Balancing reactions algebraic method limitations

As per the OP's request, here is a solution. The equation is:

$$\ce{a XeF4 + b H2O  -> c XeO3 + d Xe  + e O2  + f HF  \tag 1}$$

So atom balance yields:

$$4a=1f     \tag 2$$
$$2b=1f     \tag 3$$
$$1a=1c + 1d     \tag 4$$
$$1b = 3c + 2e     \tag 5$$

The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer greater than zero, so I proceed as follows:

Assume $a = 1$, which is the minimum non-zero value it might have. Then, from equation $(4)$, $1 = a = c + d$, which is impossible because either $c$ or $d$ would have to be zero.
Assume $a = 2$. Then $b = 4$ because equations $(2)$ and $(3)$ show that $b = 2a$. But then equation $(5)$ shows that $4 = b = 3c + 2e$. Then if $c = 1, e = 0.5$, which is impossible. Obviously, $c$ cannot equal $2$, because that would give a negative value for $e$.
Assume $a = 3$. Then $b = 6$, because $b = 2a$, as above. Then equation $(5)$ shows that $6 = b = 3c + 2e$. Then if $c = 1, e = 1.5$, which is impossible. If $c = 2$, then $e = 0$, which is impossible. As before, $c$ cannot be greater than $2$, because that would give a negative value for $e$.
Assume $a = 4$. Then $b = 8$ and $f = 2b = 16$. Then equation $(5)$ shows that $8 = b = 3c + 2e$. If $c = 1$, then $2e = 5$, so $e = 2.5$, which is impossible. If $c = 2$, then $2e = 2$, so $e = 1$. Then equation $(4)$ shows that $4 = 2 + d$, so $d = 2$. This is a solution. Lastly, $c$ cannot equal $3$, because that would give a negative value for $e$.

Almost final result: $a = 4, b = 8, c = 2, d = 2, e = 1,$ and $f = 16$. Hence

$$\ce{4 XeF4 + 8 H2O  -> 2 XeO3 + 2 Xe  + 1 O2  + 16 HF  \tag 6}$$

For a = 5, the solution is a = 5, b = 10, c = 2, d = 3, e = 2 and f = 20. This is the same as Oscar Lanzi's example solution. For a = 6, the solution is a = 6, b = 12, c = 2, d = 4, e = 3 and f = 24.

For a = 7, a double solution occurs: a = 7, b = 14, c = 2, d = 5, e = 4 and f = 28 and also a = 7, b = 14, c = 4, d = 3, e = 1 and f = 28.

There are an unlimited number of valid balanced equations, as expected. So the a = 4 solution is simply the first solution in the infinite series and it is the minimum coefficient solution, since the total number of atoms is 11a.

Bonus: Mithoron provided a link to a problem Balancing disproportionation reactions answered by Ivan Neretin years ago. The equation that was to be balanced was:

$$\ce{a XeF2 + b H2O  -> c Xe + d XeO3  + e O2  + f HF  \tag 7}$$

So atom balance yields:

$$1a=1c + 1d     \tag 8$$
$$2a=1f     \tag 9$$
$$2b=1f     \tag {10}$$
$$1b = 3d + 2e     \tag {11}$$

The equations relate the six unknown coefficients $a$ through $f$. But each coefficient is an integer greater than zero, so I first note that equations $(9)$ and $(10)$ show that $a = b$. Then equation $(11)$ shows that $b$ must at least equal $5$. Otherwise either $d = 0$ or $e = 0$, which is impossible.

Now I proceed by assuming $a = b = 5$. Then equation $(8)$ gives rise to the following four possibilities: $c = 1$ and $d = 4$; $c = 2$ and $d = 3$; $c = 3$ and $d = 2$; $c = 4$ and $d = 1$. If $c = 1$ and $d = 4$, then equation $(11)$ yields $5 = 12 + 2e$, which is impossible. Likewise for $c = 2$ and $d = 3$ and also for $c = 3$ and $d = 2$. But if $c = 4$ and $d = 1$, then equation $(11)$ yields $5 = 3 + 2e$, so $e = 1$. This is a solution: $a = 5, b = 5, c = 4, d = 1, e = 1,$ and $f = 10$. Hence the balanced equation is:

$$\ce{5 XeF2 + 5 H2O  -> 4 Xe + 1 XeO3  + 1 O2  + 10 HF   }$$

However, as Ivan Neretin beautifully pointed out in his answer, there is also the simpler redox reaction given by

$$\ce{2 XeF2 + 2 H2O  -> 2 Xe   + 1 O2  + 4 HF   }$$

So any linear combination of these, with positive integer coefficients, is also a valid balanced equation. For example, just add the last two equations together together:

$$\ce{7 XeF2 + 7 H2O  -> 6 Xe + 1 XeO3  + 2 O2  + 14 HF   }$$


Predicting product ratio by rearrangement in Dehydration Reactions

The reaction pathway I would propose for the obtained products is as below:

Coming to the percentage of each product obtained, I would defer from the data provided in the question. I would believe that V would be the major product because it is both thermodynamically and kinetically the best possible product.
I would be the next best as it is formed from the best intermediate.
Than II and IV would be the next major products as they are formed from the second best intermediate. Actually I would assume that IV would have a slight higher percentage due to the fact that it is thermodynamically more stable.
III and VI are produced in very less quantities because to form the carbocation required to form these products it has to undergo a very unfavourable rearrangement which would happen only in very less proportion.
So according to me the order should be V>I>IV≥II and very minor quantities of III and VI.


Why do carbonyl compounds undergo nucleophilic and not electrophilic addition?

They sure do. Except that the oxygen has lone pairs, which are more nucleophilic than the $\pi$ bond. Once those electrons attack an electrophile, it further deactivates the $\pi$ bond as a nucleophile.


"Adjust pH of the solution to 5.0±0.1 with acetic acid (by potentiometry)"

I suspect this is very likely a sentence adapted from of one of the methods from the Russian State Pharmacopoeia (RSP), which has numerous entries for the adverb «потенциометрически» (Eng. “potentiometrically”).
For instance, there is a nearly identical match in the normative section for the preparation of acetate buffer solution, ОФС.1.3.0003.15-1.27 Ацетатный буферный раствор рН 5,0:

К 120,0 мл 6,0 г/л раствора уксусной кислоты ледяной прибавляют 100,0 мл 0,1 М раствора калия гидроксида и 250,0 мл воды, перемешивают. Доводят рН до 5,0 потенциометрически с помощью 6 г/л раствора уксусной кислоты ледяной или 0,1 М раствора калия гидроксида и доводят объём раствора водой до 1000,0 мл.

RSP underwent a vigorous clean-up since the Soviet era when it was primarily original, and currently it is mostly translated from the European Pharmacopoeia (EP), so I decided to take a look at the corresponding entry in  EP 8.0 (currently invalidated, but one can at least find bits of it in the internet), Section 4.1.3. Buffer solutions:

Acetate buffer solution pH 5.0. 4009100.
To 120 mL of a 6 g/L solution of glacial acetic acid R add 100 mL of 0.1 M potassium hydroxide and about 250 mL of water R. Mix. Adjust the pH to 5.0 with a 6 g/L solution of acetic acid R or with 0.1 M potassium hydroxide and dilute to 1000.0 mL with water R.

EP omits the details as to how the pH is controlled, but there are several phrases in the prior text

Adjust to pH … with …, monitoring the pH potentiometrically

referring to the section 2.2.3. Potentiometric Determination of pH, so I think the phrase that you have proposed

Adjust pH of the solution potentiometrically to 5.0±0.1 with acetic acid.

is perfectly fine and will be correctly understood as a more literate alternative to something like "the pH was determined with a pH meter."
A couple of similar textbook-usecases:

Concise Encyclopedia Chemistry [2, p. 803]:

The pH of a solution is determined experimentally with a suitable pH color indicator (see Indicators) or potentiometrically, with a pH meter (see Glass electrode).

Fundamentals of Electrochemistry [3, p. 590]:

The concentrations of hydrogen ions (solution pH) and of a number of other inorganic ions … are determined potentiometrically.

References

Council of Europe; European Directorate for the Quality of Medicines & Healthcare. European Pharmacopoeia 8.0; European Directorate for the Quality of Medicines & Healthcare, Council of Europe: Strasbourg, 2013.
Concise Encyclopedia Chemistry, English language ed., 2nd ed.; Scott, T., Eagleson, M., Eds.; de Gruyter: Berlin; New York, 1994.
Bagot︠s︡kiĭ, V. S. Fundamentals of Electrochemistry, 2nd ed.; The Electrochemical Society series; Wiley-Interscience: Hoboken, N.J, 2006.


Are there chiral compounds that don't rotate plane-polarized light?

Good question.
There's a phenomenon named cryptochirality[1] (meaning “hidden chirality”), when a compound, though chiral, has practically unmeasurable optical rotation activity.

It can happen to molecules with chiral center(s) bearing very similar substituents. (So, no tricks with bonded slightly modified enantiomeric pairs are needed.)

An example is 5-ethyl-5-propylundecane $\ce{CH3[CH2]5-\overset{$*$}{C}(CH2CH3)(CH2CH2CH3)-[CH2]3CH3}$,[2] don't call it “butyl(ethyl)hexyl(propyl)methane”, found e.g. in beans. Its specific rotation is $[\alpha] < 0.001$.

Another, more common example are fats, i.e. triglycerides, $\ce{RCOOCH2\overset{$*$}{C}H(OCOR')CH2OCOR''}$,[1] if containing e.g. only palmitic, oleic and similar long acyls, optical rotation is not demonstrable.[3]

(Related topic is chirality in polymers, see e.g. Q: Chirality on Carbon of PVC molecule.)

References:

Mislow K. & Bickart P.: An Epistemological Note on Chirality. Israel Journal of Chemistry 15, 1–6 (1976)
Wynberg H., Hekkert G.L., Houbiers J.P.M. & Bosch H.W.: The Optical Activity of Butylethylhexylpropylmethane. Journal of the American Chemical Society 87, 2635–2639 (1965)
Schlenk W.: Synthesis and analysis of optically active triglycerides. Journal of the American Oil Chemists’ Society 42, 945–957 (1965)


Isolating salt from sea water using filters

Yes, this is possible and is a process called reverse osmosis. It is based on a membrane which lets water pass but not the salt ions (and other water contaminants). From Wikipedia:

  The membranes used for reverse osmosis have a dense layer in the polymer matrix -- either the skin of an asymmetric membrane or an interfacially polymerized layer within a thin-film-composite membrane -- where the separation occurs. In most cases, the membrane is designed to allow only water to pass through this dense layer, while preventing the passage of solutes (such as salt ions). This process requires that a high pressure be exerted on the high concentration side of the membrane, usually 2–17 bar (30–250 psi) for fresh and brackish water, and 40–82 bar (600–1200 psi) for seawater

Water desalination features prominently among applications of reverse osmosis:

  The largest and most important application of reverse osmosis is to the separation of pure water from seawater and brackish waters; seawater or brackish water is pressurized against one surface of the membrane, causing transport of salt-depleted water across the membrane and emergence of potable drinking water from the low-pressure side.


Calculating pH of methanoic/formic acid (HCOOH)

The reason for this is that the hydrogen atom attached to the carbon will not be lost. Only the hydrogen attached to the oxygen will be ionized. In fact, in a solution of formic acid, not all molecules will be ionized anyways. Only very strong acids will fully ionize. Even in the case of sulfuric acid, not all molecules will fully ionize and lose the second proton. It is actually a simplification.


Which of the following hydrocarbons react with sodium in liquid ammonia?

Although this is more of a case of communication rather than actual chemistry, I'll explain exactly what the author meant.

They basically want you to ignore the Birch reduction reaction. They want you to find the compound that reacts with sodium in liquid ammonia, but not via Birch reduction.

I however, do stand by that the question is very poorly framed. It is definitely ambiguous, and relies on strange, unrealistic assumptions.


Is there a difference between energy levels and electron shells?

Introduction

The absolute energy levels of all electrons in an atom are different, and will shift subtly when electrons are added or taken away. However, it is true that, generally speaking, sets of electrons in an atom will tend to have energy levels very closely clustered together, with big differences in energy to other sets of electrons in the atoms. We say that these electrons are all in the same energy level. On the other hand, election shells have to do with the mathematics of the electrons 'orbitals', and are closely related, but are not the same. If you want to avoid 'hair-splitting' detail at this point, the difference is this -

Energy levels

Each electron in an atom belongs to an energy level in which all the electrons in that level have nearly exactly the same energy, but there are big energy differences between the levels. If you were to strip away all the electrons from an atom and then return them one by one, you would find that you could only put 2 in the lowest level and then 8 in the next. The capacity for the first five levels are 2, 8, 8, 18, 18 in that order. This corresponds to the number of atoms in each of the first five periods of the periodic table and helps explain the properties of elements in the periodic table. This makes 'energy levels' a very useful idea. So energy levels go from 1 to 5 for the first 5 periods of the periodic table, and in simple terms, full energy levels tend to be 'desired' by atoms.

Electron shells

Electron shells are slightly different, but closely related. They are numbered, like the energy levels. The first shell, 'is' the first energy level. Likewise, the second shell, is the second energy level. But after this, it becomes more complicated, and this is where the confusion can come in.
With the exception of the first shell, each shell is made up of sub-shells. These sub-shells make up each energy level, but sub-shells can belong to different numbered energy level from shell 3 onwards.
To add some detail. There are 3 kinds of shell, 3 ways of satisfying the quantum maths for bound electrons, in the first 5 periods. There are labelled, 's' shells, 'p' shells and 'd' shells. In the mathematics for these, there are 'shell numbers'. For shell number '1', the mathematics only allows for 's'-shells. For shell number 2, 's'-shells and 'p'-shells are allowed, and for shell number 3, s, p and d-shells are allowed. S-shells can take 2 electrons, p-shells 6 electrons and d-shells 10. So shell level 1 has 2 electrons, level 2, 8 (2+6=8) electrons, and level 3, 18 (2+6+10=18) electrons, giving the pattern 2, 8, 18. This is clearly different from the energy levels, and to know why, we have to know more about the sub-shells.
The sub-shells are written down with the shell number first, the shell type second, and the number of electrons actually in that shell, third, (usually as a superscript). So Hydrogen's electrons would be 1s1, and Helium's 1s2. For the second energy level, Li would be 1s2 2s1, and further across the period, Carbon, 1s2 2s2 2p2, having six electrons. At the end of the period, Neon would have two full energy levels, written as 1s2 2s2 2p6. (Often lower full shells are ignored, so Neon would be written 2s2 2p6.) So far, electron 'shell numbers' coincide with energy levels. However, the third energy level has only 3s and 3p shells, not 3d. This is because the 3d shell has an energy level between 4s and 4p. So the 3d sub-shell belongs to energy level 4. Similarly, the 4d sub-shell has an energy level between the 5s and 5p sub-shells. These extra 10 electrons in the energy levels 4 and 5 produce the transition metal block in the periodic table.

Conclusion

Shells and energy levels come from the properties of atomic electrons, but one comes from the mathematics of electon orbitals, and the other from the energy that the electrons take in these sub-shells. This means that referring to energy levels as ‘shells’ will produce confusion, when an understanding of ‘sub-shells’ is needed for a better understanding of the periodic table chemistry.


How to filter out magnesium acetate (a soluble salt) quickly and cheaply from water

Ion-exchange resin can be made with $\ce{H+}$ cations on one type of resin and $\ce{OH-}$ anions on another type of resin. These would leave the water without the any of the magnesium or acetate ions effectively without replacing the ions with other ions. This is how we deionize for general chemical use. This happens according to the reactions below:
$$\ce{2HSO3R + Mg^{2+} + 2C2H3O2- -> Mg(SO3R)2 + 2HC2H3O2}$$
$$\ce{R4N+OH- + HC2H3O2 -> [R4N+ +\ C2H3O2^{-}] + H2O}$$
In these examples, the sulfonates/sulfonic acids and ammonium cations are tethered to the resin so that they can't dissolve.

The downside to this is that if you are trying to measure something else in the system, if it is ionic, it will be removed as well.

I've never had to source ion-exchange resin myself, so I'm not sure of the easiest way to go about it. There's pleanty of deionization resin on Amazon. The mixed stuff is simple to use, just mix it with your water, agitate for a while and then filter it out and you'll have removed all the ions. That is expensive, though, because it's single use.

You can also buy regeneratable ion exchange resins. You don't want to mix the regeneratable stuff, and you have to do two separate ion-exchange steps, one with the cation-exchange and one with the anion-exchange. It's pretty easy to regenerate, though, as far as I know. You just use strong acid with the cation-exchange to redissolve the soluble cations and use a hydroxide solution to regenerate the anion exchange resin. It can't be done forever because there are some cations and anions that just won't want to redissolve, but if you're doing it a lot, might be worth it. I can't seem to find the anion stuff except in bulk.

I'm guessing the mixed type would be more appropriate for most applications. If you really need it quickly, I think you can buy the filters at Lowes/Home Depot and just run it through those or open them up and get the resin out.

Update: These filters are the only ones I saw that said they actually have the "dual ion-exchange system." Their website actually says it gets down to "000 total dissolved solids," which is pretty impressive relative to other water filters from what I've seen (depending on what "000 TDS" actually means).


Determining the rate equation from a reaction where one reactant is in excess

There is no real need to invoke a "$k^{app}$".

Let the rate be $r$, so:

$$r=k[\ce{A}][\ce{B}]^n$$

Since as $\ce{A}$ is in large excess:

$$[\ce{A}]=C(A)\gg[\ce{B}] \implies r=kC(A)[\ce{B}]^n$$

So $[\ce{A}]$ doesn't vary perceptibly during the rate measurement interval.

Plot the value of $\ln r$ versus $\ln [\ce{B}]$ in the knowledge that:

$$\ln r=\ln k + \ln C(A) + n\ln [\ce{B}]$$

So the gradient of that graph is $n$.

And the intercept of that graph is $\ln k + \ln C(A)$. Since as $C(A)$ is known, $\ln k$, and thus $k$, is also known.


What is the mechanism for the auto-dissociation of nitrogen tetraoxide

Dinitrogen tetroxide $\ce{N2O4}$ is a colourless liquid composed of planar dimers of nitrogen dioxide $\ce{NO2}$.$^{[1]\ \text{[a]}\ [2]}$

As Martin noted and is to be expected from the structure of $\ce{N2O4}$, the process

$$\ce{N2O4 <=> NO+ + NO3-}$$

is not very likely to happen. Autoionisation of $\ce{NO2}$ is more common$^{[1]}$.

$$\ce{NO2 -> NO2+ +  e^-\\ NO2 +  e^- -> NO2-}$$

Ionisation energy of $\ce{NO2}$ is $9.8\ \mathrm{eV}$ while its electron affinity equals $1.62\ \mathrm{eV}$.$^{[1]}$ Technically you could look at this as heterolytic bond dissociation $\ce{N2O4}$ but that is probably not what is happening. This autoionisation is, in a way, realised for aqueous solutions$^{[2]}$:

$$\ce{\underset{electron transfer}{NO2 + NO2} + H2O -> \underset{really just HNO3}{NO2OH} +  HNO2},$$
$$\ce{\underset{electron transfer}{NO2 + NO2} + 2KOH -> KNO3 + KNO2 +  H2O}.$$

Table 1. Melting and poiling points of $\ce{NO2}$ and $\ce{N2O4}$. Note the "equality" of melting points.$^{[2]\ [3]}$
$$\begin{array}{|c|c|c|}
\hline
\mathbf{Molecule} & \mathbf{mp\ (\deg  \mathrm{C})} &  \mathbf{bp\ (\deg  \mathrm{C})}\\
\hline
\ce{NO2} & -11.2 & 21 \\
\ce{N2O4} & -11.2 & 22 \\
\hline
\end{array}$$

Table 2. Percentage of  $\ce{NO2}$ and $\ce{N2O4}$ at various temperatures (pressure $1 \ \mathrm{atm}$).$^{[2]}$
$$\begin{array}{|c|c|c|c|}
\hline
\mathbf{Molecule} & \text{below}\  \mathrm{mp}\ (\mathrm{\%}) &  \mathbf{40\ \deg  \mathrm{C}\ (\mathrm{\%})} & \text{over}\mathbf{\  150\deg  \mathrm{C}\ (\mathrm{\%})}\\
\hline
\ce{NO2} & 0 & 31 & 100 \\
\ce{N2O4} & 100 & 69 & 0 \\
\hline
\end{array}$$

$\text{[a]}$ To the best of my knowledge, the enthalpy $-55\ \mathrm{kJ}$ applies at $20 \deg \mathrm{C}$ and $1\  \mathrm{atm}$ for both components in the gas phase. NIST gives $-57.12\ \mathrm{kJ}$ at $25 \deg \mathrm{C}$ and $1\  \mathrm{bar}$. See $\ce{N2O4}$ and $\ce{NO2}$.

$[1]$ N. N. Ahmetov. Anorgaaniline keemia. (1974) (pages 375, 376)

$[2]$ H. Karik, Kalle Truus. Elementide keemia. (2003) (pages 424, 426)

$[3]$ Wikipedia:

Nitrogen dioxide
Dinitrogen tetroxide


Working with typical glass: What temperatures would I be dealing with?

The softening point of glass depends on its composition. For example,

1650°C for fused silica (pure $\ce{SiO2}$)
821°C for Corning Pyrex 7740 borosilicate glass
713°C for Wheaton 900 soda-lime glass

Since you want to use glass found at home, it might be soft lead-glass (e.g. "crystal" tumbler), soda-lime glass (e.g. bottles), or borosilicate glass (e.g. coffee carafe). Melting together different types of glass is usually not successful because unless perfectly mixed, the difference in thermal coefficient of expansion will cause the object to shatter on cooling. One way to test the coefficient of expansion is to combine two different types of glass into a thread to see if it bends on cooling. This can be used to make a glass thermometer, similar to the bimetallic strip in oven thermometers. Calibration is left to the student.

If you start with fairly pure silica sand, then, you need at least 1650°C to get a glass. Although an inexpensive air-propane torch (not oxy-propane) can theoretically reach 1,995 °C, my own experience is that it takes a long time to melt incompletely just a few grams of sand.

Adding other material to silica lowers the softening point. Soda-lime contains alakali metal oxides, $\ce{CaO}$, etc. Rather than using the caustic oxides, cheaper and safer carbonates (or bicarbonates, e.g. $\ce{NaHCO3}$) can be used, though water of hydration and $\ce{CO2}$ must be driven off by heat. You could try a small amount of very fine sand mixed with baking soda, which would eventually melt to make a lower-temperature glass... but be patient, because the glass is so viscous that it takes a long while to dissolve together.

Lead oxide is used to make lead "crystal", with a lower softening point and higher index of refraction than soda-lime glass, but working with lead in a home lab is definitely not advised!


Why does the cathode have to be made out Cu in a Zn-Cu voltaic cell?

You could certain have a zinc-copper cell in which copper was not the cathode material.  Graphite or silver or whatever could be the electrode material instead.   But once you close the circuit:

Current flows
Cu is deposited as a metal solid on the electrode surface, whatever it is made out of.
Copper metal is, as it always is, conductive.
As a result of steps 2 and 3, there is now a copper metal cathode instead of a cathode made of another material.

So even if you start without a copper cathode you wind up with one.


How to retrieve Dalton value from m/z value?

IUPAC is wrong

Martin's answer is useful for its pointer to the IUPAC Gold Book, which is one of the most authoritative sources for chemical nomenclature around.

However, on this particular question, the Gold Book is laughably wrong.

  The abbreviation $m/z$ is used to denote the dimensionless quantity formed by dividing the mass number of an ion by its charge number.

This sentence is wrong.  The $m$ in $m/z$, as used by nearly every practicing chemist in the world (except IUPAC), is the ion mass, not the mass number.  The mass number is the sum of protons and neutrons in a nucleus.  Therefore it is always an integer.  Thus, according to IUPAC, ions with a charge of $\pm 2$ would always have $m/z$ values that are either integers (if $m$ is even) or half-integers.  Thus, according to their definition, saying the $m/z$ value of $\ce{C7H7^2+}$ is 45.52683903 would be wrong.  However that is exactly what most mass spectrometrists would say it is.

Here is some evidence:

A lipid maps database entry for $\ce{C7H7}$ lists $m/z$ values for a hypothetical $\ce{C7H8+}$ ion that have four decimal places.
I happen to have the latest print edition of the Journal of The American Society for Mass Spectrometry in front of me, and paging through the contents I see things such as:

  ...an intense signal for the $\ce{Cu^2+ - His_3}$ complex at $m/z$ 245.6, instead of the $\ce{[His_3 + 2H]+}$ at $m/z$ 215.1...

  ...MS spectra of standard solutions containing (a) lactose ($m/z$ 365.11, $\ce{[M + Na]+}$; $m/z$ 381.309, $\ce{[M + K]+}$), (b) cytidine ($m/z$ 266.218, $\ce{[M + Na]+}$, ...

These excerpts are not using the (integer) mass number in their calculation of $m/z$, they are using the ion mass, measured in Daltons.  (If they were, the only numbers after the decimal point would be .0 or .5).  That means that $m/z$ values are not really dimensionless, which makes the sentence wrong again.  In fact a unit called the Thomson has been proposed for $m/z$ values in honor of JJ Thomson. However this nomenclature has not really caught on widely.  Nonetheless, $m/z$ values are not dimensionless.

The Gold Book goes on...

  It has long been called the mass-to-charge ratio although $m$ is not the ionic mass nor is $z$ a multiple or the elementary (electronic) charge, $e$.

This sentence is wrong because $m$ is the ionic mass (in Daltons) as explained above.  It is additionally misleading because $z$ is always a multiple of the elementary electronic charge, with the understanding that charge must be measured in multiples of $e$, not in Coulombs.

  The abbreviation $m/e$ is, therefore, not recommended.

Finally something I agree with IUPAC on!

  Thus, for example, for the ion $\ce{C7H7^2+}$, $m/z$ equals 45.5.

Most folks would find that to be an OK approximation to the true $m/z$, which as I said above should be 45.52683903.

For the purposes of your question, Martin was right when he said

  For all intents and purposes the mass number may be seen as the mass of the ion in u or Da...

One thing to keep in mind is that this definition of "mass number" is at odds with IUPAC and nearly every other chemical text, which restricts "mass number" to be integers.

One additional complication

It's a bit unclear in your question if you want to determine the mass of an ion or the mass of the parent (usually neutral) molecule that gave rise to that ion.  If you want to determine the mass of an ion, then the other answer is right and you can just multiply the $m/z$ by $z$, the (absolute value of the) reduced electric charge.

If you want the mass of the parent neutral molecule, you will (a) need more information, and (b) have to do a bit more work. The extra information you need is the ionization mechanism.  In electron ionization mass spectrometry, ionization is usually by electron gain or loss.  So to get the neutral molecule mass, you will have to add back the missing electrons (or substract the gained electrons) from the calculated ion mass to get the molecule mass.  For example, if $\ce{C7H7^2+}$ arose by the loss of two electrons, the $m/z$ would be 45.52683903, the ion mass would be 91.05367806, and the molecule mass would be 91.05477522.

Other types of mass spectrometry involve ionization by gaining or losing protons, not electrons, and so the difference between ion mass and neutral molecule mass is much more significant.


How can I produce hydrogen gas efficiently and cheaply?

Electrolysis of water involves the splitting (lysis) of water molecules using electricity.

$\ce{2H_{2}O(l) \rightarrow 2H_{2}(g) + O_{2}(g)}$

From the above equation, we see that the production of hydrogen will occur twice as fast as oxygen (on a volume basis). The electrode materials must be inert (non-reactive with water or its constituents). Platinum is good but very expensive, so titanium and/or nickel alloy catalysts are sometimes used. The current flows into the anode where oxygen gas is produced, whereas hydrogen gas is produced at the cathode. Generally, the higher the current, the higher the rate of production of hydrogen gas.

Since this is not a spontaneous reaction, energy is required from an external source in order to drive it. This is provided by a direct current (DC) electrical power supply, such as a battery or solar cell. The amount of energy required can be determined from the Gibbs energy of formation of water, which is -237.2 kJ/mol (at 298.15 K or 25 °C).

The theoretical minimum voltage needed for the reaction to occur is the standard electrode potential of oxygen, which is +1.23V. In practice, a slightly higher voltage (about 1.48V) is needed because of an over-potential of about 0.25V which is used up as heat.

Pure (demineralised) water can be expensive and has very low electrical conductivity, so some impurities in the water are desirable. Salt (sodium chloride) dissolved in the water will increase its conductivity dramatically, thereby reducing the current necessary to produce a produce hydrogen at a given rate. However, the chlorine gas can be produced at the anode (instead of oxygen) which is highly corrosive and hazardous. Minerals such as magnesium or calcium (present in sea water) will form insoluble precipitants on the cathode which could build up a 'passive' layer, reducing the rate of hydrogen production. Some acids or bases such as sulfuric acid or sodium hydroxide are therefore sometimes preferred.


The hydrophobic force does not exist. It is a phenomena resulting from the following considerations.
In principle, liquids should be like gases, perfectly miscible in all proportions. This is the case for non polar molecules, like hydrocarbons. This is not the case for liquids where molecules are hold together by hydrogen bonds, like water. In these liquids, adding solutes requires to break hydrogen bonds. It is acceptable if the solute is able to form new hydrogen bonds, like with small alcohols, polyols and sugars. But if mixing with another substance will break too many H-bonds, Nature refuses to make up this mixture, as if there was a hydrophobic force. Of course, this force does not exist. It is simply a consequence that mixing very different substances would cost too much energy for breaking all H-bonds of one liquid, without forming back new ones.


In general substances whose molecules have a strong dipole tend to dissolve better in water than compounds which do not. If we consider for example diethyl ether and THF. The two are almost identical the only difference is that THF is held in by ring into one posiition. This makes THF have a dipole which is more permenent. The side groups on diethyl ether can move about thus allowing the molecule to adopt different conformations.

THF is totaly soluble in water, it is infinitely miscible with water. This means any combination of THF and water will mix perfectly. On the otherhand diethyl ether has a limited solubility in water. Only a very small amount of diethyl ether will dissolve in water.

If we look at the solvent accessable surface of THF where the atomic colours represent their partial charges we can see that THF has a strong permenent dipole. Here in these pictures the more blue an atom is the more negative and the more red the more positive it is.

While ether has less of a permennt dipole as the effects of the right hand side partial positive charge is canceled out by the left hand one.

Do be careful as more than dipole strength will determine the solubility of a compound. A better test for liquids would be to measure the relative dielectric constant at different frequencies. If we have a substance whose molecules have a dipole then at a low frequency the relative dielectric constant will be higher than at a high (often microwave) frequency. The reason is that at a low frequency the molecules have time to rotate to line up. At a high frequency this does not occur becuase the molecules have too little time to rotate. Below is shown a random array of dipoles and on the right some dipoles which have been aligned by an electric field.


Why is lead(II) present in two groups of cation analysis?

Simply because lead(II) fits both analytical groups:

it forms lead(II) chloride $\ce{PbCl2}$, a poor soluble chloride, which puts it in the 1st group;
it also forms lead(II) sulfide $\ce{PbS}$, insoluble in water and diluted acids, which also puts it in the 2nd group.

One reason for putting lead in both groups is that the chloride is somewhat of a borderline case in terms of being insoluble enough to appear in Group I.  Group II represents a more complicated analysis but is more sensitive for lead.


Why is the Van der Waals constant of water lesser than benzene?

The parameter $a$ from the van der Waals equation

$$\left(p+\frac{an^2}{V^2}\right)\left(V-nb\right)=nRT$$

is not to be taken only as the measure of strength of intermolecular bonding.

Rather, it is overall measure of significance and frequency of intermolecular interaction.

It may be based on purely elastic electrostatic interactions, where only the weakest van der Waals force may ( or even may not ) be involved.

The probability 2 benzene molecules interact is bigger than for 2 water molecules.

Notes also that for n-hexane is $\pu{a = 24.71 L^2\cdot bar mol^{2-}}$. It has bigger value than benzene, in spite of the weaker interaction due the lack of $\pi$ bonding. But its linear molecule is stratched with all the hydrogen atoms in many directions, so collisions are more probable.


How to distinguish between benzyl chloride and chloro benzene?

When benzyl chloride is treated with $\ce{NaOH}$ and $\ce{AgNO3}$ , a white precipitate is formed.
Chlorobenzene does not form white precipitate reacting with $\ce{NaOH}$ and $\ce{AgNO3}$.


Can hot food ever emit x-rays or gamma rays?

In theory, yes, you can heat objects to a high enough temperature to emit x-rays or gamma rays. You cannot do this to food, and you certainly cannot do this in your kitchen (or probably any kitchen).
Let's take the lowest energy x-ray out there and see what it would take. X-rays range in frequency from $30 \times 10^{ 16}$ to $30\times 10^{10}$ hertz. The energy of one photon of 30 petahertz radiation is:
$$E=h\nu = \left(6.626\times 10^{-34}\mathrm{\ J\cdot s}\right)\left(30\times 10^{16}\ \mathrm{s^{-1}}\right) = 1.988 \times 10^{-16}\ \mathrm{J}$$
This is not a lot of energy! However, a single photon is boring. Let's consider a mole of photons. This will also ease comparison with other phenomena, whose energies are listed per mole of events.
$$1.988 \times 10^{-16}\ \mathrm{J} \times 6.022\times 10^{23}\ \mathrm{mol^{-1}}=1.197\times 10^8\ \mathrm{J\cdot mol^{-1}}$$
In theory, if you could pump that much energy into something, you should get some high energy photons out. In practice, it does not work that way. Other stuff happens first. To simplify our example, let's just consider 1 mole of water (18.0 grams) and heat it up. The fate of basically any other matter will be the same, but the energy required will vary a bit.
First, adding energy heats the water. If we start at room temperature $\left(20\ ^\circ\mathrm{C}\right)$, it takes $80\ ^\circ\mathrm{C}\times 18\ \mathrm{g}\times 4.184\ \mathrm{J\cdot g^{-1}\cdot ^\circ C^{-1}}=6025\ \mathrm{J}$ to heat that water to boiling. It takes 40.66 kJ to convert the water into gas. Neither of these puts a big dent in our energy. It takes further energy to heat the water vapor again, but let's see how far we need to take it.
Once we get enough energy into our sample of water, the molecules start to fall apart.
$$\ce{ H2O(g) -> 2H(g) + O(g)} \ \Delta H^\circ =+920\ \mathrm{kJ\cdot mol^{-1}}\ \Delta S^\circ =0.202\ \mathrm{kJ\cdot mol^{-1}\cdot K^{-1}}$$
By fixing $\Delta G=0$ at equilibrium, we can solve for a temperature at which this reaction becomes spontaneous:
$$T=\dfrac{\Delta H}{\Delta S}=\dfrac{+920\ \mathrm{kJ\cdot mol^{-1}}}{0.202\ \mathrm{kJ\cdot mol^{-1}}\cdot K^{-1}}=4596\mathrm{K}$$
We need to heat our water vapor up an additional 4218 K, which takes $18\ \mathrm{g}\times 1.996\ \mathrm{J\cdot g^{-1}\cdot K^{-1}}\times 4218\ \mathrm{K}=151.5\times 10^3 \mathrm{J}$.
So, we now have pumped nearly 200,000 J into our water sample, atomized it, and heated it to approximately 5000 K. We are now close to the temperature of the outer layers of the sun! Surely we have enough energy at this temperature to produce x-rays. Nope. At 5000 K, we produce minimal x-rays. Most of the radiation is in the visible, UV, and IR (think about what we get from the sun). Below is a plot of black-body radiation as a function of temperature (image by Wikipedia user Darth Kule and released into the public domain):

Okay, so we are far beyond the reality of what can happen in a conventional oven (or almost any reasonable heat source used for food). At this temperature, we can use the Planck Law to calculate the power output ($I$) of x-rays at the temperature. We can also do this at some normal temperatures and for gamma rays. This model is a little goofy, since food is not a black body, but we will at least calculate the max x-ray and gamma ray output.
Rather than grinding through all the maths, I'll just put in a table of some temperatures and watts. 1 watt is not a lot of power. Most lightbulbs produce light in the kilowatts.
$$\begin{array}{|c|c|c|c|}\hline
\mathrm{T\ (K)} & \mathrm{P_{x-ray}\ (W)} & \mathrm{P_{gamma}\ (W)} & \mathrm{notes} \\ \hline
378 & \approx 0 & \approx 0 & \text{boiling point of water} \\ \hline
550 & \approx 0 & \approx 0 & \text{approximate common highest temperature on residential ovens}\\ \hline
700-800 & \approx 0 & \approx 0 & \text{temperature range for wood-fired ovens, tandoors, etc.}\\ \hline
5770 & 4.26\times 10^{-129} & \approx 0 & \text{temperature of the photosphere of the sun}\\ \hline
1.57\times 10^7 & 10.4 & 7.87\times 10^{-54} & \text{estimated temperature of the center of the sun} \\ \hline
\end{array}$$
So, if you could heat your food to the temperature of the sun, it would produce minuscule x-ray radiation. It would also no long resemble food.


Chemical change - decomposing single element?

It is a chemical change

The aspect of this that is probably confusing you is that both the product and the result in the case of an element are still a form of the element.

But elements can exist in different chemical forms and moving between them is still a chemical process. A relevant example is the reaction occurring mostly in the upper atmosphere that converts dioxygen (O2) to ozone (O3). This is a specific example of what happens when you add energy (in this case from UV light) to an oxygen molecule in the presence of other oxygen molecules: the split-apart oxygen atoms react to form ozone (which is more stable than a free oxygen atom). This is clearly a chemical change as the molecular structure is different, though ozone is still elementary oxygen so the element has not changed.


Is the dimethylglyoximate anion ambidentate or not?

I'm not sure if this example qualifies as an example for dimethylglyoximate acting as an ambidentate ligand, because it, for me, is more like bidentate. Nonetheless, I'm presenting this example here only because both nitrogen and oxygen are donating lone pairs within the same molecule. See trinuclear copper complex below (ref: doi.org/10.1107/S160053680301331X):


As far as I can tell there are two material types used for this - pure ceramics or ceramic mixtures that are designed to break and be replaced, and composite metal + ceramic materials that are designed to be able to handle a couple of rounds without losing too much effectiveness. There are also techniques involving purely metal layers in overlapping structures with fibers of materials like Kevlar or ultra-high-molecular-weight polyethene (UHMWPE) (of which Spectra is a brand).

Shopping around, it seems the best materials are boron and silicon carbides (extremely hard materials, though not quite as hard as diamond), which can be formed into plates with multiple levels of structure by starting with hexagonal plates and forming larger plates from those. In addition to personal armour, the same materials and techniques are used for military vehicle (including aircraft) armour. In the ceramic/metal composite materials, you get ceramic plates formed up within a metal sheet, where a ceramic element will shatter to take most of the impact, while the rest will be absorbed by the metal matrix. The metal matrix then remains intact (but may be deformed) so that the armour is still useful but weakened around the area of impact.

Good places to start from are the Wikipedia pages for Chobham armour (the common name for the tank armour that includes composite metal/ceramic matrix materials), and Small Arms Protective Insert.


What are some of the simplest chiral molecules in biological systems

Don't get hung up on the "left-handed" and "right-handed" designations for amino acids and sugars. Those are just conventions for distinguishing one mirror image (enantiomer) of a chiral molecule from another. It just so happens that based on a particular way of drawing (Fisher projection) amino acids and sugars that it appears that the configurations are "left-handed" or "right-handed."

As for how simple chiral biological molecules get, the simplest ones that I can think of are alanine or lactic acid. I'm not sure what you consider "intelligent", but these are likely to be found in all life (as we know it).


What is the mechanism of ring contraction of 6-bromo-7-methoxy-2,3,4,7-tetrahydrooxepine from seven to five?

It's in scheme 3 of the paper you cited! The 7-membered acetal undergoes ring-opening to the α,β-unsaturated oxocarbenium ion, then conjugate addition to give the 5-membered tetrahydrofuran. A bit of arrow pushing and addition of solvent gives the other THF product

Not sure the legality of posting screenshots of papers, so I redrew the scheme:


Does fluorine ever form a double or triple bond?

The compound $\ce{NF}$, which is isoelectronic to $\ce{O2}$ is known, has been isolated in matrices and characterised and subjected to calculations. Like for dioxygen, three different states of this molecule are known: one triplet and two singlet states. Without performing any sophisticated analysis of its orbitals, we can expect a bond order of 2, and thus a double bond.

Harbison performed calculations on this compound. He came to the conclusion that the most stable triplet state is best described using only a single $\ce{N-F}$ bond. The two singlet states which require full electron pairing, however, display a much shorter $\ce{N-F}$ bond distance and are thus better described by an $\ce{N=F}$ double bond. Adding formal charges would lead to:
$$\ce{\overset{-}{N}=\overset{+}{F}}$$

Reference:

G. S. Harbison, J. Am. Chem. Soc., 2002, 124, 366–367. DOI: 10.1021/ja0159261.


How to calculate the pH of the neutralisation of HCN with excess of KOH?

While you've generally done a good job, you've made an important error. The hydrolysis of the cyanide ion should be written as an equilibrium reaction:

$$\ce{CN-(aq) + H2O(l) <=>  HCN(aq) + OH-(aq)}$$

That's because $\ce{HCN}$ is a very weak acid and its conjugated base a weak base (homework: look up the $K_b$ of $\ce{CN-}$).

Now, as you duly noted, there's excess $\ce{KOH}$ in the solution, which dissociates completely. The $\ce{OH-}$ resulting from that dissociation now push the equilibrium:

$$K_b=\frac{[\ce{HCN}]\times[\ce{OH-}]}{[\ce{CN-}]}$$

to the left. As a result, the contribution of the $[\ce{OH-}]$ coming from the hydrolysis of the cyanide is negligible and can be ignored. Only the $[\ce{OH-}]$ from the excess $\ce{KOH}$ is to be counted. So $[\ce{OH-}]\approx 0.06\ \mathrm{mol/L}$.

We can confirm that numerically by assuming $[\ce{OH-}]\approx 0.06\ \mathrm{mol/L}$ and with:

$$\frac{[\ce{HCN}]}{[\ce{CN-}]}=\frac{K_b}{[\ce{OH-}]}\approx 0.0027$$

So neglibibly little cyanide is present as the acid $\ce{HCN}$.

As regards the matter of adding molarities, in some instances that is allowed.

Say we add $n_1$ moles of $X$ to $V\ \mathrm{L}$, that would give a molarity $M_1=\frac{n_1}{V}$. At a later stage we add $n_2$ moles of $X$, that molarity would be $M_2=\frac{n_2}{V}$.

The total molarity would be:

$$M=M_1+M_2=\frac{n_1}{V}+\frac{n_2}{V}=\frac{n_1+n_2}{V}$$

But if we were to mix volumes of solutions it would be:

$$M=\frac{M_1V_1+M_2V_2}{V_1+V_2}$$

Now  they are no longer simply additive.


The Compressibility of a substance, dependent on

If you apply an external pressure on an object and analyze the deformational response of the object (say using Hooke's law in 3D), you find that the state of stress and strain within the object are completely uniform and independent of direction within the object (if the object has direction-independent properties to begin with), irrespective of the shape of the object.  This is sort of the same kind of thing that happens in thermal expansion, where the object strains uniformly in all directions, and the strains are independent of the object shape.


Stirring the reaction between hydrochloric acid and a powdered substance

To minimize thermal coupling between the stirring apparatus and the solution, the best choice is a small, slowly stirred, PTFE or equivalent coated magnetic stir bar. Heat generated by friction should be negligible.

A potential source of heat transfer into the system from this configuration is heat from the stirrer motor. To mitigate this, the surface of the stir plate should be covered with a strip of Styrofoam as thermal insulation. Furthermore, the reaction vessel could be physically separated from the stir plate by using a ring stand and clamp or equivalent. It may be necessary to adjust the thickness of the Styrofoam and distance of physical separation in order to keep the stir bar magnetically coupled to the stir plate magnet.

The setup should first be tested by doing several full runs without adding the acid or the powders. This is essentially an analytical blank. If the temperature changes can be considered negligibly small throughout the duration of an experiment, then you are ready to go. If necessary, any temperature change measured from the blank runs can be subtracted from the actual experimental measurements.


Cyclohexene is very symmetric. At first approximation, you can draw two planes of symmetry: one containing the double bond and one orthogonal, bisecting the double bond.

These two planes of symmetry mean, that both sides of the double bond are equal as are both carbon atoms. Therefore, the hydrogen atom protonating the double bond in $\ce{HBr}$ addition cannot sense any difference between either of the four possible attack pathways and chooses one at random. The bromide then has to choose the other carbon, but can again choose the plane from which it attacks (‘above’ or ‘below’ — in inverted commas because they are equivalent!) at random.

Even after the reaction, the product bromocyclohexane is highly symmetric. You can again draw a plane of symmetry through it; this time, the plane must include the $\ce{C-Br}$ bond and also bisect the carbon atom on the other side of the ring. The presence of a plane of symmetry means that the molecule cannot exist in enantiomeric forms, and since you cannot say on which side the hydrogen was previously added (they, too, are all identical), you only get one possible product.

We might consider a more complicated case and decide to add $\ce{DBr}$ instead of $\ce{HBr}$. In this case, we would indeed observe more than one type of product since a deuterium atom cannot be mapped onto a hydrogen atom by symmetry. However, we would obtain all symmetry-equivalent products at equal ratios because of the high symmetry of the starting compound.

In general, if you can find planes of symmetry then both sides are equally likely to react and you cannot predict a preference.


Choice of solvent during thin layer chromatography

Chromatography is really an empirical science. The fluid mechanical side and the adsorption phenomenon is somewhat well understood but choosing mobile phases is essentially based on trials, sadly to date. If someone has given you a ratio, say 8:2, it means someone must have tried it and they found that this mobile phase works the best. The main criterion for choosing the mobile phase is that all the mixture components must be soluble in it. Thus if glycine did not move, this means that the mobile phase has no affinity for this solute (it is perhaps insoluble in it) and it is a poor choice. One then starts to tune the polarity of the solvent mixture. Think about a petroleum sample, which literally has thousands and thousands of different molecules. Now think yourself: will they all dissolve in 100% hexane? You do need an additional solvent.

There is no theoretical restriction in chromatography to use multiple solvents or a single solvent. For example in gas chromatography, you always use a pure gas. The reason for using a mixture of solvents in liquid chromatography originates from the concept of "general elution problem of chromatography" which basically says that there is no single adsorbent or a single mobile phase which can separate everything. Not only separation scientists use a single mixture composition of mobile phase (isocratic), they can also continuously change the mobile phase composition as a function of time. This is called gradient elution. You cannot do that with thin layer chromatography without using fancy stuff.

People are trying to implement machine learning, so that one can enter the molecular structures contained in the mixture, available stationary phases, and with a lot of statistics and tons of experimental data, the computer can tell you to try a certain mobile phase composition.


Is the shorthand electron configuration of Br- just [Kr]?

You could render bromide ion that way.  But the outermost electrons in the "krypton core" are potentially reactive, valence electrons; therefore $\text{[Kr]}$ is not an accurate description of how the electrons really act.  Only the argon core acts like a noble gas, so you are better off rendering the bromide ion as $\text{[Ar]3d}^{10}\text{4s}^2\text{4p}^6$.

Cations stripped to the noble gas core, however, really are stripped to unreactive electrons, so calcium ion can accurately be rendered as $\text{[Ar]}$.


Are there known compounds of caesium with oxidation number >1?

As noted in the answer to the other question, alkali metal can exist in higher oxidation state when bonded with polycylic multidendate ligands like cryptands etc. This is an excerpt from an eBook:

  The chemistry of group 1 elements have been dominated by +1 oxidation
  states. However, there have been indications that caesium might form
  higher oxidation species. Thus, electrochemical oxidation of
  $\ce{[CsL]PF6}$ L=18-crown-6 or cryptand-[222] gives evidence for
  $\ce{Cs^2+}$ and $\ce{Cs^3+}$. Compounds containing caesium in higher
  oxidation state is yet to be isolated.

Also, compounds of form $\ce{CsF_n}$ is known to contain stoichiometric amount of caesium in higher oxidation state($\ce{Cs^2+,Cs^3+,Cs^4+,Cs^5+}$). Experimental calculations have confirmed the presence.(See here and here).

Furthermore, it has also been observed by Moock and Seepelt, Angew, Chem. Intl. Ed. Engl. 28, 1676(1989) that $\ce{Cs+}$ can be oxidised in acetonitrile solution to $\ce{Cs^3+}$ at a potential of +3.0 V($\ce{E°}$). - (See this eBook)


Why does hydrogen burn spontaneously in fuel cell at room temperature?

You guessed it right that hydrogen and oxygen mixtures will remain stable. Someone in Harvard waited for >30 years and found very little water if any. As you said, there is a energy barrier. What type of materials can lower this barrier? Catalysts.

The electrode materials require some catalysts such as platinum or nickel. Of course, making cheaper and sustainable catalysts is an extensive area of research. If a person discovers a better catalyst which is far more common than Pt, then that person certainly deserves a big award. Copper does not act as a catalyst here.


Why is the boiling point of homologous primary > secondary > tertiary alcohols whilst their solubility in water is the reverse?

The key concept here is that for solubility, you need to compare alcohols with the same number of carbon atoms. For example, the solubility order of isomers of butanol is as follows:
$$\ce{CH3CH2CH2CH2OH < (CH3)2CHCH2OH < (CH3)3COH}$$

This difference in solubility follows the order 3°>2°>1° and can be explained by the branching concept that you have used in your question. However, If you compare alcohols with different number of carbon atoms, say butanol and methanol, then the alcohol with a smaller hydrocarbon part will be more soluble in water.

The boiling point of alcohols increases with the increase in the number of carbon atoms, as stronger van der waals forces start acting in between the molecules. For example, Phenol boils at around 180°C whereas methanol boils at around 60°C. For alcohols with the same number of carbon atoms, a more exposed hydroxyl group will result in stronger intermolecular H-bonding.

TL;DR: your comparision can only be applied for alcohols with the same number of carbon atoms


What is the order of stability of the resonance structures of furan?

A quick post so you can get started on thinking about this.

Remember, resonance is just a model. The electrons are delocalized all the time; the furan does not morph back and forth among the various structures.

You can do a "formal" evaluation of which resonance forms have the most contribution to the structure by looking at formal charges, and there are rules to help you decide, from the formal charges, which resonance forms are most important.

If you're looking at reactivity, the property you're interested in may not be that the one that is "more stable is less reactive". What you might look at instead, is the consequences of attack at the 2 and 3 position, look at the stability of the intermediates formed...a more stable intermediate would tend to drive the reaction along that path.

After looking at these ideas, feel free to come back with specific questions.


Separate precious metals from catalytic converter

Let me try to talk you out of it ;-)

The precious metals in a catalytic converter sit in a honeycomb ceramic that contains a lot of synthetic Cordierite. This material isn't only heat-resistant, but also quite hard.

Industrial recycling of catalytic converters typically starts with

dismantling (decanning) the ceramic monolith using guillotine shears
breaking down the ceramic to smaller pieces in hammer mills connected by a closed tubing system
milling the material to dust in a ball-mill blender

You might want to wear a filter mask during these steps.

Industrial recycling almost exclusively uses pyrometallurgical tretment (melting the metals) then.

Having ruled that out, there are two other options:

hydrometallurgical treatment
In order to separate the metals from the ceramic, you can either try to

dissolve the ceramic in hot alkaline solution in a closed reactor under pressure (to reach temperatures above 100 °C). This leaves the metals intact.
leach the metals out by using strong oxidizing acids, such as mixtures of concentrated nitric and hydrochlorid acid. You have the metals in solution now.

What are your plans for personal protection during these steps and for the waste water management?
biomobilization
There is some ongoing research to replace the unpleasant conditions of hydrothermal treatment with microorganisms that can leach the solid metals and store them. As an example, have a look at an article by Helmut Brandl, Stefan Lehmann, Mohammad A. Faramarzi, and Daniel Martinelli, published in  Hydrometallurgy, 2008, 94, 14-17. The authors examined, whether metals could be converted to cyanido complexes, such as $\ce{[Pt(CN)4]^{2-}}$, by cyanide-producing microorganisms. A PDF of their article, titled Biomobilization of silver, gold, and platinum from solid waste materials by HCN-forming microorganisms is available at no cost from ZORA (Zurich Open Repository and Archive). In the case of platinum, however, the results were rather disappointing.


Will galvanic corrosion happen on a ring made of gold and bronze?

Moist air, rich in an electrolyte (salt particles) or human contact, providing both NaCl and H+ may supply the reagents needed for galvanic corrosion, with dissimilar metals in direct contact. This usually proceeds, albeit slowly, over time.

Note, exposure to fruit juices could be especially problematic, resulting in a matter of days of continuous contact in the presence of moisture and oxygen, in some visible signs of disfiguring pit corrosion.

Nevertheless, I recommend wearing the ring and enjoying it, but perhaps placing it in fresh distilled water (remove O2 exposure) at night after washing it with mildly alkaline soap (to clean off salts and neutralize body acids). If the vessel use is made of Aluminum, a very active anodic metal, the ring should do just fine.


Are aromatic amines or amides less basic?

Measures for acid and base strength

First, you need to be clear about 'strong' and 'weak' acids.
In order to measure the strength of an acid relative to water and find out how effective a proton donor it is, you must look at the equilibrium constant for the reaction

\begin{equation}
  \ce{HA (aq) + H2O(l) <=> H3O+ (aq) + A- (aq)}
\end{equation}

The position of equilibrium is measured by the equilibrium constant for this reaction $K_{\text{eq}}$.

\begin{equation}
  K_{\text{eq}} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{H2O}][\ce{HA}]}
\end{equation}

The concentration of water remains essentially constant (at $55.56 \, \text{mol} \, \text{dm}^{–3}$) with dilute solutions
of acids wherever the equilibrium may be and a new equilibrium constant, $K_{\text{a}}$, is defined and called the acidity constant.

\begin{equation}
  K_{\text{a}} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]}
\end{equation}

Like $\text{pH}$, this is also expressed in a logarithmic form, $\text{p}K_{\text{a}}$.

\begin{equation}
  \text{p}K_{\text{a}} = -\log (K_{\text{a}})
\end{equation}

Because of the minus sign in this definition, the lower the $\text{p}K_{\text{a}}$, the larger the
equilibrium constant, $K_{\text{a}}$, is and hence the stronger the acid. The $\text{p}K_{\text{a}}$ of the acid is the $\text{pH}$ where it is exactly half dissociated. At $\text{pH}$s above the $\text{p}K_{\text{a}}$, the acid $\ce{HA}$ exists as $\ce{A^{–}}$ in water; at $\text{pH}$s below the $\text{p}K_{\text{a}}$, it exists as undissociated $\ce{HA}$.

An acid's $\text{p}K_{\text{a}}$ depends on the stability of its conjugate base.
$\ce{HCl}$ is a much stronger acid than acetic acid: the $\text{p}K_{\text{a}}$ of $\ce{HCl}$ is around $–7$ compared to $4.76$ for acetic acid. This tells you that in solution $K_{\text{a}}$ for hydrogen chloride is $10^{7} \, \text{mol} \, \text{dm}^{–3}$ whilst for acetic acid it is only $10^{–4.76} = 1.74 \cdot 10^{–5} \, \text{mol} \, \text{dm}^{–3}$. But, why are the equilibria so different? Why does hydrogen chloride fully dissociate but acetic acid do so only partially?

\begin{equation}
  \ce{HCl (aq) + H2O (l) <=>> H3O+ (aq) + Cl- (aq)} \quad \qquad \qquad K_{\text{a}} = 10^{7} \, \text{mol} \, \text{dm}^{–3} \\
  \ce{CH3COOH (aq) + H2O (l) <<=> H3O+ (aq) + CH3COO- (aq)} \qquad K_{\text{a}} = 1.74 \cdot 10^{–5} \, \text{mol} \, \text{dm}^{–3} \\
\end{equation}

The answer must have something to do with the conjugate base $\ce{A^{–}}$ of each acid $\ce{HA}$, since this is the only thing that varies from one acid to another. In both the equilibria above, water acts as a base by accepting a proton from the acid. For the hydrochloric acid equilibrium in the reverse direction, the chloride ion is not a strong enough base to deprotonate the hydronium ion. Acetate, on the other hand, is easily protonated by $\ce{H3O+}$ to give neutral acetic acid, which means that acetate must be a stronger base than chloride ion.
A nice thing to remember:

The stronger the acid $\ce{HA}$, the weaker its conjugate base, $\ce{A^{–}}$
The stronger the base $\ce{A^{–}}$, the weaker its conjugate acid $\ce{AH}$

So, the $\text{p}K_{\text{a}}$ is a good measure for the strength of an acid: the lower the $\text{p}K_{\text{a}}$ the stronger the acid.

But what about bases? To say something about the strength of a base it is convenient to take the following viewpoint: to what extent does a protonated base want to keep its proton or in other words: how big is the acid strength of its conjugate acid? For example if you want to know which is the stronger base - say, formate anion or acetylide anion - you look up the $\text{p}K_{\text{a}}$s for their conjugate acids. You find that the $\text{p}K_{\text{a}}$ for formic acid ($\ce{HCO2H}$) is $3.7$, whilst the $\text{p}K_{\text{a}}$ for acetylene is around $25$. This means that acetylene is much more reluctant to part with its proton, that is, acetylide is much more basic than formate. This is all very well for anions - you simply look up the $\text{p}K_{\text{a}}$ value for the neutral conjugate acid, but what if you want to know the basicity of ammonia? If you look up the $\text{p}K_{\text{a}}$ for ammonia you find a value around $33$ but this is the value for deprotonating neutral ammonia to give the amide ion, $\ce{NH2^{–}}$.
If you want to know the basicity of ammonia, you must look up the $\text{p}K_{\text{a}}$ of its conjugate acid, the ammonium cation, $\ce{NH4+}$, protonated ammonia. Its $\text{p}K_{\text{a}}$ is $9.24$ which means that ammonia is a weaker base than hydroxide - the $\text{p}K_{\text{a}}$ for water (the conjugate acid of hydroxide) is $15.74$.
In conclusion: the higher the $\text{p}K_{\text{a}}$ of the conjugate acid the stronger the base.

The answer to your question

The $\text{p}K_{\text{a}}$ of the conjugate acid of aniline is $4.6$. So, since this value is much lower than the $9.24$ you had for ammonia, it is much less basic than ammonia, as is expected since the lone pair of aniline is delocalized into the phenyl ring. But for amides the $\text{p}K_{\text{a}}$ values of the conjugate acid are typically smaller than $0$, so an amide is much less basic than aniline. This is to be expected because the carbonyl group contains an electronegative oxygen atom and thus has a lower lying LUMO than the phenyl ring which can interact more efficiently with the nitrogen's lone pair, thus spreading the negative charge more evenly among the conjugated groups and leading to a higher stabilization of the molecule.


How to identify which species will undergo Oxidation and which will go Reduction

You have to look at the possible oxidation states of the atoms involved.
For instance, if you were given the molecule $\ce{F2}$, you should consider that fluorine has two possible oxidation states: 0 (in the $\ce{F2}$ molecule) and -1.
Hence, $\ce{F2}$ can only undergo reduction.
For the other molecules, you should use the same strategy: try yourself!


Effects of atmospheric gases on colour of aurora

There is good information at Glowing Gases - Aurorae

There are many factors that need to be considered.

Once an atom or molecule is excited, it can lose the energy by collision or by emission of light.

The longer the lifetime of the excited state, the higher the altitude is required to make radiation vs. collision the way energy is lost.

The atomic oxygen transition with green light has a lifetime of about 1 second, while the transition for red light is 110 seconds, so a higher altitude is required for the red transition to become significant.

For molecular nitrogen, radiative lifetime is only $4 \times 10^{-8}$ seconds according to Electronically excited molecular nitrogen and molecular oxygen in
the high-latitude upper atmosphere Ann. Geophys., 26, 1159–1169, 2008

Also, human eyes are more sensitive to green light than red light.


How can I attempt to inhibit the hydrolysis/degeneration of aspartame?

Aspartame is hydrolysed in both acid and alkaline solution with the methyl ester being the first site hydrolysed, so control of the pH of the aqueous solution is important. Addition of buffer solutions would enable control of the pH, keeping it close to neutrality.

https://www.chemguide.co.uk/physical/acidbaseeqia/buffers.html


A sample of copper(II) sulfate pentahydrate (CuSO4·5H2O) contains 0.360 g of water. What is the total number of atoms in the compound

You have to calculate the amount of substance with the mass of water.

$n[H_2O]=\frac{m}{M}=\frac{0.360\,g}{18\,g} \cdot mol = 0.02\, mol$

$1~ n[CuSO_4\cdot 5 H_2O] = 5~n[H_2O]$

$1~ n[CuSO_4\cdot 5 H_2O] = 0.004\, mol$

In 0.004 mol copper sulfate pentahydrate you have 0.36 g water.

1 molecule of copper sulfate pentahydrate is composed of 21 atoms and the Avogadro constant is $N_A = 6.022 \cdot 10^{23}\, mol^{-1}$.

Now you can simply multiply all values.

$\text{#}Atoms = 0.004\, mol \cdot 6.022 \cdot 10^{23}\, mol^{-1} \cdot 21 = \underline{5.06 \cdot 10^{22}}$


Normalization of a free particle in 1 dimension

To answer the question, over and above the errors noted in the other answer the integral is just not well defined - it is infinite. There are two ways to see this

The "common sense" method. Consider
$$ \int^\infty_{-\infty} \left[ A\sin(kx) + B \cos(kx)\right]^2dx$$
The quantity under the integral is positive semi-definite - it is always greater than or equal to zero. It describes oscillations all of which are positive. Each oscillation bounds a finite positive area between it and the x axis. As the bounds of the integral are all space there are an infinite number of oscillations. An infinite number of positive finite areas evaluates to an infinite area. Thus your integral is not well defined.

Maths - assuming A and B are real (but removing this restriction makes no difference to the final conclusion)

$$\left[ A\sin(kx) + B \cos(kx)\right]^2 = \left[A^2\sin^2(kx) + 2AB\sin(kx)\cos(kx) +B^2\cos^2(kx)\right]$$
The way to evaluate the integral is to use the identities
$$ 2sin(\theta)cos(\theta) = sin(2\theta) $$
$$ sin^2(\theta) = {{(1-cos(2\theta))}\over2}$$
$$ cos^2(\theta) = {{(1+cos(2\theta))}\over2}$$
Thus we can write the integral as
$$
\int^\infty_{-\infty} \left[A^2{{(1-cos(2kx))}\over2}+B^2{{(1+cos(2kx))}\over2}+ABsin(2kx)\right]dx
$$
The first term contains
$$
\int^\infty_{-\infty} \left[ A^2\right]dx
$$
This is infinite. Similarly the second term for $B^2$. Thus the integral is not well defined [yes this is slightly hand-waving but avoids a discussion of $cos(\infty)$ and similar]. So what you are doing wrong is trying to evaluate an integral that doesn't have an answer.
What you are running into is the fact that not all wavefunctions are normalisable. See Normalizing the solution to free particle Schrödinger equation in the physics stack exchange forum for more details


Comparing the basicity of benzamide and acetamide

Where do amides accept protons?

As Mithoron aptly pointed out in the comments, it is first important to establish which atom in the amide molecule receives the proton. Due to the strong delocalisation present in the amide, the nitrogen lone pair becomes very weakly basic and it is the lone pair on the carbonyl oxygen which is donated to form bonds with $\ce {H^+}$. Not only does the delocalisation diminish the basicity of the nitrogen lone pair, the donation of the lone pair from the nitrogen atom towards the carbonyl group effectively raises the energy of the lone pairs present on the oxygen atom.

This is more clearly illustrated in the molecular orbital picture, where we consider the interactions between the lone pair orbital on $\ce {N}$ and the $\ce {C=O}$ $\pi$ and $\pi^*$ MOs. The interaction with the $\pi$ MO helps to increase the energies of the electrons on the $\ce {O}$ atom while the interaction with the $\pi^*$ MO helps to lower the energy of the lone pair on the $\ce {N}$.

Let's look at some data

I personally find that a theoretical approach to this problem without looking at any data is non-trivial and may potentially give a false conclusion since many factors are at play. In fact, my initial thoughts without looking at any data was that benzamide would be more basic since the benzene ring can donate into the $\ce {C=O}$ group via resonance (in the same way as how a benzyl carbocation is stabilised by the donation of $\pi$ electrons from the ring) and this would certainly enhance the basicity of the carbonyl oxygen since the electron density at the oxygen atom would have increased significantly.

Cox et al. (1980) have determined via empirical measurements the $\ce {pk_a}$s of the conjugate acids of acetamide and benzamide to be $\ce {-0.73}$ and $\ce {-1.54}$ respectively, with an uncertainty of $\ce {0.02}$. This tells us that acetamide is actually more basic than benzamide. In the discussion of their results, the authors considered the factors which they thought underlied this basicity difference. These included the following:

Effect 1: inductive withdrawal due to the more electronegative $\ce {sp^2}$ hybridised carbon atoms of the aromatic ring, decreasing basicity of benzamide
Effect 2: resonance stabilisation of the positive charge in the conjugate acid of the benzamide molecule, increasing the basicity of benzamide
Effect 3: more negative entropy change of ionisation for benzamide, decreasing basicity of benzamide

The authors suggested that perhaps, effects 1 and 3 were more significant combined than effect 2, resulting in the observed difference. Effect 3 was not well-explained by the authors. I suppose effect 3 arises due to the difference in size of the methyl and phenyl substituents and the consequent difference in the extent of their solvation. The conjugate acid of benzamide may require more solvation and hence, a greater ordering of water molecules, giving rise to a more negative entropy change. This is my rationalisation of effect 3.

Further insights provided

In addition to determining the acidity constants of the conjugate acids, the authors also calculated the deviation from coplanarity of the the benzene ring of the various N-substituted benzamides. They observed that as the deviation from coplanarity increased due to the presence of large and bulky substituents on the $\ce {N}$ atom, the $\ce {pk_a}$ value of the N-substituted benzamides became more "acetamide-like". They then went on to say:

  This shows that whatever the nature of the effect which causes benzamide to be less basic than acetamide, the degree of coplanarity in the molecule is important
  to it.

Perhaps, there is intramolecular hydrogen bonding?

I believe that the effect at play here that was not mentioned by Cox et al. may be the presence of intramolecular hydrogen bonding in the benzamide molecule. To highlight the significance of this effect, I will now discuss the computational findings of Muthuraja et al. (2017). The following image (the grahical abstract of their study) shows the subject of the computation study and it also shows the computed Mulliken charges on the different atoms. Notice that most of the aromatic $\ce {H}$s have much more signficant positive charge (i.e. in excess of $\ce {+0.1}$ compared to the other protons not bound to the aromatic ring. Additionally, note the large negative charge value on the carbonyl oxygens.

The authors posit high possibility of intramolecular hydrogen bonding between the carbonyl oxygen of the lactam with the neighbouring aromatic proton. They cite the following evidence justifying their claim:

Firstly, the $\ce {O...H}$ distance ($\ce {2.18}$ angstroms) is shorter than their van der Waal distance ($\ce {2.72}$ angstroms)
Secondly, the carbonyl stretching frequency is red-shifted from the normal stretching frequency

From this computation study, I simply wish to point out that there is a significant chance that intramolecular hydrogen bonding may be present in benzamide due to the proximity of the carbonyl oxygen to the aromatic proton. This interaction decreases the availability of the lone pair on the carbonyl oxygen for donation to acids. Thus, benzamide is less basic. Furthermore, this agrees with the additional insight provided by Cox et al. (1980): As deviation from coplanarity increases, there is less and less intramolecular $\ce {H}$ bonding and thus, greater and greater availability of the lone pairs on the $\ce {O}$ for donation to acids.

References

Cox, R. A.; Druet, L. M.; Klausner, A. E.; Modro, T. A.; Wan, P.; Yates, K. Protonation acidity constants for some benzamides, acetamides, and lactams. Can. J. Chem 1980, 59, 1568-1573.
Muthuraja, P.; Shanmugavadivua, T.; Joselin  Beaulab, T.; Bena Jothy, V.; Dhandapani, M. Influence of intramolecular hydrogen bonding interaction on the molecular properties of N-p-tolyl-5-oxo pyrrolidine-3-carboxylic acid: A theoretical and experimental study. Chem. Phys. Lett. 2017, 691, 114-121.


Why is the pressure correction added in the Van der Waals equation?

You may consider the vdW equation as the deviation from the ideal gas state equation $pV_\mathrm{m} = RT$.

If you look at the vdW equation

$$\left(P+\frac a{V_m^2}\right)(V_m-b)=R T$$

you can easily see why the corrections have  their signs.

The pressure term correction has the positive sign,
as the real gas has  due molecule cohesion  smaller volume than  an ideal gas at the same pressure. Like if it was under bigger pressure than it really was.

Increasing $p$ term by a positive correction leads to lower predicted volume than for ideal gas, at the same pressure, because of cohesive force.

$$V_m=\frac {R T}{\left(P+\frac a{V_m^2}\right)}$$

The correction simulates this bigger pressure the ideal gas would need to get the same vol.

The volume term correction has the negative sign,
as the gas has at the given volume
higher pressure due molecule own volume.

Decreasing $V$ term by a negative correction leads to higher predicted pressure than for ideal gas, at the same pressure, because of own molecule volume.

$$P=\frac {R T}{(V_m-b)}$$


Can copper sulfate react with terpenes (such as limonene and terpinenes) at room temperature?

For the record, after M. Farook's agreement I am making my comment an answer.
Most of the evidence you present, together with solubility data from Wikipedia, seems consistent with just precipitating the copper sulfate. Note that because of spectrochemical effects, we would expect the complex to change color if organic ligands were to displace water.


A simple question about the direction of a chemical reaction

Before I talk about the chemistry, I will use an analogy from physics. If I drop a piece of chalk, it falls down. It does not at some point later "fall up" again. However, the atoms inside the piece of chalk do go up and down a bit because of thermal vibration. The chalk does not start jumping up because these vibrations are random rather than synchronized.

  Why does not the reaction spontaneously reverse once again forming pure Zn and H2SO4 molecules?

If you are asking why at no time the mass of elemental Zn (metal) increases again, it is because there is a strong driving force toward turning the reactants into products for this specific reaction. If you are asking about individual particles, it turns out every now and then, the reverse reaction does occur (this is true for all reactions and is called micro-reversibility).

For some reactions, this back and forth results in the reaction not going to completion. Instead, these reactions may reach an equilibrium where reactant and products are present, and the forward and reverse reactions go on at the same speed so nothing seems to change (this state is called a dynamic equilibrium).

In the reaction you are asking about the driving force forward is too large, and one of the products continuously "escapes" from the reaction mix, so the reaction goes on until one of the reactants is used up (goes to completion rather than reach equilibrium).

In any reaction, reversible or not, once the reaction reaches equilibrium or goes to completion, it does not spontaneously go in the opposite direction unless something changes (add materials, change temperature, do work).


This is a Lewis acid, not a Brønsted acid. Like boric acid $\ce{B(OH)3}$, which form an adduct $\ce{B(OH)3(H2O)}$ which then lose a proton from the water molecule of the adduct: $$\ce{B(OH)3(H2O) -> B(OH)4- + H+ }$$

Above molecule will do this at each B atom, and so lose 2 $\ce{H+}$ in total, not 4. IE will have:
$$\ce{B2(OH)4 + 2H2O -> B2(OH)6^{2-} + 2H+}$$
as the overall reaction.
Note $\ce{B(OH)3}$ acts as a Lewis acid due to vacant B p orbital, which can accept lone pair of $\ce{H2O}$ and so form an adduct. From this adduct lose proton, and hence acidic.


Suppose one of the doublets of one C=O bond moves to one of these two atoms. It will not be to the Carbon atom. The Oxygen atom is a better choice. And now this Oxygen atom would be negatively charged. As a consequence the Carbon atom is positively charged and has a vacant orbital able to accept a lone pair, for example from an $\ce{OH-}$ ion or from water. This will produce the ion $\ce{HCO3^-}$ (from $\ce{OH-}$) or $\ce{H2CO3}$ (from $\ce{H2O}$).


How many crystal forms of HCl have been identified - where to find the unit cells/lattice constants?

From the first picture depicted, I would refrain from stating there are five HCl molecules per unit cell.  One thing, while there is some choice in the definition of a unitcell (there are several definitions possible, even under the constrain "it should be the simplest and still complete representation possible"), lower multiplicities occur more often; and more frequently seen values are $Z = 1, 2, 4$.

Second remark, again to the first picture: no, it does not look so as there were five molecules per unit cell.  Here, the projection of the $(a,b)$-plane is depicted (the little green and red axis emerging the origin O bear a label).  To graphically determine the actual multiplicity or the number of molecules present in a unit cell, it is better to "walk around" the unit cell; for example by accessing either *.cif (typical for small molecule structures) or *.pdb (more typical for protein structures) in a appropriate program, like CCDC's Mercury (likely used in the first drawing), or Jmol (used for the second drawing), or Avogadro.  Each of the three programs may be deployed under Windows, Linux, or MacIntosh; and in terms of displaying structures and to perform measurements of distances and angles available for free (as free beer).

It is possible, yet not in every case necessary to go from *.pdb or *.cif via Blender to prepare a dataset for 3D print though.  The "just do it" software that comes to my mind for this purpose is Werner Kaminsky's cif2vrml (windows only).  If you have Mercury (which actually may read both *.pdb, *.cif, beside *.mol, *.mol2, *.xyz, too), there is a built-in file -> print 3D menu, too; allowing you to chose between *.vrml or (then colourless) *.stl file.  Doing so found entry in publications, where examples in the Journal of Chemical Education like DOI: 10.1021/acs.jchemed.5b00597 and DOI: 10.1021/acs.jchemed.6b00293 come pretty close to what you intend to do. (As you are not afraid of larger molecules, consider the dedicated NIH 3D exchange as a valuable source of inspiration, too.)  Both jmol as well as Avogadro (referring to 1.2.0; not to the branch of Avogadro2) equally offer VRML export, the former including a dedicated how-to here.

For Jmol (the second figure presented), there are (at least) two options to read *.cif data.  The one by the GUI typically opens the *.cif, representing both the crystallographic motif and the unit cell (the later may be switched off, if necessary / useful).  Note that the GUI in the top row differs slightly from the one accessible by right-mouse-button click inside the window.  Via the toprow gui, you may access the CLI console, too -- allowing you to access many other functions, too.  For example, if you decompressed the *.zip archive, placed your *.cif into the same directory where jmol.jar lives, typing

  load cell test.cif {1 1 1}

will load the data set, test.cif, and will fill the complete unitcell till indices {1 1 1}. For one of the data found at COD, Jmol provided me this representation:

Where to get the *.cif? Literature databases like ACS' Scifinder and Elsevier's Reaxys indicate / link to such primary literature.  Depending on the journals policy, these *.cif data may be freely available in the SI (ACS journals, for example), or not (Tetrahedron family ...).  Clearly, a dedicated database ones like CSD (organic, metalorganic; preferentially small molecules than proteins) is an option.  For the service and potential insight offered, an annual academic subscription is very well invested money (was once around US$ 300-400 per year and full time equivalent).  If you encounter their structure code in a literature reference, but not the *.cif, you may request the data free of charge here; regardless if you / your institution holds a subscription, or not.

For mineral compounds, for example ISCD or the one by AMS.  The baltic COD is much younger than the others (CSD started more than 50 years ago), so the data coverage is not yet this complete, but searchable by keyword and structure, too; yielding by the former (allowing only H and Cl as the only elements present) 5 entries for HCl.

Where to find all the *.cif of HCl? It is not really organic / metallorganic (CSD), nor really a mineral (ISCD).  Perhaps via Scifinder and Reaxys one may retrieve a more round answer.

Addition

One reference found doi:10.1038/213171a0 is based on powder diffraction studies of HCl and DCl, listing these structures known to the authors back in 1967:

It continues with further powder diffraction studies, presenting an orthorhombic model, with four molecules per unit cell:

and a projection very similar to your first figure:

From these two it may be possible to construct (manually) a *.cif file which is then readable to Mercury others.  For this nature paper, Thomson Reuter's Web of Science alone knows about 139 citations, the most recent one in 2015 10.1021/acs.jpcc.5b01653 about predicted supraconducting polymorphs of HCl and HBr in the monoclinic $C2/m$.

Furthermore, a cif file based the Italian 1933 structure report the English wikipedia refers to is accessible in the COD if searching by keyword (H and Cl elements present, excluding O and C, minimal = maximal number of elements = 2).


Why do we integrate square of wave function over N-1 electron coordinates?

$\newcommand{\el}{_\mathrm{e}}
\newcommand{\dif}{\mathrm{d}}
\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}$Consider an $n$-electron system in an arbitrary electronic state represented by the electronic wave function $\psi\el(\vec{q}_{1}, \vec{q}_{1}, \dotsc, \vec{q}_{n})$, where $\vec{q}_{i}$ stands for joint spin-spatial coordinate of $i$-th electron, $\vec{q}_{i} = \{ m_{si}, \vec{r}_{i} \}$.

Integrating the square of the absolute value of the electronic wave function over the spatial coordinates of all the electrons except for electron-one and summing up over the spin coordinates of all of them without any exception gives the probability density of finding electron-one with arbitrary spin at $\vec{r}_{1}$ while having the remaining $n - 1$ electrons with arbitrary spins at arbitrary positions,
\begin{equation*}
    \rho_{\mathrm{e-1}}(\vec{r}_{1})
    =
    \sum_{m_{s1}} \sum_{m_{s2}} \dotsi \sum_{m_{sn}} \iint \dotsi \int
    \left| ψ\el(\vec{q}_{1}, \vec{q}_{2}, \dotsc, \vec{q}_{n}) \right|^{2}
    \dif \vec{r}_{2} \dif \vec{r}_{3} \dotsc \dif \vec{r}_{n} \, .
\end{equation*}
And yes, if we "integrate" over all coordinates in this expression we'll just get an inner product of the wave function with itself,
$$
    \braket{\psi\el}{\psi\el}
    =
    \sum_{m_{s1}} \sum_{m_{s2}} \dotsi \sum_{m_{sn}} \iint \dotsi \int
    \left| ψ\el(\vec{q}_{1}, \vec{q}_{2}, \dotsc, \vec{q}_{n}) \right|^{2}
    \dif \vec{r}_{1} \dif \vec{r}_{2} \dif \vec{r}_{3} \dotsc \dif \vec{r}_{n} \, ,
$$
which is usually equals to 1 since the wave function is usually normalized to unity.

Electrons are, of course, indistinguishable, so that their labeling is arbitrary and can be changed any time.
Thus, the expression above should work equally well for calculating the probability density of finding electron-two, electron-three, etc., or, in other words, it gives the probability density of finding one particular electron, no matter which, with arbitrary spin at $\vec{r}_{1}$ while having the remaining $n - 1$ electrons with arbitrary spins at arbitrary positions.

The electron density, defined as the probability density of finding any of the $n$ electrons with arbitrary spin at $\vec{r}_{1}$, is evidently $n$ times as large as the probability of finding one particular electron,
\begin{equation} \label{eq:electron_density}
    \rho(\vec{r}_{1})
    =
    n \sum_{m_{s1}} \sum_{m_{s2}} \dotsi \sum_{m_{sn}} \iint \dotsi \int
    \left| ψ\el(\vec{q}_{1}, \vec{q}_{2}, \dotsc, \vec{q}_{n}) \right|^{2}
    \dif \vec{r}_{2} \vec{r}_{3} \dotsb \dif \vec{r}_{n} \, .
\end{equation}
Note here that, for a wave function that is normalized to unity, if we "integrate" over all coordinates in this expression we'll just get the number of electrons $n$, which is one of the well-known properties of the electron density,
\begin{equation}
    \int \rho(\vec{r}_{1}) \dif \vec{r}_{1} = n \, .
\end{equation}


Phase subscript for supercritical reagent?

The accepted if not standardised notation is Sc or sc - I've seen both.  Capital "S" followed by a lower case "c", e. g.  ScCO2 is the one I've seen most often in chemistry related papers.
EDIT - apologies, this is a prefix and not a subscript. Hopefully you get the point


A problem regarding solubility and concentration

It is supposed addition of solid $\ce{Na2CrO4}$ would cause negligible volume change, so we can afford to consider the same $[\ce{Sr^2+}]$ as the initial one.
The task is rather theoretical, supposing the time point when $\ce{[Sr^2+][CrO4^2-]}$ just reached $K_\mathrm{sp,\ce{SrCrO4}}$, while $K_\mathrm{sp,\ce{SrCrO4}}$ and $\ce{[Sr^2+]}$ are given.
The proper computation procedure is already presented in you solved task. What  else is unclear ?


Addition of Methyl Grignard and use of HClO4 in this question

Consider the resonance structures of the pyranone. MeMgCl attacks the structure shown second. The perchloric acid workup hydrolyses the enol ether to give the diketone shown bottom right. NaOH promotes the internal aldol cyclisation to give 3,5-dimethylphenol.


What is the equation for the reduction of water?

I get that this makes sense, but I have no idea how to derive it myself. Any help would be appreciated!

Good question because such questions were pursued by prominent physical chemists and Nobel laureates (Ostwald/ Nernst and Hoffmann).
The reduction of water can be tested visually and the apparatus to demonstrate it indeed named after Hoffman. Basically, it is an electrolysis of water experiment. What will see at the side connected to the negative terminal of the battery that it produces a gas, which is twice the volume as compared to the side connected to the positive terminal of the battery.
This gas which is very light, burns with a blue flame and it forms water again- so this must be hydrogen. Secondly, another interesting effect is that electrode region which is producing hydrogen becomes highly alkaline. If you put phenolphthalein indicator, the negative terminal becomes pink indicating an excess of OH(-) ions.

So three things can convince us with the following:
$\ce{2H2O + 2e- -> H2 + 2OH-}$
a) The electron rich electrode is the cathode in electrolysis cell. Water can gain electrons here.
b) The cathode becomes highly alkaline during electrolysis
c) The cathode produces hydrogen gas, which is twice in volume as compared to oxygen.
d) You can even calculate the number of electrons used to produce a given volume of hydrogen from the measured current and the time.


Why triiodide ion does not form ionic bond with cation having +2 charge?

Why triiodide ion does not form ionic bond with cation having +2 charge

It does. Specifically, quick googling for "bis-triiodide" gives out of the blue links like this: poke, poke, so the question is not 'why no bivalent triiodides', but 'why only super-large bications can form triiodides.'

The answer, I suspect, lies in charge density of the cations and their ability to form strong ionic bonds.

It is well known, that energy of ionic crystal lattice strongly increases with decrease of the size of ions involved. Triiodide ion, being quite large, is not very good for forming strong ionic lattices, forming only very weak ionic bonds. Iodide ion, on the other hand, is much better for that.

The energy of formation of triiodide anion from iodine molecul and iodide anion is very low. Thus, triiodie will break to form a crystal lattice with higher formation energy at first chance. So, for a solid triiodide to form, an ion with very low charge density is required, so the energy of crystal lattice would not benefit significantly from triiodide anione dissociation. This rules out any 'naked' cation with charge +2 or more and actually most 'naked' cations with charge +1. On the other hand, by covering dication in a shell of organic molecules strongly bound to the cation, we gain a very large cation, and a solid triiodide can form.

A very similar effect guards temperature of decomposition of solid carbonates, for example, as the decomposition involves breaking of a larger carbonate dianion into a much smaller oxide dianion.


What primary standard acid can be used to standardise NaOH solutions by titration?

Oxalic acid dihydrate and KHP are the most comfortable acids for standardization, as their content per a unit of measurement (in this case it's obviously grams, as they are solid) hardly changes. When you have liquid acids, besides that what you've already mentioned, they can absorb water and other compounds from air. As they are liquid, people usually measure amount of those compounds by volume, and volume is a variable which depends on temperature, pressure.


Looks like it's neither of the three; ortho-diethynylbenzene dianion has the highest calculated proton affinity.

From Wikipedia

according to calculations, ortho-diethynylbenzene dianion is the strongest base, with a proton affinity of $\pu{1843 kJ \cdot mol^{−1}}$.

This base claimed the title of the strongest base 2016 ish, another base may have grabbed the spot but I haven't been able to find it.

B L J Poad et al, Chem. Sci., 2016, DOI: 10.1039/c6sc01726f


The reaction to consider is the transformation of chemical energy into visible radiation (light) and a little of heat.  A glowing stick consists of at least two compartments which -- on purchase -- a supposed to be kept separate.  The snap only breaks the wall between the two so that chemicals may mix with each other to start chemoluminescence (work principle), a principle similar to a firefly in late summer, too.
To decide if the reaction is exothermic, endothermic; exergonic or endergonic you have to compare the energy of initial state (when the chemicals are separated, no emission of light) with the one after the reaction.  There may be an energetic barrier to trigger the reaction, equally known as activation energy, but in its current form, this detail is not relevant to answer the question.
Take an analogy employing potential energy you might recall from the physics lecture: imagine you ate in a restaurant on a mountain and now walk back home.  It is not relevant that some energy was required to get up from the table, it is only about the comparison between
$$\mbox{state A «being in the restaurant»} \ce{->} \mbox{state B «back in the valley» } + \mbox{energy}$$
Even the pathway, straight downhill, or passing some smaller hills to reach the destination is not important here.  The naming then is, by convention, a formalism.


Why don't minerals evaporate from seawater along with the water?

In order to get a measure of how much a compound is prone to evaporation one can check boiling points. Boiling point of water is 100 C in atmospheric pressure, for molecular NaCl, sodium sulfate and Magnesium chloride (main minerals in seawater) boiling point is above 1400 C. For ionic salts dissolved in water there is a different story since they are ionized. The ions are surrounded by water molecules in liquid phase, while if vaporized they would not have a chance of having water molecules closeby so they prefer to stay in the liquid phase. All of that being said, one can envision a certain temperature, pressure and concentration (extremely high for three) at which ions would somehow get out to the vapor phase. Since sea level temperature is quite low, mineral or salt evaporation is practically zero.

Side note on why would water evaporate at 4 C instead of 100 C:
Although it take 100 C of kinetic energy (molecule's velocity) to have "all" water molecules to escape from the intermolecular attractive forces, still "some" of them end up having the energy to escape at 4 C (sea temperature average). It is because kinetic energy has distribution over all the molecules, and you will always find some that have the necessary energy. At the same time "some" of the water molecules in the air end up not having enough kinetic energy to remain in the air, and so they end up being attracted back to the liquid. These two numbers (The molecules of water in air that condense back and the water molecules in the ocean that evaporate) are equal so we have thermodynamic equilibrium. The net number of molecules that escape the sea depends on kinetic energy distribution which is exemplified in temperature.


What are the maximum number of electrons in each shell?

Shells and orbitals are not the same. In terms of quantum numbers, electrons in different shells will have different values of principal quantum number n.

To answer your question...

In the first shell (n=1), we have:

The 1s orbital

In the second shell (n=2), we have:

The 2s orbital
The 2p orbitals

In the third shell (n=3), we have:

The 3s orbital
The 3p orbitals
The 3d orbitals

In the fourth shell (n=4), we have:

The 4s orbital
The 4p orbitals
The 4d orbitals
The 4f orbitals

So another kind of orbitals (s, p, d, f) becomes available as we go to a shell with higher n. The number in front of the letter signifies which shell the orbital(s) are in. So the 7s orbital will be in the 7th shell.

Now for the different kinds of orbitals
Each kind of orbital has a different "shape", as you can see on the picture below. You can also see that:

The s-kind has only one orbital
The p-kind has three orbitals
The d-kind has five orbitals
The f-kind has seven orbitals

Each orbital can hold two electrons. One spin-up and one spin-down. This means that the 1s, 2s, 3s, 4s, etc., can each hold two electrons because they each have only one orbital.

The 2p, 3p, 4p, etc., can each hold six electrons because they each have three orbitals, that can hold two electrons each (3*2=6).

The 3d, 4d etc., can each hold ten electrons, because they each have five orbitals, and each orbital can hold two electrons (5*2=10).

Thus, to find the number of electrons possible per shell

First, we look at the n=1 shell (the first shell). It has:

The 1s orbital

An s-orbital holds 2 electrons. Thus n=1 shell can hold two electrons.

The n=2 (second) shell has:

The 2s orbital
The 2p orbitals

s-orbitals can hold 2 electrons, the p-orbitals can hold 6 electrons. Thus, the second shell can have 8 electrons.

The n=3 (third) shell has:

The 3s orbital
The 3p orbitals
The 3d orbitals

s-orbitals can hold 2 electrons, p-orbitals can hold 6, and d-orbitals can hold 10, for a total of 18 electrons.

Therefore, the formula $2n^2$ holds! What is the difference between your two methods?

There's an important distinction between "the number of electrons possible in a shell" and "the number of valence electrons possible for a period of elements".

There's space for $18 \text{e}^-$ in the 3rd shell: $3s + 3p + 3d = 2 + 6 + 10 = 18$, however, elements in the 3rd period only have up to 8 valence electrons. This is because the $3d$-orbitals aren't filled until we get to elements from the 4th period - ie. elements from the 3rd period don't fill the 3rd shell.

The orbitals are filled so that the ones of lowest energy are filled first. The energy is roughly like this:

$$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s$$

An easy way to visualize this is like this:


Let's imagine that your transport box has a decently closing lid. It you place the herbs in the box, close the lid , stuff the box into a vacuum sealer bag, remove the air from the bag until it is shrunk around the box and finally seal it, you may consider it sealed.

This means:

According to "household standards", it is air-tight to the outside. No air from the environment will enter the bag, unless it is damaged.
There is however still air in the box. Whatever degradation, loss of flavour, etc. may happen to the herbs during transport still can, since there is oxygen.

However, if you vaccum-seal the box, or replace the air in the box with nitrogen, you will minimize (or fully prevent) degradation of the herbs.


Is it safe to condense water into a liquid nitrogen trap?

One use of a liquid nitrogen cold trap is to collect organic and water vapors that are removed from a system under vacuum.  In this case, the vapors are collected as ices (solids).  See this reference which has some useful information on Schlenk line safety and use.

Where you want to be cautious is when your collection is finished.  You do not want to keep a liquid nitrogen cold trap exposed to the atmosphere because oxygen will condense and cause an explosion hazard.  Typical procedures would include shutting of the vacuum pump, dismantling of the vacuum apparatus and allowing the trap to warm to room temperature (in a fume hood if organic vapors were collected).

Some additional references might be helpful, including this one that not only describes proper Schlenk line technique but also provides some good references; and you can see how liquid nitrogen traps are used for several industrial purposes here and here.