This is not an answer to the question "can the sulfoxide be synthesized in significant yield" (and under which conditions). Instead, this is a "naive" prediction based on the data you provide and two assumptions:

The predicted thermodynamic values in the articles you cite are accurate at $\pu{298 K}.$
The heat and entropy of formation of (2) and (3) from (1) are constant over $T.$

From the references we can build a data table of energies and entropies relative to "the ground state" $\ce{HSOH}$ assuming the $\Delta G^\circ$ values correspond to $T = \pu{298 K}:$

$$\begin{array}{lccc}
\hline
\text{Compound} & \Delta G/\pu{kcal mol-1} & \Delta H/\pu{kcal mol-1}^* & \Delta S/\pu{cal mol-1 K-1} & K_\mathrm{eq}\\
\hline
\ce{H2SO} & 25 & 15.8 & -31 & 10^{-19} \\
\ce{H2OS} & 34 & 38.2 & 14 & 10^{-25} \\
\hline
\end{array}$$

I include the order of magnitude of the equilibrium constants $K_\mathrm{eq}$ only to illustrate their ludicrously small values (and these are upper estimates!).

Could you find a temperature that would make $\Delta G^\circ<0$ for the reaction to $\ce{H2SO}$? The answer is no.  To see why we turn to the enthalpies. Since the reaction is endothermic you might hope that an increase in $T$ would help, but the accompanying entropy change is negative, so there is no way you can sufficiently increase the temperature to drive the reaction forward. The highest equilibrium constant you could hope for is $K_\mathrm{eq} = \pu{1.7E-7}$.

Things are not so bleak for $\ce{H2OS}$: the reaction is also endothermic but the entropy change is positive. At about $\pu{2800 K}$ $K_\mathrm{eq} = 1,$ and if we increase $T$ sufficiently, we can crank up $K_\mathrm{eq}$ to any arbitrarily large value.

This is of course not remotely exhaustive (therefore "naive") but hopefully a start to answering your question.

* Quoting your second reference:

  At the CCSD(T)/CBS limit and including corrections for scalar relativistic, spin orbit and core-valence correlation effects, the estimated enthalpies of formation are $\pu{−28.1 ± 1},$ $\pu{−12.3 ± 1},$ and $\pu{10.1 ± 1 kcal/mol}$ for $\ce{HSOH},$ $\ce{H2SO}$ and $\ce{H2OS},$ respectively.


Why are fluoride ions (and ions in general) reactive?

Fluoride activity, or more properly danger, comes from the fact that formation of some salts of fluoride is energetically highly favorable for aforementioned metals.

Calcium for instance, rather than in calcium carbonate, is more stable in $\ce{CaF2}$ form. Then, consequently, it will etch your bones eventually if there is a dermal contact with $\ce{HF}$. Other acids like sulfuric acid make a superficial effect and with a copious amount of water you mostly get rid of it. $\ce{HF},$ on the contrary, is a weak acid and slowly seeps through your epidermis to your bone.

$\ce{HF}$ is also reactive with glass, through silicon. It may also etch the beaker or any other container through time so you don't want to store $\ce{HF}$ for a long time in glass containers, actually same with bases to a lesser degree.

$\ce{F}$ can even react with noble gases, you should just look at the last product and compare the overall stabilities of the current form and the final form. Fluoride simply more stable in $\ce{CaF2}$ solid than in solution.


After ionization occurs, the carbonyl donates its free radical into its own double bond.  This causes the bond adjacent to the carbonyl to split into two electron s, one with spin antiparallel to that of the carbonyl radical and one of parallel spin.  The one of antiparallel spin combines with the free radical to form another pi bond.  This creates a carbon oxygen triple bond and carbon atom with a free radical (alkyl radical).  The carbon atom with the radical is the atom alpha to the carbonyl, hence alpha cleavage.  The oxygen bears a positive formal charge.  Alpha cleavage basically makes substituted carbon monoxides, and side chain radicals.  You will see the "carbon monoxide derivative" on the spectrum.

To do this quickly, just take the original molecule, draw lines showing fragmentation on both sides of the carbonyl, and then draw the carbon oxygen triple bond structure.


If I understand correctly, what you described can be seen as a form of fractional freezing.

This is done in order to make alcoholic beverages more alcoholic (Eisbock is an example): the beverage is left to partially freeze and the ice is removed.
Since ethanol has a lower melting point (-114 °C) than water (0 °C), this effectively removes more water than ethanol.

But in your case, since acetic acid has a higher melting point (16 °C) than water, the solids will have a higher concentration of acetic acid and you would want to keep that instead of the liquid.


How do I derive the Nernst equation in the case in which R is absent?

Thank you for the comment Philipp. With this clarification, the derivation is simple:

$E=E_0+\frac{RT}{nF}ln\frac{C_O(x=0)}{C_R(x=0)}$ (1.4.12) where

$C_O(x=0)=\frac{i_l-i}{nFAm_O}$ (1.4.11) and $C_R(x=0)=\frac{i}{nFAm_R}$ (1.4.13).

Than, combining equations (1.4.11) and (1.4.13) to (1.4.12), we obtain:

$E=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{nFAm_O}}{\frac{i}{nFAm_R}}=E_0+\frac{RT}{nF}ln\frac{\frac{i_l-i}{i}}{\frac{nFAm_O}{nFAm_R}}=E_0+\frac{RT}{nF}(ln\frac{i_l-i}{i}- ln\frac{nFAm_O}{nFAm_R})$=

$E_0-\frac{RT}{nF}ln\frac{m_O}{m_R}+\frac{RT}{nF}ln\frac{i_l-i}{i}$


Why does reduction always accompany oxidation and vice versa?

If something is to be reduced then there must be a reducing agent present. Also if something is oxidized then there must be an oxidizing agent present.

You must have a reducing agent or oxidizing agent present in order for a redox reaction to take place. An element can't just dislocate it's electrons. There must be an outside force present in order to do that.


"Exactly Equal" and "At Least" in electron excitation

When you ionize or break something, the resulting particles (an electron and an ion, or maybe two atoms, or whatever) would fly away from each other with arbitrary speed. Their kinetic energy would absorb any excess energy of the photon. This is the "at least" case.

On the other hand, when you excite some particle (molecule or whatever) in such a way that it remains in one piece, then you can't send it flying, because the momentum conservation forbids that. This is the "exactly equal" case.

Surely this is a simplification; in fact, a photon may excite a molecule using only a part of its energy, while keeping the rest of it to itself (Raman scattering), but that's another story.


How to use a modified Peng-Robinson equation to calculate a density?

According to this source:

  The popular EoSs such as Soave-Redlich-Kwong (SRK) [1] and Peng-Robinson (PR) [2] predict liquid density with an average absolute error of about 8%, much higher than several good density correlations. This large magnitude of error is not acceptable by industry; therefore, they are not used for this purpose.

Which is what I first thought when I thought of your question. Liquid density is a hard issue, specially in compounds with hydrogen bonding like ethanol. I wouldn't predict a gas-oriented (as in its primary use) EoS to perform just as well for liquids.

The article further suggests

  In order to overcome this deficiency, volume translated methods have been developed.

So that's what I'd recommend you to look into.


Gattermann-Koch Reaction with Phenol Substrates

You have outlined that the formyl chloride is generated as the reactive, electrophilic intermediate under these conditions.

You are planning to perform an electrophilic aromatic substitution, but in a phenol, there's high electron density on the $\ce{O}$ atom. Isn't it conceivable that your main reaction product is phenylformate here, resulting from $\ce{O}$-acylation?


Making Lyophilized Cake Lookalike using household ingredients

You may want to consider whey. Looks like Karen Smith, Dairy Processing Technologist at the Wisconsin Center for Dairy Research, already did some of the work for you. The result depends on the specific type of whey (a high score of 4 or 5 on the caking test means the material cakes readily, forming a gummy crust):

  Whey  – (Scored 2-5) –  whey exhibited a wide range of caking scores. How the whey is processed has a very large effect on the tendency of the resulting powder to cake as evident in this result. Clearly, two of the samples had large amounts of amorphous lactose and without the presence of significant amounts of protein the samples readily caked.

Conclusion: choose a whey with less protein and more amorphous sugar (such as lactose). In general you are looking for a sticky substance. The amorphous sugar is sticky (it does not form a crystalline powder) and glues the material together. Bodybuilding whey may contain too much protein. Alternately milk powder (which contains plenty of lactose) may do the trick. I'm going to the market now to get supplies and run a test.

Update: I have run a home kitchen caking test on dried skim milk. The dry solids consist of 56% sugar, 40% protein, and ~1 g fat. The milk contains 11% water.

The test was similar to that described in the linked document and consists of placing ~20 g of milk powder in a small (~3 inch diameter) aluminum baking dish and allowing it to float on hot water (~$\pu{50^\circ C}$) in a closed container for about an hour.

The caking test is of course the opposite in a sense of lyophilization, for the solid is exposed to a humid atmosphere to encourage coalescence. The following image illustrates the result:

This confirms the observation in the linked document that NFDM (lower left corner of the image array) "cakes upon storage due to presence of large amounts of amorphous lactose. The protein present limits the hardness of the cake of powder."

This verifies that milk powder cakes. However I do not have a lyophilizer at home so unfortunately cannot verify that the dried milk will remain caked once freeze dried. I might yet put it in the oven to see how this affects the structure.


Methods used to compare two spectra in order to determine the effects of irradiation of sample

The table below illustrates a possible screen, values computed displayed are rounded to four decimals.  The suggest is to use the observations prior the intended irradiation ($x_1, x_2$) separate from the observations after the intended irradiation ($y_1, y_2$).  Per wavelength

compute for $x$ and $y$ the arithmetical mean value
determine the standard deviation of this sample (e.g., $x_1$ and $x_2$), the one pocket calculators sometimes label with $\sigma(n-1)$.  The table doesn't include it, but you may compute the confidence interval with a reasonable $t$ value.  Because there were only two realizations, the degree of freedoms ($f = n - 1$) equates to 1, and an understandably high $t_{1, 0.95} = 12.71$ to map the interval of

$$\bar{y} - \frac{t \cdot \sigma}{\sqrt{n}} \le \bar{y} \le \bar{y} + \frac{t \cdot \sigma}{\sqrt{n}}$$

subtract the arithmetical mean values from each other to determine the effect of the irradiation.  Expect positive as well as negative values to occur.
compute the standard deviation for this effect.  Because the effect is computed as a difference of $(\bar{x} - \bar{y})$, the error propagation of is a sum of the corresponding standard deviations. Or add the corresponding halfes of the confidence intervals about $x$ and $y$ instead.

I have no hands-on experience with R.

| lambda |     x1 |     y1 |     x2 |     y2 | mean_x | stdev_x | mean_y | stdev_y | mean_x - mean_y | stdev_x + stdev_y |
|--------+--------+--------+--------+--------+--------+---------+--------+---------+-----------------+-------------------|
|   3999 | 0.0124 | 0.0132 | 0.0122 | 0.0113 | 0.0123 |  0.0001 | 0.0123 |  0.0013 |          0.0000 |            0.0015 |
|   3998 | 0.0125 | 0.0130 | 0.0122 | 0.0116 | 0.0123 |  0.0002 | 0.0123 |  0.0010 |          0.0001 |            0.0012 |
|   3997 | 0.0122 | 0.0131 | 0.0122 | 0.0113 | 0.0122 |  0.0000 | 0.0122 |  0.0013 |          0.0000 |            0.0013 |
|   3996 | 0.0121 | 0.0136 | 0.0122 | 0.0114 | 0.0122 |  0.0001 | 0.0125 |  0.0016 |         -0.0003 |            0.0016 |
|   3995 | 0.0124 | 0.0139 | 0.0122 | 0.0122 | 0.0123 |  0.0001 | 0.0130 |  0.0012 |         -0.0007 |            0.0013 |
|   3994 | 0.0125 | 0.0141 | 0.0122 | 0.0129 | 0.0123 |  0.0002 | 0.0135 |  0.0008 |         -0.0011 |            0.0011 |


E1 and E2 elimination reactions of isotopically labeled bromocyclohexanes

Under E1 conditions (typically hot, polar protic solvent), one of the bromines will leave as bromide to give a carbocation. Ignoring the isotope effect, there's about an equal propensity for either of the isotopic bromines to leave, giving about a 50/50 ratio of the two carbocations. Deprotonation at the beta-position gives an alkene. I've only shown the most substituted possible alkene since that will be the most stable and major product.

(Note, the carbocation could rearrange, but I'm ignoring that for this level of analysis.)

Under more strongly basic conditions an E2 reaction is more likely. There is a strong preference for an anti-periplanar arrangement of the leaving group, carbons where the alkene will form, and the deprotonated hydrogen. This stereochemical requirement is more easily visualized in the chair forms, which are not equivalent. The structure on the left is more stable since the methyl group is placed equatorially. In this conformation, there are two hydrogens that are anti-periplanar with the bromine-81 atom (green). A bulky base would react with the least hindered hydrogen (purple) and eliminate the bromine-81 to give the disubstituted alkene shown. A smaller base would be more likely to produce the most stable product, which results from deprotonating the hydrogen in blue.

Just for completeness I've shown the elimination product of the less stable chair conformation. In this case, there is only one anti-periplanar arrangement. If the orange hydrogen is deprotonated, then the bromine-79 atom can be eliminated to give the cyclohexene shown.


In short

$\ce{CH3CH2NH3Cl}$ is acidic because it is formed from a strong acid ($\ce{HCl}$) and a weak base ($\ce{CH3CH2NH2}$).

Elaborated

Notice that this compound is ionic, comprising $\ce{CH3CH2NH3+}$ cation and $\ce{Cl-}$ anion.

I mentioned that $\ce{CH3CH2NH2}$ is weak, meaning that this reaction is reversible:

$$\ce{CH3CH2NH2 + H2O <=> CH3CH2NH3+ + OH-}\tag1$$

In an acidic environment, we can rewrite it as:

$$\ce{CH3CH2NH2 + H3O+ <=> CH3CH2NH3+ + H2O}\tag2$$

Therefore, when $\ce{CH3CH2NH3Cl}$ dissolves in water, there would be initially an equal amount of $\ce{CH3CH2NH3+}$ and $\ce{Cl-}$ in terms of number of moles.

Using equation $(2)$, we know that $\ce{CH3CH2NH3+}$ will react with water reaching an acidic equilibrium.

Note that there is no $\ce{OH-}$ to start with (very little in reality), so equation $(1)$ is not applicable.

Therefore, the following species would be present:

$\ce{CH3CH2NH3+}$
$\ce{CH3CH2NH2}$
$\ce{H2O}$
$\ce{H3O+}$
$\ce{Cl-}$

Now it is apparent that $\ce{H3O+}$ makes it acidic.


I don't have an excellent understanding of the topic, but I will try to write something up.

Polycations are found with the Hg(I) oxidation state. The mercury cation, $\ce{Hg+}$, has a $\mathrm{5d^{10}6s^1}$ configuration. The mercurous cation $\ce{Hg2^2+}$ can then be considered to simply be formed by overlap of the two $\mathrm{6s}$ orbitals on two separate $\ce{Hg+}$ ions. In this sense, the bonding is qualitatively very similar to that in $\ce{H2}$. Note that the $\mathrm{5d^{10}6s^0}$ Hg(II) cation has no chance of forming such covalent bonds.

Generally, going down a group, orbitals become larger, more diffuse, and the overlap between them decreases. As a consequence, $\sigma$-bonding tends to become weaker going down the group. A simple example is provided by the alkali metal dimers:

$$\begin{array}{c|ccc}
\text{Element} & \ce{H} & \ce{Li} & \ce{Na} \\  \hline
D_\mathrm{0}(\ce{E-E})\text{ / }\pu{kJ mol-1} & 432 & 99 & 71 \\ \hline
\text{Reference} & 1 & 2 & 3
\end{array}$$

These dissociation energies are extremely small. For the analogous $\ce{Hg2^2+}$ cation, the situation is not quite as bleak. The Group 12 metals in general have more low-lying and contracted ns orbitals than the corresponding alkali metals, due to the presence of ten poorly-shielding (n-1)d electrons. However, given that zinc does not form similar cations, it is odd that mercury with its larger 6s orbitals should do so.

The most obvious explanation is simply that Zn does not commonly exhibit the +1 oxidation state. The existence of the +1 oxidation state for Hg is directly related to the relativistic stabilisation of the 6s orbital, which increases the first two ionisation energies4 of mercury.

$$\begin{array}{c|ccc}
\text{Element} & \ce{Zn} & \ce{Cd} & \ce{Hg} \\ \hline
I_1\text{ / eV} & 9.393 & 8.992 & 10.44 \\ \hline
I_2\text{ / eV} & 17.96 & 16.90 & 18.76 \\
\end{array}$$

However, the question still remains as to why Hg(I) chooses to form polycations instead of existing simply as $\ce{Hg+}$. In fact, relativistic effects have also been implicated in this for a long time: Pyykkö and Desclaux wrote in 1979 that5

  The relativistic contraction of the half-filled valence s AO makes $\ce{Au2}$ much more strongly bound than $\ce{Ag2}$. Inasmuch as $\ce{Hg2^2+}$ and $\ce{Cd2^2+}$ are isoelectronic with these two molecules, the remarkable stability of the mercurous ion [compared to the Cd analogue] can also be understood.

Therefore, this may most simply be explained as the relativistic contraction of the 6s AOs leading to stronger-than-expected overlap to form a covalent bond. As suggested by the above paragraph, similar effects are seen in the neutral dimers of Group 11 metals:6

$$\begin{array}{c|ccc}
\text{Element} & \ce{Cu} & \ce{Ag} & \ce{Au} \\  \hline
D_\mathrm{0}(\ce{E-E})\text{ / eV} & 1.95 & 1.65 & 2.29 \\
\end{array}$$

The decrease going from Cu to Ag is expected on the basis of weakened overlap, and the dissociation energy of $\ce{Au2}$ would be even lower in the absence of relativistic effects: calculations indicate that relativity contributes roughly $\pu{1 eV}$ to the dissociation energy of $\ce{Au2}$.6 Presumably, the effect in $\ce{Hg2^2+}$ is of a similar significance.

References

Frenking, G.; Bickelhaupt, F. M. The EDA Perspective of Chemical Bonding. In The Chemical Bond: Fundamental Aspects of Chemical Bonding; Frenking, G., Shaik, S., Eds.; Wiley: Weinheim, Germany, 2014; p 128.
Velasco, R.; Ottinger, Ch.; Zare, R. N. Dissociation Energy of Li2 from Laser‐Excited Fluorescence. J. Chem. Phys. 1969, 51 (12), 5522–5532. DOI: 10.1063/1.1671979. Free version provided by the Zare group here.
Jones, K. M.; Maleki, S.; Bize, S.; Lett, P. D.; Williams, C. J.; Richling, H.; Knöckel, H.; Tiemann, E.; Wang, H.; Gould, P. L.; Stwalley, W. C. Direct measurement of the ground-state dissociation energy of Na2. Phys. Rev. A 54, R1006(R). DOI: 10.1103/PhysRevA.54.R1006.
Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed; Oxford UP: Oxford, U.K., 2014; pp 836–7.
Pyykko, P.; Desclaux, J. P. Relativity and the periodic system of elements. Acc. Chem. Res. 1979, 12 (8), 276–281. DOI: 10.1021/ar50140a002.
Pitzer, K. S. Relativistic effects on chemical properties. Acc. Chem. Res. 1979, 12 (8), 271–276. DOI: 10.1021/ar50140a001.


Mass of strontium iodide contained in a 350 mL solution

The mistake is that you didn't write "mol" after $1.93 \times 10^{-6}$.

Some teachers/professors might let that go, others wouldn't, depending upon what they are really trying to teach you.

Units and dimensional analysis are important.


How is Standard Gibbs free energy measured?

Generally you measure the equilibrium constant $K$ and then convert it to $\Delta G^\circ$ (note that is change in standard free energy)

$ \Delta G^\circ = -RT \ln K $

In the case of redox reactions, this can be done with electrodes as you mention. For other reactions you usually use some spectroscopic technique (NMR, IRC, UV, etc) that can distinguish reactants from products, but calorimetry can also be used in certain cases.

Here is a video describing one example using UV.


Why is copper (II) oxide used in quantitative analysis of carbon?

You got the crux of it: low cost.  And it works, regardless of what the starting compound was.  And it's not just low cost to purchase, it is easy to regenerate the used cupric oxide multiple times by heating it under oxygen.


How does the entropic force arise within the hydrophobic effect?

Whenever a non-polar substance is put into water, water molecules will organise themselves around it in a cage-like formation. The molecules exposed to the non-polar substance at any given time will orient themselves to form as many hydrogen bonds with the rest of the solution as possible.
(Image from here)

The way I understand it intuitively is that these water molecules are restricted in the orientations they can assume, becoming more ordered to maximise hydrogen bonding, so have an overall lower entropy than if they were free to tumble like in the bulk solution. Similarly, the aggregation of non-polar molecules also allows more free movement within the non-polar 'blob'. If they were distributed evenly, there would be a higher surface area and therefore amount of 'cage' water molecules and non-polar solutes would be more restricted in their movement (trapped in a polar environment). This is the entropic factor that drives the hydrophobic effect.
It is also favourable enthalpically; hydrogen-bonding interactions and non-polar interactions are maximised respectively.
These two approaches demonstrate why it is thermodynamically favourable for non-polar substances to aggregate, not disperse. To clarify, it is still thermodynamically unfavourable to mix the two in the first place, that's why they separate; this is just the better option if they were to mix.
It is also worth noting that oil 'blobs' in water tend to be very round, due to the fact that a sphere is the shape with the smallest surface area to volume ratio. This supports the above reasoning, minimising the number of 'cage' water molecules whilst maximising movement of both water and the hydrophobe.


Solubility of dimethyl ether in aliphatic hydrocarbons

Since dimethyl ether is a very uncommon solvent, it's not a big surprise that you have hard time finding its information just using Google. However, I found two relatively important parameters of dimethyl ether: Its Hildebrand Solubility Parameter is 7.3 and Kauri-Butanol index is 90 (INVENTEC). For comparision, Hildebrand Solubility Parameter of diethyl ether is 7.62 (Solubility Parameters: Theory and Application). According to this reference, if two solvents have about the same Hildebrand values, theoretically, such solvents should have solubility behaviors similar to each other.

Thus, dimethyl ether should behave similar to diethyl ether (in liquid phase, I believe).  Keep in mind that most of hydrocarbons dissolve in diethyl ether in all ratios. For example, we run column chromatography using such solvent systems (diethyl ether/hexanes) to isolate close nonpolar organic compounds from complex mixtures.


Lower Heating value calculation from chemcial combustion equation of Butane

I see three flaws in your reasoning.

First, butane is $\ce{C4H10}$. The formula $\ce{C4H8}$ refers to several isomers of butene. I will keep using your formula, but you should specify whether you mean butane or $\ce{C4H8}$.

Second, you know the relationship between your variables $n_1$ and $n_2$ because you have the volume percents of $\ce{C4H8}$ and $\ce{CO2}$ in the exhaust. Thus, you have fewer variables.

$$\dfrac{n_1}{n_2} = \dfrac{0.75}{14.95}$$

Third, I am not sure why $\ce{N2}$ is in the equation. It does not participate in the reaction, so it should not be there. Likewise, you do not need to add $\ce{CO}$ and $\ce{H2}$ because you know that none is formed. You can balance the equation and determine the stoichiometric coefficients of $\ce{CO2}$ and $\ce{H2O}$ from the conservation of mass:

$$\ce{C4H8 + 6O2 -> 4CO2 + 4H2O}$$

To include one of your variables, we assume that the reaction does not run to completion, and that some amount $n_1$ of $\ce{C4H8}$ remains unreacted. Since each 1 equivalent of $\ce{C4H8}$ consumes 6 equivalents of $\ce{O2}$, if $n_1$ of $\ce{C4H8}$ is not consumed, then $6n_1$ of $\ce{O2}$ is not consumed. Likewise, $4n_1$ of $\ce{CO2}$ and $\ce{H2O}$ were not produced.

$$\ce{C4H8 + (6-6n_1)O2 -> n_1 C4H8 + (4-4n_1)CO2 + (4-4n_1)H2O}$$

Now, I have rendered your problem down to one variable: $n_1$. Your other variables can be written as equations involving $n_1$ by comparing your chemical equation to mine (they are equivalent). You have enough equations in $n_1$ and $n_2$ to solve for $n_1$, $n_2$, and eventually $\phi$, which then leads to the values you really want.


Is the structure of a crystalized protein the native one?

I'll use quotes from B. Rupp, Biomolecular Crystallography (p. 7-8) to answer.

Generally the structure is similar...

  Comparison of many nuclear magnetic resonance (NMR) solution structure
  ensembles with crystallographic structure has shown that the core
  structure of protein molecules remains unchanged compared with the
  solution state during crystallization. In addition, enzymes packed in
  crystals even maintain biological activity.

with some parts not visible in the experiment...

  The maintenance of the core structure and of enzymatic function shows
  that crystal structures are a very good approximation of the native
  protein solution structure. Nonetheless, highly flexible or mobile
  regions, frequently the amino- or carboxyl-termini of the protein chain or
  flexible loops connecting secondary structure elements, can be
  poorly defined or even absent in the electron density and thus can be
  modeled only with limited confidence.

but beware...

  In certain situation flexible and dynamic regions of a protein
  molecule can be rigidly fixed in a specific conformation as a result
  of crystal packing interactions. In most cases this represents just a
  snapshot of one possible conformation out of many and it must be
  understood that such a specific conformation may not locally represent
  the protein structure in solution. A simple safeguard against
  misinterpretations -- which is usually assignment of certain
  biological relevance to regions where that is de facto not warranted --
  is to display all neighboring, symmetry-related molecules in the crystal
  structure and examine if any intermolecular interactions are
  present that are a result of crystal packing. Such packing induced
  artifacts can also hamper for example drug discovery by altering or
  blocking binding sites and thus preventing an otherwise active
  substance from binding.


Odd Acidity Trend in Hexaaqua Metal Complexes

To start off with, there are two graphs on the site you linked, the second one being this one:

This, as the author of your site also states, is a much nicer graph in that it explains all but two ions ($\ce{Cr^{II}}$ and $\ce{V^{II}}$) well.

I also went around and checked different crystal structures of hexaaquametal(II) complexes for their $\ce{M-O}$ bond length to see if that may have any influence. The results are:

Vanadium: $212~\mathrm{pm}$ (octahedral)
Chromium $203~\mathrm{pm}$ (axial) and $245~\mathrm{pm}$ (equatorial; Jahn–Teller distorted octahedron)
Manganese: $214$ to $217~\mathrm{pm}$
Iron: $211~\mathrm{pm}$ (almost octahedral)
Cobalt: $208$ to $213~\mathrm{pm}$ (distorted octahedron)
Nickel: Did not find data
Copper: $200~\mathrm{pm}$ (axial) $220~\mathrm{pm}$ (equatorial)

On the lower end of this spectrum we find copper and chromium which indeed have rather low $\mathrm{p}K_\mathrm{a}$ values. This could be due to a shorter $\ce{M-O}$ bond meaning that the water molecules experience the charge stronger. (This is also why I favour the graph I put in this answer: It puts copper closer to ‘where it should be’ in ionic radii terms.)

A second thing to consider is the ligand field stabilisation enthalpy LFSE. This is strongest ($-12~\mathrm{Dq}$) for vanadium since that ion only has three electrons which all occupy the lower energy levels. This would mean that ligand exchange for $\ce{[V(H2O)6]^2+}$ is kinetically strongly inhibited especially for vanadium, since any loss of a ligand would need to overcome the stabilisation of $-12~\mathrm{Dq}$. Hence, the water molecules are strongly bound to vanadium and experience its charge all the time. On the other hand, copper has rather weakly bound ligands exemplified in how simple it is to create $\ce{[Cu(NH3)4(H2O)2]^2+}$, the deep blue tetraammincopper(II) complex.
(Note: I did not find any papers confirming or disputing this hypothesis, but I only spent a short time looking.)

Right after vanadium we encounter chromium. It only has an LFSE of $-6~\mathrm{Dq}$ — which is still pretty good, considering it all derives from unpaired electrons. Overall, I think this ion can be considered to be almost like vanadium(II) albeit a little smaller, with an extra electron and a shorter metal-water bond length which would explain its position.

Next up is manganese which looks as if it were the exception. But it actually isn’t. First off, its LFSE is $0$, thus water molecules can diffuse on and off as they see fit, therefore overall weakening the effect the ion’s charge has on them.

From iron onwards, we can again consider the water ligands bound slightly more tightly since there is again LFSE to be gained. However, remember that while the $\mathrm{t_{2g}}$ orbitals are non-bonding with respect to metal–ligand σ interactions, the $\mathrm{e_g}$ level is antibonding. We have a half-filled $\mathrm{e_g}$ level which will weaken the overall $\ce{M-O}$ bonds. This could not only explain the larger $\ce{M-O}$ distances of manganese and iron when compared to chromium but also their weaker acidity. (A weaker bond would also mean weaker charge transfer.) I suspect that this does not yet really matter for chromium due to its Jahn–Teller distortion being somehow favourable overall, and that chromium’s acidity stems from the closer water molecules.

From then on, the explanation seems rather simple, since the ion radii and the $\mathrm{p}K_\mathrm{a}$ are almost in a linear correlation. The question is how the ionic radii were deduced (which is none-trivial) and why they differ so strongly with the two sources the author of your linked site used.


Final concentration of stock solution upon dilution

You are correct.  You might find it more intuitive to think of it this way:

In the end you will have $\pu{1 mL}$ of solution. In that end solution you have $\pu{0.1 mL} \times \frac{\pu{1mg}}{\pu{mL}}$ of stock, or (cancelling the $\pu{mL}$s) $\pu{0.1mg}$ of the chemical in question.  Therefore, final concentration is $\frac{\pu{0.1mg}}{\pu{1mL}} = 0.1 \frac{\pu{mg}}{\pu{mL}}$.


What is the mechanism of chlorosulfonation of benzene?

Rob, IMO this is a difficult problem, so don't feel bad about not "seeing" the solution right off.

Chlorosulfonic acid (or chlorosulfuric acid) is both difficult and hazardous to work with so it's reactions aren't discussed all that frequently.  It can also give rise to different products at different temperatures.  At higher temperatures it generates $\ce{SO3}$ which will behave as an electrophile and react with aromatic compounds to produce sulfonation products.

image source

At lower temperatures (your case), chlorosulfonic acid reacts by way of the following equilibrium (who knew this?)

$$\ce{3 ClSO2(OH) <=> SO2Cl^{+} + 2 SO3Cl^{-} + H3O^{+}}$$

The equilibrium generates $\ce{SO2Cl^{+}}$ which is the electrophile of interest in your reaction.  $\ce{SO2Cl^{+}}$ will react with an aromatic nucleus to produce the corresponding sulfonyl chloride by way of a typical electrophilic aromatic substitution reaction.

If you'd like to read more about the complexities of reactions involving chlorosulfonic acid see here starting around page 11.

The difficult part here was identifying the source of the electrophile. Looking at the final product you get an idea of what the electrophile is, but how is it produced?  Sometimes you have to do a little googling to sort things out.

Again, IMO this was a tough one.


Why enthalpy changes in phase transition when there is no change of temperature?

For a single phase, enthalpy is a function of temperature and pressure, and, at constant pressure, $\mathrm dH = C_p\,\mathrm dT.$

However, at a change of phase, enthalpy (per unit mass) undergoes a jump change, even at constant temperature and pressure.  If the phase change is from liquid to vapor, for example, this jump change in enthalpy is called the heat of vaporization.


With a partition coefficient how do you decide which solute concentration is the numerator?

The proper way to express the partition coefficient of compound X distributed between an aqueous and organic phase, for example, is:

$$\mathrm{(K_P)_X = \frac{[X](org)}{[X](aq)}}$$

This unambiguously expresses the mathematical fraction to be used in calculating the partition constant for a solute, X between two solvents, aq and org.


Toxicity of Mercury (II) Iodide and dicopper (I) tetraiodomercurate (II)

Following the @Nilay Ghosh recommendation, I found an SDS sheet for $\ce{Cu2[HgI4]}$ and there is SDS for $\ce{HgI2}$.
For $\ce{Cu2[HgI4]}$:

Acute toxicity estimate Oral (Expert judgment): 5,1 mg/kg
Acute toxicity estimate Inhalation (Expert judgment) 4 h: 0,051 mg/l
Acute toxicity estimate Dermal (Expert judgment): 5,1 mg/kg

For $\ce{HgI2}$:

LD50 oral rat: 18 mg/kg
Acute toxicity estimate Inhalation (Expert judgment) 4 h: 0,051 mg/l
Acute toxicity estimate Dermal (Expert judgment): 5,1 mg/kg


“Likewise the equatorial Ph3P groups would both only see one axial Ph3P group, hence only a singlet would be observed” - I think you’ve just made a little mistake. The equatorial groups can see one phosphorus other than themselves, so (2nI + 1) = 2, not 1!


Advice on forming a 10% solution of Maltol

Maltol (3-Hydroxy-2-methyl-4-pyrone) appears to be most soluble in ethanol.
https://pubchem.ncbi.nlm.nih.gov/compound/maltol#section=Melting-Point gives solubility as
1 to 10 mg/mL at 72° F in water, 10900 mg/L at 15 deg C (about 1%); one gram dissolves in 82 mL water, 80 mL glycerin, 21 mL alcohol, 28 mL propylene glycol; Freely soluble in hot water, chloroform; sparingly soluble in benzene, ether, petroleum ether; soluble in alkali hydroxides giving yellow solutions

Your fluffy floating clumps suggest something like water, which would reduce the solubility of maltol and could clump it up. or maybe there is some other impurity that complicates things.

You could try is to make a 5% solution in warm dry ethanol (apparently no problem...) and then evaporate off excess solvent. If you can get 10% when hot, but the solution drops crystals on cooling, you will have a saturated solution. But maybe it won't precipitate...


How is it possible that meta xylene exists?

Friedel Crafts alkylation of toluene is far less regioselective than you might think.
The meta-substituted products are thermodynamically favoured. At higher temperatures, the alkylation yields more 1,3-dialkyl substituted benzenes.
On a larger scale, Dimethylbenzenes are typically not synthesized by alkylation of toluene, but stem from catalytic reforming processes using alkanes as feedstock.


Why does alcoholic KOH prefer elimination whereas aqueous KOH prefers substitution?

In alcoholic solution, the $\ce{KOH}$ is basic enough ($\mathrm{p}K_{\mathrm{a}} =15.74$) to deprotonate a small amount of the alcohol molecules ($\mathrm{p}K_{\mathrm{a}}= 16–17$), thus forming alkoxide salts ($\ce{ROK}$). The alkoxide anions $\ce{RO-}$ are not only more basic than pure $\ce{OH-}$ but they are also bulkier (how much bulkier depends on the alkyl group). The higher bulkiness makes $\ce{RO-}$ a worse nucleophile than $\ce{OH-}$ and the higher basicity makes it better at E2 eliminations.


How does a Pt and reference electrode measure the cell potential in a potentiometric titration?

It is a very good question. Your question has nothing to do with current being drawn or not or a reference electrode etc. Imagine a single Pt electrode dipped in an Fe(II) solution and forget about the presence of cobalt. Imagine that a metal is a source (sea) of electrons.

The moment you dip the electrode, instantaneous exchange of electrons takes place between the metal and the reducible or oxidizable ions. Electrons can be released by Fe(II) to form Fe(III) and similarly electrons can be gained by Fe(III) to form Fe(II). This exchange rate is dependent on the concentration ratios of Fe(II) and Fe(III).

As a result of this exchange, the Pt metal develops a charge. It is very surprising that you only need to remove a very very very small number of electrons to develop a measurable surface charge. Hence, there is no perceptible change in the concentration of Fe ions. If platinum electrode develops a charge, the interface (solution electrode boundary) develops an opposite charge until equilibrium is achieved.

Now given the fact that Pt electrode is charged i.e., if we connect Pt to another electrode, we can sense the potential difference between the Pt electrode and another electrode, which can be any other electrode. Now that you wish to have a stable. For that purpose you need an electrode whose potential is fixed i.e. a reference electrode.


What do these mineral classification symbols mean?

In addition to being a non-practicing chemist, I'm also a non-practicing geologist, so look no further for an answer you can trust.

Your minerals are classified as follows: C is for color (sample, gross), S is for streak (color of powdered mineral, essentially, on an unfinished streak plate), H is hardness on the Mohs scale, SG is specific gravity, L is luster, and CL is cleavage (how the sample breaks along planes dictated by the crystal habit of the mineral). See this reference for a printable chart.

The color-coding is certainly used to mark the samples in case they get jostled or mixed up and need to be matched to their appropriate slots: those codes are not related to the mineral classification.


How to evaluate the heat of formation with computational chemistry?

Your approach is quite correct, but as Jan already pointed out, it is incomplete. What you calculated is the difference in the electronic energy of the reaction $$\ce{H2 (g) + 1/2 O2 (g,\,{}^1\Delta_{g}) -> H2O (g)}.$$

Let's turn this into some kind of a tutorial and a little exercise, that you can try to reproduce at home. For a more detailed description on how it works in Gaussian, you can read the paper by Joseph W. Ochterski. I will use Gaussian09 rev. D for the calculations involved.

The enthalpy of formation of water from the standard state of the elements involved, is basically the reaction enthalpy of the combustion of hydrogen and oxygen. You are in luck, because most of the quantum chemical programs shall provide you with the necessary data for reactions straight away. What you do need to do is calculate the thermal corrections to the reaction. Essentially this is done by calculating the Hessian, i.e. a frequency calculation. The second derivatives only have a interpretable meaning if they are performed at a stationary point. Hence they must be performed with the same method, that you used to optimise the geometry.
Jan already pointed out, that HF/6-31G(d) is probably not a well enough fit for geometries and frequencies. I'd always consider some level of theory with at least some coverage of correlation energy. I often prefer BP86 or M06L and the def2-SVP basis set as a first anchor point. The smaller the system, the more accurate the methods you can use (or afford).
Jan also pointed out, that especially for oxygen, the spin state is important. (I did a quick check and am almost 100% certain you calculated it in the singlet state; see above equation.)

For this exercise I have matched your suggestion with using M06/def2-QVZPP as it is still feasible. Keep in mind, that most of the times density functional approximations have to be calibrated. It's all about what you can or cannot afford. You should perform calculations to the best of your abilities.

Okay, let's dive into some mathematics. First, we consider the reaction equation
$$\ce{H2 (g) + 1/2 O2 (g,\,{}^3\Sigma_{g}^{-}) -> H2O (g)}.$$

And we would like to know the enthalpy of formation for water at standard state (which we will use as $298.15~\mathrm{K}$ and $1~\mathrm{atm}$)
$$\begin{align}
\Delta_{f}H^\circ(\ce{H2O}) &= H^\circ(\ce{H2O})
                               - (H^\circ(\ce{H2})
                               +\frac12 H^\circ(\ce{O2}))\\
\Delta_{f}H^\circ(\ce{H2O}) &=
   [E_\mathrm{el}(\ce{H2O}) + H_\mathrm{corr}(\ce{H2O})]\\
&\qquad
 - \left([E_\mathrm{el}(\ce{H2}) + H_\mathrm{corr}(\ce{H2})]
 +\frac12 [E_\mathrm{el}(\ce{O2}) + H_\mathrm{corr}(\ce{O2})]\right)\\
\end{align}$$

The data in the expanded form we can directly take from the output of the frequency calculation:

Oxygen  mo6qzvpp.freq.log/.com
 SCF Done:  E(UM06) =  -150.322374247     A.U. after    1 cycles
 Zero-point correction=                           0.003897 (Hartree/Particle)
 Thermal correction to Enthalpy=                  0.007203
 Thermal correction to Gibbs Free Energy=        -0.016049

Hydrogen mo6qzvpp.freq.log/.com
 SCF Done:  E(RM06) =  -1.17221845130     A.U. after    1 cycles
 Zero-point correction=                           0.009902 (Hartree/Particle)
 Thermal correction to Enthalpy=                  0.013206
 Thermal correction to Gibbs Free Energy=        -0.001590

Water mo6qzvpp.freq.log/.com
 SCF Done:  E(RM06) =  -76.4321360023     A.U. after    1 cycles
 Zero-point correction=                           0.021635 (Hartree/Particle)
 Thermal correction to Enthalpy=                  0.025415
 Thermal correction to Gibbs Free Energy=         0.004010

Therefore we will obtain
$$\Delta_{f}H^\circ(\ce{H2O}) = -236.6~\mathrm{kJ/mol}$$
which I think is reasonably close to the experimental value.

Appendix
The following input files for the computation of above values.

%chk=mo6qzvpp.chk
#p M06/def2QZVPP
opt
int(ultrafinegrid)

oxygen

0 3
O
O 1 1.2
 (blank)

%chk=mo6qzvpp.freq.chk
%oldchk=mo6qzvpp.chk
#p M06/def2QZVPP
freq
geom=check guess=read
int(ultrafinegrid)

oxygen freq

0 3
 (blank)

%chk=mo6qzvpp.chk
#p M06/def2QZVPP
opt
int(ultrafinegrid)

hydrogen

0 1
H
H 1 1.2
 (blank)

%chk=mo6qzvpp.freq.chk
%oldchk=mo6qzvpp.chk
#p M06/def2QZVPP
freq
geom=check guess=read
int(ultrafinegrid)

hydrogen freq

0 1
 (blank)

%chk=mo6qzvpp.chk
#p M06/def2QZVPP
opt
int(ultrafinegrid)

water

0 1
O
H 1 1.2
H 1 1.2 2 109.0
 (blank)

%chk=mo6qzvpp.freq.chk
%oldchk=mo6qzvpp.chk
#p M06/def2QZVPP
freq
geom=check guess=read
int(ultrafinegrid)

water freq

0 1
 (blank)

Have a lot of fun.


Szabo/Ostlund exercise on spin orbital orthonormality

Consider the spin-orbital integral
$$ \langle \psi_i^{\alpha} \alpha
(\omega) | \psi_j^{\beta } \alpha (\omega) \rangle  = \langle\psi_i^{\alpha}  |  \psi_j^{\beta } \rangle
$$
which is not necessarily orthogonal.

Yes, it is not necessarily orthogonal, but the thing is that you don't need to consider such integrals in the exercise. Read more carefully: spatial orbitals from the first "alpha" set can be multiplied only by the $\alpha$ spin function while spatial orbitals from the second "beta" set can be multiplied only by the $\beta$ spin function. Thus, there are only three types of integrals you need to consider:
$$
\langle \psi_i^{\alpha} \alpha | \psi_j^{\alpha } \alpha \rangle \, ,
\quad
\langle \psi_i^{\alpha} \alpha | \psi_j^{\beta } \beta \rangle \, ,
\quad
\langle \psi_i^{\beta} \beta | \psi_j^{\beta } \beta \rangle \, .
$$


Remember your general chemistry. In an acid-base reaction, the equilibrium favors the side of the reaction with the weaker acid/base pair. Strong bases have weak conjugate acids. Strong acids have weak conjugate bases. Weak bases have strong conjugate acids. Weak acids have strong conjugate bases. Consider the following:

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

If $\ce{HA}$ is a stronger acid than $\ce{HB}$, then $\ce{A-}$ must be a weaker base than $\ce{B-}$.

Here is an example with real compounds:

$$\ce{HCl + NH3 <=> Cl- + NH4+}$$

Even though the right side of the reaction contains ions instead of neutral molecules, the right side is favored because $\ce{HCl}$ is a strong acid and $\ce{NH4+}$ is a weak acid.

When we study organic chemistry, we are given some rules of thumb to help us compare relative acidity:

$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y}$ is more electronegative than $\ce{Z}$ and they are in the same period.
$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y}$ is larger than $\ce{Z}$ and they are in the same group.
$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more resonance stabilization than $\ce{Z-}$ and if $\ce{Y-}$ and $\ce{Z-}$ are otherwise similar.
$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more inductive stabilization than $\ce{Z-}$ and if $\ce{Y-}$ and $\ce{Z-}$ are otherwise similar.
$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more $\ce{s}$-character than $\ce{Z-}$ and $\ce{Y-}$ and if $\ce{Z-}$ are otherwise similar.
$\ce{YH2+}$ is always more acidic than $\ce{YH}$, though it is hard to compare $\ce{YH2+}$ and $\ce{YZ}$ or $\ce{YH}$ and $\ce{YZ2+}$.

With these rules in hand, it is sometimes challenging to remember that we also have an experimental measure of acid strength, and that there are only a limited number of "strong" acids (those acids which are stronger than $\ce{HSolvent+}$).

The $K_\mathrm{a}$ of $\ce{HCl}$ is not easily determinable, since it is more acidic than most protonated solvents. However, the Evans pKa table and other sources often estimate it at -7 or -8, with only $\ce{HBr}$, $\ce{HI}$, and the various "superacids" including such things as $\ce{HSbF6}$ being stronger.

The Evans table lists the $\mathrm{p}K_\mathrm{a}$ of various protonated phosphines. For example

$\ce{CH3PH3+}$ is 2.7 in DMSO ($\ce{HCl}$ is 1.8 in DMSO).
$\ce{Et3PH+}$ is 9.1 in DMSO

Would we expect $\ce{ROPCl2H+}$, which is the intermediate in your reaction, to be more or less acidic than the two reference phosphonium ions above? Both $\ce{RO}$ amd $\ce{Cl}$ are electron withdrawing by induction.

$\ce{Cl-}$ maybe could attack $\ce{H}$ and not $\ce{C}$, but the $\ce{HCl}$ that forms would be so acidic in comparison to everything else in the reaction that something else would take that proton away again, regenerating $\ce{Cl-}$. Once $\ce{RCl}$ forms, there is no other good nucleophile present that is capable of displacing the chloride group.


store alcohol tinctured herbs in polypropylene for extended time

The alcohols should not be an issue, it is used for storing often.  Some oils will affect it.

Here is a random guide that includes a few common oils.


Why does concentration of solutions of alkali metals in liquid ammonia change the magnetic properties?

I always suggest students to try Google Scholar (scholar.google.com) when a simple Google search fails. I just searched three keywords : alkali metals ammonia solutions and the third result is highly relevant.

When your book talks about "in concentrated solution", it means more alkali metal in liquid ammonia. This paper, which you should search in Google Scholar, is titled A Revised Model for Ammonia Solutions of Alkali Metals, Journal of the American Society, Year 1962, volume 84, pg 2264). It talks about the formation of dimers of metals of the type $\ce{M2}$, because as you increase the metal conc. in ammonia solution conductivity decreases.
The authors write, raising objections to older models as completely wrong show the following evidence:

  "We picture the » species as a quadrupolar ionic assembly of $\ce{e- + 2M+}$ in which there is little distortion of either the ammoniated electrons or ammoniated metal ions. Presumably the electrons and ions are held in a square or rhombic configuration."

In short the picture is not that simple an no wonder the authors of a grade 12 book never discuss more details. You can start from here and search more.


Is it possible to have negative bond orders?

No, it's not possible, theoretically either.

If you see the Molecular orbital theory, maximum number of anti bonding electrons = number of bonding electrons.

And bond order is (Number of bonding electrons - number of anti bonding electron)/2

So, the minimum value that you can get is 0 and not negative, as anti bonding electrons can't exceed the number of bonding electrons.


Hybrid orbitals, photoionization and time dependence

There are really two parts to your question - one is to interpret the wikipedia entry, the other is the more general question about the transition of methane from the parent molecule to the ionized $\ce{CH4+}$.

For the first question, I believe the author has muddled the explanation a bit (though it's of course just as likely that I'm the one who is muddled). It sounds to me like they are explaining the reconciliation of molecular orbital (MO) theory and valence bond (VB) theory for $\ce{CH4}$ rather than for the cation or the ionization process itself. Here is what I think they are trying to say:

Electronic structure of $\ce{CH4}$

MO theory
If we carry out an analysis of the delocalized molecular orbitals of methane, we find that the four highest energy occupied orbitals are spread between two energy levels, with one orbital at the lower energy level and the other three at the higher level. The orbitals meet the quantum mechanical requirements of being orthonormal and matching the symmetry operations of the molecule. The energy levels are eigenvalues of the wave function energy operator, so we'll call them eigenstates.
This representation is supported by the experimental data from PES, which shows two ionization energies, with a relative intensity of 3:1.
VB theory
In contrast, the valence bond view of the molecule has four orbitals of equal energy, each one corresponding to a specific C-H bond. These orbitals can be constructed as linear combinations of the MO theory orbitals and are orthonormal to each other. As such, they form an alternate basis set for the same space as described by the MO theory orbitals (that is, any linear combination of the MO theory orbitals can be described by a linear combination of the VB theory orbitals). However, they do not match the symmetry of the molecule, and their energy levels are not eigenvalues for the energy operator.
This representation seems to be in disagreement with the experimental evidence, since it seems to suggest that only one ionization energy should be observed. However, it is not incorrect, as I will explain.
Reconciliation
The reconciliation between the VB description and the experimental result comes from realizing that "a linear combination of MO theory orbitals" is just a different way of describing a superposition of states. All quantum systems exist as a superposition of states. In the mathematical edifice that is quantum mechanics, we require that when we make a measurement of an observable of a superposition of states, the result must still be an eigenvalue for one of the states in the superposition. The probability of a specific eigenvalue appearing is a function of the contribution of the corresponding state to the superposition.

The simplest example of this is measuring the spin of an electron. Unless a constraint is imposed, we expect to measure ($+\frac12$) 50 per cent of the time and ($-\frac12$) 50 per cent of the time. In neither case do we conclude that the electron was fixed in that spin state prior to the measurement. It was in a superposition of both spins until we made the measurement.

Getting back to the methane molecule, we say that all 8 electrons in the valence orbitals can be viewed as existing in superpositions of states, with the only requirement being that within each individual molecule, the electrons are in states that are orthonormal to each other. The VB and MO theory representations are just two of an infinite number of possibilities. When we do the PES experiment, the results must always be eigenvalues, and the intensities will always have a 3:1 ratio, because the average contribution of the eigenstates to the population of orbitals across the population of methane molecules must remain 3:1.

Ionization of $\ce{CH4}$
With the above in mind, we can move on to the harder general question about time evolution of states during and after ionization. Since the measurement in the PES experiment returns an eigenvalue, it would seem that it must force the molecule into an eigenstate, just as measuring the spin of a free electron forces it into a single spin state.

We know that initially the molecule retains the tetrahedral geometry of $\ce{CH4}$, but within a short time it relaxes to a $C_{2v}$ symmetry. This change in symmetry must be accompanied by a change in the wave function and the energies (eigenvalues) of the states. I would assume that this time-evolution of the wave function also allows for scrambling of the eigenstates back to a superposition of states, but I don't know that process well enough to give a clear description. Perhaps someone else can jump in with that.

Conclusion
Referring to your specific questions -

  1.Am I wrong in thinking that superposition states do not necessarily have to obey the stated requirement about molecular symmetry?

As I described, I would say you are correct. Superpositions do not need to meet the symmetry requirement. The only requirement is that they are linear combinations of the symmetry-consistent basis set and that as a group they form an equivalent orthonormal basis set.

  And is my picture of a time-dependent molecular wavefunction immediately after ionization viable?

I think so, but I'm not certain.

  What is the justification for the assumption that the MO energies are not observables?

I don't understand this question. Energy is definitely an observable, and the PES experiment measures it. An observable is anything that can be measured.

UPDATE: Rereading your updated question, I think the confusion is between "not having a fixed value" and "not being an observable". Just because the electrons do not have defined energies does not mean that their energies are not observables. In chemistry, we use language that refers to fixed values of observables like energy, but a fundamental concept of QM is that observables do not have fixed values until they are measured. The measurement process itself collapses the wave function to a specific eigenstate. I tried to clarify that with the example of the electron spin. Position is another common example. Ideally, we would preface lots of statements with "if you were to measure x. . " but that would be really cumbersome.

  And how is it justified from the perspective of the Wikipedia article that we have to obey the symmetry of the molecule in the ionized product state, but for the hybrid orbitals in the ground state, we can for some reason disobey the same principle? Isn't this highly inconsistent?

As I tried to explain, I believe the Wikipedia article was mixing up the parent molecule and the ionized state. The immediate effect of the ionization (ie measurement) is to force the molecule into an energy eigenstate. However, that is an eigenstate of the immediate product, which still resembles the methane parent in symmetry and orbitals. There is a time evolution to the ground state of the cation, which has a different symmetry ($C_{2v}$) and different eigenstates. Since we have performed no measurements on this molecule, it is not fixed in any of those eigenstates. Since it has evolved away from the initial $T_d$ symmetry, it is also no longer an eigenstate of the parent.

To summarize, the only point in the process in which the electrons are in a fixed eigenstate is immediately after ionization (before relaxation), and this eigenstate is described by the set of symmetry-allowed MO theory orbitals that correspond to the $T_d$ symmetry of $\ce{CH4}$, not the $C_{2v}$ symmetry of $\ce{CH4+}$.


Wieland's method for triphenylamine bromination

From the reference you supplied the Wieland paper is Chem. Berichte 40 (1907) 4278

The bromination method (3 eq of Bromine in chloroform slowly added to a chloroform soln of triphenylamine) in the experimental section of the reference you supplied appears to be Wieland's method - it is referenced as such


If you want to determine the total content of oxalate you would need to extract it from the matrix with hot HCl 1M, for instance, since it will dissolve calcium oxalate crystals. Afterwords you can precipitate oxalate again at pH=4-5 with calcium nitrate, thus forming white crystals of calcium oxalate. You can redissolve them by adding sulfuric acid and identify oxalate by taking advantage of its reducing properties - if you add potassium permanganate it will be reduced from Mn +7 (purple colored) to Mn +2 (brown colour) and oxalate will be oxidized to carbon dioxide. But of course you start with a complex matrix, the easiest thing would be to use chromatography to separate and identify oxalate.


Yes, Charlotte Emma Moore and her tables really exist(ed).

See Atomic Energy Levels, 2nd edition.


Tabulation table such as the ones you reference, describe 'typical' bonds when using generic groups such as R.  You are right in your discomfort that the numbers should be somewhat different, but that would require a thermochemical measurement for every known molecule, which is difficult to achieve in practice.

Calculations such as the one you are performing are 'estimates', and while good ones, they are never as credible as an actual experiment.


When there is a situation of multiple equilibria, all of them must be satisfied. If this is the case, the whole system will be at equilibrium.

For your example and as you report, the maths let you write that

$
K_{3}=K_{1}\times K_{2} \quad (1)
$

This final equation, beyond being the mere mathematical combination of two initial equations, is obtained by considering that Gibbs free energy is a function of state: iIt describes "quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state" (Wikipedia). One may get to equilibrium via an unique straight step, or through several steps. But the final destination will be the same.

This definition allows us to write:

$
\Delta G_{3}^{o}=\Delta G_{1}^{o} + \Delta G_{2}^{o}  \quad (2)
$

Equation (2) can be converted into equation (1), by remembering that $
\Delta G^{o}=-RT\text{ln} K$.

We can first of all say that $K_{1}$ and $K_{2}$ relate to the two "pieces" of the reaction path one has to cover to reach the overall equilibrium. Knowing them is essential, if one wants to obtain the quantities for all the chemical species, at equilibrium.

$K_{3}$, on the other hand, relates to the "shortcut" (the unique "ideal" step connecting the initial and final states of a system) and is linked to $K_{1}$ and $K_{2}$, through equation (1).

Using the examples you provide and assuming initially 1 mole for $A$ and 1 mole for $B$, we write:

$
K_{1}=\frac{(y-z)y}{(1-y)^2}
$

$
K_{2}=\frac{z}{y-z}
$

We could more generally write:

$
K_{1}=f(y,z)
$

$
K_{2}=g(y,z)
$

The two constants are different functions: but they both depend on the same parameters, $y$ and $z$, which are temperature and pressure (for gases) dependent. If temperature changes, both constants will accordingly change.

To determine the equilibrium quantities, at given temperature and pressure, we'll need the values for $K_{1}$ and $K_{2}$, since knowing only $K_{3}$ won't be enough.

Beyond all this, kinetics will affect the time needed to reach system equilibrium: but that's another story.


Why does the excitation and emission spectrum of a fluorescent molecule have overlap?

Energy-Level Diagram
There is overlap because between the excitation and the emission spectra because there exists a Range of orbitals at different energies. You'll always notice that the energy for excitation is always higher, or the lambda value is always lower (these say the same thing since E=planck's constant * frequency).

I've attached a HE-Ne laser schematic from Verdeyen's Laser Electronics book. For a better visual. What you need to know is let's say your gain medium, in this case He-Ne was excited at it's lowest wavelength (we'll call it Lambda1) then it's emission wavelength may be only 2nm higher (or red shifted), Lambda2.

However, as you'll notice in this schematic there are many wavelengths at which this gain medium (He-Ne) can emit light. These other possible excitation orbitals can have lower energy levels that correspond to higher wavelengths and thus overlap on the spectrum.


The answer is hard-soft acid/base theory (or HSAB theory). Essentially, some Lewis acids and bases are hard and some are soft, with a few borderline cases.

  Hard acids and bases tend to have

Small atomic/ionic radius
Highly charges (or highly positive/negative oxidation state)
High electronegativity for bases and low electronegativity for acids
Low polarizability

  Soft acids and bases tend to have

Large atomic/ionic radius
Low or zero charge / oxidation state
Moderate electronegativity
High polarizability

According to HSAB theory, matches that are hard-hard and soft-soft form strong interactions, where hard-soft matches are weaker interactions.

$\ce{Ag+}$ is a soft Lewis acid. $\ce{Cl-}$ is a hard Lewis base, while $\ce{I-}$ is a soft Lewis base. Therefore, we would predict that $\ce{Ag+}$ and $\ce{I-}$ will have have a stronger attraction than than $\ce{Ag+}$ and $\ce{Cl-}$.

We can see this difference in attraction in the solubilities of $\ce{AgCl}$ and $\ce{AgI}$ in water. $\ce{AgCl}$ is much more soluble in water than $\ce{AgI}$ (realizing that both are nearly insoluble). See the solubility constant data:

$\ce{AgCl}\ \ \ k_{\text{sp}}=1.8\times 10^{-10}$
$\ce{AgI}\ \ \ \ \ \ k_{\text{sp}}=8.5\times 10^{-17}$


Kinetics of interstellar chemistry: Applying steady state to formation of H₂

OK, let's step through your questions one at a time.

  First off, what do they mean by steady-state number here?

Intermediates formed in reactions usually exist in low concentration.  The steady-state assumption assumes that the concentration of this (low-concentration) intermediate will not change appreciably over the course of the reaction once steady state has been established.  The rate of its formation is assumed to be equal to the rate of its consumption.  Making the steady state assumption allows us to recast the previous statement as a mathematical equation as we'll see below.

  In this case why must the rate of absorption be equal to the rate of desorption

Just like in a chemical equilibrium where it takes a finite time for the equilibrium to establish itself, it takes a finite time before the steady state is established. Once there is a finite, non-zero concentration of the intermediate and we have arrived at the steady state, then the rate of absorption and desorption will be equal and the non-zero concentration of the intermediate(s) will remain constant – a steady state.

Applying the steady-state assumption to the problem at hand allows us to state that for all of the intermediates (dust particles with 1 or 2 hydrogen atoms on them), the rate of hydrogen atom adsorption on an intermediate must be equal to the rate of hydrogen atom desorption from that intermediate. This can be written as

$$k_\mathrm a [\ce{H}] = k_\mathrm d N$$

and then rearranged and solved:

$$\begin{align}
N &= \frac{k_\mathrm a}{k_\mathrm d}[\ce{H}] \\
&= \frac{\pu{1.4 * 10^-5 cm3 s-1}}{\pu{1.9 * 10^-3 s-1}} \cdot \pu{10 cm^-3} \\
&= \pu{7.4 * 10^-2}
\end{align}$$

  Why is it $k_\mathrm d N$?

Because $k_\mathrm d N$ represents the total rate of desorption of hydrogen atoms from dust particles, the right side of our verbal equation. It represents desorption from $x_1$ and $x_2$ - the only two species that can desorb, and it is equal to

$$k_\mathrm d N = k_\mathrm d x_1 + 2k_\mathrm d x_2$$

the factor of "2" arising because there are 2 hydrogen atoms on each dust particle in $x_2$, and both are viable candidates for desorption.

  Further, why is the rate of desorption not simply the reverse of the absorption reaction and thus the same rate but with a negative sign?

It is the same rate, that's what $k_\mathrm a [\ce{H}] = k_\mathrm d N$ means.  Rates are always a positive number, we are multiplying a rate constant times a concentration, both positive numbers.  However, when we write an equation and insert the rate expression into it, we can insert a minus sign before the rate expression if the rate expression represents a pathway that is causing the concentration of a product, reactant or intermediate to decrease.

  How did they come up with those rates?

Adsorption rate: $r_\mathrm a = \sum_m k_\mathrm a [\ce{H}] x_m = k_\mathrm a [\ce{H}]x_0 + k_\mathrm a [\ce{H}]x_1$
This is just the typical kinetic expression – adsorption rate constant multiplied by the concentration of hydrogen atoms available to adsorb times the concentration of the various dust particles that can adsorb a hydrogen atom, which are only $x_0$ and $x_1$.
Desorption rate: $r_\mathrm d = \sum_m k_\mathrm d m x_m = (k_\mathrm d \cdot 0 \cdot x_0) + (k_\mathrm d \cdot 1 \cdot x_1) + (k_\mathrm d \cdot 2 \cdot x_2)$
Here $m$ is a concentration term analogous to $[\ce{H}]$.  It represents the number of hydrogen atoms on a dust particle, and can be 0, 1 or 2.  Note how it "removes" $x_0$ from the desorption equation as it should since a particle with no hydrogen atoms cannot desorb.  Notice too how it adjusts $x_2$ for the fact that there are two adsorbed hydrogen atoms in this case, so desorption is statistically twice as likely to happen.  It differs from $[\ce{H}]$ in its units: $[\ce{H}]$ is in $\pu{atoms/cm3}$ while "m" is in units of atoms/dust particle.
Reaction rate: $r_\mathrm r = \left(\frac{1}{2} \cdot k_\mathrm r \cdot 2 \cdot 1 \cdot x_2\right) \left(\frac{1}{2} \cdot k_\mathrm r \cdot 1 \cdot 0 \cdot x_1\right) = k_\mathrm r x_2$

EDIT 1:

  Also, after your explanation I understood how they got the absorption and desorption rates, but I'm still confused on how they got the reaction rate $\left(\frac{1}{2} \cdot k_\mathrm r \cdot 2 \cdot 1 \cdot x_2\right) \left(\frac{1}{2} \cdot k_\mathrm r \cdot 1 \cdot 0 \cdot x_1\right)$. Could you please elaborate further on where all the numbers came from?

Note we have a factor of $1/2$ because it takes 2 hydrogen atoms to produce 1 hydrogen molecule.  The $2$ and the $1$ come from the concentration terms $m(m-1)$.  In the "Reaction" case we have to combine two hydrogen atoms.  It is a bimolecular reaction and just like in the bimolecular reaction $\ce{A + B -> C}$ where the rate is proportional to $[\ce{A}][\ce{B}]$, here the analogous concentrations are $m(m-1)$.  Initially the concentration is $m=2$, but after that first hydrogen atom is "selected" the remaining concentration from which to pick our second hydrogen atom has been decreased to $m-1$.  What we're saying is that when we pick the first hydrogen we have two atoms to choose from, when we pick the second hydrogen atom we have one atom to choose from. If we had 5 hydrogen atoms on a particle and were still going to make 2 picks the equation for our two picks would become $5(5-1)=20$.  We would expect this case to proceed $(20/2)=10$ times faster than the case with only 2 atoms on the dust particle.

There are 2 ways to form $\ce{H2}$ in our problem. Label the hydrogen atoms A and B, then we can either pick via an AB or BA sequence. $m(m-1)$ is exactly analogous to $[\ce{H}][\ce{H}]$, it's just not a square because there is a significant difference between $2\cdot 2$ and $2\cdot 1$. When dealing with a mole of atoms ($N = 6 \cdot 10^{23}$, or Avogadro's number) and they react with one another in a bimolecular reaction, there is no need to write $(6\cdot 10^{23})(6\cdot 10^{23} - 1)$, but in our case with such small numbers the difference matters.

  Why does absorption have $[\ce{H}]x_m$ but desorption have $mx_m$?

As noted above, $m$ is a concentration term analogous to $[\ce{H}]$, but with different units.

  Finally, for the reaction rate, where did they get $m(m−1)$?

See above.

  why does the $k_\mathrm a$ term have $[\ce{H}]$ while the other terms do not?

The other terms use $m$ which, as explained above, is analogous to $[\ce{H}]$ in that they both reflect concentrations, $[\ce{H}]$ for hydrogen atoms in space and $m$ for hydrogen atoms on dust particles; they use different units as explained above.

  What happened to the $mx_m$ that was mentioned above?

Since the full rate expressions have been written out, they have been replaced with integers, 0, 1, or 2, representing the concentration of hydrogen atoms on a dust particle.

  Further, how did they derive the rate of change for $x_1$ and $x_2$?

The rate of change of $x_1$ is the rate of formation minus the rate of destruction.

The rate of formation is given by $k_\mathrm a[\ce{H}]x_0$ (the rate of a bare dust particle picking up a hydrogen atom) plus $2k_\mathrm d x_2$ (the rate of a dust particle with two hydrogen atoms losing one).

The rate of destruction is given by $k_\mathrm a[\ce{H}]x_1 + k_\mathrm d x_1$ (the rate of a dust particle with one hydrogen atom either picking up one more, or losing it).

For the time being, I'll leave the rate of change of $x_2$ as an exercise for you to test yourself on what you've learned.


What is the most accepted mechanism for the reaction of ketones with m-dinitrobenzene?

In the Janovsky reaction, an aldehyde or ketone which contains at least one $\alpha$ methylen group reacts with 1,3-dinitrobenzene and a strong base (like $\ce{NaOH}$) to form a colored Meisenheimer complex (source). The enolate acts as a nucleophile which attacks the aromatic ring that is electron-deficient due to the presence of multiple electron-withdrawing nitro substituents.

image source

As the reaction mechanism relies on the formation of an enolate, non-enolizable aldehydes like, for example, formaldehyde or benzaldehyde, do not resond to this test. Likewise, non-enolizable ketones (like benzophenone) cannot react in this way. The color of the formed complex varies with the number of nitro substituents of the benzene ring, with 1,3-dinitro compounds giving blue complexes, while trinitro compounds yield a red coloration (source).

This paper contains a list of the colors produced by various aromatic nitro compounds when reacted with acetone and $\ce{NaOH}$. It is reported that most mononitro compounds do not produce any color at all, with the exeption of 3-nitro-4-aminotoluene (orange) and 3-nitro-4-aminotoluene benzoate (orange-red). Some examples for the colors obtained from the reaction of dinitro compounds are: m-dinitrobenzene (purplish blue), 2,4-dinitrotoluene (blue), 2,4-dinitrophenol (yellow-orange), 2,4-dinitroresorcinol (brownish-green) and p-dinitrobenzene (green-yellow).


Is there a way to counteract the problems with weighing hot objects?

Depending on geometry, mass and temperature (hot is pretty vague), you might be able to use an inertial balance.

If the device used to hold the sample is rigid, you could calibrate its resonant frequency with and without the sample and with known weights.

See this high school lab and a video of the procedure.


Trend in the melting point down the group

There's a counter effect in play here: van der Waals forces. As you become larger and larger, induced dipole-induced dipole interactions become stronger and stronger.

You'll notice that, in halogens (where abstracting an electron is a fairly tall order), the trend is that, the bigger your atom, the higher your boiling point, regardless of where you are in the group. The boron group loses the conflict of ease of abstraction versus van der Waals forces at around the gallium/indium leap, and for the carbon group, the conflict is lost to van der Waals forces only between tin and lead.


Problem with synthesis of calcium formate

You probably did get your anticipated reaction, but unlike carbon dioxide or hydrogen ammonia is highly soluble in water.  Unless your reactants are more highly concentrated than is possible with the limited solubility of calcium hydroxide, the ammonia remains in solution instead of giving off bubbles or a smell.
There are various ways to render this reaction visible:

Heat the solution, driving out any dissolved ammonia.  If you generate a smell of ammonia, or better if the solution emits a gas that turns moist red litmus paper blue, you found the ammonia. You should compare with a blank where you just have the unreacted ammonium formate.

If you used solid calcium hydroxide, double-check how much you
dissolved.  Calcium hydroxide has only a solubility of 1-2 grams
per liter, so if
you dissolved more than that it must have reacted.  If you were
using magnesium hydroxide the solubility difference is even more
evident, for magnesium hydroxide is also strongly basic enough to
displace ammonia but its solubility without the reaction is very
low.

Check the pH.  Ammonia is a weaker base than the calcium hydroxide that displaces it, so you get a lower pH than you would have gotten just adding the same amount of calcium hydroxide to plain water (which in this case you would do for comparison).

Chemistry is an experimental science.  Therefore experimental design is essential for chemistry.


Does the field of organic chemistry cover all group 14 elements or only carbon?

The other group 14 elements don't have similar chemistry. In general, the elements in the 2p block are "unusual" compared to the 3p/4p/5p blocks, just as how the elements in the 3d block are quite different compared to the 4d/5d blocks, and just as how hydrogen ("1s block") is very different from the alkali metals.

The obvious difference in chemistry comes from the unparalleled ability of carbon to catenate, or in other words, to form chains of single bonds. Silicon does not do this to any appreciable extent, mainly due to bond energetics (the Si–Si bond, for example, is much weaker than C–C).

Many of the functional groups familiar to organic chemistry have little parallel in silicon chemistry. Carbonyl groups are ubiquitous, but Si=O compounds are barely stable (see Wikipedia), which is a result of the poor π overlap between Si (valence 3p orbitals) and O (valence 2p). The same applies for C=C and C≡C bonds. Silicon analogues are known, but are very difficult to make and generally need to be stabilised using a variety of tricks (see: How come there aren't any silicon analogs of alkenes or alkynes? and links within).

So, traditionally, the chemistry of Si/Ge/Sn/Pb is within the realm of inorganic chemistry. Of course, it is never a clear-cut distinction, since these elements can coexist with carbon in the same molecule. But that is a story for another time.


What element has good conductivity but is also brittle?

Whether a "metalloid" is both brittle and a good conductor depends on what you mean by "metalloid", which varies in literature, what is a "good" level of (electrical) conductivity, and even what exactly you mean by "brittle" versus "ductile".  Silicon, for instance, is only a semiconductor, with much less electrical conductivity than metals, when pure.

Wikipedia identifies beryllium and magnesium as brittle metals, although the latter can be made ductile with minor alloying.  Zinc is called "slightly brittle".


Copper (II) chloride reduction with Hydrogen gas

There are better ways to do it.  Copper (II) can be reduced to copper (I) with a variety of reagents, especially in the presence of a soft-base halide ion such as chloride which precipitates or complexes with the copper (I) (which is a good way to prevent "overshooting" to metallic copper).  Some specific approaches can be seen here.


Is the surface of the planet Mercury made of copper minerals?

I think you probably fell into a trap concerning color-coding.

In reality, mercury is dark grey and mostly covered in silicate rocks/dust.

From a scientific point of view, the "real color" of an object like mercury is of little interest. What's more interesting, is it topography etc.
To make it easier to interpret data, man often uses color-coding (since our color-vision is quite ok) to communicate different information about such things.

Or ist just about beauty:
When you see pictures of nebulae like the pillars of creation you notice they are in a lot of different colors. This is because a lot of the radiation is not in the visible spectrum, and you can "color-code" it to make it interesting to look at.


You may have just experienced supercooling.

Ordinary crystallization upon reaching the freezing point of a liquid requires nucleation sites - places for crystals to form. In absence of nucleation sites, water can be cooled down to lower temperatures until crystal homogeneous nucleation occurs. In water, this can occur as low as $224.8\ \mathrm{K\ (-48.3\ ^\circ C,\ -55\ ^\circ F)}$.

Nucleation can also be induced by disrupting the system, which happened when you shook the bottle. Even so, the fact that it only became slush may attest to the presence of impurities. A $\mathrm{19\ ^\circ F,\ 10\ ^\circ C}$ freezing point depression is not unreasonable depending on the amount of saliva that might have ended up back in the bottle.


Gaussian calculation - PCM model for IRC compared to gas phase?

Option 2 is the most scientific and rigorous. You should re-run the IRC in solvent again because the reaction path can be different from what you got in the gas phase.
Now, will it be different in solvent? Unless you are dealing with some exotic system, different reaction path with implicit solvation is unlikely (provided you have the same TS and minima with PCM). However, the reaction barriers may significantly change, which can cause two subsequent steps to become one single step (the TS between them forming a shoulder in the IRC). I have seen this happen.
What matters though is that different reaction paths can happen, so you should run the IRCs in solvent if you want to be sure of your result (also because you want to publish it).


The formula

$$c_1\cdot V_1=c_2\cdot V_2$$

is correct.

Since your stock solution has a concentration of $c=0.1\ \mathrm{mol\ l^{-1}}$, the value for the initial concentration $c_1$ is always

$$c_1=0.1\ \mathrm{mol\ l^{-1}}$$

Since you want the final volume of each solution to be $V=30\ \mathrm{ml}$, the value for the final volume $V_2$ is always

$$V_2=30\ \mathrm{ml}$$

You want four individual solutions (labelled $\mathrm a$, $\mathrm b$, $\mathrm c$, and $\mathrm d$) with four different final concentrations ($0.1\ \mathrm{mol\ l^{-1}}$, $0.05\ \mathrm{mol\ l^{-1}}$, $0.025\ \mathrm{mol\ l^{-1}}$, $0.0125\ \mathrm{mol\ l^{-1}}$. These are the individual values for the final concentration $c_2$:

$$\begin{align}
c_{2,\mathrm{a}}&=0.1\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{b}}&=0.05\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{c}}&=0.025\ \mathrm{mol\ l^{-1}}\\[6pt]
c_{2,\mathrm{d}}&=0.0125\ \mathrm{mol\ l^{-1}}
\end{align}$$

Now you can rearrange the above-mentioned equation to solve for the initial volume of the stock solution $V_1$:

$$\begin{alignat}{2}
V_1&=\frac{c_2\cdot V_2}{c_1}\\[6pt]
V_{1,\mathrm a}&=\frac{0.1\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=30\ \mathrm{ml}\\[6pt]
V_{1,\mathrm b}&=\frac{0.05\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=15\ \mathrm{ml}\\[6pt]
V_{1,\mathrm c}&=\frac{0.025\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=7.5\ \mathrm{ml}\\[6pt]
V_{1,\mathrm d}&=\frac{0.0125\ \mathrm{mol\ l^{-1}}\times30\ \mathrm{ml}}{0.1\ \mathrm{mol\ l^{-1}}}&&=3.75\ \mathrm{ml}
\end{alignat}$$

You measure these volumes of the stock solution and transfer them into four separate vessels.

Finally, you have to fill up each solution with pure water to the final volume of $V_2=30\ \mathrm{ml}$. The required additional volume of water $V_{\ce{H2O}}$ is approximately

$$\begin{alignat}{2}
V_{\ce{H2O}}&\approx V_2-V_1\\[6pt]
V_{\ce{H2O},\mathrm a}&\approx30\ \mathrm{ml}-30\ \mathrm{ml}&&=0\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm b}&\approx30\ \mathrm{ml}-15\ \mathrm{ml}&&=15\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm c}&\approx30\ \mathrm{ml}-7.5\ \mathrm{ml}&&=22.5\ \mathrm{ml}\\[6pt]
V_{\ce{H2O},\mathrm d}&\approx30\ \mathrm{ml}-3.75\ \mathrm{ml}&&=26.25\ \mathrm{ml}
\end{alignat}$$


Your calculation is correct. The standardized definition of the change in internal energy $U$ for a closed thermodynamic system is

$$\Delta U=Q+W$$

where $Q$ is amount of heat transferred to the system and $W$ is work done on the system (provided that no chemical reactions occur). Therefore, heat transferred to the system is assigned a positive sign in the equation
$$Q=750\ \mathrm J$$
whereas work done by the system on the surroundings during the expansion of the liquid is assigned a negative sign
$$W=-200\ \mathrm J$$
Thus, the change in internal energy is
$$\begin{align}
\Delta U&=Q+W\\
&=750\ \mathrm J-200\ \mathrm J\\
&=550\ \mathrm J\\
\end{align}$$

However, the question is a bit flawed since the given values are not typical for a liquid. By way of comparison, realistic values for water are shown in the following table.

$$\textbf{Water (liquid)}\\
\begin{array}{lllll}
\hline
\text{Quantity} & \text{Symbol} & \text{Initial value (0)} & \text{Final value (1)} & \text{Change}\ (\Delta) \\
\hline
\text{Amount of substance} & n & 1.00000\ \mathrm{mol} & 1.00000\ \mathrm{mol} & 0\\
\text{Volume} & V & 18.0476\ \mathrm{ml} & 18.0938\ \mathrm{ml} & 0.0462\ \mathrm{ml} \\
 & & 1.80476\times10^{-5}\ \mathrm{m^3} & 1.80938\times10^{-5}\ \mathrm{m^3} & 4.62\times10^{-8}\ \mathrm{m^3} \\
\text{Pressure} & p & 1.00000\ \mathrm{bar} & 1.00000\ \mathrm{bar} & 0 \\
 & & 100\,000\ \mathrm{Pa} & 100\,000\ \mathrm{Pa} & 0 \\
\text{Temperature} & T & 20.0000\ \mathrm{^\circ C} & 29.9560\ \mathrm{^\circ C} & 9.9560\ \mathrm{^\circ C} \\
 & & 293.1500\ \mathrm{K} & 303.1060\ \mathrm{K} & 9.9560\ \mathrm{K} \\
\text{Internal energy} & U & 1\,511.59\ \mathrm{J} & 2\,261.58\ \mathrm{J} & 749.99\ \mathrm{J} \\
\text{Enthalpy} & H & 1\,513.39\ \mathrm{J} & 2\,263.39\ \mathrm{J} & 750.00\ \mathrm{J} \\
\hline
\end{array}$$

When $1\ \mathrm{mol}$ of water with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ is heated with $\Delta H=Q=750\ \mathrm J$ at a constant pressure of $p=1\ \mathrm{bar}$, the resulting expansion is actually only
$$\begin{align}
\Delta V&=V_1-V_0\\
&=18.0938\ \mathrm{ml}-18.0476\ \mathrm{ml}\\
&=0.0462\ \mathrm{ml}\\
&=4.62\times10^{-8}\ \mathrm{m^3}
\end{align}$$

The corresponding pressure-volume work is
$$\begin{align}
W&=p\Delta V\\
&=100\,000\ \mathrm{Pa}\times4.62\times10^{-8}\ \mathrm{m^3}\\
&=0.00462\ \mathrm J
\end{align}$$
which is clearly below the value given in the question $(W=200\ \mathrm J)$.

The values given in the question are appropriate for a gas. For example, realistic values for nitrogen are shown in the following table.

$$\textbf{Nitrogen (gas)}\\
\begin{array}{lllll}
\hline
\text{Quantity} & \text{Symbol} & \text{Initial value (0)} & \text{Final value (1)} & \text{Change}\ (\Delta) \\
\hline
\text{Amount of substance} & n & 1.00000\ \mathrm{mol} & 1.00000\ \mathrm{mol} & 0\\
\text{Volume} & V & 24.3681\ \mathrm{l} & 26.5104\ \mathrm{l} & 2.1423\ \mathrm{l} \\
 & & 0.0243681\ \mathrm{m^3} & 0.0265104\ \mathrm{m^3} & 0.0021423\ \mathrm{m^3} \\
\text{Pressure} & p & 1.00000\ \mathrm{bar} & 1.00000\ \mathrm{bar} & 0 \\
 & & 100\,000\ \mathrm{Pa} & 100\,000\ \mathrm{Pa} & 0 \\
\text{Temperature} & T & 20.0000\ \mathrm{^\circ C} & 45.7088\ \mathrm{^\circ C} & 25.7088\ \mathrm{^\circ C} \\
 & & 293.1500\ \mathrm{K} & 318.8588\ \mathrm{K} & 25.7088\ \mathrm{K} \\
\text{Internal energy} & U & 6\,081.06\ \mathrm{J} & 6\,616.83\ \mathrm{J} & 535.77\ \mathrm{J} \\
\text{Enthalpy} & H & 8\,517.87\ \mathrm{J} & 9\,267.87\ \mathrm{J} & 750.00\ \mathrm{J} \\
\hline
\end{array}$$

When $1\ \mathrm{mol}$ of nitrogen with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ is heated with $\Delta H=Q=750\ \mathrm J$ at a constant pressure of $p=1\ \mathrm{bar}$, the resulting pressure-volume work is

$$\begin{align}
W&=p\Delta V\\
&=100\,000\ \mathrm{Pa}\times0.0021423\ \mathrm{m^3}\\
&=214.23\ \mathrm{J}
\end{align}$$
The corresponding enthalpy balance
$$\begin{align}
\Delta H&=\Delta U+W\\
750.00\ \mathrm{J}&=535.77\ \mathrm{J}+214.23\ \mathrm{J}
\end{align}$$
is quite similar to the values of the question $(\Delta H=Q=750\ \mathrm J,$ $\Delta U=550\ \mathrm J,$ and $W=200\ \mathrm{J}).$


Reaction of hydrogen peroxide with sodium carbonate

The impossibility of a reaction producing simultaneously $\ce{Na_2O_2, H_2O_2}$ and $\ce{CO_2}$ can be proved by the following arguments, drawn from the Merck Index, under the reference "$8728$. Sodium Peroxide".

1) $\ce{Na_2O_2}$ is produced by burning metallic sodium in a current of air, from which carbon dioxide has been removed.

2) $\ce{Na_2O_2}$ reacts so easily with $\ce{CO_2}$ that it is used for rendering air charged with $\ce{CO_2}$ respirable as in submarines and diving bells.

3) $\ce{Na_2O_2}$ absorbs water from the air forming sodium hydroxyde and hydrogen peroxide, the latter quickly decomposed into Oxygen and water.

Ref.: Maryadele O'Neil, Ann Smith, et al., The Merck Index, An Encyclopedia of Chemicals, 13th Ed. 2001, Merck & Co Inc., Whitehouse Station, NJ, USA.


How would you make/produce sulfuric acid with tools only the romans had?

There are several naturally occurring sulfate minerals such as copper(II) sulfate (pentahydrate mineral name: chalcanthite) that will decompose into $\ce{SO3}$ when heated to a few hundred degrees, which produces sulfuric acid when dissolved in water.
These compounds were known as vitriols due to their glassy appearance and sulfuric acid was known as oil or spirit of vitriol since it was derived from these minerals.
Chalcanthite seems to be a natural oxidation product of copper deposits.
Copper(II) sulfate in particular was known as Roman vitriol, but I don't know if this means that that was the primary source for the Romans.

In any case, ancient civilizations did not have any means of mass producing sulfuric acid and industrialized processes involving sulfur dioxide from the combustion of elemental sulfur arose much later.


Are there any common and identifiable hazards to being a Chemistry Teacher?

I have been teaching chemistry at the college level (if you include my years as a TA in graduate school) since 1999.  While there are certainly dangers involved in dealing with hazardous chemicals, I would not say that the job is unusually dangerous if you follow safe practices and use decent equipment.  I have known many chemistry teachers and professors who lived well into their 80's and beyond.  You're probably going to find certain workplaces which, for whatever reason, are more dangerous than most.  I have heard of (but not verified) an incident at a local college where five female biology professors were diagnosed with uterine cancer, an event which is extremely improbable to occur by chance.  Obviously, there is a substantial need to investigate such a situation and see if there is a common carcinogen present which is causing the cancer.  Such a situation could conceivably occur in the vicinity of a chemistry lab, although it seems unlikely if all standard precautions are followed.

The greatest source of danger is likely to be the students themselves.  When you have many young people together who are (1) easily distracted, (2) not particularly aware of their surroundings, and (3) lacking in manual dexterity (e.g. clumsy), they are far more likely to be a source of injury than the chemicals on their own.  So long as you are willing to invest the time in enforcing safety in your labs on a consistent basis, I think you're going to be just fine.


Iodine flasks differ from normal conical (Erlenmeyer) flasks in three main ways:

The ground glass joint.
The modified stopper.
The funnel shaped top.

Iodine itself, and iodine related molecules produced during the analysis are, to a certain extent, volatile (you can see this if you just put a bead of iodine in a sealed vial and watch the vial fill with vapour). The ground glass joint of the iodine flask prevents these vapours from escaping, improving the accuracy of your analysis.

As for the funnel, its simply there to help get your stuff into the flask in the first place, preventing loss through spillage and preventing the need for a separate glass funnel. Anything that ends up on the tapered neck is simply rinsed directly into the flask.


Predicting orbital angular momentum effects on magnetic moments

When do exceptions to this behaviour occur ? For example I have found that $\ce{Co(bipy)_{3}(ClO4)}$ has a lower experimental magnetic moment than the spin only formula predicts despite being a $\ce{Co(I)}$ (d8) compound with a $^{3}\!A_{2g}$ ground state, but I do not understand why.

According to ADVANCES IN INORGANIC CHEMISTRY AND RADIOCHEMISTRY, Volume 12, page 193 "The tris(bipyridal) cobalt(I) cation has a magnetic momment of 3.2-3.4 B.M. which is independent of temperature from 70K to 340K".

Since there are two unpaired electrons, the spin only value is $\sqrt{2(2+2)} =2.83$.  So your experimental data are inconsistent with the literature, as 3.2-3.4 is greater than the spin only value, 2.83.

  Is the deviation from the spin only formula always positive for $T$ states ? If not can we predict its sign similarly ?

No, the deviation from the spin only value is not always positive for complexes with T ground states. The main characteristic of complexes with T ground states is that the magnetic moment is temperature dependent, because the spin-orbital coupling splits the T state into states that differ in energy by an amount close to kT.

For example $\ce{Cs2VCl6}$ has a magnetic moment of 1.4 at 80K and 1.8 at 300K, with the spin only formula predicting 1.73.  So deviation can be negative, and isn't necessarily positive or negative even for the same compound, depending upon temperature.

See section 4.3 of Dr. Mark D. Spicer's magnetism lecture for more information.


In a candle, the wax first melts, then turns into a vapor above the wick and the oxidation region, where the vaporized fuel mixes with air, surrounds the central, cooler region. You can demonstrate this by extinguishing a candle and then relighting the smoke above the candle... the flame will flash back to the wick. This "flash-back" can be hazardous in a lab using ether, since a nearby flame can ignite the fumes which will burn back to the ether can.

Incense, smoldering cotton and similar combustion relies on thermal degradation of a solid, producing a narrow flame-front on the surface of the solid fuel. Since oxygen cannot penetrate the solid, the rate of combustion is limited. For that reason, cotton bales burn slowly but are very difficult to extinguish, since the interior remains hot and water cannot quickly penetrate. On the other hand, stretching a cosmetic cotton ball changes the burning from smoldering to an intense flame.


Which of the these SN1 solvolysis of two alkenyl bromides reactions is faster?

[...] SN1 solvolysis [...]

Let's recall what we know about these. We have a two-step reaction, consisting of

heterolysis of the starting material
association of a nucleophile $\ce{Nu}$

We remember that the heterolysis is not assisted by $\ce{Nu}$, there is no backside attack of $\ce{Nu}$, pushing out the leaving group. We also remember that the association step (addition of $\ce{Nu}$ to the carbocation intermediate) is usually pretty fast. Consequently, heterolysis, i.e. the formation of a carbocation is the rate-determining step.

Thank's to good old Arrhenius, we know that the rate is determined by the activation energy of this step.

The activation energy is lower when the resulting cation is more stable.

The cation is more stable when the delocalization of the charge is better.

Distribution of charge on the vinyl cation is better when it's actually a
styryl cation.

What is the geometrical requirement for this interaction and what might prevent it?


Why are acids characterized by giving hydrogen and bases characterized by receiving it when acid and base reactions are double displacement reactions?

$\ce{KOH}$ does not exist as a molecule.  In water it is completely dissociated as $\ce{K+}$ and $\ce{OH-}$.  It does not make sense to write $\ce{KOH_2^+}$ for this reason.


This difference in Gibbs Free Energy is only due to entropy right?

This is mostly true.  (I think it is 100% true when concentrations can be used in the reaction quotient $Q$, but am not sure if it is strictly true when solutions are not ideal and activities have to be used instead.)

  Because enthalpy is a constant?

Enthalpy is a "constant" only because of the way electrochemical standard states are defined.

  So obviously, entropy significantly
  interferes in the process of electrolysis.

Have no idea what this means.  How does it "interfere"?

  And yet, it is ignored in
  virtually all spheres of chemistry.

False.

  I'm beginning to doubt my chemical
  intuition because of entropy effects, and faith in it can only be
  restored if entropy really was negligible. So it is negligible or not?
  If not, why is it always ignored?

It isn't negligible and is not ignored.


Why does N₂ react with O₂ to Form NO at high temperatures?

$\Delta G = \Delta H - T \Delta S$

In the case of the $\ce{N2 + O2 -> 2NO}$ , $\Delta H$ and $\Delta S$ are both positive, so the reaction is thermodynamically favorable at high temperature (such as in lightning) but not at low temperature.

If the temperature drops to room temperature after NO is formed, it is thermodynamically favorable for NO to decompose to nitrogen and oxygen.

However, that NO is unstable at room temperature tells us nothing about the rate of the decomposition reaction.  In fact there was an interesting 40 year study showing very little decomposition of NO sealed in glass tubes over that time period.  The authors' calculations show that without a catalyst the timescale of decompsition could be $10^{29}$ years!


What is the link between electrode surface areas and EMF?

If by EMF you mean the ideal, thermodynamically determined, voltage then the answer is that increasing the surface area does not affect voltage/EMF. If by EMF/Voltage, the questions means operational voltage then it will increase because increasing the surface area of electrodes decreases the overpotential losses (Tafel equation) by essentially decreasing the resistance due to the kinetics of the reaction thus increasing operational/real voltage of a cell.


"What technologies are available currently to detect a broader range of particles in a closed system" a LOT of technologies allow this: spectrophotometry, NMR, GC-MS, etc. These technologies give a lot of information about the molecular structure (but not about the molecular  function).

The hard part in your idea is trying to replicate the sensivity of a nose, which is, at the moment, impossible as far as I know.

Trying to deduce via a robotic nose the smell of a novel compound is a task that could, theorically, be made if that "nose" had exactly the same receptors and "circuits" of our nose, since receptors interact in a unique way with molecules, and that is the source of all those unique smells that we sense.

A more approximate approach, doable with current technologies, could be a database-based approach: a lot of "smelly" molecules could be categorized and described in terms of smell.
The analyzer (one of the machines mentioned earlier, for instance) could identify the molecule(s) involved, and the database could provide information about the smell in that environment.

The problem? A sistem like this doesn't work like a nose at all! Any unknown molecule couldn't be "smelled".

Maybe in the future, with the improvement of technology, a molecular docking study will be performed in seconds, and that, when all the proteins associated with smell will be known, could allow us to know the exact way a molecule affects our receptors, our brain and our perception, hence creating a working "digital nose".

And finally, let me add a note: you said "particles from the environment have a custom signature which we associate as smell of a banana or orange". One might argue wether that "signature" is in the molecule or in the nose itself. While many compounds are sensed in a similar way within a population, there are genetic mutations between species and even individuals. Puppies are known to be appealed by cat feces, for instance: is that just a behavioural thing, or their perception of the smell of cat's feces is different from ours?
Some people like the smell of truffles, others hate it: it's (at least partially) because of the individual ability of detecting molecules.

The point is not that of detecting a "signature", but in detecting molecules and interpreting them via our specific "signature".
If olfactory proteins didn't exist, smell wouldn't exist: it's not a physical property of the molecule, but the result of an interaction with who "smells".

Don't be fooled: that's not a philosophical sentence: light wouldn't cease to exist if we had no visual receptors. But for sure colours wouldn't exist, since also colours exist only because we are evolved to detect light as a mixture of three different chemical responses (each associated with one of the RGB colours).


This is the Skattebøl rearrangement, the first step is of which is a lithium-halogen exchange to form a carbenoid. A mechanistic study of the rearrangement of monosubstituded lithium bromocyclopropylidenoids found that allene formation can occur either in a concerted (Scheme 2) or step-wise (Scheme 3) fashion depending on the ability of the substituent to stabilize the intermediate carbene$^{[1]}$.

The study found that electron donating groups (like the methyl groups in your example) lower the barrier to allene formation. For several EDG-substituted species, the free carbene structure 14 could not be found as an energy minimum and all attempts to locate these structures lead directly to allenes. This means the stepwise mechanism does not play a role in the Skattebøl rearrangement for EDG-sbustituded species$^{[1]}$.

$[1]$ Azizoglu, A.; Balci, M.; Mieusset, J.-L.; Brinker, U. H. The Journal of Organic Chemistry 2008, 73 (21), 8182–8188.