversion at the end of the reactor, $\nu_i$ is the stoichiometric coefficient of $i$ and $\xi$ is the extent of reaction.

Mass balance:
$$\ce{IN + PROD - CONS = OUT + VAR}$$
Therefore,
$$
\begin{cases}
F_{A,in} + 0 - arV = F_{A,out} \\
F_{B,in} + 0 - brV = F_{B,out} \\
F_{C,in} + crV - 0 = F_{C,out} \\
\end{cases}
$$
As $$\xi = -rV = -X\frac{F_{i,in}}{\nu_i}$$ we have for $\ce{A}$:
$$\begin{alignat}{}
F_{A,out} &= F_{A,in} - arV \\
& = F_{A,in} - a(-\xi) \\
& = F_{A,in} - a\left(X\frac{F_{A,in}}{a}\right) \\
& = \left(1 - X \right)F_{A,in}
\end{alignat}$$
For $B$ with can find as well:
$$F_{B,out} = \left(1 - X \right)F_{B,in} = \frac{b}{a}\left(1 - X \right)F_{A,in}$$

My problem with C:
Now if I follow my calculus for $\ce{C}$, here is what I get:
$$\begin{alignat}{}
F_{C,out} &= F_{C,in} + crV \\
& = F_{C,in} + c(-\xi) \\
& = F_{C,in} + c\left(X\frac{F_{C,in}}{c}\right) \\
& = \left(1 + X \right)F_{C,in}
\end{alignat}$$
Which states that if there is no $C$ at the beginning, there is no $C$ produced at all! Which is wrong. Can you tell me where I am wrong?

NB: I have no clue, which tags fit best for this question.

Sure about the $F_{B, out}$ expression?

@EashaanGodbole well I guess I am not wrong but please tell me what you think is wrong? $$\begin{alignat}{}
F_{B,out} &= F_{B,in} - brV \
& = F_{B,in} - b(-\xi) \
& = F_{B,in} - b\left(X\frac{F_{B,in}}{b}\right) \
& = \left(1 - X \right)F_{B,in}
\end{alignat}$$
As $$\frac{a}{b} = \frac{F_A}{F_B}$$ it gives the result:$$F_{B,out} = \left(1 - X \right)F_{B,in} = \frac{b}{a}\left(1 - X \right)F_{A,in}$$

Oh, then it's no issue. The $\dfrac{a}{b} = \dfrac{F_A}{F_B}$ condition isn't mentioned in the question.

I can't tell you precisely where your reasoning starts to be wrong, but here is another way to think about the problem.
Notice that for a specific reaction as you gave here, the term $rV$ is always the same independently of the molecule you choose to determine it. It does vary for sure over time though (don't read I said it was a constant).
As you wrote:
$$\begin{cases} F_{A,in} + 0 - arV = F_{A,out} \\ F_{B,in} + 0 - brV = F_{B,out} \\ F_{C,in} + crV - 0 = F_{C,out} \\ \end{cases}$$
Thus, $$rV = \frac{F_{A,in} - F_{A,out}}{a} = \frac{F_{B,in} - F_{B,out}}{b} = \frac{F_{C,in} - F_{C,out}}{-c}$$
You can write what happens for $C$ in terms of $A$: $$\frac{F_{A,in} - F_{A,out}}{a} = \frac{F_{C,in} - F_{C,out}}{-c} \Leftrightarrow F_{C,out} = F_{C,in} + \frac{c}{a} \left(F_{A,in} - F_{A,out}\right)$$
Which boils down to, $$F_{C,out} = F_{C,in} + \frac{c}{a} X F_{A,in}$$
And if you write equations in function of a particular molecule, $A$, as you did here, make sure that the conversion you wrote $X$, is to be more precise, the conversion of $A$ at the exit of the reactor. Otherwise, it doesn't make sense.

EDIT:
Note as well that your equation for $B$ is wrong as well, using the same reasoning as before you should find: $$F_{B,out} = F_{B,in} - \frac{b}{a} X F_{A,in}$$ which make sense because the flow of $B$ must be less at the end of the reactor if this reaction really occurs.


text:
Why would the first Tg be higher than a rescan Tg of an epoxy by DSC?
I have some small cured epoxy shafts I am cutting segments out of for checking Tg by DSC. First scan gave a Tg of 118 °C, follow on was 109 °C. Scans were repeated on a new sample of the shaft with similar results.
250 °F curing dicy based epoxy system. Neat resin before cure has exotherm onset at 127 °C and the curve's peak is 142 °C by DSC.
Any idea why the first scan would be higher than the rescan?

Because, while it was cured, it was far from being thermally inert. One could say, you cured it a little bit more with another heating. It's a bit like with cooking - reheated may taste better ;)

250°F is 121°C ("C" is Culoumb, btw., and "F" is Farad). Yea, you probably re-cured your system. However I would expect the $T_g$ of an epoxy resin to go up, not down. What you saw in the first run was likely not the glass transition but the actual curing.Without seeing the actual curve, or at least the temperature profile, that's just guessing however.

What temperature range are you running, at what rate? What does the packaging of your epoxy say about curing time? And a peak indicates an endo-/exothermal process, like a reaction. Is there still a small peak in the second run?

related https://www.sciencedirect.com/topics/engineering/dicyandiamide-dicy


text:
Can I normalize my data in order to compare it?
I have three chromatograms of the same eluted protein but at different concentrations, since it got diluted during purification process) and hence peak maximums at very different absorbances.
The highest is around 100 AU and lowest 20 AU.
I want to present these and look at the difference in peak shape (because I want to look at aggregation). Can I simply multiply the data that generated the peak max at 20 AU by 5? Because then it is easier to compare the shapes of the peaks when they  overlay each other.
Is this something one would do or is it totally forbidden?
Thanks in advice.

1 kilogram times 5 is 5 kilograms, which is a totally different thing. But 1 arbitrary unit times 5 is another arbitrary unit, which is pretty much the same thing.

Where readings such as absorbance 120  (AU) come from? There is nothing arbitrary in absorbance which is a well defined quantity withno units per se.

You can conveniently normalize the peak height to unity, if your sole purpose is to compare the peak shapes. It is not forbidden at all. However you should not attempt any quantitation.
You can read the paper here
Total peak shape analysis: detection and quantitation of concurrent fronting, tailing, and their effect on asymmetry measurements

I assume you have a UV/Vis absorbance based measurement. Then the absorbance should follow Beer´s Law which states that absorbance is proportional to concentration. Therefore you should be fine with your normalization approach. You could check if you are in the linear range by performing a calibration.

Strictly speaking: the fact that data are from UV/Vis measurements doesn't imply that the samples obey Beer law. Moreover absorbance is the peak integral not the height. Checking linearity could be done by diluting the samples but it is not inherent to the question nor necessary. All that is required is comparing shapes, and this is easier at the same scale and can be done.  What could be the actual shape of one of the two features at a different concentration is right matter for discussion. And it seems what the OP want to do.

Agree. But linearity is absolutely necessary in order to be able to compare peak shapes directly because otherwise the shape depends also on the peak witdth.


text:
Why potential energy is neglected in this problem for calculating the energy difference between two stationary Bohr Orbits?

Question
An electron in a hydrogen like atom makes a transition from a state in which its de-Broglie wavelength is $\lambda_1$ to a state in which its de-Broglie wavelength is $\lambda_2$, then wavelength of photon generated due to this transition will be ___________.

The solution for this problem as given in my book is as follows:

In the first step, it is given that the energy difference is just equal to the kinetic energy difference. How is this possible? Why have they neglected the potential energy in calculating the energy difference since the two stationary orbits in Bohr's model have different potential as well as kinetic energies.
Please clarify my doubt.

There is a statement known as virial theorem. Roughly speaking, it says that in systems like this, the potential energy equals (-2)*kinetic energy. No, the potential energy is not neglected at all.

@IvanNeretin, Thank you for your reply. I haven't heard about virial theorem, but I forgot that total energy is (-1)*kinetic energy while asking this question, which I recalled only after seeing the answer. From your comment, I learnt that the name of the property in "virial theorem". Thank you.

@GuruVishnu Let's talk in this room.

My solution...
Hope this helps!!


text:
Which electronic transition of the hydrogen spectrum corresponds to "the third line from the red end"?

Question
In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inner-orbit jumps of the electron for Bohr orbits in an atom of hydrogen?
(A) $3\to2$
(B) $5\to2$
(C) $4\to1$
(D) $2\to5$
Only one option is correct.

My approach
I just drew the electron transition diagram like this:

Then, I counted the third line from the red end, i.e., from the left side starting from the first line of the particular series. I then counter checked with the given options. Surprisingly two options matched, i.e., option B as well as C. I don't know further how to decide which is the correct answer out of B and C, since only one answer is correct. There is no other details mentioned in the question and thus I am confused.

4-1 and 5-2 fit the question as you suspect.

It is kind of an odd question. However since the question mentions a red line, I'd assume that it is referring to the Balmer series which is in the visible range. Thus the answer would be (B).

Since only one answer is correct, we must consider the transitions in the visible region only i.e., the Balmer Series. If multiple answers are correct then it is wise to choose both the options B and C.
The transition $n_2 \to n_1$ can be easily calculated by using the following formula and it is not necessary to draw the transitions, which might be difficult for higher transitions where drawing the transitions would be time-consuming.
$$n_2=n_1 + n$$
where $n$ represents the $n^{th}$ line in a particular series. Since here we are talking about Balmer series, $n_1=2$ and for the third line $n=3$. Substituting the values
$$n_2=2+3=5$$
Hence the required transition is given by option B.


text:
Ratio of total energy to kinetic energy in hydrogen atom when the reference is changed

If in the hydrogen atom potential energy at $\infty$ is chosen to be $\pu{13.6 eV},$ then what is the ratio of total energy and kinetic energy (with the sign) for the first Bohr orbit?

I think we don't consider the reference frame, and I answered with the relation
$$E_\mathrm{tot} : E_\mathrm{k} : E_\mathrm{p} = -1 : 1 : -2,$$
but the answer given is zero. Why?

What would be the potential energy of the 1st orbit in this reference frame? Then what would be the total energy?

P.E = $-13.6$eV in this frame. But would the K.E be still the same?

Of course K.E. will be still the same, but P.E. will not. In particular, it won't be -13.6 (it was like that when we had $0$ at $\infty$, which is not the case anymore).

Potential energy would be $-27.2$eV in the normal case. In this reference, it would be $-13.6$eV.

Sure. My fault, I confused it with the total energy. Well, anyway, K.E. stays the same as P.E. changes; what will happen to the total energy and the ratio?

T.E would be zero! So, the ratio would be zero.

So, is the ratio of all those quantities true only when we consider the potential at $\infty$ to be zero?

If you are talking about the virial theorem, then yes, it only applies when we have $0$ at $\infty$.


text:
Cleaning of chlorbutyl plungers
This is my first post, so please bear with me.
At work I have some chlorobutyl plungers that I need to clean from any residual material that might be left after production.
A colleague of mine told me that he has seen a GMP (Good Manufacturing Practice) description of how to clean chlorobutyl plungers, but was unable to find it. It relates to pharmaceutical industry and the standard approved ways of working. The plungers I am referring to are commonly used in penfill cartridges for subcutaneous injections.
Any advice on how to clean the chlorobutyl plungers (e.g. with what chemicals, under what conditions) to ensure that there are no residual solutions/oils/materials on the plungers?


text:
Measuring change in concentration of hydrogen peroxide
I am conducting an experiment to find the rate constant of the decomposition of hydrogen peroxide, $\ce{H2O2}.$ For this, I was planning on using the integrated rate law:
$$\ln(\text{final concentration}) = -kt + \ln(\text{initial concentration})$$
However, I am not sure how I can measure the change in the concentration of $\ce{H2O2}.$ I would appreciate it if someone can propose a method.

Typically you measure the volume of evolved O2.

After that, what should I do?

Then calculate the concentration of H2O2 from these data.

Alternatively, if you want to work with liquid, you could do titrations by sampling the solution over time. http://www.titrations.info/permanganate-titration-hydrogen-peroxide


text:
Role of THF in oxymercuration of alkenes
I am trying to self learn solomon's organic chemistry book, but there is no explanation over why we should use THF in oxymercuration. After a quick search I found out that THF help stabilize in hydroboration, but is it the same for oxymercuration? and I still don't understand why and how THF can help stabilize in hydroboration.

It's the solvent.

$\ce{THF}$ is a polar aprotic solvent, that is, it can stabilize positive charges but not negative ones since it won't form hydrogen bonds with the help of any of its own hydrogen atoms.
Thus the carbocationic intermediate, that is the mercurinium ion is still stabilized, allowing the reaction to proceed.
And in Hydroboration, $\ce{THF}$ acts as the solvent as well but plays the additional role of shifting the equilibrium between diborane and borane to the side of the latter. As you might know, the Boron atom of $\ce{BH3}$ is electron deficient, which plays a key role in the formation of the trialkylborane intermediate of the Hydroboration reaction. Formation of diborane quenches this electron deficiency of borane by the formation of two 3-centred, 2-electron bonds in the dimer.

so without THF, the mercurinium ion will be reactive and won't react as intended pathway?

@AegisNine It won't necessary be reactive but the lowering of energy of an intermediate is always desirable. It will solvate the intermediate.

@FᴀʀʜᴀɴAɴᴀᴍ, What if we use a polar protic solvent? It still stabilises the mercurinium ion right.

@Intellex Yes. Only anionic intermediates are destabilized as such by solvent effects and that is in the case of a polar aprotic solvent.


text:
What is meant by pure solids or pure liquids?
In by book, under the topic Heterogeneous Equilibrium it is given

For pure solid and pure liquid, active mass is taken to be unity, i.e., 1 as they remain constant throughout the reaction.

Could you please explain what are the pure solids and liquids being mentioned here?
I read some related answers to similar question but none addresses which compounds are pure which are not. Or, if I haven't noticed properly, kindly give me the link to clarify my doubt.

Pure substances here refer to substances other than solutions in the liquid/gas phase, I think. For instance, Bromine in its pure state is a liquid but liquid HCl is more often than not an aqueous solution.

Pure refers to the elemental state. But, https://chemistry.stackexchange.com/questions/24082/reference-states-for-activities pretty much answers it.

Can you please give the name of that book?


text:
Is the 1,4-adduct of carbonyl compounds ever a major product over the 1,2-adduct?
If, say, propenal undergoes nucleophilic addition of hydrazine, will the 1,2- adduct or the 1,4-adduct be preferred? Why?
There is an obvious reasonace structure with an electrophilic beta carbon so under proper conditions will the 1,4-adduct dominate?

Organocuprates have much higher tendency to add in a 1,4- fashion. Selectivity can be tuned as a function of the reagent (usually).

Google for 'Michael addition' and you'll find that in fact plenty of nucleophiles add 1,4-.

@orthocresol I found out about the Michael addition after posting this question but I can't find anything about hydrazine adding 1,4 to an enone. My textbook mentions a similar addition reaction producing a 1,2-adduct. So I'm confused when I should rule out the 1,4- adduct.

That's because hydrazine is special: it has two NH2 groups, so one can condense with the carbonyl group to form an imine, and the other one can add to the double bond.

@orthocresol So does that mean the product wouldn't be a hydrazone as one might expect?


text:
How to extract melanin from human hair
I am working on a project wherein I am studying various properties of human hair and its utilization..I would like to know method by which we can extract melanin from human hair. Can anyone suggest some cost effective method?


text:
Why the formation of PH4+ is difficult compared to NH4+?
My textbook asks this as an objective question.
What I had in mind was that the lone pair on phosporous was more distributed as compared to nitrogen(since nitrogen is more electronegative and smaller in size ) and hence the statement. But the option given in the textbook says the answer is lone pair of phosphorous resides in almost pure s orbital.
How do you know if the lone pair is placed in pure s orbital or pure p orbital

https://chemistry.stackexchange.com/questions/71803/why-is-nh3-a-stronger-base-than-ph3

The bond angle in $\ce{PH3}$ is about $92-93^\circ$, and by the percentage $s$ character formula i.e $\cos(\theta) = s/(s-1)$ you can calculate the $s$ character and that comes out to be 6%. And that where drago rule holds true , that is the orbitals involved in bonding are pure $p_x$ ,$p_y$, $p_z$ and as a character is high the lone pair is present in pure s orbital and highly penetrated towards phosphorus. Hence acts as a poor Lewis base.

Sorry that is P character 94 percent and 6 percent s character

You can click on "edit" to modify your answer.

regarding "Drago's rule": https://chemistry.stackexchange.com/questions/64613/what-is-dragos-rule-does-it-really-exist

how does this differ from "Bent's rule": https://en.wikipedia.org/wiki/Bent%27s_rule ?

We think that both P and N contain lone pair of electrons ,but this is not true if it comes to you donating that lone pair to hydrogen ion AKA H+
Here comes back bonding into action it simply means that if an atom has got lone pair ,it would distribute it to a group to which it is bonded if that group contains (or is) an atom  which has its orbital unfilled or empty or has an incomplete octet
First two reasons have got almost same meaning.
P's lone pair get distributed over its own 3D orbital hence availability of lone pair decreases in its pure orbital (lone pair can only exist in pure orbitals)so bond to H + is well ..... prohibited.

This explanation has been interpreted from Concise Inorganic Chemistry by J.D. Lee

No, phosphorus’ high-energy unpopulated d orbitals have no contribution whatsoever and there is no relevant backbonding. Not to mention that there is certainly no hybridisation between phosphorus’ s lone pair and any of the 3d orbitals. I hope Lee doesn’t state that in their book, because if they do it would be better described as highly outdated.

@Jan After doing a keyword search and going over the relevant parts of the book here I can confirm that Lee makes no direct claims of such lone pair "distribution" as stated in this answer. This is entirely the answerer's version of interpretation of said book


text:
Is it possible to predict release of acid fumes in certain acid reactions?
Upon mixing both acids, the reaction between Formic Acid and Sulfuric Acid, besides producing Carbon Monoxide and Water, releases a certain amount of acid fumes.
Does anyone know if the fumes, in that specific reaction, are of only one of the acids or both of them?
Also, is there a good way to predict release of acid fumes in similar reactions? Is it a characteristic of certain acids?
Thank you.

It is probably just hot water vapor and some entrained tiny droplets of the acids. It may not always be easy to predict when significant fuming is likely, but it is always possible to predict that using a working fume hood is a great idea.

Note that formic acid  releases carbon monoxide even spontaneously, what is reason it should not be stored in a gas tight containers.

Partly related: Making carbon monoxide gas from formic acid and sulfuric acid? and Vaporized sodium hydroxide?

Formic acid is volatile.
Any volatile liquid, including water, if heated ( externally or spontaneously in mixture ) forms fumes, if temperature in gas above the liquid drops below its dew point.
Remember water "fumes" over a boiling kettle.
Additionally to these physical-chemical reasons, if there is boiling (or other strong mechanical disturbance)  instead of plain  evaporation, the non volatile sulphuric acid can pass  to fumes by mechanical way.

Perhaps boiling kettle? :-)

@Ed V Sure, kettle. :-D As non native, even if I know both words, I may confuse them as almost homophones.  :-)


text:
Which of these reactants will not generate this racemic amine?

I understand that in E there is no electrophile the amine can attack, however, I do not understand why A is an answer as well. The amine in A can perform an intramolecular attack on the aldehyde and form the product below. Why can it not?
Also, how does the oxygen leave anyway?

This is an example of reductive amination, which generally proceeds through two main steps: (i) condensation reaction between the amine and carbonyl to generate an imine; and (ii) reduction of the imine to the corresponding amine.

Clearly (e) can't do this reaction since there is no carbonyl. The reason why (a) is also an answer comes down to the wording of the question - it specifically states that the product is racemic. The stereocentre in (a) is (R) in the starting material, and is not involved in the reaction so the product will exist as a single (R) enantiomer.

In the question, all the amides are secondary. Secondary amines can't form imines, thus how is an imine formed?

Fair point, they will just form iminium ions rather than imines, but the mechanism remains exactly the same.

Ah, ok. That explains the -1 charge then in the picture.


text:
Ethyl acetate produces red colour on ceric ammonium nitrate test
When I did CAN test on ethyl acetate red colour was seen. I thought water may get added with Ethyl acetate to produce
this compound which in turn gave red coloration but carbon should not make three bonds with oxygens easily.
What could be the possible explanation to this phenomena?

Water hydrolyses the ester to form an alcohol which responds to the CAN test.

@FᴀʀʜᴀɴAɴᴀᴍ  could you please tell that alcohol's structure fromed in this process? I had thought of one, as mentioned in the question, but that seems unlikely.

Since the ester is ethyl acetate, on hydrolysis, ethanol and acetic acid will be formed.

I had also thought about that but the solution of ethyl acetate with water did not change colour of blue litmus paper neither it gave effervescence on adding sodium bicarbonate.

In that case I think I might be wrong. Ester hydrolysis does require an acid catalyst which is absent here.

It could be that the ester gets reduced to two alcohols but I doubt how that would proceed either.


text:
Numerical simulation and calculations of pH
a. Reaction of $\ce{CO2}$ addition in water is as follows:
$\ce{CO2}$ + $\ce{H2O}$ $\ce{->}$ $\ce{H2CO3}$
b. Then, pH of $\ce{H2CO3}$ solution is increased to 12 by addition of 1M $\ce{NaOH}$.
So, the reaction would be:
$\ce{2NaOH(aq)}$ + $\ce{H2CO3(aq)}$ $\ce{->}$ $\ce{Na2CO3(aq)}$ + $\ce{2H2O}$
c. This new solution at pH 12 is added to $\ce{CaCl2}$ solution; the final reaction of the system would be:
(  $\ce{Na2CO3(aq)}$ + $\ce{2H+}$ + $\ce{2OH-}$  ) + $\ce{CaCl2}$ $\ce{->}$ $\ce{CaCO3(s)}$ + $\ce{2 NaCl}$ + $\ce{2H2O}$
My questions are:
1. Is the final chemical reaction correct?
2. By setting the pH of step c equal to 12, which is ( $\ce{Na2CO3(aq)}$ + $\ce{2H+}$ + $\ce{2OH-}$ ), and addition of 50 ml of that to $\ce{CaCl2}$ solution, is it possible to make some sort of pH calculations?
$\ce{[H+]} = 10^{-12} M$
$\ce{[OH-]} = 10^{-2} M$
$\ce{[CO3]}$ = assuming 0.05 M in 50 ml
3. Overall, is it possible to do a numerical simulation (e.g., MATLAB) of change in the overall solution pH? OR, predict the final pH by calculations?
Thanks and Best Regards,

This can't bu done precisely, as long as you don't know how much CO2 was dissolved. Still difference between cases should be negligible as long as you add surplus of CaCl2.

As an answer to your overall question: it is indeed possible to do numerical simulations of pH in different solutions or mixtures, and there is a number of programs or spreadsheets available if ou search the net. My personal favorite is Curtipot made by Prof. Gutz: http://www.iq.usp.br/gutz/Curtipot_.html


text:
Why is activation energy drawn in a potential energy diagram in reactions?

I learned in chemistry class that activation energy is the av. KE that the colliding particles need in order to successfully react. However, they are always drawn as a "hill" in a PE diagram.
I do not understand why they are classified as PE, when they are KE.

Related: https://chemistry.stackexchange.com/questions/86212/is-the-kinetic-energy-turning-into-activation-energy-when-the-reaction-takes-pla/86218#86218

The activation energy is not the average kinetic energy that the colliding particles need in order to successfully react. Instead, it is the energy barrier between reactants and products. Often, the energy required to cross the barrier comes from the kinetic energy of collisions.
To describe the rate of a reaction there are two models, the Arrhenius model and the more sophisticate Eyring model:
The Arrhenius model uses the activation energy, defined as the energy required to reach the transition state; the "energy" here is of various nature, not only kinetic an includes the chemical potential.
The Eyring model uses the Gibbs energy of activation for the transition state that includes an enthalpy and an entropy term ($ΔG^‡= ΔH^‡ – T ΔS^‡$). In comparison, for a unimolecular, one-step reaction, the activation energy can be approximately expressed as $E_a = ΔH^‡ + RT$.
There are different y-axis in these diagrams as explained in a related answer:
Y-axis of the reaction co-ordinate graph. Depending on the model and reaction you want to study you can use the Gibbs energy (Eyring model), energy in general (Arrhenius model) or enthalpy in case you want to show the heat absorbed or released in a reaction at constant pressure.
Why does your graph show potential energy?
In simple terms, potential energy is stored energy while kinetic energy is the energy of motion. Potential energy is converted into kinetic energy, therefore you can account for the latter including it in the overall potential energy.

The energy of the products is in the ground state of all states (vibration, rotational, electronic)?


text:
Exact weights for a chemical reaction og Silver and Nitric Acid
I'm an amateur analog photographer and want to react Nitric Acid (62%) and silver to Silver nitrate. Silver nitrate is the basis for almost all black and white photography.
I'm trying to figure out the correct ratio of silver to nitric acid. The reaction is $$\ce{3Ag(s) +4HNO3(aq) -> 3AgNO3(l) +NO(g) +2H2O (l)}$$
Can I calculate the ratio of silver to acid with the formula $3 \times (\text{molar weight of Ag}) / 4 \times (\text{Molar weight of HNO3}) \times 0.62$ ?

This is just a bad idea. Molar proportion's your last concern with this reaction.

I'm perfectly aware of the reaction and it's bi product. Thanks for your concern!

Every 3 moles of Ag reacts with 4 moles of Nitric acid.
Therefore the mass ratio is:
 $$ \frac{m(Ag)}{m(HNO3)} = \frac{3}{4}\frac{MM(Ag)}{MM(HNO3)} = 1.28 $$
MM is molar mass(g/mol)

I have to multiply  4 MM(HNO3) by a factor of 0.62 to take the diluted acid into account, right?

So you want the ratio:  silver mass/volume HNO3 solution?

Silver mass / mass of dilluted HNO3 preferably.


text:
Cauchy-Schwarz integral screening different inequalites
Eq. (9.12.23) in [1] states that:
$$\sum_{abcd}c_{ab}c_{cd}g_{abcd}>0\tag{MEST 9.12.23}\label{91223}$$
with $c_{ab}=c_ac_b$ for $c_a$ being some distribution coefficients and $g_{abcd}$ being the two-electron integrals in AO basis. Now for $c=a$ and $d=b$ giving:
$$c_{ab}c_{ab}>0$$
Which seem obvious as long as $c$ are real numbers.
It can thus be seen that to fulfill Eq. (\ref{91223}) that:
$$g_{abab}>0 \tag{MEST 9.12.24}$$
Which fulfill the requirements for the Cauchy-Schwarz inequality, thus giving:
$$\left| g_{abcd} \right| \leq \sqrt{g_{abab}}\sqrt{g_{cdcd}}\tag{MEST 9.12.25}$$
Now since $c_{ab}=c_{a}c_{b}$ this implies that:
$$c_{aa}c_{bb}>0\tag{1}$$
which in turns further implies that:
$$g_{aabb}>0\tag{2}$$
Which surely also fulfill the requirements for the Cauchy-Schwarz inequality, giving:
$$\left| g_{abcd} \right| \leq \sqrt{g_{aabb}}\sqrt{g_{ccdd}}\tag{3}$$
Now further doing some index juggleling will lead to the following six different inequilities:
$$\left| g_{abcd} \right| \leq \sqrt{g_{abab}}\sqrt{g_{cdcd}}\tag{Ineq. 1}\label{ineq1}$$
$$\left| g_{abcd} \right| \leq \sqrt{g_{acac}}\sqrt{g_{bdbd}}\tag{Ineq. 2}$$
$$\left| g_{abcd} \right| \leq \sqrt{g_{adad}}\sqrt{g_{cbcb}}\tag{Ineq. 3}$$
$$\left| g_{abcd} \right| \leq \sqrt{g_{aabb}}\sqrt{g_{ccdd}}\tag{Ineq. 4}$$
$$\left| g_{abcd} \right| \leq \sqrt{g_{aacc}}\sqrt{g_{bbdd}}\tag{Ineq. 5}$$
$$\left| g_{abcd} \right| \leq \sqrt{g_{aadd}}\sqrt{g_{ccbb}}\tag{Ineq. 6}$$
I think I have ever only seen Eq. (\ref{ineq1}) described before, surely this could be because the rest are obvious extensions, that I just haven't thought of as being obvious before.
Or that I have made some flaws in my reasoning.
In practise are all of the inequilites used?
If not anyone got clues about why not?
Especially Ineq. 4-6 is somewhat different from Ineq. 1-3.
[1] : Molecular Electronic-Structure Theory, Trygve Helgaker, Poul and Jørgensen Jeppe Olsen

It's not clear what any of these quantities mean. Please provide more context.

Quantities should now have been clarified @Zhe

As you have already stated in the question, (Ineq. 1) is the standard inequality used in Cauchy-Schwarz integral screening. (Ineq. 2) and (Ineq. 3) are not valid inequalities, which is clear if you select atomic orbitals such that a & b are spatially disjoint from c & d and the proposed upper bounds vanish. (Ineq. 4), (Ineq. 5), and (Ineq. 6) follow from a different use of the Cauchy-Schwarz inequality where the inner product is interpreted as being over the six spatial integrals and requires the Coulomb kernel to be pointwise non-negative so that you can take its element-wise square root,
$$ \begin{align} |g_{abcd}| &= \left| \int d\vec{r} d\vec{r}' \phi_a(\vec{r}) \phi_b(\vec{r}) V(\vec{r},\vec{r}') \phi_c(\vec{r}') \phi_d(\vec{r}') \right| \\
&= \left| \sqrt{V}\phi_a \phi_c \cdot \sqrt{V}\phi_b \phi_d \right| \\ & \le \| \sqrt{V}\phi_a \phi_c \| \| \sqrt{V}\phi_b \phi_d \| = \sqrt{g_{aacc}} \sqrt{g_{bbdd}} \end{align}. $$  (Ineq. 4) is not useful for integral screening because it is strictly weaker than (Ineq. 1) since $g_{abab} \le g_{aabb}$ (for real orbitals). Whereas $g_{abab}$ vanishes as orbitals a & b become spatially disjoint, $g_{aabb}$ only decays algebraically in the average distance between a & b. (Ineq. 5) and (Ineq. 6) are not strictly weaker than (Ineq. 1), but they have the same problem of being weakly decaying bounds that don't screen very much. For example, they are tight and thus tighter than (Ineq. 1) when a=b and c=d, but this is a class of integrals that will only be screened out for extremely large spatial separations between a & c.


text:
Question based on finding impurity of a sample of Sodium bicarbonate
12.5g of commercial  sodium bicarbonate gives 3 g of carbon dioxide. What is the percentage in purity in the sample?
$$\ce{2NaHCO_3 -> Na_2CO_3 + CO_2 + H_2O}$$
I have solved this question, but I just need my solution verified, as it’s not given. I have solved it as follows
Molar mass of sodium bicarbonate=84
Molar mass of carbon dioxide=44
expected yield of carbon dioxide=3.27g, obtained using stoichiometric calculations.
But obtained product=3g
%yield =$\frac{obtained product}{expected product}.100$
Therefore %yield =91.74%
The answer given is 8(I don’t know how that’s possible), so I would appreciate if someone would let me know if my solution was right. Thanks!

I haven't checked the numbers, but the calculations are set up correctly. Are you sure that you are not asked about the percentage of impurity? If that is the case, 100-91.74 is quite close to 8 :D

Yup, it’s asked as percentage purity


text:
Electronic configuration of Co³⁺
Cobalt (atomic number 27) has electronic configuration $[\ce{Ar}]\mathrm{(4s)^2 (3d)^7}.$
I have seen two different configurations and explanations for $\ce{Co^3+}$ from a variety of different sources:

either it loses two $\mathrm{4s}$ electrons and becomes $[\ce{Ar}]\mathrm{(3d)^6}$ in accordance with aufbau principle;
or it loses one $\mathrm{4s}$ electron and two $\mathrm{3d}$ electrons to become $[\ce{Ar}]\mathrm{4s (3d)^5}$ which is supposedly more stable due to its half filled orbitals and lack of electron pairs.

I think the one with more stability or lesser energy will be the one which is correct, but how can I find this out without experimental data? And what is the right answer?


text:
Why does ammonia have higher melting point but lower boiling point than HF?
I guess the reason is hydrogen bonding, but shouldn't both the trends be similar in that case?

Boiling point is simple to predict. Melting point is tricky.

@IvanNeretin I think even the boiling point of ammonia should be more than HF as it can form 4 hydrogen bonds while the later only 2

You think wrong. Ammonia is no different from HF in this regard.

@Stan That doesn't quite make sense because ammonia molecules can only accept 1 hydrogen bond. You have them all form 3 hydrogen bonds via their N-H's because there aren't enough acceptors.

You are comparing pears and apples. Two very different molecules, two very different substances. There isn't one reason why their melting and boiling points are what they are.
At the very qualitative level you are looking at it, their Mp is the same. And the Bp of HF is significantly higher because its hydrogen bonds are much stronger.


text:
Symmetry Space Group Diagram
I am doing an introductory course to Crystallography and I have found in the following web page: http://pd.chem.ucl.ac.uk/pdnn/symm3/sgmonoc.htm a guide for the elaboration of the diagrams of the spatial groups.
In this tutorial, they say the following: "The symmetry operator for the c-glide plane is of the form x,T-y,1/2+z, while the symmetry operator for the two-one screw axis is of the form T′-x,1/2+y,T′′-z. It is possible to show (using matrices) that the only values for T, T′, and T′′, which permit the symmetry operators to form a closed group, are 1/2, 0, and 1/2, respectively"
They could tell me how to calculate the values of T, T' and T'' using the matrices.

Why, that's simple. Apply each operator twice and see where this gets you.

Sorry, but I don't understand you

They say the symmetry operator has a certain form. What does that mean, if anything? I'll tell you what. It means that the operator grabs any point (x,y,z) and relocates it to certain other coordinates. Now I suggest looking at what happens if you apply the operator twice.

But, I would like to know how to obtain the values of T, T', T'' through the use of matrices.

Take a huge whiteboard marker and write "Matrices" on your chair where you will be sitting while doing this. I think that will count as the use of matrices. Seriously, what is the point of using a specific instrument you don't really need?

I know how to perform operations with matrices and I know the mathematical definition of group. So don't worry about it. Thank you.

Then perform the glide plane operation with matrices, if you insist so, and do that twice.

But, so I can calculate the values of T, T', T''. In the manual they insist on the existence of unique values of T, T', T'' for which a closed group is formed and that these values can be easily deduced with matrices. But, they do not deduct it simply name that with matrices could be done.

Do the operation twice and see what happens. That would be some symmetry operation too. An operation of what kind?

It's just that this is the only way I think I'm going to be able to understand for space groups. For example, in the space group Pnma how do I know that there are three helical axes (which are located at 1/4) or that the sliding plane of type "a" is also located at 1/4. I'm not going to be able to understand this.

Or, for example, why does a diagonal plane (type "n") appear in the spatial group Ccc2?

You are missing an important part in the quote (the first sentence):

Fixing the point of inversion at the origin implies that one of the symmetry operators is -x,-y,-z. What are the symmetry operators for the glide plane and screw axis? (The answer is clearly listed, but suppose it was not given here!) The symmetry operator for the c-glide plane is of the form x,T-y,1/2+z, while the symmetry operator for the two-one screw axis is of the form T′-x,1/2+y,T′′-z. It is possible to show (using matrices) that the only values for T, T′, and T′′, which permit the symmetry operators to form a closed group, are 1/2, 0, and 1/2, respectively. This produces the symmetry operators listed on the right-hand side of the space-group diagram shown above.

Once you fix the origin, T, T' and T'' are indeed fixed. Otherwise, they are not. So to show that they are, you have to combine a glide plane or screw axis with the inversion operation for the chosen origin, and then make an argument.
These arguments usually go as in this example: Two symmetry operations combined give a pure translation. This translation has to be compatible with the crystal lattice, otherwise you have a contradiction.

And, this can be done mathematically. In other words, without knowing these arguments, I can deduct them.

For example, in the space group Pnma. Can I know from matrices that the helical binary axis is at 1/4? Or that the "a" plane is too?

Or, for example, why does a diagonal plane (type "n") appear in the spatial group Ccc2?


text:
Why does the dipole moment order of group 15 hydrides increase down the group?
The order given in my book is

$$\ce{NH3} > \ce{SbH3} > \ce{AsH3} > \ce{PH3}$$

Phosphorous is more electronegative than arsenic and antimony. Then shouldn't the dipole moment be more in case of $\ce{PH3}?$
And why does the anomaly of $\ce{NH3}$ occur?

Geometry/symmetry  also has a role in determining the dipole, for example, if the molecules were planar  as is $\ce{BF3}$  the dipole would be zero.

@porphyrin could you convert this into a proper answer for this question

The query you have is a common mistake, and this is aksed as a very common question in examinations, well, I wouldn't blame you as the reason is simple but not easy to think of.
What you said about E.N. of phosphorous being more than arsenic and antimony is absolutely correct. But what you missed, is that lower down the group the direction of dipole itself reverses.
E.N. of $\ce{P}$ is $2.19$, while that of $\ce{As}$ is $2.18$ and of $\ce{Sb}$ is $2.05$
Whereas H as an E.N. of $2.20$. So hence now you can see clearly that your order is justified, as in actuality, H becomes the negative end of the dipole.
Another factor that further widens this gap is that the dipole vector in chemistry notations is from + to-, which means all vectors add up in case of $\ce{As}$ and Sb

Why should I not use doubt? @The_Vinz ?


text:
Synthesis of copper selenide
If I react selenium powder with copper chloride powder above the melting point of selenium (221 °C), will I end up with a chunk of copper selenide, with the chlorine given off as gas, or will I lose some or all of the Se as as selenium chloride gas?

Sounds like a bad idea to me. Chlorine certainly won't go off without a fight.

And selenium compounds can be very foul smelling. The smell also persists for a long time.

The eutectic in the Cl-Se binary system is at -52C (yeah, surprised me too). All copper-selenium phases have melting points above that of selenium, the most selenium rich are stable in contact with liquid selenium making it hard to predict what would happen.

There are many ways to synthesize copper selenide. It generally has two forms $\ce{Cu2Se}$ and $\ce{CuSe}$ but actually it is a non-stoichiometric compound. They are grown as nanoparticles for various applications like making semiconductor, optoelectronic components, sensors etc.

By direct combination of copper and selenium vapour / direct displacement of cupric salts with hydrogen selenide.(1)(2)
Using copper(II) chloride dihydrate ($\ce{CuCl2.2H2O}$) and sodium selenite ($\ce{Na2SeO3}$) in ternary oil-in-water (o/w) emulsion system. The reducing agent, sodium borohydride ($\ce{NaBH4}$) used to reduce sodium selenite source
at room temperature(3)
Using thiol ligands(4)
Using copper acetate dihydrate and hydrazine hydrate(5)

As for your reaction, chlorine will not get displaced that easily. It needs a counter ion for chloride to make bond. That's why instead of just selenium, hydrogen selenide is used. It can be better if any displacement is avoided and just direct combination of elemenys is done. See method 1.
References

http://copper.atomistry.com/cuprous_selenide.html
http://copper.atomistry.com/cupric_selenide.html
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.452.6282&rep=rep1&type=pdf
https://pubs.acs.org/doi/abs/10.1021/acsami.5b01085
http://www.asianjournalofchemistry.co.in/User/ViewFreeArticle.aspx?ArticleID=25_10_57


text:
Why does the image of cyclo[18]carbon look like a nonagon?
The $\ce{C18}$ allotrope cyclocarbon has been synthesized and imaged.[1] Science has most details behind a paywall, but this discussion includes an image:

In this octakaidecagonal molecule, each $\ce{\color{blue}{C}}$ is bonded viz. $\ce{C-\color{blue}{C}#C}$. Yet the above image looks like a nonagon. Why does every other carbon atom stand out as an obvious vertex while the others don't? My thoughts:

The internal angles are close to $180^\circ$, making vertices hard to see. However, I wouldn't expect the internal angles of this octakaidecagon to differ much.
I wonder if this molecule has a delocalised ring analogous to the one in benzene. On this idea, each of the nine visible edges could alternate between the states $\ce{C-C#C}$, $\ce{C#C-C}$. But this wouldn't explain why "odd" vertices have one appearance while "even" ones have another. I'm not convinced of this idea anyway, because it would average out to $\ce{C=C=C}$, unlike the $1.5$-bonds in benzene.

References:

Kaiser, K.; Scriven, L. M.; Schulz, F.; Gawel, P.; Gross, L.; Anderson, H. L. An sp-hybridized molecular carbon allotrope, cyclo[18]carbon. Science 2019, eaay1914. DOI: 10.1126/science.aay1914.

Related: How small is the smallest known carbon ring containing only double bonds?

The first thing to say is that I'm not sure where that image is taken from; it's neither in the original article nor in the supporting information to the article. Therefore, it appears to be more of an "artist's impression" rather than an actual atomic force microscopy (AFM) image, which is what was reported in the paper.
Nevertheless, the actual AFM images of $\ce{C18}$ are in Figs. 3Q and 3R. They are referred to as "AFM far" and "AFM close" respectively because of the height of the probe ($\Delta z$):

One can indeed see that there is 9-fold symmetry (technically, $D_\mathrm{9h}$). This implies that $\ce{C18}$ has a 'polyyne' structure in which there are two different types of bonds $\ce{-C#C-C#C-\phantom{}}$, rather than a 'cumulene' structure in which every bond is equivalent $\ce{=C=C=C=C=}$ (prior to this, computational studies had been equivocal as to which form was more stable).
The bright spots within the ring do not correspond to carbon atoms, but rather to carbon–carbon triple bonds. This is consistent with the AFM images obtained for other similar intermediates in the synthesis of cyclo[18]carbon. In the authors' own words:

Assigning the bright features in the “AFM far” images to the location of triple bonds, we observed curved polyyne segments with the expected number of triple bonds: 5 in $\ce{C22O4}$ and 8 in $\ce{C20O2}$. At small tip height, we observed sharp bond-like features with corners at the assigned positions of triple bonds and straight lines in between. This contrast was explained by CO tip relaxation, in that maxima in the potential energy landscape, from which the tip apex was repelled, were located above the triple bonds because of their high electron density. In between these maxima, ridges in the potential landscape led to straight bond-like features.

(The two bright spots outside the ring are due to individual $\ce{CO}$ molecules.)


text:
H and H- atomic and ionic radius
A hydrogen anion has a bigger radius than neutral H . Why is this happening ? I mean both electrons exist at the same shell so this is not it. Both H- and H have 1 proton so this is not it as well . Both electrons of the hydrogen ion have the same energy level and they don't shield the charge of the core from one another. What is it then? Pauli repulsion?

Same shell all right, but they feel each other, don't they?

and if that was the case the H- couldnt exist since it would feel a total charge of 0 meaning it could be anywhere...

Not quite. Imagine the impossible situation where one electron was right next to the proton. Then the other electron would experience zero net attraction/repulsion. This would be like having a neutron and one electron. But the inverse square law implies that there is a difference between 1) when the 2 electrons are on the "same side" of the H atom and 2) being on "opposite sides" of the H atom. Make sense?

Ed V first of all electrons dont have a precise location until measured .

And second they experience attractive magnetic force due to having opposite spins.

You have so many misconceptions, I don't think one question could cover them. Most important thing for this question is that you seem to think that data for ionic or atomic radii are for bare atom or ion which is not true. These are just calculated values for particular types of compounds.

I'm honestly at a loss what you're trying to achieve here. The 1s of a hydride is not of the same energy as that of a hydrogen radical, for starters.

tell me why it isn't.

Electrostatic repulsion between electrons is the key to understanding $\ce{H-}$. The electrons might be delocalized, but they are also correlated - they avoid each other.

Both electrons of the hydrogen ion have the same energy level and they don't shield the charge of the core from one another.

The Aufbau and Pauli exclusion principles (and the assumption of hydrogenic orbitals) are insufficient to explain the structure and stability of $\ce{H-}$. In particular, it helps to drop the assumption that the electrons occupy the same spatial orbital and differ only according to their spin quantum number [1]:

As  pointed  out  in  the  introduction  Bethe  [5]  and Hylleraas [6] were the first authors to prove the stability of the negative hydrogen.  But it was Chandrasekhar [1]who first introduced a clever wave function to describe the H− system which leads to a beautiful physical picture. The key concept introduced by Chandrasekhar was to break the symmetry between the two electrons, which is a way to introduce implicitly the electron correlation.  [...]
We conclude that the negative hydrogen ion is a special  atomic  system  whose  stability  depends  completely on the electron correlation. [...] mean field theories and other approximations do not take into account correlation, preventing the possibility of predicting stability for the negative hydrogen ion.

Reference
[1]. Ignacio  Lopez  de  Arbina  de  Frutos. Stability of the Negative Hydrogen Ion:  Variational Approach with Electron Correlation.

Would have preferred to include a reference to a peer-reviewed article or textbook but the linked article does a good job of explaining the pertinent details.

what about the attractive magnetic force?

due to the opposite spins?

@Engineer It's not clear what you are asking. Maybe you are talking about the exchange interaction? https://en.wikipedia.org/wiki/Exchange_interaction ? Quote: "Indeed, in general, the direct magnetic interaction between a pair of electrons (due to their electron magnetic moments) is negligibly small compared to this electric interaction."


text:
Partial dissolution of non-polar solutes in polar solvents
I have learnt that in order for dissolution to occur, the solute-solvent intermolecular forces must be strong enough to overcome the solute-solute and solvent-solvent forces. However if this is the case, why are non-polar substances such as methane able to partially dissolve in polar solvents such as water? Aren't the dispersion forces always too weak to break the pre-existing hydrogen bonds present in water?

In solids, reorientational and especially translational molecular motion are highly hindered. The motion of individual molecules is constrained about mean fixed positions within a regular lattice. Such regularity is absent in liquids, where significant thermal energy is associated with rearrangements in molecular position and orientation.
This is of course one defining property of a liquid (and more generally of fluids). The corollary of this structural irregularity is that it leads to a greater number of potential "defects" in the network of interactions within the substance.
Consider water. The wikipedia explains that in crystalline ice, water forms hydrogen bonds with 4 other water molecules. Meanwhile,

[from] TIP4P liquid water simulations at 25 °C, it was estimated that each water molecule participates in an average of 3.59 hydrogen bonds. At 100 °C, this number decreases to 3.24 due to the increased molecular motion and decreased density, while at 0 °C, the average number of hydrogen bonds increases to 3.69.[35]

The net result of this structural irregularity is that even molecules that interact weakly through dispersion interactions (or, in principle, not at all) with water have a significant if small chance of persisting within the liquid. Therefore, it is less a matter of dispersion interactions "breaking" hydrogen bonds, and more of apolar solutes occupying "defects" in the liquid.
References in wikipedia article
[35] Jorgensen, W. L.; Madura, J. D. (1985). "Temperature and size dependence for Monte Carlo simulations of TIP4P water". Mol. Phys. 56 (6): 1381.


text:
What does a comma signify in inorganic chemistry?
$\ce{(Ca{,}Co)CO3}$
I have never ever seen this before. What does this mean exactly?

It just means that relative amounts of calcium and cobalt are  indeterminate.

That's a mixed carbonate, similar to https://en.wikipedia.org/wiki/Mixed_oxide

Quoting [1]:

Comma. In accord with general usage, the comma "," should be used to
  indicate continuous substitution, partial as well as complete, if
  there is no change of CN. Vacancies (symbol $\square$) are treated here just
  like atoms. This usage is restricted to chemical formulas for atoms on
  a specific site. The comma is not appropriate for structure site
  formulas because different symbols indicate a change in the
  coordination number. Examples for proper use of the comma include
  forsterite-fayalite $\ce{[(Mg{,}Fe)2(SiO4)]}$ and hornblende $\ce{\{(Na{,}\square)Ca2(Mg{,}Fe{,}Al)_5[(Si{,}Al)4O11]2(OH)2\}}$. In structure type formulas, the
  comma is used to mean "or" where substitution may occur. An example is
  in the zeolites where H2O and M units may both occur.

Comma representation are often used in writing chemical compositions of minerals as the elements and its concentration vary to a certain extent. See minerals.net:

A number of minerals contain a varying amount of two or more elements like Aurichalcite, which has a chemical formula of
  $\ce{(Zn{,}Cu)5(CO3)2(OH)6}$ contains an unspecific varying amount of
  zinc (Zn) and copper (Cu). This is indicated by comma separating the
  Zn from the Cu. If a chemical formula with two elements in
  parenthesis is separated by a comma, the number of those elements
  vary. Aurichalcite has a variable amount of zinc of copper where the
  combination of both these elements totals five. The more dominant
  element is usually listed first. (Emphasis added.)

For more information, see this previous Chemistry.SE question and see here to find steps regarding how to write such type of formulae using chemical markups.
References:

Smith, D. K.; Roberts, A. C.; Bayliss, P.; Liebau, F. A systematic approach to general and structure-type formulas for minerals and other inorganic phases. Am. Mineral. 1998, 83 (1-2), 126–132. DOI: 10.2138/am-1998-1-212. Online version: http://database.iem.ac.ru/mincryst/fst.html

When quoting, please make sure to quote correctly, or mark your edits within the quote.

A short “obligatory” reference from IUPAC Recommendations [1, p. 21], section IR-2.2.3.1 regarding use of parentheses in formulae:

(d) In solid-state chemistry, to enclose symbols of atoms occupying the same type of site in a random fashion. The symbols themselves are separated by a comma, with no space.
Example:

K(Br,Cl)

Related question on chemistry typesetting in $\mathrm\LaTeX$: Comma in chemical formula.
References

IUPAC “Red Book” Nomenclature of Inorganic Chemistry, 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; IUPAC Recommendations; Royal Society of Chemistry: Cambridge, UK, 2005. ISBN 978-0-85404-438-2. PDF

I have seen this nomenclature used as a shorthand to indicate a solid solution of two different ionic compounds that share a common anion. For example:

Wang, Q.; Grau-Crespo, R.; de Leeuw, N. H. Mixing Thermodynamics of the Calcite-Structured (Mn,Ca)CO3Solid Solution: A Computer Simulation Study. J. Phys. Chem. B 2011, 115 (47), 13854–13861. DOI: 10.1021/jp200378q.
González-López, J.; Ruiz-Hernández, S. E.; Fernández-González, Á.; Jiménez, A.; de Leeuw, N. H.; Grau-Crespo, R. Cobalt incorporation in calcite: Thermochemistry of (Ca,Co)CO 3  solid solutions from density functional theory simulations. Geochim. Cosmochim. Acta 2014, 142, 205–216. DOI: 10.1016/j.gca.2014.07.026.

Since it's a solution of two different compounds, its composition can be variable.


text:
what are the cons of converting plastic to fuel
I have been trying to read how to fix plastic problems, there seems to be 3 outstanding solutions

Re-use
Recycle
Decompose to Fuel

Recycling is expensive, requires lots of water and the quality degrades on each recycle. I can partially understand why companies won't recycle
However I am unable to understand why would we not decompose to fuel. I have tried reading following articles:

Can plastic be used as fuel for vehicles?
https://www.power-technology.com/comment/plastic-to-fuel/
http://www.anthropocenemagazine.org/2019/02/new-technique-converts-plastic-waste-to-fuel/
http://www.deccanchronicle.com/science/science/180717/hyderabad-techie-recycles-plastics-into-fuel.html

It seems the primary way is Anhydrous pyrolysis requires no water, limited sorting and seems to convert 80% of plastic to fuel. I am unable to understand why would everybody not do this ? seems a viable solution and a profitable business

Perhaps better is to use plastics as fuel directly, in high capacity, high temperature facilities, providing electricity or hot water.

Following the previous comment, what kind of "fuel" are you looking for? Not all fuels are made equal. Wood is useful as a fuel, but so is ethanol.

@Poutnik using plastics as fuel directly has more pollution impact than converting and using it.

@BuckThorn Looking at petroleum fuel - diesel, petrol etc

@Gaurav Shah I would say the opposite. There is environmental impact of conversion and small local device combustion has worse parameters.

makes sense that way. Thanks @Poutnik


text:
How does cross-conjugation destabilizes a molecule?
Wikipedia article on ortho effect suggests that o-toluic acid is more acidic than benzoic acid. This fact is rationalized by suggesting that steric effect causes COOH group to "twist" out of the plane and which increases it's acidity as it inhibits "cross-conjugation"that it was previously taking part in.
My instructor told me that cross-conjugation is two parallel conjugated sets of atom both terminating on a common atom. He says this parallel character of such conjugation makes conjugation on common atom less effective.
My question is -

How accurate is it to point out cross conjugation as the reason for acidity of o-toluic acid ?

How correct is the explanation of cross conjugation by my instructor?

Edit #1
There are some valid concern raised by Alchimista. It seems like I have not made my question clear to understand. I would rephrase it a bit ->
I am not advocating the usage of cross conjugation to explain acidity of o-toulic acid. But merely seeking clarification for its usage to explain acidity of o-toluic acid by the wikipedia article(linked here for convenience -https://en.wikipedia.org/wiki/Ortho_effect#General_explanation_for_first_ortho_effect). To rephrase the question, Wikipedia article states that cross conjugation "destabilizes" conjugate base of benzoic acid but never justifies the statement.
My main concern is not the justification for acidity of o-toluic acid as there are plenty of answers here on that(Link->https://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases). What I want to know is, if cross conjugation destabilizes benzoic acid as mentioned in the article and if yes then how?

It seems the use of the term criss-conjugation is used differently in your description. Also in the example of o-toluic ac. is steric hindrance that prevents effective conjugation, and this should make the conjugated base less stable → the acid less strong. But perhaps I misunderstood. Also note that conjugation is merely topological and shouldn't be use as synonymous of delocalisation.

@Alchimista -> 1."Also in the example of o-toluic ac. is steric hindrance that prevents effective conjugation, and this should make the conjugated base less stable→ the acid less strong". I am not able to comprehend your statement, are you implying that o toulic is weaker than benzoic? pKa of o toulic is 3.91 while that of benzoic is 4.19. Sincere apology if I have incorrect understanding of your statement. English is not my first language.

2."conjugation is merely topological and shouldn't be use as synonymous of delocalisation" completely agreed. Sorry for misusing terminology.

3.3."It seems the use of the term criss-conjugation is used differently in your description" I assume it was a typo and you are referring to cross conjugation and not something I am unaware of (I believe this is reasonable as "o" is pretty close to "I" in keyboard if this is not the case, it would be very kind of you to post some relevant article). I am completely sure in my understanding of term cross conjugation. Part (1/2)

I agree that probably I am not able to explain it properly as definition given in my post is a mixture of my own and some translated thoughts expressed by my instructor in another language. Part(2/2)

The Wikipedia article that I referred to in my post ->  https://en.wikipedia.org/wiki/Ortho_effect#General_explanation_for_first_ortho_effect

Yes the only thing I overlooked is about the fact that substituted benzoic ac.s are stronger as it is the delocalisation within the conj.acid that is inhibited.  But cross conjugation is not the context here.  We are dealing with delocalisation M effects and steric hindrance  it suffice to say that the ring and the carboxylic gr. are not in plane.

Look at this figure it might clarify the situation https://qph.fs.quoracdn.net/main-qimg-61d6d230e0462107958eefa8fa369677

@Alchimista->1."But cross conjugation is not the context here." I am not advocating the usage of cross conjugation to explain acidity of o-toulic acid. But merely seeking clarification for its usage to explain acidity of o-toulic acid by the wikipedia article (linked again for convenience ->https://en.wikipedia.org/wiki/Ortho_effect#General_explanation_for_first_ortho_effect). To rephrase the question in the post, Wikipedia article(not me) states that cross conjugation is responsible for "destabilization" of benzoic acid but never justifies the statement ,I am asking if article is correct (P1)

I do not attribute 'twisting' of COOH to cross conjug. and understand that it is consequence of steric interaction. My main concern in not the justification for acidity of o-toulic acid(which I believe you are assuming) as there are plenty of answers here on that. I am only concerned by the aforementioned Wikipedia article's claim that cross conjugation destablisies benzoic acid. I edit the post to make it clearer. :)

Lovan I get confused. Basically the reason is that the +M of the phenyl ring is inhibited or suppressed. I do not see why cross-conj. is mentioned in Wikipedia.  It might be a further matter of analysis if the substituent would be a group with its own mesomeric effect but certainly is just adding an unnecessary word in o-toluic acid.


text:
Finding out more stable conformer via molecular orbital interaction

Out of these two, which one is more stable? I'm unable to imagine the orientation of the lone pairs of O and ABMO of C and Cl bond.
Any tips to tackle these kind of problems?

Try writing the lone pairs of the two heteroatoms as if they were attached to the atoms by bonds, in a tetrahedral geometry. What structure minimizes LP-LP interaction? :)

@The_Vinz Based on your idea, I would say Cl in axial position is favorable. I also want to look at it another way. Can we consider the pi type overlap between lone pair of O and sigma asterisk bond of C and Cl?

Yep, the answer is that: chlorine in axial position minimizes the so-called 1-3 diaxial interaction between the lone pairs, and is thus more stable (from a substantially qualitative AM1 calculation, around 15 kJ/mol lower in energy). As to the molecular orbitals, I confess that personally I couldn't extimate that interaction "on paper"

Search for "anomeric effect", you will probably find a number of websites explaining this in some detail.


text:
Number of stereoisomers of polysubstituted cubane

The molecule given below is a substituted cubane.

a) How many stereoisomers exist for this molecule?
b) How many pairs of enantiomers are possible?

How do I calculate the number of stereoisomers? Since each carbon is linked to four different groups, can I claim that all are chiral and, therefore, $2^8$ stereoisomers are possible?
If that is true, since there is no element of symmetry, can I say that $2^4$ enantiomeric pairs exist?
Edit: Looking at the comments clears things up but brings forth another question. If, say, iodine was present at any other position that would not lead to a tetrahedron, what would happen?

True, but they don't exist. Try to draw some and you'll see. Only 2 isomers are possible.

The molecule u have asked about is very similar to a substituted adamantine...check this out https://chemistry.stackexchange.com/questions/58727/chirality-of-substituted-adamantanes

A tetrahedron may be inscribed in a cube,

I hope I am not too late.
All carbons are $\mathrm{sp^3}$-hybridised and three bonds of every carbon are fixed (one corner shares three bonds), so fourth one has only one place in space to maintain an $\mathrm{sp^3}$ tetrahedron. This one is optically active, so this and its enantiomer are the only two possible isomers.
Now you get why we don't use the $2^n$ formula. And similar thing happens in a bridgehead position.


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Finding the isotherm with given minima

Get the temperature of the isotherm for water for which the local minimum is at $\pu{100 Pa}.$ Use the values of $a$ and $b$ of water.

My approach
In the van der Waals equation, set
$$\frac{\mathrm dP}{\mathrm dV} = 0$$
and get the value of $T.$ Plug this value in the initial van der Waals equation and and solve for $V$ from the biquadratic.
The calculations get very messy and the values come out to be unrealistic. (For example, temperature of the order of $10^6$)
My doubt is whether the $\pu{100 Pa}$ as provided by the problem setter a realistic value?

I got 546.0 K, but I did the calculations in Mathematica.  What's the answer supposed to be?

I do not have the answer. 546.0 K seems like a reasonable answer, did the problem boil down to pV^3 -aV + 2ab = 0.

Using Mathematica, I obtained $T=546 K$, as follows.  One of the nice things about using Mathematica for physical calculations is that it has the ability to understand/keep track of/cancel out/convert units.
The volume is non-physically small -- about 750,000 x smaller than what we'd expect for a mole of, say, ideal gas at $P=100 Pa, T = 546 K$).  One does expect a non-physically small volume, since this is at a non-physically low point on the isotherm, but I would not have predicted it would be this much smaller.
If you were doing it by hand, you'd probably want to use a different approach:

Thanks, you seem comfortable with Mathematica, could you solve the same problem but this time for local maxima? Is the answer 1.78 K in this case?

@StabReberie  Yes, that's easy enough; I just change (in the Solve statement at the end) dPdV2 > 0 to dPdV2< 0, and then rerun the notebook.  When I do this, I  get T = 1.79 K [essentially the same as yours; you may have a round-off error; note that Mathematica uses R =8.31446 J/(mol K) = 0.0820574 L atm/(mol K)], and V = Vm = 0.0744 m^3.