Does deprotonation of a hexaaquaion continue until neutral charge

Let's look at an example: Hexaaqua aluminium, or $\ce{[Al(OH2)6]^{3+}}$

It is acidic and will undergo deprotonation of a ligand water molecule spontaneously:
$$ \ce{[Al(OH2)6]^{3+} <=> H+ + [Al(OH)(OH2)5]^{2+}} $$

You can even get it to deprotonate twice:
$$ \ce{[Al(OH)(OH2)5]^{2+} <=> H+ + [Al(OH)2(OH2)4]^{+}} $$

But, as indicated, both reactions are equilibrium reactions. So if the concentration of protons in solution is high enough, the hydroxo ligands will be reprotonated and become neutral aqua ligands.

If you add some hydroxide to the solution, you can precipitate aluminium hydroxide, since it is not soluble in water. While this might be a driving force for the deprotonation, it doesn't seem to be the case here.

So we have shown that the process does not necessarily continue until the charge of the complex ion is neutralized.

Now, concerning your question why this doesn't happen until all aqua ligands have lost a proton:

Again, a part of the solution is the equilibrium idea brought forward in the upper part. But there's another way to think about it: If you have a negatively charged ion, you also need a positively charged counterion. The only counterion that exists in the solution of water and the aluminium compounds is the freshly abstracted proton, which happily will bind back to the hydroxoligand.

There are also $\pi$-donor effects that are stronger in the hydroxo than the aqua ligand. So it might be favourable to have no more than three hydroxo ligands in an octahedral complex so no two $\pi$-donors are across each other.


How to capture the gas produced by combusting alcohols

If you look at the overall reaction, there is actually no gas produced, but there is a consumption of gas:

$\ce{CH3(CH2)_n-OH + \frac{3}{2} (n + 1)O2 -> (n + 1)CO2 +(n + 2)H2O}$

So 1.5 mol of oxygen are consumed, while only 1 mol of carbon dioxide is formed, so the volume will be less if the temperature is maintained.

To perform this experiment, you must collect the carbon dioxide in a solution of alkali and then titrate the solution.
The carbon dioxide will be converted to carbonate:

$\ce{2NaOH + CO2 -> Na2CO3 +H2O}$

A constant flow of air must be put into the system and the gas that comes out must be bubbled with a diffuser through the alkaline solution.
The air flow must be slow, so that all the carbon dioxide is washed out of the gas stream.

In the titration there will be two equivalence points, the first one of the excess of alkali and the second one of the carbon dioxide.
This one has a much smaller pH change.

Be sure to make a careful blank of the alkali, because it normally has some carbonate that will add to the value of carbon dioxide.


How to give an evidence in support of the arenium ion mechanism?

You say you've already drawn the energy profile diagram for your proposed mechanism , now you want to make sure that your reaction mechanism proceeds through the most stable transition state , ( Least activation energy ) and therefore , on the energy profile diagram , the peak occuring at the transition state should not be too high .Also you wanna make sure that your final products have energies lesser than the  reactants , because in that case , your $\Delta H$ would be $-ve$ and from the equation $\Delta G=\Delta H-T\Delta S $, there would be a greater chance your reaction moves forward.

You also should have a look at this


Can vapour pressure be used to 'generate' work for free?

I believe the vapour produced in the step 3 of you experiment is due to the particles with high order speed in the Maxwell distribution of speeds curve that escape because of their higher kinetic energies. When escaping, they will reduce the average kinetic energy of particles in the water sample, which will then reduce the net temperature of sample.
This will then prompt the system (water sample) to take energy, in the form of heat, from the surroundings to reestablish the thermal equilibrium. Hence taking back the work done by system on surrounding via expansion of the bottle through heat to maintain its temperature. Although, as you can see in the Maxwell distribution of speed, the particles with enough kinetic energy to escape from the water surface are very less at low temperatures.
So, if you take a system with just enough temperature and manage to isolate it from the surroundings, I think you should see a visible drop in the temperature of water proportional to the work done in expanding the water bottle.
Mathematical description of the above Explanation:
Let's say the total energy change taking place after the particles escape water surface is $\Delta U$ out of which $W$ is used to do work on the surroundings. The remaining energy ($\Delta U - W$) is used to establish an equilibrium temperature inside the bottle and hence is given back to the air-water system inside the bottle.
This step may be happening simultaneously with the next steps I am writing but I am assuming that the air- water system inside the bottle will reach an equilibrium first before any work is done. This is purely to make the math easier and because thermodynamic is the study of equilibriums rather than processes so it won't affect the results as long as this equilibrium is established even when we are studying the final change taken place.
So the net change in temperature inside the bottle is $$\frac{-W}{C_{system}}$$ where $C_{system}$ is the heat capacity of the system.
Net temperature difference between system and surrounding $$\Delta T=\frac{W}{C_{system}}$$
So the system will absorb energy in the form of heat to bring the temperature back to equilibrium $$Q=C_{system}\Delta T=>Q=C_{system}\biggr[\frac{W}{C_{system}}\biggr]=>Q=W$$
So we can see mathematically that the system will take back all the energy it gave to the surroundings by doing work in the form of heat.


Is cyclopropyl methyl carbanion also exceptionally stable?

Sigma resonance only works on cyclopropyl methyl carbocation (and its cyclopropyl derivatives). So I don't think that cyclopropyl methyl carbanion will be exceptionally stable.


Are artificial ingredients better than natural ones because they are made in laboratory conditions?

The question is by no means restricted to vanilla. It refers to any natural product that can be reproduced artificially. Ditto for the answer.

Sure, the artificial product might have the right chemical formula. Also, it can be made as pure as you want. The problem is that the natural product is not pure. It is a bunch of various compounds, and they all contribute to our perception of the flavour. You identify the compound that makes up 90%, investigate its structure, develop the synthetic route, and finally you hold it in your hands, and it feels just about right, but something is amiss. So you look for the next component, which constitutes 9% and hence is a lot harder to isolate, and after a long and tedious procedure you identify it, and repeat everything, and finally you end up with the product that feels totally right. Well, almost so. Is it the remaining 1% that is responsible for the difference? You never know until you try.

So it goes.


What's the explanation for the chelate effect?

The chelate effect is generally agreed to be a thermodynamic effect caused by the change in entropy upon binding of a bidentate ligand.

General description of the chelate effect:

Consider a coordination complex,  [M]L2, where L = a standard monodentate ligand. If we now add a bidentate ligand, two monodentate ligands are released.

The ligand substitution reaction in the forward direction is therefore generally favourable:  ∆S, the entropy term is positive due to two molecules reacting to become three, with this favourable entropy term having the effect of tending to make ∆G negative (negative ∆G is favourable, i.e. the reaction likely to proceed).[*]

Once a bidentate ligand is bound, it then becomes less favourable to go in the other direction (the entropy term would be negative due to entropy decreasing in the system, which in turn would tend to make ∆G positive, and hence the reaction unfavourable).

Example and thermodynamic data:

To give a more concrete example demonstrating the significance of the entropy term, consider the following reaction and associated thermodynamic data (taken from Physical Inorganic Chemistry, S. F. A. Kettle, 1996):

Clearly, the entropy term dominates, causing ∆G to be negative in both cases (hence causing the forward reaction to be favoured in both cases).

[*]: The caveat here is that we're assuming that enthalpy really isn't contributing a whole lot, this is clearly a vast over-simplification, and there are cases where enthalpy becomes important (i.e. when the complex is significantly more stable than the starting materials). We're also ignoring the fact that the act of binding a bidentate ligand itself has some unfavourable entropy, since we're containing the ligand and removing some conformational flexibility


Are allylic three degree alcohols oxidized by the Jones' reagent via a classical carbocation intermediate?

It makes sense. Jones's oxidation occurs in presence of a strong acid. Hence, the oxygen can get protonated and leave as a water molecule, giving rise to a tertiary carbocation. The charge can then delocalize, the secondary alcohol get formed, and this alcohol then undergo oxidation. The oxidation will probably drive the equilibrium towards the product.


Is it possible to predict the color of the flame test result based on the element's structure?

The question is why would you want to predict that? All elements have been screened for allowed and forbidden transitions ranging over the complete visible and non-visible spectrum. You can just look it up there. There are also multiple possible transitions which lead to the often complex spectra.

And then there is the problem with the actual jump. In a simple shell-model where you only add electrons in shells this jump is just to the next level. And you can introduce Rydberg-atoms and get element specific Rydberg-constants. But I remember to have tried that for a Lithium atom and if I didn't do a mistake back then it would have required quite a high energy level to get the red emission while in reality the principal quantum number (n) doesn't even change for lithium. You can see this pretty easily if you look at the energy diagram for the famous yellow sodium emission line. It's a jump from the $3p$ to the $3s$ orbital.

But in the end, as I said, they have all been calculated and confirmed. You can just look at lists on the net where every possible transition is given with the orbitals, energy, wavelenght and something like an intensity.

And engines like WolframAlpha even use that. You can just search there for let's say sodium emission spectrum and it will give you the spectrum. And there are also websites where all spectra are given.

I'm a bit confused with your question. It sounds like you don't have such a list. I guess, as you are talking about flame colors and not spectra you are only looking for the flame color itself. This should be mentioned in most analytical chemistry books and I think there was even a Wikipedia article with pictures. But at some point this will become quite hard. Lithium vs Calcium vs Strontium is such a case where I wouldn't be too sure if I could distinguish them by their flame color without looking at spectra. In the end the observed emission will be mixture of many weak and strong emissions.


Using acetone instead of acetonitrile on reverse-phase prep-HPLC

If you want to go completely green, you should use ethanol.
In many sectors it is considered the only green solvent (except for water of course).
Then, to be really green, you should make sure that it comes from fermentation, not from synthesis.
This quality exists.

The only problem with ethanol is the same as with acetonitrile, the price.
In this case it is due to the taxes that it normally has.
However, if you are working for some institution, it is probable that you can buy it tax free.

Another alternative is to use methanol.
It is not so green, but you can call it wood alcohol.
Methanol works very well in HPLC.
The only difference with acetonitrile is that the pressure in the system will be higher.
This is normally not a problem because the equipment is designed for it.

I prefer not to use acetone because it can give reaction with some compounds, or with itself to form isophorone or other condensation products.
As Waylander correctly points out in his comment, the UV cut off of acetone is also a problem in case you are using UV detection.

                 UV cut off (nm)
 acetonitrile        190
 methanol            205
 ethanol             210
 acetone             330

To change from one solvent to another you will have to change the gradient.


Explanation of the Selection rule in light of IR spectroscopy

The operator for interaction with infrared light is the permanent dipole of the molecule, $\mu$.  I don't know anything about representation theory, but in group theory, you decompose the molecule by its symmetry elements to find the eigenmodes.  The infrared active modes are those modes in which the permanent dipole changes over the course of a vibration.  An excellent reference is Symmetry and Spectroscopy by Harris and Bertolucci which should be available in your library.


Disproportionation Reactions of the Halogens: Writing balanced net ionic equations for four reactions. See details

(answer to original question)

No, you are not on the right track.

If bromate is a starting material, it should not appear as a product in reaction 1. Instead, consider how bromide and bromate can react with each other.

If iodate is a starting material, it should not appear as a product in reaction 2. Instead, consider how iodide and iodate can react with each other.

For reactions 3 and 4, a hint would be "disproportionation".


Calculating surface area from BET results

Your problem is not the unit conversion, but the units you have used. The adsorption cross-section of $\ce{N2}$ ($s$) has been found using BET method is $\pu{0.162 nm^2}$ (Langmuir). Yet, you have used $\pu{0.162 nm^3}$, more than a factor of $1 \times 10^9~\mathrm{nm/m}$.


When calculating the distance between molecules in a solution, is it reasonable to approximate a sphere around each molecule?

The distribution of molecules in solution is not uniform but has a distribution. The calculation of nearest neighbour distances from one molecule to another has been worked out a long time ago see,  S. Chandrasekhar, Rev. Mod. Phys. 1943, v15, p1.
Let $w(r)$ be the probability that the nearest neighbour occurs between distance  $r$ and $r+dr$. This must be the probability than no molecules exist closer to the victim molecule that one occurs at a distance $r$ and that this molecule exists in the shell $r$ to $r+dr$. Thus,
$$ w(r)=\left(  1-\int_0^r w(r)dr \right)4\pi r^2 n$$
where $n$ is the average number of molecules / unit volume. (1 mol/m3 $\equiv 10^3N_A/10^{27} = 0.6023 $ molecules / nm$^3$).
This looks hard to solve because $w(r)$ is also inside the integration, so it is necessary to isolate the integral first (by dividing both sides by $4\pi r^2 n$) and then differentiating both sides to find
$$\displaystyle \frac{d}{dr}\left(\frac{w(r)}{4\pi r^2 n}\right)= \frac{d}{dr}\left(1-\int w(r)\right) =-4\pi r^2 n \frac{w(r)}{4\pi r^2 n}$$
and the last step we effectively multiply by 1. The result is
$$ w(r)=4\pi r^2n\exp(-4\pi r^3n/3)$$
The average distance between molecules is
$$\langle r \rangle =\int_0^\infty rw(r)dr =\Gamma (4/3)/(4\pi n/3)^{1/3} =0.55396n^{-1/3}$$
where $n$ is in number of molecules/volume then $n^{-1/3}$ is in units of length.


NaHCO3 decomposition dynamics and extent vs Temperature

The paper doesn't tell how fast will it be.
As for the rest, have this:


Rate of Reaction for Elementary Reactions

Intuitively, a reaction between A and B to give C will go twice as fast when you double the concentration of A (because collisions between A and B will occur twice as frequently). Similarly, doubling the concentration of B will also double the reaction rate. But what about the specific case where A and B are the same substance (A + A -> C)? When you double the concentration of A, the rate is quadrupled; when you triple the concentration of A, the rate goes up by a factor of 9, etc.


What does the prefix "dihydro" in the systematic name of luminol refer to?

On the left is 5-amino-2,3-dihydrophthalazine-1,4-dione and on the right is the molecule which the name is based on, phthalazine. The 2,3-dihydro refers to the fact that at the 2 and 3 positions (the two nitrogens) hydrogens have been added (hydro is the prefix for hydrogen).


How do we explain mutations, in the deterministic world of chemical reactions?

The topic is much larger than a scope for a single answer. But staying at general, low level chemical principles, it can be pinned down to several mechanisms leading to possible mutations:

Chemical reactions are not deterministic. Generally, on molecular level, there is less than 100% probability the desired and expected reaction happens, and more than 0%  probability it does not happen, or other reaction happens.
A molecule may react with other molecule in more than one way, with alternative reactions having their probability to form  "by-products".
A molecule may react with similar, but still native molecule, that it is not supposed to react with to form desired product.
A molecule may react with a molecule that should not be present, coming e.g. from pollution, medical treatment or poisoning.
Some foreign molecules may not react directly, but can affect DNA replication or synthesis of RNA or proteins.
A molecule may be broken by ionizating irradiation like UV, X-ray or  $\gamma$ photons, high energy subatomic particles or ions from the natural body radioactivity ($\ce{^{14}C}$, $\ce{^{40}K}$), from natural radioactive background ($\ce{U}$, $\ce{Ra}$, $\ce{Rn}$), from pollution (radioctive elements from coal burning ), medical treatment, radiation incidents or even poisoning.


Can you make deuterium-depleted water through a freezing method?

Not easily. For all practical purposes, $\ce{D2O}$, $\ce{DHO}$ and $\ce{H2O}$ behave the same -- including complete miscibility. Though some small fraction of the liquid that freezes first would be enriched in $\ce{D2O}$, and to a smaller extent in $\ce{DHO}$, it would still be composed largely of $\ce{H2O}$.

According to Wikipedia, the Girdler process is the most energy efficient means of separation.


How can I choose an indicator for determining the concentration of a base using its pKb values?

Since:
\begin{align}
K_\mathrm{w} &=K_\mathrm{a} \cdot K_\mathrm{b}\\
K_\mathrm{a} &=\frac{K_\mathrm{w}}{K_\mathrm{b}}\\
\log{K_\mathrm{a}} &=\log K_\mathrm{w}-\log K_\mathrm{b}\\
K_\mathrm{w} &=1 \times 10^{-14}\\
\log{K_\mathrm{a}} &= -14-\log K_\mathrm{b}\\
-\log{K_\mathrm{a}} &=14+\log K_\mathrm{b}\\
-\log{K_\mathrm{a}} &=\mathrm{p}K_\mathrm{a}\\
-\log{K_\mathrm{b}} &=\mathrm{p}K_\mathrm{b}\\
\mathrm{p}K_\mathrm{a} &=14-\mathrm{p}K_\mathrm{b}
\end{align}

Using the above formula, the three $\mathrm{p}K_\mathrm{a}$ for the conjugate acid of the conjugate base are: $13.5~(\mathrm{p}K_\mathrm{a_1})$, $10.3~(\mathrm{p}K_\mathrm{a_2})$, and $6.3~(\mathrm{p}K_\mathrm{a_3})$, where $\mathrm{p}K_\mathrm{a_1}=14-\mathrm{p}K_\mathrm{b_1}$, etc.

An important equation to note is that the $\mathrm{pH}$ at the equivalence point is given by the equation: $$\mathrm{pH_\mathrm{eq}}=\frac{\mathrm{p}K_\mathrm{a1}+\mathrm{p}K_\mathrm{a2}}{2}$$

This means that two of the equivalence points exist at $$\mathrm{pH_\mathrm{eq_1}}=\frac{13.5+10.3}{2}=11.9$$ and $$\mathrm{pH_\mathrm{eq_2}}=\frac{10.3+6.3}{2}=8.3$$

The $\mathrm{pH}$ range for methyl orange is $3.0~-~4.4$ and the $\mathrm{pH}$ range for phenolphthalein is $8.2~-~10.0$. Phenolphthalein would be useful for detecting the second equivalence point $(\mathrm{pH_\mathrm{eq_2}})$, since this shift occurs within phenolphthalein's range.

The final equivalence point cannot be calculated from the information given, but given that methyl orange's color transition occurs in an acidic medium, and also the relatively basic $\mathrm{p}K_\mathrm{a_1}$, methyl orange can be eliminated as a possible indicator.


Radioactivity is not at all dependent on the atomic or molecular structure: it a purely nuclear phenomenon. The only thing that is relevant is the structure of the nucleus.

The nucleus contains only protons and neutrons and the distinguishing feature that makes nuclei radioactive is, crudely, the ratio of the two. Some arrangements are unstable for a variety of reasons a full understanding of which requires fairly advanced quantum theory. What you need to remember is that some ratios or arrangements of protons and neutrons are inherently unstable and spontaneously decay in a variety of ways to more stable nuclei. For example, 39K (which contains 19 protons and 20 neutrons) is stable but adding a neutron makes 40K which is slightly unstable and spontaneously decays in several possible ways. You can tell this isn't a simple phenomenon, though, as 41K, with two extra neutrons is stable again.

It is also important to note that radioactive decay is essentially uninfluenced by anything outside the nucleus. For any given nucleus there is a probability that it will decay but that event occurs randomly and we can't tell when it will happen (but the rate of decay is stable and not influenced by temperature, pressure or the chemical compounds the nuclei are present in).


Why does no antitoxin for botulinum type H exist yet?

The premise of this question is based upon outdated information.

According to Study: Novel botulinum toxin less dangerous than thought

  Researchers from the Centers for Disease Control and Prevention (CDC) and the University of Wisconsin (UW), writing in the Journal of Infectious Diseases (JID), said it turns out that the novel toxin can be blocked by available antitoxins after all. That suggests that it isn't really a novel toxin, the authors said

(Attention grammar police: The above is an exact quote from the linked reference.  If you feel the compulsion to keep editing this again and again, please do it in a way that makes clear it is no longer an exact quote.)


Why can't the N- end of the nitrile anion attack the alkyl halide in an SN2 fashion during an alkylation reaction when OH- can?

It can.
According to this JACS paper here steric effects are a major factor in whether the ketenimine is formed or the C-alklyated product - the more hindered the nitrile, the more N-alkylation.


No, you should not do the problem your way.
Given
$$(0.08−x)(0.09−x)=10^{−14}$$
You can solve for $x$ here, but its value will be quite close to $0.08$. This is the only way to make the product a small number, by making the one of the multiplicands very small.
However, $x$ will be so close to 0.08 that will have trouble figuring out the value of $0.08 - x$, which is what you want. Therefore, the right answer is obscured by your method of solving the problem.
In this problem, you can work around that by calculating the pH value from the hydroxide ion concentration, which, by the argument above, will be roughly $0.01\ \mathrm{M}$, consistent with a pH of 12. However, in the most general case, you may not be able to do that to determine what the final equilibrium concentrations are.


No, you do not have to have as many π electrons as p-orbitals or any type of direct relationship. As long as you have a continuous, circular arrangement of p-orbitals and an electron count $4n+2$ (for monocycles), you can expect your molecule to be aromatic.

Pyrrole, furan and related molecules are aromatic. They have 6 π electrons like benzene, but only five p-orbitals.
The cyclobutadienyl dianion has been synthesised as a dilithium salt. The authors claim evidence for aromaticity.[1]
The dianion of cyclooctatetraene has also been synthesised and examined by ESR as early as the 1960’s. The authors state:

  The assumption that both the dianion and the radical-anion are planar […] is consistend with [their equation] within experimental error.

Thus it can be assumed that $\ce{cot^2-}$ is also an aromatic system.[2]
Disubstituted dications of cyclooctatetraene have been synthesised a decade later by Olah et al. The published proton NMR data implies that the protons are more deshielded than a simple triene moiety would imply. The summary the authors give states the following:

  The $\ce{^{1}H}$ and $\ce{^{13}C}$ NMR spectroscopic characterization of the reported new cyclooctatetraene dications indicates these ions possess those characteristics expected of a $\ce{C8}$ 6 π aromatic system and hence further extends the evidence for the great predictive utility of the $4n+2$ Hückel rule.

$\ce{cot^2+}$ is thus also an aromatic system.[3]

These last two bullet points must be the ones you are looking for. The cyclooctatetraene dianion, a cyclic $\ce{C8}$ species with 10 π electrons ($n=2$) is aromatic, as is also the cyclooctatetraene dication, the same $\ce{C8}$ species albeit with 6 π electrons ($n=1$).

For most systems, aromaticity is either already given (as they have e.g. six π electrons in their neutral state; compare benzene etc.) or they are only one electron away (cyclopentadienyl, cycloheptatrienyl or tropylium). The way these molecules obtain aromaticity is relatively fixed since greater charges destabilise molecules.

Only for those systems starting with $4n$ π electrons such as cyclooctatetraene, which are antiaromatic by the Hückel definition, can there be a choice. Both the dication and the dianion are generally unfavourable, but they are similarly unfavourable thus the system can be pushed into either direction.

It is highly unlikely that a tetraanion or tetracation of benzene exist outside of a $4~\mathrm{K}$ matrix, much like it is very unlikely for a cyclopentadienyl-trication or a cycloheptatrienyl-trianion to exist outside of similar conditions due to their great charge density.

References:

[1]: A. Sekiguchi, T. Matsuo, H. Watanabe, J. Am. Chem. Soc. 2000, 122, 5652. DOI: 10.1021/ja0004175.

[2]: H. L. Strauss, T. J. Katz, G. K. Fraenkel, J. Am. Chem. Soc. 1963, 85, 2360. DOI: 10.1021/ja00899a004.

[3]: G. A. Olah, J. S. Staral, G. Liang, L. A. Paquette, W. P. Melega, M. J. Carmody J. Am. Chem. Soc. 1977, 99, 3349. DOI: 10.1021/ja00452a027.


Reaction of hypohalous acids with alkenes

Let us deconstruct our reagent:

Hypochlorous acids, more commonly, 'bleach', is an oxidant.  Additionally, it is not a great acid, pKa 7.40.

And let us deconstruct our product:

Additionally the chloride is on the least substituted carbon (anti-Markovnikoff) without any indication of added radical source.

In this context it is unwise to view this as an analogous to HCl addition.

Your product could be readily explained by an epoxidation-type reaction followed by a nucleophilic substitution with residual chloride ion.  The regioselectivity of Cl on the less substituted carbon would seem to support that, as well as the absence of a O-Cl bond in the product.


Need to dissolve glue placed intentionally in a lock

To paraphrase the link you mentioned,
Without knowing the specific glue, you should try different solvents to see if it dissolves the glue:

Start with a mild, non-polar solvent like paint thinner.
If that doesn't work, try a polar solvent like acetone.
If that doesn't work, try tetrahydrofuran (PVC pipe solvent)

All of these solvents are flammable and pretty nasty if you get them on you. To minimise risk, use small quantities at a time, minimise the time that the lid is off the container. The fumes shouldn't be too much of a problem if you're outside, but still take care.


Do all double displacement reactions lead to precipitates?

No, recall the definition of a precipitate "insoluble ionic compound". Take this reaction into consideration $$\ce{ NaOH~(s) + HCl~(l) <=> H2O~(l) + NaCl~(aq)}$$ none of the products are insoluble. You'll learn more when you go over gas evolution reactions.


Modeling perturbations of a quantum mechanical system

I don't understand exactly what is your question:

If you start in an eigenstate of $H_p$, say $\varphi_j$ and then your Hamiltonian changes like $H(t) = H_u + f(t) (H_p - H_u)$ where $f(0) = 1$ and $f(T) = 0$ -$T$ being the final time of the perturbation- then in order to follow the evolution of $\psi(t)$ all you need to do is to solve the time dependent Schroedinger equation
$$i\hbar \partial \psi(t) / \partial t = H(t) \psi(t)$$
where $\psi(0) = \varphi_j$. There are many numerical procedures to do so. For instance if you are interested, as it seems, in following the dynamics in the basis of the eigenstates of $H_u$, then you expand $\psi(t) = \sum_k a_k(t)
\phi_k$, where $\phi_k$ are eigenstates of $H_u$, and you solve a set of ordinary differential equations.

If $f(t)$ is very fast ($T << (\Delta E)^{-1}$, where $\Delta E$ is a characteristic energy of the system, say the energy difference between adjacent eigenstates) then you can use the sudden approximation; if $f(t)$ is very slow you can use the adiabatic approximation, if $H_p - H_u$ is small you can use first order perturbation theory, etc.

In any case it is important to realise that in Quantum Mechanics there is no relaxation unless you couple your initial state to a continuum and there is no return to equilibrium unless you do the same (couple to continuum) and work with the proper formalism that allows you to define the (thermal) equilibrium, which are the density matrices, unless by equilibrium you are referring to the initial state. Again, in ordinary Quantum Mechanics $\psi(t)$ will oscillate and will not relax to the initial state $\varphi_j$ (or $\phi_0$); at most, it will return to the initial state as part of an oscillation, called revival.


Are highly nitrated cyclopentadiene compounds possible?

Mono-nitrated cyclopentadienes and their benzo-analogues (indene, fluorene) have been prepared in the context of studies on nitro-aci-tautomerism.

As far as higher substitution is concerned, there seems to be at least a crystal structure of potassium dinitrotrichlorocyclopentadienide reported in Acta Cryst., B

In the abstract, the authors (Y. Otaka, F. Marumo, and Y. Saito) mention the preparation of this compound by Haga in 1969, (sorry, reference is behind paywall) via

  nitration of hexachlorocyclopentadiene with alkali nitrite in an aqueous or ethanolic solution

NOTE: Hexachlorocyclopentadiene once was a favoured starting material in the synthesis of insecticides via Diels-Alder reaction. However, it is a potent teratogen and the "mother" of least five members of the dirty dozen (persistent organic pollutants) nowadays banned for good reason.

You might want to read the MSDS and then decide that you'd rather prefer to work with something else.


Entropy of solid, liquid, and gas at triple point of water

Triple point defines a situation of simultaneous equilibrium between the solid, the liquid and the gas phases.

For such an equilibrium, you simply write:

$$
\Delta S= \frac{\Delta H}{T}
$$

that rises from the fact that $\Delta G=0$.

Considering water molar ($m$) enthalpies for each phase transition at $273\,K$:

$$
\Delta H_{melting,m}=6.01\,kJ\,mol^{-1}
$$
$$
\Delta H_{vaporisation,m}=45.05\,kJ\,mol^{-1}
$$
$$
\Delta H_{sublimation,m}=51.06\,kJ\,mol^{-1}
$$

one realises that

the entropy of the gas phase is higher than the entropy of the liquid phase.
the entropy of the liquid phase is higher than the entropy of the solid phase

In this respect the triple point has no peculiar behaviour compared to other points where a two-phase equilibrium is established.


Raoult's law is an approximation for ideal solutions.

Those links have more detail, but the basic idea is that when you assume that the interactions between solute and solvent molecules don't depend on the identity of the molecules, but the vapor pressure does, then it makes sense that the vapor pressure of the mixture would be a linear combination of the vapor pressure of the molecules in the mixture.

Think about it like a bunch of slightly sticky spheres bouncing around in a bucket. A certain percentage will randomly bounce out of the bucket - that's the vapor pressure. If you have a different type of sphere with less mass, more will bounce out in the same amount of time - the vapor pressure will be higher. If you mix them, the total vapor pressure is given by the total number of both types that bounce out. The more volatile ones leave more easily, so the total increases.

Since ideal gases don't interact with each other and since ideal mixtures have negligible enthalpy of solvation, all that is left is entropy terms. If you are looking for a connection to entropy, that is it.

In cases where the ideal approximation is not realistic (water and a volatile solute, for example), Raoult's law doesn't really hold, but it is still a reasonable approximation when the concentration is low simply because most functions look approximately linear for small deviations. It also tends to work reasonably well because most things that will form a solution tend to have similar interaction energies - otherwise they are less likely to mix in the first place.

Of course, there are many, many practical cases where Raoult's law doesn't work, and so we have corrections for that: fugacity and activity.


The most important question is why do we need low pressures? If you recall kinetic theory of gases, there is a concept of mean free path. This quantity roughly tells us the distance a particle would travel before colliding with another particle. If you want the electrons to travel from one end to the other, you need to provide a large mean free path to them. How can one obtain that? Simply by removing "matter" from the cathode ray tube. Practically this is achieved by vacuum pumps. Although nobody gives importance to vacuum pumps, it was this technology which allowed great progress in the field of cathode ray experiments and hence understanding matter.

Now, the "glow" phenomenon is rather a spectroscopic problem. Electron don't glow green/pink etc, it is the matter which starts to glow when it interacts with electron. The glow can originate from several sources such as recombination phenomenon, fluorescence of the glass, a fluorescent zinc sulfide screen placed inside, or even Xrays. When the pressures are moderate, other phenomena like which you quote as striations, appear. This arises from accelerated electrons exciting the gas particles and then losing energy (dark space). This video explains some of it. Cathode ray striations. However it is not a trivial phenomena. It took ages to understand it.
The observation of striations indicate that there is enough gas (low vacuum) in the discharge tube.


The reaction energetics

Using the method of the superposition of configurations, the electron affinity of $\ce {Na}$ was theoretically determined to be $\ce {+0.54 eV}$ $\ce {^1}$, that is, around $\ce {-52.1 kJ/mol}$. The gas-phase process

$\ce {2Na (g) -> Na^+ (g) + Na^- (g)}$

has been determined to be endothermic by $\ce {4.54 eV}$ $\ce {^2}$ while the solid-state process

$\ce {2Na (s) -> Na^+ .Na^- (s)}$

has also been estimated to be endothermic by $\ce {0.8 eV}$ $\ce {^2}$.Thus, your assertion regarding the large exothermicity of the reaction is questionable. However, it is worthy to note that the $\ce { \Delta H_f}$ and $\ce {\Delta G_f}$ for $\ce {Na^+.Cry Na^-}$, where $\ce {Cry}$ = [2.2.2] cryptand, are $\ce {-10 kJ/mol}$ and $\ce {+28 kJ/mol}$ respectively $\ce {^3}$. Your large exothermic enthalpy may likely refer to the lattice energy, i.e. for the process

$\ce {M^+.Cry (g) + M^- (g) -> M^+.Cry M^- (s)}$.

For $\ce {M = Na}$, the $\ce {\Delta H}$ and $\ce {\Delta G}$ for the above process are $\ce {-323 kJ/mol}$ and $\ce {-258 kJ/mol}$ respectively $\ce {^3}$.

Preparation of the alkalide

$\ce {Na^-}$, $\ce {K^-}$, $\ce {Rb^-}$, and $\ce {Cs^-}$ anions are stable both in suitable solvents and in crystalline solids$\ce {^3}$. The latter can be prepared either by cooling a saturated solution $\ce {^4}$ or by rapid solvent evaporation.

The principal difficulty in preparation of crystalline salts containing alkalide ions by the method of cooling a saturated solution is the low solubility of these alkali metals in the amine and ether solutions $\ce {^3}$. Without a sufficiently large concentration of the metal dissolved in solution, precipitation of the solid upon cooling would be insignificant. This difficulty was resolved by the use of crown-ether and cryptand complexes, such as those of [18]crown-6 and [2.2.2] cryptand] $\ce {^3}$. The complexating agent complexes with $\ce {M^+}$ , shifting the equilibrium (1) far to the right, significantly increasing the concentrations of the dissolved metal ions.

(1) $\ce { 2M (s) -> M^+ (sol) + M^- (sol)}$

(2) $\ce { M^+ (sol) + Cry (sol) -> M^+.Cry}$

This technique of using complexating agents was also what Dye et al. used in their synthesis in 1973 $\ce {^4}$. As reported by Dye et al., a sufficiently concentrated solution of sodium metal (in excess) dissolved in ethylamine with [2.2.2] cryptand was first prepared. The solution is then cooled to dry ice temperatures, giving a gold-coloured crystalline solid precipitate. Through thorough analysis, this precipitate was then determined to be $\ce {Na^+.Cry Na^- (s)}$ with $\ce {Cry}$ being the [2.2.2] cryptand.

References

Weiss, A. W. Theoretical Electron Affinities for Some of the Alkali and
Alkaline-Earth Elements. Phys. Rev., 1968, 166 (1), 70-74
Tehan, F. J.; Barnett, B. L.; Dye, J. L. Alkali anions. Preparation and Crystal Structure of a Compound which contains the Cryptated Sodium Cation and the Sodium Anion. J. Am. Chem. Soc., 1974, 96 (23), 7203–7208
Dye, J. L. Compounds of Alkali Metal Anions. Angew. Chem., 1979, 18 (8), 587-598
Dye, J. L.; Ceraso, J. M.; Lok, M. T.; Barnett, B. L.; Tehan, F. J. A Crystalline Salt of the Sodium Anion (Na-). J. Am. Chem. Soc., 1974, 96 (2), 608-609


Has the half-life of an isotope ever been accurately predicted before it could be measured?

A quick perusal of the post-WWII literature (particularly Glenn Seaborg) pulls up papers like The New Element Californium (Atomic Number 98) where comparisons of the observed alpha decay half life are said to be in line with predictions. Such predictions follow along from papers such as Systematics of Alpha-Radioactivity.

So, yes, it would seem that predictions of radioactive half lives were part and parcel of the creation of new elements in the laboratory.


Substitute molten salt bath with temperature range from 400–500 °C

I wasn't sure what other conditions you needed for your salt bath, yet I listed few for your convenience together with the salt bath of $\ce{KNO3/NaNO2}$ (bath 5) for comparison:
$$
\begin{array}{ccc}
\text{Bath#} &\text{Composition} & \text{Approx. melting poing} & \text{Working temp. range} \\\hline
1 & \ce{NaCl}, \ce{KCl}, \ce{BaCl2}, \ce{CaCl2}  & \pu{400 ^\circ C} & \pu{500-800 ^\circ C} \\
 & (10\!-\!15 \!:\!20\!-\!30 \!:\!40\!-\!50 \!:\!15\!-\!20) &  &  \\
2 & \ce{NaCO3}\, (45\!-\!55), \ce{KCl}\, (55\!-\! 45)  & \pu{450 ^\circ C} & \pu{550-900 ^\circ C} \\
3 & \ce{NaCl}\, (20), \ce{KCl}\, (30), \ce{BaCl2}\, (50)  & \pu{540 ^\circ C} & \pu{570-900 ^\circ C} \\
4 & \ce{NaNO3}\, (100)  & \pu{370 ^\circ C} & \pu{400-600 ^\circ C} \\
5 & \ce{KNO3}\, (50\!-\!60), \ce{NaNO2}\, (50\!-\!40) & \pu{135 ^\circ C} & \pu{160-550 ^\circ C} \\
\end{array}
$$

I also listed $\ce{NaNO3}$ bath in case you won't mind using it since it is not potassium salt and working range is ideal for your requirement. As indicated, you can change composition as much as you wanted to to get right melting point and the working temperature range you wanted to.


A possible reaction of heating Terephthalic acid

Following are the Statement from “On Terephthalic Acid and its derivatives” By Warren De la Rue and Hugo Muller February 7, 1861:

On heating, terephthalic acid sublimes without previously
fusing(melting). The sublimate, which is indistinctly crystalline, has
the same composition and properties as the original acid, and
therefore, unlike other bibasic acids, terephthalic acid cannot be
converted into an anhydrite by merely heating it

Terephthalic acid when pure is white opaque powder, insoluble in water and it sublimes at $\pu{300^{\circ}C}$ without melting.
When heated to decomposition it emits acrid smoke and irritating fumes. $\ce{FeOOH}$, nickel hydroxide and nickel oxide are suitable catalyst for thermal decomposition of terephthalic acid (at low temperature) but since nickel compounds are expensive so $\ce{FeOOH}$ is generally used. At higher temperature (in absence of catalyst) Terephthalic acid$\ce{_{gas}}$ is decomposed to carbon dioxide,benzene and benzoic acid.
For further reading refer to this more in depth paper
"Determination of the Thermal Decomposition Products of Terephthalic Acid by Using Curie-Point Pyrolyzer" by Mehmet uluturk


Potassium permanganate to analyze iron compounds

The Standard Methods for the Examination of Water and Wastewater 3500-Fe is a method for determining total $\ce{Fe}$, $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ in water using a colorimetric detection technique.  As in your titrimetric method, this method can only directly measure $\ce{Fe^{2+}}$.  In order to measure $\ce{Fe^{3+}}$ (the only other reasonably stable oxidation state of $\ce{Fe}$) you must first reduce the $\ce{Fe^{3+}}$ to $\ce{Fe^{2+}}$.  I believe you could use the same reduction technique to prepare a solution of $\ce{Fe^{3+}}$ for your titration method.

The procedure sited above states that:

  "[the $\ce{Fe}$ in the sample is] reduced to the ferrous state by boiling
  with acid and hydroxylamine."

The hydroxylamine solution is prepared by dissolving 10 g $\ce{NH2OH⋅HCl}$ in 100 mL water.  50 mL of the water sample is acidified with 2.0 mL of concentrated $\ce{HCl}$, and 1 mL of the $\ce{NH2OH}$ solution is added to the acidified water sample.

As $\ce{Fe^{3+}}$ species are frequently insoluble, the procedure described above calls for boiling the solution down to 15-20 mL to get all of the iron both dissolved and reduced before bringing the solution back up to 50 mL for analysis.  If the solution were prepared from a soluble $\ce{Fe^{3+}}$ salt like $\ce{FeCl3}$ for example, simply boiling the solution for 5 minute or so would likely be sufficient.

So, I suspect you could measure the $\ce{Fe^{3+}}$ in your solution using your titration method if you first reduced the $\ce{Fe^{3+}}$ to $\ce{Fe^{2+}}$ using the hydroxylamine reduction described above.


Diffusion of carbon from ferrite to cementite in the eutectoid transformation

Diffusion is perhaps more complicated than initially taught. Yes, all things being equal you will get diffusion from regions of high concentration to regions of low concentration.

But, in the case you mention, all things are not equal - there is a phase change occurring, and that means that there are chemical potential gradients as well as concentration gradients. Now you have an additional driving force for motion of impurities as they prefer to be in the phase with a lower Gibbs free energy.


I could say explosives are things that blow up, but this would not a helpful answer.

It is also not that helpful to describe explosives as compounds with highly exothermic decomposition. This characteristic alone does not define an explosive.

What matters is rate of the reaction in the bulk compound. This rate defines two key types of explosive. If the reaction, once started, proceeds through the bulk compound faster than the speed of sound we call it detonation. If the reaction is slower than the speed of sound we describe it as deflagration.

This distinction matters because of the different uses explosives can be put to. Propellants, such as the explosives used to drive bullets, tend to deflagrate (as you want a relatively slow, controlled reaction to drive the shell out of the barrel without blowing up the barrel). But if you want something to really go bang and create a mess (as the military do) or if you want something to break up rock (as quarriers and miners do) then you want something that will detonate.

The structures of some well-known explosives are shown below:

Most well known real world explosive compounds are made from these compounds plus some added materials to make them easier to handle. C-4, a well known military explosive is mostly RDX with plasticisers to make it easy to mould like plasticine. The favourite terrorist plastic explosive is Semtex, similar to C-4 but made from a mixture of PETN and RDX. TNT is often used in high-explosive military shells (partially because it can be melted by steam and poured into shell casings). The first modern explosive, dynamite, is based on nitroglycerine mixed with diatomaceous earth (which makes it much easier to handle than the dangerously shock-sensitive pure compound).

The chemical reactions typically involve the production of CO2, H2O and, sometimes, N2.

Another characteristic that matters is how easy the reaction is to set off. Some compounds are very shock sensitive like mercury fulminate or nitroglycerin. This is only useful in detonators (in small quantities) or for chemists who have a death wish. Many bulk explosives require a detonator to trigger an explosion, making them much safer to handle and transport. C-4 in small quantities will actually burn slowly (and was sometimes used by soldiers in the Vietnam war to warm rations).


Purity of calcium carbonate in limestone

We know that 2 moles of $\ce{HCl}$ are required to react completely with 1 mole of $\ce{CaCO3}$. We find out the number of moles of $\ce{HCl}$ in $\mathrm{150ml}$ $\mathrm{0.1M}$ $\ce{HCl}$ ($\mathrm{0.1N} \ce{HCl} = \mathrm{0.1Ml} \ce{HCl}$), which is $\mathrm{0.015}$ moles. Now the amount of $\ce{CaCO3}$ reacted is ($\mathrm{\frac{0.015}{2}}$) = $\mathrm{0.0075}$ moles.

Amount of $\ce{CaCO3}$ in gms = $$\mathrm{0.0075~moles\times 100~}\text{(mol.wt of CaCO3)} = \mathrm{0.75g}$$

...implying the purity is 75%.


Formation of 2,4-Dichlorophenol by free chlorine mediated oxidation of Triclosan

I'm not sure if this qualifies as an answer since it's more of a guess. Catechols (1,2-dihydroxybenzene derivatives) can be oxidized to ortho-quinones. Applying that transformation here (chlorine in water is certainly oxidizing conditions) gives the oxonium of the ortho-quinone, which can be hydrolyzed to 2,4-DCP.


Why is the standard enthalpy of formation of elements in their native forms zero?

It is purely a matter of definition. A standard enthalpy of formation describes the change in enthalpy during the formation of 1 mol of a target compound by reacting the (pure) elements it consists of, whereby each element is expected to be in its most stable modification for the given temperature. Now take the case of oxygen as an example: At standard conditions $\ce{O2(g)}$ is the most stable modification of oxygen. That means at standard conditions the reaction

$$\ce{2O <=>> O2(g)}$$

is defined in such a way that each of the elemental $\ce{O}$s are defined to be exactly the same state as an oxygen atom that is part of an $\ce{O2(g)}$, i.e. their enthalpy $2H^{\circ}(\ce{O})$ matches exactly that of $\ce{O2(g)}$: $H^{\circ}(\ce{O}) = \frac{1}{2}H^{\circ}(\ce{O2(g)})$. This gives you a reference point for $H^{\circ}(\ce{O})$ which is otherwise nearly impossible to measure much as the Standard Hydrogen Electrode is used to set up a reference system for the electrode potential.

A table showing some standard enthalpies of formation can be found here. There it is also indicated which phases are used as reference form.


Since this got no answers, I'm posting an answer from Reddit user midnight-cheeseater, which can be found here.

  There are two main possibilities here, depending on how the ions are adsorbed:

  If the solid only adsorbs a type of ion, either positive or negative. Such as the example quoted of alumina adsorbing fluoride ions. The alumina will become negatively charged at the specific points where the fluoride ions stick to it. These negative charges will attract anything nearby which is positively charged. So the $\ce{Na+}$ ions will associate themselves with the negatively charged areas on the alumina.
  As a solid adsorbs one ion, it may release another. This process is commonly seen in ion-exchange resins. Such as those which are used to produce "de-ionized" water. These adsorb positively charged metal ions and release $\ce{H+}$ ions at the same time. They can be regenerated by running fairly concentrated nitric acid through them.

  The counter-ions are typically chloride, sulfate or nitrate, as well as some other halides, carbonate and phosphate. So this essentially converts a dilute solution of metal salts into a dilute solution of hydrochloric, sulfuric and nitric acids. Which explains why "de-ionized" water is rather acidic, as opposed to distilled water which is neutral.

  Another common example found in dishwashers and water softening systems is an ion-exchange resin which adsorbs $\ce{Ca^2+}$ or $\ce{Mg^2+}$ ions while releasing $\ce{Na^+}$ ions at the same time. This converts dissolved calcium or magnesium salts (usually bicarbonates or sulfates) into sodium salts. These are regenerated by running saturated $\ce{NaCl}$ solution through them, which is why dishwashers and water softeners require a supply of salt.


Shaving cream is a special type of mixture called a colloid, consisting of tiny air bubbles dispersed in liquid. This particular type of colloid can further classified as a foam. Since the mixture itself consists of two distinct phases, it doesn't really make sense to classify the entire mixture as one phase. If you had to pick one though, it is most nearly a liquid since that is the dispersion medium.


Easy method of analysing iodine content in supplements?

Take some potato starch dissolved in water (you can easily get it from the water when you boil potatoes), and put a tablet into the water. If the water becomes bluish black, there is iodine in the tablet.

Yes, there are testing labs that will determine the amount of iodine in the tablets which will cost somewhere around \$100 to \$200.


Are we using an assumption in calculating the ratio of first state excited to unexcited chemical species using the Boltmann distribution?

The expression for the Boltzmann equation is:

$\eta_{i} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}$,

wherein $\eta_{i}$ the probability is to encounter the ensemble in state $i$ is in the ensemble, $g_{i}$ the degeneracy of state $i$, and $\epsilon_{i}$ its energy (I reckon the other variables are familiar to you).

To calculate the ratio between two states, you can simply divide the above expression for two different states by which you obtain:

$\dfrac{\eta_{i}}{\eta_{j}} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)} \cdot \dfrac{\sum_{i} g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{g_{j} \exp \left( \dfrac{-\epsilon_{j}}{k_{b}T} \right)} = \dfrac{g_{i} \exp \left( \dfrac{-\epsilon_{i}}{k_{b}T} \right)}{g_{j} \exp \left( \dfrac{-\epsilon_{j}}{k_{b}T} \right)}$

The expression above shows that because states $i$ and $j$ are of the same ensemble, you do not have to use (or even know) the energy levels of the other states in that ensemble if you are only interested in the probability ratio between the two states.


Is it possible to generate electricity by using two inert electrodes in a galvanic cell?

A non-chemical way to generate electricity using two inert electrodes in salt water is through magnetohydrodynamics. If the salt water moves past the electrodes in a magnetic field, it generates an electric current without consuming the electrodes or electrolyte.

This could make an interesting "magic trick" or puzzle: use a moving magnetic field, e.g. a magnet attached to a motor, in a box below a beaker filled with salt water, and insert two carbon electrodes connected to an AC current meter. Of course, make sure the motor is very quiet.


Entropy change in an adiabatic expansion

"One mole of ideal gas initially at a pressure of 1 atmosphere and T = 298 K, is expanded into a volume 50% larger adiabatically."

The question doesn't have sufficient information for a solution, because we don't know if the adiabatic expansion is reversible or irreversible.
If reversible:
If the expansion is done reversibly, then we know entropy of the universe can't be changing (because reversible), and that heat flow from the gas to the surroundings is zero (because adiabatic).  Without heat flow, the only way the entropy of the universe can stay constant is if both the entropy of the universe and of the gas stay constant.  In this case you can apply both of your equations.  But keep in mind that your equation for $dS$ is only true for ideal gases.
$dS = \frac{\delta q_{rev}}{dT} = 0$
$dS = 0 = \frac{C_V}{T}dT + \frac{R}{V}dV$
$\frac{dT}{T}=(-\frac{R}{C_V})\frac{dV}{V}$
This is now a differential equation you can solve for temperature if you know volume.
As the other answers have indicated, a reversible expansion will involve the gas doing work on the surroundings.  One way to think about why this is so is that for the process to be reversible, the potential of the gas to do work must not be lost, and so if this potential for doing work doesn't remain in the gas, it must go to the surroundings.  A loss in the potential to do work (of the universe) is an increase in entropy.  And in a reversible process that can't happen.
Think of gas in a piston, where the piston is held down by a pile of sand.  The reversible expansion involves slowly removing sand, grain by grain, from the piston.  As the grains are removed, the piston rises and the gas lifts the remaining sand, doing work against gravity.
If irreversible:
...then we still don't have enough information because there are many ways the process can be irreversible.  But one particular, extreme case of irreversibility is if the adiabatic expansion occurs without transfer of any energy, then we can solve the problem.  In this extreme scenario, neither heat nor work are not transferred.  You can thus infer that $dU=0$ for the gas, so
$dU=-TdS + p dV=0$
$TdS = p dV$
Using your definition for $dS$ for an ideal gas,
$T \frac{C_V}{T}dT + T \frac{R}{V}dV = p dV$
Keep in mind that $\frac{RT}{V}$ for an ideal gas is $p$, so you should get that
$C_V dT = 0$
Which is also a differential equation that you can integrate to get the temperature change.  (The obvious solution is that the temperature does not change in this case.)
In this second case, now that we know that $T$ is constant, we know $dT$ is zero, and thus using your equation for $dS$, we get that $dS = \frac{R}{V}dV > 0$, i.e., the entropy of the gas increases, which makes sense because the process was irreversible.  The potential to do work has been lost.
Think of gas confined to one half of a cylinder, and the other half of the cylinder is empty (i.e. vacuum).  If the divider between the filled and unfilled halves is suddenly removed, that would be a case of adiabatic, isoergic expansion.


Continuous chemical reaction network versus chemical reaction network - difference?

I suspect that the distinction is that a CRN considers discrete species concentrations whereas a CCRN considers continuous species concentrations: "The CCRN model is continuous, dealing with real-valued concentrations of species...". More precisely, whereas a CRN deals with concentrations $x(t), y(t) \in \mathbb{Q}$, a CCRN deals with concentrations $x(t), y(t) \in \mathbb{R}$.

On a microscopic level, it's easy to see why $x(t), y(t)$ should be elements of $\mathbb{Q}$: a microscopic concentration is simply a ratio of the number of one molecule to the total number of molecules in the system, and the numerator and denominator must thus be integral.

For macroscopic systems, however, since the denominator---the total number of molecules in the system---is so large, it becomes a very good approximation to take $x(t), y(t) \in \mathbb{R}$. Moving to $\mathbb{R}$ is convenient, since then by physical considerations $x(t)$ and $y(t)$ are continuous, and we can, under mild assumptions, take derivatives and return to a rate-law formalism involving differential equations rather than difference equations.

Remark. I'm not familiar with this field, and I don't know how common the CRN/CCRN distinction is. Indeed, the Wikipedia article assumes differentiability of $x(t), y(t)$ and hence that $x(t), y(t) \in \mathbb{R}$.


What simple formal systems are there that have behavior similar to actual chemistry?

The Lewis model (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3285721/) meets those criteria.


There is nothing weird in here. What you've been told is wrong, or maybe you have misunderstood it. Regions like this exist because stuff are miscible. (ss) stands for solid solution, meaning that the solid crystal structure can accommodate some amount of the other component (Al2O3 in this case).

The diagram is a bit misleading, because naming something as MgO(ss) is not entirely correct. Adding the (ss) implies that it is not pure MgO. A more correct way to name the fields in the diagram would be to rename MgO to "periclase" and MgAl2O4 to "spinel", which are the mineral names for these compounds. They will not imply pure compositions of the solids, and conceptually allow the solid solutions to occur.

Some applications of the lever rule: For example, take a look at 10% Al2O3 and 2000 degrees. You basically have periclase, which in pure form as the composition MgO. However, it has 10% of Al2O3 in it. It is still periclase though. Let's take a more complicated example: At 40% Al2O3 and 1800 degrees. You have two solid phases, periclase and spinel in roughly equal proportions. The periclase has 10% Al2O3 in it, and the spinel has 65% Al2O3 in it. Note that it's less than the ideal spinel should have!

Also, notice that Al2O3 in the diagram does not have (ss) in it. This means that it's always going to be Al2O3, with no MgO in it at all.

The curves that separate solid solution (ss) fields are appropriately called solvi (singular: solvus). They mark phase changes that are purely in the solid state, without any melting. For example, let's say you are at 70% Al2O3 and 2000 degrees. You have only one phase: a spinel. By cooling down the system, you will cross the solvus line and periclase will be added to the system, on the expense of spinel. You will be able to microscopically see it as patches or stripes or something like that of periclase in the spinel. Wikipedia has an image to demonstrate that (although it's built for eutectic compositions, the morphology is similar).


Ethanol clearly can be used as a fuel.

But there are several reasons why it isn't.

First is simple economics. Ethanol costs more to produce than gasoline: oversimplifying only slightly, fermentation is more complicated than simple distillation. Those countries that do use mixtures of ethanol in gasoline do so for "environmental" reasons (which are not always well thought through as being made from plants doesn't make something good if associated production costs are high or alternative uses are better). And "environmental" causes are a good excuse to subsidise favoured groups, like US or Brazilian farmers. In the absence of subsidy, there would be very little use of ethanol as a fuel.

But those are not the only reasons. Ethanol carries less energy than the same volume of gasoline. In other words your mileage will be lower; your fuel efficiency will be lower.

Worse, though, ethanol doesn't have exactly the same characteristics as gasoline. Engines may need to be adjusted so the alternative fuels burn equally well, which is annoying but fixable easily in electronically controlled engines. In addition, Ethanol may be more corrosive to some of the components used in fuel pipes (nothing like as bad as methanol, but clearly worse than gasoline). Ethanol is also much more hygroscopic than gasoline, absorbing significant amounts of water from air. This is another potential cause of contamination in fuel systems and may make corrosion worse.

In short, there are no fundamental reasons why ethanol can't be used. But there are a range of practical issues that make it less economical or inconvenient.


What is the difference between these equations for the Contact process?

In principle, your sources are correct and not contradictory.

In order to avoid the formation of problematic sulfuric acid mist, sufur trioxide is not directly absorbed into water. Instead, it is absorbed in sulfuric acid of 98 %–99 % concentration, which approximately corresponds to an azeotrope where the partial pressures of sufur trioxide, sulfuric acid, and water are relatively low.

Sulfuric acid reacts with the absorbed sulfur trioxide to form disulfuric acid:

$$\ce{H2SO4 + SO3 <=> H2S2O7}$$

The disulfuric acid can be converted to sulfuric acid by reaction with added water:

$$\ce{H2S2O7 + H2O <=> 2 H2SO4}$$

Thus, the overall net reaction is as follows:

$$\begin{align}
\ce{H2SO4 + SO3 \;&<=> H2S2O7}\\
\ce{H2S2O7 + H2O \;&<=> 2 H2SO4}\\
\hline\ce{SO3 + H2O \;&<=> H2SO4}
\end{align}$$


How to inhibit Mn2+ autocatalysis in the oxalate-permanganate reaction?

At first the reaction is slow. But when we heat up, the reaction gets catalyzed due to formation of $\ce{Mn^2+}$ ions. It happens due to the fact that $\ce{Mn^2+}$ gets oxidised to $\ce{Mn^3+}$ ions. (Source).

$$\ce{4Mn^2+ + MnO4- + 8H+ -> 5Mn^3+ + 4H2O}$$

So, we have to inhibit the formation of $\ce{Mn^3+}$ ions so as to inhibit the autocatalysis reaction. For this,  we have to use  phosphonate and carboxylate based chelating agent that solubilize (hydr)oxide-bound $\ce{Mn(III)}$ . The following is some relevant information from the site:

  phosphonate-based chelators that solubilize (hydr)oxide-bound
  Mn(III) via ligand-promoted and/or reductive dissolution at
  circum-neutral pHs.

  pyrophosphoric acid (PP), methylenediphosphonic acid (MDP), and
  phosphonoacetic acid (PAA), ligand-promoted dissolution is
  predominant: from pH 6−8, initial dissolution rates and plateau
  concentrations for dissolved Mn(III) decrease in the order PP > MDP >
  PAA, and at pH 5, MDP reacts equally well (with birnessite) or more
  efficiently (with manganite) than PP, and PAA remains the least
  reactive chelator. For manganite reacting with an excess concentration
  of aminophosphonate/carboxylate-based chelators, the
  aminophosphonate-containing iminodimethylenephosphonic acid and
  glyphosate yield appreciable amounts of dissolved Mn(III), but the
  aminocarboxylate-based methyliminodiacetic acid yields solely
  dissolved Mn(II) via Mn(III) reduction.


Calculating Heat Release from Bomb calorimeter

Heat of combustion is the amount of heat (enthalpy) liberated when 1 mole $\ce{CH4}$ reacts (combusts) with oxygen.

In other words, what is the heat of reaction in this case?

$\ce{CH4(g) + 2O2(g) -> CO2(g) + 2H2O(l)}$

Then you must ask yourself, what happens after the heat is produced i.e. where does it go? It quickly turns into an accounting problem...

Take a look at: Constant Volume Calorimetry


Why do chalcogens (Group VI) stink so badly?

I think that there are two reasons: the sulfur is quite common in biochemistry and at the same time a lot of sulfur compounds are toxic. The bio-degradation of sulfur compounds mostly leads to some sulfides which very strongly bind transition metals, i.e. they can destroy/block Metalloenzymes. If you add some millions years of evolution it's a v.good feature to smell them (that would also work for small phosphines and amines). Your chances for food-poisoning are lower, chance for survival higher (Long Live "The selfish gene").


The good news is there is no good reason to add vinegar to your aquarium.  You are correct that if you did, the addition of hydroxide to maintain proper pH would then neutralize acetic acid to acetate.

You can dissolve about 4.5 g of aspartic acid into a liter of neutral water. I assume you try to keep your saltwater tank at a pH of around 8.3 however, which should significantly increase it's solubility as aspartic acid is acidic, having a pKa of 3.9. This is the reason you could not dissolve the aspartic acid with 5% acetic acid, which still pretty acidic with a pKa of 4.8.

So the solution is to just use the slightly basic aquarium water itself, plus a dose of patience, to dissolve the aspartic acid. You could place it in a tea-bag next to your aerator or something if you wanted to isolate it and see how quickly it dissolves.


Deposition of saturated solution and affect of evaporated water

Unfortunately, your calculations are a little bit wrong because of the poor choice of the total initial mass. It is $200 g$, and not $220~ g$. So here are the right calculations.
As $100 g$ water dissolves $109g$ $\ce{KNO3}$ at $60$°C, producing a total mass of $209g$, you may deduce that $200g$ of your initial solution contains $(200/209)·109g = 104.3 g$ $\ce{KNO3}$ and $(200/209)·100g = 95.7g$ water.
Now you loose $20 g$ water by evaporation. The amount of $\ce{KNO3}$ is still $104.3 g$. Remaining water is $95.7g - 20g = 75.7g$ water. At the final temperature $20$°C, $100g$ water can only dissolve $32g$ $\ce{KNO3}$. As a consequence, $75.7$ g water can dissolve $(75.7/100)·32g = 24.2 g$ $\ce{KNO3}$. The rest is deposited in the flask. It is $104.3 g - 24.2 g = 80.1 g$.
Please note that the mass has a unit, the gram. It is incorrect to say "the solubility of $\ce{KNO3}$ in $100g$ of water is $32$ at $20°$C and $109$ at $60°$C". You should have said that "the solubility of $\ce{KNO3}$ in $100$g of water is $32 g$ at $20$°C and $109$ g at $60$°C


Determining overtones bands of degenerate modes

The issue with this approach is that the symmetric direct product doesn't commute, but it is a bit more subtle than that. Summarizing in somewhat less mathematical terms the answer given to my question on MathSE, the issue stems from the fact that once you move past doing a single direct product, there is no longer just a symmetric and an antisymmetric part, but rather $n+1$ parts, where $n$ is the number of direct products being taken. So, trying to drop any antisymmetric terms that arise at each step will still leave terms that aren't in the symmetric part of the representation. (To say this in mathematical language for those interested, the representation for $n$ direct products of a particular irreducible representation of a group can be decomposed into $n+1$ parts by projecting onto the symmetric group $S_{n+1}$).

Physically, the reason for needing only the symmetric part rather than the whole direct product is that we are filling quanta into one particular mode. For example, if we wanted to find a combination band composed of two different $\Pi_u$ modes, we could just take the direct product to find the representation. But, filling two into a single $\Pi_u$ mode limits the number of possible distributions of the quanta to only those given by the symmetric direct product.

As to why the recursive formula gives the symmetric direct product, I give a summary of the argument from Wilson, Decius, and Cross's Molecular Vibrations for the case of a doubly degenerate mode.

Let $Q_a$ and $Q_b$ be the normal coordinates of a doubly degenerate mode. We choose these coordinates such that for an operation $R$      $$Q_a \xrightarrow{R}R_aQ_a$$
$$Q_b \xrightarrow{R}R_bQ_b$$
While the same coordinates will not diagonalize each $R$ in the group, we can find a set of coordinates for each $R$ that will. The character of $R$ for the $(v-1)^{\mathrm{th}}$ overtone, $\chi_v(R)$ can then be determined by summing the contribution of $R$ acting on each possible distribution of quanta between the modes $$\chi_v(R)=R_a^v+R_a^{v-1}R_b+...+R_aR_b^{v-1}+R_b^v$$
We can see that recursion formula holds by looking at the form of the terms on the right hand side
$$\chi(R)\chi_{v-1}(R)=(R_a+R_b)(R_a^{v-1}+R_a^{v-2}R_b+...+R_aR_b^{v-2}+R_b^{v-1})=R_a^v+2R_a^{v-1}R_b+...+2R_aR_b^{v-1}+R_b^v$$
and
$$\chi(R^v)=R_a^v+R_b^v$$
Averaging these two terms, we obtain the recursion formula from the question.
$$\chi_v(\hat{R})=\frac{1}{2}[\chi({R})\chi_{v-1}({R})+\chi({R}^v)]$$

By a similar argument, we can obtain a recursive expression for triply degenerate states
$$\chi_v({R})=\frac{1}{3}\bigg[2\chi({R})\chi_{v-1}({R})+\chi({R}^v)+\frac{1}{2}\big(\chi({R^2})+[\chi({R}]^2\big)\chi_{v-2}({R})+\chi({R}^v)\bigg]$$


Is there a paper glue that is NOT dissolved by acetone?

Something you can try to use is tape to put under the glue. This may stop the acetone from seeping through. Not quite sure but maybe if you really need it done. Also, Elmer's glue does offer a spray on version regardless of what some say on your comments.


Why is ammonia more basic than acetonitrile

p orbitals are higher in energy than s orbitals.
sp3 orbitals with 75% p-character are therefore higher in energy than sp orbitals with 50% p-character.
This means that the lone pair in ammonia is higher in energy than that in acetonitrile.
Higher energy lone pair corresponds to greater basicity.

No need to bring electronegativity into the picture.

For a similar question see Relative acidities of alkanes, alkenes, and alkynes


Dependence of mass transfer coefficient on solubility

No.

In all but highly exotic situations, diffusive and convective fluxes of solutes are linear in the solution-phase concentration of the species:

$$
\begin{align}
J &= -D {\mathrm dC\over \mathrm dx}  & \mathrm{(diffusion)} \\
N_c  &= k_c\left(C_\infty - C_\mathrm{surface} \right)  \quad & \mathrm{(convection)}
\end{align}
$$

Whatever the dissolved concentration $C$ (or concentration gradient ${\mathrm dC\over \mathrm dx}$) of the species is, that is the value you use in both of the above equations.  Whether that concentration (gradient) arises from a species with low, intermediate, or high solubility doesn't matter at all. I suspect your book is using the term environment to refer to things like the geometry, temperature, and fluid-flow characteristics of the system.

The main way I can think of where the solubility might affect the overall mass transport achievable in the system is in its effect on the maximum possible dissolved solute concentration at the surface.  If solubility limitations constrain $C_\mathrm{surface}$ to a low value, then the magnitude of $N_c$ will necessarily be low.  The low solubility is typically assumed not to appreciably change the value of $k_c$, however.


Why is SnCl2 produced when tin reacts with hydrochloric acid?

First, let's consider possibility, that indeed some cation is more stable than others. Let's look at ionization energies of tin (in eV)

\begin{array} {|r|r|}
\hline
1 &7.34 \\
\hline
2 &14.63 \\
\hline
3 &30.50 \\
\hline
4 &40.73 \\
\hline
5 &72.28 \\
\hline
\end{array}

As you can see, the ionization energies increase monotonically, meaning that each subsequent step to higher charged cation is progressively harder. This means, that if only cation stability (in vacuum) was considered, first all available tin should produce tin (I) compound and only then move further. This is, obviously, not the case.

Still, if you are lazy, it is perfectly OK to go with notion of 'stable' valence states, that are quite easy to memorize. Standard electrode potentials are sometimes quite puzzling from uninvolved point of view and it is easier to use the tables without much thoughts about them.

Still, if we want a reasonably acceptable explanation. We can see, that difference between 2nd and 1st ionization energies (~7 eV) and between 3rd and 4th (~10) is lower than between 2nd and 3rd (~15). This suggests that (II) state should oxidize disproportionally harder. Since cations are stabilized by ion-dipole or ion-ion interactions that strengthens linearly with charge, this suggests extra stability of tin (II) ionic compounds comparing to (I) and (III).

The oxidation state (IV) of tin is a completely different matter. Cations of charge $3+$ and higher very rarely truly bear said charge. Typically, they are involved in Lewis acid-base interactions reducing the charge considerably. Tin (IV) compounds have Tin bound covalently. In this case stability of compound is guarded by 8 (18 for d-elements) electron rule.

Mind you, explanation above are not very convincing from quantum mechanical POV, but it is good enough for most cases
.


Proving a relation for fugacity of component i in term of independent variables (T,V)

I haven't been able to work out all the details of the mathematics yet, but the derivation of this equation must start out from the following equation:

$$nRT\mathrm{d}\ln{\phi} = (V - V^{ig})\mathrm{d}P,$$

where $\phi$ is the fugacity coefficient of the mixture, $n$ is the total moles of the mixture, $V$ is the volume of the mixture $znRT/P$ and $V^{ig}$ is the volume of the mixture under ideal gas conditions: $nRT/P$.  Once the equation is integrated to get $RTn\ln{\phi}$, we get the fugacity of the $i$th component of the mixture by evaluating the following partial derivative:

$$RT\ln{\phi_i} = RT\frac{\partial (n\ln{\phi})}{\partial n_i}$$

Hope that this helps a little.

ADDENEUM

If we write, $$d(nV)=\left(\frac{\partial (nV)}{\partial n_i}\right)_{p,n_k}dn_i+\left(\frac{\partial (nV)}{\partial p}\right)_{ni,n_k}dp$$

then it follows that:  $$\left(\frac{\partial (nV)}{\partial n_i}\right)_{p,n_k}=-\left(\frac{\partial (nV)}{\partial p}\right)_{ni,n_k,V}\left(\frac{\partial (p)}{\partial n_i}\right)_{n_k,V}$$

The rest is easy.  The subscript $n_i$ in the OP equation is a typo.  It should be $n_k$, representing all other species being held constant.


Bond Angles and intermolecular interactions

When people talk about bond angles in molecules, they are usually talking about the equilibrium bond angle in isolation. This can be derived from the energy minimized structure of the molecule in isolation.

As you point out, the angles in a real, non-isolated molecule will fluctuate due to interactions with neighboring molecules. However, the actual angle formed by three successive atoms in a molecule will naturally fluctuate even if the molecule is isolated and has some kinetic energy. Everything's always jiggling around.


Intuitive understanding of standard electrode potential

The absolute electrode potential, i.e. the actual volts between electrode and electrolyte, is incredibly difficult to measure accurately.    Instead we simplify: declare the hydrogen electrode potential to be zero, when actually it's roughly +4.4V @25C deg.    Then measure other electrode potentials using a standard hydrogen electrode to contact the electrolyte.


Do carboxyl groups behave like bases and amino groups behave like acids when peptide bonds are created?

Yes, an amino acid consists of a carboxyl group, commonly called carboxylic acid, plus the amino group. And yes, a carboxyl group can act as an acid, i.e. release its proton to give a carboxylate. Conversely, an amino group can act as base, its free electron pair accepting a hydrogen to give an ammonium derivate. That is in fact what happens in aquaeous media or in the solid state of amino acids: the carboxylate protonates the amino group creating a zwitterion.

But these are only one type of reactions both groups are capable of performing. In the case of the amino group, as long as it is not protonated (i.e. still has an unbonded electron pair) it can attack nucleophilicly. Maybe you have heard of $\mathrm{S_N}2$ type reactions, that are rather similar.

The nucleophilic attack of the nitrogen could happen on the carboxylate’s proton — but then we wouldn’t be gaining anything, since it would just be the same acid-base reaction. (And this wouldn’t be considered a nucleophilic attack, usually.) Instead, the nitrogen needs to go to the carbonyl group — which happens to be the most positive carbon in an amino acid, so should be a great target for a nucleophile.

The initial result, as I tried to show in the scheme below, is a tetrahedral carbon with a negative charge on the ex-carbonyl oxygen and a positive charge on the nitrogen. If one of the nitrogen’s protons jumps across to the ex-hydroxy oxygen, that one gets positively charged, and then electron pairs move, we release water and gain our peptide bond.

(The picture is really only the basic mock-up and no real effort put into it. It neither exactly matches the biological peptide bond mechanism, nor does it match the mechanism that a chemist would use in the lab. But it conveys the basic picture.)

So while you did technically use the basicity (or postivie-charge loving) of the nitrogen, you see that the acidity of the carboxyl group plays no role in the reaction itself.


Detection limit for log-log calibration curves

I can add some points for the OP which are too long to serve as comments. I have talked to a couple of leading persons who work with a lot a of LODs. The first lesson, which is important, that there is no "the limit of a detection" of an analytical procedure. Second, some analytical chemists also state that it should be abolished because of the same reason- there is no the limit of detection. Limit of detection varies from method to method of calculation. You will find a couple of common things in legally accepted methods for LODs as the German DIN-32645, and ISO-BS 11843-2:2000 (Incorporating corrigendum October 2007). They all deal with linear calibration curves and no one want to mess around with quadratic or log transformed calibration curves. In the literature, in my opinion, LOD is the most incorrectly reported number and the experts agree with this opinion in personal conversations.
Now quadratic calibration curves are completely kosher. They are being used in atomic absorption spectrophotometers for ages (~ 40 years). Similarly, another HPLC detector based on light scattering does a small manipulation that makes the calibration curve non-linear. These detectors are used in pharmaceutical industry all the time. So, if your sole purpose is to do quantification, within 1 to 500 mg/L, this calibration curve should serve you well but do not calculate the LODs using the formula 3.3xsigma/slope. The problem is the determination of sigma, and the multiplying factor. When Kaiser developed these formulae (mostly in German papers), he never bothered to say anything about nonlinearity because nobody wanted to work with non-linear calibration curves before the computers.
The log transformation as suggested in the comments changes the noise behavior. If you really wish to determine a number called LOD for this instrument you can do the following (for the satisfaction of a novice reviewer of your paper or another higher up who wants a LOD number from you).
P.S. Do not round off concentrations and areas, the instrument readings look simulated to me because they are only changing in units of 50 and 100. Round off is a bad idea, it should only be done at the end.
a) Check the linearity range of your method. I plotted the data up to 50 mg/L. It is linear.
b) Obtain the readings in the linear range in triplicate or quadruplicate. You already have this data.
c) Look at the standard deviation of peak area vs. concentration curve. As you can see it is not a flat line (slope of 0). Hence the Kaiser's formula fails right here. You cannot apply 3.3*sigma/slope formula. The noise is heteroscedastic. Kaiser's formula is valid for homoscedastic noise.
Suggestions to obtain an LOD of your method.
d) Make about 7-8 standards in the range 0.1 mg/L to 20 mg/L. Repeat the process a to c. What we need is a standard near the estimated LOD.
e) If you still see a non-zero slope for the peak area standard deviation vs. concentration, apply a weighted least squares method. ISO-BS 11843-2:2000 has all the solved examples.


You are correct in thinking that urea and sucrose would have van't hoff factor (i) = 1, since they are non-electrolytes and don't undergo disassociation(or ion pairing).

What you are left with are the following: $\ce{NaCl}$ , $\ce{BaCl_2}$ and $\ce{[Cr(NH_3)_4Cl_2]Cl}$.

Let us begin with the the two chloride salts: $\ce{NaCl}$ , $\ce{BaCl_2}$. we assume 100% disassociation for the two (strong electrolytes) and we have i = 2 and i = 3 respectively for both.

Now, the complex would indeed disassociate but not completely (the extent would be determined $K_f$ for the complex). Thus 1 < i < 2.

Based on this, I believe you can order them on the basis of increasing osmotic pressure.


The reasoning stated is partially correct, but the final relation you arrived to is incorrect. I will try to explain why and write the N dependence explicitly for completeness. The crucial thing is that when one writes an expression such as $$\left(\frac{\partial H}{\partial T} \right)_{P,N}$$
what one really means is "take the partial derivative of $H$ written as a function of $T$, $P$ and $N$ with respect to $T$". When you take the partial derivatives in equation (2) then, you should take them considering $H$ as a function of $T$, $P$ and $N$. You have to consider then the expression
$H = H(T,P,N) = S(T,P,N)T + \mu(T,P) N$. If you differentiate that equation with respect to T and P the result is:

$$\left(\frac{\partial H}{\partial P} \right)_{T,N} = \left(\frac{\partial S}{\partial P} \right)_{T,N} T + \left(\frac{\partial \mu}{\partial P} \right)_{T} N~~~;~~~ \left(\frac{\partial H}{\partial T} \right)_{P,N} = \left(\frac{\partial S}{\partial T} \right)_{P,N} T + S + \left(\frac{\partial \mu}{\partial T} \right)_{P} N$$
If you replace those two relations in your equation (2) the result is:

$$ \mathrm{d}H = \left(\left(\frac{\partial S}{\partial P} \right)_{T,N} T+ \left(\frac{\partial \mu}{\partial P} \right)_{T} N \right) \mathrm{d}P + \;\left(\left(\frac{\partial S}{\partial T} \right)_{P,N} T + S+ \left(\frac{\partial \mu}{\partial T} \right)_{P} N\right)~\mathrm{d}T.$$

You can indeed equate this with your equation (1), which is what you ask in your main question. This is the same thing one ordinarily does when expressing a scalar as a function of different sets of coordinates, for instance $f = f(x,y) = x^2 + y^2$ and $f = f(r,\theta) = r^2$ this means, equating both, that $x^2 + y^2 = r^2$, which is a relationship that must hold if you want both $(x,y)$ and $(r,\theta)$ to refer to $f$ (this last sentence may be a bit tautological, I hope what it means is clear, note that it is certainly not "always" true that $f(x,y)$ and $f(r,\theta)$ are equal, since they are two different functions, albeit expressed with the same letter, for instance, $f(x=1,y=0) = 1$ but $f(r=2,\theta= \pi) = 4$, there must be a specific relation between the coordinates for these to be equal). If you do this you obtain:

$$\left(\left(\frac{\partial S}{\partial P} \right)_{T,N} T+ \left(\frac{\partial \mu}{\partial P} \right)_{T} N \right) \mathrm{d}P + \;\left(\left(\frac{\partial S}{\partial T} \right)_{P,N} T + S+ \left(\frac{\partial \mu}{\partial T} \right)_{P} N\right)~\mathrm{d}T = T~\mathrm{d}S + V~\mathrm{d}P.$$

Note that you would get the same expression even if you considered processes in which $\mu \mathrm{d}N$ wasn't cero, cause both terms would cancel out. If one remembers that $T\mathrm{d}S = T\left(\frac{\partial S}{\partial P} \right)_{T,N}\mathrm{d}P + T\left(\frac{\partial S}{\partial T} \right)_{P,N}\mathrm{d}T$ then this simplifies to:

$$\left(\left(\frac{\partial \mu}{\partial P} \right)_{T} N \right) \mathrm{d}P + \;\left(S+ \left(\frac{\partial \mu}{\partial T} \right)_{P} N\right)~\mathrm{d}T = V~\mathrm{d}P.$$

Dividing through by N:

$$\left(\frac{\partial \mu}{\partial P} \right)_{T} \mathrm{d}P + \;\left(\bar{S}+ \left(\frac{\partial \mu}{\partial T} \right)_{P} \right)~\mathrm{d}T = \bar{V}~\mathrm{d}P.$$

This is true if and only if:

$$\left(\frac{\partial \mu}{\partial P} \right)_{T} = \bar{V}~~;~~\left(\frac{\partial \mu}{\partial T} \right)_{P} = -\bar{S}.$$

These are correct relations and can also be deduced from the Gibbs-Duhem equation: $N\mathrm{d}\mu -V\mathrm{d}P + S\mathrm{d}T = 0$.


Charge and spin multiplicity change in IRC search

In other words, does the IRC search still occur on the doublet PES or has it (somehow) gone to the singlet PES?

You could check that by looking at the output. Right at the end of each and every SCF optimization Gaussian prints a block that starts with SCF Done and contains information about energy of the state to which SCF converged and its total spin. For instance, below is one such block in the results of test170 calculations from Gaussian 09 test set:

 SCF Done:  E(UHF) =  -40.1568456632     A.U. after   10 cycles
             Convg  =    0.6432D-08             -V/T =  1.9978
 <Sx>= 0.0000 <Sy>= 0.0000 <Sz>= 0.5000 <S**2>= 0.7832 S= 0.5165

Now, regarding GaussView, interestingly, it also wrongly shows that the state is singlet and not a doublet for that test job. Besides, GaussView reports the following when opening the output:

  Unresolvable inconsistency between charge and multiplicity and orbital occupancies. Orbital data will be ignored.

Honestly speaking, I suspect that is a bug, because while reading the output I could not see any inconsistencies. I've tried few other IRC jobs on radicals form the Gaussian test set (test171, test452, test453) and GaussView always reports that a state is singlet when in fact it is a doublet. Moreover, in some cases GaussView even failed to read all the steps of a normally terminated IRC calculation. So, the only inconsistency I've noticed so far is that GaussView could not always correctly read Gaussian outputs of IRC calculations on states that are not singlets.


This answer uses electronegativities for the calculation of oxidation states as proposed in the Expanded Definition of the Oxidation State by Hans-Peter Loock in 2011.

Comparing Electronegativities

The following table shows an excerpt from Pauling electronegativities ($\chi_{\mathrm{Pauling}}$):

$$\begin{array}{cc}
\hline
\text{Element} & \chi_{\mathrm{Pauling}}\\
\hline
\ce{H} & 2.20\\
\ce{N} & 3.04\\
\hline
\end{array}$$

Looking at $\ce{N-H}$ Bond (heteroatomic)

The table shows, that the nitrogen atom is more electronegative than the hydrogen atom. It will therefore attract the electrons more than hydrogen does, and thus polarize the bond. Imagining the gedankenexperiment of an hypothetical heterolytical bond cleavage - presumed by applying pulling force to the atoms, until they get separated - it is assumed, that the electrons are allocated to the nitrogen atom during separation.

Looking at $\ce{N-N}$ and $\ce{N=N}$ Bonds (homoatomic)

Looking again at the previous gedankenexperiment, now with two identical Atoms bound to each other. Since they have, of course, the same electronegativity, there is no polarization of the bond. It is assumed, that separation would therefore result in an homolytical bond cleavage, allocating always one electron to each of the two nitrogen atoms for each bond respectively.

Summing up

Now, having a look on the complete structure of the molecule in question, and applying the previously states rules:

Last thing to do is calculating the atoms hypothetical charge after separation, which is to be equatable with the oxidation state:

$$ \text{Oxidation state} = N_\mathrm{i}(\ce{e-}) - N_\mathrm{f}(\ce{e-}) $$

With $ N_\mathrm{i}(\ce{e-}) $ representating the number of electrons in a free atom, and $ N_\mathrm{f}(\ce{e-}) $ the one after separation (One should not forget the lone pairs).

One will end up with the following oxidation states for the different (nitrogen) atoms, with the last entry representating the mean oxidation state of all nitrogen atoms:

$$\begin{array}{cccc}
\hline
\text{Atom} & N_\mathrm{i}(\ce{e-}) & N_\mathrm{f}(\ce{e-}) &\text{Oxidation state}\\
\hline
\ce{H} & 1 & 0 & +1\\
\hline
\ce{N-1} & 5 & 5 & 0\\
\ce{N-2} & 5 & 5 & 0\\
\ce{N-3} & 5 & 6 & -1\\
\hline
\ce{N} & - & - & -\frac{1}{3}\\
\hline
\end{array}$$

And the structure with assigned oxidation states:


What to do about a un-dissolved substance?

I think what you forgot is that the solubility of boric acid (or sodium chloride.. or most solids) is lower at room temperature.

So when you dissolved the boric acid in boiling water, there's no problem at first. As it cooled, you will have a supersaturated solution until eventually the boric acid will recrystallize or crash out.

I'd create a new mixture using room temperature water and add only a small amount of boric acid (as mentioned by @K_P, about 5g / 100 mL of water). Let it dissolve slowly while you mix.

Then if you spray, you shouldn't leave large amounts of boric acid around.

Updated Answer

Yes, if there's only ~10% water, you will want to add more (room temperature) water. This will dissolve more boric acid. Stir/shake to dissolve as much as possible, wait a bit and then you can spray again.

As mentioned, you don't want to add water to a hot acid. In this case, adding more room temperature water to a room temperature acid will probably produce heat, but add slowly and carefully.


It should be N-methyl-2-butanamine. Amine is treated as a functional group attached. So we write it as alkanamine as it is the only functional group here.

Image source: ChemSpider


It is ambiguous. 2500 can either be 2.5E3 or 2.500E3. The number 2506 would clearly be 2.506E3. Changes of last two digits being zeros randomly is 1%. If this is a home question or on a test then work the answer to 4 figures state that the problems seems ambiguous and that your assumption is that there are only two significant figures in the given temperature and round your answer to 1.6E3.

I'll point out that your answer has the same ambiguity. Is 1600 really 1.6E3 or 1.600E3?


Acidic nature of boric acid when its concentration high

In the equation $\ce{H2O + H3BO3 -> H+ + B(OH)4-}$, one negative charge on the anion is shared among 4 electronegative oxygens. Yes, one positive charge on the proton is also shared with some water molecules, so the orthoborate ion probably shares some negative charge with some waters, but let's keep it simple.

Boric acid is known to complex with sugars to generate higher acidity. If boric acid complexes with itself, as in $$\ce{3 H3BO3 -> H+ + [B3O3(OH)4]- + 2 H2O}$$, the one negative charge on the anion can be considered to be distributed among 7 electronegative oxygens, or maybe still distributed mostly on the OH groups, but since they will be separated more, the stability of the ion will be greater, and the tendency to lose a proton will be greater: greater acidity. The electronegativity of boron is less than that of hydrogen (2.02 vs 2.2, electronegativity, Wikipedia), so when the atoms in the ions are counted, the first ionization equation has one hydrogen and one boron as the metallic atoms getting the positive charge, while the second equation has one hydrogen and 3 borons donating electrons - more sharing of the positive charge here, tho the hydrogen still gets the positive charge.

But this structure may be only a suggestion. Well-known polyborate anions include the triborate(1-), tetraborate(2-) and pentaborate(1-) anions. The condensation reaction for the formation of tetraborate(2-) is as follows Wikipedia

$$\ce{2 B(OH)3 + 2 [B(OH)4]− ⇌ [B4O5(OH)4]2- + 5 H2O}$$

The italicized text mentions an observation of pH that is consistent with the complexation of boric acid with itself. It doesn't prove the existence of the triborate ion only, and perhaps is only a way to keep the text shorter without ignoring complexation completely. Then along comes a student who looks into the sentence and sees that something was omitted or not explained thoroughly. (Keep asking questions!)


Is cobalt(II) in the hexaamminecobalt(II) complex high spin or low spin?

Theoretically, you cannot predict a priori whether a compound is high- or low-spin. For some reason, a lot of people seem to think that it depends only on the ligand and that it is possible to unambiguously use the position of the ligand in the spectrochemical series to figure out whether a complex is high- or low-spin. This is a very narrow viewpoint and leads to lots of mistakes: for example $\ce{[Co(H2O)6]^3+}$ is low-spin although $\ce{H2O}$ is fairly low on the spectrochemical series. In truth it depends on (at least) the ligand, the metal, as well as the oxidation state, and there is no magic formula or rule that allows you to combine all three factors.

However, as a general rule of thumb, most 3d metal complexes are high-spin. If you return to the fundamental criterion for high- vs low-spin, i.e. the relative magnitudes of $\Delta_\mathrm o$ and the pairing energy, you will find that for many 3d metals, $\Delta_\mathrm o$ is small due to the poor overlap of the 3d orbitals with ligand orbitals. $\ce{[Co(NH3)6]^2+}$ falls squarely into this category.

The main exceptions occur when you have ligands that are very high on the spectrochemical series. These are nearly always π-acceptors, such as $\ce{CN-}$, bpy, phen, etc. Even so, it should be noted that there are some 3d π-acceptor complexes that are still high-spin, such as $\ce{[Co(bpy)3]^2+}$, so this shouldn't be taken as a rule but rather a rough generalisation. The other big exception is when you have high oxidation states, mainly +3 or higher. So, for example, Co(III) is nearly always low-spin except in $\ce{[CoF6]^3-}$. But there are plenty of high-spin Mn(III) and Fe(III) complexes, so even there it is not a clear-cut rule.


Calculate the amount of substance of an isotope based on the percent composition

If you have given an accurate account then your teacher is mistaken. We've all gotten confused while trying to talk and think at the same time.
\begin{align}
\pu{42.3 g} \ce{Cl}\cdot 0.24\frac{m(\ce{^37Cl})}{m(\ce{Cl})}
&= \pu{10.2 g} \ce{^37Cl} \\
\implies
\frac{\pu{10.2 g}~ \ce{^37Cl}}{\pu{36.97 g//mol}\ce{^37Cl}} &= \pu{0.276 mol}
\end{align}


Are there any "catalysts" that decrease the rate of reaction

Well, such a "negative catalyst" cannot increase the activation energy of already existing path.
An alternative with the higher activation energy would not be significantly taken. You can detour a river making it flow easier by a side channel, but not if you lead it across a hill.
It could act differently, as an  inhibitor, blocking the low activation energy path by reaction with the key intermediate product or existing catalyst, or by competitive inhibition.
Many redox reactions, including food deterioration, are catalyzed by traces of transient metals acting as redox catalysts, typically copper. An additive, forming metal complexes like EDTA, in large extend inhibits these catalyzed reactions.


An automated approach to visualize several xyz structures

You can use VMD to do this.
Note that the following is only tested on Linux with VMD version 1.9.2, i don't know how it works out on Windows or Apple OS.
The following is from a script that i used to render a trajectory. The first thing to do is to load your xyz file as trajectory. I.e. in a format so that vmd allows you to "slide" through the structures in its normal view. Then execute the script in the tlc console. The script iterates over the steps in incremenets of 2 and assumes that you have 2000 structures. The iteration variable is i. The actual command that changes the frames is the animate goto $i command. You would need to adjust that to fit your requirement. I used the tachyion renderer and saved the output figures in tga format. You will most likely have to adjust the path to the renderer binary which is the path /software/vmd-1.9.2/lib/vmd/tachyon_LINUXAMD64 in my case. VMD saves first a .dat file that contains the scene and then applies the renderer to this dat file. This is why i am setting two path variables, first fin for the .dat file and then fout for the actual tga file.
for {set i 0} {$i < 2001} {incr i 2} {
animate goto $i
set fin [format "pic_%04d.dat" $i]
set fout [format "pic_%04d.tga" $i]
render Tachyon $fin "/software/vmd-1.9.2/lib/vmd/tachyon_LINUXAMD64" -aasamples 12 $fin -format TARGA -o $fout -res 1920 1080
}

The nice part about this is that you can implement other commands during your loop, like a zoom or a rotation. All you have to do, is find out what the tcl command in vmd is.


Explain volume contraction in mixtures of alcohol and water

Liquid water can be thought of as having an average configuration that tends to be an open, almost crystalline structure, somewhat like that for hexagonal ice crystals $\mathrm{^{[1,2,3]}}$.  This structure has been confirmed by both x-ray and neutron diffraction experiments.  Below is a depiction of the hexagonal structure of ice for reference:

Source: Boundless. “The Structure and Properties of Water.” Boundless Chemistry Boundless, 26 May. 2016. Retrieved 06 Feb. 2017 from boundless.com

This phenomenon of the ice-like structure of liquid water is also illustrated in the observation that at temperatures below $\pu{4^oC}$, the density of water decreases with decreasing temperature as it becomes more ice-like prior to actual solid crystalline ice formation.

So, if we envision this average crystal-like configuration for liquid water, then two effects come into play regarding the volume contraction observed when water is mixed with ethanol.

Firstly, some fraction of the hydrocarbon tails of the ethanol molecules is able to fit within this lattice, filling some of the voids within this water structure.

Secondly, the hydroxyl moiety of ethanol engages in hydrogen bonding with water.  As ethanol only has a single hydrogen bond donor, however, the crystal-like structure of the liquid water cannot be fully maintained, and thus it somewhat collapses.

Also, please take a look at these answers regarding the general concept of volume contraction of liquid mixtures, not specific to the case of ethanol and water.

1 http://www.diss.fu-berlin.de/diss/servlets/MCRFileNodeServlet/FUDISS_derivate_000000003174/2_Chap1.pdf;hosts=
2 http://crystal.med.upenn.edu/sharp-lab-pdfs/sharp_EncLifeSci.pdf
3 "Water structure and changes in thermal stability of the system poly(ethylene oxide)–water", J. Chem. Soc., Faraday Trans. 1, 1981,77, 2053-2077