What is the molecular structure of xenon hexafluoride?

TL;DR Xenon hexafluoride has a fluxional structure in the gas phase, with multiple rapidly interconverting conformers. The three most important conformers have $C_\mathrm{3v}$, $O_\mathrm{h}$, and $C_\mathrm{2v}$ symmetries. The minimum energy conformer is probably $C_\mathrm{3v}$.

Experimental evidence
The structure of xenon hexafluoride ($\ce{XeF6}$) has always been controversial; it is a famously strong oxidising agent, so experimental studies of it have been hampered by difficulties in the isolation and storage of pure samples.
Early studies indicated that the structure of $\ce{XeF6}$ showed deviations from octahedral symmetry. Gas-phase electron diffraction (dating from 1968) showed that the structure was approximately, but not exactly, octahedral: the experimental diffractogram differed from the theoretical diffractogram calculated for a hypothetical octahedral molecule.1 The authors considered a wide range of possible geometrical distortions, and found that $C_\mathrm{2v}$ or $C_\mathrm{3v}$ structures were relatively consistent with the spectrum obtained, with the latter being the most probable configuration.2 These could be interpreted in terms of a stereochemically active lone pair:

If we start with an undeformed octahedron we can characterize the deformation to the equilibrium structure in terms of the site of the repulsive lone pair [...] Symmetry considerations would seem to dictate that the lone pair be centered on the corner, face, or edge in the equilibrium structure. [...] If the lone pair protrudes through the center of a face, a $C_\mathrm{3v}$ structure results [...]  If the lone pair points toward an edge a $C_\mathrm{2v}$ configuration is induced.

This approach has stood the test of time, and is reproduced in multiple recent inorganic chemistry textbooks.3–5
At roughly the same time, experiments with molecular beams of $\ce{XeF6}$ were carried out by Falconer et al. These beams were not deflected in an inhomogeneous electric field, and the authors derived an upper bound of $\pu{0.03 D}$ for the electric dipole moment of a rigid structure.6
This was occasionally presented as evidence for an undistorted structure, but the calculations did not take into account the possibility of a highly fluxional process, and later Bernstein and Pitzer showed that depending on the barrier height, dipole moments up to $\pu{0.6 D}$ could be plausible.7 On top of that, the distorted conformers were not expected to have large dipole moments, as the dipole moment of the lone pair would oppose the net dipole moments of the $\ce{Xe-F}$ bonds.
Vibrational spectroscopy offers a unique possibility to eliminate, or at least reduce, the influence of dynamic processes, since the timescales involved are extremely short. In 1972, Claassen et al. carried out infrared (IR) and Raman analysis of $\ce{XeF6}$ at various temperatures.8 A group theoretical analysis leads to the prediction that an octahedral molecule should possess two IR-active bands, one $\mathrm{t_{1u}}$ stretch and one $\mathrm{t_{1u}}$ bend. However, the IR spectra at both room temperature and $\pu{4 K}$ (in an argon matrix) show at least three stretches alone. Similar inconsistencies with an octahedral structure were observed in the Raman spectra. However, the extremely broad peaks precluded a conclusive assignment to a specific distortion.
Based on all the experimental data, Pitzer and Bernstein concluded that $\ce{XeF6}$ could be described by a pseudorotational model with the potential minimum most likely being a $C_\mathrm{3v}$ structure.9 Following in the footsteps of Bartell and Gavin,2 they attributed this deformation to a second-order Jahn–Teller effect, which we will now turn to.

Theories of bonding
Clearly, traditional VSEPR theory cannot satisfactorily explain the structure of $\ce{XeF6}$. Often, the structure is described as having a stereochemically active but fluxional lone pair, which is generally accurate; however, VSEPR cannot be used to predict the region of space which the lone pair occupies.
We could reject the 2c-2e bonds utilised in VSEPR theory, and move instead to 3c-4e bonds, which works for the lower fluorides $\ce{XeF2}$ and $\ce{XeF4}$, and is briefly explained here. Essentially, each $\ce{F–Xe–F}$ interaction is modeled as a linear 3c-4e interaction. However, this model predicts a regular octahedral structure, which is again incorrect.
As far as simple bonding theories go, the best explanation is given by qualitative molecular orbital theory. A MO scheme for octahedral $\ce{XeF6}$ can be constructed from Xe 5s, Xe 5p, and F 2p orbitals (neglecting π-type interactions). It is similar to the one for $\ce{SF6}$ (see here), except that there are two more electrons:

This opens up the possibility of a second-order Jahn–Teller distortion, and I have described the mathematical basis of this elsewhere. By distortion to a lower symmetry, the HOMO and LUMO are allowed to mix, which leads to a stabilising interaction. The distortion should be mediated by a vibrational mode with symmetry $\mathrm{A_{1g} \otimes T_{1u} = T_{1u}}$, which does indeed lead to a $C_\mathrm{3v}$ structure. The Jahn–Teller effect, as well as the application to the case of $\ce{XeF6}$, is described in the textbook by Albright et al.10 I have reproduced one diagram from the book (p 364):

Equation 14.4 shows the HOMO-LUMO mixing upon lowering of symmetry; equation 14.5 depicts the particular linear combination of the traditional $2\mathrm{t_{1u}}$ orbitals that transforms as $\mathrm{a_1}$ in the lowered symmetry; and equation 14.6 shows the interaction of the $2\mathrm{a_{1g}}$ orbital with this linear combination.

Computational studies
Computational studies of $\ce{XeF6}$ have generally been plagued by a strong dependency on the level of theory and on the basis sets used, due to the very small energy differences between the relevant conformers. There is still a lack of agreement on the exact details of the potential energy surface.
Kutzelnigg et al. provided the first thorough in silico investigation of xenon hexafluoride, along with a number of valence-isoelectronic species.11 They concluded that the balance between an undistorted $O_\mathrm{h}$ structure and a distorted $C_\mathrm{3v}$ structure is governed by competing effects. Distortions are found to be present when nonrelativistic Hartree–Fock theory is used, and the addition of polarisation functions (e.g., Xe 4f) to the basis set favours the $C_\mathrm{3v}$ structure. However, relativistic effects as well as electron correlation both favour the $O_\mathrm{h}$ structure. In the case of $\ce{XeF6}$, these lead to very small differences between the two structures.12
Very sophisticated coupled-cluster calculations (CCSD(T)-F12b) were carried out by Peterson et al., and led to the conclusion that the $C_\mathrm{3v}$ structure was the ground state.13 The eight equivalent $C_\mathrm{3v}$ structures (an octahedron has eight faces) underwent interconversion via $C_\mathrm{2v}$ transition states $\pu{1.08 kcal/mol}$ higher in energy. The perfect $O_\mathrm{h}$ structure was found to be a saddle point $\pu{1.80 kcal/mol}$ above the ground state, with a threefold degenerate imaginary vibrational frequency.
However, the accuracy of these calculations was called into question by Cheng et al., who found that the $O_\mathrm{h}$ structure was in fact a local minimum (i.e. no imaginary frequencies).14 According to the authors, the imaginary frequency found by Peterson et al. arises due to basis set effects in the CCSD(T) triples correction.
All in all, computations broadly support the fact that $\ce{XeF6}$ is dynamically distorted from $O_\mathrm{h}$ symmetry. However, the exact details have yet to be ironed out, and given that the energies involved are so close to each other, "an unbalanced calculation may fortuitously give the right answer" (in Kutzelnigg's words11).

Notes and references
1. Gavin, R. M., Jr.; Bartell, L. S. Molecular Structure of XeF6. I. Analysis of Electron‐Diffraction Intensities. J. Chem. Phys. 1968, 48 (6), 2460–2465. DOI: 10.1063/1.1669470.
2. Bartell, L. S.; Gavin, R. M., Jr. Molecular Structure of XeF6. II. Internal Motion and Mean Geometry Deduced by Electron Diffraction. J. Chem. Phys. 1968, 48 (6), 2466–2483. DOI: 10.1063/1.1669471.
3. Housecroft, C. E.; Sharpe, A. G. Inorganic Chemistry, 4th ed.;  Pearson Education, Inc.: Upper Saddle River, NJ, 2012.
4. Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed; Oxford University Press: Oxford, U.K., 2014.
5. Miessler, G. L.; Fischer, P. J.; Tarr, D. A. Inorganic Chemistry, 5th ed.;  Pearson Education, Inc.: Upper Saddle River, NJ, 2014.
6. Falconer, W. E.; Büchler, A.; Stauffer, J. L.; Klemperer, W. Molecular Structure of XeF6 and IF7. J. Chem. Phys. 1968, 48 (1), 312–318. DOI: 10.1063/1.1667921.
7. Bernstein, L. S.; Pitzer, K. S. Electric field deflection of molecules with large amplitude motions. J. Chem. Phys. 1975, 62 (7), 2530–2534. DOI: 10.1063/1.430833.
8. Claassen, H. H.; Goodman, G. L.; Kim, H. Spectral Observations on Molecular XeF6: Raman Scattering and Infrared, Visible and Ultraviolet Absorption in the Vapor and in Matrix Isolation. J. Chem. Phys. 1972, 56 (10), 5042–5053. DOI: 10.1063/1.1676986.
9. Pitzer, K. S.; Bernstein, L. S. Molecular Structure of XeF6. J. Chem. Phys. 1975, 63 (9), 3849–3856. DOI: 10.1063/1.431880.
10. Albright, T. A.; Burdett, J. K.; Whangbo, M.-H. Orbital Interactions in Chemistry, 2nd ed.; John Wiley & Sons, Inc.: Hoboken, NJ, 2013.
11. Kaupp, M.; van Wüllen, Ch.; Franke, R.; Schmitz, F.; Kutzelnigg, W. The Structure of XeF6 and of Compounds Isoelectronic with It. A Challenge to Computational Chemistry and to the Qualitative Theory of the Chemical Bond. J. Am. Chem. Soc. 1996, 118 (47), 11939–11950. DOI: 10.1021/ja9621556.
12. Interestingly, $\ce{RnF6}$ is predicted to be octahedral even at the SCF level, i.e. ignoring electron correlation. This is a result of the strong relativistic effects. On the other hand, $\ce{ClF6-}$ is octahedral simply because of electron correlation effects; the energy gained by distortion is insufficient. Together, these clearly show that even the second-order Jahn–Teller approach described earlier is incomplete, as it fails to explain other species valence isoelectronic with $\ce{XeF6}$.
13. Peterson, K. A.; Dixon, D. A.; Stoll, H. The Use of Explicitly Correlated Methods on XeF6 Predicts a C3v Minimum with a Sterically Active, Free Valence Electron Pair on Xe. J. Phys. Chem. A 2012, 116 (39), 9777–9782. DOI: 10.1021/jp3084259.
14. Cheng, L.; Gauss, J.; Stanton, J. F. Relativistic coupled-cluster calculations on XeF6: Delicate interplay between electron-correlation and basis-set effects. J. Chem. Phys. 2015, 142 (22), 224309. DOI: 10.1063/1.4922112.


How is it thermodynamically possible to obtain energy from a piece of coal?

Coal in open air can and does sponstaneously combust.

See for example Spontaneous Heating and Fire in Coal Mines and Coal’s Spontaneous Combustion Problem.

At low temperature the change in Gibbs free energy of coal oxidation is still strongly negative, just like at high temperature.

But at low temperature, the reaction rate is low.

Still, if the rate at which heat is dissipated is less that the rate at which heat is produced, temperature increases and the reaction rate increases until there is appreciable combusiton.


Determining ratio of competing fragmentation processes in mass spectrometry

Your three photons represent the total energy added to the system:

$$E_\mathrm{photon} = 3 \cdot \frac{hc}{\lambda} = \pu{13.98 eV}$$

That, less the sum of ionization energy and kinetic energy of the removed electron, leaves the change of the internal energy of the cation:

$$E_\mathrm{int} = \pu{13.98 eV} - \pu{9.10 eV} - \pu{1.4 eV} = \pu{3.48 eV}$$

Finding the value of $\log k$ for both curves at this energy will give the relative rates of simple cleavage vs rearrangement.


Why "RCA clean" solution has to be prepared every time (and not stored long-term)?

Hydrogen peroxide is not a very stable substance. Over time, and however you store it or however carefully you prepare it, it will decay. So solutions of a given concentration need to be prepared fresh when you want to use them. You could try to store them but you won't know whether the concentration is still suitable for whatever you are going to do with it at some later time. And, the decay may well generate gas and other side-products of oxidation what may interfere with the use and may be unsafe.

Hence you would make the mixture fresh every time when you want to use it.


Lithium carbonate has poor water solubility.  Cesium triphenylcyanoborate is insoluble in water.


Do ion-exchange resins soften water through a kinetic equilibrium process?

(1) Water softeners use an ion exchange material that is reversible, so that after it has been used a while and the ion exchanger becomes loaded with calcium ions, it can be regenerated to the sodium form by flushing with salt (sodium chloride) solution. This would suggest that the affinity of sodium and calcium ions for the resin do not differ that greatly. I would doubt that either kinetic or steric factors count much here either, although a divalent ion like calcium might bind more tightly by being held electrostatically by more than one anionic group on the resin.

(2) I think the primary reason water softeners work is that they typically use the ion exchange resin in the form of a vertical column: you put the hard water in the top and soft water exits at the bottom. When hard water is introduced at the top of the column, a local equilibrium is established between ions in solution and ions bound to the resin, with a somewhat lower concentration of calcium ions and a corresponding higher concentration (at a 2:1 ratio) of sodium ions. As the water moves down the column, this equilibrium is reestablished again, but this time, the concentration of calcium is lower still. This repeats a large number of times, until all the calcium ions are replaced with sodium ions. I have heard this effect referred to as "mass action".


Electrolysis of water for demonstration: ideal electrolyte for safety, stability, conductivity, and ease of maintenance

I suggest $1\%$ w/w $\ce{NaOH}$. Point-by-point:

$1\%$ w/w is $\sim\!10~\mathrm{g\over L}$, which for $\ce{NaOH}$ at $40~\mathrm{g\over mol}$ is $\sim\!0.25~\mathrm{M}$, corresponding to a $\mathrm{pH}$ of $13.4$. Standard personal protective equipment (coats, gloves, goggles) should suffice for observers and maintainers.
$\ce{NaOH}$ is extraordinarily soluble, so you will have no difficulties with precipitation.  You will want to keep the system covered when not in use, if you're not already, to minimize evaporation.  You may get some riming on the interior, but rinsing with DI water should take care of it.
According to Schweitzer$^\dagger$, the materials in your system have the following resistances to $10\%$ (not $1\%$) $\ce{NaOH}$:

Plexiglas (acrylic): Resistant, to $80~^\circ\mathrm F$
Brass: Good (less than 0.02" penetration per year), to $200~^\circ\mathrm F$
Silicone elastomer: Resistant, to $80~^\circ\mathrm F$
ABS: Resistant, to $140~^\circ\mathrm F$
Platinum: ..... darn well better hold up! (Though, if there are any gaps in the coating, the substrate might be exposed to attack.)

Per the first chart here (PDF link), the conductivity of $1\%~\ce{NaOH}$ should be about $50~\mathrm{mS\over cm}$, which is over ten times the $\sim\! 4.5~\mathrm{mS\over cm}$ of your current $4\%$ citric acid recipe.
I salute any common microorganisms able to grow in a $\mathrm{pH}~13+$ solution. (See, e.g., Table 3-5 here, where the maximum survivable $\mathrm{pH}$ for any species listed is $11$.)

Obviously, if you should switch, monitor the health of the system materials to be sure the above literature chemical resistance data actually holds in reality.  There's always the chance some small component of the system is of a material you haven't listed, too.

(To note, I re-thought the vinegar-plus-sodium-bisulfate approach after discovering that both brass and acrylic/Plexiglas are poorly resistant to vinegar, and that the bisulfate is not likely to provide as much of a conductivity boost as I was thinking.)

$^\dagger$ P.A. Schweitzer. "Corrosion Resistance Tables, Part C, P-Z." 4th Ed. New York: Marcel Dekker, 1995, pp. 2685-2688. (Amazon link)


Initially one would assume that cumene (isopropyl benzene) would be the major product. However, kinetic control does have a say in this reaction, according to Gilman and Means.1

At - 2°, n-propyl bromide with benzene and aluminum chloride gave n-propylbenzene, identified as its sulfonamide. Genvresse (6) obtained both n-propylbenzene and isopropylbenzene by conducting the reaction at reflux temperature. Konowalow (7) found that below $0^\circ$, n-propyl chloride gave n-propylbenzene, while from $0^\circ$ to reflux temperature it gave mixtures of n-propylbenzene and isopropylbenzene. More recently it has been shown that at $-6^\circ$, 60% of the monopropylbenzene was n-propylbenzene, and 40% was isopropylbenzene. At $35^\circ$, this ratio was reversed.

a low temperature would facilitate the production of larger amounts of n-propylbenzene due to kinetic control. However, this is still not a very efficient method of preparing n-propylbenzene, and acylation followed by Clemmensen/Wolff-Kishner reduction would be more suitable here.
References:

Gilman, Henry, R. N. Meals, “Rearrangements in the Friedel-Crafts Alkylation of Benzene” Journal of Organic Chemistry 1943, 08(2), 126–146. doi:10.1021/jo01190a003.


I am assuming you are in grade school.
The problem is, as was mentioned before, that photosynthesis is very complicated.
The formula you know of is extremely simplistic and skips over 100 reactions on the way. you can find plenty of information in Wikipedia, or if that is too much you can use the Simple Wikipedia.

In high school we took some kind of water plant or algae and put it in an upside down test tube filled with water, as oxygen was formed it filled the end of the test tube. This is a fairly simple way to monitor photosynthesis and you can use it to study photosynthesis. for example: how does the temperature effect photosynthesis, how does the color of light affect it etc.


Assuming that by air bubble you're referring to the pocket of air now sealed in the bottle above the water surface:
Immediately after sealing the bottle, the air pocket is not at all compressed: it's at the same pressure as the surrounding environment (the air outside the bottle).

However, because you have a closed container, some water will evaporate, until an equilibrium is reached between the water vapor in the air pocket and the water in the liquid phase. This occurs when the chemical potential of the water in the two phases are equal.

Due to that vaporization, the total pressure in the air pocket (which now contains a higher amount of water vapor than the surrounding atmosphere outside the bottle) will be larger than the surroundings. This is a prominent effect when it's hot outside (have you ever left a sealed water bottle in a hot car?).

While this effect is usually small, as gases mix very well, it is non-zero and so yes, the air pocket is slightly compressed due to presence of water vapor.

As for whether the air pocket can mix with water: Assuming our container can tolerate such pressures and we were to externally compress it: liquids are much less compressible than gases, and so the air would (given high enough pressure) condensate, mixing into the liquid phase (but you wouldn't achieve this by hand!)

To address your final point: let's say we invert the bottle, so the air pocket travels as a bubble through the liquid phase to re-emerge at the other end of the bottle. One could ask, "why doesn't the air bubble fragment into little bubbles as it progresses upwards?"

To be fair, you do see a few bubbles splinter off, but for the most part, the bubble is intact. The air bubble stays largely intact due to the viscosity of water (the intermolecular attractions between water molecules is larger than the entropic benefits of mixing with the air bubble). So, to answer your question, the fact that the air bubble doesn't mix is not due to its intermolecular attraction, but rather it's due to water's intermolecular attraction.


Equilibrium Concentrations of Products/Reactants

Adding a reactant increases the concentration of products at equilibrium, but it doesn't necessarily mean that the product concentrations will be larger than the reactant concentrations at equilibrium.

Suppose you have a reaction $\ce{A <=> B}$. Then
$$K = \frac{[\ce{B}]}{[\ce{A}]}$$
where $[\ce{A}]$ and $[\ce{B}]$ are the concentrations at equilibrium and $K$ is a constant. The ratio $[\ce{B}]/[\ce{A}]$ is constant at equilibrium, so if you add some $\ce{A}$ to the system, some $\ce{A}$ must react and become $\ce{B}$ to maintain the ratio. The equilibrium concentration of the products will increase.

But that increase in product concentration won't necessarily mean that the equilibrium product concentration becomes higher than that for the reactants. If $K$ is bigger than one, the equilibrium concentration of the products will be greater than that for the reactants; if it's less than one, the equilibrium concentration of the reactants is bigger.


Why do ionic substances with higher lattice energies tend to be less soluble in water than substances with lower lattice energies?

When an ionic compound dissolves in water, the water molecules surround the individual ions. Since water is very polar, when water surrounds an ion there is a decrease in overall energy. However, to pull the ions apart in the first place, there has to be an energy increase to overcome the lattice energy. If the lattice energy (cost) is larger than the energy of solvation of the ions (gain) the compound won't dissociate, and it will be insoluble.

There is a large grey area where lattice energy and solvation energy are close, and for those compounds temperature has a big influence on solubility.


Synthesis of 3-Methylenecyclopentene from 1-methylCyclobutene

Here is my try.

Ozonolysis
Aldol condensation using a hindered base (like LDA)
Wittig reagent to convert $\ce{C=O}$ to $\ce{C=C}$

Step 1:

Step 2:

I used $\ce{LDA}$ which abstracts a proton from the least hindered carbon near to a carbonyl group. (So that the resulting carbanion is stable).
I am not drawing the enolate ion formation etc. You can find the reaction mechanism for aldol condensation here and the reaction using $\ce{LDA}$ here.

After protonation, you can heat the compound to remove $\ce{OH-}$ which is a simple dehydation. There are two possible alkenes. One of them (the one given below) is more stable because of conjugation.

Step 3:

We are almost done. To replace $\ce{C=O}$ with $\ce{C=C}$, use wittig reagent.


How can compounds like XeO4 form tetrahedral structures despite having double bonds?

$\ce{XeO4}$ has a tetrahedral structure as you noted, but $d$-orbitals are not involved in explaining its geometry or hybridization. The central xenon atom in $\ce{XeO4}$ is best described as simply $\ce{sp^3}$ hybridized.  We have 8 xenon electrons so we need 4 orbitals to put them in.  The 4 equivalent orbitals are formed by mixing the $s$- and $p$-orbitals (just as we do with carbon) to form the equivalent $\ce{sp^3}$ orbitals pointing towards the vertices of a tetrahedron (which minimizes electron-electron repulsions between the 4 bonds).

The molecule can be represented by several resonance structures and I've drawn two of them (A and B) below.  I've only drawn the lone pairs on one of the oxygen atoms for each resonance structure, but they are the same for all oxygens in each resonance structure.  So the resonance structures tell us that the xenon-oxygen bonds in $\ce{XeO4}$ are some mix of single and double bond character.  In resonance structure A, the xenon and oxygens are neutral; while in resonance structure B, the xenon has a charge of +4 and each oxygen has a charge of -1.

The $\ce{Xe-O}$ bonds in $\ce{XeO4}$ are very weak. In resonance structure A the pi overlap is very poor due to the different atomic sizes of xenon and oxygen. In resonance structure B we just have a simple single bond connecting the oxygen and xenon and it contains 2 electrons, both from xenon and shared by oxygen, oxygen does not contribute any electrons to this bond. The single bond resonance structure B probably contributes more to the description of this molecule than resonance structure A. The bonds are so weak that $\ce{XeO4}$ is unstable and explosively decomposes around -36 C.

In summary we can say that

the xenon in $\ce{XeO4}$ is $\ce{sp^3}$ hybridized
the $\ce{Xe-O}$ bond is a partial single-double bond with the single bond probably being a more accurate description, and
the $\ce{Xe-O}$ bonds are extremely weak because of the poor pi overlap in resonance contributor A; and oxygen does not contribute any electrons to the xenon-oxygen bond in resonance contributor B.


Which ceramic-like material is used in halogen MR16, GU5.3 and so on bulbs?

Actually, aside from heat resistance, a primary qualification is that the adhesive cement thermal expansion coefficient match that of the quartz (or glass or alumina) envelope over the temperature range of storage and operation, to avoid cracking the envelope (the metal fixture is more flexible and "forgiving").
There are a number of cements in use, such as, "aluminum phosphate refractory cement... a mixture of aluminum phosphate... phosphoric acid [and] alumina". Glassbond sells a variety of adhesives, such as a phenolic/melamine adhesive for glass lamps with operating temperature below 300°C. Sauereisen has a long history making high-temperature adhesives, including those of silicate and of magnesium oxysulfate composition. One could try making these cements at home or in the lab, though they're readily available commercially. You could experiment with refractory furnace cement, though it likely would have a different coefficient of expansion, cracking the envelope.
BTW, I remember, in school, attaching thermocouples to incandescent lamps with a variety of cements... and watching the glass envelope form small star-shaped cracks at each joint, for many of those adhesives, as the lamp warmed. Sorry, but I don't remember which formulations actually succeeded past a few on/off cycles.


What is the net ionic equation of sodium hydroxide when it dissolves in water?

When you write down the net ionic equation, you have to consider that the educt and the products are in different phases, a solid and a liquid phase. Undissolved $\ce{NaOH}$ is a solid, denoted by the index $\ce{(s)}$, while the dissolved $\ce{Na+}$ and $\ce{OH-}$ ions are solvated in the aqueous phase (index $\ce{(aq)}$ in the equation):

$$\ce{NaOH(s) ~->~ Na+(aq) + OH^{-}(aq)}$$


Finding the amount of nitrous oxide produced in the reaction between lead sulfide and nitric acid

The first step is to calculate the number of moles of $\ce{PbS}$:
$$n(\ce{PbS}) = \frac{\pu{4.7 g}}{\pu{239 g mol-1}} = \pu{0.0197 mol}$$
The next step is to calculate how many moles of $\ce{NO}$ will be produced based on the stoichiometry of the reaction. As stated in a comment, it looks like this is the step that was left out of your calculations. Because only $\pu{2 mol}$ of $\ce{NO}$ are produced for every $\pu{3 mol}$ of $\ce{PbS}$, only $2/3$ of $\pu{0.0197 mol}$, or $\pu{0.0131 mol}$ $\ce{NO}$ will be formed.
At this point we know all of the variables required to invoke the ideal gas law:
$$
\begin{align}
V(\ce{NO}) &= \frac{nRT}{p}\\
 &= \frac{\pu{0.0131 mol} \times \pu{0.0821 L atm mol-1 K-1} \times \pu{300K}}{\pu{1.10 atm}}\\
 &= \pu{0.29 L}
\end{align}
$$
And this gives us the correct answer of a. Again, as your answer was off by a factor of $2/3$, it is pretty clear that the second step above regarding the stoichiometry of the reaction is the only step that you left out of your calculations.


The taste of potassium chloride is not close enough to sodium chloride so the salt substitutes usually contain a mix of potassium and sodium chloride.  This is healthier than pure sodium chloride.


Determine molecular weight of biomass based on ultimate analysis

As a general rule, it is not a good idea to treat complex materials like wood as something you can calculate a mole of. You can of course analyze the material and get elemental rates, but that does not mean they are bonded in such way that this rate forms a minimum fraction which represents the whole thing.

When you calculate the number of moles of each atom in 100g, what you have is an estimation of the expected quantities in that sample. Undoubtedly, if you took different samples of the wood you would find different values, specially if sampling from different parts of the material.

As an exaggerated example, think about the elemental constitution of the human body, which is mainly 65% Oxygen, 18% Carbon, 10% Hydrogen and 3% Nitrogen in mass (source). With that in mind you can calculate a mole ratio of these elements and create a "representative formula". Obviously, though, there's no such thing as the human molecule, and you can't say that someone's body is X or Y moles of human.

Yet, like I originally answered, if you define what the representative formula is, you can calculate it's molar mass.

$$\ce {M}=4.0841*12.0107 + 5.77*1.00794+2.81*15.9994+0.0086*14.0067 = 99.947 \ \ce{g/mol}$$

Knowing ${\ce M}$, the number of moles is ${\ce 700,000/99.947 \approx 7004}$, and the concentration of biomass is $7004\ \ce{mol/m^3}$


What's special about the purine scaffold?

It is important to remember that the purine scaffold is one of the most fundamental organic structures in all lifeforms. From an evolutionary standpoint, this means that there must have been a simple, effective way for this scaffold to be prepared in the pre-biotic earth. After all, the first organisms to contain purine scaffolds did not have "access" to complex enzymes, proteins, etc.
Take a look at the molecular formula for adenine, a simple purine derivative. C5H5N5. We could rewrite this as (HCN)5. As the molecular formula alone might lead you to conclude, adenine can be made from only one chemical: hydrogen cyanide. Interestingly, hydrogen cyanide was a very abundant chemical in the earth's primordial atmosphere. Roy et al.1 replicated the atmospheric conditions of the primordial earth and found that substantial quantities of adenine could be produced! The mechanism for this reaction is quite beautiful, and I will include Peter Jacobi's rendition of it below. The mechanism for the formation of adenine also explains why other constitutional isomers are not produced. We must consider the alternating nature involved in these nucleophilic-electrophilic interactions (e.g. two nitrogen atoms will never "decide" to couple).

So, to conclude, the purine substructure is "special" because it not only has the unique ability to hydrogen bond and form a base pair for DNA replication but was also incredibly easy to synthesize in the primordial earth's atmosphere. Although life forms do not use HCN to synthesize the purine substructure anymore, its synthesis is still incredibly simple.
1 Roy, D.; Najafian, K.; Schleyer, P. V. R. Proc. Natl. Acad. Sci USA 2007, 104, 17272-17277.


Why does menthol (e.g. peppermint) feel cool to the tongue?

Menthol it self gives a cold feeling in the mouth because it is active at the same receptor (an ion channel) on the tongue that cold temperature triggers. Interestingly, although they act at the same receptor, they act at different sites, so that provides the intensified response when eating a mint and then drinking water. This reference gives an excellent detailed answer, with references to the original papers, which I'll  summarize here.

Menthol acts at the TRPM8 protein which forms an ion channel that allows $\ce{Na+}$ and $\ce{Ca^2+}$ ions to flow into cells and this sends a signal saying "cool" to the brain. (As an aside, this protein monitors temperature across the body and not just on the tongue.) Cold temperatures actually change the confirmation of this protein, which allows the ions to flow more freely, and sends the signal to the brain. Menthol, on the other hand, stabilizes the open channel (allowing ions to flow even more freely) and also...

  “menthol shifts the voltage dependence of channel activation to more negative values by slowing channel deactivation”.  This is very significant to my question because it supports a claim made by the first web page I visited which stated that menthol acts on the receptors, leaving them sensitized for when the second stimulus is applied (i.e. cold water) resulting in the enhanced sensation.  This mechanism of binding is very clearly different from the mechanism of cold affecting the TRP channels.  This is why the sensation is increased when both stimuli are applied, yet is not affected after addition stimulation from the same stimuli (i.e. eating another mint).

All in all, pretty cool.


How can I free base cystamine dihydrochloride?

Maybe the MgSO4 is absorbing your product. You may want to try another drying agent Try Na2SO4, zeolites, or azeotropic distillation (e.g. with benzene). Maybe NaOH will work, but it may react with cysteamine.


How would one calculate the entropy change for an adiabatic irreversible process?

I agree with you over here.

  $\underbrace{\Delta U}_{\text{state function}} = \underbrace{\Delta W}_{\text{path function}}$

  And $W_{\text{reversible}}\neq W_{\text{irreversible}}$

But this part seems to be wrong.

  If two states $A$ and $ B$ are connected by a reversible path, they can never be connected by an irreversible path during an adiabatic

Just because the process is adiabatic irreversible it doesn't mean that the reversible path needs to be adiabatic as well.

The only requisite for calculating the entropy for an irreversible process is simply that the path taken must be reversible.

So until now we have established that the heat transfer does not have to be 0. For an adiabatic process.

Even though $W_{\text{irrev}} \neq W_{\text{rev}}$  the extra parts will be balanced by the heat transfer in the reversible process.

For example let's take adiabatic expansion.

$$W_{\text{irrev}} > W_{\text{rev}}$$

Numerically speaking in an absolutely hypothetical situation.

\begin{align}
\ce{ ∆U &-> 10~J}\\
\ce{ W_{irrev} &-> 10 ~J}\\
\ce{ W_{rev} &-> 5~J}\\
\end{align}

So the possible proposed reversible path will automatically have a transfer of heat into the gas . Whose value will be 5J.
 .


What is happening in this video of solvated electrons donated from sodium in ammonia?

I'll try an answer to this question because I watched this video a while back and did a bit of reading on it at the time and I think I understand the big picture. The problem is that these solvated electrons are very complicated things, and do not lend themselves to the traditional ways that chemists would like to think about things. For that reason, there is quite a lot of literature coming from physicists which is very complicated but nonetheless valuable. Additionally, solvated electrons show up in more than just this rather unusual metal-liquid ammonia solution. For instance, a recent paper in Nature chemistry by Sieferman et. al. [1] showed that there are transient solvated electrons at the surface of water which can have major implications for natural electron transfer reactions. So, understanding these systems is truly quite important.

The Blue Color:

In order to get at the deep blue color which is seen in this sodium-ammonia solution, we must understand what it is that is absorbing (reddish) visible light and hence leading to the transmission of a lot of blue light.

As it happens, at low concentrations of sodium (I'll address this in a bit), the solvated electrons exist in a bound state. This is not too dissimilar to ordinary electronic energy levels, except that there is not really a molecule to be found. Rather, the electron exists in a bound state which it creates for itself. That is, the electron polarizes the surrounding solvent such that it is contained in a bound state. Sometimes people refer to this electron as being a "polaron". One important difference between this and energy levels in a molecule is that electronic excitations generally take place from one bound state to another bound state. In the case of a solvated electron, however, the transitions are from a bound state to a continuum state.

Obviously one question that would be nice to answer is why is it such a deep blue solution? Well, this sort of question is always hard to address because the color of things is more or less an accident. The fact that it is blue just tells you that the electron is relatively weakly bound because it is excited by reddish light.

For more details on the bound-continuum transition, see Aulich's paper [2] which uses photo-electron emission to study solvated electrons as a function of concentration and photon energy. For a theoretical treatment of the bound states at dilute concentrations of metals, see Jortner's paper [3].

The Transition to a Bronze Color:

One reason that people have been very interested in this system is because of this transition that takes place as the concentration of metal increases. Let me first try to give some intuition as to why this transition takes place at all. First, going off of our model that the electrons polarize the solvent and exist in a low-energy, bound state, we should expect that as more electrons are added to the solution (i.e. as more sodium is added), these cavities which the electrons occupy will become closer and closer to each other. This means that electron-electron interactions become quite important. This means two things: first, we have to begin worrying about satisfying Pauli exclusion, and second, we have to be concerned that at some point the electrostatic repulsions will become larger than the binding energy of this electron.

These two points have been studied by some physicists. Alavi and Frenkel's paper [4] develops a model and then performs simulations on ideal fermions (I honestly don't know what makes a fermion ideal and don't understand this paper), demonstrates that the low-concentration bound states are very unstable as the concentration of electrons increases. Thus, some transition must take place.

The best paper on the theory of this transition can be found here [5]. The reason this transition is so interesting is that it is a "metal to non-metal" transition. Remember when you were first introduced to metals and the characteristic description is that they are a "sea of electrons"? That is, the electrons are free to move around, and this leads to high conductivity and other observable properties, one of which is that they absorb and reflect nearly all wavelengths of light, which leads to the silvery color of most metals.

Well, the exact same phenomenon is being observed here, but it's very peculiar because this is all happening in a liquid. For instance, Schroeder et. al. wrote a paper [6] which measures the conductivity of several metal-ammonia solutions as a function of concentration of the metal. I unfortunately couldn't add the relevant figure, but the conductivity basically increases linearly with concentration of the metal until it hits a plateau and does not increase any further.

This gradual increase explains what is going on when we see a separation into two phases. The explanation goes like this. There is only so much space which allows for the existence of the bound states we described above because these bound states occupy a cavity of relatively large volume in the solvent. As more metal is added, more electrons are free in the solution, but the solution is already saturated with these bound electrons. Thus, the electrostatic and exclusion effects become such that any additional electrons added can only exist in a metallic state. This is peculiar because this metallic state is in the liquid phase and is actually fairly dense (I read this somewhere but can't remember the reference... Will update later). If one continues adding electrons, they always become incorporated into the metallic state because the bound states are saturated. Eventually, enough electrons are present that the destabilizing effects due the presence of other electrons is large enough that no possible bound state can exist and the whole system becomes metallic. Hence the plateau in the conductivity plot I described above. At this point, I believe the sodium ceases to dissolve and just stays as plain old metal sodium.

I do not have a good explanation for why the metallic phase is specifically bronze colored as opposed to silvery. I suspect this is a very subtle feature and would certainly be temperature dependent. I haven't described the temperature dependence of all of this, but there quite a few papers that address only this point and it seems to be quite complicated.

I also don't know why the metallic phase forms on the top. I suspect this may just be an artefact of the experiment in that video. That is, it is entirely feasible that solid would sink to the bottom if it didn't just freeze to the side of the test tube. I could be wrong about that though.

References:

Siefermann, K. R., Liu, Y., Lugovoy, E., Link, O., Faubel, M., Buck, U., ... & Abel, B. (2010). Binding energies, lifetimes and implications of bulk and interface solvated electrons in water. Nature chemistry, 2(4), 274-279. DOI
Aulich, H., Baron, B., Delahay, P., & Lugo, R. (1973). Photoelectron emission by solvated electrons in liquid ammonia. The Journal of Chemical Physics, 58(10), 4439-4443. DOI
Jortner, J. (1959). Energy levels of bound electrons in liquid ammonia. The Journal of Chemical Physics, 30(3), 839-846. DOI
Alavi, A., & Frenkel, D. (1992). Grand‐canonical simulations of solvated ideal fermions. Evidence for phase separation. The Journal of chemical physics, 97(12), 9249-9257. DOI
Jortner, J., Cohen, M. H. (1976). Metal-nonmetal transition in metal-ammonia solution. Physical Review B, 13 (4), 1548-1568.
Schroeder, R. L., Thompson, J. C., & Oertel, P. L. (1969). Conduction in Concentrated Solutions of Several Metals in Liquid Ammonia. Physical Review, 178(1), 298. DOI
Thompson, J. C. (1968). Metal-nonmetal transition in metal-ammonia solutions. Reviews of modern physics, 40(4), 704.
Schroeder, R. L., Thompson, J. C., & Oertel, P. L. (1969). Conduction in Concentrated Solutions of Several Metals in Liquid Ammonia. Physical Review, 178(1), 298.


Compare ONO bond angle in nitrogen oxides

You are right, that's $\mathrm{NO_2^+}$, since it is linear, and whoever says otherwise is wrong.

Also, those (except for NO2) are not compounds, but particles.


Are all hydrolysis reactions non-redox reactions?

Hydrolysis refers to any compound reacting with water, including cases where the water acts as an oxidizing or reducing agent.  Magnesium silicide, for instance, reduces water leaving both hydrogen and oxygen from the water in negative oxidation states:
$$\ce{Mg2\overset{-4}{Si} + 4\overset{+1}{H}_2O -> 2Mg(OH)2 + \overset{+4}{Si}\overset{-1}{H}_4}$$


How to distinguish two molecules with same formula but different geometry with vibrational spectroscopy

The starting point is to look at the potential point groups of the two molecules and work out if there are common vibrational mode symmetries between IR and Raman spectra, assuming that you are able to obtain both. In the 3rd and 4th columns of the point group are the $x, \,y, \,z$ and $xy, \,xz$ etc operators. The $x, \,y, \,z$ operators transform as does a dipole (so IR transitions) and  act as an alias for the symmetry species.

For example in the $C_{2v}$ point group each  IR transition must be one of $A_1,\,B_1 ,\, B_2$ species and as there are 12 transitions  there must be several of the same symmetry but different frequency.

The Raman transitions correspond to product terms , $x^2,\, xy$ etc. (because Raman spectroscopy depends of the polarisability which is proportional volume and so area in cross-section ) and these have also $A_1,\,B_1 ,\, B_2$ symmetry species as well as $A_2$ species (in $C_{2v}$) so we do expect to see some different transition frequencies between IR and Raman spectra. We do not necessarily know what the symmetry species are for these transitions, just that there is a difference.

In the case of $D_{2h}$ point group as there is a centre of inversion, there are no common transitions between IR and Raman spectra. Look at the point group and you will see this.

In more detailed analysis the polarisation of Raman signals can be used to determine which transition is $A_1$ etc. Spectroscopy books give details of this.

look at Understanding group theory easily and quickly for an explanation of point groups.


Is this gas in thermodynamic equilibrium?

I don't think "equilibrium" is a good word to use in your question.  We have no way of knowing if the gas is at equilibrium from the data you present.  A highly non-ideal gas that was exploding might give the same relationship.  A gas undergoing a rapid unimolecular reaction wouldn't really be at a true chemical equilibrium but could behave as an ideal gas from a pressure / temperature standpoint.

Did you really mean to ask if the gas was ideal?  An ideal gas follows $PV = nRT$, which can be re-written in several ways:

molar density $\frac{n}{V}$ as a function of pressure $P$: $$\frac{n}{V} = \frac{1}{RT}\;P$$, which is the equation of a straight line through the origin (bottom graph).
molar volume $\frac{V}{n}$ as a function of pressure $P$:  $$\frac{V}{n} = RT \frac{1}{P}$$, which is the equation of a hyperbola (top graph).

Your data is consistent with the gas being ideal (and isothermal).


How to calculate the number of geometrical isomers in allenes?

If we have one double bond in a hydrocarbon compound we have an olefin or alkene.  Ethylene is the simplest example of this class of compounds.  The carbons in the double bond and the 4 atoms attached to them lie in the same plane.  One pair of cis-trans isomers is possible in compounds with a single double bond.

If we add a nother double bond directly on to the end of the double bond in ethylene such that an sp hybridized carbon is created in the process, then we have formed an allene.  The 4 substituents at the end of the double bonds in allene lie in planes that are oriented 90° to one another.  Look at the allene (bottom line) in the following drawing.  One set of substituents (R3, R4) are located in the plane of the screen; the other set of substituents (R1, R2) are located in a plane perpendicular to the screen.  Allenes cannot generate cis-trans isomers, but they can generate enantiomers.

Now let's add one more double bond to allene such that we now have 2 sp hybridized carbons (top line in the above drawing).  We see that, like in ethylene, the 4 atoms connected to the double bond lie in the same plane as the double bond and one pair of cis-trans isomers is possible.

Any compound with 3 or more cumulative double bonds is a member of the cumulene family. Any cumulene with an odd number of double bonds is geometrically structured like ethylene (the 4 atoms connected to the double bond lie in the same plane as the double bond) and is capable of having one pair of cis-trans isomers. Cumulenes with an even number of double bonds are structured like allene (the 4 atoms connected to the double bonds lie in perpendicular planes) and cannot display cis-trans isomerization, but can have enantiomers.


What chemical available in 18th century would explosively react with phosphorus?

I think if you just add oxygen you'll just end up with a lump of burning phosphorus. I would suggest having a timed explosive charge at the centre of a phosphorus payload to essentially achieve something like a white phosphorus bomb.

Since white phosphorus is pyrophoric, you only need to disperse it for it to ignite. To do this, you could either use an automated striking mechanism to ignite a fuse to something like black powder, or my Improvised Munitions Handbook suggests a combination sulfuric acid/potassium chlorate or perchlorate (note: not chloride) + sugar device, with the mechanism keeping the two components separate, would be suitable.

The main problem with your design specifications as I see them is having an excuse to have the external surface of phosphorus exposed to the air: white phosphorus will only glow from the air-exposed surface, so if you put it behind glass, for example, it won't glow. A fine wire mesh as your external covering might do the job, but I'm not sure exactly what effect that has on reliability or effectiveness (and may cause the device to spontaneously and accidentally catch fire, especially if not packed properly).

Near explosion + being covered in burning white phosphorus = much sadness, probable death. See: White Phosphorus - Effect on People.

(Usual disclaimer: I am not a munitions designer.)


Among Benzyl carbocation (Ph-CH2+) and Methyl methoxy carbocation (CH3-O-CH2+), why is the latter more stable?

let's see the resoance forms of benzyl carbocation and methoxy carbocation:
-Benzyl carbocation:

You are right, benzyl carbocation have multiple ressonance forms which means that It must be very stable. In fact, benzyl carbocation is one of the most stable carbocations.
-Methoxy carbocation:

In this case, we only observe 2 resonance forms.
so the question is, why methoxy carbocation is more stable despite of having less resonance forms??
Well, the key is the stability of this resoance forms...
In benzyl carbocation, none of its resoance forms have a complet octet, wich mean that the resonance forms are not very stable and therefore the positive charge is not very delocalized.
However, in the case of the methoxy carbocation, one of the resonance form have a complet octet, which means that the resoant form are pretty stable and therefore the positive charge is quite delocalized between the O and de C
In conclusion, number of resonance forms are important, but their stability is also a really important factor.


Finding equilibrium concentrations in solution from initial reagent concentrations and equilibrium constant

No need for the cubic equation. Just firstly assume that $\ce{A}$ and $\ce{B}$ reacted completely, then $\ce{[C]}=\pu{2M}$, $\ce{[D]}=\pu{1M}$, and from $K_c$ you can calculate $\ce{A}$ which is the same as $\ce{B}$.
That's the approximation usually used with this kind of problem.
The answer should be $\pu{2E-4M}$.


Batch conversion of Waters mass spectrometry file formats

Waters Databridge actually allows the conversion of multiple files at the same time, but somehow it is not very intuitive to use (not a native UI) and also does not really work in batch mode. Still, it is somewhat possible to quickly export Waters .raw files using Databridge, and it is the best solution I found so far. From there, a simple python script allows me to average the mass spectra (that are still chromatograms), so that it at least doesn't take hours to export every single mass spectrum by hand.

For other people in my position, having to convert hundreds of Waters .raw files, I hope that helps at least somewhat.


Can I electrolyze saturated sodium hydroxide?

At $T=25\ \mathrm{^\circ C}$, saturated sodium hydroxide solutions contain about $50\ \%$ sodium hydroxide. Hence, the concentration of $\ce{Na+}$ is about $c(\ce{Na+})=19\ \mathrm{mol\ l^{-1}}$ and the concentration of $\ce{H3O+}$ corresponds roughly to $\mathrm{pH\approx15}$, i.e. $c(\ce{H+})\approx10^{-15}\ \mathrm{mol\ l^{-1}}$ (neglecting activity coefficients).

The positive ions $\ce{Na+}$ and $\ce{H3O+}$ are attracted to the cathode.
At the cathode, the redox potentials $E$ imply that, even under the given alkaline conditions, reduction of $\ce{H+}$ to $\ce{H2}$ is still preferred over reduction of $\ce{Na+}$ to $\ce{Na}$:

$$\begin{alignat}{3}
\ce{Na+ + e- \;&<=> Na}\qquad &&E^\circ = -2.713\ \mathrm V\qquad &&E&&=E^\circ+0.05916\ \mathrm{V}\times\log_{10} [\ce{Na+}]\\
& && && &&=-2.713\ \mathrm V+0.05916\ \mathrm{V}\times\log_{10} 19\\
& && && &&=-2.64\ \mathrm V\\[6pt]
\ce{2H+ + 2e- \;&<=> H2}\quad\quad &&E^\circ = +0.000\ \mathrm V\qquad &&E&&=E^\circ+0.05916\ \mathrm{V}\times\log_{10} [\ce{H+}]\\
& && && &&=0.000\ \mathrm V+0.05916\ \mathrm{V}\times\log_{10}(10^{-15})\\
& && && &&=-0.89\ \mathrm V
\end{alignat}$$

Therefore, the thermodynamically preferred reaction is the reduction of water to hydrogen according to
$$\ce{2H+ + 2e- -> H2 }$$
or in this case better (in view of the given alkaline conditions)
$$\ce{2H2O + 2e- -> H2 + 2OH- }$$

Nevertheless, under the given alkaline conditions, the concentration of $\ce{H3O+}$ is very low. The predominant positive ions are $\ce{Na+}$, which migrate to the cathode. Thus, especially when applying a high current density, the major part of the current of positive charges that reach the cathode is carried by $\ce{Na+}$. Therefore, $\ce{Na+}$ actually can be discharged at the cathode:
$$\ce{Na+ + e- -> Na}$$
The ratio of discharged $\ce{Na+}$ to $\ce{H+}$ also depends on the cathode material. For some materials, the reduction of $\ce{H+}$ to $\ce{H2}$ is kinetically inhibited, which results in a significant overpotential.

However, any generated $\ce{Na}$ immediately reacts with water (except when a mercury cathode is used):
$$\ce{2Na + 2H2O -> 2Na+ + H2 + 2OH- }$$

Therefore, the net reaction is again the above-mentioned reduction of $\ce{H2O}$ to $\ce{H2}$. The $\ce{Na+}$ ions act as electrolyte and might be considered to act as a catalyst:

$$\begin{align}
\ce{2Na+ + 2e- &-> 2Na}\\
\ce{2Na + 2H2O &-> 2Na+ + H2 + 2OH- }\\
\hline\ce{2H2O + 2e- &-> H2 + 2OH- }
\end{align}
$$


Yes, you are on the right track, let's look at the situation in more detail.  In the haloacid equilibrium $$\ce{HX <=> H+ + X-}$$ anything that stabilizes HX will push the equilibrium to the left and make the $\mathrm{p}K_\mathrm{a}$ more positive. Anything that stabilizes the proton and the halogen anion will push the equilibrium to the right and make the $\mathrm{p}K_\mathrm{a}$ less positive. Here's a table that summarizes some of the haloacid data:

$$\begin{array}{cccc}
\text{X} & \text{H-X bond strength / kJ mol}^{-1} & \text{Electronegativity of X} & \mathrm{p}K_\mathrm{a}(\ce{HX}) \\
\hline
\ce{F} & 565 & 4.0 & 3.1 \\
\ce{Cl} & 427 & 3.0 & -7.0 \\
\ce{Br} & 363 & 2.8 & -9.0 \\
\ce{I} & 295 & 2.5 & -11.0
\end{array}$$

There are 3 main factors that influence this equilibrium: bond strength, electronegativity and polarizabilty.

In the "Bond Strength" column we see that the $\ce{HX}$ bond strength decreases as we move down the column. This is because, as you noted, overlap between the hydrogen $\mathrm{1s}$ orbital and the halogen orbital is most effective with the fluorine $\mathrm{2sp^3}$ orbital (in fact, overlap is so good in the fluorine case that the bond has a significant covalent nature) and the effectiveness of the overlap decreases as we move down the column. Therefore, $\ce{HX}$ bond strengths should make the $\mathrm{p}K_\mathrm{a}$ decrease (less positive, more negative) as we move down the halogen column.

Your analysis of electronegativity is correct. The more electronegative the halogen, the more stable the halide anion. This factor should cause the $\mathrm{p}K_\mathrm{a}$ to increase (more positive) as we move down the halogen column.

The larger the atom, the more polarizable it is. The more polarizable an atom the more stable is its corresponding anion because we have our charge spread out over a larger volume which is a stabilizing feature. This factor should cause the $\mathrm{p}K_\mathrm{a}$ to decrease (more negative) as we move down the halogen column.

So we have two factors, bond strength and polarizability, that work together to push the $\mathrm{p}K_\mathrm{a}$ lower as we move down the halogen column; and one factor, electronegativity, that works in the opposite direction. Based on the observed $\mathrm{p}K_\mathrm{a}$'s, the first two factors win out over the electronegativity.

BTW, I like your $\sigma$, $\sigma^*$ reasoning. Here's how I think of it in those terms.  As you bring two atomic orbitals closer together, they will split and form two new molecular orbitals. The degree of splitting is dependent upon the overlap of the two atomic orbitals.  Overlap is a measure of electron density between the atoms.  An electronegative atom will share its electron less readily decreasing the splitting.  If there is a strong bond, then the electrons are well shared and the splitting will be larger.


How to Protect one -OH group in 1,4-benzenedimethanol using TBDMS

The product is symmetrical so it doesn't really matter which alcohol gets protected, you'll end up with the same product.

There are many examples of doing this kind of mono protection of simple substrates. Quite often a fairly large excess of the diol is added to try and ensure no bis protection takes place. In your case the starting material should be easily separable from product by column chromatography.

A potentially better method is to formally deprotonate the diol with 1 eq of a base such as sodium hydride or butyl lithium then trapping with a solution of the silyl chloride. This often works well in cases where bis protection may occur, as energetically its unfavourable for both oxygens on the molecule where there's molecules that are neutral


It's much simpler than all that.

Pure gas
$$\mu_g(T,P)=\mu^0(T)+RT\ln P$$
where $\mu^0(T)$ is the molar free energy of the pure gas at temperature T and pressure 1 Bar, and P is the pressure expressed in bars.

Pure liquid

If $P_{sat}(T)$ is the equilibrium vapor pressure at temperature T, then, at pressure $P_{sat}(T)$, we know that the free energy of the liquid is equal to the free energy of the vapor.  So, $$\mu_l(T,P_{sat})=\mu^0(T)+RT\ln P_{sat}(T)$$If we neglect the effect of pressure on the chemical potential of the liquid (because the specific volume is very small), then, for a pure liquid at temperature T and pressure P:$$\mu_l(T,P)=\mu^0(T)+RT\ln P_{sat}(T)$$

Ideal gas mixture

For species A in an ideal gas mixture, the chemical potential of species A is the same as the chemical potential of the pure species, except evaluated at the partial pressure of A in the gas mixture:$$\mu_{Ag}(T,P,y_A)=\mu^0_A(T)+RT\ln p_A=\mu^0_A(T)+RT\ln (Py)_A=\mu_A(T,P)+RT\ln y_A$$
where $p_A$ is the partial pressure of A (bars), $\mu^0_A(T)$ is the free energy of pure A at temperature T and pressure of 1 bar, P is the total pressure of the mixture (bars), $\mu_A(T,P)$ is the free energy of pure A at the temperature and  pressure of the mixture, and $y_A$ is the mole fraction of A in the mixture.  So the chemical potential of A in the mixture is equal to the chemical potential of pure A at the same temperature and pressure as the mixture plus the correction term for mixing $RT\ln y_A$.

Ideal liquid mixture

By analogy to an ideal gas mixture, the chemical potential of species A in an ideal liquid solution is given by:

$$\mu_{Al}(T,P,x_A)=\mu_{Al}(T,P)+RT\ln x_A=\mu^0_A(T)+RT\ln P_{A,sat}(T)+RT\ln x_A$$
where $\mu^0_A(T)$ is the free energy of pure A at temperature T and pressure of 1 bar, $P_{A,sat}(T)$ is the saturation vapor pressure of species A at temperature T (bars), and $x_A$ is the mole fraction of species A in the liquid.


How to create a 3 M solution of ammonia from 25% ammonia

$\text{Produce a solution of }3\;\text{M}\;0.5\;\pu{dm^3}\;\ce{NH3}$

$\\\text{Ammonia}\;25\%\;\text{(mass-percent)}\\\rho_{ammonia}=0.82\pu{\frac{g}{cm^3}}$

Suppose 100ml

$$0.82\times 100\times 0.25=20.5\;\pu{g}\;\ce{NH3}\text{ per 100 ml of solution}$$

$$n_{ammonia}=\frac mM=\left(\frac{20.5}{17.031}\right)\pu{moles}$$
$$C_{ammonia}=\frac{n_{ammonia}}{100\times10^{-3}}\approx 12.037 \;\pu{M}.$$

To decide $V_1$, which is needed in order to prepare the solution of $0.5 \;\pu{dm^3}$ $3\;\pu{M}$ ammonia,

$$V_1=\frac{C_2V_2}{C_1}\approx0.1246 \pu{dm^3}=124.6\; \pu{ml}.$$


Explaining the colour change in the potassium permanganate titration of iron(II) ions

I think this is because $\ce{MnO4-}$ ions have a very distinct colour. At the end of the reaction, the solution is colourless. I think that the $\ce{Fe^3+}$ ions give the solution a brown colour, but it is in a very low concentration. The brown $\ce{Fe^3+}$ ions are too few to give the whole solution a brown colour.


Major product formed when 1,1,1-trichloropropane is treated with aqueous KOH

Figure: 1,1,1-trichloropropane

${\color{green}{\checkmark}}$ The above compound involves $\color{red}{\ce{C-H, C-C, C-Cl}}$ bonds, whose bond energies are $413, 348, 328 kj/mol $ respectively. So, initially $\color{orange}{\ce{C-Cl}}$ bond must break.

${\color{green}{\checkmark}}$ It may be noted that generally alcoholic $\color{blue}{\ce{KOH}}$ causes elimination (dehydrohalogenation reaction) in the molecule of haloalkane, while aqueous $\color{blue}{\ce{KOH}}$ leads to substitution. As in your reaction, aqueous $\color{blue}{\ce{KOH}}$ is used, substitution reaction is most preferable.

${\color{green}{\checkmark}}$ As primary carbon and strong $\color{orange}{\ce{C-Cl}}$ (compared to other $\ce{C-Br, C-I}$ bond) is involved in 1,1,1-trichloropropane, $S_{N2}$ mechanism is preferable.

From the above thought process, the most preferable reaction would be:
$$\ce{Cl_3C-CH_2-CH_3 + 3KOH(aq) ->[\text{heat}] (OH)3C-CH_2-CH_3 + 3KCl} $$

${\color{green}{\checkmark}}$ It may be noted that more than one $\color{blue}{OH}$ group cannot be present on the same carbon atom. In such a case, the compound will be extremely unstable and will readily lose a water molecule to form more stable aldehyde, ketone or carboxylic acid.

$$\ce{(OH)_3C-CH_2-CH_3 -> (OH)CO-CH_2-CH_3 + H_2O}$$

You are right, propionic acid along with water molecule is going to be the product.


When can a Secondary standard solution be used for standardising something instead of a Primary?

If you are doing pH titrations, and your primary standard is an acid - say - potassium hydrogen phthalate, you can't use that to standardize an acid directly. Instead, you would make a secondary standard of - say - sodium hydroxide, titrating against your primary standard. Then, you are ready to standardize the acid you want to use in your titration.

Secondary standards are also used if the primary standard is expensive. Hydrochloric acid solutions are cheap and stable, but you can't make them by weighing in a solid. So instead, you make an expensive primary standard and a lot of cheap hydrochloric acid of unknown concentrations, and determine the titer of the hydrochloric acid. Then, you can use it as a secondary standard.


How thick is the colored oxide layer on heat treated stainless steel?

Not very thick. Carbon steel develops the same colors at much lower temperature/ times . Gold to pink to blue to purple , stainless shows the colors better because the base metal is brighter . It is caused by the thickness of oxides , a matter of a few layers of oxide molecules ; unfortunately I cannot give you a number in microns and can't remember which book it is in.


Is Internal Energy = (3/2)nRT for a ideal monoatomic gas?

Molecules have internal energy due to intermolecular interactions, as well as translational kinetic energy, rotational energy, vibrational energy, electronic energy (and if you care to include them, nuclear energy and the mass-energy of the protons, neutrons and electrons themselves).

By saying "ideal", energy due to intermolecular interactions is eliminated.

By saying "monoatomic", energy due to rotation and vibration is eliminated.

So only translational kinetic energy and electronic energy remains.

(3/2)nRT is the translational kinetic energy, and since almost all atoms are in the ground electronic state at low temperature, it is a good expression for internal energy as long as the temperature is low enough that essentially all atoms are in the electronic ground state.


If this is in regard to mechanisms with organometallic complexes or coordination compounds, there are a set of specific mechanisms that are well defined. These are:

Ligand dissociation / association
Reductive elimination / oxidative addition
Intramolecular insertion (e.g. CO migratory insertion) / elimination (e.g. $\beta$–hydride elimination)
Metallacycle processes (e.g. $\sigma$-bond metathesis)
Nucleophilic / electrophilic addition and abstraction are also possible, similar to organic chemistry.

These mechanisms should be in inorganic textbooks but can also be viewed on the first page here. The later slides in this pdf cover more of the details of each specific mechanism.

In many reaction mechanisms with organometallic complexes, it is a process of a number of these smaller mechanisms occurring, leading to the overall chemical transformation that is stated in a reaction equation. These overall mechanisms can be complicated and need to be evaluated for each case. Knowing the basic underlying mechanisms will certainly help clarify processes that seem quite complex at first glance.

There are subtleties in all of these mechanisms to consider beyond the basic types above. Significantly more detail can be found in these textbooks that I have previously utilized which you may also find informative:

Collman, J.; Hegedus, L.; Norton, J.; Finke, R. Principles and Applications of Organotransition Metal Chemistry. 1987. University Science Books. Mill Valley, California.

Spessard, G.; Miessler, G. Organometallic Chemistry. 1997. Prentice-Hall. Upper Saddle River, New Jersey.


Why it is only carbons that have hydrogens attached (and not just hydrogens directly attached) to the charged carbon which result in hyperconjugation?

Hyperconjugation happens because of an overlap of orbitals.

Consider this: in (uncharged) 2-methylpropane, the central carbon has sp3 hybridisation (and a hydrogen bound to it).

When the carbocation of the image is formed, the hydrogen leaves, the hybridization becomes sp2 (hence, planar), and now there is a free pi orbital available.

Here is what causes hyperconjugation: pi orbitals are quite larger than how they are drawn in books, and they can overlap partially with the orbitals of the adjacent hydrogens, and get part of their charge.

In the example of butane, the thing is the same: when the carbocation is formed, the carbon and the two hydrogens attached become planar (sp2), and a perpendicular pi orbital is formed. It cannot interact with the hydrogens attached to the charged carbon because they are on its nodal plane. Instead, it overlaps partially with the orbitals of the hydrogens on the adjacent carbon.


Why is dialysis tubing impermeable to sucrose when it should retain >12,000 Da?

There exist many types of semi-permeable membranes (the ones used for dialysis tubing), with various pore sizes. One of the very common lab experiment on the topic of osmosis is using a sucrose solution (sucrose is cheap) and small-pores membrane, such that water and small ions (typically Na+ and Cl–) can pass, but not sucrose. The one which you link to simply happens to be a variety with much wider pores.


Why several of isomers of octane such as 2,3-dimethylhexane and 2,4-dimethylhexane doesn't have a value for melting point from any databases?

I recommend to include more than one (secondary) reference and to widen the querry -- if there is access -- to general chemistry data bases like Chemical Abstracts (by the American Chemical Society), or Reaxys (by Elsevier), and Springer Materials, too.

In addition, do not stick to one chemical name, and equally include in your querry unique identifiers like the CAS registry number -- applicable regardless how the organic nomenclature is applied (including databases in other languages) -- rather than one chemical name.  By doing so, you may rapidly find addtional data, for example among chemical suppliers like Sigma-Aldrich, that lists 2,3,4-trimethylpentane (CAS-RN 565-75-3) in its catalog with a melting point of −110 °C and a boiling point of 113-114 °C.

Beside electronic ressources, consider additional ones available in print, like volumes in Landolt-Bornstein dedicated to condensed matter (now included in Springer Materials).


How does PEDOT:PSS work (mechanism) in Organic Electrochemical Transistors?

This is about the use of well known words in semiconductor physics in the relatively new context of organic electronics. This is not always strictly justified, and at least confusing in many occurrences.

I will try not to slip in any deep debate about communities. Just let me say that obviously general physics applies both in classical semiconductors and organic semiconductors*, in particular when the output of a device - rather than its microscopic structure - is considered. It is fine to me to use whatever approach and treat these systems up bottom (semiconductors people do that) or vice versa, but keeping in mind that the constituents are covalent molecules.

Reality is in the middle, and organic semiconductors are likely to be compared to amorphous inorganic semiconductors even in the case of the firsts having rather crystal like structures. Molecules will be always therein, and a kind of hopping processes will be more or less crucial to conduction anyway.

PEDOT:PSS is in fact an ionic compound, a "poly(salt)", let me say. I want to say more than general ionomer, I do NOT want to propose a new name. It will be clear as you read. Be careful of italic.

For those not familiar with the material here is the link to it in Wikipedia: https://en.m.wikipedia.org/wiki/PEDOT:PSS

I would have placed a posive charge already in the structure, sketching a bit more than one EDOT ring.

Let's start with a neutral PEDOT chain itself. It is poly(ethylene-dioxythiophene), a conjugated polymer with quite low energy gap and low oxidation potential.

No drastic oxidative process, and even photoinduced electron transfer process (transfer to whatever available e-acceptor within the sample, purposefully added or not) lead to relatively stable radical-cations delocalised within the pi system along the polymer chain (there is an adapted or adopted name for these too, polarons).

If the anionic species formed are relatively stable too, then long living charges are formed.

In presence of very stable ionic species, either added or formed by the same process, a polimeric salt is obtained. In this polimeric salt charges of one sign are in chain, not in pending functional groups which is a much trivial case for which coincidentally polystyrene sulphonate PSS is an example.

PEDOT:PSS is right  one example of the polimeric salts that I have just introduced. The double dot here means a certain ratio depending on formulation, as the PPS providing the counterions (neutrality) can be in excess. The degree of oxidation can also varies depending on formulation.

You said " As at least I understand it, the PEDOT chain is rich with holes and then is doped with PSS which has sulfonate anions that compensate these holes."

Is the use of "doped", or alternatively "then", that confuses you.

The oxidized PEDOT chain is rich with holes in the sense that is rich of positive radical-cations and they behave as the carriers. However the chain is far to be neutral, is effectively positively charged, as you know.

Actually there are positive charges (positive carriers) by virtue of the presence of the external compensating PSS counterions. The latters are strictly necessary for the so called holes to exist within the PEDOT chains.
This is not the case for inorganic semiconductors, e.g. doped silicon is inherently neutral.

PSS stabilize the charges that in its absence will be either transient or prone to further chemical reactions as for radicals are generally very reactive.

I think this description should help you in getting the subtleties of this field and those of its dictionaries.

I could have simply written that

in conjugated pi system the so called doping is a redox process, transferring electron between the system and a second material

(generally from - to respectively, with notable exceptions of which fullerenes are probably the best known examples).

The carrier in oxidized PEDOT are positive and their motion is the opposite motion of electrons residing on the highest singly occupied MO, that ideally spread all along a perfect chain. In reality along a chain their will be a distribution of segments with different effective conjugated length, each with its corresponding  singly occupied orbital.
Positive charges move i.e. electrons move between these in chain or inter-chain SOMOs, which can be regarded as a band. In such a sense these carriers are called holes.  They move by a deficit of electrons in the normally occupied levels.
As for the convenience of calling  these carriers holes or not is up to you.
The point is that PEDIT:PSS can be regarded as a whole neutral material with mobile positive charges.

Then you said " Then of course when you apply a gate voltage, cations compensate the PSS anions instead which according to literature makes the PEDOT:PSS less conductive, but I would have thought this would make it more conductive as it liberates the holes in the PEDOT backbone. What do I seem to be missing here?".

I do not know what device structure you refer to. The conductivity of PEDOT:PSS is high enough to shorts two metals or doped Si - metal contacts, and to work as antistatic layer.

I believe that the behaviour of your diode will be dictated mostly by a second material with conductivity/mobility compatible to a transistor function.

PEDOT:PSS is a conductor tout court (between metals) or is in general applied*** as a selective contact when its SOMO levels mentioned above match (or mismatch) the relevant conduction levels of other materials.

In a sense the term semiconductor should be reserved to neutral PEDOT (although its bulk conductivity will be rather that of an insulator) while PEDOT:PSS has metal like bulk conductivity but a hole** type conduction mechanism as described above.

*it is clear what one mean saying "organic semiconductors", but I would prefer "extended pi conjugated systems". The same for "doped", I would say "oxidised (or reduced)". And so on. And not, there are not dampling bonds in molecules, even when it happens they are organic semiconductors.

**this should clarify that sometimes an adopted nomenclature is the first solution at hand ;)

***PEDOT:PSS is also reported as smoothing, adhesion promoter, etc all uses not pertinent to this Q.

EDIT: I realised you refer specifically to electrochemical transistors. Well, the above is still answering. Indeed, upon negative bias electrolyte cations infiltrate the material thereby neutralising the PSS anions. For the above discussion this reduces back the PEDOT, diminishing the available amount of carriers and as such the drain current.


Easiest/Fastest way to obtain a good quantity of fine Iron powder?

A less laborious path to very fine iron dust (in fact, likely  pyrophoric iron) perhaps follows from first purchasing or preparing iron(II) oxalate (a path is detailed in a reference provided below). Next, thermally decompose the salt and finally, stabilize the nano-iron product with say mineral oil.
The details on the thermal decomposition in an atmosphere of hydrogen is discussed in this paper: Pyrophoric Nanoparticles and Nanoporous Foils for Defense Applications. See also this discussion which includes a suggested preparation for Iron(II) oxalate together with a video here.
Lastly, follow a similar process as is outlined in this US Patent 3,520,676 which claims, to quote:

Pyrophoric powder of metals (such as iron, cobalt, nickel and alloys thereof) is stabilized to prevent its spontaneous ignition. Such stabilization is achieved by Wetting the metal particles with a high-boiling organic liquid (such as an ester of carboxylic acid, mineral oil, silicone oil or fatty acid) and holding the wetted particles in the presence of oxygen until a thin oxide layer is formed on the surfaces thereof.

in order to stabilize the very fine iron product for your intended use.
A particular advantage to this approach is the low cost, and actually, not a great need for a large amount of the nano-particles for your demonstration purposes.


How to test for and remove lead(II) sulfate from a sample of 1 M sulfuric acid?

To answer your question on how to test for Lead sulfate, I refer you to this paper, to quote:

  In 1929 Fischer [1] introduced a sensitive reagent for small quantities of lead which has found interesting applications in widely diverse fields. Diphenylthiocarbazone, or dithizone as it is commonly called...

And further:

  We have made no study of this reagent, except to find means of using it in our problem of measuring the solubility of lead sulfate. Dissolved in chloroform, dithizone imparts a noticeable green tint to the solution, provided the solution is not too concentrated. The lead-dithizone complex, however, has a bright cherry-red color.

Also of import;

  Dithizone is known to react with metals other than lead. The precautions taken to avoid interference by other metals are discussed on page 58. These precautions, together with the spectroscopic analysis of the lead salt and the redistillation of water and acid, make it highly improbable that the results were affected by other metals. Confirmation of this is found in the analysis of solutions of known lead content.

If Pb is present, dilution appears to be the best path due to solubility.

[EDIT] Here is an theoretical path to the removal of heavy metals that does not require dilution. I am suggesting UV light therapy employing TiO2 photocatalyst to act on cold H2SO4 containing heavy metal impurities. Heavy metals are expected to precipitate.

With respect to the mechanics of the photolysis, I referred, for example, to this source. Per cited Equation [R1]:

$\ce{ hv -> e- + h+ }$

Per Equation [R3], expected reaction with an electron hole:

$\ce{H2O + h+ -> .OH + H+ }$

$\ce{H2SO4 = H+ + HSO4- }$

$\ce{H+ + e- = .H }$

where the hydrogen radical behaves as a e-/H+ pair, see Hydrometallurgy 2008: Proceedings of the Sixth International Symposium, p. 818, a commercial reductive leaching equation, to quote:

  $\ce{ PbS + 2 •H = Pb + H2S }$ (5)

In the current case:

$\ce{ .H + PbSO4 -> Pb (s) + HSO4- }$

However, the associated transient sulfate radical is also created:

$\ce{ .OH + HSO4- -> H2O + .SO4- }$

Leading to the persulfate ion:

$\ce{ .SO4- + .SO4- -> S2O8(2-) }$

So, the sulfate ions is not alone. Possibly more advantageous, further  pump air/oxygen into the solution during the UV light treatment. Then per Equation [R3], expect the superoxide radical anion formation:

$\ce{O2 + e- = .O2- }$

which becomes, in acidic conditions, the $\ce{.HO2}$ radical, which may similarly react with Lead sulfate as follows:

$\ce{ .HO2 + PbSO4 -> PbO2 (s) + HSO4- }$

or, just oxidize any liberated Pb.

I would even suggest you try this process out by creating a test solution (with at least added Lead) and heat in H2SO4 to form a heavy metal impurity and then apply the UV light/TiO2 (and also added oxygen) therapy.


Circular dichroism of a hydrazone boron complex

On the origin of chirality, you should read something about atropoisomerism. This sort of chirality arises due to a hindered rotation around a chiral axis.
It is most notably encountered in biaryls, because by introducing substituents you can make rotation more difficult and observe a CD. However, it is also present in very simple molecules. Think about H2O2: every conformation where the dihedral angle is between 0° and 90° is chiral. Probably your teacher didn't point this out when explaining butane conformations, but gauche conformers exist in enantiomeric forms. The problem with these small molecules is that to freeze rotation you should basically go to very low temperatures, and then separate the enantiomers. You would observe CD. If you can do that, you got material for a pretty awesome paper in my opinion!

In this case, think about rotating the phenyl ring around the N-Ph bond. Draw the structures, and you will notice that they are chiral. The claim of the authors is that in a high viscosity solvent this rotation is hindered at room temperature, thus you will observe CD (see more on pg. 46 of the Supp. Info).
Concerning the racemic mixture, I'd say you are right. It is not clear to me why they see CD without separating the enantiomers, or maybe they did but they didn't explain it thoroughly. They refer to some previous work, so maybe you could find more details in other papers.

Why is the maximum band so high?
High with respect to what? 150mdeg really doesn't mean anything if you don't have the concentration and cell length and you can't work out the delta epsilon and the g-factor. CD people never use mdeg because it is not an absolute unit. The authors do not report concentration and cell path, so you can't really say whether it is strong or not, also because absorbance is reported in arbitrary units, so you can't really work out the g-factor. Clearly, they're not CD people, they use it as tool in their research, so the information they provide is not really helpful to answer this question.

Concerning the signals, have a look at figure 64 in the Supp. Info. Think about spectra as a series of lines to which you apply some lineshape function (usually Gaussian or Lorentzian). Now, two isolated bands are exactly superimposable if they have the same intensity and the same broadening. If they are not superimposable, one of these two things is not right, or the bands are not isolated. Since you are speaking about two different spectra, you can always normalise them to the same intensity. Additionally, they are recorded in the same conditions (possibly simultaneously), so there is really no reason why the broadening should be different. The only other possiblity is that there is more than one transition under that signal.
The "dark state" means basically that, in UV such state has a low absorption because of some reason. However, some transition that is dark in UV can have a pretty significant CD signal, because maybe it is only magnetically allowed (think about the n-pi* transition in carbonyl groups, or some charge-transfer state).

I tried my best, however the most interesting point (why do they see CD) still remains obscure to me, mostly because it's weekend and I am too lazy to read the paper! If I remember, I'll maybe read it next week.

Edit:
One thing that I didn't mention is that the sp3 N atom is chiral as well, so actually here, when you put in the axial chirality, you're really speaking about diastereoisomers, if you can fix the chirality of the N atom. This happens in very particular cases (see Cinchona alkaloyds), but since we're in a high viscosity medium, who knows, this might be the case, and explain why only one of the two possible configurations of the chiral axis is preferred.


Stability of ferrous ions and iron(II) hydroxide in acidic conditions

The stability of iron(II) hydroxide J. Chem. Educ., 1957, vol. 34 pages 178-179 shows that the book isn't entirely correct.

  ...Iron(II)
  hydroxide would thus appear to be unstable in neutral
  solutions and to possess increasing stability as the
  basicity of the medium increases.

  The foregoing reasoning led to the attempt to prepare
  iron(II) hydroxide from the chloride under such
  conditions that the hydroxyl ion concentration was not
  allowed to fall below approximately one molal until the chloride ion had been removed. The hydroxide was precipitated from strongly basic solution by the addition of dilute iron(II) chloride to a saturated solution of
  sodium hydroxide in an air-free system with vigorous
  stirring. The precipitate was repeatedly washed with
  molal sodium hydroxide (oxygen and carbonate-free)
  until a test for chloride ion with silver nitrate could
  no longer he obtained. Five additional one-liter washings with normal base completed the preparation of the
  iron(II) hydroxide used for the solubility studies.
  This technique led to the formation of a coarse-grained
  precipitate, white when viewed by reflected light and
  slightly greenish tinged by transmitted light. Dilutions
  of the material were made to one-tenth molal base
  without evidence of darkening when maintained under
  an oxygen-free atmosphere.


Can sensitive nuclei like 1H be exploited to obtain 13C spectra?

We are able to record 1D heteronuclear experiments using these techniques, and absolutely do make use of this behaviour as a commonplace method for measuring 1D spectra of low sensitivity nuclei, such as $\ce{^13C, ^15N}$ and $\ce{^29Si}$. However, in order to get frequency information for the insensitive nucleus, we still need to use what is called a direct-detect method, where the nucleus of interest is actually the acquisition nucleus.

The most common method for sensitivity enhancement uses the general class of experiments called INEPT experiments; Insensitive nuclei enhanced by polarization transfer. The DEPT series (DEPT45, DEPT90 and DEPT135) are a sub-class of the INEPT experiment.

The signal of some nuclei can be improved by taking advantage of low power 1H decoupling during the relaxation period to allow nOe transfer to occur. This, as you mention, is typically how $\ce{^13C}$ spectra are recorded, and is why the $\ce{CH3}$ peaks ar much bigger than quaternary peaks. This method cannot be used for nuclei that have a negative magnetogyric ratio, such as $\ce{^15N}$ and $\ce{^29Si}$, as the nOe will actually generate a diminished signal. For these nuclei, the INEPT method is the preferred experiment for acquisition.

The INEPT magnetisation transfer step is at the core of most common 2D sequences, such as the HSQC. It provides a much greater signal enhancement than the nOe, although is strongly dependent on correct calibration of pulses.

The theoretical maximum sensitivity gain for nOe is $\displaystyle1+\frac{\gamma_\mathrm{S}}{2\gamma_\mathrm{I}}$
The theoretical maximum sensitivity gain for INEPT is  $\displaystyle\left|\frac{\gamma_\mathrm{S}}{\gamma_\mathrm{I}}\right|$

For $\ce{^13C}$, this equates to $\approx 3\times$ for nOe and $\approx 4\times$ for INEPT. For $\ce{^15N}$, it is $\approx -4\times$ for nOe (yep, negative) and $\approx 25\times$ for INEPT.

Glenn Facey at University of Ottawa has a couple of nice examples of INEPT and DEPT that are well worth looking at.

Both of these classes of experiment (nOE and INEPT) typically rely on interaction with nearby $\ce{^1H}$ signals: nOe is a through space interaction, and INEPT is a through bond interaction. Other nuclear pairs can be used, and many examples such as INEPT using $\ce{^19F-^29Si}$ have been published. The detection of insenstive nucleus environments with no directly attached sensitive nucleus (such as $\ce{^13C}$ quaternary centres) still requires long range coupling, and there are many published long-range INEPT methods.

The 2D HSQC method is an indirect detection method — we get information about the insensitive nucleus by acquiring another nucleus. Equivalent 1D variants of these 2D experiments do exist (so, a 1D HSQC), but they will only give you information about those $\ce{^1H}$ signals that have coupling to $\ce{^13C}$ — they don't provide any information about the $\ce{^13C}$ frequencies, unless you do a selective variant. So, a sel-HSQC will only irradiate a small band of the $\ce{^13C}$ frequency window, and therefore only give $\ce{^1H}$ signals that correlate to a $\ce{^13C}$ nucleus in this band. This is sometimes useful to probe a specfic question of structure, but a very tedious method of stepping through frequency bands of the $\ce{^13C}$ region (which is what a 2D experiment does for you).

For what it is worth, there are a number of cost-effective ways to improve sensitivity without having to fork out the big bucks for an 800. Choice of probe (X-observe, cryoprobe, microprobe) will offer improvements over increase in magnetic field at a fraction of the cost, and Shigemi tubes and micro cell tubes will also provide significant boosts in  sensitivity.


Reaction of cinnamic acid with Br2/Na2CO3

The sodium carbonate is just there to mop up any $\ce{HBr}$. The $\ce{Br2}$ adds across the double bond to give cinnamic acid dibromide.
The treatment of cinnamic acid dibromide with $\ce{KOH}$ at elevated temperature eliminates 2 eq. of $\ce{HBr}$ to give Phenylpropiolic acid. This is known to decarboxylate through the intermediate acetylene anion.


How to create .CIF file from DRX data in a paper, for compound not found in crystallographic database

You can't find this entry in CCDC's CSD, because the scope of this database are small organic, or organometallic molecules' crystal structures.  (For large organic molecules, analogue studies' data may be found, e.g., on wwpdb.org.)  Because your compound is considered as inorganic*, you'll find it in the ICSD, the inorganic crystal structure database.
Note, the paper includes the statement:

«Further details of the crystal structure investigation may  be  obtained  from  Fachinformationszentrum Karlsruhe, 76344 Eggenstein-Leopoldshafen, Ger-many (fax: +49-7247-808-666; e-mail: crysdata@fiz-karlsruhe.de,   http://www.fiz-informationsdienste.de/en/DB/icsd/depotanforderung.html) on quoting the depository number CSD-417100.»

Because CCDC and ISCD established in 2018 a search interface in common, and entries' labels in both databases differ from each other, it is easy to find the record here.  It is not as comfortable as their own GUI and API interfaces (example ICSD), but it is possible to download the record's .cif.
Note that complementary to these two databases, there are others like e.g., the American Mineralogist Crystal Structure Database and the COD.  Entries 2100155, 2100156, and 2100157 of the later for example point to the .cif about $\ce{Sr(OH)2 ·8H2O}$ of a study by Ricci et al. at 20, 100 and 200 K, published 2005 in Acta Cryst. B.

*) A simplified definition of «organic compound» is that it is molecular and contains at least one carbon atom excluding carbon oxides ($\ce{CO, CO2}$, and others), their hydrates, and simple metal carbonates ($\ce{H2CO3, HCO^-_3, CO^{2-}_3}$).


rust caused by salt solution & stainless steel

Chloride will pit/rust 304 ,not quite as much rust/pits on 316/317. Many things make it worse such as limited air access ( under deposit or crevice corrosion ).Oxygen can help rebuild the passive film. For your long term exposure ceramics and glass containers is the best fix. For metal containers ,much higher alloy is needed such as Incoloy 825.


Do 1s & 2s orbitals overlap of different atoms while forming sigma bond? I

Orbital combinations involving inner shell electrons can indeed interact with outer-shell counterparts having the same symmetry.  But, the inner shell orbitals are so deeply buried inside their respective atoms that the amount of overlap, and therefore the effect on bond/orbital energies, is very small.  We find that in general, molecules are well modeled, without unneeded complexity, by including only outer shell orbitals.  Moreover, any overlap involving inner shells, or even with outer shells, usually does not involve just $s$ orbitals.  There can be $p$ orbital combinations and even $d$ orbital combinations having the same symmetry as the $s$ orbital ones, so the higher angular momentum orbitals also mix in.  Only hydrogen and helium atoms with just $1s$ electrons can avoid that.

It isn't just our description of chemical bonding that exploits this property of inner shell electrons.  We sometimes identify elements in a sample by hitting the samples with X rays that knock out inner shell electrons, and measuring the spectrum of absorbed energies.  The energy required to knock out inner shell electrons from an atom is a characteristic of that element, and lack of strong overlap with other atoms makes that energy essentially invariant.


What is this glassware for a distillation apparatus called?

The "A" on the glassware gives it away as an Aldrich glassware component;
and Ron is correct calling it a Kugelrohr bulb. It is actually a single bulb ball tube.

This specific item can be found in the catalogue at Aldrich's website


An old German book about chemical records (Quadbeck-Seeger, H.-J., Ed. Chemie Rekorde; Wiley-VCH: Weinheim, 1997) gives the Cullinan Diamond as the largest found natural molecule and a synthetic diamond of 38.4 carat as the largest made artificial molecule.


Why would Auger electron detection have potential for higher spatial resolution than EDX

I think this is a good question. The problem is more fundamental. In both cases, you would be using electron beams on the sample.
In EDX, you are shining a beam of electrons and detecting the X-rays. Here comes the main problem. It is hard to focus the X-rays. Mea culpa here. As Jon commented it is the depth of penetration of the electron beam and the corresponding X-ray emission that main affects spatial resolution. This picture roughly tells the story:

Auger electron electron spectroscopy, you are focusing a beam of electrons and you detect electrons. It is relatively easy to focus electrons using "lenses". Hence you can point point the exact origin where the electrons are being ejected.
Also, think about the depth profiling. EDX is not a surface technique unlike Auger. I used to have this misconception until I attended a workshop on surface analysis. The electrons can easily penetrate the sample and similarly originate from the some depth as well.


Equilibrium: Equations with complex roots

Complex roots to mass/concentration balance equations should be discarded as they have no physical interpretation. If there are only complex roots, then odds are there's some error in the manipulation of variables that lead to the equation.

Your case seems different though. Notice the very small magnitude of the imaginary part of the complex roots; the number is "almost real". This happens at times simply because of Mathematica's numerical root search algorithms, which aren't always guaranteed to reach the exact value. I'm actually a bit surprised that it got caught up in such a simple calculation.

As proof that the imaginary part of the root is nothing but an issue with the calculation, I simplified the equations until I found a better representation for Wolfram Alpha to parse, hoping the complex root would disappear.

First, let $A=1.77\times 10^{-4}$ and $B=3.3\times 10^{-4}$ for concision. Isolating $x$ in the second equation gives:

$$x=\frac{B(0.1-y)-y^2}{y}$$

Replacing $x$ for the above expression in the first equation then yields, after some tinkering:

$$(A+B)y^3+(B^2-0.1A-0.1B-AB)y^2+(0.1AB-0.2B^2)y+0.01B^2=0$$

Aha, a simple polynomial. Alpha should be able to deal with that just fine. Replacing the values of $A$ and $B$ produces the following equation, which Alpha solves with no imaginary part. The physically meaningful solution out of the three roots is $y\simeq 0.00458505$. Replacing that value in the equation for $x$ completes the problem, with $x\simeq 0.00228225$, very close to the approximate solutions.


Can you eliminate cigarette smoke odor between walls?

Getting rid of odor is a difficult task in general, particularly since molecules stick to surfaces and often stick in surfaces.

There's a good reason that "Febreze" works well. It contains a cyclodextrin ring that binds many odor-causing molecules. So the molecules are still there, but bound in a form you can't smell anymore.

(I remember an anecdote that they had to develop special fragrances to work with Febreze, but I can't find that anywhere.)

I suspect spraying paint primer inside the walls could work. This is often used to fix smoke damage from fires.


Why doesn't liquid propane in a closed coke bottle burst?

Yes, based on what we can see in the video, your guess appears to be correct: as the propane-filled bottle warmed up, just enough propane evaporated to keep the pressure inside the bottle equal to the equilibrium vapor pressure of the liquid propane.

According to the video, the ambient temperature outside at the time it was recorded was "about 45 °F", or about 7 °C.  Using the formula given here, I calculate the vapor pressure of propane at that temperature to be about 4400 mmHg, or about 590 kPa or about 5.9 bar.

Meanwhile, according to this page, the pressure inside a warm can or bottle of Coke can reach at least 380 kPa, or about two thirds of the vapor pressure of propane on a cold day.  As the bottles are certainly designed with a considerable safety margin, to make sure that they won't burst even if handled carelessly or slightly damaged, it's not surprising that they can easily withstand the pressure of the propane in the video.

BTW, this is exactly how aerosol spray cans work: they contain a mixture of the liquid being sprayed and a propellant substance (quite often propane) that has a boiling point at 1 atm only slightly below room temperature (or, equivalently, that has an equilibrium vapor pressure only slightly above 1 atm at room temperature).  Thus, as the can is drained, the partial boiling of the propellant maintains the pressure inside the can at the propellant's vapor pressure, which is high enough to propel the spray out of the nozzle, but not so high that it would require an excessively sturdy and expensive can to contain it.

As for what would happen at higher temperatures, at room temperature (i.e. 25 °C), the vapor pressure of propane would be about 7100 mmHg or 950 kPa (according to the formula, or about 7600 mmHg or 1000 kPa according to the table, which seems to be taken from a different source).  According to this random forum post, the small ½ liter Coke bottles used in the video can apparently withstand at least 180 psi, or 1250 kPa, so the propane-filled bottle probably wouldn't burst even at room temperature (unless it happened to be damaged or otherwise particularly weak).  If the temperature was raised to, say, 45 °C (113 °F, a very hot day), the vapor pressure of the propane would rise further to about 1500 kPa, which just might be enough to make the bottle fail.  Also, elevated temperatures will soften and weaken the plastic somewhat, making failure more likely.

In any case, in my personal experience, the weakest point of such bottles seems to be the relatively thin and weak cap, which would likely fail at some point before the bottle itself did.  I wouldn't be surprised if that was by design, to make the typical failure mode relatively safe and predictable.