text:
Determine heat flux from temperature profile

The heating
  element and the insulator are of equal thickness L. Heat transfer in
  the air film adjacent to the heater is assumed negligible.

I've noticed that I find these type of problem the hardest to understand, so before I ask my question, does anyone know a good video or website which explains in detail how you should interpret temperature/concentration profiles?
My answer to (a) would be that both of the points should have the same heat flux. I know that one of the relationships used for layered wall is that q=constant, meaning that it is the same through every layer, however is that the correct way of thinking for this problem? I don't think so and don't understand how I should interpret the profile.
Another way I thought about it was that the heat flux for the layers should be:
$$q_1=(k_1/L)\cdot(T_\text{rear} - T_{12})$$
$$q_3=(k_3/L)\cdot(T_{23} - T_\text{front})$$
and that:
$$k_3>k_1$$
$$(T_\text{rear} - T_{12})\gt(T_{23} - T_\text{front})$$
If I just for simplicity assume that $L=1\ \mathrm m$. Then I would get that
$$q_1=(\text{something small})\cdot(\text{something big})$$
$$q_3=(\text{something big})\cdot(\text{something small})$$
Hence another reason why I think that the heat fluxes are the same.

Your first line of thinking to part a) is correct. For conductive heat transfer through rectangular layers, the heat flux needs to be constant and equivalent through all layers for steady state, or you would have a 'build up' of heat at between one layer. This would locally increase the temperature, and we wouldn't be at steady state.
For b) your second approach to a) is useful. You know the relative sizes of the temperature gradients from the sketch, and know from a) that the heat flux must be the same for each slab. Combine these two and you will find a requirement on the relative magnitudes of the conductivities.

Starting from the following equation (the one dimensional Fourier law) for heat flux $q$: $$q=-k\frac{\mathrm{d}T}{\mathrm{d}x}$$ where $k$ is the material's thermal conductivity, I obtained the following relations:
$$|q_{2\to1}|=q_{1}=k_1\frac{\Delta{T_1}}{L}$$
$$|q_{2\to3}|=q_{3}=k_3\frac{\Delta{T_3}}{L}$$
$$|q_{2\to1}|=k_2\frac{\Delta{T_{12}}}{L_{12}}$$
$$|q_{2\to3}|=k_2\frac{\Delta{T_{23}}}{L_{23}}$$
where $$\Delta{T_1} = T_{12}-T_{rear}$$ $$\Delta{T_3} = T_{23}-T_{front}$$
$$\Delta{T_{12}} = T_{max}-T_{12}$$
$$\Delta{T_{23}} = T_{max}-T_{23}$$
For instance, $q_{2\to1}$ is the heat flux within layer 2 in the direction of 1 (away from the maximum).
Now if I assume that $L_{23}=L_{12}$ (that the T maximum in layer 2 is positioned halfway in the layer), then
$$\frac{|q_{2\to1}|}{|q_{2\to3}|}=\frac{\Delta{T_{12}}}{\Delta{T_{23}}}$$
Since
$$\Delta{T_{23}}>\Delta{T_{12}}$$
I conclude that
$$|q_{2\to3}|>|q_{2\to1}|$$
This answers (a). But the accuracy of this comes down to the assumption $L_{23}=L_{12}$ and I suspect there may be another way to solve this part.
The result of (a) can now be used, together with
$$\frac{|q_{2\to1}|}{|q_{2\to3}|}=\frac{|q_1|}{|q_3|}$$
and expressions above, to show that
$$k_3\Delta{T_3}>k_1\Delta{T_1}$$
$$\frac{k_3}{k_1}>\frac{\Delta{T_1}}{\Delta{T_3}}$$
Now since
$$1<\frac{\Delta{T_1}}{\Delta{T_3}}$$
we have that
$$\frac{k_3}{k_1}>1$$
which answers part (b).


text:
Electrolysis rust removal - side products - identify lead
my kids want to restore an old hand tool made of mostly iron.
I've choosen electrolysis as method for rust removal. We used Natron to enhance the conductivity of the water. The process worked so far and we get as expected some chemical side products swimming on top of the solution. We have brown stuff (iron) and from the wires some blue (silver or copper), and we have back stuff. The solution was pretty black afterwards as well as some scum swimming on top of it.
I don't want my kids to come in contact with (too much) lead. Is there an easy way to determine if the hand tool has lead in it?

there are consumer lead testing kits available at most hardware stores

Add the information regarding the composition regarding the electrodes. However lead is very heavy it would not like to float on the top. Secondly, lead compounds are mostly white with an exception of oxides (yellow/ red) and iodide (yellowish). Most likely the black floating scum is carbon in iron. Wear gloves, and wash your object with a mild abrasive such as a baking soda, and it should be fine. After all it was meant as a hand tool. It will not become "toxic" after a brief electrolysis.

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text:
Decrease of volume by half in equilibrium - with specific numbers- how to solve?

$$\ce{CO (g) + Cl2(g)<=>COCl2}(g)$$
  We have
  2 mols of $\ce{CO}$, 1 Mol of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$
  contained in 1 L.
  The system is currently in equilibrium.
We then decrease the volume by half to 0.5 L
What will be the new concentration in equilibrium of CO?

I tried calculating at first the constant of concentration;
$Kc=\frac{[\ce{CoCl2}]}{[\ce{CO}]\times[\ce{Cl2}]}$
I found it to be $4.8825$
then I calculated the concentration with the new Volume:
$[\ce{CO}]= \pu{2mol}/\pu{0.5L}= \pu{4 M}$
$[\ce{Cl2}]= \pu{1mol}/\pu{0.5L}= \pu{2 M}$
$[\ce{CoCl2}]= \pu{9.765mol}/\pu{0.5L}= \pu{19.53 M}$
obviously this wasn't in equilibrium so I went to do this:
$\ce{CO + Cl2<=>COCl2}$
$4M-x$, $2M-x$,  $19.53+x$
(as the number of mols in the products is smaller and the system wants to fix the increase in pressure)
When I calculated all of this in the Kc formula it came to me that the Concentration of CO increased, and is now larger than 2 M ,
and in the textbook answers it is stated that it is supposed to be less than 2M, how is it possible?
I would be very glad for help with calculations and technique of solving as well!

What have you tried so far? Do you have thoughts on how to approach the problem? You should edit the question to include these so it doesn't get closed as homework. @RonAvraham

I've edited the question to use some math and chemistry formatting, let me know if I messed anything up. @RonAvraham

HINTS - Is the system a gas or an aqueous solution? How should the equilibrium change and why?

the system is gas, and the equilibrium is supposed to move to the products as they have less mols.

After getting to values 4M−x, 2M−x, and 19.53+x, I assume you solved the quadratic equation? So long as x is a positive real number, then the equilibrium will shift to favor the product and [CO]<4

What did you get for x?

I found X to be 0.73 and 5.74 obviously the second one is not viable, though the first soliution seems logical the textbook states that [CO] will be smaller then 2 M so I guess that it is just a mistake in the book..

Maybe they meant to say after dilution the reaction goes in the reverse direction (for the reasons you stated), so you expect a concentration less than 2 M.

I solved the quadratic equation by Wolfram Alpha

The solution indicates that $\ce{[Cl2]}=\pu{1.26926M}<\pu{2M}$,but$[\ce{CO}]=\pu{1.26926M}<\pu{2M}$, if we have
1 mol of $\ce{CO}$, 2 Mols of $\ce{Cl2}$,and 9.765 Mols of $\ce{COCl2}$
contained in 1 L.

@MaxW I appreciate to provide me feedback  about the answer

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text:
why do cone cells have broad responsivity spectra peaks?
My understanding of vision is that, although the cone cells in our eyes can detect three colors, there is enough overlap in the responsivity spectra (as shown here: https://en.wikipedia.org/wiki/Spectral_sensitivity) that we can see any wavelength in the visual range.
What is the physical reason for the broadness of the responsivity spectra peaks? Is it interactions with water molecules, interactions with the protein that the retinal molecule is attached to, or is it broad because many vibrational states of the retinal can be excited?

They aren't particularly broad afaik.

It is a question I am thinking since a while. I suppose that are proteins deforming the conjugated chain.  We'll see, perhaps in a while you can try biology. ...


text:
Molecular orbitals used with CCSD(T) geometries
Suppose I am given some geometry data (say, of a water dimer) from CCSD(T).  If I were to do a single-point energy calculation to generate orbitals for the system, wouldn't CCSD give something more representative of the correlation in the system than DFT?
I see "CCSD(T) orbitals don't exist - used DFT orbitals instead" kind of statements in papers (quotations below), and wasn't sure why they don't even mention CCSD.  Perhaps it's just due to speed/availability, but that would lead to another question "why is it not worth it to use CCSD orbitals with CCSD(T) geometry".
The same question can be asked for DFT orbitals being used with MP2 geometries, but MP2 perturbs about HF ... so clearly DFT is needed to capture correlation.
[edit: more context]
What I'm trying to do is "export" these orbitals to a diffusion quantum Monte Carlo (DMC) calculation which doesn't do geometry optimization by itself but augments the imported orbitals with explicit correlations (eg. something that depends directly on electron pair distances).  Usually the final DMC energy is not very sensitive to whether the orbitals are from CC or DFT (I have outputs from both), but I was checking here to make sure I wasn't missing any "deeper" reason.

From ChemPhysChem 2018, 19 (15), 1886–1894:

Since MP2 orbitals are nonexistent, the charge-transfer interactions were computed using the B3LYP functional with the aug'-cc-pVTZ basis set at the MP2/aug'-cc-pVTZ geometries, so that at least some electron correlation effects could be included.

and J. Phys. Chem. A 2017, 121 (49), 9544–9556:

Due to the nonexistence of CCSD(T) orbitals, ΔE(del) was calculated with ωB97XD/aug-cc-pVTZ.

I'm guessing that a lot of times, this comes down to the necessary methods being very costly and/or not implemented. The methods that then consume these orbitals are considered to be not even semi-quantitatively correct by many, so DFT orbitals are probably sufficient most of the time.

I thought I had a straightforward answer to this, but I believe the difficulty is associated with the fact both these methods are perturbative in nature. In principle, one should be able to generate a new set of orbitals, but I believe in practice the orbitals from perturbative QM methods are actually somewhat arbitrary as the energy is only minimized with respect to excitation amplitudes out of the reference determinant which does not guarantee the orbitals are stationary. Thus, properties in these methods are computed via linear response rather than as expectation values over the orbitals.

The real question is "what do you mean by orbitals?", which is partially related to "what do you want to look at them for?" Certainly all of these methods have natural orbitals which can be obtained by diagonalizing the corresponding 1-particle density matrix (which will have non-integer occupation numbers), and also certainly there are reference orbitals used to compute these methods. Those cited statements are, to some extent, meaningless without the answers to those questions.

And to clarify, if you run a single-point, you have generated an energy, not orbitals. All orbitals are an implementational tool to obtain this energy and, in that sense, no orbitals "exist", in that they are not observable and can be replaced with an alternative method to obtain the same energy. So in order to obtain orbitals you need to clarify what you mean by that.

I also don't know what you mean by using "X orbitals with Y geometries". All of these calculations are done with the Born-Oppenheimer approximation. You fix the nuclei, then solve for the electronic energy. How you obtained the geometry has nothing to do with what method you subsequently use to compute energies or properties.

Thanks for the comments.  They could have constituted answers on their own.  I added a few more sentences to the question for more context.

Pulling together some of my comments into an answer:
The principal question at issue is "what do you mean by orbitals?", which is partially related to "what do you want to look at them for?" You can obtain an corresponding 1-particle density matrix from all of these methods, which can be diagonalized to yield natural orbitals (which will have non-integer occupation numbers. There are also certainly reference orbitals used to compute these methods (usually Hartree-Fock, but there are methods like orbital optimized-MP2 (OO-MP2) and Brueckner doubles (an orbital optimized CC variant)). Those cited statements are, to some extent, meaningless without the answers to those questions.
If you run a single-point, you have generated an energy, not orbitals. All orbitals are an implementational tool to obtain this energy and, in that sense, no orbitals "exist", in that they are not observable and can be replaced with an alternative method to obtain the same energy.
For your purpose, since you are just exporting to DMC, yeah wouldn't expect the results to vary too much by orbital, the DMC procedure just needs a mostly reasonable starting point. CC natural orbitals would probably produce a more compact wavefunction and therefore maybe run faster than DFT orbitals, but they are enough more expensive to compute that it's probably not worth it. Probably the orbitals you got from the CC calculation you did are just the HF reference orbitals. It's also worth pointing out that natural orbitals are probably not so different from the DFT orbitals unless there is a lot of correlation (in which case, DFT method will probably produce poor orbitals). For systems where the reference state is qualitatively wrong (e.g. diradical system that might suffer from a lot of spin-contamination in HF), OO-MP2 would improve that and those orbitals would probably be substantially better and that would be a case where the cost would probably be justifiable.

The orbitals used for the coupled clusters variants that you mentioned are (almost always) those from the Hartree-Fock method, which is performed as a first step. Note that the latter does not account for correlation energy. The case of MP$n$ is exactly the same.
In the case of DFT the orbitals are obtained from the Kohn-Sham method. I suppose that they are preferred due to this.


text:
Direction of mass transfer in two film theory

In an extraction process compound A is transferred from one solvent to another solvent. The two solvent may be, for example, water and dichloromethane. These two solvents are immiscible and when mixing them together, one of the solvents, say solvent 1, forms droplets that are dispersed in solvent 2, as depicted [...].

Fig. 4 shows the concentration profile across the interface between the two solvents. For this concentration profile, answer the following questions:
(a) In which direction is the mass transfer, from solvent 1 to solvent 2 or vice versa? Motivate your answer by a short explanation (2 pt).

How can you tell which direction the mass transfer will go?
I would assume that because solvent 1 forms droplets in solvent 2, then solvent 1 should be the organic phase. Would the mass transfer be from solvent 2 to solvent 1? Because the aqueous phase would contain compound A.
Edit
Just to see if I have understood it correctly, if I am given the concentration profile below. Then mass transfer would occur from the aqueous phase to the organic phase because the aqueous phase has a higher concentration.

Just keep in mind: the driving force of mass transfer is a concentration gradient. Nature tries to 'equilibriate' a difference in concentration between two phases.
As such, the mass transfer in that sense will be from the phase of high concentration (S2) to the phase of lower concentration (S1).
Two film theory is only a simplification of the mass transport directly at the interface, in which we assume there are two boundary layers on each side of the interface in which the concentration gradient is constant, giving us linear concentration profiles, and the concentration directly at the interface is determined by Henrys Law. This makes dealing with such problems much easier.
Mass and Heat Transfer are often taught in one course as they have many similiarities, but Heat Transfer offers a more intuitive introduction to the principles of transport phenomena in my view:
In heat transfer the driving force is the temperature gradient. We intuitively know that heat transfer will be from the hotter system to the colder system, and formally know that from the 2nd law of thermodynamics. Can you see the similarity to mass transport?

So whenever I am given a concentration profile and asked to determine the direction the mass transfer, I should look at which side has the higher concentration at the interface and then I know that it will be directed from that layer to the one with the lower concentration?

If you don't mind, could you have a look at the edit I made and tell me if I have understood the concept? Thank you!

yea that should be fine, as long as you are not talking about osmosis.


text:
Why is phosphorus (v) oxides molecular formula P4O10 and not P2O5
I understand that the name derives from the empirical formula; however, I do not understand why it does not exist as $\ce{P2O5}$. Is there a compound that exists with this as its molecular formula?

See Wikipedia's articlePhosphorus pentoxide. Essentially $\ce{P4O10}$ is a molecule and $\ce{P2O5}$ is not.

Just out of curiosity why do you think isolated P2O5 molecules would be stable? As far as I know only 1 element (Iodine) forms a pentoxide for which the most stable form in the solid state is an isolated X2O5 molecule, all other pentoxides have more extended structures. Why should phosphorous be different from the norm?

@Karl not really, it could look like N2O5 does.

Unfortunately, "exists" is not well-defined enough.

@ian there is also dibromine pentoxide, which however requires low temperature to prepare and store.

There is no clear rule about how to name a compound which is actually existing as a dimer, a trimer... or an octamer, as long as it does not make a difference in a reaction equation, which is usually the case.
It does not matter in an equation if you count sulfur as a single atom entity (S) or as its true form (S8). You will find the same quantity in both cases.
The same stands for P2O5/P4O10: the true form is usually P4O10 (depending on the solvent), but the 2 forms give the same result at the end.

The short answer is probably that in any case of stoichiometry it doesn't matter. Much as SteffX explained it above. But while we have it already. There are cases where I think it has at least some benefit if we use the not shortened formula. This would probably be one of these cases. As you may know, the $\ce{P4O10}$ is formed when white Phosphorus $\ce{P4}$ reacts with air. The reason for this is that in the $\ce{P4}$-tetrahedron we have smaller bond angles (60°) than we would expect from a head-on bond formation of three p-type orbitals (90°). Therefore the orbitals cannot overlap perfectly and there is some considerable bond strain.
My former Professor for Inorganic Chemistry is a famous phosphorus chemist and he explained it in much more details back then. But if it's about angles in $\ce{P-P}$-bonds, triangular shapes are some of the worst things that can happen. And for the $\ce{P4}$-tetrahedron (white phosphorus) we have a polyhedron that consists of triangular faces only.
So if we introduce oxygen into the system it could enter between every $\ce{P-P}$-bond to increase the angle and therefore reduce the strain. If you draw this on some paper you will see that a $\ce{P4O6}$ results (sometimes called $\ce{P2O3}$). And what you should be able to see is that you can still connect all $\ce{P}$-atoms to get back the original tetrahedron. So the overall shape did not change. We call this a 'topotactic oxidation', oxidation, where the original shape remains, we simply add something inbetween. And in the final step, the phosphorus is at $\ce{P^3+}$ now, we can even oxidize it up to it's highest possible oxidation state $\ce{P^5+}$ by adding more oxygen. It will attack the terminal positions, so we get four additional oxygens and a final $\ce{P4O10}$. This is a relationship many people often forget about. So it's quite nice to remind them of the original $\ce{P4}$-tetrahedron by not shortening the formula down to $\ce{P2O5}$. And as many suggested above it's just the unit you will see.


text:
Does mixing calcium dichloride and magnesium sulfate induce a reaction, and how does it affect the hygroscopy of the sample?
I have a problem with a sample made from $\ce{CaCl2*2H2O}$ and $\ce{MgSO4*7H2O}$ in about equal amounts. I took the two salts, pressed them together, and the result was a weird brine that was very wet and almost seemed to turn into water before my eyes.
I am wondering what happened here, and was hoping that someone could walk me through it. As you can tell by the title, I believe that a reaction might be involved that I did not consider when making the sample.
As far as I know, the two salts are both very hygroscopic, so it could be that they simply adsorbed water and that is why the sample became wet. However, the effect seemed more severe after mixing them together, so I am thinking that a reaction might have been involved.
Could it be that the two salts reacted this way?
$$
\ce{CaCl2*2H2O + MgSO4*7H2O -> CaSO4*2H2O + MgCl2*6H2O + H2O}
$$
And if so, what happens with the leftover water that is not bound? Especially since $\ce{MgCl2*6H2O}$ is even more hygroscopic than any of the other salts, could it be that the water was attracted by the $\ce{MgCl2*6H2O}$, which then turned the sample into a brine?
I'm just trying to understand what happened, I did not add any heat, but I did add a lot of pressure.

You really need what will be a very complicated phase diagram to figure this out. It would seem obvious that enough water came from the original salts to create a "brine". If there were any solids left then four possible salts with a number of different hydrates.

You are correct.
In the system $\ce{Ca^2+*Mg^2+*Cl^-*SO4^2-*H2O}$, the least soluble phase is $\ce{CaSO4*2H2O}$.
Once you're mixing $\ce{CaCl2*2H2O}$ and $\ce{MgSO4*7H2O}$, your $\ce{CaCl2*2H2O}$ starts absorbing atmospheric moisture and then dissolving in it. Then you have a brine which dissolves the $\ce{MgSO4*7H2O}$, speeding up the kinetics of the thing. The solid that you precipitate by the recation has less $\ce{H2O}$ per mole of salt, so the excess $\ce{H2O}$ just ends up in the brine (the "water"). Pressing it speeds it up because there is more surface area of the two salts in contact, which is where the reaction happens.


text:
AP Chem Question about Reaction Rate - Confused?
I found this question in the AP Chem book, and I understand why B is the answer (as the answer key says). However, I don't understand why D is not also correct.
The question is:
A multistep reaction takes place with the following elementary steps.

Step I. A + B <-> C
Step II. C + A -> D
Step III. C + D -> B + E

Why would increasing the temperature make the reaction rate go up?

(A) it is an endothermic reaction that needs an outside energy source
to function
(B) The various molecules in the reactions will move faster and
collide more often
(C) The overall activation energy of the rxn will be lowered
(D) A higher fraction of molecules will have the same activation
energy

I understand that B is correct based on collision theory, but why isn't D correct as well? Don't a greater % of molecules have the same activation energy at higher temperatures? (Or am I misinterpreting answer choice D?)
Thank you!

“A higher fraction of collisions will have an energy higher than the activation energy” is a correct statement. This is the main reason reaction rates increase at higher temperature. They also increase due to higher frequency of collisions. I don’t understand the statement in choice D.

If you strike out the word same in (D) then the statement would be true. With same however the statement is weird. I think same was thrown in to trap a careless reading of the statement.

Tangential to MaxW's point, at any temperature, all the molecules have the same activation energy, but at higher temperature more molecules will have energy in excess of that activation energy.


text:
Behavior beyond the critical pressure
I have seen some phase diagrams that look like this:
https://www.learner.org/courses/chemistry/images/text_img/phase_diagram_water.jpg
And also some like this:
https://en.wikipedia.org/wiki/Phase_diagram#/media/File:Phase-diag2.svg
My confusion is related to the Critical Point. The first picture doesn't really explain where the gas/supercritical fluid boundary is, and the right angle indicating the supercritical fluid boundary at the critical point seems unusual to me. I have some general questions about behavior beyond the critical temperatures and pressures.
I think my specific question is best summarized by this picture.

More specifically, suppose you have an arbitrary compound X with a Liquid/Gas critical temperature of 320 K and critical pressure of 40 Bar which follows the general shape of the above phase diagrams. I have the following questions
1) At 330 K  and 10 bar (basically beyond the critical temperature but below the critical pressure) would X be a gas? or would it be a supercritical fluid? Would we need more information to be able to decide?
2) As a more general part of 1), is it possible that a gas exists beyond its critical temperature?
3) At pressures above its critical pressure, a liquid can still exist so long as it's temperature is lower than the critical temperature correct?

All that really happens at the critical point is that the liquid and vapor have the same properties. I prefer T-v diagrams myself, then you can see clearly that at the critical point (and above) the densities of the liquid and vapor converge and only one "phase" exists. It is not accurate to call it a liquid or a vapor ... it would be closest to a dense vapor I guess. I don't like your diagram as it seems to insinuate there are phase boundaries above the critical point. Dotted lines would be better to distinguish the supercritical fluid from the vapor or compressed liquid phases.

I like the second image you linked, that describes how I think of the supercritical region.

To answer the questions on your diagram, the vapor in that region is still a gas, but is often referred to as superheated. The liquid is still a liquid but is referred to as a compressed liquid. On the phase line, there is just enough pressure to form a liquid... if pressure was decreased some of the liquid would become a vapor.. In the compressed region there is so much pressure that you could decrease the pressure, and it would still stay a liquid... this is a (overly)simple way of thinking about it.

Often, the term used past the critical point is ‘fluid’.

Possible duplicate of Why must both the critical temperature and pressure be exceeded to achieve the supercritical phase?

In order to be in the "critical state", both temperature (x-axis) and pressure (y-axis) must be above the critical point. So answers to your questions become quite obvious:

Very hot but below the critical pressure, it is a gas. You don't need any more information than the temperature and the pressure.
Yes if its pressure is less than its critical pressure
Correct.


text:
what is the basic idea for making bioplastic?
I have the following questions about making bioplastics:

Why is only lactic acid used in making bioplastic?
Are there any other methods used in making bioplastic without converting
starch into lactic acid?
What are the features of lactic acid that makes it suitable for
making bioplastic?
Are there any alternatives for making good bioplastic?

This question as it stands is too broad please ask the 4 questions separately

A general answer to your broad question is that poly(lactic acid) is used because fermentation of sugar to lactic acid is easy and has high mass yield, and polymerization to the polyester is straightforward. But it's not the only bioplastic. There are many others in use or in development. For example, ethanol from fermentation can be dehydrated to ethylene which is polymerized to polyethylene.
So yes, there are other methods and other alternatives to PLA. If you are interested in a particular application, I would recommend researching possible materials for that application and asking a more specific question about the comparable qualities of materials for that application.


text:
Comparison of chemical shift of methyl group attached to alcohol and carboxylic acid
I understand that hydrogens can be deshielded by electron-withdrawing groups, and the more powerful the group, the greater the chemical shift.
However, I ran into a problem when qualitatively compare the chemical shifts of the hydrogens in the methyl group of methanol and ethanoic acid: the carboxylic acid group is stronger than the alcohol group, the oxygens of the carboxylic acid group isn't directly attached to the methyl carbon, unlike the alcohol oxygen.
How would I qualitatively compare the chemical shifts of the methyl groups of methanol and ethanoic acid?

That carbon is part of the carboxylic group ($\ce{-C(O)OH}$)

@NightWriter thanks, I've edited to better reflect what I mean

It's a good idea to start with some data. The table below shows relevant $\ce{^1H}$ shifts in $\pu{ppm}$ for three compounds in various deuterated solvents (shift values are from Cambridge Isotope Labs): methanol (MetOH), formic acid (FormAc) and ethanoic or acetic acid (AcAc).

$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\text{Compound} & \ce{CDCl3} & \ce{(CD3)2SO} & \ce{C5D5N}\;\text{or}\;\ce{C5H5N}  & \ce{C6D6}\;\text{or}\;\ce{C6H6} & \ce{D2O} \\ \hline
 \text{Acetic acid} & 2.13 & 1.95 & 2.13  & 1.63 & 2.16\\ \hline
 \text{Formic acid} & 8.02 & 8.18 & 8.54 & 7.24 & 8.22\\ \hline
 \text{Methanol} & 3.48 & 3.20 & 3.57 & 3.09 & 3.35\\ \hline
\end{array}
$$
The protons in methanol are more deshielded than those in ethanoic acid because in methanol an electron withdrawing oxygen atom is directly bonded to the carbon, while in ethanoic acid there is an intervening carbon.
If you remove that intervening carbon things get interesting, as the case of formic acid illustrates. The carbon atom in a carboxylic group is $\mathrm{sp^2}$ hybridized, which makes it more electrophilic than the $\mathrm{sp^3}$ hybridized carbon in methanol. Of course the carbon in a carboxylic group is also bound to two electron withdrawing oxygen atoms, rather than one, which results in greater deshielding. The formic acid proton is also subject to deshielding magnetic anisotropy associated with the delocalized electrons of the carboxylic group. The anisotropy effect is less pronounced in ethanoic acid because the methyl group is further away from the carboxylic group. The ultimate result is that the formic acid proton is very strongly deshielded compared to protons in the other compounds.

In NMR spectroscopy, the chemical shift value of active nucleus is telling us the strength of the magnetic field that particular nucleus feels.  Nuclei in a molecule are not separated from charged particles such as electrons; therefore, the electrons will always have an impact on the nuclei. A nucleus feels a field that doesn’t exactly equal to applied external magnetic field. The difference between the external magnetic field and the field felt by the nucleus is very small and we call this small difference, shielding. For example, if the nucleus is a hydrogen atom (In vocabulary in NMR spectroscopy, it is called a proton) in the molecule, it feels different magnetic fields depending upon its location and the neighbor it directly attached to. To answer your question of this difference properly, there needs to be a few words in NMR vocabulary to define:
Shielding (Upfield shift): Shielding is a barrier made of inner-shell electrons and it decreases the nucleus’ pull on the outer electrons. In most books, shielded is defined as a nucleus whose chemical shift has been decreased due to addition of electron density, magnetic induction, or other effects. Hence, the nucleus feels weaker magnetic field than the applied field.
Deshielding (Downfield shift): Deshielding is the opposite of shielding. When we say that a nucleus is deshielded, what we mean is that nucleus’ chemical shift has been increased due to removal of electron density (e.g., a barrier made of inner-shell electrons), magnetic induction, or other effects. Hence, the nucleus feels stronger magnetic field than the applied field.
Magnetic Induction: In NMR spectroscopy, magnetic induction (e.g., magnetic anisotropy) is the phenomenon that an external magnetic field (applied field) causes $\pi $-electrons in an electron cloud to circulate. As these electrons move around, their own electric charge causes them to create their own magnetic field (induced field). Possibly the best example of magnetic induction occurs with the benzene molecule, in which its 6 $\pi $-electrons are delocalized and free to move around the aromatic ring (6 $\mathrm{sp}^2$ carbons), inducing a magnetic field at the center such a way that it is opposite to the applied field (right hand rule). Thus, outside the ring feels more than applied field thus benzene protons are deshielded (see picture below). On the other hand, protons inside the ring (molecule in the RHS of the picture) feel less magnetic field and hence, shielded.

Magnetic anisotropy is in effect when delocated $\pi$-electrons are in nearby such as in alkenes, alkynes and carbonyl groups (See picture below). As Night Writer pointed out, electronegativity difference between neighboring nuclei plays a bigger role in methanol and other compounds such as $\ce{CH3X}$ (see the Table below as well).

$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\ce{CH3X}  & \ce{CH3F} & \ce{(CH3OH} & \ce{CH3Cl} &  \ce{CH3Br} & \ce{CH3I} &  \ce{CH3H} & \ce{(CH3)4Si} \\ \hline
 \ce{X} & \ce{F} & \ce{O} & \ce{Cl} & \ce{Br} & \ce{I} & \ce{H} & \ce{Si} \\ \hline
 \text{Electronegativity of}\; \ce{X} & 4.0 & 3.5 & 3.1 & 2.8 & 2.5 & 2.1 & 1.8 \\ \hline
 \delta \; \text{in ppm} & 4.26 & 3.4 & 3.05 & 2.68 & 2.16 & 0.23 & 0 \\ \hline
\end{array}
$$
The above picture and the data of the table are directly from University of Calgary page.
Keep in mind that this chemical shifts values can vary by the solvents used (see Night Writer's Table), operating temperature, pH, etc. (Ref.1). Some values listed in Ref.1 are displayed in following Table for your convenience:
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\text{Compound} & \ce{C-H} & \ce{CDCl3} & \ce{(CD3)2CO} & \ce{(CD3)2SO} & \ce{C6D6} & \ce{CD3CN} & \ce{CD3OD} & \ce{D2O} \\ \hline
 \text{acetic acid} & \ce{CH3} & 2.10 & 1.96 & 1.91 & 1.55 & 1.96 & 1.99 & 2.08 \\ \hline
\text{acetone} & \ce{CH3} & 2.17 & 2.09 & 2.09 & 1.55 & 2.08 & 2.15 & 2.22 \\ \hline
\text{acetonitrile} & \ce{CH3} & 2.10 & 2.05 & 2.07 & 1.55 & 1.96 & 2.03 & 2.06 \\ \hline
\text{DMSO} & \ce{CH3} & 2.62 & 2.52 & 2.54 & 1.68  & 2.50 & 2.65 & 2.71 \\ \hline
\text{nitromethane} & \ce{CH3} & 4.33 & 4.43 & 4.42 & 2.94 & 4.31 & 4.34 & 4.40 \\ \hline
\text{ethyl acetate} & \ce{CH3CO} & 2.05 & 1.97 & 1.99 & 1.65  & 1.97 & 2.01 & 2.07 \\ \hline
\text{ ethyl acetate} & \ce{OCH2} & 4.12 & 4.05 & 4.03 & 3.89  & 4.06 & 4.09 & 4.14 \\ \hline
\text{ methanol} & \ce{OCH2} & 3.49 & 3.31 & 3.16 & 3.07  & 3.28 & 3.34 & 3.34 \\ \hline
\text{ ethanol} & \ce{OCH2} & 3.72 & 3.57 & 3.44 & 3.34  & 3.54 & 3.60 & 3.65 \\ \hline
\text{2-propanol} & \ce{OCH} & 4.04 & 3.90 & 3.78 & 3.67  & 3.87 & 3.92 & 4.02 \\ \hline
\text{toluene} & \ce{CH3} & 2.36 & 2.32 & 2.30 & 2.11 & 2.33 & 2.32 & - \\ \hline
\end{array}
$$
Keep in mind that chemical shift of regular alkane $\ce{-CH3}$ is around $\pu{0.9 ppm}$, yet the downfield shifts shown in entries 1-3 and 6 are due to magnetic anisotropy induced by carbonyl group ($\ce{-C=O}$). Similarly, relatively large shifts in entries 4 and 5 are due to $\ce{-S=O}$ and $\ce{-N(O)=O}$ groups, respectively. The similar effect on last entry is duo to nearby double bond ($\ce{-CH=C-CH3}$).
References:

H. E. Gottlieb, V. Kotlyar, A. Nudelman, “NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities,” J. Org. Chem. 1997, 62(21), 7512–7515 (DOI: 10.1021/jo971176v).

In brief:
Indeed is not exactly O (of the carboxyl group) that  withdraws electrons from the methyl group.  but the carboxyl itself. Not surprisingly we call this -M effect. So it does not matter much that there is a carbon in between.


text:
Stability of alkadienes

Which of the following is the most stable?
(i) Conjugated alkadiene $(\ce{CH2=CH-CH=CH2})$
(ii) Isolated alkadiene $(\ce{CH2=CH-CH2-CH=CH2})$
(iii) Cumulated alkadiene $(\ce{CH2=C=CH2})$
(iv) All are equal.

I am not able to understand the meaning of "isolated" here.
I am thinking the answer here to be (ii), as it will show more hyperconjugation, but answer is (i).

Well, it's simple: conjugated double bonds feel great, isolated double bonds feel normal, and cumulated bonds feel terrible. Hyperconjugation is a small and unsignificant thing compared to that.

"Isolated" means that the double bonds do not interact and it is, in this example, the opposite of "conjugated". The key here is to see if the double bonds and single bonds are alternating like double-single-double-etc.

KM: Make an effort to justify your answer of ii. That way the question will not be marked as homework and dumped. Otherwise you  will not get help.

It’s better to compare dienes with the same total chain length and thus the same molecular formula, for example the following pentadienes, so that the differences in enthalpy of formation (or as well in enthalpy of hydrogenation) directly indicate the effects of the arrangements of the double bonds.
Isolated double bonds
 – penta-1,4-diene ($\Delta_\mathrm f H^\circ=105.7\ \mathrm{kJ/mol}$)
Conjugated double bonds
 – (3Z)-penta-1,3-diene ($\Delta_\mathrm f H^\circ=81.4\ \mathrm{kJ/mol}$)
 – (3E)-penta-1,3-diene ($\Delta_\mathrm f H^\circ=76.1\ \mathrm{kJ/mol}$)
Cumulated double bonds
 – penta-1,2-diene ($\Delta_\mathrm f H^\circ=140.7\ \mathrm{kJ/mol}$)
 – penta-2,3-diene ($\Delta_\mathrm f H^\circ=133.1\ \mathrm{kJ/mol}$)
(All values taken from “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.)
Double bonds are isolated if they are separated by two or more single bonds so that they cannot interact with each other. The enthalpy of hydrogenation is close to the sum of the enthalpies of hydrogenation for the individual double bonds.
Double bonds are conjugated if they are separated by just one single bond. Because of the interaction between the double bonds, systems containing conjugated double bonds are more stable than similar systems with isolated double bonds.
Successive double bonds with no intervening single bonds are called cumulated double bonds. Systems containing cumulated double bonds are less stable than similar systems with isolated double bonds.

As a suggestion, a short sentence rationalizing why cumulated double bonds are less stable would make this an even better answer.

A graphic, which is not to scale, of @Loong's explanation and @Zhe's request is offered below. [All values are in Kcal/mol (NIST). Values adjacent to structures are heats of formation and heats of hydrogenation are adjacent to arrows.] Although the order of stability is conjugated>isolated>cumulative, the heats of formation of 1,4-pentadiene and 1,3-butadiene are in the wrong order although 1,2-propadiene (allene) is OK. Hydrogenation of these three dienes give different alkanes. The heats of hydrogenation show the least amount of heat is liberated from the most stable, conjugated diene and the most heat liberated from the least stable, cumulated diene. As Loong has pointed out, comparing C5 compounds is appropriate. Now (E)-1,3-pentadiene (in red) is more stable than 1,4-pentadiene which is more  stable than 1,2-pentadiene (in red). Hydrogenation results in the formation of pentane in all three cases. The more stable the diene, the less heat is liberated on hydrogenation.


text:
Temperature profile at steady state

Consider a solid wall consisting of three layers where the layers are of equal
  thickness but made out of different materials: $\ce{Layer 1}$ is made out
  of copper (high thermal conductivity), $\ce{Layer 2}$ is made out of a
  highly porous ceramic (low thermal conductivity), and $\ce{Layer 3}$ is
  made out of concrete (medium thermal conductivity).
How does
  the steady state temperature profile inside the wall look like?

I'm not sure if this question is asked in a way to mislead me into drawing the T-profile as (2) when at steady-state it actually should look like (1).
I apologize if this question is seen as off-topic or simple, but I would appreciate it if someone would explain if I am correct in saying that the T-profile at steady state should be linear.
Thank you!
(1)
 (2)

No, the stationary profile won't be linear. Why would it?

If you would provide your thoughts on why you think it might be one or the other it would be on topic.

No, the temperature gradient is not uniform across all three layers.
To see why, start from the Fourier law for heat flow, shown in my solution to this problem.
Since at steady state the heat flux $q$ must be equal for all layers, we have that
$$q_{1}=q_{2}=q_{3}$$
And since the layers are equal in thickness,
$$ k_1\Delta{T_1}=k_2\Delta{T_2}=k_3\Delta{T_3} $$
Clearly, if the $k$ are not equal, the $\Delta{T}$ cannot be equal either, so the profile will not be that in (1).


text:
Definition of an "equivalent"
What does it mean to add "1 equivalent" of something? I looked it up on google and couldn't really find a good explanation for it.
Thanks!

An equivalent is an equivalent, it doesn't change its meaning in chemistry. What you are asking about is probably molar equivalent, but the name is also self-explanatory...

For example, read: https://en.wikipedia.org/wiki/Equivalent_%28chemistry%29 and http://www.chem.ucla.edu/~harding/IGOC/O/one_equivalent.html

For detailed explanation for gen. chem, read: http://dl.clackamas.edu/ch105/lesson6acid_base_solutions.html

An equivalent is the factor of the number of molecules you need, at minimum, to carry out a reaction. The number of equivalent depends on your choice of what you consider is your main reactant.
Simply put, if you take a chemical equation where A + B → C, you need 1 eq. of B compared to A, i.e. you need one molecule of B for each molecule of A.
But sometimes, you can get something like 2A + B → C, and you want to know the "number of equivalent of B you should add to A". The quantity of B is one half that of A. If you consider your main reactant to be A, then you need to add only 1/2 (aka 0.5) equivalent of B.


text:
how to calculate the initial concentration of NaOH in a mixture?
I have doubts about how to find the equation for the calculation of the initial concentration of $\ce{NaOH}$. I have the following information:
 in the kinetic study of a reaction,
We measure $\ce{20.00 ml}$ of $\ce{CH_3COOC_2H_5}$ solution of 0.0400 $\ce{mol/L}$ and also $\ce{20.00 ml}$ of $\ce{NaOH}$ solution of 0.0400 $\ce{mol/L}$
We mix both substances in one Erlenmeyer.
The question is:  find the equation to calculate the concentration of $\ce{NaOH}$ at $\ce{t=0}$.
It is a bit weird to me because a concentration of the $\ce{NaOH}$ solution is already given.
The best (and the only) that I can think is this:
if the concentrations of the reactives are equal, then:
$$\ce{C_iV_i = C_fV_f}$$
then we have,
$$\ce{C_i=C_fV_f/V_i}$$
$\ce{C_i}$ is the initial concentration of $\ce{NaOH}$, $\ce{V_i}$ is the initial volume of $\ce{NaOH}$, $\ce{C_f}$ is the final concentration of the whole mixture ( which I don't know how to find), and $\ce{V_f}$ is the final volume of the whole mixture
Probably this is wrong but I don't have any other ideas.
 any clues?
Thanks.
(Note: apologies for my poor English)

Right equation. The gist here is that you must assume that the volumes are additive. That isn't strictly true, but since both solutions are fairly dilute the assumption is  "good enough" for this problem.

@MaxW but how do I find the total concentration of the reaction? I don't think I can just sum up both concentrations. I'm confused.

? I said "you must assume that the volumes are additive." So the total volume is 20+20=40 ml

As @MaxW pointed out, if you assume that the volumes are additive implies that your total volume is 40 mL. Now, remember the definition of molar concentration:
$$C=\frac{\text{moles solute}}{\text{total volume}}=\frac{n}{V}$$
That means that the concentration and the volume are inversely proportional, thus if initially, you have a $\text{0.0400 } \frac{\text{mol}}{\text{L}}$ in 20 mL and then you increase the volume twice, then your concentration decrease to the half of initial value. In numbers:
$$\frac{\overset{\text{this are your moles of solute}}{\overbrace{\frac{\text{0.0400 mol}}{\text{1 L}}\times{\text{20 mL}}}}}{\underset{\text{this is your final volume}}{\underbrace{\text{40 mL}}}}=\text{0.0400 }\frac{\text{mol}}{\text{L}}\times\frac{1}{2}=\text{0.0200 }\frac{\text{mol}}{\text{L}} $$


text:
Reaction between phenylamine and halogens
I understand that phenylamine is considerably more reactive than benzene, meaning that it is able to react with bromine water to give 2,3,6-tribromophenylamine. But I was then wondering why I haven't seen anywhere mention reactions with other halogens. So will phenylamine react with other halogens to produce a similar compound, I see no reason why it wouldn't happen? Do text books just use bromine water as it is convenient?
Will  $\ce{C_6H_5NH_2 + 3Cl_2 -> C_6H_2NH_2 Cl_3 + 3HCl}$ happen?

Why would you get N,N-Dichloroaniline instead? and would iodine do the same? Why isn't bromine behaving in this way?

@user55119 Yeah, redox would happen, but this product is unlikely.

It's aniline's ring that is easily oxidated, not nitrogen.

Chlorination of aniline with chlorine is rapid and gives 2,4,6-trichloroaniline according to the references cited in this paper here. The earliest reference cited is 1845 by AW Hofmann so it has been around a long time. More modern approaches use N-Chlorosuccinimide as the reaction is more controllable

Why is this the case then?

In general researchers don't want trichloroaniline, with NCS it is easier to isolate the desired mono or di-chloro aniline products.

No, why does the reaction of chlorine result in a different product and no chloro aniline?

I'm not sure I understand you. The reaction of chlorine directly with aniline gives chlorination of the benzene ring with the same substitution pattern as Bromine does.

Ok, I see now, I didn't read it carefully enough. would iodine behave in the same way then if both chlorine and bromine behave the same?

As far as I can see iodine is not reactive enough to tri-iodinate. It appears to only go in once to give 4-iodoaniline.

Do we only get mono-sub due to the deactivating nature of the iodine, thus making it harder to get a di-iodo product?

It could be that, though iodine is a fairly weak deactivator. Sterics are against it too


text:
Determining the reduction potential of a half cell reaction at a given pH

$E^\circ_\ce{Cu^2+|Cu} = \pu{0.34 V}$.
  What will be reduction potential at $\mathrm{pH} = 14$ for the same couple?
  Given that $K_\mathrm{sp}$ of $\ce{Cu(OH)2}$ is $10^{-19}$.

My Attempt
I think that this problem might be solved by using the Nernst equation
$$E = E^\circ - \frac{0.0591}{n}\log_{10} K,$$
but I don't know how do I apply that.
We have the $E^\circ$ and the number of electrons participating (which is equal to 2), but I don't know how do I determine the equilibrium constant.
Any help or hint would be appreciated.

Hint: try to link solubility product and $K$; start by writing the reaction of dissolution of copper(II) hydroxide.

Okay. But then what's the role of pH?

pH is going to affect that equilibrium.

I am getting $K = K_{\text{sp}}$. Which doesn't yield the correct answer.

I am so sorry. That does give me the correct answer. Made a calculation error. Thanks anyways.

If you think you figured it out, you can post an answer to your own question:)

The gist is that the standard potential is for a copper solution in its standard state, ie 1 molar $\ce{Cu^{2+}}$. Since the concentration of copper is so much less the reduction potential of a copper solution at pH 14 is going to be much greater. I'd guess so much greater than you'd have water hydrolysis instead of copper plating.

@MaxW but I am getting a negative answer :/ I don't understand how are you predicting a higher reduction potential. Please correct me if I am wrong anywhere.

The K you refer to relates to the overall cell reaction and should be Q, the reaction quotient. This is only equal to K when the potential difference between the two 1/2 cells has fallen to zero.
This question does not refer to a full cell, but to a 1/2 cell. The form of The Nernst Equation you must use, assuming we are at 298K, is:
$\rm E=E^\circ+\dfrac{0.0591}{z}log\dfrac{[oxidised\,\, form]}{[reduced\,\, form]}$
Where z is the number of moles of electrons transferred which, in this case = 2.
This becomes:
$\rm E=E^\circ+\dfrac{0.0591}{2}log[Cu^{2+}]$
The 1/2 cell in question is:
$\rm Cu_{(aq)}^{2+}+2e\rightleftharpoons Cu_{(s)}\,\,\,\,\,\,\,\,\,\,E^\circ=+0.34\,V$
At such a high pH the copper(II) ions will precipitate.
$\rm pH=14\,\,\,so\,\,\,pOH=0$
So $\rm c[OH^-]=1\,\,mol/l$
This will precipitate out the copper(II) ions as the hydroxide.
So we have:
$\rm Cu(OH)_{2(s)}\rightleftharpoons Cu_{(aq)}^{2+}+2OH_{(aq)}^-$
$\rm K_{sp}=[Cu^{2+}][OH^-]^2=10^{-19}$
$[\ce{OH-}]$ is 1 (pH is 14), so $[\ce{Cu^2+}]$ is equal to $\pu{e-19}$.
As you can see this is a very low concentration. Applying Le Chatelier's Principle to the 1/2 cell couple we would predict that reducing the concentration of Cu(II) will drive the position of equilibrium to the left. This would push out electrons and make the electrode potential of the 1/2 cell more negative.
Putting in the numbers into the Nernst Equation:
$\rm E = 0.34+\dfrac{0.0591}{2}\log \pu{e-19}$
$\rm E = 0.34- 1.12\,\,\,\,\,V$
$\rm E = -0.78\,\,\,\,V$
As predicted, it has been reduced considerably.


text:
Nitration of m-bromochlorobenzene

1) What could be the most plausible explanation to the very low percentage of the nitro group attacking the site between bromine and chlorine?
- is it solely because of steric hinderence
- or is it because of the inefficient pi-electron overlaps by the larger chlorine and bromine atoms suppressing resonance in the region between them
- or is it inductive effect
2) why is the first product (given above) more formed than the second. Is it because of the more electronegativity of Cl over Br which could lessen the stability of compound 2 by negative induction effect( which destabilised nitro group which itself is electron withdrawing).

Inductive effects scale down with increased distance from the electronegative atom. What's more acidic? trifluoroacetic acid or 3,3,3-trifluoropronanoic acid?

Zhe, yeah I got that point fine but my doubt is if it even the same for aromatic compounds? Or do the compounds by any chance distribute the positive charge over the ring to maintain homogeneousity?

Positive charge in what sense? The localization of charge into the $\pi$ system is probably fairly even, but the inductive effect is much stronger at the 2-position than the 4-position, relative to the halide.

Ok so is induction the sole reason for the cause? Is induction that strong and effective to cause only 1% of the last product? Thank you.

No, that's probably also steric. But steric effects should be similar between products 1 and 2.

Yeahhh but still there's a large difference between the percentage of 1 and 2 too.

That's from the inductive effect. Product 3 has double the steric interactions.


text:
Preference for tin or iron in the reduction of nitrobenzene
My NCERT book(Class 12) said the following for reduction of nitrobenzene using $\ce{Sn/HCl}$ and $\ce{Fe/HCl}$ :

Now, I have been following this mechanism for this particular reduction, and it seems to me that the metal is getting oxidized to +2 state wherever it is shown that 2 electrons are being transferred. If that is the case, then:
•Why is $\ce{Fe + HCl}$ being preferred?
•Because I looked it up on Wikipedia and it says that $\ce{SnCl2}$ also gets hydrolysed to liberate $\ce{HCl}$ as follows:
$\ce{SnCl2 (aq) + H2O (l) ⇌ Sn(OH)Cl (s) + HCl (aq)}$
Or have I interpreted the mechanism incorrectly?

Probably preferred for cost reasons, iron being cheaper than tin

@Waylander But the way the book has stated it,it seems that this regeneration of HCl takes place only with Fe. So is there a problem with the yield in case of Sn+HCl?

I can see no problem with your. My guess is that the preference the authors express is for Fe/HCl over catalytic hydrogenation.

The book's reasoning is correct. Fe/HCl gives $\ce{FeCl2}$ on reaction and it gets hydrolyzed by steam vapors to produces more hydrochloric acid and hydrogen to push the reaction forward thus making the reaction self-sustaining to produce more aniline from nitrobenzene.
$$\ce{Fe + 2HCl ->[160°C,CH3OH] FeCl2 + H2}$$
$$\ce{3FeCl2 + 4H2O(steam) ->[Hydrolysis] Fe3O4 + 6HCl + H2 }$$
whereas $\ce{SnCl2}$ gives $\ce{Sn(OH)Cl}$ and 1 mole of hydrogen chloride which is insufficient and it gets exhausted after sometime.
The other thing is cost factor as @Waylander said. Iron scrap is way cheaper than tin.
Both Sn/HCl and Fe/HCl reduction(Bechamp reduction) is replaced by catalytic hydrogenation due to the latter being more efficient and more productive in industial production. Fe/HCl reduction is now an academic interest and is studied as a route to produce iron oxide based pigments.


text:
How can the ocean act as a buffer?
I was reading up on the buffering effect of the ocean and I was quite confused about how the buffer is able to do anything. The buffer I will refer to is below:
$$\ce{CO3^{-2} +H2O <=> HCO3 + OH^{-}}$$
$$\ce{HCO3^{-} +H2O <=> CO3^{-2} + H3O^{+}} \tag{2}$$
When we have an excess of $\ce{CO2}$ this reaction (between sea water and carbon dioxide) will shift to the right:
$$\ce{CO2 +H2O <=> H2CO3}$$
Thus it should also protonate
$$\ce{H2CO3 + H2O <=> HCO3^{-} + H3O+} \tag{1}$$
However the products of this ionization produce species that are on the opposite sides of Equation 2. Therefore how can the equilibrium shift away to counteract this increase as both sides would have had an increase? I've heard that what should happen is that Equation 2 shifts to the left and decreases the concentration of carbonate ions but I don't understand why it shifts.

I took the liberty of numbering your equations so that they can be referred to. Now to the point: equilibrium (2) is not as important as (1).

But without (1) wouldn't there be no buffering action - which would defeat calling it an "ocean buffer"?

That's right, (1) is important. (2) isn't.


text:
Why other benzene derivatives do not undergo nitrosation
Solomon and Fryhle $11 th$ ed :
Most other benzene derivatives except phenols and tertiary aromatic amines do not undergo C- nitrosation reaction.
I agree that if there is a primary or secondary aromatic amine, the reaction will follow other routes but in case of other activating substituents EAS is the only route left for the electrophile $\ce{-^+N=O}$ for example with anisole.
Also it seems to contradict  ron's answer
So which one is correct?

I don't see any contradiction.

Anisole can be nitrosated in good yield reference here.  This, and other references, note that the PhOMe group is unstable to the reaction conditions being cleaved to give phenol.
m-Xylene can be nitrosated in good yield, o-xylene and toluene in modest yield in TFA or sulfuric/acetic acid mix under nitric oxide reference here
Nitrosonium tetrafluoroborate will also nitrosate a range of electron-rich aromatics, including anisole and xylenes reference here
This suggests to me that the authors of the textbook quoted are in error, as two of the references predate the publication date of the textbook.


text:
Is the 12-6 Lennard Jones potential a good description of van der Waals interaction energy?
It seems to me that the Lennard Jones potential oversimplifies the more complex nature of van der Waals forces. However, in most of the MD simulation package, this is generally accepted.

Of course LJ is not precisely accurate, but it is computationally cheap. This makes it good enough for many applications.


text:
Volume change in positive deviation from Raoult's Law
Let us take a mixture of ethanol and acetone as an example. When ethanol is mixed with acetone, the hydrogen bonding between ethanol molecules gets disturbed as actone molecules get in between the ethanol molecules. This results in $\Delta_\text{mix} H\gt0 $ as heat must be supplied to make the solution from its constituents (as attraction is less) but I am not sure about $\Delta_\text{mix} V$. This page says it is greater than zero because volume expands on mixing. But I think it should be less than zero because the observed vapour pressure is more than that predicted on the basis of theoretical calculations, so a greater amount of solution will be present in the form of vapour, hence volume of solution will be less (than the sum of the volumes of its constituents). Where am I wrong?

An intuitive explanation can be found on the page you referenced. The lessened attraction between the constituents effectively means that the average "intermolecular distance" would be greater after mixing than it was before, hence the increased volume. The vapor pressures before and after mixing are not directly relevant here.

I agree with the increased intermolecular distance, but more vapour pressure means more amount of vapour over solution. As total amount of solution + vapour is constant, this means less solution, hence less volume of solution, so there are two conflicting processes which one dominates?

@drake01 By the volume change is meant the total volume after mixing, at the same temperature versus sum of volumes of the same amount of liquids before mixing. It should be obvious that evaporation changes the volume,  as the there is less of liquids.

I'm a little late to the party here, but wanted to chime in.
The link you posted isn't working anymore, but you appear to be making too many assumptions about the system being measured.  Nothing about the measurement of the volume change implies there is necessarily an open space above the liquid for vapor to form.  Imagine an open beaker of ethanol with a small disk floating on top so that no molecules can escape upwards, but it can move up and down with the change in liquid level.  This would also keep the pressure of the liquid constant (although it doesn't really matter since the liquid is essentially incompressible) .
If you were able to inject a certain amount of acetone into this beaker without opening it to the atmosphere for very long, the resulting volume of liquid in the beaker would be greater than the sum of its parts. Hopefully it makes sense that if there is no space for liquid to form, the positive volume change of mixing can show itself.
As far as the idea that more vapor would form to decrease the volume of liquid, realize that evaporation is a time-dependent process, so likely the mixed liquid would reach an effective equilibrium state where it's volume change could be observed before the extra liquid has time to evaporate.  It is true that it would evaporate faster if the effective vapor pressure (technically "fugacity" or "chemical potential") were higher, as the driving force for mass transfer would be greater, but this is also very dependent on the way the system is specified.  Technically if you left the container open in a large room forever, it would all evaporate in either case.

but I am not sure about $\Delta_{mix}V$. [...] I think it should be less than zero because the observed vapour pressure is more than that predicted on the basis of theoretical calculations

$\Delta_{mix}V$ refers to the change in the volume of the liquid with mixing, compared to the pure components. The vapour plays no role in this computation. Note also that the lower vapour pressure relative to the pure components means that the free energy of the liquid mixture is lower than the arithmetic average over the free energies of the pure components. This can certainly be interpreted as being due to stronger association between the components in the mixture than in the average of the pure liquids, but does not necessarily imply a specific direction of change (positive or negative) in volume.


text:
Why is it vital that atoms in the same sample are all given the same charge? Mass spectrometry
When being tested in the spectrometer why is it vital that the atoms in the same sample are given the same electrical charge e.g all are c+1 or all are +2 etc

As far as I know, the degree of ionization of particles in mass spectrometry depends on the energy supplied. In other words, it's tunable. However, you would want to use your electron beam (let's say it's electric ionization) as an analytical tool, not a sledgehammer. It makes the MS result analysis more feasible if the ionization energy is just enough to remove as few electrons as possible. Simplifies the spectrum, so to speak. Is that what you're asking about?

This is a misconception that during ionization, each ion must get a +1 or +2 charge. It is desirable to have +1 so m/z corresponds to the molecular mass of the ion or its fragment. The state of ionization heavily depends on what you are using to ionize your target analyte. For example if you use electron impact, you will see fragments mostly with +1 charge. However in other techniques called matrix assisted laser desorption (MALDI), the state charge can be as high as +12 and so on. They are detected as such. In electrospray ionization, +2 is also common.
There is no way a mass spectrometer "knows" a priori the state of charge of an ion. All charge ratios are detected. However it can distinguish the sign. It depends on the user whether negative or positively charged ions reach the detector. Some ions are better detected in positive mode and some in negative mode.

In mass spectrometry, the position of each peak you observe depends on the value m/z (i.e. mass over charge). If you have only z=1, then the position will depend only on the mass (m/1 = m).(1)
In a mass spectrometer, you will generate a mixture of charges at first but ions with a charge greater than 1 are filtered out.
(1) Obviously, it also works with a -1 charge.


text:
Determination of solubility equilibrium using galvanic cell reactions
This is the problem:
We can determine the solubility equilibrium for silver bromide using cell:
Ag (s) | AgNO3 (aq) || KBr (aq) | AgBr (s) | Ag (s)
We know that:
$\text{AgBr} + \text{e}^- \rightarrow \text{Ag} + \text{Br}^- \ \ \ \text{E}^0=0.095 \text{V}$
$\text{Ag}^+ + \text{e}^- \rightarrow \text{Ag}\ \ \ \text{E}^0=0.799 \text{V}$
First of all isn't that cell diagram incorrect? As far as I understand it is the silver reducing from silver nitrate to metallic silver thus being cathode. And vice versa for the silver bromide that is being produced more as silver is oxidating. Shouldn't the cell diagram be like this instead:
KBr (aq) | AgBr (s) | Ag (s) || Ag (s) | AgNO3 (aq)
Anyways, if we consider the Nernst equation:
$$E = E_0-\frac{RT}{nF}\ln{k}$$
and assume that the galvanic cell is in equilibrium when there is no voltage (i.e. $E=0$). We can then write:
$$\ln{k}=\frac{E^0nF}{RT} \\ \leftrightarrow k=e^{\frac{E^0nF}{RT}}$$
Then plug in the numbers:
$$\ln{k}=\frac{(-0.095+0.799)\ \text{V}\cdot 1 \ \text{mol}\cdot 96485.31 \ \frac{\text{C}}{\text{mol}}}{8.31451\ \frac{\text{J}}{\text{mol}\cdot \text{K}}\cdot 293.15 \ \text{K}} \\ \leftrightarrow
k= 1.26754\cdot 10^{12}$$
But the right answer is:
$$1.26 \cdot 10^{-12}\ \text{mol}^2/\text{dm}^6$$
Any ideas what am I doing wrong?

The leftmost and rightmost parts of the galvanic cell notation are supposed to be electrodes, not electrolytes (no KBr (aq) and the likes of it there). What is more, the double vertical line represents the interface where the junction potential has been eliminated ((c) Atkins), i.e. salt bridge, which in this case is the interface between electrolytes. The left hand side of the cell is anode, where the oxidation occurs, and the silver ions are reduced at the cathode, which is written on the right hand side. It seems to me that the correct notation is:
Ag (s) | AgBr (s) | KBr (aq) || AgNO3 (aq) | Ag (s)
In turn, the cell reaction is $\ce{Ag+ + Br- -> AgBr}$, and the EMF of this reaction is +0.704 V. It is not surprising that the equilibrium constant of this reaction is much larger than unity, otherwise silver bromide wouldn't be so stable and so insoluble. What you're missing is that the solubility product, which you want to calculate, is an equilibrium constant of the inverse process - not the formation, but the dissolution of $\ce{AgBr}$. So take the inverse of the equilibrium constant which you obtained from the galvanic cell EMF, and you'll get $7.9\cdot10^{-13}$, which is pretty close to the reference values.

Consider the net reaction you are looking at:
$$\ce{Ag+  + Br- -> AgBr}, E^{\circ} = 0.704\ \mathrm{V}$$
The equilibrium, not surprisingly, lies to the right, which we can tell from the standard potential and from knowledge about the solubility of silver bromide.
This means that you expect the concentrations of reactants to be quite small. When they then go into the equilibrium constant, I expect a very large equilibrium constant. In other words, your answer is much more consistent with the chemical situation than the expected answer, based solely on the magnitude of the two answers.


text:
Inorganic chemistry handbook with reaction lists
There is a handy handbook on common inorganic substances' properties called "Reactivity of Inorganic Substances" (eds. R. A. Lidin, V. A. Molochko, and L. L. Andreeva). It was originally written in Russian and then translated into English (see the review here). The entries in this handbook go like this (sorry for the Russian, couldn't find the English version):

You see that the compound's name is followed by the brief summary of its physical properties (color, solubility in water), general notes on its stability under ambient atmosphere, and then the list of the chemical reactions which were considered by the authors to be the most important or illustrative for this particular substance.
The question is - are there any reputable handbooks that are similar to this one, with not only the summary of the physical properties, but the summary of possible chemical reactions as well, or is this book one-of-a-kind?

There was much good chemistry written in other languages than English, and to some extent still is.  Skipping sources in Russian were like omitting important ressources, and not only from Russia itself.

In the 1970s, learning Russian/German was one of the requirements to do a PhD in chemistry. If one had to become an electrochemist, Russian was highly recommended. Still many top universities in USA require a foreign language exam for a PhD in mathematics. It is sad that most of the scientists don't bother to learn other languages now.

There is Holleman-Wiberg, an originally German (the very first edition was actually dutch, in 1898) compendium of biblical size, including a ~5-20 page chapter on every main group and side group element, with properties and reactions for most all simple inorganic compounds. And of course the CRC Handbook of Chemistry and Physics.

Indeed Holleman Wiberg is another giant contribution. It was quite recently translated into English. Most of the people are also not aware of Mellor's Treatise on Inorganic Chemistry from 1920-30s. It is at least 9-10 volumes by a single author- element by element. One of the best ever.

@M.Farooq They translated it again? There was an English edition already in the first quarter of the last century. I love these books which, to avoid making them in two or even three volumes, are printed on this very fine, thin paper, like telephone books, or bibles.

I meant 2001. I have seen it in the library. However I was not aware of the early Holleman's translation.

An Intro in Russian edition of Reactivity of Inorganic Substances by Lidin et al. claims its uniqueness:

Настоящее пособие носит информационно-справочный характер, не имеет аналогов в отечественной и зарубежной химической литературе, охватывает все разделы неорганической химии, изучаемые в химических вузах.
(English) This reference handbook has no analogues among domestic and foreign chemical literature, covers all inorganic chemistry sections studied in university chemistry course.

However, Turova's Inorganic Chemistry in Tables comes pretty close.
It doesn't only list reactions per element groups, but also arranges them in visual blocks; available both in Russian and English [1, 2]:

References

Турова, Н. Я. Неорганическая Химия в Таблицах; Высший химический колледж РАН: Москва, 1997.
Turova, N. Inorganic Chemistry in Tables; Springer: Heidelberg ; New York, 2011. ISBN 978-3-642-20486-9.

I am not aware of a list like this however, the biggest database on Earth of inorganic materials/reactions is the Gmelin Handbuch der Anorganische Chemie. Obviously in German since late 1770s and specific to element by element. The later editions started in English since 1970s. It is descriptive unlike the reference you have mentioned. Reaxys (Elsevier) has taken over Gmelin Handbuch and its organic counter part Beilstein. All in English.


text:
N-heterocyclic carbene metal complexes. Question about the delocalization of the nitrogen lone pairs
I am wondering why nitrogen lone pairs are delocalized in NHC metal complexes. What I can think of is, when complex is formed, plus charge is formed on the carbon atom and therefore delocalization from the nitrogen atoms. But there should be then "plus" sign like in the picture below.

But these kind of carbene complexes are always depicted without the plus sign. Can anyone please tell me what I am missing out?

They're just depicting the merged resonance structures. Just draw out one of the resonance structures, and it should be perfectly clear. This is basicially a N,N-disubstituted imidazoline.

Imidazolium ion is depicted normally (with the plus sign) but NHC metal complexes are not. As I said I think NHC metal complexes should be written with plus sign, but normally they are written without it and I was wondering why.

To my knowledge that is correct. You only care about the overall charge of the organometallic complex anyway, and you're just polluting the structure with charges. Should there be a charge on the chloride? They're X-type ligands but the phosphine is L-type. Nah, just draw brackets around the whole thing as needed and write the total charge.

Note this exact issue in this example. With the formal positive charge in that imidazoline, what is the overall charge on the complex? If one answers correctly with neutral, then why, in the name of all that is holy, is there a positive charge with no corresponding negative charge? The answer is that it's a carbene, and there's a hidden negative from that, but again, we've chosen not to show the negatives so why bother with the positive?

Some parts of this answer probably border a bit on personal opinion, but I will at least try to justify having such an opinion.
The first thing to do is to actually examine the heterocycle. What kind of ligand is it?

This is a carbene. It donates two electrons via the lone pair on the carbon in plane with the ring. Notice that there is a smear of positive charge from the delocalization of the $\pi$ system between the N-C-N system. That is balanced by the negative chage on the carbon, however. Overall, this ligand is neutral, so it is an L-type ligand.
What's the overall charge on this metal complex? This is a ruthenium(II) ion (2 chlorides, everything else neutral), so this is a $d^{8}$ metal ion. You have 2 electrons from the NHC, 2 from each chloride (4 total), 2 from the phosphine ligand, and 2 from the carbene (the double bond is back-bonding). That gives 10 electrons from ligands, so this is a fine-looking neutral 18 electron complex.
For a complex that's overall charge zero, why is there a plus charge hanging out by itself? Well, it turns out that we're extremely sloppy with charges. For example, the chlorides are X-type ligands but the phosphine is not. Do we stipulate this difference in the structure? No we do not, though we use it during electron counting. The authors here felt that the smear of positive charge of the carbene, which is how you would draw the structure in isolation, should be shown but chose not to show the negative charge of the carbene, which is the case by convention. That seems somehow inconsistent to me because the overall charge of this ligand is neutral.
Look at these other representations:

Mendeleev Communications, 2019, 29 (1), pp 38-40. (10.1021/om3009298)

Organometallics, 2012, 31 (23), pp 8355–8359. (doi: 10.1016/j.mencom.2019.01.011)
Both representations of the NHC are used: without explicit $\pi$ system and with.
Note that neither of them say anything about the positive charge. This seems more consistent to me, and these authors, at least, agree with me.

In the first NHC structure, there should be a C, so that it's not confused with $\ce{CH2}$.


text:
What is the velocity of de Broglie wave?
Here is my calculation:
 note: $ν$ – frequency, $V$ – velocity of wave, $v'$– velocity of object.
$$
\begin{align}
λ &= h/mv' \tag{1}\\
mc^2/h &= ν \tag{2}\\
λν &= V \quad \text{(wave equation)} \tag{3}
\end{align}
$$
Substituting 1 and 2 in 3 we get:
$$h/mv' \cdot mc^2/h = V$$
Simplifying we get:
$$c^2/v' = V$$
However, according to the book, the answer is $hv/mc$.
What is wrong with my answer?

What is equation (2)? This appears to be a mass up of mass-energy equivalence and computing the energy of a photon.

Some de Broglie waves are not (in any meaningful sense) moving: they are standing waves (like electron orbitals in an atom) where it doesn't make sense to talk about how fast the wave moves.

I think the problem arises partially from the confusing notations.
"Velocity of wave" $V$ is essentially $c$, "velocity of object" $v'$ is usually denoted with the speed of particle $v$ and the energy of photon $E$ is for some reason expressed via mass–energy equivalence $E = mc^2$ when it should be just kinetic energy of the particle $E = mv^2/2$ (but the latter is anyways irrelevant here).
So, your equation turns into this:
$$c^2/v' = V$$
$$c^2/v = c$$
$$c = v$$
which just states that the velocities of both wave and particle are equivalent.
I suggest you just combine the de Broglie equation
$$λ = \frac{h}{mv}$$
and the relation between wavelength $λ$ and frequency $ν$
$$λ = \frac{c}{ν}$$
so that
$$λ = \frac{h}{mv} = \frac{c}{ν} \quad \implies \quad v =\frac{hν}{mc}$$


text:
How do we determine the proportions of hydrogen to helium (to other elements) that make up the Sun given the absorption spectra?
I am reading about the composition of the Sun and how it is made up of mostly hydrogen (~90%) and helium (>10%) and other elements with much smaller quantities. The explains that different elements give different absoprtion spectra (different lines) yet I do not understand how the proportions were calculated. Can someone explain this to me?

Lines differ not only in terms of position, but also their intensities.

Also I think that such an analysis should assume many things as well as modelling the star. I think the only access is to what directly radiates, photosphere and the exterior

I'm voting to close this question as off-topic because it would be better on Astronomy.SE

Spectral lines and how they work is clearly a part of chemistry even when applied to the sun. After all one of the elements–helium–was first discovered in the spectral lines of the sun.

I agree is chemistry but as I said before there is no way to know the composition of  the sun in a straightforward way. It would be like to look at an almost opaque  bottle to decide its content. Still more direct evidence can be attained for the photosphere, chromosphere and corona.

@Mithoron the question is on spectroscopic methods...just because it is not your preferred area of study does not mean it is off-topic. I could have just asked how can the proportions of an element be identified through line spectroscopy giving the Sun as an example..


text:
Calculate pH of a buffer solution

Calculate the pH in a buffer prepared from $\pu{50 mL, 0.30 M}$ formic acid ($\ce{HCOOH}$) and $\pu{30 mL, 0.40 M}$ sodium formate ($\ce{HCOONa}$).

My way of solving:
$n_{\ce{HCOOH}} = 0.05 \times 0.3 = \pu{0.015 mol}$
$n_{\ce{HCOONa}} = 0.03 \times 0.4 = \pu{0.012 mol}$ (which means $\ce{NaOH} = \pu{0.012mol}$)
$\ce{HCOOH + NaOH <=> HCOONa + H2O}$
After doing a BCA-table, I got that the remaining moles of HCOOH = $\pu{0.003 mol}$
and HCOONa = $\pu{0.012 mol}$
Calculating the new concentration:
[HCOOH] = $\ce{0.003 / 0.08 = 0.0375M}$
[HCOONa] = $\ce{0.012 / 0.08 = 0.15M}$
Using the Henderson–Hasselbalch equation:
pH = pKa + log [HCOONa]/[HCOOH] = $\ce{3.75 + 0.60 = 4.35}$
So my answer is that the pH of the buffer is $\ce{4.35}$
Answer key
However, in our answer key, they found the new concentration of HCOOH and HCOONa by simply taking:
[HCOOH] = $\ce{(0.05 *0.3)/0.08 = 0.1875M}$
[HCOONa] = $\ce{(0.03 *0.4)/0.08 = 0.15M}$
and when using those concentrations in the Henderson–Hasselbalch equation they get the answer pH = $\ce{3.65}$
Question
Is the answer key correct? If we are adding a base, then the pH should increase to become more basic, not decrease as in the answer key. If the answer key is correct, why does the pH decrease?

The answer key's method is certainly valid given that the system is buffered. I suspect that your calculations for final concentrations are incorrect. For one thing, you should be doing those calculations on concentrations, since the equilibrium expression all use concentrations, not amount of substance.

@Zhe But shouldn't the pH increase and become more basic?

@katara - No look at your calculation of the moles. There are 0.015 moles of formic acid and only 0.012 moles of sodium formate. Thus you would expect the solution to be more acidic than the pKa value.

@katara - you shot yourself in the foot by going off on a tangent and thinking about NaOH. It is true that if you started with 0.027 moles of formic acid and 0.012 moles of NaOH then you would end up with 0.015 moles of formic acid and 0.012 moles of sodium formate, but why go "backwards" to solve the problem?

Is the answer key correct?

Yes, they have a very good approximation of the pH, and their way of solving it is straightforward and conventional.

After doing a BCA-table, I got that the remaining moles of HCOOH = 0.003moles

The sum of formic acid and formate has to remain 0.027 mol. You could start out with 0.027 mol of formic acid and turn some of it into formate by adding NaOH. Or you could just mix them as described in the problem. However, you are claiming that there is a total of 0.015 mol of formic acid plus formate, so you lost some atoms there. As the comments state, there is no reason to think about where they got the sodium formate (stock room?).

If the answer key is correct, why does the pH decrease?

Decrease or increase depends on your frame of reference. If you compare it with a 1:1 buffer where the pH = pKa, it decreased because there is a higher concentration of acid than conjugate base.
If you compare it to the pH of 0.3 M formic acid (roughly pH = 2.1), then it increased because you added conjugate base (neutralized and diluted the acid).


text:
Is the differential, dp, exact or not?

Consider the differential $$\mathrm dp=\frac RV\,\mathrm dT+\left(\frac{2a}{V^2}-\frac{RT}{V^2}\right)\,\mathrm dV$$ (where $a$ is a constant value)
(a) Determine whether the above differential, i.e. $\mathrm dp$, is exact or not. Show all your steps and evaluation of the appropriate partial differentials!

I have no idea on how to even start this. As far as I can tell the differential is exact, but I don't know how to prove or show it. I'm really struggling to show the steps involved. I would appreciate any advice and thank you very much in advance.

I'm voting to close this question as off-topic because it is a mathmatics question.

Not all chemistry is done in a vial.

Your function $p$ is a function of the independent variables $V$ and $T$ i.e., $p(V,T)$. The other variables, $a$, and $R$ are constants.
Let me rewrite the differential as
$$\mathrm{d}p = A(V)\,\mathrm{d}T + B(T,V)\,\mathrm{d}V$$
where
$$A(V) = \frac{R}{V}$$
and
$$ B(T,V) = \left( \frac{2a}{V^3} - \frac{RT}{V^2} \right) $$
A differential is exact if
$$\left( \frac{\partial A}{\partial V} \right)_T = \left( \frac{\partial B}{\partial T} \right)_V $$
Note that subscripts denote that the variable is being held constant during the partial differentiation. The problem already told us what our $A$ and $B$ terms are, and I wrote them out explicitly above. Now, we just need to take the appropriate partial differentials and compare. If they are equal, then $\mathrm{d}p$ is exact!
Writing out the formulas took awhile, it seems a shame to stop now... A little bit of differentiating, remembering to hold the appropriate variables constant during differentiation leads to
$$\left( \frac{\partial A}{\partial V} \right)_T = -\frac{R}{V^2}$$
$$\left( \frac{\partial B}{\partial T} \right)_V = -\frac{R}{V^2}$$
If the two partial derivatives are the same, the differential is exact. I will let you be the judge.

I don't think you should basically have revealed the full solution.

In my experience people mess up the differentiation and then think they did everything wrong

Full solutions are fine! We encourage them: https://chemistry.meta.stackexchange.com/questions/4287/hint-answers-revisited

@orthocresol Well, in this case it's rather there shouldn't be an answer as I don't see how this question wouldn't violate HW policy.

@Mithoron, that is a fair point and I will leave it to the community to decide whether to close it. However, given that there is an answer, I would much prefer that it is good (like this) and not "HINT: $show$ $\partial^2p/\partial T\partial V = \partial^2p/\partial V\partial T$".

No way around it being a Homework question, but I answered it because there was a time when I had the same question. I tried to provide a means for understanding the solution. The final answer is not that important, but the concept... I hope I showed the concept. If I did, then it is a worthwhile question to have, and solution to have.

@orthocresol Hmm, such answer wouldn't do, but a comment... Oh, well; unfortunately now it went to HNQ and question is rather undeservedly getting  upvotes :(

For a given function, $F(x,y,z,...)$, it's differential $\text{d}F$ is given by:
$$
\text{d}F = \left(\frac{\partial F}{\partial x}\right)_{y,z} \text d x +\left(\frac{\partial F}{\partial y}\right)_{x,z} \text d y  + \left(\frac{\partial F}{\partial z}\right)_{x,y} \text d z  +\; ...
$$
To say that a differential is an exact differential is to say that is if the differential of a function and hence is of the form given about.
For the case given:
$$
\mathrm dp(T,V)=\frac RV\,\mathrm dT+\left(\frac{2a}{V^2}-\frac{RT}{V^2}\right)\,\mathrm dV
$$
If $\mathrm dp$ is an exact differential, that would mean that:
$$
\left(\frac{\partial p}{\partial T}\right)_{V} = \frac RV\ \text{and } \left(\frac{\partial p}{\partial V}\right)_{T} = \left(\frac{2a}{V^2}-\frac{RT}{V^2}\right)
$$
There are two equivalent way to determine whether this is true, you can integrate the partial derivatives of $p$ to recover a form for $p$, or you can differentiate each term once more to so that both give identical values for the mixed second derivative.
Integration
Taking indefinite integrals of the suspected derivatives:
$$\int \frac RV\ \mathrm d T  = \frac{RT}{V} + g(V) \\
\int \left(\frac{2a}{V^2}-\frac{RT}{V^2}\right) \mathrm d V = \frac{-2a+RT}{V} + h(T) \\
\to p(T,V) = \frac{RT-2a}{V} + c
$$
Differentiation
It is typically easier to compare the suspected derivative by differentiation.  If $p$ is  a true function of $T$ and $V$, by the symmetry of mixed derivatives:
$$
\frac{\partial }{\partial T }_V \left(\frac{\partial p }{\partial V }\right)_T = \frac{\partial }{\partial V }_T \left( \frac{\partial p }{\partial T } \right)_V
$$
Assuming:
$$ \left(\frac{\partial p}{\partial T}\right)_{V} = \frac RV\ \\
\to \frac{\partial }{\partial V }_T \left( \frac{\partial p }{\partial T } \right)_V = -\frac{R}{V^2}
$$
Assuming
$$ \left(\frac{\partial p}{\partial V}\right)_{T} = \left(\frac{2a}{V^2}-\frac{RT}{V^2}\right) \\
\to \frac{\partial }{\partial T }_V \left( \frac{\partial p }{\partial V } \right)_T = -\frac{R}{V^2}
$$
Both terms of the differential $\mathrm dp$ imply the same mixed second derivative, hence  $\mathrm dp$ is an exact differential.


text:
How to determine steady state of multi-step reaction with both reversible and irreversible steps?
I am trying to determine the relative concentrations of $\ce{U1}$, $\ce{U2}$, and $\ce{E}$ when the following reaction is at steady state. Note that $k_t$ is the zero order rate constant of new $\ce{U1}$ entering the system and $k_d$ is the first order rate constant of substances leaving the system.
$k_1$ and $k_2$ are the first order rate constant for the forward and reverse reaction, respectively, of $\ce{U1}$ turning into $\ce{U2}$, and $k_e$ is the first order rate constant for $\ce{U2}$ reacting to $\ce{E}$. First order reactions are all first order in the respective reactant.
All the individual rate constants are known but the concentrations are not.
Can anyone point me in the right direction for how to determine the relative concentrations at steady state? I'm imagining an equation or set of equations such that when values for each rate constant are plugged in, the concentrations of the products are determined.

I find this rather confusing. I'm a chemist not a chemical engineer. It seems however that "k" is being used for rate constants and flow rates. So kt and kd seem to be flow rates and k1, k2, and ke seem to be rate constants. // I'm imaging some kind of reaction chamber with an input and output assuming complete mixing at any instant. // I don't get how kd relates to U1, U2, and E.

In other words I'm thinking that kt is a flow rate like 1 liter per minute where U1 has a concentration of 0.100 mole per liter. Hence the flow rate of kd must be 1 liter per minute also, and for the output [U1] + [U2] + [E] = 0.100 moles per liter since all of the reactions are 1:1.

Thanks to Karsten's comments clarifying the differences between equilibrium and steady state, I found this great resource:
https://www.youtube.com/watch?v=hCE_-GNnnMA
Although the math is slightly more complex than in the example video, the logic is the same:
at steady state, $dX/dt = 0$, so each concentration $U_1, U_2$, and $E$ can be calculated.
$\frac{dU_1}{dt}=k_t-U_1 k_1+U_2 k_2-U_1 k_d = 0$
$\frac{dU_2}{dt}=U_1 k_1-U_2 k_2- U_2 k_e-U_2 k_d = 0$
$\frac{dE}{dt}=U_2 k_e- E k_d = 0$
Solving this system of three linear equations for the three unknowns gives:
$U_1=(k_t+\frac{k_1 k_2 k_t}{k_1 k_e+k1 k_d+k2 k_d+k_e k_d+k_d^2})\times\frac{1}{k_1+k_d}$
$U_2=\frac{k_1 k_t}{k_1 k_e+k_1 k_d+k_2 k_d+k_e k_d+k_d^2}$
$E=\frac{k_1 k_t k_e}{k_1 k_e k_d+k_1 k_d^2+k_2 k_d^2+k_e k_d^2+k_d^3}$

Now that you got your answer, could you edit the question to tell us the context of the problem? The question was closed, and if it remains closed, it will be deleted and others won't be able to learn from your thought process.

Karsten, it looks like my original wording has been pretty extensively rehashed and generally made more clear. What would make this more clear?

I think the problem is clear, and your answer is clear. What is not clear is what the context is. Is this a transport problem or a series of chemical reactions? What do U1, U2 and E represent - conformations of the same molecule, different molecules, different compartments? Why is the rate for the decay of U1 identical to that of U2 and E? If you add the specifics why your are interested in this model, it would be more relevant and easier to see what the assumptions are that led you to the model. It's great that it also contains the abstract description to use it on other systems.

Can anyone point me in the right direction for how to determine the relative concentrations at equilibrium?

This system will not reach equilibrium as long as there are irreversible reactions removing U2. If someone claims to have found an equilibrium state, the rates associated with k1 and k2 would be equal, but the [U2] would decrease because of those irreversible reactions.

I know that kt=3kd must be true for the whole system to be in balance, but I have been struggling to figure out how to relate the other rate constants.

Usually, lowercase k is used for rate constants, with the rate of reaction given by a rate law that involves k, but also concentrations of the reactants and perhaps other species. The way you are talking about it, all reactions would be zero order. For zero order reactions, there is no equilibrium (a melting process has no equilibrium - if the temperature is above the melting point, everything melts, and if it is below the melting point, everything freezes).

I have read that the Keq for multi-step reactions is the product of each single equilibrium constant, but I'm not sure how to apply this with the irreversible reactions in the system.

You can't. There is the concept of steady state, though, where concentrations of intermediates remain constant while the forward and reverse rate of a given reaction are different. In this case, the rates increasing the concentration of an intermediate are equal to the rates decreasing the concentration of that same intermediate, but there is a net reaction. This is typical for living organisms (they can, however, reach equilibrium after they die).

RE: This system will not reach equilibrium as long as there are irreversible reactions removing U2. ... // Not really true. An equilibrium can be established between k1 and k2 assuming that kd and ke are negligible. // Also in reaction chambers it is typical to have a quasi-stable steady state even though there is no true equilibrium.

Sure, if kt, kd, and ke are zero, U1 and U2 will be at equilibrium.

Re: if kt, kd, and ke are zero, U1 and U2 will be at equilibrium. Not necessarily. It seems to me that this problem is about a reaction chamber that reaches some sort of quasi-steady state, not a true equilibrium.

Karsten, thanks for the thoughtful reply. You've helped me realize that I'm muddling the concepts of equilibrium and steady state in my mind. I now am clear on the differences and am looking for concentrations at steady state.


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Nucleophilic addition - Propanal vs Benzaldehyde
According to my school textbook, carbon atom of the carbonyl group benzaldehyde is less electrophilic than that of propanal because polarity of the carbonyl group is reduced in benzaldehyde due to resonance.
I can't understand how. If I draw the resonating structures, the oxygen atom takes the delocalised electron and not the carbon atom. So, resonance shouldn't affect the electrophilicity of the carbon atom. Moreover, benzene being a -I group, induces a partial positive charge on the carbon.
What is the correct reason?

Observe the correct resonance structures properly. See this https://i.stack.imgur.com/fK7LM.jpg

@SoumikDas Okay, but doesn't the carbon attached to the carbonyl get a positive charge in the last resonating structure? So, its electrophilicity should increase? Honestly, I don't exactly understand how resonating structures affect electrophilicity or nucleophilicity, so should I change my question to that?

I also didn't see the last structure exactly,instead of charge seperation like that,it will be a $\ce{C=O}$ bond there. So, forget about increase of electrophilicity, it will be less definitely.Now, see what resonating structures are basically delocalised $\pi$-electron density structures. If you have more resonating structures concerning the carbonyl carbon, in the resonance hybrid the $\pi$-electron cloud will be more towards the carbonyl, and due to such enhanced $\pi$-cloud, it will be difficult for the Nucleophile to approach resulting in decrease in electrophilicity.

@SoumikDas Oh I got it. I didn't know I was supposed to think in terms of $\pi$ electron cloud shift. Btw, is there a way to justify this decrease in electrophilicity just the way increase in electrophilicity in say, phenol, is justified (the lone pair of electrons of oxygen are delocalised in the ortho and para positions, making those positions more electrophilic)?

Firstly, the ortho and para positions of Phenol become rich in electron density and become more nucleophilic (not electrophilic). So, first clarify this. And the easiest way to look for enhanced or diminished electrophilicity is to observe the structure of the molecule. If there is more $\delta +$ charge on any atom in native state or in resonance hybrid, that centre is electrophilic, and if the charge on some other species is $\delta - $, it tends to become nucleophilic.So, simply, decrease of $\delta +$ charge or increase in $\delta -$ results in decrease in electrophilicity.

@SoumikDas Sorry, I got muddled while typing the last comment. I meant more nucleophilic for the example of phenol. Referring to the second part of your comment, in my example, the carbon of the carbonyl doesn't get any decrease in $\delta +$ or increase in $\delta -$ in the resonance hybrid, so it's electrophilicity shouldn't decrease? Or is the increase in $\delta -$ because the carbon is sp2 hybridised in resonating structures, causing it to become more electronegative? I have understood the logic in terms of shift in $\pi$ electron cloud

In the resonance hybrid the carbon becomes more enriched in s character, that's true.. And increase in s character results in increase in electron density due to increase in electronegativity...