text:
What is the correct definition of hydration enthalpy and why is it always negative?
The first definition of hydration enthalpy I found on the net is:

The hydration enthalpy is the enthalpy change when $1$ mole of gaseous
  ions dissolve in sufficient water to give an infinitely dilute
  solution. Hydration enthalpies are always negative.

Another definition I found in a book is:

Hydration enthalpy is the energy released when $1$ mole of a compound is
  dissolved in sufficient water to give an infinitely dilute solution.

Now I have three questions:

Which of the two definitions is the correct one? Why?
Why exactly are hydration enthalpies always negative? Suppose I have hydrated copper(II) sulphate which is mostly stable as $\ce{CuSO4.5H2O}$. I convert it to $\ce{CuSO4.6H2O}$ by adding more water forcefully. Shouldn't the hydration enthalpy be positive in this case as it will make the final product less stable than the initial one?
Which definition of hydration enthalpy applies for $\ce{CuSO4}$? I do not think $\ce{CuSO4.5H2O}$ is "infinitely" dilute by any extension of logic. Please explain this if possible.

P.S: I am a novice in thermochemistry. It would be helpful if you could keep the explanations as simple as possible.

I'd just like to note that both definitions are commonly called "enthalpy of hydration", so if you see it in a book, make sure to check the context and figure out which one they're referring to. It's very common to talk about the "heat of hydration of so-and-so ion", which refers to the first of your two definitions.

Could you please answer "Suppose I have hydrated copper(II) sulphate which is mostly stable as $\ce{CuSO4.5H2O}$. I convert it to $\ce{CuSO4.6H2O}$ by adding more water forcefully. Shouldn't the hydration enthalpy be positive in this case as it will make the final product less stable than the initial one?" @orthocresol

Hydration is defined as the following process (with any salt but using copper(II) sulphate as an example):
$$\ce{CuSO4 + 5 H2O -> CuSO4 . 5 H2O (s)}\tag{1}$$
This reaction will only happen if the associated enthalpy $\Delta_\mathrm{hyd} H^0$ — or more precisely, the associated Gibbs free energy $\Delta_\mathrm{hyd} G^0$ is negative. For some compounds, that is not the case; for example sodium chloride does not form hydrates: reaction (2) does not occur spontaneously.
$$\ce{NaCl + n H2O} \nrightarrow \ce{NaCl . n H2O}\tag{2}$$
Thus, we cannot measure a hydration enthalpy. If we can measure it, the process must be spontaneous and thus the enthalpy negative. Thus, all measurable hydration enthalpies are negative.
Sometimes, salts can form multiple hydrates. However, not every hydrate is always possible. Hydration enthalpies only exist for those hydrates which are possible.

Your definitions are basically identical only that they do not measure hydration enthalpies but solvation enthalpies. That is the process as shown in equation (3).
$$\ce{CuSO4 (+ H2O) -> Cu^2+ (aq) + SO4^2- (aq) (+ H2O)}\tag{3}$$
For that process, the enthalpy does not have to be negative. In fact, quite a few ionic compounds dissolve endothermicly, e.g. $\ce{NH4Cl}$.

Even the process $\ce{Na^+(g) \rightarrow Na^+(l)}$ is called hydration. Is the enthalpy for it always negative?

FYI "Hydration energy (also hydration enthalpy) is the amount of energy released when one mole of ions undergo hydration which is a special case of solvation. It is a special case of dissolution energy, with the solvent being water."  So the last example you gave falls under hydration only. See  https://en.wikipedia.org/wiki/Hydration_energy

I don't agree with "If we can measure it, the process must be spontaneous and thus the enthalpy negative." Spontaneous is about Gibbs energy, not enthalpy. I guess the entropy is expected to be negative (constraining water molecules to the crystal lattice) but unless you say that explicitly, the enthalpy could be positive or negative for a spontaneous reaction or process.

@KarstenTheis Of course, it's not easy to think oneself back into a five year old answer ... However, I think I wrote that because starting material (ionic compound) and product (hydrate of the ionic compound) are both solids and somehow I recall that in those reactions the entropy term would become zero and only the enthalpy ends up determining $\Delta G$ and thus the spontaneity? It's been a long time and I could have drawn wrong conclusions from prior knowledge.

Definition 2 sounds like the definition for enthalpy of solution.
In that case, both are correct: they measure different things.
Hydration enthalpies are always negative because hydrating an ion would always stabilize it. For hydration enthalpies, note that the waters of hydration are absent: you are hydrating the gas phase ions.
You can also measure both for $\ce{CuSO4}$. There is no hydration enthalpy for the hydrate.

No, $\ce{CuSO4}$ does not have a hydration enthalpy. Look at the definition again. You need to dissolve the component ions to infinite dilution. That preludes the ability to "hydrate" the hydrate.

To me both definitions are correct. The complex cation hexaaquacopper(II) cannot undergo further hydration and therefore it is infinitely dilute given that all 6 ligands have been provided. From the definition, since hydration enthalpy is a measure of energy released from a mole of compound then the final compound possesses less energy than the initial one. enthalpy final minus enthalpy initial in this case will always be negative. in terms of stability the pentaaquacopper(II) complex should be unstable because 5 is not a common coordination number for the copper metal ion so the complex should be less stable and more energetic than the hydrated one.


text:
What is the difference between hexanes and petroleum ether
I'm trying to use hexanes as the nonsolubilizing recrystallization solvent in a multisolvent recrystallization (benzene is the solubilizing solvent) that requires an oxygen free atmosphere and hexanes is present in the glovebox. I am using it in place of light petroleum ether (not in glovebox) and was wondering what the differences between hexanes and light petroleum ether were to determine if I can adequately substitute the two.

My understanding is that they're fairly similar, but hexanes is purified to be mostly hexanes, whereas for petroleum ether, you get a mixture of hydrocarbons that boil in a given range. For recrystallization purposes, the substitution is probably fine. I guess it depends on how finicky your recrystallization is.

Petroleum ether:
This is a mixture of a lot of different compounds (aliphatic hydrocarbons in this case) that as a mixture have a boiling point within a specified range. The contents are dealt with as having low amounts of aromatic compounds and high amounts of this and that, not one specific formula. It's like a blackbox to me and can be used for crystallisations that are usually easy to do.
Hexane:
In the synthesis laboratory I worked in, hexane was dealt with as we would deal with petroleum ether, since it was a mixture the many possible constitutional isomers and not pure n-hexane. We bought the technical grade hexane in our chemical store and purified it with a solvent purification system to have it dry and compatible for glove box use. Since they appeared so clean in our Schlenk flasks I thought in the beginning that it was pure n-hexane but a simple 1H NMR spectrum taught me otherwise.
The hexanes mixture was used for medium difficulty crystallisations.
Pentane:
The pentane we bought and purified, however, was n-pentane and was used for difficult crystallisations that were not easy to reproduce. Already among organometallic compounds there are examples in which obtaining crystals at all is an enormous struggle and having single crystal quality is simply a matter of hope.
Therefore, by going from petroleum ether to hexane to n-hexane (or n-pentane) is a matter of going from a wide mixture to a compound to a pure chemical.

Any rationale for this finding, that stuff crystallises better from pentane than hexane or petrol ether? Btw., the use of n-hexane is frowned upon, because it is supposed to be rather more toxic than e.g. heptane. A mixture named "isohexane" seems to be widely used as substitute, where cyclohexane is regarded as too expensive.

Ah, thank you! I knew about benzene being frowned upon but not n-hexane.

The difficulty in crystallisation comes from highly symmetric clusters, large solvent spaces in the crystal lattice, or sometimes the compound crystallises only once and cannot be recrystallised because it decomposes. To find the optimal crystallisation conditions, it's best to rely on solvents with high purity and always the same properties. This cannot be found in petroleum ether. Petroleum ether is different in laboratories all over the world and pure compounds have a higher chance of leading to a reproducible experimental procedure.

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text:
Why can't chlorine atoms form hydrogen bonds?
Why can't chlorine atoms form hydrogen bonds even though they have very similar electronegativity to nitrogen, which can?
Electronegativities are as follows:
$\ce{Cl} - 3.16$
$\ce{N} - 3.04$
Is it because of the size ?

The existence of hydrogen bonds involving chlorine atoms was disputed a while ago, but experimental data show that chlorine atoms can form hydrogen bonds, and are in fact very common. See C.B. Aakeröy et al., The C–H···Cl hydrogen bond: does it exist?  New J. Chem., 1999,23, 145-152. DOI: 10.1039/A809309A

I really hate to do that, but quora actually has a good answer on that. (a as in one, only one good answer, i.e. the one by by Mario Barbatti) find it here

also related: http://chemistry.stackexchange.com/q/5503/4945

and here is real answer http://chemistry.stackexchange.com/questions/21773/why-is-hydrogen-bonding-generally-defined-to-include-only-three-period-two-eleme

If I wanted to close, I'd rather close as the latter. @Mith Anyway, I will let you guys decide...

Yeah, probably second one is better.

This discussion is really dubious. I would be careful about saying that chlorine engages in hydrogen bonds and argue with e.g. Cl-H-Br hydrogen bonds. Hydrogen bonds are mainly interesting in a biological context and then there are no chlorine hydrogen bonds. Even if there are chlorine bromine hydrogen bonds when you stir fuming HCl in a magic acid mixture at 400°C this doesn't mean that they occur in real world scenarios as well. Show me one protein where a Cl h.-bond occurs and I am convinced. $\mathrm{Cl^-}$ is just not basic enough to engage in significantly strong hydrogen bonds. Period.

@ketbra Seems you're not accepting the existance of weak H-bonds. All this N,O,F stuff are moderately or very strong examples but there can be weak or even very weak ones and they are also important, for example in protein folding.

@Mithoron Than could you please give an example where Cl is important in protein folding?

General stuff about weak H-bonds: http://pubs.rsc.org/en/content/articlepdf/2000/CC/A900221I Halogens in hydrogen bonds: http://pubs.acs.org/doi/pdf/10.1021/cg015522k#page=8&zoom=auto,-99,758

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text:
Word for a specific enzymatic process?
Foreword:
I am developing a variable-processing program for my Computer Science research project.
One particular part of the program mimics the behavior of biological enzymes, except instead of catalyzing break downs and compositions of chemicals, its catalyzing break downs and compositions of variables (specifically arrays).
The process in question is a 2 part process:

initiating process:

(in terms of enzymes:) breaks down certain compounds into smaller compounds
(in terms of program:) fragments certain variables (certain arrays) into fragments

finalizing process:

(in terms of enzymes:) combine the smaller compounds into larger compounds different from the originals (like mixing and matching the parts)
(in terms of program:) combine the fragment-variables into larger variables different from the originals (like mixing and matching the parts)

The process in question all together:

(in terms of enzymes:) breaks down certain compounds into smaller compounds,then recombines them into larger compounds that are different from the original large compounds (like mixing and matching the parts)

(in terms of program:) fragments certain variables (certain arrays) into fragments then recombines them into larger variables that are different from the original large variables (like mixing and matching the parts)

What is the process in question called?
Afternote:
I'm looking for a name, so the word could be a new word constructed from recognizable Latin or Germanic roots, and that would be an acceptable answer.
To get full coverage of an answer, I asked this question the English Language & Usage SE site, Stackoverflow SE site, and this SE site.

the purpose of asking on the different sites is to get different types of feedback, so if you want to provide an answer on each site, please be considerate and don't dupe your answers. Answer with a different light and perspective. I am trying to get a full-coverage answer

Cross-posting questions on multiple sites is generally frowned upon. It would probably be better if you decided which perspective suited you the best and asked it on one single site.

With that said, I'll give you two chemical options: rearrangement and isomerisation.

"Digestion" and "ligation" sound good to me.  So do "lysis" and "agglomerization".   Calling the combined process a "recombination" or even a "fermentation" seems apt.

In the broadest sense, "catabolic" for breakdown and "anabolic" for reconstruction might be suitable. Not sure what the two together would be, though.

@hBy2Py Amphibolic describes that. // general comment: OP is looking for a word to describe the entire process, not the individual break down and build up components. It's very common to say that "XXX enzyme catalyses the isomerisation of ABC to DEF". This class of enzymes even have their own name: isomerases. Hence my original proposal.

To add onto the existing comments, "catamorphism" and "anamorphism"  are existing programming terms for functions that do something like what you describe. While we could inappropriately transplant terms and call your process a "amphimorphism," I'm pretty sure you'd just be better off just describing exactly what you're doing. If you use a single word that's  unknown or with an ambiguous meaning, your documentation isn't of much use to anyone.


text:
How to calculate the magnitude of magnetic field of an orbital?
I read that according to QM  the electron in the ground state orbital produces no magnetic field, if that is true, can you explain why so and in what states there is a magnetic field and how that is produced? Why the difference?
Anyway, I am interested in learning how the classical model calculated it, since I read different values in different papers: for example hyperphysics says its value is .4 T, elesewhere I found .5 and .6, and If I use the classical formula I get .125 T
Does 0.4 Tesla refers to the exact value of the magnitude of the magnetic field od 1S state?

Orbitals don't "produce" magnetic field, only electrons. Where your original statement about ground states come from?

The hyperphysics page is referring to the 2p state, not the ground state.

@DavePhD,thanks,  do you know how they got .4 T and what would the value of ground state be? is it 1/8 T?

@user104372 hyperphysics means that given that the splitting between the two different 2p spin-orbit states is 0.000045 eV, and considering that this is due to the energy of the states being +/-
μB, where μ is the magnetic moment of the electron, B is 0.4T. For the ground state, 1s, there is no orbital momentum. The magnetic moment of the electron is only in the field of the proton, which is a much weaker field. The magnetic field would be equation 77 here: http://www.pha.jhu.edu/~rt19/hydro/node9.html

Hyperphysics means that the energy difference between the 2p J=3/2 state and the 2p J=1/2 state is as if the magnetic moment of the electron were aligned with and against a magnetic field of 0.4T.
The energy of the two states, relative to the ground state, is (according to NIST)
2p J=3/2 : 10.19885142904 eV
2p J=1/2 : 10.19880606470 eV
The difference is 0.00004538204 eV or $7.271004 \times 10^{-24}$ J
The electron magnetic moment is $−9.28476462 \times 10^{-24}$ J/T
So the energies are as if they are raised and lowered by being in a field of 0.3915556 T.
For the ground state, 1s, there is no orbital angular momentum.  The 1s state is only split by interaction with the proton.  This is sometimes referred to as hyperfine structure. The energy difference is $5.87433 \times 10^{-6}$ eV.

@tvb orbitals that are not s-orbitals have angular momentum and this creates a magnetic field.  A classical formula for the magnetic field is equation 42 here http://www.pha.jhu.edu/~rt19/hydro/node4.html#SECTION00031000000000000000 , but that is as if there are circular orbits, which is not true

Thanks, Dave, very helpful  answer and comment. Can you briefly say why s-orbital have no L while other do have, or should I ask a separate question?

@tvb "s" means l=0, "p" means l=1, "d" means l=2, etc.  Also, the angular momentum operators involve derivatives with respect to angular coordinates, while the wavefunctions for s orbitals are independent of angle, so such derivatives are zero.  For more see http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect8.pdf especially at the bottom of page 78.


text:
Why is there a difference in available energy in saturated vs. unsaturated fatty acids?
On Wikipedia it states:

In cellular metabolism, unsaturated fat molecules contain somewhat less energy (i.e., fewer calories) than an equivalent amount of saturated fat.

Can someone explain why that is true in simple words? (that is, if possible, in a way understandable by someone who is not conversant with biology nor chemistry)

I think the 'energy' referred to here is Gibbs free energy. One way to think of things, admittedly an over-simplification,  is as follows: all of the ATP generated by oxidative phosphorylation is generated by electrons 'held' in C-C and C-H bonds. None comes from water or C-O bonds.  Count the total number of C-C plus C-H bonds in the saturated and unsaturated fatty acids. Saturated fatty acids have the potential to 'make' more ATP compared with their unsaturated equivalents, and so far as cellular metabolism is concerned  may be thought of as a better 'source' of free energy

The total energy stored in macromolecules is commonly quantified by the amount of $\ce{ATP}$ produced, as $\ce{ATP}$ is used throughout the body to couple reactions and make them thermodynamically favorable.
First, as the oxidation of a double bond in an fully saturated fatty acid requires a reducing agent, you here produce $\ce{FADH2}$ from $\ce{FAD+}$ which is equivalent to a $\ce{1.5 ATP}$ gain.[1]
Also, unsaturated fat has double bonds in cis positions that are unfavorable to beta oxidation.[2] An enzyme coupled reaction is needed for the double bond shift. Therefore, an enzyme coupled reaction that requires energy (in the reductant form $\ce{NADPH}$) is needed to make the molecule a trans fat with a shift in bond position in order to continue oxidizing the molecule completely. $\ce{NADPH}$, usually is defined as $\ce{2.5ATP}$ equivalents[1] and this extra energy expenditure in shifting the double bond is why less energy is ultimately produced.  Note, this $\ce{NADPH}$ is only used when unsaturated fats have an even numbered bond position.[2] This happens with polyunsaturated fats when the first bond is shifted.
For example:
18:2 linoleic acid will yield $1.5 + 1.5 + 2.5 = 5.5$ less $\ce{ATP}$ than 18:0 stearic acid.

References:
[1]: http://www.saylor.org/site/wp-content/uploads/2013/04/BIO101A-6.2.3-CellularRespiration.pdf
[2]: https://en.wikipedia.org/wiki/Beta_oxidation

Yes, it is a fact that saturated fats contain more gross energy due to higher H+ concentration. However, unsaturated fats are more digestible and have higher absorption rate, and therefore the metabolizable energy that unsaturated fats provide is higher than that provided by saturated fats.

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text:
Birch Reduction Of Anthracene

Which ring of anthracene is reduced during Birch Reduction? What will the final product look like? Why?

My teacher said that the more electron deficient ring should undergo the reduction but I couldn't figure it out still. Any ideas?

There are two ways of rationalising this. One way is to look up or qualitatively determine the frontier orbitals (HOMO and LUMO) of anthracene, the other is to think about the stability of various radical anions.
But first, a quick note on the Birch reduction and its mechanism. As you may know, the $\ce{Na/NH3}$ system effectively works as a single electron transfer reagent. The electron is added to the π system’s LUMO, and the resulting radical anion is trapped by a proton (donated from a proton source; e.g. added ethanol) to give a radical. This radical is still electron-deficient and thus susceptible to another single electron transfer reduction resulting in a second anion which again is captured by a proton resulting in the reaction product.
Frontier orbital analysis
A great resource for frontier orbitals is the WOrbit page of Hans-Ulrich Wagner from the LMU Munich. Among the systems whose orbitals you can look at is, luckily, anthracene. For posteriority, I have included the HOMO and LUMO pictures in the image below.

Figure 1: LUMO and HOMO of anthracene as calculated on the WOrbit page linked above.
The LUMO is of importance; after all, we are adding an electron. We see that the MO is most localised on carbons 9 and 10 of the central ring — thus, this is where the first electron will be added in a localised picture, and this is where the first proton will capture.

Scheme 1: First reduction sequence.
As I drew in scheme 1, this gives us the radical 3, of which I have only shown one resonance structure. Applying the same analysis for the second step will tell us that the second electron will be added exactly at that position where I have drawn the radical (data not shown). Thus, we find that the central ring is reduced as depicted in scheme 2.

Scheme 2: Second single electron transfer and subsequent reduction of radical 3 to 9,10-dihydroanthracene 5.
Stability of radical anions
If we consider the $+I$ effect on substituted carbons, we see that we only need to consider the three structures shown in scheme 3 below; reversing charge and radical in structure 2 would result in the negative charge being on the tertiary carbon which is much less favourable when compared to the secondary one.

Scheme 3: Possible radical anions of anthracene.
At first, it looks like 6 and 7 might be better stabilised structures: 6 effectively represents a methylnaphthalene radical and 7 a naphthylallyl radical. However, if we take a closer look at 2 we see that it can be rearranged in a manner shown in scheme 1 above. If you consider the above drawn structure 3 as a radical anion (i.e. before protonation), notice that the radical is now a diphenylmethyl radical. This diphenylmethyl radical wins in favourability against both naphthyl derivatives.
As a general rule, if you can preserve the aromaticity of a benzene ring the structure is ‘better’ than one where you can’t. And two benzene rings as are present in structures 3 and 4 (and 5) are better than the naphthyl rings of structures 6 and 7. Thus, it is most favourable to reduce the central ring.
Note, by the way, that depending on the conditions employed, 5 is not inert towards further Birch reduction. Mehda, Sen and Ramesh report a triple Birch reduction of anthracene leading to the very interesting tricyclic triene 8 in $80~\%$ yield (scheme 4).[1]

Scheme 4: Synthesis of 1,4,5,8,9,10-hexahydroanthracene.[1]

Reference:
[1]: G. Mehta, S. Sen, S. S. Ramesh, Eur. J. Org. Chem. 2007, 423. DOI: 10.1002/ejoc.200600610.


text:
How is phosphate stable disobeying octet rule?

Doesn't the phosphate now have 10 electrons mutually shared between the oxygen atoms? I am unable to understand how the phosphate bond would be formed if phosphate does not obey the octet rule.
Thank you!
Reference: http://www.chemguide.co.uk/organicprops/aminoacids/makedna.gif

The octet rule is just a rule. Once you leave the 2nd row of the periodic table, the octet rule doesn't apply. Further reading: https://en.wikipedia.org/wiki/Hypervalent_molecule

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text:
Is the Ebullioscopic constant temperature-dependent?
In tables, the ebullioscopic constant is usually written as a single value.
But, when the value is not available, it can be calculated by applying the formula: $$K_{\mathrm b} = RT_{\mathrm b}^2/\Delta h_\text{vap}$$
I was wondering: since in this last definition the ebullioscopic constant does depend on temperature, being the value of $\Delta h_\text{vap}$ temperature-dependent (its value is different if calculated at room temperature or at the boiling point @1 atm) what is, in real life, the best approximation of the behaviour of $K_b$?
In other words, does it behave like a constant or its value changes at two different temperatures? (eg: boiling point elevation for a solution @atmospheric pressure vs boiling point elevation for the same solution @reduced pressure)

I'd say that Kb can be described as constant in the same sense as the enthalpy of vaporization can be considered constant; it's approximately constant over a small T range or when T << Tcritical. In other words, for Kb to be exact at a given temperature, the latent heat of vaporization would have to be corrected for temperature differences.

Short answer
As long as external pressure is constant, it does not make sense to discuss $K_b$'s temperature dependence. If external pressure is changed, boiling point of solvent changes and hence also the ebullioscopic constant.
Longer version
Condition for equilibrium (skip if familiar)
Boiling is a reversible equilibrium process. By the second law of thermodynamics,
$$\mathrm{d}S_\text{universe}=0.$$
Assuming constant pressure,
$$\mathrm{d}S_\text{universe}= \mathrm{d}S-\frac{\mathrm{d}H}{T}.$$
Provided additional constant temperature,
$$-T\mathrm{d}S_\text{universe}= \mathrm{d}H - T\mathrm{d}S = \mathrm{d}G_\text{P,T}.$$
We also know that $G=G(P,T,n_1,\ldots, n_k).$ Thus
$$\mathrm{dG}=V\mathrm{d}P-S\mathrm{d}T + \sum_{i=1}^{k}\mu_i\mathrm{d}n_i,$$
and therefore
$$\mathrm{d}G_\text{P,T} = \sum_{i=1}^{k}\mu_i\mathrm{d}n_i=0.$$
Currently, we have two phases: a gas and a liquid. Boiling is a dynamic equilibrium, the differentials are non-zero, $\mathrm{d}n_1 = -\mathrm{d}n_2$, and so
$$\mathrm{d}G_\text{P,T} = 0 \iff G_\text{liquid} = G_\text{gas} \iff \mu_\text{liquid} = \mu_\text{gas}.$$
Clausius$-$Clapeyron equation (skip if familiar)
Suppose a closed system with initial equilibrium, constant temperature and pressure. We saw that
$$G_\text{liquid} = G_\text{gas}.$$
Now change the temperature. To achieve a new equilibrium in a closed system, the pressure would also need to change. At this new equilibrium, $G'_\text{liquid} = G'_\text{gas}$, which is why
$$\mathrm{d}G_\text{liquid} = \mathrm{d}G_\text{gas},$$
or equivalently,
$$V_\text{liquid}\mathrm{d}P - S_\text{liquid}\mathrm{d}T = V_\text{gas}\mathrm{d}P - S_\text{gas}\mathrm{d}T.$$
Now
$$\frac{\mathrm{d}P}{\mathrm{d}T}=\frac{S_\text{gas}-S_\text{liquid}}{V_\text{gas}-V_\text{liquid}}.$$
Note that the RHS depends again only on the initial and final conditions of phase equilibrium. Thus we are allowed to use $\Delta S = \Delta H/T$ to arrive at
$$\frac{\mathrm{d}P}{\mathrm{d}T}=\frac{\Delta H}{T(V_\text{gas}-V_\text{liquid})}.$$
Because $V_\text{gas} >> V_\text{liquid}$, and due to the ideal gas law
$$\frac{1}{P}\frac{\mathrm{d}P}{\mathrm{d}T}\approx\frac{\Delta H}{RT^2}.$$
Assuming $\Delta H$ is not a function of temperature,
$$\ln \frac{P_2}{P_1} \approx -\frac{\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right).\tag{Clausius–Clapeyron}$$
NOTE. There is a small but important detail to keep in mind here. The assumption that $\Delta H$ is independent of temperature will only make our equations more inaccurate but it will not affect our answer. See the definition further below.
Boiling point elevation
We have
$$\ln \frac{P^\text{s}}{P^\circ} \approx -\frac{\Delta H_{vap}(T^\circ_\text{b})}{R}\left(\frac{1}{T^\circ_\text{b}} - \frac{1}{T^\text{s}_\text{b}}\right)\tag1$$
where

$P^\text{s}$ – vapor pressure of solution at initial boiling temperature $T^\circ_\text{b}$,
$P^\circ$ – saturation pressure of solvent, by definition of boiling, $P^\circ = P_\text{ext}$,
$\Delta H_{vap}(T^\circ_\text{b})$ – entalphy of vaporisation at initial boiling temperature $T^\circ_\text{b}$,
$T^\text{s}_\text{b}$ – new boiling temperature at some external pressure $P_\text{ext}$,
$T^\circ_\text{b}$ – boiling temperature of solute at the same external pressure $P_\text{ext}$.

From Raoult's law
$$P^\text{s} \approx P^\circ X_\text{solvent} = P^\circ(1 - X_\text{solute}),$$
and thus from $(1)$
$$\ln (1 - X_\text{solute}) \approx -\frac{\Delta H_{vap}(T^\circ_\text{b})}{R}\left(\frac{1}{T^\circ_\text{b}} - \frac{1}{T^\text{s}_\text{b}}\right).$$
From first-order Maclaurin series, and since boiling point elevation is usually small for our equations to be a good approximation,
$$\Delta T_\text{b} \approx \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}X_\text{solute}.$$
Granted molar mass $M$ is in units $\mathrm{g/mol}$, and molality $c_m$ has units of $\mathrm{mol\ /\ 1\ kg\ solvent}$, then
$$\Delta T_\text{b} \approx \underbrace{\frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}}_{K_\text{b}}c_m \equiv K_\text{b}c_m$$
or
$$K_\text{b} \equiv \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}. \tag2$$
It does not matter whether our previous equations are mere approximations, this is now a definition of the ebullioscopic constant. It is evident that it is in no manner dependent on external temperature. It is only dependent on the boiling temperature of a pure solvent at some well-defined external pressure $P_\text{ext}$. To change this temperature, we would have to change the external pressure, or pick a different solvent altogether.
If the external pressure is transformed, $T^\circ_\text{b}$ will not remain the same. There is no reason to believe that $(T^\circ_\text{b})^2$ at this new pressure $P'_\text{ext}$ will cancel with the change in vaporisation entalphy. And thus, the ebullioscopic constant is, in fact, not constant.
Conclusion

If external pressure remains the same, it makes no sense to talk about temperature dependence. This is because the ebullioscopic constant is defined as $$K_\text{b} \equiv \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}$$
which is dependent only on the original boiling temperature of the pure solvent at some well-defined external pressure.
If external pressure is changed, the pure solvent's boiling point $T^\circ_\text{b}$ will change to a new value. There is no cause to believe that the square of the boiling temperature will cancel with the altered entalphy of vaporisation. Hence, ebullioscopic constant will change, and is, in this sense, dependent on temperature. The exact function will vary with the solvent at hand but is most likely approximated by polynomials with empirical coefficients if need be.

I skiped all your answer because I was familiar with it all. Shall I downvote it ?

@Hexacoordinate-C Haha. One cannot comprehend the answers by those who have more rep. That's not how the system works

In short: depending on the solvent in question, the temperature of ebullioscopic determination and the accuracy you aim for, the temperature dependence of the ebullioscopic constant may be negligible.
Alexander Apelblat published with Springer a book Citric Acid (ISBN 978-3-319-11232-9), partially visible in the book view by Google Canada.  In chapter 2, entitled "Properties of Citric Acid and Its Solutions", he provides an example for water's ebullioscopic constant and its temperature dependence. Page 28 starts with this:

The page in question may be accessed for example from here, then jumping to chapter 2 / page 13, and manual browsing to page 28.  (As a side note, the preview does not allow to see page 29.)


text:
What are the decay products of HTO?

Tritium, symbol $\ce{T}$, is hydrogen with a mass of $3\ \mathrm{amu}$. It is radioactive and undergoes $\beta$ decay. Which of the following could be the only products after a quantity of $\ce{HTO}$ undergoes decay?

$\ce{HeOH}$
$\ce{H2O}$, $\ce{O2}$ and $\ce{He}$
$\ce{H2O}$, $\ce{H2}$ and $\ce{He}$

The answer is 2 only, however I am finding it difficult to understand how I could have worked this out. Tritium undergoes beta decay, so: $$\ce{^3_1H+ -> ^3_2He^{2+} + e- + \bar {\nu}_e}$$
I had assumed it wouldn't be possible for the electron emitted in beta decay to become an orbital electron. I have seen the following question: What does HTO decay into?, but it doesn't answer my specific query. I could state that the helium ions produced gain 2 electrons from the oxygen ions to become helium, with these oxygen atoms then reacting to form $\ce{O2}$ and some of the hydrogen combining with oxygen to form water, leading to answer 2, but this feels somewhat arbitrary. What approach could be taken to solve this problem?

You walk to a local market with 50$$$ in your pocket, then you check again and find nothing (0$$$). Were you robbed? You didn't notice anything suspicious. Well, supposedly, you could compare 0 and 50 and do the arithmetic, but this feels somewhat arbitrary... No, it is not arbitrary at all. This is the conservation law, arguably the most reliable thing in chemistry. There is no going around it.

I don’t consider your working to be arbitrary at all. Let’s first consider the β decay including the associated $\ce{OH}$ part of $\ce{HTO}$:
$$\ce{_1^3THO -> _2^3He^2+ + OH- + e-^ + \bar{\nu_e}}\tag{1}$$
As you suggested, the electron created in equation (1) won’t stick around, it is emitted into the environment, leaving us with a very reactive $\ce{He^2+}$ ion. Obviously, that is a strong oxidant as it wants to get reduced itself. The closest electron sources are $\ce{OH-}$ and $\ce{HTO}$, so we can consider a number of other reactions. However, it makes sense to organise it in a simple redox reaction. The reduced component is obviously going to be $\ce{He^2+}$ which only leaves $\ce{HTO}$ being the oxidised component — and since both hydrogen and tritium are in the $\mathrm{+I}$ oxidation state in $\ce{HTO}$, it must be oxygen which is oxidised to either peroxide or elemental oxygen. Indeed, peroxides are not stable and typically disproportionate into oxygen and water anyway, so we might as well just consider the formation of oxygen. Here goes:
$$\textbf{Reduction:}\\
\begin{gather}\ce{He^2+ + 2 e- -> He^{\pm 0}}\tag{2}\end{gather}$$
$$\textbf{Oxidation:}\\
\begin{align}\ce{2HTO^{-II} &-> O2^{\pm0} + 4 e-}\tag{3.1}\\
\ce{2 HTO & -> O2 + 4 e- + 2 H+ + 2 T+}\tag{3.2}\end{align}$$
$$\textbf{Redox:}\\
\begin{gather}\ce{2 He^2+ + 2 HTO -> 2 He + O2 + 2 H+ + 2 T+}\tag{4}\end{gather}$$
Equation (3) and (4) would be similar if I had chosen to oxidise only to $\ce{H2O2}$ and then would have been followed by an equation (5) to show the disproportionation of hydrogen peroxide:
$$\ce{2 H2O2 -> O2 + 2 H2O}\tag{5}$$
In the end, you will arrive more or less at the same substances.
Note that to end up with hydrogen gas, you would have needed a reductive environment capable of reducing $\ce{H^{+I}}$ to $\ce{H2^{\pm0}}$ — not the case if you have a severly electron-desiring ion like $\ce{He^2+}$ around.

The reactions are impressive looking but mostly bogus.

@MaxW Are they now? On what basis? Suggest better ones that lead to the final products? Or suggest entirely different final products altogether?

First off HTO doesn't exist as a pure compound. H and T would exchange so that you's have a mixture of $\ce{H20}$, $\ce{HTO}$, and $\ce{T2O}$. The electron from the decay is energetic enough to cause all sorts of weird ionizations. $\ce{H2O^{2-}}$ could only exist for a fleeting moment until it would grive an electron to anothyer water molecule to form two $\ce{H2O^-}$.

@MaxW Where do you see $\ce{H2O^2-}$? I also decided to completely ignore the hydrogen/tritium exchanges, since on average $\ce{HTO}$ should be the most prevalent compound by statistics. You are probably correct about the electron, though.

The answer really can't be deduced by an understanding of the various chemical reactions, but I think we can all agree that answer (1) is obviously wrong. So the answer must be either (2) or (3).
Since the decay of tritium is consuming a species of hydrogen the solution will be hydrogen deficient, so (3) is out.
Thus (2) is the answer and it is more from logical deduction rather than an understanding of all of the chemical equations.


text:
Are alkenes with bulky groups attached (like the one shown in the picture) chiral or achiral?

Is the compound in the picture chiral or achiral?

I think it should be achiral for the following reasons:

There is free rotation. I know, but for every configuration produced by free rotation there is an exactly opposite configuration.
Also at certain point of time all the groups can come to the same plane.

This question is from my textbook exercise. And the answer says it is chiral. But I've no idea why. Also is it optically active or inactive?
EDIT:
@K_P gave a very useful article related to these type of compounds:

Is there free rotation around the bond between the phenyl ring and the double bond?

I think so @K_P tell me if I am wrong

related http://chemistry.stackexchange.com/questions/37794/when-can-a-molecule-be-considered-freely-rotating-at-room-temperature

Have you sudied atropisomerism? It is a very important topic and your exercise is an example of this

also http://chemistry.stackexchange.com/questions/24901/do-molecules-with-axial-chirality-have-stereogenic-units

@K_P  I read it now. But how to understand whether the group is bulky enough for atropisomerism ?

Your picture is not right, you need to draw the methyl groups (and the bromine atom) on the benzene ring so it will be more clear why there is hindered rotation. If this is useful at all, I was not sure if there is atropisomerism, I assumed there was, because of the answer in your textbook and then did some googling that confirms it: article

Related: http://chemistry.stackexchange.com/a/35515/14857

Is the compound in the picture chiral or achiral?

The answer in practice is: it depends. You say that there would be free rotation along the phenyl–vinyl bond. However, every adjacent carbon atom — the 2- and 6-positions of the phenyl ring and the 1-position of the double bond — are substituted. For free rotation to occur, these substituents must pass each other. Since they are all bulky, though, free rotation is very effectively hindered. A planar rotamer with full π conjugation between the phenyl ring and the double bond is not populated at any extend which can be shown by the NMR shifts of the corresponding carbon atoms (and could also be shown if the methyl group on the double bond were a proton, because it’s shift would be vastly different from one in styryl conjugation).
Under low enough temperatures, the compound you have been given will indeed form to separable enantiomers due to the bromine atom which destroys the molecule’s symmetry. These are examples of axial chirality; if the chlorine atom is above the phenyl ring’s plane, it is (aR), if it is below, the compound is (aS). The two enantiomers can be separated, e.g. by chiral HPLC, and they will show optical activity.
At higher temperatures, thermal energy will be enough to overcome the rotation barrier and an optically active single enantiomer will racemise. This temperature is typically above room temperature for a system such as your’s but well below the temperature required for alkene (E/Z) isomerisation.
This concept is called atropisomerism and is also the reason why BINOL and BINAP are chiral.

Do you know of any web resource which has listed down all known atropisomers ?

@S007 All known is wa~y too broad; I don’t think that even all bisphenyl-based atropisomers are listed anywhere.  Unless of course you’re talking about different conceptual building blocks but then again I still don’t know. (But at least the list would be manageable.)


text:
Do electrons in an atom revolve around the the nucleus clockwise or counterclockwise?
Do electrons in an atom revolve around the the nucleus clockwise or counterclockwise?
Is there any rule to determine?

No. Electrons don't revolve around anything at all. You have a choice of two inappropriately-termed words to describe the electronic state you're interested in: spin up and spin down. Neither term refers to spin in any sense of the word, not to mention up or down, which are even less appropriate. In short: electrons have a property called spin which can have values of up or down (more confusingly notated as 1/2 or -1/2).

What would be the spin dependence of time?

Come to think of it, even ordinary things in 3D do not rotate clockwise or counterclockwise. (Also, electrons are not ordinary things, and they do not rotate at all.)

Even in an orbital the electrons in an atom so not rotate, they just have a probability of being at some position. The spin of an electrons as mentioned above does not literally mean that they 'spin' but rather that they have a property that obeys equations describing angular momentum. Yes it is all very odd compared to everyday experience, but its what experiments teach us.

http://chemistry.stackexchange.com/questions/36987/does-the-shared-electron-in-covalent-bonds-revolve-around-nucleus

Lets make this clear:

Electron is elementary particle.
Like other particles it can be also described as wave (wave–particle duality).
It may be a point particle, but it possibly has finite, but very small radius.
Electrons do move around nuclei. Corpuscular nature of electron completely excludes the possibility that electrons in atoms don't move. They also may or may not have true orbits depending on which interpretation of quantum mechanics is really "true". So, they move and are very fast; their velocities can be even relativistic (see this answer) and in such cases even Schrödinger equation doesn't describe atoms properly.

As Ivan pointed out in comments, orbits of electrons in atom, which wasn't disturbed in any way, can't be seen, which is true for any system (observer effect). Still even single electron during ionisation of an atom was observed using attosecond laser; a film of movement of electron cloud was created (see also).
As for OP's question "clockwise" or otherwise is relative to observer and therefore there's no point in asking that. Circular motion in original Bohr model was only an approximation and electrons move in unpredictable, chaotic way. Even on short timescales single atoms can be considered spherically symmetrical - they don't have left side or up side.

The question does pose itself (and I am in no way qualified to understand that paper … I gave up quickly) whether electrons can be seen to circulate the nucleus or whether they just move randomly in its vicinity.

@Jan Bohr modelled orbits with circels Sommerfield improved with ellipses, but the movement is more or less chaotic https://en.wikipedia.org/wiki/Old_quantum_theory#Limitations_of_the_old_quantum_theory

@Mithoron You are confusing people for no reason. These papers do not say that an electron residing on an orbital moves in any literal sense. The second paper describes an excited electron which oscillates between certain states (each state being a probability cloud), and the first one deals with an ionized atom and separated electron. By skipping these technicalities, you fuel the false impression which many students already have, that electrons on (say) 1s orbital sort of fly around the nucleus, like planets around the Sun. This is quite wrong. Sorry, but you aren't making this clear.

@IvanNeretin Well, for starters electrons do not reside on orbitals. Orbitals are only models, by no means a perfect ones. I felt the need to make this post because it's you who is wrong and I'll try to convince you.

Yeah, sure, if we look deeper, the orbitals are but abstractions and all that. Now what's about the electrons which are "moving"? In one case, an electron is oscillating between two probability clouds, which probably can count as "moving", but mentioning that without further explanation still does create a wrong impression. In the other case, an electron gets wrestled free from an atom and is moving in a somewhat classic fashion; that's OK. Now to the point: what about an electron in a ground-state atom, is it "moving"?

@IvanNeretin See Dave's answer - http://chemistry.stackexchange.com/questions/13581/i-am-trying-to-picture-how-electrons-move-around-in-atomic-orbitals

The problem with the interpretations of quantum mechanics, and also the reason why there are so many of them, is that they are all equivalent when it comes to the observables. Bohmian trajectories, even if they exist, aren't and can't be seen. Your answer implies that the true trajectories of electrons in undisturbed atoms were actually observed. This is quite an overstatement.


text:
Why doesn't a C4H8 alkane exist?
This structure:
     H   H
     |   |
 H - C - C - H
     |   |
 H - C - C - H
     |   |
     H   H

Which would be $\ce{C4H8}$, like butene, but an alkane, because it's functional group is C - C.

Um....cyclobutane is a commercially available liquefied gas. Am I missing something here?

With the missing bond added, your structure is fine and part of a family known as cycloalkanes.

https://en.wikipedia.org/wiki/Cyclobutane

It does exist. The formula $\ce{C_{n}H_{2n+2}}$ is only applicable to straight chain alkanes. They can't have any degrees of unsaturation, which is defined as a multiple bond or a ring.


text:
Why aren't methanal and ethanal also ketones?
Why aren't the aldehydes methanal/formaldehyde and ethanal/acetaldehyde also ketones? Ketones are defined by replacing any H - C - H in an alkane with C=O, which is true for both of these: methanal replaces two hydrogen atoms in methane with C=O, while ethanal does it with ethane.

They are not ketones because the carbonyl group is at a terminal position. In a ketone, both neighbouring atoms of the carbon  in $\ce{C=O}$ need to be carbon atoms. In the aldehydes, one of the neighbours is a hydrogen atom.

That definition of ketones is wrong!

Aldehydes deserve a different name because they have different properties, for example being easily oxidized to acids, as in the Benedict test which distinguishes "reducing" sugars like glucose (an aldehyde) from non-reducing ones like fructose (a ketone).  The relation between aldehydes and ketones is like the relation between alcohols and ethers, both of which can be described as R-O-R, but the properties are significantly different when one of the R's is a hydrogen.

According to the Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995), ketones are defined as follows:

ketones:
  Compounds in which a carbonyl group is bonded to two carbon atoms: $\ce{R2C=O}$ (neither $\ce{R}$ may be $\ce{H}$). NOC Rule C-311.1. (…)

The mentioned nomenclature of organic chemistry (NOC) Rule C-311.1 refers to the obsolete 1979 IUPAC recommendations:

311.1 – The generic name “ketone” is given to compounds containing an oxygen atom doubly bound to a single carbon atom with the carbonyl group $\ce{>C=O}$ joined to two carbon atoms. (…)

The corresponding rule in the also obsolete 1993 IUPAC recommendations reads:

R-5.6.2.1 Ketones. The generic term “ketone” refers to compounds containing a carbonyl group, $\ce{>C=O}$, joined to two carbon atoms. (…)

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) refers to the definition of the 1995 glossary of class names:

P-64.1.1 Ketones are defined classically as compounds in which a carbonyl group is bonded to two carbon atoms: $\ce{R2CO}$ (neither $\ce{R}$ may be $\ce{H}$). (…)

All definitions include the criterion that the carbonyl group is bonded to two carbon atoms. Since in methanal (formaldehyde) as well as in ethanal (acetaldehyde) the carbonyl group is not bonded to two carbon atoms, these compounds are not ketones.

Functional groups are identified based on their characteristic reactions. Aldehydes and ketones are structurally similar, and reactivity is certainly closely related for both groups. However, there are reactions where the outcome is very different for aldehydes than for ketones. Here is one of them:

Oxidation

Aldehydes react with a variety of oxidizing agents to form carboxylic acids (formaldehyde is often over-oxidized to carbon dioxide).
$$\ce{RC(O)H ->[\ce{CrO3}][\ce{H2SO4}] RCO2H}$$
Ketones, on the other hand, are hard to oxidize. Acetone (dimethyl ketone, 2-propanone) is sometimes used as a cosolvent in the chromic acid oxidation (above).
$$
\ce{RC(O)R ->[\ce{CrO3}][\ce{H2SO4}]} \text{no reaction}
$$
That is not to say that ketones cannot be oxidized, but different reagents are needed:
$$\ce{RC(O)R ->[\ce{RCO3H}] RCO2R}$$


text:
Are the $+\leftrightarrow+$ and $-\leftrightarrow-$ methanol rotational transitions dipole-allowed?
I am quite confused about the origin and justification of the rotational selection rules of methanol, and I would appreciate any help or references. As far as I understand, there are three irreducible representations of the symmetry group of methanol ($A_1$, $A_2$,and $E$) and the two $A$-type are linked by dipole-allowed radiative transitions. Now, I think the literature also calls these ($A_1$ and $A_2$) $A^+$ and $A^-$ representations, I suppose because the parity of the eigenfunctions which generate the $A_1$ and $A_2$ are $+1$ and $-1$, respectively. Is this correct?
From what I have studied, because the electric dipole generates the $A_2$ representation, usually the allowed transitions follow the selection rule $A_1\leftrightarrow A_2$ (because $A_2\times A_2=A_1$), and I have seen this in many references. However, a lot of strong $A$-type methanol rotational lines are $++$ or $--$. I have noticed that all of these are associated with transitions with $|\Delta J|=1$ and all of the $+-$ and $-+$ transitions are associated with $\Delta J=0$. In fact, I have seen this statement as a selection rule in some papers (in this reference, they say that the $+-$ refers to K-type doubling ... ?), but it is not at all clear to me where does these rules come from.
I assume that the symmetry group of methanol is $C_{3v}$ (in books sometimes they call it $G_6$), and the representations above are usually the three which appear in the character tables. I think the group $C_{3v}$ represent permutations and permutation-inversions involving the protons of only the methyl group, because the other proton is too far and cannot be easily swapped with the others. The $J$ above represent the total rotational angular momentum of the molecule. Apparently, the usual approach is to express the wavefunctions as linear combinations of the product of symmetric top rotor wavefunctions and the wavefunctions  associated with the (hindered or not) rotation of the methyl group around the CO axis. I am interested in transitions at the torsional ground state ($\nu_t=0$) so I guess we can assume that the rotational potential is large. The $K$ I mention briefly above is the projection of the angular momentum in the main principal axis of the molecule, which is roughly aligned with the CO bond. Since methanol is asymmetric, there are two energy levels with the same $J$ and $K$ in the rotational spectrum, which is what I think they call the $K$ doubling.
I am in no way an expert on this, so if you want to point out inaccuracies in what I have just wrote I would be thankful.

crosspost of http://physics.stackexchange.com/questions/291290/are-the-leftrightarrow-and-leftrightarrow-methanol-rotational-transiti

It would help is you state  what you consider the point group to be, C1, Cs or something else. Are you worried about the restricted rotation of the OH group or do you ignore this, in which case the molecule is C3v. Have a look at http://chemistry.stackexchange.com/questions/58609/understanding-group-theory-easily-and-quickly/58613#58613  which may help in identifying transitions.

Thank you for your comment. I include a paragraph with more details.


text:
What constituents in soft drinks are the cause their strong pungent smell?
Wikipedia and PubChem tells us that $\ce{CO2}$ is odorless in lower concentrations but can have a mildly pungent acidic smell at higher concentrations.
However, there is no pungent smell from chemical reactions such as the combustion of cooking gas, effervescence from carbonate upon the addition of an acid, and the sublimation of a piece of dry ice.
But many soft drinks and soda waters (including Sprite, 7 up, Pepsi, Coca-Cola, and Thums Up) emit, open opening, a strong burning smell that may cause severe irritation to eye and nose. The same burning is experienced during the burping (or belching) that comes out after consuming a glass of such drink
So, I want to know, is this due to $\ce{CO2}$? Maybe the manufacturers pump too much $\ce{CO2}$ into their drinks that it reacts in a different way than $\ce{CO2}$ from LPG-oven flames or carbonate/acid reactions? Or maybe that 'burning effect' is due to some mysterious additive?

Yes, it is $\ce{CO2}$ and not some additive. It is just that we can't feel it in low concentrations, exactly as Wikipedia says.

http://chat.stackexchange.com/transcript/message/33376929#33376929 there is phosphoric acid. http://chat.stackexchange.com/transcript/message/33377067#33377067  links here

Yes there is; so what? It is not like you pour Coca-cola down into your nose to smell the solution. Phosphoric acid per se is not very volatile. Sure, some of it might get to your nose in the form of tiny droplets, but those are responsible for a minor component of the smell at best. I can attest to that. I did smell pure $\ce{CO2}$ out of curiosity.

and that pure CO2 was just same as cold-drinks? or more? or less (approx)?

Yes, pretty much so.

@IvanNeretin you could write an answer.

CO2 doesn't irritate. I believe burning is caused by acidic droplets. You can prove/disprove this by arranging experiment with some membrane to retain the droplets.

@sa7 Have you ever tried breathing nearly pure CO2?

Nope, I think I would loose consciousness very soon.

It is definitely $\ce{CO2}$, nothing else.

Yes, that strong smell is mostly $\ce{CO2}$, possibly with a faint tint of volatile aromatizers from the beverage.
Our sense of smell works in surprisingly non-straightforward ways. Indeed, the smells of the same compound in high and low concentrations are sometimes perceived as quite different, to the point of being totally unrecognizable. A classic example is skatole, which may produce either a pleasant flowery aroma or a fecal stench, depending on the concentration alone. Looks like $\ce{CO2}$ is just a somewhat less extreme example of the same effect.
Air with 5% of $\ce{CO2}$ doesn't smell much of anything (not that it is OK to breathe it for long), which leads most people to conclude that $\ce{CO2}$ is odorless. This is quite wrong. If you try to inhale some pure $\ce{CO2}$ (which I do not recommend or endorse), that's when the smell kicks in. Your breathing just stops instinctively, as if you were trying to inhale water. By that moment, though, some $\ce{CO2}$ reaches your olfactory nerves. I am at a loss of words as to what it feels like. Being slapped across your face with a shovel seems like a moderately adequate approximation. Besides, you feel that familiar refreshing sharp odor of an open Coca-cola bottle.
So no, this is not due to some mysterious additive.

+1'd for "Our sense of smell works in surprisingly non-straightforward ways".  It is surprising fact that scatole, which is one of the bad-smelling component of stool; produce a pleasant aroma when the concentration is low!. Same has been told about indole, too!. However It is tough to accept because fecal-odor smells bad even-when it comes from distance. And after purging; when the bathroom-is cleaned, the remaining trace of smell is never felt pleasant.


text:
Calculating concentration of dissolved gas from its mole fraction
I came across the following (solved) problem in my textbook:

The Henry's law constant for $\ce{O2}$ dissolved in water is $\pu{4.34 \times 10^4 atm}$ at $\pu{25 ^\circ C}$. If the partial pressure of oxygen in air is $\pu{0.4 atm}$, calculate the concentration (in terms of molarity, $\pu M$) of dissolved oxygen in water, in equilibrium with air at $\pu{25 ^\circ C}$.

The following is my attempt at a solution to the above problem:
From Henry's law, we have that
$$p{\ce{O2}} = K_\mathrm H \cdot x_\ce{O2}$$
where $x_\ce{O2}$ is the mole fraction of oxygen dissolved in the solution, $K_\mathrm H$ is the Henry's law constant, and $p_{\ce{O2}}$ is the partial pressure of oxygen over the liquid.
Given that the Henry's law constant for a mixture of water and some amount of dissolved oxygen is $\pu{4.34 \times 10^4 atm}$, this gives us
$$x_\ce{O2} = p_{\ce{O2}}/K_\mathrm H = 0.4/(4.34 \times 10^4) = 9.2 \times 10^{-6}$$
To calculate the concentration of oxygen in this mixture in terms of its molarity, I need to know the number of moles of oxygen present in the solution and also the number liters of water in which the oxygen is dissolved.
To calculate the number of moles of dissolved oxygen, I can use the following equation:
$$x_\ce{O2} = \frac{n_\ce{O2}}{n_\ce{O2} + n_\ce{H2O}}$$
However, no information regarding the number of moles of the solvent, water, has been provided.
Hence, I checked the solutions provided by my textbook and found that they have arbitrarily assumed that we are dealing with $\pu{1 L}$ of water.
My question is this: are we allowed to arbitrarily take the volume of water as $\pu{1 L}$ even when absolutely nothing has been mentioned about this in the problem?

The gist is that as long as you have a specified mole fraction, the concentration of oxygen does not depend on the amount of water present. This is what is called an "intensive property". Intensive properties don't depend on how much stuff there is, and therefore, you can arbitrarily specify any volume you like. In order to make the calculations simpler, this specified volume is usually some unit quantity, like $\pu{1 L}$.
It is somewhat analogous to determination of an empirical formula, in that it doesn't matter how many grams of the compound is present, you will always end up with the same empirical formula. A common tactic is to assume $\pu{100 g}$ of compound is present, such that if there is $x\%$ of carbon by mass, there is $x\ \pu{g}$ and therefore $x/\pu{12 mol}$ of carbon.

Here, if you have $\pu{2 L}$ of water, then $n_{\ce{H2O}}$ is double of that if you only have $\pu{1 L}$. But since you have twice as much water, to maintain the mole fraction (that you determined via Henry's law), you have to dissolve twice as much oxygen - otherwise the mole fraction won't be the same. So $n_{\ce{O2}}$ is double of that if you only have $\pu{1 L}$. The factor of 2 will cancel out in the equation you gave:
$$\begin{align}
x_{\ce{O2}} &= \frac{2n_{\ce{O2}}}{2n_{\ce{O2}} + 2n_{\ce{H2O}}} \\
&= \frac{n_{\ce{O2}}}{n_{\ce{O2}} + n_{\ce{H2O}}}
\end{align}$$
If you are really uncomfortable with assuming the volume, then simply assign an unknown value: $k$ litres, for example. You should find that $k$ cancels out and you are left with the exact same value for the molar concentration of $\ce{O2}$.

There is, perhaps, a more "elegant" way to prove this. Note that
$$\begin{align}
\frac{1}{x_{\ce{O2}}} &= \frac{n_{\ce{O2}} + n_{\ce{H2O}}}{n_{\ce{O2}}} \\
&= 1 + \frac{n_{\ce{H2O}}}{n_{\ce{O2}}} \\
&= 1 + \frac{V_{\ce{H2O}} \times [(\pu{1000 g L-1})/(\pu{18 g mol-1})]}{n_{\ce{O2}}} \\
&= 1 + \frac{\pu{55.56 mol L-1}}{c_{\ce{O2}}}
\end{align}$$
where $V$ is the volume and $c = n/V$ the molar concentration. You can see that the molar concentration is uniquely determined by the mass fraction $x$ and the exact value of $V$ doesn't matter at all.


text:
Relationship between surface area and vapor pressure
How is vapor pressure independent of the surface area and volume of a liquid, although rate of evaporation depends on surface area?
Please someone help me I am totally confused.

The rate of evaporation is proportional to surface area. The rate of condensation is proportional to surface area. The vapor pressure is the equilibrium pressure where the rate of evaporation is equal to the rate of condensation. Since the scaling factor is the same, the vapor pressure is independent of the surface area.

Sorry but why is rate of condensation proportional to surface area?

@samjoe Because it's the microscopic reverse of evaporation.


text:
How does a Lewis acid differ from an oxidizing agent; Lewis bases from reducing agents?
A Lewis acid is defined as an electron acceptor. An oxidizing agent is also defined as an electron acceptor (in the electronic theory of oxidation).
So then what is the basic difference between an Lewis acid and an oxidizing agent?
Similarly;
A Lewis base is defined as an electron donor (Lewis theory). A reducing agent is also defined as an electron donor (electronic theory). Therefore, what is the difference between Lewis bases and reducing agents? Are they exactly same thing?
There is a similar question which, however, only considers bases and nucleophiles and does not go into detail about oxidation and reduction.

For Lewis acid and base theory, the electrons must be in a pair. Also, we don't think of them as being transferred. The donation or acceptance of the pair creates a bond. Whereas for oxidation/reduction, you transfer some number of electrons, not necessarily 2.

@Zhe make it an answer.

Could I write "all lewis acids are oxidising agents but all oxidising agents are not lewis acid"? similarly, "all lewis bases are reducing agents, but not all reducing agents are lewis bases"? If my understanding is too poor, please forgive me but I prefer to clear confusion.

An acid and an oxidising agent are very different concepts that apply to different phenomena — although a single compound can be both at the same time, e.g. $\ce{H2SO4}$. However, a compound can also be an acid and a reducing agent, e.g. $\ce{HI}$.
The Brønsted acid/base theory is clearly distinct from any type of redox reaction, so I suppose that the confusion lies with the extension of said theory that the Lewis theory provides. But first a short word on the Brønsted theory: therein, an acid is a compound that can transfer a $\ce{H+}$ ion onto another compound.
The Lewis theory realises, that the acidic transfer reagent need not be hydrogen. Instead, anything with an unoccupied low-energy orbital can function as a Lewis acid; most notably boron compounds and metal cations. Upon creation of a Lewis acid-base pair, a coordinate bond is formed. Basically, this means that the energy of the lone pair of the Lewis base is lowered because it interacts with the empty orbital of the Lewis acid whose energy is raised. This can be depicted as $\ce{LB\bond{->}LA}$. The key concept is that the interaction is fully reversible and that upon dissociation the Lewis base keeps its lone pair.
When discussing oxidising or reducing agents, however, the electrons are irreversibly transferred. They actually move from orbitals of the reducing agent (which will be oxidised) into orbitals of the oxidising agent (which will be reduced). This transfer is thermodynamically favourable and thus the electrons won’t just ‘go back’.
To exemplify this, let’s consider two reactions of $\ce{H2SO4}$; one where it is an oxidising agent and one where it is an acid (Brønsted type; the proton of $\ce{H2SO4}$ is actually the Lewis acid).
$$\text{Acid-base:}\\
\ce{H2SO4 (l) + NH3 (g) <=>> NH4+ HSO4- (s)}$$
Note that ammonia, our base, is a gaseous compound at room temperature. If you employ appropriate conditions (i.e. temperature and low pressure), the equilibrium will be drawn to the left, releasing ammonia as ammonia ($\ce{NH3}$ molecules) including its lone pair. $\ce{H2SO4}$ will stay behind.
$$\text{Oxidising agent:}\\
\ce{2H2SO4 + Cu -> CuSO4 + SO2 ^ + 2 H2O}$$
$\ce{SO2}$, being a gas, will diffuse away. However, even if you perform this in a pressurised vessel, there will be no significant back-reaction because the electrons have been fully transferred from copper to sulfur.

The argument for Lewis bases/reducing agents is the same in reverse.

Thanks, but I am still confused. Could you add a brief statement about "if this, then Lewis base, and if this, then reducing agent" (similarly for acid and oxidising agents)


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Does the carboxylate group show a positive or a negative inductive effect?
In the benzoate ion, does the carboxylate group show a $-I$ or a $+I$ effect?
I learnt that the carboxylate group is a $-I$ group, but according to a website the carboxylate group in benzoate shows a $+I$ effect.

So, which one is correct? Does the carboxylate group show a positive or a negative inductive effect?
Please neglect mesomeric effect when answering.

To the close voters, I think this question shows enough thought and reasoning not to close it. To the OP, please make sure to [edit] and cite your source.

the carboxylic acid group always shows -I effect due to the presence of electronegative oxygen atoms, however the carboxylate group(COO-) shows +I effect due to excess negative charge.

The carboxy group is electron-withdrawing both by inductive and by mesomeric mechanisms. I have drawn two depictions in the scheme below. The upper one is how the mesomeric effect of the carboxy group can be explained. This effect is only relevant if the $\ce{COOH}$ fragment is in the same plane as the ring.

In the case of 2,6-di(tert-butyl)benzoic acid, the bulky tert-butyl groups prevent the mesomerism depicted above. The carboxy group is perpendicular to the ring. Yet, this group still exercises an electron-withdrawing inductive effect. It can be explained by a mesomeric structure of the $\ce{C=O}$ double bond with charge separation:
$$\ce{C=O <-> C^+-O^-}$$
This positive charge is, naturally, electron-withdrawing and can be used to explain the $-I$ effect of carboxy groups. No matter which mesomeric depiction you choose, you can never get a negative charge on the carboxy carbon, thus it will always remain partially positive. That also makes sense considering it is bonded to two electron-withdrawing, strongly electronegative oxygen atoms.

I believe he's asking about the carboxylate group not the carboxy group


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Is the inductive effect always measured relative to hydrogen?
Wikipedia has defined the Inductive effect thus:

"In Chemistry and Physics, the inductive effect is an experimentally observable effect of the transmission of charge through a chain of atoms in a molecule, resulting in a permanent dipole in a bond."

Recently, I learned from a teacher that the inductive effect of any atom/group is always measured with respect to hydrogen.
Under this definition, we would ideally attach the atom/group with an atom of hydrogen and observe the direction of the dipole moment caused by the difference in electronegativity between hydrogen and the atom/group in question. If the electrons-initially positioned at the middle of the bond-are pushed toward hydrogen, the atom/group is termed as a "$+I$", electron-donating group. On the other hand, if the electrons are pulled toward the atom/group, it is termed as a "$-I$", electron-withdrawing group. Thereafter, this atom/group will always exhibit this type of inductive effect. Even if a situation does arise, in which it appears as though an atom/group that exhibits the $-I$ effect with hydrogen is exhibiting the $+I$ effect with another atom/group, this must be the cause of other dominating effects such as the mesomeric effect and in those cases also, it is said that the atom exhibits the $-I$ and not the $+I$ effect.
To illustrate the clear distinction between the definition put forward by the teacher and the one on Wikipedia, let us consider a molecule of $\ce{Cl2}$.
Under the definition on Wikipedia, it is clear that $\ce{Cl2}$ is a non-polar molecule (i.e. No permanent dipole moment has been experimentally observed in this molecule). Hence, no inductive effect is in play and neither of the two $\ce{Cl}$ atoms show either the $+I$ or $-I$ effect.
On the other hand, if we work with the definition put forward by the teacher, we observe that an atom of $\ce{Cl}$ exhibits the $-I$ effect when attached to an atom of hydrogen, being more electronegative than it. Therefore, in the $\ce{Cl2}$ molecule, no permanent dipole exists as a result of the $-I$ effect of both chlorine atoms cancelling away from opposite directions.
It appears that only one of these two definitions is definitely correct and I am confused as to which one that is.

Why should one Cl atom exert inductive effect on the other? They are both equally electronegative, so they will pull the shared electron pair with equal 'force' and so there would be no inductive effect.

I think this is in principle a problem in terminology. You will always have an (a net) inductive effect when you have different electronegativities. The terms +I/-I should only be used in organic chemistry, and there you would indeed compare them to the standard CH bond.

Your definition is falling apart because you are over-applying it. In the definition, you can see that it says:

...transmission of charge through a chain of atoms in a molecule...

It doesn't make sense to apply this in the case of a two-atom molecule, because the inductive effect is used to measure how one bond effects the electron density of other bonds in the molecule. You can assign any bond a dipole moment, but inductive effects are used specifically to explain phenomena observed in organic chemistry, where chains of molecules are (exceedingly) common.
To answer your other question: yes, inductive effects are always measured relative to hydrogen. Linear free-energy relationships are a quantifiable way of measuring a reaction's sensitivity with respect to a certain parameter. The Hammett equation is used to study how sensitive a molecule is in a given reaction to a change in its subsituent's inductive and resonance effects, for which the reference subsituent in all cases is a hydrogen atom. This is perhaps because the $\ce{C-H}$ is the most common bond in all of organic chemistry.
Linear free-energy relationships are particularly useful because similar as to what you said, substituent groups will always have the same relative inductive effects no matter the molecule. A quaternary ammonium group will always have a larger $-I$ effect (and larger $\sigma$ value) than a nitro group than a trifluoromethyl group. A molecule's response to these changes can give great mechanistic insight into organic reactions and the pathways through which they proceed.

A table of the different types of Hammett relations$^{[1]}$:

Their reference reactions:
$\hspace{3.25cm}$

$[1]$ Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science: Sausalito, CA, 2006.

I'm afraid I don't quite understand the last bit completely: "Linear free energy relationships are a quantifiable way of measuring a reaction's sensitivity with respect to a certain parameter." Is it possible to simplify this so that a highschool student can also understand it?

The Hammett equation, for instance, measures how sensitive the ionization of various substances are to changing the substituent para to the carboxyl group. There are other relationships that measure things such as how leaving groups or nucleophiles are affected by changing substituents, how the choice of solvent effects the reaction... the list goes on. If you are interested you can read more in the link I added.

I myself am having a little bit of trouble wrapping my head around the Wikipedia definition which they cite from an organic chemistry textbook. I am specifically troubled by the phrase ‘resulting in a permanent dipole moment.’ According to all definitions of the inductive effect I know, a methyl group is electron-donating ($+I$). Yet, if you consider para-xylene, the molecule suddenly has a centre of inversion (point group $C_\mathrm{2h}$). Therefore, there is no permanent dipole in para-xylene. (Depending on the methyl rotamers, it might also have $C_\mathrm{2v}$, in which case the dipole would be perpendicular to the benzene ring plane — and since there is almost no vertical structure, ignoring the three aliphatic $\ce{C-H}$ bonds, it is predicted to be very small.)
There is no reason to assume that methyl groups suddenly lose their $+I$ effect just because two of them are bonded to two different sides of a benzene ring. The NMR supports the assignment of a double $+I$-substitution with the aromatic protons of para-xylene displaying chemical shifts of $7.05~\mathrm{ppm}$ (compared to $7.36~\mathrm{ppm}$ in benzene; same solvent $\ce{CDCl3}$). Obviously there is additional shielding present in spite of no observable permanent dipole, which can only be attributed to the methyl groups’ $+I$ effect. The same case can be made for other symmetrically para-substituted benzenes such as para-nitrobenzene ($8.41~\mathrm{ppm}$ in $\ce{C6D12 + C3D6O}$), para-dichlorobenzene ($7.26~\mathrm{ppm}$ in $\ce{CDCl3}$) and others. In these cases, especially dichlorobenzene, the point group is actually predicted to strictly be $D_\mathrm{2h}$ — strictly containing inversion.
Thus, when having to choose between your teacher’s and Wikipedia’s definions, I want to go with your teacher’s because it better explains what we discuss in basically every seminar.


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Reaction of carboxylic acids with primary amines

$$\ce{RCOOH + RNH2 -> RCONHR + H2O}$$

$\ce{R}$ stands for any alkyl group.
I have two queries regarding this reaction:

I understand that the reaction is an $\mathrm{S_N{AE}}$ (nucleophilic addition elimination type of reaction). But during the reaction $\ce{R-(CO)-N^+RH2}$ intermediate is formed. After this step the $\ce{OH^-}$ which was eliminated acts as base takes away $\ce{H^+}$ from  $\ce{R-(CO)-N^+RH2}$, converting it into $\ce{R-(CO)-NRH}$. But why did $\ce{OH^-}$ extract $\ce{H^+}$ from $\ce{R-(CO)-N^+RH2}$  and not $\ce{R^+}$ ? Isn't $\ce{R^+}$ always a more stable carbocation compared to $\ce{H^+}$ ?
Why does this reaction require a basic medium?

After this step the $\ce{OH-}$ which was eliminated …

No. $\ce{OH-}$ is never eliminated but formally. To eliminate the oxygen required, it needs to first accept a proton from somewhere leading to $\ce{R-C(O^-)(OH2+)-NHR}$. The hydroxide ion is simply much too bad a leaving group for any $\mathrm{S_N}$ reaction; it must always be transformed into a better one.
This leads to a severe problem in the reaction. Basic conditions hinder the reaction because they decrease the degree of protonation on that oxygen. However, under acidic conditions the amine will be protonated in solution ($\ce{RNH3+}$) and a protonated amine no longer has any lone pairs that can attack nucleophilicly. You’ll want neutral to mildly basic conditions for the reaction to work out.
Another thing to consider: the ‘elimination of a proton’ is a simple acid-base reaction. Proton displacements are typically much more rapid than any carbocation eliminations because protons, being as small as they are, can tunnel. Additionally, you should never compare the stability of a naked proton against that of a naked carbocation in these eliminations. A naked proton will not be generated, it will attach to a base immediately — and there are enough bases around: $\ce{OH-, H2O, RNH2, \dots}$ Furthermore, carbon-heteroatom bonds are pretty stable if the heteroatom side does not equal a good leaving group (e.g. $\ce{NHTs}$). Thus, these are hardly ever cleaved.
In practice, you would not use a particularly acidic or basic medium in this reaction. Rather, you would use acid activation techniques such as the Steglich conditions ($\ce{DCC, DMAP}$).


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How does a diatomic molecule vibrate in response to heat?
Do you know of any applet/animation that shows how a diatomic molecule like $\ce{O2}$ or $\ce{N2}$ vibrate in response to/ to produce heat?

Can it vibrate only on one axis, the one joining the two atoms? What happens to the electrons during the vibration?

That's the only degree of freedom available. Otherwise, you're probably coupling rotation with the vibration.

It can vibrate only by extending and contracting the bond. The more energy that is added the bigger the bond excursion. (More vibrational levels are excited).  The bond only extends by a few % of its length in lower vib levels so its pretty hard to see. If heaps & heaps of energy are added (temp say $1000$ K)  then the bond can break as it is  actually anharmonic in its motion. When this happens the atoms fly apart.  Additionally the molecule can also rotate while it is vibrating. Rotation is slower than vibration so many vibrational periods occur within a $360$ degree rotation.

This post actually contains three questions, each of which would probably have benefited from being asked in individual posts.

The diatomic can vibrate only by extending and contracting the bond. The more energy that is added the bigger the bond excursion. (More vibrational levels are excited and have bigger excursions). The bond only extends by a few % of its length in lower vibrational levels so its pretty hard to see. If heaps & heaps of energy are added (temp say $1000$ K) then in a real molecule the bond can break as it is actually anharmonic in its motion. When this happens the atoms fly apart.
Additionally the molecule can also rotate while it is vibrating. Rotation is a slower motion than vibration so many vibrational periods occur within a $360$ degree rotation.
At any given constant temperature in a sample of molecules the energy moves between vibrations, rotations and translations so that each type of energy is in equilibrium.
Some of the energy can  also be transferred as  radiation, in the infra-red (for vibrations hundreds to a few thousand $\pu{cm^{-1}}$) and microwave region ( in the range few $\pu{cm^{-1}}$) for rotational transitions but only if the diatomic molecule has a permanent dipole, i.e. is not homonuclear. If these molecules are in equilibrium with surrounding then these surroundings must also emit/absorb radiation to keep the total energy constant. If you were to suddenly isolate your block of (heteronuclear diatomic) molecules, say in space, then the radiation emitted would over a few hours remove energy so that the molecules will end up in their zero point levels.
The figure shows the potential energy profile of HI. The the average length for the for the lowest level (quantum number $0$) is approx $106.4 $ pm and the chnage due to vibration is about $\pm 10 $ pm and clearly greater for higher vibrational levels. The vibrational period is $\approx 14 $ fs.

In a hetero-nuclear diatomic molecule microwave radiation can start a molecule rotating by causing it to move from one rotational state to another, it can also loose energy by emitting photons. But in homo-nuclear diatomic molecules there is no dipole and so radiation cannot influence it (other than to electronically excite it if the radiation has enough energy, i.e. visible/uv photons). Rotation  and translational energy levels can be populated or depopulated in all types of molecules by energy transfer during a collision.


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What is the major hydrolysis product of 2,2,2-trifluoroethyl 2-acetamido-3-phenylprop-2-enoate?
In the below reaction we have to find the major hydrolysis product of 2,2,2-trifluoroethyl 2-acetamido-3-phenylprop-2-enoate.

My doubt is why the major hydrolysis is not when $\ce{OH-}$ at the $\ce{C}$ of $\ce{CF3}$ as it is more electropostive. However the major product is given by the following reaction, why?

That's a not a hydrolysis product. Are you just asking for the "hydrolysis product" because of the reaction conditions?

@Zhe which one is not hydrolysis product . The one which I thought or the other one

The other one. How is that hydrolysis?

Also, C-F bonds are notoriously hard to hydrolyze. You should have supposed that hydrolysis would simply break the ester, that would be at least logical.

@IvanNeretin why it is hard to hydrolyse

Plain hydrolysis would just replace trifluorexthoxide with hydroxide.
But you might want to consider that your staring material in equilibrium with another tautomer:

Deprotonation of the $\ce{OH}$ in the right structure, followed by an intramolecular reaction (transesterification/lactonization) in which $\ce{CF3CH2O-}$ is kicked out should give rise to an oxazolone.

How can we know which reaction can produce major product

Does the second reaction is also hydrolysis

@J.Doe Hydrolysis is an extremely broad term. I'd rather call this a cyclization, which is still broad but describes the outcome of the reaction much better. On a second look, this looks a bit an intramolecular variant of the Erlenmeyer-Plöchl reaction.

That reaction is not hydrolysis, it's a transesterification.

@Variax Yes, just as I wrote in my answer ;)


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Oxidation of guanine with singlet oxygen
I'm studying the oxidation of guanine using singlet oxygen and one step of the reaction is not very clear to me.

My problem is with the first step: it is a 4+2 Diels–Alder cycloaddition, and Fukui's rules suggest that this kind of reaction doesn't happen photochemically, only thermally.
However, since the reaction is between guanine and singlet oxygen, the oxygen has been photochemically excited, and I don't think it would happen thermally with oxygen at the ground state because otherwise we would have constant oxidation.
So what exactly is happening there?

You're right that ground state $\ce{^3O2}$ doesn't undergo this reaction.
There is no direct excitation of oxygen either.
The key in these processes is the electronic excitation of tetraphenyl porphyrine (TPP). The excited singlet state of the dye undergoes intersystem crossing to a triplet state. The latter interacts with ground state oxygen and generates singlet oxygen.
$$\ce{TPP ->[h\nu] ^1TPP^*}$$
$$\ce{^1TPP^* ->[ISC] ^3TPP^*}$$
$$\ce{^3TPP^* + ^3O2->[sens] TPP + ^1O2}$$
Singlet oxygen then acts like the typical dienophile in a Diels-Alder reaction.


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Why do alpha-hydroxy ketones respond positively to Tollens', Fehling's, Benedict's, Schiff's and HgCl2 tests?
I learnt from a certain textbook that aldehydes and formic acid respond positively to Tollens', Fehling's, Benedict's, Schiff's and HgCl2 tests.
However, in another textbook I read that even alpha-hydroxy ketones respond positively to these tests. What is the reason behind this? I thought ketones are not oxidizable (unless under drastic conditions).
Fructose is an alpha-hydroxy ketone, but all alpha-hydroxy ketones are not fructose. In fructose there is a terminal hydroxy group which facilitates Tollens' test, etc. in basic medium. But why do alpha-hydroxy ketones like benzoin give these tests even though there is no terminal hydroxy group? That part is still unclear to me. Could you please answer that part?

Related: Why does fructose reduce Tollen's reagent and Fehling's solution?

I think $\alpha$-hydroxy ketone is being oxidized to a diketone... Still thinking about mechanism of action.

I think you should be able to find your answer here: http://www.acadblock.com/organic-chemistry/tollens-reagent-l4b2/

There is a nice answer in this link: https://www.quora.com/Why-is-Fehlings-solution-able-to-oxidise-%CE%B1-Hydroxy-ketone-but-not-ketones

Duplicate?: https://chemistry.stackexchange.com/questions/94620/in-what-way-could-benzoin-give-tollens-test/94630#94630

I answered a similar question with proposed mechanism : https://chemistry.stackexchange.com/questions/156199/do-all-%ce%b1-hydroxy-ketones-give-tollens-test/156371?noredirect=1#comment321838_156371

Its because of two tautomerizations happening, the ketone becomes en-ol, and then another tautomerization happens with the original hydroxy group (which is terminal), making it an aldehyde group. All three forms exist in equilibrium, so it responds to Tollens' test.

Not all alpha hydroxy ketones can undergo double tautomerization to form an aldehyde.


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Schweizer's reagent from/with CuO?
While attempting to make Schweizer's reagent in household conditions, using $\ce{CuSO4 + NaOH + 25\%  NH4OH}$ route (all at least 95% pure, either technical or lab grade purity), I've stumbled upon a problem. While trying to precipitate $\ce{Cu(OH)2}$, the temperature of the solutions rose to about 50 Celsius, and instead of expected blue precipitate, I got an almost black (with a hint of violet) one instead, which I assumed is $\ce{CuO}$. However, I was able to dissolve some of the precipitate in $\ce{NH4OH}$ slightly (a suspension was made from about 10 g of precipitate in 200 ml ammonia), resulting in dark blue solution with only traces of suspended black particles ($\ce{CuO}$?), appearing very like the expected Schweizer's reagent itself.
The resulting substance was able to dissolve regular cotton & cellulose samples overnight, however the dissolution was quite slow and unimpressive, with some of the sample material remaining almost intact; a chalk-coated paper was quite resistant to it (although it also dissolved somewhat eventually).
I don't have prior experience with Schweizer's reagent, so I don't know if the rate of dissolution I got here was the proper, expected behaviour here or a sign of dud.
Since, from what I know, $\ce{CuO}$ doesn't react nor dissolve in aq ammonia, does this experiment suggest a trace amounts of $\ce{Cu(OH)2}$ present in the (assumed) $\ce{CuO}$ precipitate, later forming a low-quality Schweizer's reagent, or am I missing something here?
As side questions:

was the precipitate really $\ce{CuO}$ (if so, how to avoid creating it, i.e. what are the necessary conditions for $\ce{Cu(OH)2}$ creation here), and what is the easiest way to check for $\ce{CuO}$ presence in general?
how efficient should be a properly done homemade Schweizer's reagent (using the aforementioned $\ce{CuSO4 + NaOH + 25\%  NH4OH}$ route) in dissolving various celluloses in practice?

Since, from what I know, $\ce{CuO}$ doesn't react nor dissolve in aq ammonia, does this experiment suggest a trace amounts of $\ce{Cu(OH)2}$ present in the (assumed) $\ce{CuO}$ precipitate, later forming a low-quality Schweizer's reagent, or am I missing something here?

Well in general this is true, but because the fresh precipitate can be partially hydrated and amorphous, which does allow the ammonium hydroxide to dissolve it.

was the precipitate really $\ce{CuO}$ (if so, how to avoid creating it, i.e. what are the necessary conditions for $\ce{Cu(OH)2}$ creation here), and what is the easiest way to check for $\ce{CuO}$ presence in general?

Your precipitate was likely a hydrated copper oxide. Heat causes the copper hydroxide to lose its water.  To prevent $\ce{CuO}$ from forming, you want to keep the solution cool, perhaps with an ice bath. The easiest way to verify this is by visual observation. Copper oxide is black and copper hydroxide is blue, there is no false positive/negative to this for your preparation.

how efficient should be a properly done homemade Schweizer's reagent (using the aforementioned $\ce{CuSO4 + NaOH + 25\%  NH4OH}$ route) in dissolving various celluloses in practice?

This is not easy to answer and depends greatly on the concentration of your reagents, temperature, pH and so on. I've seen videos on YouTube however showing about 2g of cotton balls dissolving in 500mL solution as a reference