Is the density of the solution significantly different from the density of the solvent?

This depends a great deal both on the solute and the concentration.

At the concentrations that we normally drink coffee at, you can almost certainly ignore the difference in density of the solution. But ccording to your question you are trying to dissolve 0.1m3 coffee (I assume the minus sign is an error) in 0.5m3 water. That's 100L of coffee in 500L of water, a 100L/600L=16.667% volume/volume solution! A lot of very strong coffee!

As a general rule, the density rho is given as:

rho = (mass component 1 + mass component 2) / (volume component 1 + volume component 2 )

This assumes (correctly) that masses can be added. Strictly speaking, the volume of a solution is not equal to the sum of the volumes of its components, as there is a slight volume change on dissolution, but this is normally insignificant.

The hard part is to know the solid density of coffee, because it cannot be readily measured, as coffee comes in powder form. If you are not using instant coffee there is the problem that the husks of the grains will not dissolve. But in your case, you are told the density of the coffee is 1500kg/m3

So you can easily work out the masses of water and coffee, and then the density of your enormous vat of extra strong coffee.

Hint: it's going to be more than water, but not different enough to make any difference regarding the correct navier stokes model to follow.

EDIT: on re-reading, it seems you are assuming that only 20% of your 0.1m3 of coffee will dissolve: that is 0.02m3 in 0.5m3 of water, an approximate concentration of 4%. That's not quite as absurdly strong as before. I was a little confused because 0.1m3 is 20% of 0.5m3. However I'm not sure how you decided that you would reach equilibrium with 20% of the available coffee dissolved.


What is the mathematical basis behind the Jahn-Teller effect?

1. Perturbations

As already mentioned, the Jahn–Teller effect has its roots in group theory. The essence of the argument is that the energy of the compound is stabilised upon distortion to a lower-symmetry point group. This distortion may be considered to be a normal mode of vibration, with the corresponding vibrational coordinate $q$ labelling the "extent of distortion". There is one condition on the vibrational mode: it cannot transform as the totally symmetric irreducible representation of the molecular point group, as such a vibrational mode cannot bring about any distortion in the molecular geometry (it may lead to a change in equilibrium bond length, but not in the shape of the molecule).
$\require{begingroup} \begingroup
\newcommand{\En}[1]{E_n^{(#1)}}
\newcommand{\ket}[1]{| #1 \rangle}
\newcommand{\n}[1]{n^{(#1)}}
\newcommand{\md}[0]{\mathrm{d}}
\newcommand{\odiff}[2]{\frac{\md #1}{\md #2}}$

In the undistorted geometry (i.e. $q = 0$), the electronic Hamiltonian is denoted $H_0$. The corresponding unperturbed electronic wavefunction is $\ket{\n{0}}$, and the electronic energy is $\En{0}$. We therefore have

$$H_0 \ket{\n{0}} = \En{0}\ket{\n{0}} \tag{1}$$

Here, the Hamiltonian, wavefunction, and energy are all functions of $q$. We can expand them as Taylor series about $q = 0$:

$$\begin{align}
H &= H_0 + q \left(\odiff{H}{q}\right) + \frac{q^2}{2}\left(\frac{\md^2 H}{\md q^2}\right) + \cdots \tag{2} \\
\ket{n} &= \ket{\n{0}} + q\ket{\n{1}} + \frac{q^2}{2}\ket{\n{2}} + \cdots \tag{3} \\
E_n &= \En{0} + q\En{1} + \frac{q^2}{2}\En{2} + \cdots \tag{4}
\end{align}$$

In the new geometry (i.e. $q \neq 0$), the Schrodinger equation must still be obeyed and therefore

$$H\ket{n} = E_n \ket{n} \tag{5}$$

By substituting in equations $(2)$ through $(4)$ into equation $(5)$, one can compare coefficients of $q$ to reach the results:

$$\begin{align}
\En{1} &= \left< \n{0} \middle| \odiff{H}{q} \middle| \n{0} \right> \tag{6} \\
\En{2} &= \left< \n{0} \middle| \frac{\md^2 H}{\md q^2} \middle| \n{0} \right> + 2\sum_{m \neq n}\frac{\left|\left<m^{(0)} \middle|(\md H/\md q)\middle|\n{0} \right>\right|^2}{\En{0} - E_m^{(0)}} \tag{7}
\end{align}$$

The derivation of equations $(6)$ and $(7)$ will not be discussed further here.1

Distortions that arise due to the $\En{1}$ term are called first-order Jahn–Teller distortions, and distortions that arise from the $\En{2}$ term are called second-order Jahn–Teller distortions.

2. The first-order Jahn–Teller effect

Recall that

$$E_n = \En{0} + \color{red}{q\En{1}} + \cdots \tag{8}$$

Therefore, if $\En{1} > 0$, then stabilisation may be attained with a negative value of $q$; if $\En{1} < 0$, then stabilisation may be attained with a positive value of $q$. These simply represent distortions in opposite directions along a vibrational coordinate. A well-known example is the distortion of octahedral $\ce{Cu^2+}$: there are two possible choices, one involving axial compression, and one involving axial elongation. These two distortions arise from movement along the same vibrational coordinate, except that one has $q > 0$ and the other has $q < 0$.

In order for there to be a first-order Jahn–Teller distortion, we therefore require that

$$\En{1} = \left<\n{0}|(\md H/\md q)| \n{0}\right> \neq 0 \tag{9}$$

Within group theory, the condition for the integral to be nonzero is that the integrand must contain a component that transforms as the totally symmetric irreducible representation (TSIR). Mathematically,

$$\Gamma_{\text{TSIR}} \in \Gamma_n \otimes \Gamma_{(\md H/\md q)} \otimes \Gamma_n \tag{10}$$

We can simplify this slightly by noting that the Hamiltonian, $H$, itself transforms as the TSIR. Therefore, $\md H/\md q$ transforms as $\Gamma_q$, and the requirement is that

$$\Gamma_{\text{TSIR}} \in \Gamma_n \otimes \Gamma_q \otimes \Gamma_n \tag{11}$$

In all point groups, for any non-degenerate irrep $\Gamma_n$, we find that $\Gamma_n \otimes \Gamma_n = \Gamma_{\text{TSIR}}$. Therefore, if $\Gamma_n$ is non-degenerate, then

$$\Gamma_n \otimes \Gamma_q \otimes \Gamma_n = \Gamma_q \neq \Gamma_{\text{TSIR}} \tag{12}$$

and the molecule is stable against a first-order Jahn–Teller distortion. Therefore, all closed-shell molecules ($\Gamma_n = \Gamma_{\text{TSIR}}$) do not undergo first-order Jahn–Teller distortions.

However, what will happen if $\Gamma_n$ is degenerate? Now, the product $\Gamma_n \otimes \Gamma_n$ contains other irreps apart from the TSIR.2 If the molecule possesses a vibrational mode that transforms as one of these irreps, then the direct product $\Gamma_n \otimes \Gamma_q \otimes \Gamma_n$ will contain the TSIR.

In a rather inelegant article,3 Hermann Jahn and Edward Teller worked out the direct products for every important point group and found that:

  stability and degeneracy are not possible simultaneously unless the molecule is a linear one...

In other words, if a non-linear molecule has a degenerate ground state, then it is susceptible towards a (first-order) Jahn–Teller distortion.

Take, for example, octahedral $\ce{Cu^2+}$. This has a $\mathrm{^2E_g}$ term symbol (see this question) - which is doubly degenerate. The symmetric direct product $\mathrm{E_g \otimes E_g = A_{1g} \oplus E_g}$. Therefore, if we have a vibrational mode of symmetry $\mathrm{E_g}$, then distortion along this vibrational coordinate will occur to give a more stable compound. Recall that the vibrational mode cannot transform as the TSIR, so we can neglect the $\mathrm{A_{1g}}$ term.

What does an $\mathrm{e_g}$ vibrational mode look like? Here is a diagram:4

It's an axial elongation, which happens to match what we know of Cu(II). However, there is a catch. The vibrational mode is doubly degenerate (the other $\mathrm{e_g}$ mode is not shown), and any linear combination of these two degenerate vibrational modes also transforms as $\mathrm{e_g}$. Therefore, the exact form of the distortion can be any linear combination of these two degenerate modes. It can also involve negative coefficients, i.e. it might feature axial compression instead of elongation; there is no way to find that out using arguments purely based on symmetry.

Therefore, in this case, it is simply a coincidence that the $\mathrm{e_g}$ mode displayed is exactly the same as the form of the distortion seen in Cu(II). Nevertheless, it is encouraging to see that the axial elongation indeed transforms as $\mathrm{e_g}$ - which lends credence to the group theoretical analysis above.

On top of that, there's also no indication of how much distortion there is. That depends on (amongst other things) the value of $\En{1}$, and all we have said is that it is nonzero - we have not said how large it is.

This is what is meant by "impossible to predict the extent or the exact form of the distortion".

3. The second-order Jahn–Teller effect

Pearson has written an article on second-order Jahn–Teller effects.5

For the second-order term, the energy correction is of the form

$$E_n = \En{0} + q\En{1} + \color{red}{\frac{q^2}{2}\En{2}} \cdots \tag{13}$$

Here, the $q^2$ term means that $\En{2}$ has to be negative if we want to see a second-order Jahn–Teller distortion. Unlike the first-order case, if $\En{2} > 0$, there will not be any distortion.

The second-order correction to the energy comprises two terms. The first term, $\left<\n{0}|(\md^2 H/\md q^2)|\n{0}\right>$, is always positive. (The proof of this is left to the reader. Hint: use the fact that $\md H/\md q$ is hermitian.) It may be interpreted as a restoring force that tries to bring the nuclei back to their original positions, and it is related to the fact that if the electronic state remains unperturbed (i.e. $\ket{n} = \ket{\n{0}}$), the unperturbed nuclear positions represent the most stable nuclear configuration.

The second term has the form

$$\sum_{m \neq n}\frac{\left|\left<m^{(0)} \middle|(\md H/\md q)\middle|\n{0} \right>\right|^2}{\En{0} - E_m^{(0)}} \tag{14}$$

which may seem like a slight monstrosity, but it is actually much easier to analyse than it looks. The summation over $m$ indicates that we are going to count every single electronic state $\ket{m}$ that is not the ground state $\ket{n}$. Since the numerator is a square modulus, it is either zero or positive, and since $\En{0} < E_m^{(0)}$, the denominator is always negative; therefore, this term is necessarily either zero or negative.

If $\En{2}$ is to be negative, then we need the second term to dominate the first. For this to occur, there are two prerequisites:

The denominator must be small, i.e. $\ket{m}$ must be a low-lying excited state such that the energy gap $\Delta E = \En{0} - E_m^{(0)}$ is small;
The numerator must not be zero, i.e. there must be a low-lying excited state of the appropriate symmetry $\ket{m}$ such that $\Gamma_m \otimes \Gamma_q \otimes \Gamma_n$ contains $\Gamma_{\text{TSIR}}$.

The first condition usually means that it will suffice to consider the first few excited states. In many of the examples I know of, the excited state that mixes with the ground state is the first excited state. In such cases one can even simply get rid of the sum and set $\ket{m}$ to be the first excited state.

These symmetry requirements are much less restrictive than previously, and second-order Jahn–Teller distortions tend to be much more widely seen than first-order distortions. A small selection of compounds in which second-order Jahn–Teller distortions are important are: p-block hydrides, $\ce{PbO}$, $\ce{Hg^2+}$, $\ce{WMe6}$, $\ce{R2Sn=SnR2}$, and anti-aromatic compounds such as cyclobutadiene. Let us use octahedral $\ce{Hg^2+}$ as an illustration.

In undistorted $O_\mathrm{h}$ symmetry, $\ce{Hg^2+}$ has a closed-shell $\mathrm{d^{10}}$ configuration and therefore its electronic ground state is $\Gamma_n = \mathrm{A_{1g}}$. However, upon excitation of one electron from the 5d orbitals (specifically the $\mathrm{e_g}$ set) to the 6s orbital (which transforms as $\mathrm{a_{1g}}$), the term symbol changes to

$$\Gamma_m = \mathrm{E_g \otimes A_{1g} = E_g} \tag{15}$$

Therefore, a vibrational mode transforming as $\mathrm{E_g}$ will facilitate a second-order distortion, since

$$\Gamma_m \otimes \Gamma_q \otimes \Gamma_n = \mathrm{E_g \otimes E_g \otimes A_{1g}} \tag{16}$$

contains the TSIR. Again, there is no way of knowing the exact form or the extent of the distortion; we only know that it transforms as $\mathrm{E_g}$. In the case of Hg(II), the distortion is manifested as an axial compression to give a "2 short, 4 long" coordination geometry, which is often described as "linear".

A second factor that favours the distortion is the extremely small 5d–6s gap in mercury, due to relativistic 5d destabilisation and 6s stabilisation. To see the importance of the small $\Delta E$, consider $\ce{Zn^2+}$, which has a larger 3d–4s gap; linear Zn(II) compounds are rare, while linear Hg(II) compounds are the norm.

Most of the time, the relevant excited state $\ket{m}$ arises from promotion of an electron from the HOMO to the LUMO. It is easy to show that if this is the case,

$$\Gamma_m \otimes \Gamma_n = \Gamma_{\text{HOMO}} \otimes \Gamma_{\text{LUMO}} \tag{17}$$

The second-order Jahn–Teller distortion can then be viewed as a reduction in symmetry, such that the HOMO and the LUMO, which transformed as different irreps in the undistorted geometry, now transform as the same irrep and therefore mix with each other. In the case of $\ce{Hg^2+}$:

This interpretation using the symmetry of individual orbitals, however, only works when the relevant excited state $\ket{m}$ is derived by excitation of an electron! In some (admittedly very rare) cases, it is possible that both $\ket{n}$ and $\ket{m}$ are derived from the same electronic configuration. This is the case for cyclobutadiene, and the Jahn–Teller effect in cyclobutadiene cannot be rationalised using orbital mixing. $\endgroup$

Notes and references

(1) For more details, look up perturbation theory in your quantum mechanics book of choice. In such treatments, the perturbation is usually formulated slightly differently: e.g. $H$ is taken as $H_0 + \lambda V$, and the eigenstates and eigenvalues are expanded as a power series instead of a Taylor series. Notwithstanding that, the principles remain the same.

(2) There is a subtlety in that the symmetric direct product must be taken. For example, in the $D_\mathrm{\infty h}$ point group, we have $\Pi \otimes \Pi = \Sigma^+ + [\Sigma^-] + \Delta$. The antisymmetric direct product $\Sigma^-$ has to be discarded.

(3) Jahn, H. A.; Teller, E. Stability of Polyatomic Molecules in Degenerate Electronic States. I. Orbital Degeneracy. Proc. R. Soc. A 1937, 161 (905), 220–235. DOI: 10.1098/rspa.1937.0142. n.b. Considering that I don't have a more elegant proof for it, I don't have much of a right to call it inelegant.

(4) Albright, T. A.; Burdett, J. K.; Whangbo, M.-H. Orbital Interactions in Chemistry, 2nd ed.; Wiley: Hoboken, NJ, 2013.

(5) Pearson, R. G. The second-order Jahn–Teller effect. J. Mol. Struct.: THEOCHEM 1983, 103, 25–34. DOI: 10.1016/0166-1280(83)85006-4.


Comparing energies of isoelectronic species, but one protonated

I think they can and should be compared. Consider the $\ce{H_2^+}$ molecular ion. When dissociating this correctly (that is, including symmetry breaking), one should eventually end up with $\ce{H}+\ce{H^+}$. One can then extract the dissociation energy of the bond from the calculations. This is equivalent to the proton affinity of atomic $\ce{H}$. By applying this logic to larger species, one reaches the conclusion, that yes, $\ce{H+}$ by definition has zero energy and should enter calculations of relative energies as such.

As an aside, the calculation of negatively charged species in the gas-phase can lead to erroneous results, but any argument made for $\ce{A-/AH}$ will certainly hold for $\ce{B/BH^+}$.


How to calculate concentration of unknown saturated stock solution from diluted solution whose concentration is known?

Unless I misunderstood (please give more info), you've already done the most difficult part of the job by determining the concentration of your diluted solution!
And you just diluted your stock solution ten times, so you'd only have to multiply by ten the value that you found for your diluted concentration...


Is there a simple way to separate deuterium oxide from tap water?

It depends how pure you want the $\ce{D2O}$ to be and what you consider simple :)

Electrolysis of water strongly favors "H" being converted to hydrogen gas rather than "D", by a factor of about 8 to 1 (depending upon the electrodes).

Inexpensive Equipment for the Preparation and Concentration of Pure $\ce{D2O}$ Ohio Journal of Science volume 41, number 5, pages 357-365 (1941) explains in great detail how to carry out the electrolysis method.  The article says the process is suitable for "those who have occasion to prepare or concentrate heavy water for research, demonstration or
advanced student laboratory use".

In a first stage, $\ce{NaOH}$ solution (initially $\pu{0.5 M}$) is subject to electrolysis, until only 1/32 of the original volume remains.  This increased the original concentration of deuterium by a factor of 12.

By then the $\ce{NaOH}$ is too concentrated, and $\ce{CO2}$ is added (as dry ice) to neutralize the solution, and the first stage solution is distilled to get rid of the sodium carbonate. About $5~\%$ of the $\ce{NaOH(D)}$ concentrate would be saved and added to the distillate to provide electrolyte for the next stage.

Then a second stage of electrolysis is performed. By the end of the second stage they reached about $10~\%$ deuterium. The process can be repeated to achieve a desired deuterium level.  They used three stages.  Yield can be improved by capturing the deutrium/hydrogen gas of the late stages, burning it to make water, and feeding this back to an earlier stage.


You are correct in your assumption. For these types of problems it is usually safe to assume like dissolves like and choose the answer with the most similarity in terms of intermolecular bonding. For a general description you can check here.

When you get into questions with two options that have similar intermolecular forces it gets a little hairy. For example, I think a question like, "which is more soluble in carbon tetrachloride: n-Pentane or n-Hexane?" would be fairly challenging to answer without experimental or computational data to compare.


Stereospecificity of the bimolecular electrophilic substitution reaction

The $\mathrm{S_E2}$ mechanism is most commonly found using organometallic reagents ($\ce{R-Li, R-MgBr}$ etc.) in which the metal is the electropositive element react with electrophiles.
Unlike with $\mathrm{S_N2}$, there are multiple stereochemical outcomes, which can make it difficult to determine whether the reaction is indeed stereospecific. This is described in Modern Physical Organic Chemistry (emphasis mine):

One of the most interesting aspects of these reactions is stereochemistry. Varying results are obtained depending upon the metal, the solvent, and the R-group.
Source: Modern Physical Organic Chemistry, Ansyln and Dougherty [1]

Criterea for a stereospecific reaction
Before going any further, it may be instructive to define what we mean by stereospecific:

stereospecific is used for those reactions where the configuration of the starting material and the configuration of the product are related in a mechanistically constrained way
Source: Molecular Orbitals and Organic Chemical Reactions (Reference Edition), Fleming [2]

What this means in practice is that if one enantiomer of a starting material gives one enantiomer of product, then the opposite enantiomer of starting material must give the opposite enantiomer of product. What stereospecific does not mean is high selectivity, many reactions are highly stereoselective but in no way mechanistically stereospecific (and equally stereospecific reactions are not always $100~\%$ stereoselective).
The empirical evidence
Unlike the $\mathrm{S_E2}$ reaction, where many examples have been studied, it has been considerably more challenging to examine the stereochemical cause of the $\mathrm{S_E2}$ reaction. In general, chiral alkyl-metal species are prone to racemisation (chiral Grignards, as an example, are barely known), and those chiral species that are stable and usually such for steric reasons (such as a constrained ring system).
Many of the experiments that have been conducted are described in March[3], and the overwhelming majority of them go with retention (note that this doesn't necessarily imply a stereospecific process in itself and it has often been hard to study both enantiomers, for various reasons).
The orbital explanation
If we consider the orbitals involved during an $\mathrm{S_E2}$  reaction, it becomes clearer why there are multiple stereochemical outcomes, consider first $\mathrm{S_E2}$  at a saturated centre, such as $\ce{MeLi}$ attacking an electrophile :

Source: Molecular Orbitals and Organic Chemical Reactions (Reference Edition), Fleming [2]
In this reaction, the electrophile is quite able to interact in such a way as to give rise to overall inversion or overall retention, in order to distinguish, these two electrophilic substitutions are labelled $\mathrm{S_E2}_\text{back}$ and $\mathrm{S_E2}_\text{front}$ respectively, though both are formally designated DEAE if the IUPAC system is followed.[4]
This differs from the $\mathrm{S_N2}$ reaction in which inversion is almost always observed due to an anti-bonding interaction in the transition state if the nucleophile approaches in such a way as to give retention:

Source: Molecular Orbitals and Organic Chemical Reactions (Reference Edition), Fleming [2]
Conclusion
The short answer is that it doesn't appear that the reaction is stereospecific. Orbitally speaking, there are two pathways leading to two different products (front or back attack), which may be observed in poor selectivity.
It appears, with the reactions studied so far, that the outcome is far more likely to be down to sterics/electronics as to whether the electrophile approaches from one side or the other

References and Notes:
[1]: Ansyln, E.V.; Dougherty, D.A.  Modern Physical Organic Chemistry; University Science Books:California, 2004
[2]: I. Fleming  Molecular Orbitals and Organic Chemical Reactions (Reference Edition); Wiley:Chichester, 2012
[3]: March, J.; Smith, M.B.  Advanced Organic Chemistry; Wiley:New Jersey, 2013
[4]: IUPAC Recommendations for the Representation of Reaction Mechanisms: Acc. Chem. Res. 1989, 22, 343


Why is the boat form of paraldehyde not theoretically possible?

A $\ce{CH_3}$ group is quite big. Steric hindrance is the reason.

Its not theoretically impossible, but it is energetically a lot less favourable than (1) and (2).

You wouldn't find it in nature so to speak.

to 2: A $\ce{OH}$ group is sterically a lot smaller than a $\ce{CH_3}$ group.


What is a chemical daily life experience that can be modelled by superior mathematics?

Diffusion is one example from the daily life. For example, the absorption of ethanol from the small intestine to blood can be modeled using differential equations. The extraction of caffeine from grounded coffee is also a diffusion phenomenon.


Creating a Battery using azobenzene in High School

Azobenzenes are among the most widely studied molecular photoswitches, yet, I'm not sure this is a project capable of doing in a high school lab. It probably need sophisticated instruments (e.g., azobenzene needs to be covalently bound to carbon nanotubes (CNTs)) and has done previously, notably by Department of Materials Science and Engineering in Massachusetts Institute of Technology (MIT) (Ref.1). Yet, I'd encourage you to consult people who know the field (such as Prof. Timothy J. Kucharski at MIT and Harvard) and get some advice how to do it in a laboratory environment or do a oral presentation (that would be cool as well). I'll attached some figures of Ref.1 for your convenience here:

Figure 1: Synthesis and characterization of azobenzene-functionalized SWCNTs (Adapted from reference 1. Copyright @ 2014 Nature Publishing Group).

Figure 2: Photochemical charging and thermal energy release (Adapted from reference 1. Copyright @ 2014 Nature Publishing Group).

Figure 3: Photochemical and thermal cycling (Adapted from reference 1. Copyright @ 2014 Nature Publishing Group).

Figure 4: Effect of packing photochemical switches in the solid state (Adapted from reference 1. Copyright @ 2014 Nature Publishing Group).

Also read Ref.2 for more information on molecular solar thermal energy storage using chemicals other than azobenzene, which also included.

References:

T. J. Kucharski, N. Ferralis, A. M. Kolpak, J. O. Zheng, D. G. Nocera, J. C. Grossman, “Templated assembly of photoswitches significantly increases the energy-storage capacity of solar thermal fuels,” Nature Chemistry 2014, 6, 441–447 (DOI: 10.1038/NCHEM.1918).
A. Lennartson, A. Roffey, K. Moth-Poulsen, “Designing photoswitches for molecular solar thermal energy storage,” Tetrahedron Letters 2015, 56(12), 1457–1465 (https://doi.org/10.1016/j.tetlet.2015.01.187).


As titration is based on the relationship of molar quantity ($n$), volume  ($V$) and concentration ($c$) of the reagents

$$ c = \frac{n}{V}$$

and as your titration is to determine the concentration of NaOH in a known volume of solution of analyte, you would need to know both the volume as well as the concentration of the $\ce{H2SO4}$ solution deployed by addition via the burette.  The later is missing in your list.

Knowing the stoichiometry of the neutralisation is equally of importance, since each equivalent of $\ce{H2SO4}$ reacts with two equivalents of $\ce{NaOH}$. (This differs from a determination of, say, of / with $\ce{HCl}$, or $\ce{H3PO4}$.)  Your current list neither accounts for a chemical equation, nor this stoichimetric factor.


Why is an isothermal process not unique in nature?

(1) An isothermal process is one during which the system temperature $T=constant$. For an ideal satisfying $pV=RT$ this means that $pV=constant$ irrespective of how you achieve it. You can go up and down, back and forth on the same isothermal, but it is still an isothermal process having the same temperature. So it is not a unique process between two fixed points.

(2) The uniqueness of the adiabatic process is true only for a simple system, that is one whose state can be described by temperature and one additional "mechanical" parameter, e.g., for a gas $p = p(T,V)$, or a magnet $M=M(T,B)$, etc. It is definitely not true for more complicated systems, for example a gas with magnetic properties $p=p(T,V,B)$. Instead what is true is that through every equilibrium point there is a unique adiabatic surface (not a curve) determined by a pair of variables say $T,V$ or $V,B$ or $T,B$ such that any curve lying in that surface is an adiabatic process. That such adiabatic surfaces exist is the basis of one of the mathematical proofs of the existence of the entropy function and this quite famous proof is due to Caratheodory.


You can indeed find the conversion (not the absolute concentrations), even without calibrating the system.

At time t = 0, take an exactly measured volume of reaction mixture (e.g. 1 mL) and dilute it in a volumetric flask with water and/or acetonitrile to an exact volume (e.g. 100 mL).

Inject an exact volume (e.g. 50 uL) of the diluted solution in the LCMS and note down the (UV-based) peak area of the limiting reagent. This is your ${A_0 = \epsilon \cdot C_0 \cdot k}$, where ${C_0}$ is the concentration of your limiting reagent at this time, and ${k}$ is a constant related to the sample size and dilutions.

At any time t, do exactly the same as above, and you get an area ${A_t = \epsilon \cdot C_t \cdot k}$. Under these conditions, ${k}$ is the same, and of course so is ${\epsilon}$.

As you know, the conversion is ${y = 1 - \frac {C_t}{C_0}  = 1 - \frac {A_t}{A_0} }$.


Anything that stabilize one of the species in an acid-base equilibrium influences the free energy difference between the reactants and products, which alters the equilibrium constant:

$$\ce{HA <=> H+ + A-}$$

$$\Delta_r G^\circ = -RT\ln K_{eq}$$

You forget the solvent. Acidity and basicity depend on the interaction between the acid/base and the solvent. While steric hindrance will decrease the rate solvent access, it more importantly also decreases solvation of the acidic or basic site. More steric hindrance prevents additional solvent molecules from reaching the site to stabilize an ionic conjugate acid or base.

Consider the difference in $\mathrm{p}K_a$ between methanol, ethanol, 1-butanol, and tert-butanol. These are equilibrium constants, so they most certainly derived from thermodynamic phenomena!

Methanol $\ce{CH3OH}$ has a $\mathrm{p}K_a = 15.5$ (in water)

Ethanol $\ce{CH3CH2OH}$ has $\mathrm{p}K_a = 15.9$ (in water)

1-Butanol $\ce{CH3CH2CH2CH2OH}$ has $\mathrm{p}K_a = 16.1$ (in water)

t-Butanol $\ce{(CH3)3COH}$ has $\mathrm{p}K_a = 17$ (in water) (I've also seen 18 as its $\mathrm{p}K_a$)

The difference between methanol, ethanol, and 1-butanol (all primary alcohols) is small, while t-butanol has a larger pKa. If sterics had no effect, then both butanol isomers should have similar acidities since both isomers have similar inductive destabilization of the conjugate base by electron donating alkyl groups.

However in methanol, ethanol, and 1-butanol, more water molecules can solvate the alkoxide anion conjugate base, stabilizing that conjugate base more than the conjugate base of tert-butanol, which due to sterics cannot have as much solvation of the anion.

Now, gas phase acidities, in which there is no solvent, should have no solvent and solvation effects, and thus no (or at least reduced) steric effects.


Plastic membrane that lets water in from the outside only?

The comments above have gently hinted that it would be a violation of thermodynamics for a passive membrane to force a substance in a single direction. Molecules have an equal probability of moving through the membrane in either direction, unless you apply a force to make them move in a specific direction, such as electro-osmosis or active water transport in plants.

Consider that if such a membrane as you posit existed, you could use it to pump water from a lower level to a higher, and then power a water-wheel: perpetual motion. Sadly, Maxwell's demon is getting a bit arthritic to shuttle water molecules, and the dampness only aggravates its condition.


Role of tautomerism and its difference from hyperconjugation

Tautomerism basicially means that you have different isomers that are able to convert into each-other. This is not really similar to hyperconjugation which is an electronic effect in a given molecular geometry. The main difference between them is that hyperconjugation is thought to be a property of a given molecular geometry while tautomerization is an equlibrium between separate, distinct, stable configurations.

There are many types of tautomerizations, as explained for example on the Wikipedia. Probably the most well-known type of tautomerisation is the keto-enol tautomerisation, which is an equilibrium between a ketone molecule and an enol (an alcohol connected to an $sp^2$ carbon). The equilibrium between two isomers is the explanation why enols are only stable under special scenarios: they tautomerise into ketons! Another important example for tautomerism is the equlibrium between the chain and the ring isomers for sugars.

So again: tautomerism is an equilibrium between different (real, existing) isomers as opposed to hyperconjugation which is just defined within a single structure.


What is it about the relationship between the Fe²⁺ and Fe³⁺ in magnetite that makes it diamagnetic?

Magnetite has a spinel structure with two types of $\ce{Fe}$ sites: octahedral and tetrahedral, respectively in green and orange below.

(in this structure, oxygen atoms are the small red balls).

This structure is called a spinel, with $\ce{Fe^{2+}}$ ion in tetrahedral coordination and $\ce{Fe^{3+}}$> ions in octahedral sites. The coupling between atoms from these two lattices are superexchange interactions, which result in an antiparallel alignment of spins. However, because the magnetic moments of the two lattices are not equal, their addition results in a net magnetic moment, even though they are of opposition alignment. This is called ferrimagnetism, and in practice most of its macroscopic properties are similar to ferromagnetism, though they stem from a completely different microscopic magnetic ordering.


What really is an equation of state (EOS) and how are they developed?

They are merely brief and approximated ways to describe observed relations among state parameters. They do not need to have a theoretical background, but many times they have. It helps to create them and provides some reliability. The theory used to guide the construction of a EOS is anyone that turns to be useful.

how would he/she start and which script should he/she follow?

It is a very broad question. In thermodynamics there are few state parameters used normally. Thermodynamic theory is a macroscopic theory of somewhat general applicability, but EOS can be very specific, and this specificity can not be provided by thermodynamics, so, it turns to be a wide search.
In cases where simple general relations exist (mostly already discovered), it can be found just by fitting a data to a function.


Tips for maximizing yield, purity and crystal size during recrystallization

When I grew crystals of aluminium potassium sulphate I filled a jar with boiling water and placed it in a cooking pot that was also full of water. Doing this helps to maintain a constant temperature. I set the cooker to a low heat and began dissolving the alum until the solution was saturated. At this point I removed the jar containing the solution and suspended a piece of string for a crystal to grow on. This would be my seed crystal. I left the jar on a window sill to cool and waited.

A few days later a crystal had formed so I lifted out the string and set it aside. Then I reheated the solution to dissolve any crystals that had formed on the jar as they would compete with the suspended seed crystal, reducing its size. It is also a good idea to filter the solution at this point. Once I had finished this step I placed the string with the seed crystal back into the jar and placed it on the window sill again. After a week I moved the jar to a refrigerator (around 3 degrees C)  to further cool the solution. This left me with a pretty big, well shaped crystal.

I didn't do this but you could try evaporating off the water after the previous step to grow the crystal further.


For the half reaction HSO3- -> S2O6^2- why is it that 2e- are on the right hand side when balancing and not one electron?

Why is $2$ placed in front of the $\ce{HSO3}$?

This is, of course, the elemental law of the conservation of mass. You have $x$ atoms that you put into your equation, you need to get $x$ atoms out again. While hydrogen and oxygen can be taken care of with water, you need to account for the sulphur immediately. Hence you need two sulphurs.

  The oxidation number of S on the left is +IV and +V so when adding electrons to balance the oxidation number why is $\ce{2e-}$ added to the right instead of one?

The oxidation numbers are, of course, correct, but remember that in the first step you wrote a $2$ before the sulphur. Therefore you now have two sulphurs that you need to treat with the appropriate number of electrons — either by stealing or by gifting.

Since you have two sulphurs and they each lose one electron, you need to add $\ce{2 e-}$ to the right.

Note: The OP already understood what went wrong through hints in the comments. Unfortunately, OP decided not to write up a self-answer.


Assuming the "4-fold" is not referring to "use four equivalents of X per molecular function Y to obtain the intended chemical reaction", it refers more often to a specical class of star shaped polymers, called dendrimers.  Commonly, around a nucleous, you have several times the same chemical group, as, for example, drawn here:

In this example, instead of a linear chain of polyethylene glycol (PEG), four branches are stitched together; the fourth example may relate the closest to your question.

Source of the picture: Zhu, Biomaterials 31 (2010) 4639-4656, doi: 10.1016/j.biomaterials.2010.02.044


Ivan's answer is indeed thought-provoking. But let's have some fun.

IUPAC defines oxidation as:

  The complete, net removal of one or more electrons from a molecular
  entity.

My humble query is thus - what better way is there to remove an electron than combining it with a literal anti-electron? Yes, my friends, we shall seek to transcend the problem entirely and swat the fly with a thermonuclear bomb. I submit as the most powerful entry, the positron.

Since 1932, we've known that ordinary matter has a mirror image, which we now call antimatter. The antimatter counterpart of the electron ($\ce{e-}$) is the positron ($\ce{e+}$). To the best of our knowledge, they behave exactly alike, except for their opposite electric charges. I stress that the positron has nothing to do with the proton ($\ce{p+}$), another class of particle entirely.

As you may know, when matter and antimatter meet, they release tremendous amounts of energy, thanks to $E=mc^2$. For an electron and positron with no initial energy other than their individual rest masses of $\pu{511 keV c^-2}$ each, the most common annihilation outcome is:

$$ \ce{e- +\ e+ -> 2\gamma}$$

However, this process is fully reversible in quantum electrodynamics; it is time-symmetric. The opposite reaction is pair production:

$$ \ce{2\gamma -> e- +\ e+ }$$

A reversible reaction? Then there is nothing stopping us from imagining the following chemical equilibrium:

\begin{align}
\ce{e- +\ e+ &<=> 2\gamma} &
\Delta_r G^\circ &= \pu{-1.022 MeV} =\pu{-98 607 810 kJ mol^-1}
\end{align}

The distinction between enthalpy and Gibbs free energy in such subatomic reactions is completely negligible, as the entropic factor is laughably small in comparison, in any reasonable conditions. I am just going to brashly consider the above value as the standard Gibbs free energy change of reaction. This enormous $\Delta_r G^\circ$ corresponds to an equilibrium constant $K_\mathrm{eq} = 3 \times 10^{17276234}$, representing a somewhat product-favoured reaction. Plugging the Nernst equation, the standard electrode potential for the "reduction of a positron" is then $\pu{+2 116 413 V}$.

Ivan mentions in his answer using an alpha particle as an oxidiser. Let's take that further. According to NIST, a rough estimate for the electron affinity of a completely bare darmstadtium nucleus ($\ce{Ds^{110+}}$) is $\pu{-204.4 keV}$, so even a stripped superheavy atom can't match the oxidising power of a positron!

... that is, until you get to $\ce{Ust^{173+}}$ ...


What causes reactions to react the way they do?

If the reactants are AB + CD, it will be a double displacement (2 pairs of compounds)

An example of this might be two salts that dissociate into their respective ions in a solvent, and then the cation and anions that were not previously paired find their way to each other, if they form a salt that is not easily dissociated in this solvent, we have a new ionic compound (salt) which precipitates out as a visible solid in the mixture container. This might not always be the case, when we write net ionic equations we leave out anything that was not involved in creating the precipitate.

If we have the net ionic equation: $\ce{A^{+}_{(aq)} + B^{-}_{(aq)} -> AB_{(s)}}$ occurring when the solutions mix, then $\ce{AB}_{(s)}$ is said to be the precipitate. Notice that we leave out the other ions C and D because they were observers in the chemical equation, perhaps their attraction to each other is not strong enough to overcome the solvent. We should not always be able to tell whether this will happen or not unless we refer to experiments in which this substance had been seen to precipitate out before, perhaps even measuring its solubility in some experiments, the extent to which it will dissociate can also be calculated from experimentation.

  Why can't, for example, AB + CD synthesize to create ABCD?

If you are trying to determine if a covalent bond is compatible, one method is to use the octet rule and see what combinations might occur. For example, you have reactants of silicon and fluorine. What will form? Looking at the periodic table to see the electrons they contribute, draw a lewis structure for fluorine and silicon. Silicon contributes 4 electrons, and 4 fluorine atoms contribute the rest to give all elements full octets. Silicon tetrafluoride seems to be the most stable using this model. This is one method, but often times reactions occur the opposite way too. However, perhaps for some reason it is more likely for the reactants to be present in a given reaction, even after it seems to have stopped on a macroscopic scale (even thought it really has not stopped activity), we need to come up with more things to explain why this behaviour might happen. So as scientists, we devise new models.

This is a less than obvious and bigger topic than you might realise. A general chemistry textbook might have a chapter called chemical kinetics or reaction mechanisms which describe this ten times better than I can. Some things that will help you in understanding some models of reactions and their mechanisms are things like rate laws and how to determine rate constants in the laboratory, or how to determine the activation energy proposed by Svante Arrhenius in 1880's which basically means that two chemicals need to have at least a certain amount of energy (the activation energy) to overcome their repulsive behaviors and other limiting factors (the energy required to actually break apart a molecule's covalent bonds). For a collision of two particles to produce a different particle, there are other things that determine if a new molecule is formed from a collision of two molecules. First, like I said they must overcome repulsions with kinetic energy but they often need to be aligned spatially for a certain molecule to form in some proposed mechanisms. A mechanism is a proposition that an in-between step occurs in the reaction, but it is very unlikely that two hydrogen molecules will smash into a carbon dioxide molecule and create methane and oxygen. We must propose intermediate steps to make it sound more probable. although this would represent the reverse reaction of combusting methane, and will likely never occur because the potential energy of the system would increase, and we know how much the universe hates that kind of behaviour, if not then read on.

Some concepts that will also help you understand "why reactions happen" are some laws like the 2nd law of thermodynamics which states that the entropy of the universe is always increasing. This explains why a reaction in which a reaction of type that you describe $\ce{ABC + D -> ABCD}$ is actually quite possible. The problem is that the reverse reaction also takes place with a higher probability, these molecules might actually break apart more than they combine. This two-way speciality is why this reaction will likely be written as such: $\ce{ABC + D <=> ABCD}$ indicating that it does possibly happen, but can go both ways, and most often goes backwards. The reason for this is that there is more entropy (aka: disorder) when there are more moles of molecules present. In your examples the reactions are unlikely because the reactants ($\ce{ABC}$ and $\ce{D}$) are twice the number of moles than the products ($\ce{ABCD}$). This is a limiting factor for the reaction to commonly take place. It is useful to note that this reaction might occur once in awhile, but it is more often that molecules fluttering about will break apart than create larger and more complex molecules. Indeed, it is more likely that this reaction will be reversed and proposed as a mechanism of a larger process, if at all, where D breaks off and combines with something else, lets say E and ABC still remains along with a very unbreakable compound called ED which has a triple bond that will likely not be broken at the given temperature.

Lets talk more about entropy: You have a snowball you just rolled up in your hand (it is composed of many flakes of ice). Think of each ice particle that makes up the snowball as an atom in a larger molecule (the snowball). Chances are when this snowball starts moving (like molecules tend to do) it will run into something, and when that happens it will become many different smaller flakes and chunks (atoms and smaller molecules). This is because the universe prefers disorder and chaos. If you threw your clothes freshly out of the dryer into your closet, will they miraculously fold themselves neatley and in order? No, because that would make them more orderly and less chaotic, instead they will flail about in the air and land in a disorganized manner, this then makes sense by the phenomenon of the second law of thermodynamics.

Look back on your equations now, why can't $\ce{AB + CD -> ABCD }$ occur? Hopefully this is more obvious to you now. There is more disorder in the reactants than the products. This reaction might still occur, but normally balanced chemical equations represent the net affect that tends to occur. Perhaps occasionally someone will win at the game BINGO, but that does not necessarily mean that the same person will keep on winning at BINGO, more likely it will be someone else in the room, and then someone else, etc. With that BINGO analogy, the person who won at bingo played again, and lost. The same way ABCD might have been created, but then almost immediately was destroyed again.

Chemical engineers might desire to create your molecule $\ce{ABCD}$ and to do this what they might do is add a large amount of $\ce{AB}$ molecules into the reaction chamber. Due to the overwhelming number of these molecules, the reaction is more likely to take place. Not everything in the container will eventually be $\ce{ABCD}$ molecules, but there will be more of them that there was before. These kinds of reactions are actually called equilibriums meaning that the reaction can occur both ways. We write the equilibrium reaction as $\ce{AB + CD <=> ABCD}$ to explicitly show this. The quotient of reactants to products will not be 1:1. Sometimes the rate constant is very large for a reaction implying that it will go to completion, in which case we simply write the reaction in a single direction as you often see with chemical equations.

That is far as I am willing to explain because I am merely repeating snippets of what any chemistry book will explain in a chapter on chemical kinetics, hopefully my repetitive explanation will help you get a better idea of what is actually going on here, but there is still a lot more that you could learn or I will bother to explain.

Food for thought: In the equilibrium reaction $\ce{AB + CD <=> ABCD}$, A chemical engineer decreases the pressure of the reaction chamber, will the reaction shift more in favor of the reactants or the products? These are some questions that chemistry books will ask you after chemical kinetics are discussed. But I do not expect you to be able to answer it yet.


The molar mass $(M)$ of a mixture with known mass fractions $(w_{i})$ is calculated from:

$$
\frac{1}{M} = \sum_{i}\frac{w_{i}}{M_{i}}
$$

So in this case
$$
\frac{1}{M} = \frac{1/8}{0.17034}+\frac{7/8}{0.034} = 26.469\ldots\\
M = \frac{1}{26.469\ldots} = 0.037779872\ \mathrm{kg}/\mathrm{mol}
$$


How does methyl vinyl ketone dimerise with heating?

The dimer of methyl vinyl ketone (1) is not shown in the choices in the question. The dimer is A shown below. It is a "tail-to-tail" dimer. Product B is an intramolecular ketal as a probable mixture of epimers. Compare with this paper.

B. P. Mundy, et al., J. Org. Chem., 1978, 43, 2347.


As described by the definition of a state function, that it is
  independent of the path adopted to bring about a change in it [...]

This is not the definition of a state function, since it is missing the crucial first part of it. Here is the definition: a state function is a property of a system that depends only on the current equilibrium state of the system, not on the way in which the system acquired that state.

The point is that something is a state function if you can tell its value by knowing just the equilibrium state of a system and nothing else. As Chester Miller said in his comment: tell me the temperature and pressure (i.e. specify the state of a system), and I'll tell you the volume. It is irrelevant how the system acquired that equilibrium state, the volume is determined by just the state.

  Wouldn't volume change differently if we brought about a change in it
  by say increasing/decreasing temperature rather than say pressure?

In general, of course, when changing the system differently, its volume will also change differently. But, if by doing different changes you eventually will reach the same equilibrium state, all the state functions (and not just volume) will have the same values.


Name for a spin state with a multiplicity of 13

I have seen multiplicity 13 referred to as simply a 13-let. This #-let nomenclature is sometimes used for high multiplicities (e.g. higher than 10).


In the DNA backbone, which bonds exactly are considered ester bonds?

All of them are right, regarding ester bonds themselves. But only one of them is right, regarding DNA context. AMP/ADP/ATP/GMP are based on ribose ( see RNA ), not on deoxyribose ( DNA ).
Phosphoric acid has 2 ester bonds in DNA.
You have somehow confused an ester and an ether bond.
$\ce{R_1-O-R_2}$ is an ether bond,
$\ce{R-O-X}$ is an ester bond, where $\ce{X}$ is a carbonyl ( from carboxylic acids or carbonic acid ) or some nonmetal like $\ce{S}$ or $\ce{P}$ for esters of inorganic acids like $\ce{(CH3O)2SO2}$.
If we are to consider it as a bond or a group depends on the point of view. It is both.
It is a group, if we focus on the $\ce{CO-O}$ atom group of esters of organic acids or $\ce{O}$ for mineral acid esters.
It is a bond ( or a bridge ), if we focus on what the group connects.


In addition to Martin's answer, there is at least an old recommendation
by E. B. Sandell and T. S. West in Pure Appl. Chem., 1969, 18, 427-436 (DOI), which states:

  Buffer capacity or buffer index. The capacity of a solution to resist changes in pH on addition of acid or base, which may be expressed numerically as the number of moles of strong acid or strong base required to change the pH by one unit when added to one litre of the specified buffer solution.


Vinegar salt reaction to produce acid; does it matter what type of vinegar I use

There are two show stoppers :

$\ce{HCl}$ is strong acid, $\ce{CH3COOH}$ is weak acid. The reaction is practically completely shifted toward acetic acid and sodium chloride. Especially if acetic acid is diluted.
$\ce{HCl}$ forms azeotrope with water with the maximal boiling point near 20% $\ce{HCl}$. If you heat the mixture, that would boil water and acetic acid with minimal traces of $\ce{HCl} $ that would dissolve back in the condensate.


What is the mechanism of halogen displacement?

I'm guessing your talking about the halogen displacement reactions:

$$\ce{2NaBr + Cl2 -> Br2 + 2NaCl}$$

which is a redox reaction:

$\ce{2 Br- -> Br2 + 2 e-}$, bromine loses electrons (is oxidised)
$\ce{Cl2 + 2e- -> 2 Cl-}$, chlorine gains electrons (is reduced).

For more detail look in to the redox potentials, this explains what will reduce what.


Why are so many different organophosphates able to deactivate acetylcholinesterase?

Acetylcholine's normal function is to cleave acetylcholine's ester bond to convert it into acetic acid and choline, but certain compounds can interfere. The short answer is that these nerve agents all share similar structural components allowing them to be bound by acetylcholinesterase, where they then either react to alter the enzyme's binding site, preventing activity, or are just very slow to hydrolyze, temporarily inhibiting the enzyme until it can remove the compound.

Not all of these related acetylcholinesterase inhibitors work in exactly the same way, but if you examine their structures and compare them with acetylcholine, the normal substrate of the enzyme, you'll notice some similarities:

Acetylcholine:

Sarin:

Tricresyl phosphate:

You can see that they all have this central double-bonded oxygen (a few have sulfur instead, but are converted to oxygen metabolically), which is important for fitting into the enzyme. The ester bond holding the choline to the carbonyl in acetylcholine is relatively easily hydrolized, but phosphoesters are much slower to hydrolyze. From what I can tell, this seems to be one mechanism behind trecresyl phosphate's toxicity and probably why it's not as acutely toxic as something like sarin or VX. Sarin, on the other hand, has a pretty good fluoride leaving group. This lets it form a strong covalent bond with a serine's sidechain in ACE's binding pocket. This reaction is effectively irreversible, making sarin very dangerous as it totally disables any ACE it reacts with, meaning it takes very little to cause a significant effect.


Assigning Peaks and Determining J Values for Mata-Disubstituted Benzene

There's almost nothing you can do in this situation, unfortunately. Because of strong coupling and other issues, aromatic peaks often do not conform to the so-called $n+1$ rule, which makes it highly non-trivial to extract coupling constants. It is probably possible to "fit" coupling constants to the multiplet shape by simulating the spectrum for a given set of $J$'s and seeing which set best matches the observed spectrum.

The HSQC trace is unfortunately not useful in this regard. At first glance, it may look simpler than the 1D proton spectrum you undoubtedly have. But that is because HSQC spectra are invariably recorded with low resolution in the proton dimension. The main reason for that is because of the high-power 13C decoupling that is performed during the acquisition. Acquisition times therefore need to be kept to approximately 100 ms, or there is a serious risk of damaging the spectrometer. On the other hand, a 1D proton spectrum doesn't utilise 13C decoupling, and can have acquisition times of several seconds.

Assuming the spectral width is constant, a shorter acquisition time means poorer resolution (this is a property of the discrete Fourier transform).


Safety in handling of concentrated ammonium fluoride

In $\ce{SiO2}$ etching, there are two active fluorine-containing species, $\ce{HF}$ and $\ce{HF2-}$.

The rate of etching at 25 degrees C is proportional to :

$$9.66[\ce{HF2-}] + 2.50 [\ce{HF}] -0.14$$

according to A Study of the Dissolution of SiO2 in Acidic Fluoride Solutions J. Electrochem. Soc. vol. 118, pp. 1772-1775.

The relavent equilibrium constants at 25 degrees C and ionic strength of 1M are

$$\frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]} = 0.0013M $$

$$\frac{[\ce{HF}][\ce{F-}]}{[\ce{HF2-}]} = 0.104M $$

From the second equilibrium, it can be seen that if the fluoride concentation is 10M as in the question, the more active $\ce{HF2-}$ will be present at a concentration of about 100 times $[\ce{HF}]$.

Multiplying the two equilibrium equations together, and substituting $[\ce{F-}] = 10M$:

$$\frac{[\ce{H+}]}{[\ce{HF2-}]} = 1.4 \times 10^{-6} $$

Therefore at $pH = log (1.4 \times 10^{-6}) = 5.9$, there will be about 1M $\ce{HF2-}$, diminishing by a factor of 10 for each pH unit increase.

Experimentally, concentrated Ammonium Fluoride has a pH of 7.2 according to http://www.solvaychemicals.us/SiteCollectionDocuments/sds/P20117-USA_CN_EN.pdf (which also says to avoid glass).

At pH 7.2, [$\ce{HF2-}$] is 0.05M.

In conclusion, there will be a significant concentration of the most active etching species, $\ce{HF2-}$, present in 10M ammonium fluoride and I would be concerned about glassware.


Good question but there is no such thing as immiscible gases. Immiscibility is caused by surface energy/tension.  Gases do not have surfaces and therefore lack surface tension.


What is the chemistry behind the reaction of KNO₃, K₂CO₃ and Sulfur?

Here's just a guess, but I do have some confidence that I'm right:
$\ce{KNO3 + S -> K2SO4 + K2SO3 + K2S + K2O + SO2 + N2}$
$\ce{K2CO3 ->[\Delta] K2O + CO2}$
The first, unbalanced equation is the reaction between potassium nitrate and sulfur, producing quite some gas as product, and a variety of potassium compounds left behind, with the composition depending on your formulation of the powder. Of course there might be some sulfur trioxide produced, but the major oxide of sulfur should be the dioxide. In the second equation is the thermal decomposition of potassium carbonate releasing carbon dioxide. This is the part that I'm not very sure of, but it may be to release more gas to give your reaction more puff. However, IIRC the decomposition requires a very high temperature. In contrast, bicarbonate has a much lower decomposition temperature and also release more gas(carbon dioxide and water), but the products would likely be absorbed by potassium oxide to form hydroxide and carbonate, so  bicarbonate might not help much.


Why is the melting point of propane lower than that of both methane and ethane?

Ok, let's first look at the trend...

$$
\begin{array}{c|c|c|c}\hline
\text{Chemical} & n \text{ in }\ce{C_nH_{2n+2}} & \text{m.p. / K} & \text{m.p.}(n) - \text{m.p.}(n-1)\text{ / K} \\
\hline
\text{Methane} & 1 & 90.7 & \\
\text{Ethane} & 2 & 90.4 & -0.3 \\
\text{Propane} & 3 & 85.5 & -4.9 \\
\text{Butane} & 4 & 133\mathrm{-}139 & +50.5 \\
\text{Pentane} & 5 & 144.1 & +8.0 \\
\text{Hexane} & 6 & 177\mathrm{-}179 & +36.9 \\
\text{Heptane} & 7 & 182.2\mathrm{-}183.0 & +4.6 \\
\text{Octane} & 8 & 216.0\mathrm{-}216.6 & +33.7 \\
\text{Nonane} & 9 & 219.0\mathrm{-}220.0 & +3.2 \\
\text{Decane} & 10 & 242.7\mathrm{-}243.9 & +23.8 \\
\hline
\end{array}
$$

Using the mean for the alkanes with melting point ranges given (e.g. 136 for n-butane and 178 for n-hexane), it isn't clear there is any "great" linear trend. In fact it seems that the even number n-alkanes are on one curve and the odd n-alkanes are on another.

There is a "general trend in that with increasing MW the MP increases. However the exceptions would seem to be both methane and ethane, not propane. Also it is obvious that a simple correlation with MW alone is not sufficient to characterize the trends. All in all I think one must be careful not to read too much into such trends without a very careful analysis.


How to test for fluoride levels in drinking water?

A good reliable set-up is pricey.  Here's a link to a reliable, quality probe for fluoride ion detection in water, it costs \$740  (US).  The meter that it plugs into sells for \$790 (US) if you get the less expensive hand-held device.  There also manuals and instructions on how to actually use the gear, at the website as well.


Is the reaction involving a proton transfer a redox reaction?

Donation of a proton is not an indication of a redox reaction, rather an acid-base reaction (which may or may not be also involving a redox process). The only way to show the reduction/oxidation takes place is to demonstrate that the oxidation numbers do change:

$$
\begin{align}
\ce{\overset{+1}{H}_2\overset{-2}{S} (aq) + \overset{-3}{N}\overset{+1}{H}_3 (aq) &→ \overset{-3}{N}\overset{+1}{H}_4+(aq) + \overset{+1}{H}\overset{-2}{S}^-(aq)}\tag{1}\\
\ce{\overset{0}{Pb}(s) + 2 \overset{+3}{Fe}\overset{-1}{Cl}_3 (aq) &→ 2 \overset{+2}{Fe}\overset{-1}{Cl}_2 (aq) + \overset{+2}{Pb}\overset{-1}{Cl}_2 (aq)}\tag{2}
\end{align}
$$

As you can see, oxidation numbers only change in reaction (2), and are preserved in reaction (1).


Relationship between a temperature increase of an exothermic reaction, Keq, and kf (forward rate constant)

You know that both the forwards $k_f$ and reverse $k_r$ rate constants will increase with temperature, simply because there is more energy present and so it takes less energy to surmount the activation barrier, both from the product and reactant sides.

The question is then which is changed more by changing the temperature. The equilibrium constant is as you write $K_Q=k_f/k_r$ and now using the Arrhenius equation we get
$$K_Q=\frac{A_f\exp(-\Delta E/(RT))}{A_r\exp(-(\Delta E+E_0)/(RT))}$$

where the A's are pre-exponential factors (and are constant vs temperature) and $\Delta E$ the activation barrier for the forwards reaction and $E_0$ the extra energy for the back reaction and  is positive since the reaction is exothermic. (If you sketch the barriers hopefully this should be clear).

Simplifying gives
$$K_Q=\frac{A_f}{A_r}\exp(E_0/(RT))$$

As T  is increased $K_Q$ becomes smaller. Thus the answer is (D).


derivation of partial differential equations for changes in U

One definition of a state function such as internal energy $U$ is that its infinitesimal change is equal to an exact differential. Thus in general

$$\mathrm{d}U=\sum_{i=1}^n\left(\frac{\partial U}{\partial x_i}\right)_{x_j} \mathrm{d}x_i.$$

There are various variables with regard to which one can apply the partial derivative. Entropy, volume, temperature to name a few. Usually we assume that our thermodynamic state functions are two-dimensional but this is only if the number of moles is constant.

$$\mathrm{d}U= \frac{\partial U}{\partial T}\mathrm{d}T + \frac{\partial U}{\partial V}\mathrm{d}V +\sum_{i=1}^k\frac{\partial U}{\partial n_i}\mathrm{d}n_i \overset{\mathrm{d}n_j\ =\ 0}{\implies} \mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)_{V,n_j}\mathrm{d}T + \left(\frac{\partial U}{\partial V}\right)_{T,n_j}\mathrm{d}V.$$

A good way to think of this assumption is that other functions are also simply functions of $T$ and $V$, e.g., $S = S(T,V)$. From the equation

$$\mathrm{d}U = \left(\frac{\partial U}{\partial T}\right)_{V,n_j}\mathrm{d}T + \left(\frac{\partial U}{\partial V}\right)_{T,n_j}\mathrm{d}V$$

you can reach the other two equations by imposing $\mathrm{d}T = 0$ and $\mathrm{d}V = 0$, respectively.


To begin this short discussion one has to differ between Sn2 and addition to the carbonyl group, as the former is driven by orbital interactions while the latter is driven by coulomb interactions.
This can be explained with a closer look at the carbonyl function.
due to the highly polarized π-bond between carbon and oxygen, carbon is rendered quite positive. It results that any nucleophile bearing a negative charge will be attracted to attack it. In addition the reaction of a carbanion and a ketone or aldehyde is irreversible!
there is fundamentally no way in which the carbanion could ever be considered as a leaving group.

On the other hand Sn2 reactions appear on saturated carbon (in the best case methyl or primary) with at least one good leaving group (e.g. a halide or tosylate). Often the sigma bond connecting carbon and the leaving group is not even polarized, at least not significantly. Just consider the electronegativity of iodine, the BEST leaving group for SN2.
That is why orbital interactions are more dominant in thess kind of reactions. It can be easily noticed that a potential nucleophile has to donate its electrons into a sigma antibonding orbital in order to  react with the carbon electrophile. These kind of orbital can be considered as very high in energy as it is the counterpart of the sigma bonding orbital which is quite stable.
Therefore the best nucleophiles for nucleophilic attack on saturated carbon is usually a second or third row nucleophile like sulphur or phosphorus (Wittig reaction).

Therefore for a carbanion the irreversible addition to the carbonyl group happens much faster, resulting in a alkoxid. The new formed alkoxid is in close proximity to the chlorine substituted carbon and therefore likely to form the five membered ring .

Any references that support everything written above can be found in Clayden, Organic chemistry second edition.
Specifically the chapter nucleophilic substitution at saturated carbon will help.

Best regards,
John


Ethylene glycol was a boiling point of 193 °C using any water cooled condenser will crack. Use a air cooled condenser. It is just a glass rod. The length of tube depends on how much you want to condense and how quickly you want it to come over. Remember visible condensation should always take place in the condenser. If it goes further down the tube then some hot vapours will be lost to atmosphere if your system is open. If under vacuume the ethylene glycol will boil at a lower temperature but you should use a cold trap at minus 40 °C or below to catch vapours which come over. Remember liquid ethylene glycol is toxic and the vapours are more dangerous check the MSDS. I suggest that you should work with a closed system in a fume cupboard for this job or you might never see your grandchildren. Good luck.


Is it possible to synthesize amino acids from hydrocarbons?

Out of memory, I don't recall a method for the direct synthesis of $\alpha$-amino acids from plain hydrocarbons (including alkenes and alkynes).

The situation is different for $\beta$-amino acids. The $\ce{Pd(II)}$-catalyzed reaction of alkenes with amines in the presence of carbon monoxide, known as aminocarbonylation can furnish these amino acids or their derivates.

  […] from crude oil via a set of complex chemical transformations

however, is pretty vague and may involve anything, including the formation of aldehydes and their subsequent use in a Strecker reaction.


Why are there free electrons in a metal?

When electricity flows, the electrons are considered "free" only because there are more electrons than there should be, and because the transition metals, such as iron, copper, lead, zinc, aluminum, gold etc. are willing to transiently accept and give up electrons from the d-orbitals of their valence shell.

Transition metals are defined in part by their stability in a wide range of "oxidation states"; that is, in several combinations of having too many or too few electrons compared to protons. This is thought to be because of the d orbital in their valence shells. Compared to the s and p orbitals at a particular energy level, electrons in the d shell are in a relatively high energy state, and by that token they have a relatively "loose" connection with their parent atom; it doesn't take much additional energy for these electrons to be ejected from one atom and go zooming through the material, usually to be captured by another atom in the material (though it is possible for the electron to leave the wire entirely). This impetus can be caused by many things, from mechanical impact to chemical reactions to electromagnetic radiation (aka light, though not all of it visible); antennas work to capture radio frequencies, because the light at those frequencies induces an electric current in the wire of the antenna. Now, in the absence of a continuous force keeping the electron in this higher energy state, the electron (and the metal atoms) will naturally settle into a state of equilibrium. Electricity is generated when just such a force is acting on the metal, giving energy to the electrons in the d orbital and forcing them to move in a certain direction.

This impetus can come from many sources, as discussed, be it the movement of a magnet within a coil of wire, or a chemical redox reaction in a battery creating a relative imbalance of electrons at each of two electrodes. The end result is that the electrons, given additional energy from this voltage source, are ejected from their "parent" atom and are captured by another. The "holes" left behind by these electrons are filled by other electrons coming in behind them from further back in the circuit. Thus, the energy provided by the voltage source is carried along the wire by the transfer of electrons.

The analogy typically made is to the flow of water, and it generally holds in many circumstances; the "voltage source" can be thought of as being like a pump or a reservoir, from which water flows through pipes, and the amount of water and the pressure it's placed under (by the pump or by gravity) can be harnessed to do work, before draining back to a lower reservoir. The pipes are similar to wires in many ways; the larger the diameter, and the smoother the inside of the pipe, the more and the faster water can flow through it (equivalent in many ways to the thickness and conductivity of the metal wire), and when under enough pressure (high enough voltage), the pipes will actually expand slightly and hold more water than they would at low pressure (this is a property of wires and other electrical conductors called "capacitance"; the ability to store a charge while under voltage and to discharge it after the voltage is released).


Why is the deposition of silver more uniform from [Ag(Cl)₂]⁻ (aq) than from Ag⁺?

The uniformity of a deposition depends on more parameters, but in this case the difference is in electromigration effect. $\ce{Ag+}$ ions are atrracted to the electrode surface, but the complex $\ce{[Ag(Cl)2]-}$ is repulsed. Cyanides form  complexes as well: $\ce{[Ag(CN)2]-}$ or $\ce{[Ag(Cl)3]2-}$ and they are also  repulsed.

Silver electrodepostion form $\ce{AgNO3}$ ($\ce{Ag+}$ in solution) is without additives nearly impossible as dendrites are created. Fukunaka, Yamamoto, Kondo, Ionic Mass Transfer Associated with Dendrite Growth of Electrodeposited Silver in AgNO3 Solution, 1989 ECS.

The metal deposition from a complex molecules is described in Wilkinson, P. Understanding Gold Plating. Gold Bull 1986, 19, 75 – 81

Some chemicals, that are usually added to electrolyte, affect the deposition because they create a film as they are adsorbed at  inner Helmholz plane. (E.g. acidic Cu deposition with alcohol addition Popescu et al, Materials Chemistry and Physics 72 (2001) 332–336.)


How to control the Friedel–Crafts alkylation reaction to methylbenzene as the major product?

The solution to this problem of having chance of poly-alkylation of Benzene rings by Friedel-Crafts alkylation has a general way, and the way is to first perform Friedel-Crafts Acylation and then reduction of the introduced Carbonyl carbon to saturated carbon atom.

Now, proceeding via this way helps to avoid all the problems that arise in a Friedel-Crafts alkylation. Firstly, after the first step, in the main product, we are having one Benzene ring attached to a carbonyl carbon , which deactivates the ring for participating in further electrophillic aromatic substitution and reduces any chances of poly-acylation which upon reduction also doesn't give any poly-alkylated product. Secondly, Friedel-Crafts acylation helps to reduce the possibility of rearrangement which occurs very often in the alkylation step. That's because the positive charge is not on any primary,secondary or tertiary carbon atom but on the acyl carbon which will not undergo any changes in its carbon skeleton structure. So, we can add the unrearranged carbon chain to the Benzene ring and reduce to get less-substituted carbon chains attached to the ring.

To solve your problem, I would rather prefer that you first react Benzene with a mixture of $\ce{CO}$ and $\ce{HCl}$ in presence of anhydrous $\ce{AlCl3}$ (Gatterman-Koch Aldehyde Synthesis) to get Benzaldehyde. Then you can perform a Clemmensen reduction ($\ce{Zn -Hg / conc. HCl}$) or a Wolff-Kishner reduction ($\ce{H2N-NH2/KOH}$) to get Toluene as the exclusive product.

Please note that, you can always control the reaction conditions to get Toluene by Friedel-Crafts alkylation only, still there remains a good amount of chance of getting di-substituted product as those kinds of controls need to be optimised and difficult to carry out and you have to perform purification techniques anyways. However, if you proceed via acylation and reduction, your side products will  be in lesser amount and product can be easily obtained.


What is the number of lone pairs of central atom when it has an odd number of electrons?

This seems to be a tricky bit of chemical terminology that may not have a lot of practical relevance, so I wouldn't get too hung up on this, if I were you. Technically, I would say that you are correct in that lone pairs should be pairs of electrons, as this is how the IUPAC defines them.


Let's start with the fact that bouyancy of an object in open water is caused by displacement, but liquids that are less dense and may spread out do not displace water, they just lie on top of it. If I put a block of wood in a fish tank that is full, it displaces and spill over a mass of water equal to the mass of the block, but if I slowly pour (or rather spray) oil on a full tank of water, the oil will spill over, not the water. The definition for displacement is (per Google):

  the moving of something from its place or position.
  the occupation by a submerged body or part of a body of a volume that would otherwise be occupied by a fluid.

The first definition is what WD-40 is doing to water, the second one is what causes buoyancy. The WD-40 displaces the water in that it moves it from where it is, but how?

Looking at the ingredients from the SDS, we can get a feel for their purpose:

$$\begin{array}{|c|c|c|}
  \hline
\textbf{Material} & \textbf{Hazards} & \textbf{Purpose (not stated on SDS)}\\
\hline\text{Aliphatic Hydrocarbon} & \text{Flammable Liquid Category 3}  & \text{Thin oils and evaporate quickly}\\
 & \text{Aspiration Toxicity Category 1} & \text{}\\ \hline
\text{Petroleum Base Oil} & \text{Not Hazardous} & \text{Provide long term protection} \\ & & \text{from water and air}\\ \hline
\text{LVP Aliphatic Hydrocarbon} & \text{Aspiration Toxicity Category 1} & \text{thin oil and evaporate slowly}\\ \hline
 & \text{Simple Asphyxiant}\\
\text{Carbon Dioxide} & \text{Gas Under Pressure,} & \text{Propellant}\\
& \text{Compressed Gas}\\ \hline
\text{Non-Hazardous Ingredients }&\text{Not Hazardous} & \text{This does the real work}\\ \hline
\end{array}$$

The oils are nice because once the water is removed, the oil will remain, preventing it from returning. But it is the non-hazardous ingredients that do most of the work. Hard to say what exactly this 10% is made of but it has to get water to "unstick"/"unwet" the surface that is being treated. To do this WD-40 is likely using a non-ionic surfactant of some sort similar to what is used in shampoos to emulsify the water and allow the flow of the WD-40 spray to remove (displace) the emulsified water. $%The rest of the wD-40 components act as carrying agents to remove the water from whie$ This speculation about surfactants is further supported by the fact that WD-40 foams when sprayed which is not characteristic of pure hydrocarbons.

Additionally, there would likely be some material to adhere to the surface to prevent further interaction with water or oxygen. There is the petroleum base oil, but it needs a way to stick to the metal. Likely, the use an unsaturated fatty acid in some way whether it is pure oleic acid or boiled linseed oil or something similar to those to bond to most metals and other materials and prevent further corrosion.


Why does lead nitrate powder on heating starts turning yellow from the top rather than from the bottom?

Lead oxide is a peculiar type of salt. It's brown when it's hot and turns yellow when it's cool. The lower most part remaining white could be due to two possible reasons:

The top-powder becoming yellow indicates the region is more cooler than the salt with is present at the lower part of the test tube. The lower most part is in direct contact with the flame and the  upper part is receiving heat due to conduction and losing heat to the atmosphere; many factors could be at play here.
Another possible deduction could be the paucity of atmospheric oxygen at the bottom of the test tube than at the top and hence the rate of conversion to lead oxide to the top would be faster than at the bottom.

In case you haven't done the experiment and want to see what the OP is trying to convey, here's a youtube video that demonstrates the whole lab-based thermal decomposition of lead nitrate (it starts at 1.32  incase the timed setting of the link fails):
http://www.youtube.com/watch?v=csEnz9ZsAFM&t=1m32s


Relative magnitudes of intermolecular forces from phase diagrams

This is an interesting situation. It would be easier to answer if the mp and bp of A were both either higher or lower than for B.

But your measure of the strength of intermolecular forces will be melting point - and boiling point. And they give different answers, so how do we make a decision?

Consider the boiling point first: the higher boiling point of B suggests higher intermolecular forces, as you have concluded. Heat of vaporization is needed to put a molecule into the vapor phase, and it seems that B needs more energy to go gaseous.

But A melts at a higher temperature, suggesting that its molecules are held together in a solid phase by stronger bonds. Solid A needs more energy to turn it into a liquid; perhaps A can pack into a tighter solid structure. It would then need more heat to melt it, so it has a larger heat of fusion.

Then your answer must have two parts: A is more tightly bound in a solid framework, but after you put enough heat into both A and B to melt them, B has more intermolecular bonding capacity as a liquid than A does. So B boils higher and is more strongly bound as a liquid.


Basicity comparison of ammonia derivatives (and guanidine)

In triphenyl amine (3), the lone pair of nitrogen is in conjugation with phenyl groups and thus would be least available for protonation.

Due to resonance structures of guanidine (2), it should be the strongest base.

Although number 4 doesn't exist as it is written. Let's suppose it is $\ce{N(CH3)2}$ with a negative charge (the only possible way to have two methyl groups on a nitrogen).

If so, it is more basic than guanidine, because charged compounds are generally more reactive than neutral ones and will attract $\ce{H+}$ more quickly.

So if number 4 is charged, then the basicity will decrease in order
4 > 2 > 1 > 3


Electrons distribution of oxygen atoms in ozone

The ozone molecule is described with two resonant structures(as in figure) in which all the oxygen atoms respect the octet rule. The oxygen in the middle has a positive charge because has only 5 valence electron  and the oxygen with the single bond has a negative charge because it has 7 valence electrons.
 In the real molecule both the bonds have the same length so there isn't one single bond and one double bond. Also both the terminal oxygen atoms have the same partial negative charge instead of being one neutral and one negative(-1).
The picture below is more realistic of the situation:

In the electrostatic potential map you can see that the two lateral oxygen are equivalent:


Matching a Slater-type wavefunction with a minimal (STO-nG) Gaussian basis set

TL;DR

The procedure to represent a Slater-type orbital (STO) as a linear combination of Gaussian-type orbitals (GTO) is outlined in W. J. Hehre, R. F. Stewart and J. A. Pople, J. Chem. Phys. 1969, 51, 2657 and is a quite straight forward least square fitting.

Derivation

I'll be quoting/ rephrasing quite liberally and take the 1s orbital as an example. The STO is defined as
\begin{align}
\phi_\mathrm{1s}(\zeta_1, \mathbf{r}) &=
  (\zeta_1^3/\pi)^{1/2}\exp(-\zeta_1 \mathbf{r}).
  \tag1\label{STO-1s}
\end{align}

The general form of a Gaussian-type 1s orbital is defined as
\begin{align}
g_\mathrm{1s}(\alpha, \mathbf{r}) &=
  (2\alpha/\pi)^{3/4}\exp(-\alpha \mathbf{r}^2).
  \tag2\label{GTO-1s-general}
\end{align}

The STO is expressed as a sum of $K$ GTOs; the coefficients and exponents are obtained for a STO with $\zeta=1.0$ and the scaled uniformly. Hence we can write
\begin{align}
\phi_\mathrm{1s}'(\zeta, \mathbf{r}) &=
  \zeta^{3/2}\phi_\mathrm{1s}'(1.0,\zeta\mathbf{r}),
  \tag3\label{GTO-1s-zeta1}
\end{align}
where
\begin{align}
\phi_\mathrm{1s}'(1.0, \mathbf{r}) &=
  \sum_k^K d_{\mathrm{1s},k}\,g_\mathrm{1s}(\alpha_k,\mathbf{r}).
  \tag4\label{GTO-1s-lc}
\end{align}

In order to obtain the coefficients $d$ and exponents $\alpha$ the contracted GTO is fitted to the form of the STO via a least square method. Note that this is a quite unique approach, since most other basis sets are optimised to produce the minimal energy. It all comes down to minimising the integral
\begin{align}
I &=
  \int \left(
         \phi_\mathrm{1s}(1.0,\mathbf{r})-\phi_\mathrm{1s}'(1.0,\mathbf{r})
       \right)^2 \mathrm{d}\mathbf{r}.
  \tag5\label{min-int}
\end{align}

For $K=3$, i.e. STO-3G, the following values are obtained:
\begin{array}{llll}\hline
k & \alpha_{\mathrm{1s},k} & d_{\mathrm{1s},k}      & I\\\hline
  &                        &                        & 3.31\times10^{-4}\\
1 & 1.09818 \times10^{-1}  & 4.44635 \times10^{-1}  &\\
2 & 4.05771 \times10^{-1}  & 5.35328 \times10^{-1}  &\\
3 & 2.22766                & 1.54329 \times10^{-1}  &\\\hline
\end{array}

Implementation

Finally you need to scale the exponents to use the right exponent. From $\eqref{GTO-1s-zeta1}$ we see
$$\mathbf{r}\mapsto\zeta\mathbf{r}_{\zeta=1.0}$$
and from $\eqref{GTO-1s-general}$ we know $\mathbf{r}\mapsto\mathbf{r}^2$ and therefore the exponent
$$\alpha_\zeta\mapsto\alpha_{\zeta=1.0}\zeta^2.$$

Hydrogen for example has $\zeta=1.24$ (Tab. IX) and therefore the exponents that need to be implemented are:
\begin{array}{lll}\hline
k & \alpha_{\mathrm{1s},k} & d_{\mathrm{1s},k}      \\\hline
1 &       0.16885540       &      0.44463454        \\
2 &       0.62391373       &      0.53532814        \\
3 &       3.42525091       &      0.15432897        \\\hline
\end{array}

For Q-Chem this is explained in the manual under 7.4.3 and needs to have the form
\begin{array}{llll}
\text{\$basis}\\
\ce{X} & 0\\
L & K & \text{scale}\\
α_1  & C_1^{L_\mathrm{min}} & C_1^{L_\mathrm{min}+1} &\dots& C_1^{L_\mathrm{max}}&\\
α_2  & C_2^{L_\mathrm{min}} & C_2^{L_\mathrm{min}+1} &\dots& C_2^{L_\mathrm{max}}&\\
\vdots    & \vdots & \vdots &\ddots & \vdots \\
α_K  & C_K^{L_\mathrm{min}} & C_K^{L_\mathrm{min}+1} &\dots& C_K^{L_\mathrm{max}}&\\
\text{****}\\
\text{\$end}\\
\end{array}
where $\ce{X}$ is the atomic symbol, $L$ is the angular momentum, $K$ is the degree of contraction, $\text{scale}$ is the scaling applied to all exponents, $\alpha_i$ are the exponents, and $C_i^L$ are the contraction coefficients.

As a practical example, the STO-3G basis set shall look like this:

$basis
   H   0
   S   3   1.00
      3.42525091             0.15432897
      0.62391373             0.53532814
      0.16885540             0.44463454
****
$end

For the actual fitting procedure, i.e. minimising $\eqref{min-int}$ you should consult a more technical oriented network page. It might be on topic at Mathematics, Computational Science, or even Mathematica depending on the question that troubles you. Please check their guidelines before asking.

General remarks

The most principle shortcoming of GTOs is that they cannot describe the cusp correctly. It's a trade-off you just make for faster integral evaluation. With STO-6G you are getting the orbital quite right already and there is a reason why we usually use STO-3G and not some higher contraction. It just offers the best cost : accuracy ratio.
This can be quite neatly seen in the graphical representation. The black curve  is the original STO, purple is STO-6G, cyan is STO-4G, and red is STO-2G plotted online at the wonderful FooPlot. Hack it here (I hope that works).

Going from STO-6G to STO-7G won't give you much more accuracy, maybe not even going to STO-10G. You probably have to go more into the direction of STO-20/30G. At that point it is probably more economic to switch to a program that can handle STO natively. From Wikipedia I see two, that handle those: ADF, and MolDS. The latter seems young, but is free.

Addendum:

If the FooPlot link does not work, here are the used formulae:

(1/pi)^(1/2)*exp(-abs(x))

0.679*(2*0.152/pi)^(3/4)*exp(-0.152*x^2)+
0.430*(2*0.852/pi)^(3/4)*exp(-0.852*x^2)

0.292*(2*0.088/pi)^(3/4)*exp(-0.088*x^2)+
0.533*(2*0.265/pi)^(3/4)*exp(-0.265*x^2)+
0.260*(2*0.955/pi)^(3/4)*exp(-0.955*x^2)+
0.0568*(2*5.22/pi)^(3/4)*exp(-5.22*x^2)

0.130*(2*0.0651/pi)^(3/4)*exp(-0.0651*x^2)+
0.416*(2*0.158/pi)^(3/4)*exp(-0.158*x^2)+
0.371*(2*0.407/pi)^(3/4)*exp(-0.407*x^2)+
0.169*(2*1.19/pi)^(3/4)*exp(-1.19*x^2)+
0.0494*(2*4.24/pi)^(3/4)*exp(-4.24*x^2)+
0.00916*(2*23.1/pi)^(3/4)*exp(-23.1*x^2)


What would one have to do differently if they used NaMnO4 instead of KMnO4 for redox titrations?

You are right. Preparing sodium permanganate is more expensive. The second reason is, sodium permanganate is hygroscopic and occurs as a monohydrate. This means that sodium permanganate is more soluble than potassium permanganate. (Solubility of $\ce{KMnO_4}$ is $0.065$ g/ml and that of $\ce{NaMnO_4}$ is $0.9$ g/ml)

So if you require a larger concentration of $\ce{MnO_4^-}$ in the solution, you can go for $\ce{NaMnO_4}$.


In vibrational spectroscopy, what affect do overtones have on the spectra?

Since a spectrum, including its overtones, is essentially a property of a molecule, one cannot really say how the overtones affect a spectrum. So, I will answer this question by first describing what an overtone is, giving an example of an effect that overtones sometimes participate in, and then an example of how overtones can be useful.

Overtones are charatceristic vibrational frequencies of a molecule which take place at approximately integer multiples of the $v=0$ vibrational modes. The existence of these overtones demonstrate that molecular vibrations are anharmonic in nature. This follows from the fact that for a purely harmonic vibration, the only non-zero transitions are those for which $\Delta v=\pm 1$. That is, harmonic oscillators can only move on state at a time. Thus, the fact that we observe these overtones tell us that the vibrations are not harmonic or anharmonic.

Now, in IR and Raman spectroscopy, there is an effect called Fermi Resonance. This effect basically allows for the mixing of multiple vibrational states. If I'm not mistaken, this also only happens for anharmonic vibrations because energy needs to be allowed to be shared between the two states which is not possible if the vibrations are pure harmonic, normal modes. I bring this up because it is quite common that one observes a Fermi resonance between an overtone and a higher frequency ground state mode. To quote the Wikipedia page,

  High resolution IR spectra of most ketones reveal that the "carbonyl band" is split into a doublet. The peak separation is usually only a few cm$^{−1}$. This splitting arises from the mixing of νCO and the overtone of HCH bending modes.

Thus, if you wish, one effect that overtones have on IR spectra is their ability to participate in Fermi resonance, which leaves the appearance of a peak being split.

Sometimes it can be difficult to know if a peak being observed is in fact an overtone. Generally, overtones have low intensity, but if they participate in a Fermi resonance, they gain intensity and will look like a normal ground state vibration (depending on what mode they come from).

One place I have seen something likely similar to this is in the following paper[1]. They compute the IR spectrum of a water cluster, including the bending overtone, and show that it is likely the experimental spectrum was mis-assigned. In this example, being able to compute a spectrum independent of the bending overtone is quite useful.

Experimentally, overtones can be quite useful if there is a large number of peaks around the same frequency as a particular mode you are interested in studying. For instance, it is thought that the water dimer may play a significant role in absorption of radiation in the atmosphere and thus contribute to global warming. It is very hard, however, to determine the abundance of water dimers in the atmosphere because of the large amount of water vapor which covers up its entire infrared spectrum. Thus, ref. [2] used the fourth overtone of the $\ce{O-H}$ stretch to experimentally measure water dimers in the atmosphere.

Another example of the same use of overtones is in ref. [3]. There, the authors are interested in studying a possible intramolecular hydrogen bond, but one cannot just look at the lowest energy vibrations because they are swamped by a regular $\ce{O-H}$ stretch. Thus, the authors record infrared spectra in the regime where one would expect to find the first, second, third, and fourth overtones. It is confirmed that these are the peaks observed by simply computing the spectra from first principles. Thus, one also learns which geometric configurations cause which overtone peak.

In summary, overtones are not so much something that "affect" a spectrum, but are a necessary part of the spectrum because we live in a world where molecules vibrate anharmonically. I have also given examples of when overtones are quite a useful tool.

Just to get you thinking a bit, if and when you learn about two-photon or two-dimensional spectroscopies, imagine how overtones can then be used to gain even deeper insight into the molecules of interest.

[1]: Wang, Y., & Bowman, J. M. (2013). IR spectra of the water hexamer: Theory, with inclusion of the monomer bend overtone, and experiment are in agreement. The journal of physical chemistry letters, 4(7), 1104-1108.

[2]: Pfeilsticker, K., Lotter, A., Peters, C., & Bösch, H. (2003). Atmospheric detection of water dimers via near-infrared absorption. Science, 300(5628), 2078-2080.

[3]: Howard, D. L., Jørgensen, P., & Kjaergaard, H. G. (2005). Weak Intramolecular Interactions in Ethylene Glycol Identified by Vapor Phase OH− Stretching Overtone Spectroscopy. Journal of the American Chemical Society, 127(48), 17096-17103.


Recrystallization of two unknown compounds?

Choosing a recrystallization solvent is as much an art as a science, even when you know what the substance is. Without knowing, it is a matter of making educated guesses. You want a solvent where it is much more soluble in the hot solvent than in the cold. Given that this is a lab exercise, we can assume that these are common substances. So, the one that dissolves in base is likely a carboxylic acid--perhaps benzoic acid or something close. That substance dissolves about 10 times as much in hot water as in cold - so I would try water first.  If it is too insoluble, then you could go to methanol, ethanol, etc., moving toward more nonpolar solvents until you get it to dissolve when heated. Then, hopefully it will crystallize out on cooling.  For the other substance, it is obviously soluble in dichloromethane and is not an acid. Perhaps it is non-polar. I would start with ethanol, and see how that works - if you need to go to a more non-polar solvent, toluene might be a good choice. Good luck!


Would decomposition take place in an atmosphere without oxygen?

I think your experiment is along the same lines as one performed in 1680 by Antonie van Leeuwenhoek. His experiment is described in this Wikipedia article:

  In his letter of 14 June 1680 to The Royal Society, Antonie van
  Leeuwenhoek described an experiment he carried out by filling two
  identical glass tubes about halfway with crushed pepper powder, to
  which some clean rain water was added. Van Leeuwenhoek sealed one of
  the glass tubes by using a flame and left the other glass tube open.
  Several days later, he discovered in the open glass tube ‘a great many
  very little animalcules, of divers sort having its own particular
  motion.’ Not expecting to see any life in the sealed glass tube, Van
  Leeuwenhoek saw to his surprise ‘a kind of living animalcules that
  were round and bigger than the biggest sort that I have said were in
  the other water.’ The conditions in the sealed tube had become quite
  anaerobic owing to consumption of oxygen by aerobic microorganisms.

Obviously his experiment was not as tightly controlled as what you propose.  Also, I think the actual size of your "very large glass container" would be the key to microbial activity in your experiment, as anaerobes do require water to thrive. It sounds like van Leeuwenhoek gave his bugs plenty water for the scale of his experiment.

My guess is that there would be some anaerobic decomposition before it pretty much dried out and "mummified".


Nitrogen is actually surrounded by 3 effective pairs. But 2 of them are bonding and the third one is a lone pair. According to VSEPR theory, the structure should be  trigonal planar.

The angle is not exactly $120^\circ$, as the electrostatic  repulsion between non-bonding pair and bonding pair is higher than the repulsion between two bonding pair. So, the angle of $\ce{O-N-O}$ is slightly smaller than $120^\circ$.
The distribution of the three electron effective pairs is trigonal planar. But the molecule structure is bent (like letter V).


Why doesn't D block contraction affect the other P block groups 14 15 16 etc

As a property showing d-block contraction, the question cited atomic radius. However, d-block contraction can manifest itself in other properties as well:

$\ce{GeO}$ is described as much more stable than $\ce{SiO}$. $\ce{Sn(II)}$ compounds are described as reducing. [Riedel]
The pentachlorides of $\ce{P, As, Sb}$ show a kink in their stability: $\ce{PCl5}$ melts at 167°C, $\ce{AsCl5}$ decomposes at $-50\mathrm{°C}$, $\ce{SbCl5}$ melts at 4°C (and decomposes at 140°C) [HoWi]
The redox potentials in acidic solution for the (0)/(IV) oxyanions and oxides of chalkogens show a peak at $\ce{Se}$ [HoWi]:

\begin{align}
\ce{SO2} + \ce{4e-} + \ce{4H+} &\leftrightharpoons \ce{S} + \ce{2H2O} \quad &E^0 = +0.50 \ \mathrm{V} \\
\ce{SeO_3^{2-}} + \ce{4e-} + \ce{6H+} &\leftrightharpoons \ce{Se} + \ce{3H2O} \quad &E^0 = +0.74 \ \mathrm{V}\\
\ce{TeO2} + \ce{4e-} + \ce{4H+} &\leftrightharpoons \ce{S} + \ce{2H2O} \quad &E^0 = +0.57 \ \mathrm{V}
\end{align}

the oxidative strength (at both low and high pH) of halogen oxyacids follows the pattern $\ce{BrO4-} > \ce{H_5IO6} > \ce{ClO4-}$. [HoWi]

I think that stability of oxidation states, stability of comparable compounds, and redox potentials of similar compounds are valid examples of d-block contraction. Of course, other chemists may arrive at different conclusions.

Literature:
[Riedel] E Riedel, Anorganische Chemie, Vol. 6, de Gruyter (2004).
[HoWi] AF Holleman, E Wiberg, Lehrbuch der Anorganischen Chemie, Vol. 101, de Gruyter (1995).


Measuring or calculating the pH of 4M NaOH? I prepared the solution in Milli Q water

For $\mathrm{pH}>12$ and $\ce{Li+}$ or $\ce{Na+}$ concentrations greater than $0.1\ \mathrm{M}$, glass electrodes experience alkali error. Because the concentration of $\ce{H3O+}$ ions in solution is so low, interference from alkali metal ions (like $\ce{Na+}$) becomes noticeable and causes the electrode to read lower than the true $\mathrm{pH}$ of the solution. $\ce{K+}$ typically cause less error than $\ce{Li+}$ or $\ce{Na+}$ due do its larger size.

Since sodium hydroxide is a strong base and dissociates completely into $\ce{Na+}$ and $\ce{OH-}$ ions, though, it's actually quite simple to just calculate the $\mathrm{pH}$.

$$K_\mathrm{w}=\ce{[H3O+][OH-]} \Rightarrow$$
$$ 1 \times 10^{-14} \ \mathrm{M^2}=\ce{[H3O+] \cdot 4\ \mathrm{M}} \Rightarrow $$
$$\ce{[H3O+]}=2.5 \times 10^{-15}\ \mathrm{M}$$
$$\mathrm{pH}=–\log(2.5 \times 10^{-15}\ \mathrm{M})=14.60$$

If you are worried about whether or not you truly have a $4\ \mathrm{M}$ solution of $\ce{NaOH}$, you should try titrating it with a known concentration of acid (in the past I always used potassium hydrogen phthalate). This process is called standardization in acid-base parlance.


The only saccharides I know of that have only a single chiral centre are the trioses L- and D-glyceraldehyde, but, at least according to the Gold Book definition, anomers are specifically diastereomers of cyclic forms of sugars. Trioses are too short for a cyclic form to normally arise and therefore can't anomerize.


Fundamental vibrational modes for QM9 molecules

Only for a diatomic molecule and only assuming the vibration is harmonic. In this case, $\mathrm{ZPVE}=\omega_1/2$. If its anharmonic or has rovibrational coupling, even the expression for a diatomic doesn't not allow you to determine $\omega_1$ from just the ZPVE. With polyatomic molecules, the ZPVE has contributions from all the vibrations, so you can't just extract a single vibrational frequency.
Rotational constants are not used in computing $\omega_1$ (unless you are considering forms of rovibrational coupling).
You can determine $\omega_1$ from other electronic structure methods and even classical molecular dynamics, but assuming you don't want to compare against experimental data, you would need to repeat these DFT calculations to get the corresponding $\omega_1$. For this dataset to include ZPVE, the underlying calculations would have had to determine all the vibrational frequencies of the molecule, but they seem not to have included this data in the QM9 set.


How does B2Cl4 have two perpendicular c2' axes?

Diboron tetrachloride has the same symmetry as allene, which I will use to demonstrate the other two $\ce{C_2^{'}}$ axis. The back $\ce{CH_2}$ group is turned 90° in comparison to the front $\ce{CH_2}$ group. This conformation is most stable as it minimizes steric interactions between both groups.

In this case, a picture tells more than I thousand words I think.

The two $\ce{C_2^{'}}$ axis are turned by 45° in comparison to the C-H bonds. It therefore also has two dihedral planes, thus belonging to the $\ce{D_{nd}}$ point group.

If you have difficulties picturing this in your head that is not strange. Many people make mistakes in assigning point groups. You get a better visualization by practicing online.


Balancing REDOX reactions in a basic solution, when no H+ ions are present

The bottom equation looks balanced to me. There are the same number of atoms of each element on each side of the equation and charge is conserved. The hydrogen in the $\ce{H2O}$ is balanced by the hydrogen in the $\ce{HAsO2}$ on the other side so it looks fine.