How to achieve clean dishes and glasses?

There are several laboratory techniques to rigorously clean glass. Here are a few of the more common ones, to add to the other answers. There are probably more procedures, and it's also common to perform more than one treatment to maximize efficiency.

Practically none of these have any use at home, because they are too dangerous, too expensive, and just outright unnecessary. A soapy scrub will do just fine cleaning anything in 99% of cases, even if water still beads.

1 - Ultrasonication in pure solvents or detergent solutions.

Often, the main impurities on the surface of glass are organic substances which are present at all times in air outside of strictly clean environments, and which strongly adhere to the glass when they come into contact. By dipping them into a solution with some affinity for the organic substances, it is possible to strip them off. Common options are regular dishwashing liquid (usually sulphonic acids with large alkyl chains), solutions of quaternary ammonium salts, aqueous ethanolanime, or pure deionized water, ethanol, isopropanol, acetone, among many others.

It is much more effective, however, to couple the aforementioned compounds with ultrasonication rather than simply dipping or scrubbing. In an ultrasonicator, the liquid is vibrated at frequencies higher than $10\ kHz$. Though macroscopically the liquid doesn't seem to change much, on the microscopic level there is chaos. The high frequency vibrations locally produce very fast solvent jets and cavitation bubbles, which effectively sandblast anything closeby. Glass manages to stay mostly unaffected due to its stiffness and strong bonding, but anything not very firmly attached to its surface will be flung away.

2 - Prolonged exposure to high temperatures

Common glass is thermally stable and can be heated strongly before it softens, so it can take temperatures at which many other substance decompose, evaporate or burn (if exposed to oxygen). Different types of glass have different maximum temperatures depending on their composition. Also, the thermal expansion while heating and contraction while cooling can be problematic for delicate glassware.

3 - Soak in base bath

Strong bases react with many substances and help dissolve them off the surface of the glass. However, while very inert, glass itself will also slowly react with concentrated basic solutions at room temperature. This also helps clean because if very persistent stains manage to survive the basic conditions, the glass surrounding them will begin dissolving slightly, and after a while the stain ends up detaching from the substrate as the glass behind it will have dissolved away. The regular options are concentrated $\ce{NaOH}$ or $\ce{KOH}$ in water or ethanol.

4 - Soak in highly oxidizing solutions

Glass is composted mostly of silicon already in its most oxidized form ($\ce{SiO2}$). Hence exposure to very oxidizing conditions will not damage the surface of the glass, while anything else on it is likely to dissolve, be it organic or inorganic matter. Classic options for dipping the glass are acidic piranha solutions (a mixture of pure $\ce{H2SO4}$ and concentrated $\ce{H2O2}$), basic piranha solutions (concentrated aqueous $\ce{NH4OH}$ and concentrated $\ce{H2O2}$), sulphocromic solution ($\ce{H2SO4}$ + $\ce{H2Cr2O7}$, in disuse due to the toxicity of $\ce{Cr^{6+}}$) and sulphonitric solution ($\ce{H2SO4}$ + $\ce{HNO3}$). These are all pretty heavy-duty mixtures. For example, here's why piranha solutions earn their name.

5 - Treatment with ozone/oxygen plasma

This is a step up from even the piranha solutions. If all you want to do is oxidize organic matter on the surface of the glass, then why not cut out the middleman and directly bathe the glass in reactive oxygen? The glass sample is first put into a closed chamber containing oxygen (air works, but pure oxygen is better). If ozone is being used, then the chamber is lit from the inside by a mercury lamp, which produces lots of ultraviolet light. The light is capable of dissociating oxygen molecules and producing ozone, which then is directly exposed to whatever is on the surface of the glass. In the case of oxygen plasma, the chamber is pumped down to a low pressure of pure oxygen gas before a coil produces strongly varying electric and magnetic fields inside the chamber, generating all sorts of oxygen species such as singlet oxygen (an electronically excited form of standard, triplet oxygen), free oxygen atoms, and several ions and radicals. Here's a nice video on how this process works in practice.

After all the work you put into cleaning a glass surface, it will immediately start adsorbing impurities if outside a vacuum, and it may take as little as a few minutes before it is too dirty to use in very sensitive and specialized applications. I have found that water already beads significantly again less than 10 minutes after the final treatment.


para-dichlorobenzene - number of proton NMR signals

Be very careful with how you interpret this molecule. Although the 1H NMR spectrum appears as a single line, this is actually an AA'A''A''' system, meaning that, although chemically equivalent,  all four protons are magnetically non-equivalent, but have coincidental chemical shift. The coupling between all protons is different; JA-A''' is different to JA-A''' is different to JA'-A'' is different to JA'-A''' and so on.

So yes, the spectrum is as you thought. p-dichlorobenzene IS a single peak in the 1H NMR spectrum.

Below is a set of simulations for a generic para disubstituted benzene system, which is an AA'BB' system. A is chemically equivalent to A', but magnetically non-equivalent. A is chemically AND magnetically non-equivalent to B. Your system is just a special case example of these types of systems, with all protons being chemically equivalent, but magnetically non-equivalent. These are second order systems, which have a lot of complex additional splitting structure in their spectrum. Like all spin systems, as the chemical shift difference becomes less (the peaks move closer together), the tilting, (or tenting or roofing, depending on which text book you read) becomes more pronounced, and the wings on the main peaks become smaller and smaller. Eventually, they collapse into a single peak at very small chemical shift difference.

The fine text which you can't read (sorry about that), gives the coupling constants for the system, which are bog-standard ortho/meta/para couplings: Jortho=7Hz, Jmeta=3Hz; Jpara=1Hz. The only variable in the simulations is the chemical shift of the downfield peak, which slowly moves towards its upfield partner.


Representing conformers using SMILES and Graph representations

For the SMILES strings as defined by Daylight, you can expect these representations tell about the molecular constitution plus information about ($E$/$Z$)- and CIP's $(R)$/$(S)$-configuration. But nothing about the conformation like the boat / twist form of cyclohexane which generally is dynamic, or a faithful preservation of information like s-cis to describe the conformation of butadiene favorable for the Diels-Alder cycloaddition reaction.
Further, I speculate SMILES strings are not a faithful representation of locked conformations as in the axial $P/M$ configuration of 2,2',6,6'-substituted 1,1'-biphenyls. However, perhaps someone else may clarify this point.


How to calculate the amount of chlorine emitted yearly by a leak in the air conditioning system?

For the sake of correct formalism, here the answer as it should be given in the textbook:

The mass flow rate of the leaking substance, under the assumption of linear scaling, is
$$\dot{m}_{\ce{CHF2Cl}} = 12 ~\mathrm{kg\, month^{-1}} = 144~\mathrm{kg\, a^{-1}}\; .$$

The mass flow rate of chlorine, as contained in the leaking substance, is then calculated as follows:
$$\dot{m}_{\ce{Cl}} = \dot{m}_{\ce{CHF2Cl}} \cdot \frac{M_{\ce{Cl}}}{M_{\ce{CHF2Cl}}} = 144~\mathrm{kg\, a^{-1}} \cdot \frac{35.453~\mathrm{g\,mol^{-1}}}{86.47~\mathrm{g\,mol^{-1}}} = 59.04~\mathrm{kg\, a^{-1}}$$

For the end result to the question "how many kilograms of chlorine will be emitted into the atmosphere each year?", please note that the mass flow rate still has a dependency on time
$$ \dot{m} = \frac{\mathrm{d} m}{\mathrm{d} t}\; , $$
which means that we have to do an integration so we can calculate the mass:

$$ m(\tau) = \int_0^\tau \dot{m}\,\mathrm{d}t = \dot{m}\,\tau$$

For one year, we get
$$ m_{\ce{Cl}}(1~\mathrm{a}) = \dot{m}_{\ce{Cl}}\cdot1~\mathrm{a} = 59.04~\mathrm{kg}\; , $$
which means that your attempt at answering your own question were very good, on spot and totally valid.


Is the pyrolysis of tetrachloroaluminate(III) anion possible?

Unlike lithium tetrafluoroborate $\ce{LiBF4},$ sodium tetrachloroaluminate $\ce{NaAlCl4}$ doesn't decompose upon melting, and doesn't show any signs of decomposition or phase transition up to approx. $\pu{800 °C}$ [1, p. 688]:

FIGURE 3. Calculated $(\ce{NaCl + AlCl3})$ phase diagram at $\pu{0.1 MPa}$ (dotted lines are liquid–liquid miscibility gap boundary and monotectic at $P > \pu{0.1 MPa}).$

Molten sodium tetrachloroaluminate is used as electrolyte in ZEBRA batteries with normal operating temperature range of $\pu{270–350 °C}.$
References

Robelin, C.; Chartrand, P.; Pelton, A. D. Thermodynamic Evaluation and Optimization of the $(\ce{NaCl + KCl + AlCl3})$ System. The Journal of Chemical Thermodynamics 2004, 36 (8), 683–699. https://doi.org/10.1016/j.jct.2004.04.011.


You are correct; the answer in your book is wrong. I get (E) for both alkenes.

For the alkene on the upper right:

$\ce{COOH}$ has precedence over $\ce{CH2OH}$ because the first carbon atom is bonded to $\mathrm{ \{O,O,O\}}$ and the second is bonded to $\mathrm{\{O,H,H\}}$.
Clockwise around the ring has precedence over counterclockwise. While both initially are $\mathrm{\{C,H,H\}}$, the next carbon atom clockwise is $\mathrm{\{C,C,H\}}$ while the next carbon atom counterclockwise is $\mathrm{\{C,H,H\}}$.

For the alkene on the lower left, the nitrile groups win in both cases:

On the left: $\mathrm{\{N,N,N\}}$ versus $\mathrm{\{N,H,H\}}$
On the right: $\mathrm{\{N,N,N\}}$ versus $\mathrm{\{C,C,H\}}$


Andersen thermostat in MD with rigid bond constraints

Hünenberger’s Thermostat algorithms for molecular dynamics simulations (PDF available here) is a generally concise and accurate reference on the topic. On stochastical thermostats such as Langevin, it does not use the number of degrees of freedom (i.e. reduced by the number of constraints) but actually uses the number of atoms.

One would have to follow the proper derivation of the thermostat's target ensemble (from the original Andersen paper) to make sure that is correct, but it seems likely to fix your problem of having too low kinetic energy.


Is it true that the better the trial function, the better approximation for the energy in variational principle?

TL;DR
No this is not true. Although a "better approximation" might yield a lower energy, this is not always the case.
Full answer
There is a major issue in the formulation of your problem: You want

to relate the difference between an approximation $\phi$ and the exact
ground-state wave function ψ0 with the difference between the
approximate energy $E=\int\phi^* H\phi \mathrm{d}\tau$ and the exact energy

The second part of that statement includes a clear mathematical definition of the target property (energy $E=\int\phi^* H\phi \mathrm{d}\tau$). However, there is no clear definition of what a better approximation of the wave function actually is. How do you want to define or measure it, so it can be expressed in a mathematical way?
There a various options to define such a measure, which would allow comparing two trial wave functions and judge their quality. The most obvious and commonly used one is the total energy. But this is the property you want to compare with, so we need something else. Examples for further options are:

overlap with the exact wave function (if available) $\langle \phi|\psi_0\rangle$
expectation values other than total energy, e.g. $\langle r\rangle$ or $\langle r^2\rangle$
any molecular properties of the chosen model system (e.g. dipole moment, excitation energies, etc.)

The next problem is: the choice of the measure for a good approximation might influence the result. For example the overlap $S_i=\langle \phi_i|\psi_0\rangle$ for two different trial wave functions $\phi_1$ and $\phi_2$ (possibly each with a different variational ansatz) might yield $S_1 > S_2$ (meaning $\phi_1$ would be the better approximation), while the total energy of the second trial wave function might be lower (and therefore close to the true ground state): $E_2 < E_1$.
The different measures on how good of an approximation a given trial wave function is do not correlate, therefore there is no general answer which one is actually better. It depends on the property you want to calculate.
There is a even paper available covering this issue and including a more detailed example. Sadly, it is paywalled although it is primarily intended for educational purposes: https://pubs.acs.org/doi/10.1021/acs.jchemed.8b00959

how can I estimate the difference between the approximate energy and
the exact energy?

In essence, this is the central question quantum chemistry is concerned about. And the field developed a whole lot of different methods to achieve reasonable approximations. Thus, I am afraid there is no simple answer to this question.


In charcoal and graphite, conjugated pi systems (the molecule orbital configuration found in carbon-carbon double bonds) act as the conductive band, where electrons can move freely.

In diamond, all bonds are SP3 hybrid bonds (the molecule orbitals found in single bonds). Therefore, no pi systems and thus no conductive band exists.

Edit (after  Jon Custer's comment, thanks): Wikipedia says: Except for most natural blue diamonds, which are semiconductors due to substitutional boron impurities replacing carbon atoms, diamond is a good electrical insulator, having a resistivity of 100 GΩ·m to 1 EΩ·m[28] (1011 to 1018 Ω·m). Thus it is possible to dote diamonds with boron to make them more conductive.


Although "paradox" is not quite the right term, what you have discussed is actually a simple, yet interesting and important phenomenon.

Given the ideal situation as you have presented, your thoughts on what would happen are correct. If the system were to achieve $\pu{100\%}$ humidity with respect to the pure water, that would always be slightly over $\pu{100\%}$ humidity with respect to solution B, so water would condense onto B. This of course would cause the humidity with respect to pure water to drop, so more would evaporate. All things being ideal, eventually all of the water in A would transfer to the solution in B (plus the gas phase). The rate that would happen would depend strongly on the concentration of the solute in B.

An example of this situation playing out in the environment can be seen in the growth of water and ice particles in clouds. The vapor pressure of one particle can differ from that of another due to solute concentration, particle size (small droplets have a greater vapor pressure than larger droplets) and phase (supercooled water droplets have a greater vapor pressure than ice particles at the same temperature). Regardless of the cause of the vapor pressure difference, whether it's solute concentration or one of the other factors, the situation is the same as depicted in your question. And the result is the same; the particles having greater vapor pressure will end up evaporating and then condensing onto the particles having lower vapor pressure (of course we're talking about vapor pressure of water/ice only, not any solute). This is generally referred to as the Wegener–Bergeron–Findeisen process, although this technically only refers to water-to-ice vapor transfer.


Any structure, is, in theory, make able. The real question is whether anyone would want to waste a sufficient amount of time to make it.

The structure you've drawn doesn't seem so unreasonable. With enough time planning a synthesis, and enough hours of graduate student time I see no reason to object


$\ce{NO_x}$ can be quantitatively removed from such combustion streams by absorption/dissolution in basic aqueous solutions containing an appropriate oxidizing agent. While $\ce{NO2}$ is readily soluble in aqueous solution, $\ce{NO}$ is not. However, $\ce{NO}$ is easily converted to $\ce{NO2}$ when an oxidant is added to the solution.

The PhD thesis by Chen-Lu Yang at NJIT, "Aqueous Absorption of $\ce{NO_x}$ Induced by Sodium Chlorite Oxidation", concluded the following:

$\ce{NO2}$ is essentially quantitatively absorbed in $\ce{NaOH}$ aqueous scrubbing solutions.
$\ce{NO}$ is not absorbed by $\ce{NaOH}$ aqueous scrubbing solutions.
$\ce{NaClO2}$ containing aqueous scrubbing medium quantitatively oxidizes $\ce{NO}$ to nitrate [which is highly water soluble].

The net effect of these solubility/reactivity properties is that $\ce{NO}$ and $\ce{NO2}$ can be quantitatively removed from a gas stream using an aqueous solution of $\ce{NaOH}$ and $\ce{NaClO2}$

Additionally, according to this PubChem entry, $\ce{NO}$ is soluble in sulfuric acid, ethanol and carbon disulfide, while according to this Wikipedia article, $\ce{NO2}$ is soluble in carbon tetrachloride, nitric acid and chloroform.


How do I work out the amount of H+ ions in a sodium hydroxide and water solution?

The pH can be directly related to the concentration of $\ce{H3O+}$:

$$\text{pH}=-\log{[\ce{H3O+}]}$$
$$\Downarrow$$
$$[\ce{H3O+}]=10^{-\text{pH}}$$


It will surely proceed as a secondary carbocation and no hydride shift will take place. Because if hydride shift takes place, bridgehead carbon will acquire positive charge i.e. will become unstable.

For detailed explanation check this out:
Why are bridge bonds with negative or positive charge on bridge head unstable?


Correlation between the equilibrium constant and delta G

The first step is simple:

$$\frac{{\ln x}}{{\log x}} = \ln 10 \approx 2.303$$

thus

$$ - RT\ln {K_{{\text{eq}}}} \approx  - RT \times 2.303 \times \log {K_{{\text{eq}}}}$$

The second step is terrible because $\text{kcal}$ is used and the units are missing:

$R = 1.9872041 \times {10^{ - 3}}{\text{ kcal K}^{ - 1}}{\text{ mo}}{{\text{l}}^{ - 1}}$

$T = 310{\text{ K}}$ (my guess)

$$\begin{aligned}
  \Delta G =  & \; {-} RT \times 2.303 \times \log {K_{{\text{eq}}}} \\
   =  & \; {-}1.9872041 \times {10^{ - 3}}{\text{ kcal K}^{ - 1}}{\text{ mol}^{ - 1}} \times 310{\text{ K}} \times {\text{2}}{\text{.303}} \times \log {K_{{\text{eq}}}} \\
   =  & \; {-}1.42{\text{ kcal mol}^{ - 1}} \times \log {K_{{\text{eq}}}} \\
\end{aligned} $$


The concentration is defined as the amount of particles per unit volume. If you have a greater concentration of sulfuric acid, there are more sulfuric acid 'particles' ready to collide and react with the calcium carbonate particles. This is a simple result from the collision theory of chemical reactions.

So, increasing the concentration of the sulfuric acid should increase the amount of carbon dioxide (and other products). However, this is not always the case.

Consider the case where calcium carbonate is the limiting reagent/reactant. This basically means that all of it gets used up before the other reactant gets used up. In a case like this, increasing the sulfuric acid concentration will not affect the amount of products, simply because there's not enough calcium carbonate.

It's like asking if having more flour is going to help you bake more cake. Often, it does. But if you don't have enough of the other ingredients, it doesn't matter how much flour you have! Obviously, this is an incomplete analogy because you could probably still make a cake, albeit a nasty one.


Without addition of chemicals? Try out ultrasound (the same physical principle opticians use to clean your pair of glasses).  If you whish, this may be handled even contact less (some background and here), to mention examples.


Best ways to compress and store tons of CO2?

Poor carbon dioxide has been unnecessarily defamed as a greenhouse effect. This is not only gas in the atmosphere which is an infrared absorber. Guess what, water vapor is a major culprit as well along with methane (recall cows digestive system). Nature has a very delicate balance, carbon dioxide is plant's food as well.

I recall some people doing PhDs on carbon dioxide remediation in chemical engineering. One of options is to pump carbon dioxide into deep rocks with a hope that it will convert into carbonates, mainly calcium carbonate in deep layers. One can then study all types of phase diagram of carbon dioxide etc.

One green solution is a massive drive for the plantation of trees. This option is far better than more expensive options.


Does the triplet sigma state of a diatomic molecule experience spin-orbit coupling?

Yes I would think so because the projection of orbital angular momentum is zero in forming $\Omega=\Lambda + \Sigma$. This would seem to correspond to Hund's case (b) where $\Lambda =0$ but $S \ne 0$, so no spin orbit coupling (unlike Hund's case (a)(c)(d)). The spin angular momentum remains fixed in space and the molecule rotates under it. If $R$ is the whole body (molecular) angular momentum this couples with the spin so that each rotational level splits into $2S+1$ components and the total angular momentum $J$ takes values from $R+S$ to $R-S$. The splitting is scaled with a spin rotation constant rather than spin-orbit constant but in $^3\Sigma$ there is also a $s_1\cdot s_2$ spin-spin coupling term.


How to transform a tertiary alcohol to a carboxyl?

I am not sure that what you claim to want to do is what you actually want to do.

Cholesterol hemisuccinate is prepared from cholesterol, but not by converting the alcohol functional group into a carboxyl group. The alcohol oxygen becomes one of the oxygen atoms in the ester, but not the carboxyl oxygen. Cholesterol hemisuccinate can be prepared in the lab probably from cholesterol and succinic anhydride in the presence of a non-nucleophilic base (like pyridine or triethyl amine), which can act as a proton shuttle. An acidic workup will be needed to remove the base and protonate the carboxylic acid. The scheme below shows this reaction (sans acidic workup). The alcohol oxygen is highlighted in red.

An identical reaction could be done with your molecule, diplopterol. Dissolve both (1 eq) diplopterol and succinic anhydride (1.1 eq.) in dichloromethane or diethyl ether (or any low polarity aprotric solvent you like). Cool on an ice bath. With stirring, add pyridine (1.1 eq.) dropwise. Let the reaction warm to room temperature, probably about 2 hours. The product will likely precipitate as the pyridinium salt. Check the progress by TLC or some other means. The product will probably be more polar than the reactant. Work up by adding dilute HCl (probably 0.1 M). Extract the aqueous layer with your organic solvent of choice. Combine the organic layers. Wash with brine. Dry over magnesium or sodium sulfate. Remove the solvent by evaporation (rotary being preferred), and purify by chromatography if necessary.

Alternatively, if the product does precipitate at the pyridium salt, collect the salt by filtration, wash the precipitate with ether and then with cold ethanol (will wash away the reactants). Then, add the precipitate to the 0.1 M HCl and extract with ether if necessary. Further purification is probably not needed.


Ground state electron configuration of chromium

WebElements page about chromium (and a number of resources) agree with the comment by @Philipp:

  The ground state electron configuration of ground state gaseous neutral chromium is  $\ce{[Ar]}3d^54s^1$

Which in some resources is written as $\ce{[Ar]}4s^13d^5$

Based on the Royal Society of Chemistry article The trouble with the aufbau principle:

  it appears that the most stable configuration for atoms of chromium, copper, niobium, molybdenum, ruthenium, rhodium, silver, platinum and gold involves only moving one electron into an $s$ orbital.

Chromium is one of a handful of Transition elements that share this electron configuration.


Is cis- or trans-cycloundecene more stable?

There is experimental data for the methoxy derivative: https://webbook.nist.gov/cgi/cbook.cgi?ID=C78289131&Units=SI&Mask=8

I think that means the (Z) form is favored. I would think the situation is similar in the absence of the methoxy group.


No, it can't be the same.  If it were exactly the same, then it would mean that ATP was storing every last iota of available free energy.  Another way of saying that is that ATP synthesis would be 100% energy efficient.  That is, the free energy change $\Delta G$ in the reaction $\ce{ADP + Pi + energy -> ATP}$ would be zero.

But perfectly reversible reactions that have 100% efficiency never happen in the real world.  Even theoretically, they can only happen at infintessimally slow rates.

Actual ATP synthesis isn't 100% efficient.  It's efficiency depends on the details of the particular organism and its biochemistry, etc.  Many organisms show efficiencies of 30 - 45% in the synthesis of ATP.


What else other than zinc can reduce Cr(III) to Cr(II)?

Wikipedia describes the reduction of $\ce{CrCl3}$ to $\ce{CrCl2}$ with zinc metal, $\ce{LiAlH4}$$^{*}$, and $\ce{H2}$. You could use any of these methods and stir with $\ce{NaOAc}$ to form $\ce{NaCl}$ and $\ce{Cr2(OAc)4}$:

$$\ce{2CrCl2 + 4NaOAc + 2H2O -> Cr2(OAc)4(H2O)2 + 4NaCl}$$

I'm not sure if magnesium or manganese would work well, but I would assume they would do the reduction. The problem is that you want something that will reduce $\ce{Cr^{3+}}$ to $\ce{Cr^{2+}}$ but not any further. Using something with a more negative reduction potential could cause overreduction. These are good ideas, and I would certainly give credit to my students for proposing them, they just aren't mentioned specifically anywhere I'm seeing.

$^*$You would need to isolate $\ce{CrCl2}$ in the case of $\ce{LiAlH4}$ reduction (or kill off any excess LAH, at least) because you don't want to reduce the acetate added after the reduction.


How do technicians treat spilt mercury and bromine?

Any reputable chemical supplier will normally sell a mercury spill kit, and anyone working with mercury these days should have one. (Either a commercially prepared one, or an equivalent one that's assembled from individual supplies.) Generally speaking, they contain some sort of absorbent/adsorbent to pick up mercury, and a chemical which will react with the mercury and keep it from volatilizing. Sulfur is a common choice (recommended by the EPA), as mercury sulfide is pretty darn inert, as well as being very insoluble in water. (Also, "flowers of sulfur" does double duty as an absorbent.)

After the cleanup, the absorbent is collected (by sweeping - vacuuming can spread mercury vapors), packaged securely, and sent to a hazardous waste disposal company.

As elemental bromine is less widely used than elemental mercury, designated spill kits for bromine are less widely available, but the procedure is generally the same: spread a chemical to react with the bromine and render it more inert, then spread an absorbent to contain the liquid. Elemental bromine is rather reactive (an oxidizer), so a number of common reducing agents can be used. A 5% solution of sodium thiosulfate is a common choice (1) (2).

Again, if you're working with elemental bromine, your lab should have a spill response kit already assembled, with sufficient pre-prepared sodium thiosufate and absorbent to handle the amount of bromine you normally deal with.

The previous was for general information purposes only. If you are looking to work with elemental mercury or bromine, consult with your institution's chemical safety officer before purchasing for more information and restrictions which may apply in your area. Your institution likely also has chemical safety and spill response training, which will inform you of current best practices in handling mercury and bromine spills.

As always with accident handling, the most important thing is not to put yourself in greater danger by attempting to fix it. If the size or extent of the spill is such that you're unsure you can handle the spill safely, the best course of action is to evacuate the premises and let the hazardous materials team do the cleanup. (They'll likely use the same techniques of reacting and absorbing the material, but they'll be wearing more extensive personal protective equipment while doing it.)


Determining structure from C-NMR spectrum, is my answer correct?

It is not always a good idea to focus only on $\ce{^{13}C}$-$\mathrm{NMR}$ when you are doing structure elucidation on unknown compound using spectral data. Yes, $\ce{^{13}C}$-$\mathrm{NMR}$ gives you valuable info, so does FTIR to some extent. Yet, $\ce{^{1}H}$-$\mathrm{NMR}$ gives the most info about the compound as I depicted in following diagram:

All your analysis is correct except for the final conclusion on the compound. It is not 2,3-dimethoxybenzaldehyde as you concluded. But it is 3,4-dimethoxybenzaldehyde. How do I know that? Look at the diagram and analyze the coupling pattern (disregard the small but significant meta-coupling, which give close to $\pu{2 ppm}$ here).

If the compound is 2,3-dimethoxybenzaldehyde, the $\ce{^{1}H}$-$\mathrm{NMR}$ spectrum of it should displays a doublet, a triplet (or dd), and a doublet in aromatic region (~$\pu{6.7-8.5 ppm}$), in that order going from upfield to downfield. However, what given $\ce{^{1}H}$-$\mathrm{NMR}$ exhibit is a doublet, a singlet, and a doublet in the aromatic region (disregarding the meta-coupling). The singlet indicates the presence of isolated aromatic ring proton, and two separate doubelets also indicate the presence of two isolated but neighboring aromatic ring protons. Since three substitution groups are one formaldehylde (-$\ce{CHO}$) and two methoxy groups (-$\ce{OCH3}$), there are three possibilities: The compound would be 2,5-dimethoxybenzaldehyde, or 2,4-dimethoxybenzaldehyde, or 3,4-dimethoxybenzaldehyde:

The possibility of been 2,4-dimethoxybenzaldehyde can easily be ruled out, because it would show the upfield shift of the chemical shift of the singlet corresponding to the proton on position 3 of the ring (been ortho to two -$\ce{OCH3}$ groups), even shielded than the doublet at the $\pu{6.99 ppm}$. This is justifiable conclusion since the resonance structures of 2,4-dimethoxybenzaldehyde (using only 2-methoxy group) indicate that the ortho ($\ce{C}$-3) and para ($\ce{C}$-5) positions will be more shielded than meta positions. Also, with respect to 4-methoxy group, the ortho ($\ce{C}$-3 and $\ce{C}$-5) positions will be even more shielded than corresponding ortho positions  in 3,4-dimethoxybenzaldehyde.

However, distingushing 3,4-dimethoxybenzaldehyde from 2,5-dimethoxybenzaldehyde is tricky. Both would have a doublet at $\pu{6.99 ppm}$ (for the proton at $\ce{C}$-5 for the former and the proton at $\ce{C}$-3 for the latter) since substitution effects are the same for both. Same argument is valid for the singlet (with meta-coupling) at $\pu{7.43 ppm}$ (for the proton at $\ce{C}$-2 for 3,4-dimethoxybenzaldehyde and the proton at $\ce{C}$-6 for 2,5-dimethoxybenzaldehyde) since, again, the substitution effects are identical for both. Yet, chemical shift for the proton at $\ce{C}$-6 position of 3,4-dimethoxybenzaldehyde and that of the proton at $\ce{C}$-4 position of 2,5-dimethoxybenzaldehyde would not be identical. That is because ortho-effect of a substitution group is somewhat larger than para-effect of the same group in general (there are few examples you may find here), regardless of its sheilding or deshielding effect. Thus, the chemical shift for the proton at $\ce{C}$-4 position of 2,5-dimethoxybenzaldehyde should be more shielded than that for the proton at $\ce{C}$-6 position of 3,4-dimethoxybenzaldehyde. My reasoning:

The proton at $\ce{C}$-4 position of 2,5-dimethoxybenzaldehyde should be deshielded by para-$\ce{CHO}$ group (blue shift), while shielded by ortho-$\ce{OCH3}$ group (larger red shift).
The proton at $\ce{C}$-6 position of 3,4-dimethoxybenzaldehyde should be deshielded by ortho-$\ce{CHO}$ group (larger blue shift), while shielded by para-$\ce{OCH3}$ group (red shift).
The both protons at $\ce{C}$-6 position of 2,5-dimethoxybenzaldehyde and at $\ce{C}$-2 position of 3,4-dimethoxybenzaldehyde are deshielded by ortho-$\ce{CHO}$ group (larger blue shift), while shielded by ortho-$\ce{OCH3}$ group (larger red shift).
Comparition of reason 1 & 3 clearly shows these two chemical shifts should be switched (instead of singlet followed by doublet, it should be doublet followed by singlet).

Thus correct resonance order would be given by 3,4-dimethoxybenzaldehyde.


Why does iron oxide form hydrous iron oxide while chromium oxide does not?

TL;DR: Guignet's green is a green pigment with composition $\ce{Cr2O3.2H2O}$.
Long answer
Chromium oxide is largely used as a green pigment in various industrial and commercial purpose as it is the only green pigment which can withstand both acidic as well as alkaline environment. It is also used as permanent paints for cements, distempers and paints. It can also withstand high temperatures, hence used for coloring glasses, enamels and pottery glazes.
Guignet's green or $\ce{Cr2O3.2H2O}$ is the hydrous form of the oxide which is used for similar purpose as the anhydrous form. This pigment becomes anhydrous when heated at 200 °C, hence not suitable for glass and ceramic industries.
It is prepared by mixing 3 parts of boric acid and 1 parts of potassium dichromate which is heated in reverberatory furnace. It becomes a dull red mass initially but swells up releasing oxygen to give a nice green color. It is then boiled with water whereby chromium borate decomposes to hydrated chromium oxide and boric acid. The precipitate is dried and separated while the mixture is washed with mother liquor to evaporate the boric acid.
$$\ce{2K2Cr2O7 + 4H2B2O4 -> 2Cr2B2O6 + 2K2B2O4 + 3O2 + 4H2O}$$
$$\ce{Cr2B2O6 + 3H2O -> Cr2O3.2H2O + H2B2O4}$$
Reference

S. P. Das Gupta & H. N. Bose (1948) A Note on Preparation of Chromic Oxide Green, Transactions of the Indian Ceramic Society, 7:2, 60-63, DOI: 10.1080/0371750X.1948.10877850


How to relate the conductance of an aqueous electrolyte to its conductivity

The answer is quite simple, if you look at the equation carefully. Conductance and conductivity are related by cell constant, $\displaystyle\frac{\text{Area}}{\text{Distance}},$ where the area is the area of the electrodes and the distance is the distance between the electrodes.
$$G = \sigma\frac{A}{l}$$
In conductivity measurements, the cell constant was either given by the manufacturer or it was determined experimentally by making a certain concentration of $\ce{KCl}$, the conductivity of which was well established and standardized.
In short, you cannot convert micromho to conductivity because you do not know the cell constant. And weight by weight conversion to molarity is not that trivial. You need to know exact density of each solution.
Did you see the end pages of the manual from you posted the old graph, there is a useful chart (for future readers).


Neutralization is a reaction between acid and base which results in the formation of a salt and water.
When you have any ionic compound and dissolve it into water, dissociation occurs.
This means that you get a solution of ions.
So if you dissolve acids, bases or salts into water, you end up with a mixture of ions.
Then, more things may happen or not.

In the case of the given reaction, you will get a solution with:

$\ce{Na+ + Cl- + H+ + SO4^{2-}}$

If the system is closed, nothing will happen.
But if the system is open, then $\ce{HCl}$ will evolve because it is a gas.
Since the sulphuric acid is very polar and a liquid, water is not even required.
This is, in fact, the standard procedure to prepare anhydrous HCL in the lab. Just drop $\ce{H2SO4}$ on $\ce{NaCl}$. It must be said that $\ce{NH4Cl}$ works better.


How to know whether to add latent heat when calculating final temperature?

It is less obvious than you seem to assume (say, if water cup in your example were already at $0^\circ$C, then the ice would not melt), but still pretty obvious.

The rule of thumb is:

Assume there is no phase change.
Calculate the final temperature.
Check whether there is a phase change in any of the substances involved as they go from starting to final temperature.
If no, we are done. If yes, assume there is a phase change and go back to step 2.


What happens if you cool water in a container too small for it to freeze?

Good question.  Let's assume the container is infinitely strong, non-deformable, and constant in volume.  Let's also assume that cooling the water is an equilibrium process -- that way, we won't have any supercooling.

At equilibrium, the first tiny bit of ice that freezes will take up more volume than the water it froze from.  This will raise the pressure on the rest of the water.  Eventually the pressure may get so high that additional freezing of more water is not thermodynamically favored.

Of course, as the pressure is raised, even the solid ice compresses a bit, freeing up a bit more volume for the liquid water.  According to this paper from 2004, ice is less compressible than water, so as a starting assumption, it may be approximately true to neglect the ice compression effect.

Figure 4 from that same paper gives the freezing point depression of water as a function of pressure:

To fully answer your question, in addition to that data, an equation that gives pressure as a function of ice volume would also be needed.  If we make the assumption I was talking about above -- i.e. that ice is incompressible, then from the data point that water has a constant compressibility of 46.4 ppm per atm we can come up with a very simple version of that equation.

$\frac{\Delta V_{water}}{V_{water}}=46.4 \times 10^{-6} \times P$, where P is the pressure in atmospheres.

Before freezing of a fraction $X$ of the water:

$$V_{ice} = X V_{tot}$$
$$V_{water} = (1-X) V_{tot}$$

After freezing:

$$ V_{ice} = X V_{tot} 1.11 $$
$$ V_{water} = (1-X) V_{tot} - \Delta V_{ice} $$

Combining those equations, you can get

$$0.11 \frac{X}{k(1-X)}= P$$, where $k$ is the compressibility of water.  If even 1% of the water in the container freezes (and all our assumptions are true), then the pressure will be 24 atmospheres!  Freezing 10% of the water would mean a pressure of 260 atmospheres.  Looking at the chart above, reaching this point would require a temperature of only 271 or 272 K, i.e. only -1 °C or -2°C.  Freezing 45% of the water would reach a pressure of 2000 atm, already off the chart above -- but the temperature required to reach that point would only be 253K or -20 °C, the setting of the average home residential freezer!  ((Of course, at these extreme pressures, (i) ice is actually compressible, and (ii) the compressibility of liquid water is not constant but also a function of pressure, so the calculations would get quite a bit more complicated.))

The lesson is that for even moderate degrees of cooling, you'd need a very, very strong container.


In general, one usually doesn't work directly with single atoms, though there are ways to do that as well, usually involving scanning probe microscopy (basically poking at atoms with very sharp probes to either manipulate them or just image them).

Atoms become much easier to work with when they are are ionized and can be manipulated by electric or magnetic fields (atoms have neutral charge so the Lorentz force can't be used to control them). Even then, most techniques deal with small groups of ions rather than individual ones. Ions are relatively easy to manipulate and we can easily trap them, accelerate them into beams, and focus them. The LHC is, of course, a very impressive (and expensive) example of this. The way it works is that packets of ions are accelerated by electric and magnetic fields to form two ion beams, which are then focused into a very small area and made to intersect, allowing the ions to collide and all the cool particle physics stuff to happen.

Manipulation of ions is also used in countless other scientific applications: semiconductor manufacturing, surface analysis, medical analysis, radioisotope dating, to name a few.


Is There a Difference in the Definition of Complete and Stoichiometric Combustion?

Stoichiometric, theoretical or complete combustion, is the idealized process where a hydrocarbon is completely reduced to carbon dioxide and water, consuming oxygen (or, if other elements are present, they get all also oxidated, like sulphur compounds yielding sulphur dioxide, but as you can see, not to its maximum oxidation state, sulphur trioxide is not produced in real combustions)

It is called stoichiometric for obvious reasons, stoichiometric amounts of the hydrocarbon and oxygen / water carbon dioxide are consumed / generated. It is supposed to be a complete reaction.

But is also called theoretical, because it does not take place like that in real life. In practice, long hydrocarbon chains do not break down completely as efficiently as it's expected, subproducts are obtained, not only carbon dioxide is formed, but also carbon monoxide, unburned elemental carbon and like in every chemical reaction, it does not proceed to completion.

Also, if there is not enough oxygen (not at least an stoichiometric amount), incomplete combustion takes place, yielding almost exclusively carbon monoxide (toxic stuff).
In some cases, such as the fuels combustion processes, another process called pyrolysis, occurs before combustion. In this incomplete combustion, products of pyrolysis remain unburnt and important contaminants are produced, like the product of partial oxidation of ethanol can produce acetaldehyde, which is really toxic (Considerably small LD50) and very very flammable (flash point under 0 Celsius degrees).


From this website:

  Every substance has its own specific heat and each phase has its own distinct value. In fact, the specific heat value of a substance changes from degree to degree, ...

It would mean that a solute in water would change the specific heat from plain water.

My guess is that if the density is greater than water, then the specific heat would be greater (because you are heating the water and the solute). To be certain, you would need to determine it for yourself.


Why is Gibbs free energy more useful than internal energy?

Thanks for the edit - I see now what you meant.

The short answer for why we "need" Gibbs energy is that the internal energy for a spontaneous process at constant temperature $T$ and pressure $p$ does not necessarily decrease. Internal energy is the thermodynamic potential (wiki link) for constant $S$ and $V$ - this makes sense if you think about it:

If the internal energy decreases, then heat must be released by the system, and/or work must be being done on the surroundings. However, real spontaneous processes happen where heat is absorbed and work is not done on the surroundings - the reason is that if the net entropy increase is greater than the enthalpy decrease, the process will still be spontaneous.

The equation for $\mathrm{d}U$ includes entropy terms, but not in a way that reflects spontaneity - the equation for Gibbs free energy does.

The other reason we "need" Gibbs energy is that it also tells us how much "thermodynamically useful" work we can get out of a process.

Mathematical Explanation:

The two inequalities that you started with only represent spontaneous process under very specific conditions:

$$\mathrm{d}U < 0$$

is true for spontaneous conditions only when $S$ and $V$ are constant. The reason for this is that we define spontaneity as the direction in which total entropy of the universe increases - it is at the root of our concept of "spontaneous" vs. "nonspontaneous."

For example, to arrive at this expression for $\mathrm{d}U$, we start with the Clausius inequality:

$$\mathrm{d}S - \frac{\mathrm{d}q}{T} \geq 0$$

This is just a way of saying that the net change of the entropy of the universe must be greater than or equal to zero for any process, and that the net entropy change of the universe is given by the change for the system ($\mathrm{d}S$) plus the change in surroundings, $-\frac{\mathrm{d}q}{T}$.

To turn this into something useful, we make a restriction: constant volume. This lets us assume no $pV$ work, and we can therefore write:

$$\mathrm{d}S - \frac{\mathrm{d}U}{T} \geq 0$$

and

$$T\,\mathrm{d}S \geq \mathrm{d}U$$

since $\mathrm{d}U = \mathrm{d}q$ at constant $V$.

If we then assume no entropy change for the system, we can also write:

$$\mathrm{d}U \leq 0$$

In other words, the restriction of constant $S$ and $V$ are built-in into the inequality. You can substitute in other thermodynamic variables to get $\mathrm{d}U$ in terms of $T$ and $p$, but it doesn't tell you anything about spontaneity under conditions other than constant $S$ and $V$.

Gibbs energy is useful because the inequality $\mathrm{d}G < 0$ is true for spontaneous processes under constant $T$ and $p$ conditions, which as you mentioned, are easy to achieve in a lab.

The other thermodynamic potentials are derived in similar ways - by holding different variables constant and seeing what happens when change in entropy is greater than zero.


Carbon-13 NMR of dimethyl 3,4,5,6-tetraphenylphthalate?

Here's a numbered drawing of the title compound to help us keep track of things.  First thing we notice is that there are 13 different carbon atoms in the structure and we see 13 signals in the spectrum, that's a good start.

Let's start with the easy ones.  The upfield peak at 52.28 is in the expected position for C1, the ester methyl carbon.  Down at the other end of the spectrum, the peak at 168.71 is in the expected position for an ester carbonyl carbon (C2).

In the region for aromatic carbons we see 5 low intensity peaks between 143 and 132.  These belong to the (unprotonated) quaternary carbons 3, 4, 5, 6, 10.  Using diethyl phthalate as a model compound, I would guess that the peak at 132.19 is from C3.  The two peaks at 138.73 and 138.60 have the same height and are closely spaced suggesting very similar environments, I might guess them to be C6 and C10, just not sure which is which.  Noting that the peak for the carbon corresponding to our C5 is downfield of the peak corresponding to our C4 in the model compound, I would expect the same trend to hold here, just with each of the peaks shifted downfield due to attachment of a phenyl group.  So I would assign the peak at 143.27 to C5 and the peak at 139.29 to C4.

This leaves the protonated aromatic carbons for us to consider.  The 2 peaks at 125.97 and 126.83 are half the intensity of the other protonated aromatic signals, therefor assign them to C9 and C13 as there are only half as many of these carbons as compared to the other protonated aromatic carbons. It's not clear which peak belongs to C9 and which to C13.  This leaves us with the 4 signals due to C7, C8, C11 and C12.  Using the same resonance structure - delocalization arguments I made in your benzaldehyde cnmr question from yesterday, I would expect C7 and C11 to be downfield from C8 and C12.  Therefor, I would assign C7 and C11 to the peaks at 129.71 and 130.88, not knowing which is which.  C8 and C12 then belong to the signals at 126.88 and 127.41, not knowing which belongs to which.

Obviously there is some guess work in here and if it was an important project I'd do some further experiments and look for some additional model compounds in order to confirm these assignments and remove the discrepancies, but this is a good start.


Which group is more suitable for metal nanoparticles functionalization?

Sulfur containing groups, such as thiols, are in general more strongly attached to silver nanoobjects than amine groups. However, the problem with sulfur and silver is that the strength and structure of sulfur-silver bonds depends critically on the local topology of the silver surface, and in-layer lateral interactions between stabiliser molecules can have a strong impact on this, so there isn't a clear pattern of reactivity of sulfur-containing groups and silver nanoobjects. You can easily get irregular, disimilar or patched coverage.

Amine is much more straightforward, links itself more weakly and is routinely used in the synthesis of silver nanoparticles as a stabiliser. However, if you want permanent functionalisation of the nanoparticle (vía covalent bonding), probably amine-silver bonding will be too labile.


Reaction of citric acid and calcium chloride

This procedure is done in our laboratory class, but first you have to add sodium hydroxide until the lemon juice is somewhat alkaline. Then add the calcium chloride (use $1~\mathrm{ml}$ for every $2~\mathrm{ml}$ of the lemon juice + $\ce{NaOH}$ solution).
I think the calcium chloride would first react with the hydroxide, producing calcium hydroxide, which would then react with citric acid to form calcium citrate.

$$\ce{2 C6H8O7 + 3 Ca(OH)2 -> Ca3(C6H5O7)2.4H2O + 2H2O}$$

Filter out the salt (precipitate) formed, weigh it, then you would have to add sulfuric acid afterwards to react with the calcium citrate. This would then form citric acid (soluble) and a precipitate which is calcium sulfate.

$$\ce{Ca3(C6H5O7)2.4H2O + 3H2SO4 -> 2C6H8O7 + 3CaSO4.2H2O + 2H2O}$$

You would then have to concentrate the filtrate to obtain the citric acid crystals.


Is the ethyl cation really more stable than benzylic and allylic carbocations?

I believe this to be a simple typo in the book. In the original image you'll see that in the surrounding schemes $<$ and $\to\text{Stability}$ are given. Only in this line the relations are reversed, while the stability indicator remains the same.

However, since we're at it, we can at least try to put some values to the task. I'll adopt the same scheme used in the comparison of the t-butyl cation and the benzyl cation, or the radicals. I calculated the isodesmic reactions of the form in $\eqref{isodesmic}$ at the DF-B97D3/def2-TZVPP level of theory.
$$\ce{ R+ + CH4 -> RH + H3C+ }\tag{1}\label{isodesmic}$$

According to this we'll find the following order in decreasing stability: t-butyl, benzyl, allyl, (ethyl). Note that the ethyl cation is a non-classical cation, basically a proton coordinating to the π-bond of ethene; see also: Which carbocation is more stable, the ethyl- or 1-propyl-carbocation?.

\begin{array}{llr}
  \ce{R+}  &  \ce{RH}  &  \Delta G / \pu{kJ mol-1}\\\hline
  \ce{H3C+}           &  \ce{CH4}         &      0.0  \\
  \ce{[H2C=CH2]H+}    &  \ce{H3C-CH3}     &   -197.9  \\
  \ce{H2C=CH-CH2+}    &  \ce{H3C-CH=CH2}  &   -259.3  \\
  \ce{H5C6-CH2+}      &  \ce{H2C-C6H5}    &   -343.0  \\
  \ce{(H3C)3C+}       &  \ce{HC(CH3)3}    &   -370.9 \\\hline
\end{array}

(Side note: I was unable to remove a small imaginary mode from the isopropyl carbocation. That has almost no influence though.)

(I won't attach geometries or absolute energies this time, because that would exceed the character limit.)


What are the units that measure free energy?

There are several different free energies, but I'll tackle Gibbs and Helmholtz only here:

Gibbs free energy: $G=H-TS$, for $H=U+PV$. Hence, by dimensional analysis, Gibbs free energy has the SI units of Joule.
Helmholtz free energy: $A=U-TS$. Again, by dimensional analysis, Helmholtz free energy also has the SI units of Joule.

Since both are only "free energies", it makes sense to assign them the SI unit Joule.

That said, however, almost everywhere they are specified in terms of $\pu{kJ/mol}$ because that makes the comparison across different substances easier. Since $G$ is an extensive function, its value can vary even with the amount of substance. Your calculated $\Delta G_\mathrm f^\circ$ can differ from mine even if we take the same substance, just because we had the amounts different. Hence, to ease comparison and computations, they are "specified" in terms of $\pu{kJ/mol}$, although their correct SI unit is Joule.


Why is H2SO4 a good dehydrating agent while other strong acids are not?

As pointed out by Maurice and myself, the argument provided in the answer by iad22agp is rather incorrect. Sulfuric acid is a dehydrating agent just because it is available in concentrated form whereas rest of the common acids like HCl(aq) are not. No, it is not a concentration effect. First of all, it is not possible to have HCl(aq) more concentrated than 38% wt/wt under normal conditions. This is already a saturated solution of the gas.
Secondly, liquified HCl cannot be a good dehydrating because as everyone knows HCl g fumes heavily in the presence of moisture. It cannot fulfill the criterion of a dehydrating or dessicant. The same arguments apply to conc. nitric acid as well.
The two main reasons which make conc. sulfuric acid a good dehydrating agent are (a) it has a very low vapor pressure and (b) secondly the hydration of sulfuric acid is quite exothermic as compared to other common acids. Maybe a quantum mechanical study reveal "why" this is so. Acid strength alone cannot be used to predict the dehydrating capability. One has to think about the vapor pressure as well.
General chemistry book also preach that sugars are dehydrated by conc. sulfuric acid. It is actually a wrong notion to say that sulfuric acid extracts water out of sugar. There is no water molecule in sugar. This is a chemical reaction between sugar and conc. sulfuric which generates water and carbon. The true demo for sulfuric acid dehydration is the dehydration of hydrated copper sulfate. Adding a few drops on the blue crystals makes them white (anhydrous) copper sulfate almost immediately.


Do the π-electrons from the C=O bond in tropone contribute to Hückel 's rule?

In addition to Philipp's answer I'd like to share some orbital pictures which shall make the case rather obvious.

Let me shortly remind you of the key part of the IUPAC definition for aromaticity:

  In the traditional sense, 'having a chemistry typified by benzene'.

I will therefore follow the same approach as in my answer to "Is phenoxide aromatic?" in comparing the orbital pictures.

The following orbitals are for the π system of benzene, in increasing order of energy (DF-BP86/def2SVP). The two orbitals on the right are degenerate:

Analogously, the π system for cyclohepta-2,4,6-trien-1-one is extensively delocalised and holds eight electrons. Note that the symmetry of this molecule is only $C_\mathrm{2v}$ removing a lot of constraint from the electronic structure. This is also why the Hückel rule breaks down, since there are no more degeneracies observed.

Also note, that the LUMO of this molecule (pictured below) still has considerable bonding character and that reduction would most likely not destroy aromaticity. (The same statement also applies to oxidation.)


What material should I mix with graphite to make a brick?

Clay would work, yes. It is, after all, dirt. Adding graphite to it would only be changing the composition of that dirt.

Typically, you'd only need any sort of binding agent: If you're willing to make a weak brick, you can add some cornstarch and literally press it into a brick. You could also try things like epoxy and acrylic compounds, or things like plaster.

If you wanted to keep the composition of the brick to be mostly carbon, you could consider using chemicals like bitumen and tar, although these bricks would be more flexible (depending on the graphite to whatever ratio).


Why does mixing inks make subractive colors?

The issue is that you're thinking about inks reflecting color, when in reality you should be thinking about the inks absorbing colors.

If the function of inks were to reflect colors, then by adding more and more ink (of a single color) you'd be able to make brighter and brighter colors, as more ink means more reflection. Instead, it's the white paper (rather than the ink) that's reflecting the light, and more and more ink (even of a single color) makes things darker because more and more light is being absorbed.

That's why mixing pigments/inks is a subtractive process - as you add inks, you increase the amount of light that's being absorbed.

So let's go to your example. Let's simplify things by considering only three wavelenghts of light: red, green and blue. Your red ink absorbs the green and blue light, whereas the green ink absorbs the red and blue light. Thus when you mix the two, each absorbs the light it would in isolation, and you have very small amount of blue light, and also very little red or green light, leaving you with a very dark, perhaps slightly yellowish-brown color.

Now consider subtractive inks. You have yellow ink, which absorbs just the blue light and you have magenta ink, which absorbs just the green light. You mix the two together, and the magenta ink absorbs the green light, the yellow ink absorbs the blue light, and the red light is free to be reflected, resulting in a red colored piece of paper from Y+M.

Generally speaking, modern inks (at least the standard, non-color changing variety) involve no chemical reactions in color production: mixing inks and pigments is solely a physical change based on absorption of light.


As this is clearly a homework question, I will only point you in the right direction.

Balance the oxidation reaction:

$$\ce{S2^2- -> 2 SO2 + $n$e^-}$$

Balancing the charges will give you $n$. (I'll leave the balancing to the OP, as this is homework).

The equivalent mass of $\ce{SO2}$ in this reaction is then:

$$N=\frac{2M}{n}$$

The factor $2$ is needed because the electronically balanced reaction has two molecules of sulphur dioxide in the equation.


Explaining the result of experiment---Aluminum foil heated in boiling KMnO4 solution

Potassium permanganate solution tends to decompose with time, especially those which are not acidified. Just let an aqueous solution sit in a glass bottle and it will leave a brown film of tenacious manganese dioxide on the surface at room temperature. If water has any organic material, it will become brown and tap water is not pure water. It decomposes on the skin as well leaving a brown stain. You Al surface is no different, with a thin deposit of manganese dioxide, it may appear yellowish-golden.

Long time ago potassium permanganate was added to water wells to disinfect water. Of course it would decompose with all the organics. If you are well after three years, what is the problem?


What is the oxidation state of all the atoms in this compound, W(CH3CN)3(CO)3?

HINT:

Compound is neutral, So the sum of oxidation state of all the element should be ZERO.

In what you have got, is sum of oxidation state of all the element is zero? This way you can check whether you are right or wrong.


Does it make a difference in energy if a reaction is done in many steps or in one step?

You might want to take a look at Hess's law. It states that the total enthalpy change during the complete course of a reaction is the same whether the reaction is made in one step or in several steps. But that covers only the thermodynamic aspect. In reality, you also have kinetic barriers (activation energy) which add to the energy needed for a reaction to run (or released during a reaction). That's why catalysts are so useful: They break a reaction down into steps (involving different intermediates) with lower activation barriers than the original reaction so that the reaction as a whole can proceed at a lower temperature.


What constraints are imposed on a wavefunction by the symmetry of the system?

I use quantum chemistry about as much as a lumberjack uses math (that is, just enough not to fell a tree on one's own head), so my answer will be short and hopefully useful as a tl;dr to someone else's more comprehensive take.

In quantum chemistry, if two operators commute, their values are simultaneously observable. In other words, there is a basis of functions which are eigenfunctions to both operators at once. For an opposite example, think of coordinate and momentum; they do not commute, hence the Heisenberg uncertainty and stuff.

When our problem has some symmetry (including, but not limited to, geometrical symmetry like $T_d$), this means that symmetry operators commute with its Hamiltonian. Therefore the eigenfunctions of the latter must also be eigenfunctions of the former. In other words, the application of symmetry operator to any of our states is equivalent to a multiplication by some number. That's the constraint you were after. Since a symmetry operator of a finite point group must have finite order (by applying it a few times, you get back where you started), the said number must be a root of unity.

That's about the size of it. The numbers for different symmetry operators can be chosen in a multitude of ways, each corresponding to some representation. The character tables tell you how many functions of each representation do you have. Since the functions of different representations are orthogonal to each other and don't interact, this helps us to split a huge problem into smaller chunks.

With methane, the qualitative result could have been envisioned in a simpler way, almost without any calculations. Think of a carbon atom with four $sp^3$ orbitals. They are all equivalent and orthogonal to each other. Now bring in the four H atoms. This will produce four $\sigma$ orbitals (and $\sigma^*$ too, we just don't care about those at the moment). But wait, those $\sigma$ orbitals are no longer orthogonal, so they must interact, if only a little. And because of symmetry, each two of them overlap by exactly the same amount. So it is as simple as Hückel can be: find the eigenvalues of the matrix $$\begin{pmatrix}
 0&-1&-1&-1\\
-1& 0&-1&-1\\
-1&-1& 0&-1\\
-1&-1&-1& 0
\end{pmatrix}$$
The symmetry approach I outlined above still applies; it is just that the system became so small that one can manage without it. The thing can be solved by hand. Or it can be noted that we're dealing with a complete graph, and their spectra are well known (just remember that math people have it upside down, for they use $\mathbf1$s instead of $\mathbf{-1}$s). Either way, the eigenvalues are $\{-3,1,1,1\}$, which kinda explains the 1:3 thing.

So it goes.


How were weights of elements and compounds determined?

If you check out the Wikipedia page on Properties of Water, it has a very small history section which says Cavendish (not a chemist, arguably) determined the composition of water in 1781. Then following that lead to Cavendish's article, you'll learn quite a lot, unless you already knew it. As far as us putting "the complete picture" in front of you on a plate, yeah, that's not going to happen. We'd have to know what your background was, and even then what exactly would satisfy your concept of what the "complete" picture was. The up-side is that although satisfying yourself is up to you, the web has a lot of resources you can get help from. When it comes to a lot of scientific, mathematical, and technical subjects, though, you'll have to put in the study (and practice) time to learn and understand it, course work and textbooks will be difficult to replace with unorganized web research. OK, enough philosophy. I encourage you to put together a time-line of alchemy, natural philosophy, and science at around the start of the 19th Century. You probably know that the idea that matter is composed of small indivisible particles (atoms) was first seriously considered by the ancient Greeks, right? And that various technologists have used recipes (based on weights or volumes or both) to repeatedly create (or destroy) various materials of interest, from time immemorial. Alchemists were dedicated transmutation, so what they knew and when they knew it is important here. Proust didn't demonstrate his law of definite proportions until almost 20 years after Cavendish discovered the constituents. So first we knew what it was made of (oxygen and hydrogen, although the concept of elements was not yet firmly established.) And see also https://en.wikipedia.org/wiki/Oxygen#Early_experiments as well as the following discussion on the Phlogiston theory. Probably the reason you haven't gotten a clear answer here is because this isn't a technical question, it is a historical question. To get a clear complete "picture" you need to literally waste your time learning about all sorts of discredited theories (Phlogiston, and the earlier even more unfortunate Earth, Air, Fire and water "theory") and then learn how the various people overcame the blunders made before them to find the "right" explanations for the world around them. Needless to say, that is an ongoing effort but to recapitulate all of the dead-ends isn't something we're going to do in a short answer, frankly. Anyway, put together a time-line and you'll begin to see how the various strands of the story wove together. HTH.


How to get meniscus down to the calibration mark with a volumetric pipette

If you have to stopper the pipette with your finger, it just takes practice. I believe the proper way to do it is to just roll your finger to the side very slowly, but I've heard several strategies including rotating the pipette like you said, and rocking/sliding your finger back and forth.

It is a lot easier if you have access to pipette bulbs like these, which let you squeeze the side arm to expel liquid smoothly:

These are easy to use too - you just roll it down to expel the liquid:

Most labs should have either of these, especially university ones. Maybe less common in high schools.


Differentiating Al2O3 from SiO2 using chemical methods

Although not a chemical method, using flotation in bromine to separate your mixture will work at room temperature. Unlike the chemical methods given in other answers, this physical method of separation does not destroy or otherwise degrade the components of your mixture, so you can recover them.

The density of bromine at 20 $^\circ$C is 3.1 g$\cdot$cm$^{-3}$; the densities of $\ce{Al2O3}$ and $\ce{SiO2}$ at the same temperature are 3.95 g$\cdot$cm$^{-3}$ and 2.65 g$\cdot$cm$^{-3}$, respectively. The $\ce{SiO2}$ will float and the $\ce{Al2O3}$ will sink.

As noted in the comment thread above, this method will work unless the particle size is too small.

And as noted in the comments below, sodium polytungstate solution (with a density of 3.1 g$\cdot$cm$^{-3}$) is an alternative to bromine.

Because there are likely a large number of substances with a density in the range required to separate your mixture, the suggestions here represent a small number of those choices.


Why did permanent marker come off of glass after rubbing it with my fingers?

Glass is a very smooth surface and therefore things adhere to it less effectively such as the ink in your permanent marker. Maybe your marker is not of a very good quality or maybe you didnt leave it long enough for the ink to stabilize on the surface.

Why dont you conduct an experiment to try your various hypotheses by varying one factor while keeping the rest constant? For exaple:

1) Try different brands of markers on the same glass surface leaving them to dry for the same time and then rubbing them in the same way.
2) Use the same marker but leave it to dry for different amount of time and then rub it in the same way each time.
3) Use the same marker on different surfaces and rub in the same way.
4) Try with different amounts of dilute saline solution to simulate sweat or cooking oil to simulate skin oil etc.
etc.

Or even better: Try at glass surfaces that have been etched with different sandpaper grits.

Your experiment is reproducible though, i tried as well and small amounts come off easily.


Nicotine concentration is a serious concern, as the material is a serious poison (dose matters).  Liquid Chromatography/Tandem Mass Spectrometry is a little pricey.

http://oxfordchemserve.com/e-cigarette-refill-nicotine-quantitative-test-kit/
  Specific test kit "designed with the non-chemist in mind"
http://oxfordchemserve.com/wp-content/uploads/products_img/Oxford_Nicotine_Test.pdf
   Explanation - it is a pH titration of nicotine base.
http://www.eliquidtest.com/
   Another titration kit.  Google for the product spectrum.

Saliva or urine nicotine test kits detect nicotine persistent metabolite cotinine and are therefore irrelevant.

http://www.medpagetoday.com/Pulmonology/Smoking/44238
Warning about E-liquid refills.
http://www.veppocig.com/product_images/uploaded_images/eliquid-nicotine-strength-guide-2.png
Claimed dosage guide.  Caveat emptor.
http://www.learn.eversmoke.com/nicotine-strength.html
A second dosage opinion.

Glycerin can pyrolyze to acrolein.  This is not a good thing to inhale.  NEVER inhale butter flavor.  Butanedione destroys lungs: "popcorn lung," bronchiolitis obliterans, obliterative bronchiolitis, constrictive bronchiolitis are life-threatenting, cumulative, and irreversible.


Difference between tendency of benzene and thiophene to undergo sulfonation

In the chapter Aromatic five-membered ring heterocycles with one heteroatom in Organic Chemistry by  J. William Suggs (2002) the following explanation can be found (pp. 403-404) which summarizes the thoughts already pointed out in the comments:

  The resonance stabilization energy of benzene is greater than that of these heteroaromatic compounds.
  The order of aromaticity is benzene > thiophene > pyrrole > furan.

  […]

  All three of these ring systems undergo electrophilic aromatic substitution and are much more reactive than benzene.
  In part, this reactivity difference arises because the rate-determining step in electrophilic aromatic substitution is the first step, which breaks up the aromatic $\pi$ system.
  Since thiophene, pyrrole, and furan have less stabilization to lose than benzene, the intermediate is lower in energy and the overall reaction proceeds more rapidly.

  […]

  All three of these heteroaromatic rings undergo electrophilic aromatic substitution, preferentially at C-2.
  The reactivity order is pyrrole > furan > thiophene because of several factors, including the electronegativity of the heteroatom and the resonance stabilization of the aromatic ring.


Is it possible to use glycerol DNA/RNA drying during extraction?

if glycerol residues stay in DNA, it might interact with later PCR. Overdrying is an underestimated issue and often DNA resuspension is not only long (>3h) but also very partial even after one night resuspension. it is not possible to evaporate ethanol first but reaching the point where sample are still wet but ethanol mostly removed is the best option


formation of a compound due to photo-isomerisation

Stilbene has been extensively studied with pump-probe absorption spectroscopy as well as time resolved fluorescence.

In the ground state (S0) there is a large activation energy to go from trans  to cis and vice versa so that both species are stable at room temperature. This is not the case in the first excited state (S1)and the trans converts to the cis in a few tens of picoseconds and it is even faster for the cis excited state to convert to the trans.
The potential energy scheme is shown below (Not to scale). On the left is the trans species with a large barrier to for the cis (on the right). This has a higher energy than the trans but also has a large barrier to form the trans.
Detailed analysis suggest that the reaction occurs by a lengthening of the double bond before twisting. Exciting the trans puts it into the first singlet excited state (S1) where there is a small barrier on the pathway to cis, but as it happens there is an intersection with the ground state (by 3ps label & twisted species) and from here either the ground state trans or ground state cis can be formed. If the cis ground state is excited then there is effectively no barrier to form the trans ground state and the excited cis molecule lasts for only a few ps.

For a review or organic photochemistry calculations see Bernardi et al. Chemical Society Reviews, p321, 1996, in particular fig 1. A lot of the original experimental work was done by the Hochstrasser group, see Annu. Rev. Phys. Chem. 2006. 57:37–64 for many references.


What is the purpose of froth in washing clothes?

The froth has little or no effect on the detergent action. In fact detergent manufacturers have to add anti-foaming agents to stop excessive foam generation in automatic washing machines.

Froth/foam is generated because surfactants in the detergent adsorb at the air water interface and stop the water film that makes up the walls of the bubbles from collapsing. Note that the surfactant does not cause bubbles to form - bubble formation is always energetically disfavoured and the foam is only kinetically stabilised. The foam is caused by agitation entraining air into the surfactant solution so the amount of foam will depend on the intensity of the agitation as well as the type and concentration of the surfactant.

It is conceivable that in some cases the foam may assist with dispersal of mineral soils. This is the basis of the industrial froth flotation process. However this plays at most a very minor role in domestic washing machines. Not that it's directly relevant, but I mention it out of interest, foam is thought to play a role in cleaning teeth i.e. in toothpaste. It entrains particles of food so they can be spat out along with foam.


what is the difference between how taste and smell are detected?

Taste receptors work with water (or saliva, or whatever other liquid that happens to occur in your mouth) and the things dissolved therein.

Smell receptors work with air and airborne chemicals.

Short of that, they are similar in that both are wide-range chemical sensors, and different in that they send different signals to our brain.


$\ce{Cu2Cl2}$ is actually the dimer of cuprous(I) chloride $\ce{CuCl}$. Cuprous chloride dimerises and also trimerises and tetramerises in gas phase. The molecular structures of the species are measured with Gas-Phase Electron Diffraction and Quantum Chemical Calculations. Some information about the various species of cuprous chloride is given here:[1]

  The molecular geometry of gaseous cuprous chloride oligomers was determined by gas-phase electron diffraction at two different temperatures. Quantum chemical calculations were also performed for $\ce{Cu_nCl_n (n=1–4)}$ molecules. A complex vapor composition was found in both experiments. Molecules of $\ce{Cu3Cl3}$ and $\ce{Cu4Cl4}$ were present at the lower temperature (689 K), while dimeric molecules ($\ce{Cu2Cl2}$) were found in addition to the trimers and tetramers at the higher temperature (1333 K). All $\ce{Cu_nCl_n}$ species were found to have planar rings by both experiment and computation. The bond lengths from electron diffraction ($r_\mathrm g$) at 689 K are 2.166±0.008 Å and 2.141±0.008 Å and the $\ce{Cu-Cl-Cu}$ bond angles are 73.9±0.6° and 88.0±0.6° for the trimer and the tetramer, respectively. At 1333 K the bond lengths are 2.254±0.011 Å, 2.180±0.011 Å, and 2.155±0.011 Å, and the $\ce{Cu-Cl-Cu}$ bond angles 67.3±1.1°, 74.4±1.1°, and 83.6±1.1° for the dimer, trimer, and tetramer, respectively.

Reference

(1) Hargittai, M.; Schwerdtfeger, P.; Réffy, B.; Brown, R. The Molecular Structure of Different Species of Cuprous Chloride from Gas-Phase Electron Diffraction and Quantum Chemical Calculations. Chem. Eur. J. 2003, 9 (1), 327–333. DOI: 10.1002/chem.200390027


Why hyperconjugation can't stabilize carbanion?

In carbanion the carbon carrying negative charge contains one $\mathrm{sp^3}$ hybrid orbital which contains a pair of electrons and therefore there is no possibility that it will undergo bonding interaction with the neighbouring $\ce{C-H}$ bond, due to repulsion between the electrons.


According to my opinion i think you are wrong because the sulphate ion does not enter the other compartment. This is because the function of the salt bridge is to maintain electrical neutrality by either allowing the sulphate ion to reside in it or by allowing the cations inside the salt bridge to enter the compartment so as to balance the ions.


Given the volumes: determine the pH and the final temperature of a mixture knowing only the initial pH and the temperature of the un-mixed components

For the temperature, this method of calculation will give you a rough estimate of the final temperature, but it will not be exactly right. The reason is that mixing two solutions comes at an energy cost/gain due to the interaction of the two solution molecules. For some mixtures this enthalpy of mixing yields a temperature rise (i.e. energy is released from the molecular potential energy), for others it yields a temperature decrease (i.e. energy is taken up to 'make the mixing possible'). For many mixtures this difference is not more than a few degrees, which makes your linear estimate reasonable, but it is good to keep in mind that it will not be exact.

For the pH, the shortcut you are taking is incorrect and will at best give you a very crude estimate. The reason is that the pH scale is logarithmic, it is correlated with the concentration of protons in the solution as $\ce{pH}=-\log_{10}\; [\ce{H+}] $, where $[\ce{H+}]$ is the proton concentration. (I will make the assumption here that you have simple liquids, which are not buffered.) Therefore, you would first need to calculate the proton concentration in both solutions from this equation, yielding $10^{-3}\,\ce{M}$ and $10^{-5}\,\ce{M}$ for sols A and B resp. This means that the amount of $\ce{H+}$ in sol A is $2\cdot10^{-3}\,\ce{mol}$ and that in sol B $10^{-5}\,\ce{mol}$. Mixing the two you get 3 liter of solution with $2.01\cdot10^{-3}\,\ce{mol}$ protons, which is $6.7\cdot10^{-4}\,\ce{M}$ which works out to $\ce{pH=}3.17$. Indeed, very far off from the 3.67 predicted in your linear method.

Short discussion on the pH: because the scale is logarithmic and your 2 $\ce{pH}$ values are actually not that far from each other the linear estimate is still ok-ish. If, for example, you would have used $\ce{pH}$ 0 and 7 you would have found $\ce{pH}=0.17$ vs. $2.33$ from your estimate.


How does water freeze and crystallize on a cold superhydrophobic surface?

As acknowledged in your question, there are a large number of factors at play here. The most basic of these is that there are three fundamental mechanisms by which freezing of a droplet can initiate when in contact with a superhydrophobic surface: 1) heterogeneous nucleation where an ice freezing nucleus (IFN) is present in the water droplet, 2) heterogeneous nucleation where the superhydrophobic surface serves as the IFN, and 3) homogenous freezing of the water droplet.

An excellent demonstration of the first or third scenario above (it is not clear from the article which is actually the case) is given in video format from an article published in Nature$^1$. In this experiment, the water droplet is evaporatively cooled until freezing initiates at the water/air interface, and continues without adhering to the aluminum-based superhydrophobic surface.

Whether or not freezing will initiate at a superhydrophobic surface depends upon the degree of supercooling, the presence or absence of potential IFN within the water droplet, and the potential for the specific material (the superhydrophobic surface) to serve as IFN. It would seem that the latter of these would be unlikely, except that a surface being hydrophobic doesn't mean that it is equally icephobic. Basically, the more the surface resembles the crystalline structure of ice, the more likely it is to act as an IFN.

Please don't hesitate to ask for any clarifications in the comments below.

1) Thomas M. Schutzius, Stefan Jung, Tanmoy Maitra, Gustav Graeber, Moritz Köhme & Dimos Poulikakos, Spontaneous droplet trampolining on rigid superhydrophobic surfaces, Nature 527, 82–85 (05 November 2015) doi:10.1038/nature15738


Can a reference electrode be inserted into an electrochemical cell horizontally?

As long as there's nothing about the design of the reference electrode (RE) that will make it malfunction when oriented horizontally, it should be fine.  Obviously if there's free liquid inside the RE that will slosh away from the frit or the electrode internals (or spill out onto the lab bench!) when you tip it, then that won't work well at all. The concern about bubbles collecting on the bottom surface is very real, especially if you're not stirring very vigorously.

One practical downside to inserting the CE directly like this is that the frit of the RE will be liable to clogging/degrading much more rapidly than if you use a Luggin. Also, the composition of the internal electrolyte of the RE will change more rapidly if directly exposed to your working solution without a Luggin.

Of course, depending on your setup, this will allow you to get your reference much closer to your working electrode, reducing the $iR$ drop in your measurement.


Generating molecular graph Laplacian based on Hückel theory

You're describing simple Hückel theory. This is usually applied to $\pi$-conjugated systems to understand the stabilization in conjugated and aromatic molecules.

There are a variety of Python packages, for example:

Felix Plasser wrote one for his blog and the current location is at his homepage
Randle Taylor has a SHMO package on GitHub, including a GUI called Orbis

I believe the version from Plasser will read in files using Open Babel's python interface, so certainly you could handle SMILES.

People would perturb the $\alpha$ (atom site energies) and $\beta$ (bond interaction) parameters in various ways. It's hard to find good tabulations of these, but the best appears to be:
"A brief review and table of semiempirical parameters used in the Hueckel molecular orbital method" J. Chem. Eng. Data 1967 122 pp. 235-246

You mention the somewhat related Extended Hückel theory (EHT). This considers both $\sigma$ and $\pi$ bonds and is thus a lot more useful, although as you mention you need coordinates to calculate overlap values, etc.


Can submicron mould particles go through building structures, non permeable foil paper?

You are asking two different questions

  I read that submicron mould particles can go through walls, floors and
  ceilings. How so aren't these structures meant to be air tight?

Not true. Mould particles can not seep through concrete walls or thick wood, even though they are porous structures. The interstitial pore size is too small and tortuous to let any mold particle pass through it randomly.

  What about the lowest thickness of foil paper which is meant to be non
  permeable, can submicron mould particles go through that?

No, an aluminum foil is impervious to solids no matter what their size is. Yes, probably electrons and X-rays will pass through but nothing particulate.


Is it possible to freeze water by dissolving a salt?

It is possible, in a significantly different way that you envision. Freezing point depression via the cryoscopic constant is an example of a colligative property, which holds only for relatively dilute solutions. Once the solutions get very concentrated, intermolecular interactions become more complicated and do not generalize, meaning different compounds will affect the solvent differently.

Therefore, with enough of the right solute, it is possible to actually increase the freezing point of water. Oscar Lanzi's example of clathrates is interesting, but if you want to use salts, then I present you tetrabutylammonium hydroxide. It is known to form stable hydrates with a defined composition and containing a large amount of water, such as $\ce{(C4H9)4N+OH^-.30 H2O}$. The 30-hydrate is a solid which melts at approximately 30 °C, containing 67.6% water by weight.

This specific compound, at this specific concentration, forms a particularly stable network of solvating water molecules. You can think of it as stabilizing the structure of ice, allowing it to occur above the normal melting point of pure water. Thus, in principle, if you add enough of the anhydrous salt to water, when you reach the right ratio, the solution will "freeze".

Many compounds actually display this behaviour of forming solid hydrates at certain defined compositions, but tetrabutylammonium hydroxide is unusual in that it forms solid hydrates with a very large amount of water.


Apparent paradox in the formation of ice at room temperature

Now that you've got $K_{eq}$, you need to take a look at the expression for the equilibrium constant, which is $$K_{eq} = \frac{a_{\text{ice}}}{a_{\text{water}}},$$ where $a$ is the activity of each species.
For a pure liquid and a pure solid, the activity is defined as 1, so the expression on the right hand side of that equation (usually referred to as the reaction quotient $Q$) can only take on three values:
1 if ice and water are both present
0 if water is present but not ice
undefined if water is absent, whether or not there is ice.
So if a mixture of ice and water is put in a room temperature environment, the ice melts because the reaction quotient of 1 is greater than the equilibrium constant of 0.8. Only when all of the ice melts and it makes the step change to $Q=0$ does it get below the equilibrium constant. Then we have a situation where anytime any infinitesimally small amount of ice crystallizes, then $Q$ exceeds $K_{eq}$ and the crystal returns to liquid form.


What determines how much water of crystallisation a salt has?

The water of crystallization in a crystal of any ionic salt can be associated with the size of the cation, magnitude of charge on it, tendency for co-ordination and hydrogen bonds, its lattice structure and other factors involving the anion. But we cant actually theoretically determine why a particular salt has these man water of crystallization. We determine this experimentally in the labs.Then after finding this data for a set of salts we try and link them with the above mentioned factors and try to give a suitable explanation which justifies real data to the greatest possible extent. A theoretical explanation would be very complex and definitely involve way more parameters than what I mentioned above to be accurate for max cases.Even there isn't any 100% full-proof theory or mathematical formula that would allow us to calculate the water of crystallization in any random salt.


Do vinyl cations adopt a classical or non-classical structure?

TL;DR: there have been many theoretical investigations of the relative energies of these two forms for the parent vinyl cation with more recent work indicating that the bridged form is slightly more stable (by about 1–3 kcal/mol)4,5. This prediction has received support from recent experimental work as well6.

Taking such relatively small energy differences into account I would not be surprised if the situation is reversed with a particular substituent. Thus, answering the second part of the question ("Does it depends on what the substituents are?") requires high-level calculations for a wide variety of different substituents which seems to be absent in literature yet.

Classical structure: trigonal (sp2) or linear (sp)?

First note that (give Marko a credit for paying our attention to that), as it is mentioned in Stang's "Vinyl Cations" book, the bent $\mathrm{sp^2}$-hybridized structure with a $\ce{C=C-H}$ bond angle of 120° (OP calls it "classical") is about 50 kcal/mol higher in energy than the linear $\mathrm{sp}$-hybridized form in all of the theoretical calculations.

  These calculations confirm the organic chemist's intuitive and
  qualitative feeling that carbenium ions prefer a trigonal planar
  geometry with an empty p orbital and that it takes energy to bend the
  methyl cation from its low-energy planar form.

Thus, in the studies below the meaning of the classical structure is different from the OP's one: the linear $\mathrm{sp}$-hybridized structure is called classical in almost all the studies.

So, Stang's "Vinyl Cations" references a CI/DZP study of the vinyl cation by Weber, Yoshimine, and McLean 1 as the most recent ab initio correlated calculations published at the time the book was written (the late 70s), so I'll start a review from this study.

Weber, Yoshimine, McLean, 1976

It is was found that the linear and bridged conformations have energies equal to within 0.01 kcal/mol with the later being lower in energy. However, such a small difference is below the accuracy of the CI/DZP level of theory, and thus, the authors didn't claim that the bridged conformation is more energetically favourable, just that it can be such ("probably" is the key word in the quote from the paper below).

  While our CI calculation shows the linear and bridged conformations to
  have energies equal to within 0.01 kcal/mol, we clearly cannot claim
  this absolute accuracy. In fact, taking into account basis set
  deficiencies, both in the one- and $n$-particle spaces, along with
  optimization effects as discussed above, we interpret our calculations
  to predict the relative energies of the two structures to be within
  1–2 kcal/mol with the bridged structure probably having the lower
  energy.

Besides, note that in the study the geometries have been optimized at the SCF level rather than at the correlated while CI calculations were done at the SCF optimized geometries. Taking into account that correlation effect were found to have major impact on relative stability of the two forms as mentioned in the study, it is especially interesting to look at the results obtained by Raghavachari, Whiteside, Pople, and Schleyer2 who optimized the structures not only at the HF/6-31G* but also at the MP2/6-31G* levels as well as did more high-level correlated calculations (MP4) with a bigger basis set (up to 6-311G** of TZP quality).

Raghavachari, Whiteside, Pople, Schleyer, 1981

For the bridged structure the $\ce{C-C}$ bond length was found to increase from 1.21 Å at HF/6-31G* to 1.23 Å at MP2/6-31G* while the distance of the bridging hydrogen to the middle of the $\ce{C-C}$ bond stays the same (1.18 Å). For the classical structure the authors report no significant changes in the geometry. With respect to the relative stability it was found that use of the MP2 geometries results in a slight additional stabilization to the bridged form: the MP4(SDQ)/6-31G**//MP2/6-31G* energy difference is 1.3 kcal/mol while the MP4(SDQ)/6-31G**//HF/6-31G* energy difference is only 0.6 kcal/mol (in both cases in favour of the bridged form).

  With the larger 6-311G** basis, the bridged structure is stabilized
  further, and our final result at MP4(SDQ)/6-311G**//MP2/6-31G*
  indicates 11 [bridged] to be more stable than 10 [classical] by 3.0 kcal/mol.

The authors also mention that these final results are quite in line with the CEPA study by Lischka and Kohler3 done using a comparable basis set: the Lischka and Kohler geometries were very similar to the HF/6-31G* geometries obtained in the study and the bridged form was found to be 4.0 kcal/mol lower in energy than the classical form as well.

A further study done by the Pople group is also is in agreement with the findings quoted above: the bridged form predicted to be more stable by 3.1 kcal/mol (including zero point energies)4. А study by Lee and Schaefer5, though, reports a somewhat lower difference.

Lee, Schaefer, 1986

In a recent CI/DZP study by Lee and Schaefer with the largest CI expansion considered being CISDT the energy difference was found to be 0.68 kcal/mol with the nonclassical structure again lying lower. The authors also considered the effect of the zero-point vibrational energy corrections and found that it further lowers the nonclassical structure by 0.29 kcal/mol.

  Our best strictly ab initio estimate of $\Delta E$(classical-non- classical) for $\ce{C2H3+}$ is obtained by adding the above 0.29 kcal correction to the 0.68 kcal difference from CISDT. In this manner, $\Delta E$ is predicted to be 0.97 kcal/mol. We estimate that a proper variational treatment of quadruple excitations might increase this $\Delta E$ value by perhaps 0.5 kcal/mol.

Weber, J.; Yoshimine, M.; McLean, A. D. A CI study of the classical and nonclassical structures of the vinyl cation and their optimum path for rearrangement. J. Chem. Phys. 1976, 64 (10), 4159–4164. DOI: 10.1063/1.431986.
Raghavachari, K.; Whiteside, R. A.; Pople, J. A.; Schleyer, P. V. R. Molecular orbital theory of the electronic structure of organic molecules. 40. structures and energies of C1-C3 carbocations including effects of electron correlation. J. Am. Chem. Soc. 1981, 103 (19), 5649–5657. DOI: 10.1021/ja00409a004.
Lischka, H.; Koehler, H. J. Structure and stability of the carbocations C2H3+ and C2H4X+, X = hydrogen, fluorine, chlorine, and methyl. Ab initio investigation including electron correlation and a comparison with MINDO/3 results. J. Am. Chem. Soc. 1978, 100 (17), 5297–5305. DOI: 10.1021/ja00485a010.
Curtiss, L. A.; Pople, J. A. Theoretical study of structures and energies of acetylene, ethylene, and vinyl radical and cation. J. Chem. Phys. 1988, 88 (12), 7405–7409. DOI: 10.1063/1.454303.
Lee, T. J.; Schaefer, H. F. The classical and nonclassical forms of protonated acetylene, C2H3+. structures, vibrational frequencies, and infrared intensities from explicitly correlated wave functions. J. Chem. Phys. 1986, 85 (6), 3437–3443. DOI: 10.1063/1.451828.
Crofton, M. W.; Jagod, M.; Rehfuss, B. D.; Oka, T. Infrared spectroscopy of carbo‐ions. v. classical vs nonclassical structure of protonated acetylene c 2 h + 3. J. Chem. Phys. 1989, 91 (9), 5139–5153. DOI: 10.1063/1.457612.


Is there a physical explanation why benzene's pi molecular orbitals in a Frost circle are degenerate but pyridine's are not?

My first thoughts are along the lines of your second-last paragraph. Here is a quick sketch of how to formalise it. From simple Hückel theory, you can obtain the coefficients of the AOs in the MOs:

$$|\psi_i\rangle = \sum_a c_{n,i} |n\rangle$$

where $c_{n,i}$ denotes the coefficient of AO $|n\rangle$ in the MO $|\psi_i\rangle$.

In Hückel theory the main difference between benzene and pyridine is that one atomic orbital is shifted in energy, so one of the $\alpha$ terms turns into, say, $\gamma$. Without loss of generality we will label the nitrogen with the index $n = 1$. In theory, the resonance integrals $\beta$ involving the nitrogen should change as well, but to a first approximation we assume they are the same.

In this case therefore the Hamiltonian matrix, when cast in the AO basis, changes from $\mathbf{H}_0$ to $\mathbf{H}$:

$$\mathbf{H}_0 = \begin{pmatrix}
\alpha & \beta & 0 & 0 & 0 & \beta \\
\beta & \alpha & \beta & 0 & 0 & 0 \\
0 & \beta & \alpha & \beta & 0 & 0 \\
0 & 0 & \beta & \alpha & \beta & 0 \\
0 & 0 & 0 & \beta & \alpha & \beta \\
\beta & 0 & 0 & 0 & \beta & \alpha
\end{pmatrix} \qquad \longrightarrow \qquad \mathbf{H} = \begin{pmatrix}
\gamma & \beta & 0 & 0 & 0 & \beta \\
\beta & \alpha & \beta & 0 & 0 & 0 \\
0 & \beta & \alpha & \beta & 0 & 0 \\
0 & 0 & \beta & \alpha & \beta & 0 \\
0 & 0 & 0 & \beta & \alpha & \beta \\
\beta & 0 & 0 & 0 & \beta & \alpha
\end{pmatrix}$$

Most of the Hamiltonian is unchanged, so we could consider this within the framework of perturbation theory, where we have a perturbed Hamiltonian $\hat{H} = \hat{H}_0 + \hat{V}$. The relevant perturbation is therefore

$$\mathbf{V} = \begin{pmatrix}
\gamma - \alpha & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0
\end{pmatrix}$$

or to condense this back into operator form, $\hat{V} = (\gamma - \alpha)|1\rangle\langle 1|$.

The proper mathematical analysis is slightly annoying because the (unperturbed) eigenstates that we're interested in looking at are degenerate, which then necessitates that you cast the matrix $\hat{V}$ in the basis of the degenerate states of interest. Here, the two states we're interested in are $|\psi_2\rangle$ and $|\psi_3\rangle$. The application of Hückel theory to benzene gives the formulae for these unperturbed states as

$$\begin{align}
|\psi_2\rangle &= \frac{\sqrt 3}{3}|1\rangle + \frac{\sqrt 3}{6}|2\rangle - \frac{\sqrt 3}{6}|3\rangle - \frac{\sqrt 3}{3}|4\rangle - \frac{\sqrt 3}{6}|5\rangle + \frac{\sqrt 3}{6}|6\rangle \\
|\psi_3\rangle &= \frac{1}{2}|2\rangle + \frac{1}{2}|3\rangle - \frac{1}{2}|5\rangle - \frac{1}{2}|6\rangle
\end{align}$$

Note that this choice of coefficients is not unique, as the two states are degenerate: any linear combination of these two MOs is also a permissible MO with the same energy. In any case, the relevant matrix elements of $\hat{V}$ in the new basis $\{|\psi_2\rangle,|\psi_3\rangle\}$ are given by

$$V_{i-1,j-1} = \langle \psi_i | \hat{V} | \psi_j \rangle$$

and if you assume that $\langle n | n' \rangle = \delta_{nn'}$ (i.e. the AOs have no overlap with each other, which is a key assumption of simple Hückel theory anyway), then it becomes very simple:

$$\mathbf{V} = \begin{pmatrix} \frac{\gamma-\alpha}{3} & 0 \\ 0 & 0\end{pmatrix}$$

Perturbation theory tells us that the eigenvectors of $\hat{V}$ are the "states which are stable towards the perturbation". What this means is that, since the degeneracy is lifted by the perturbation, you also remove the free choice of having any linear combinations be a permissible MO: the coefficients of AOs in the new MOs must be uniquely determined (ignoring the phase factor). Since $\hat{V}$ is diagonal in the basis $\{|\psi_2\rangle,|\psi_3\rangle\}$, it conveniently means that both $|\psi_2\rangle$ and $|\psi_3\rangle$ are the eigenvectors of $\hat{V}$.

More interesting here is the first-order corrections to the eigenvalues of $\hat{H}_0$, which are simply the eigenvalues of $\hat{V}$. In this case, since $\hat{V}$ is diagonal, the eigenvalues are just $(\gamma-\alpha)/3$ and $0$: that is to say, the MO $\psi_2$ is shifted in energy by $(\gamma-\alpha)/3$ and the MO $\psi_3$ is unshifted in energy.