Why must the leaving group in E1cb be poor?

There is a range of elimination reactions with E1cb at one end, E1 at the other end and E2 in between. It is not uncommon for these different reaction pathways to compete with one another.  For example, in some elimination reactions the E1 and E2 pathways can operate in competition with one another.  An activation energy is associated with each of these 3 reaction pathways.  Whichever pathway has the lowest activation energy will be the major pathway followed.  By changing solvent, reaction temperature, relative strength of the nucleophile, relative strength of the base, leaving group stability, etc., we can raise or lower the activation energy for each of these 3 pathways and shift a reaction towards one side of this mechanistic range or the other.

  Typically the leaving group for E1cb is poor (like -OH or -OR) but why
  must this be the case?

The E1 mechanism involves ejecting a leaving group in its first step, while the E1cb mechanism involves removing a proton in its first step. Let's consider how changing the leaving group can shift an elimination reaction towards one pathway or the other.  To a first approximation, changing a leaving group will not affect how hard or easy it is to remove the proton. So it is reasonable to assume that the activation energy for the E1cb process doesn't change as we vary our leaving group.  If we use a better leaving group (make our leaving group more stable) that means that we have made ejecting the leaving group a lower energy pathway and the E1 process will become more favorable relative to the other elimination mechanisms.  If we change our leaving group to one that is an extremely poor leaving group (make our leaving group less stable), then ejecting it becomes a higher energy process and the E1 reaction becomes less competitive with the other reaction pathways. Saying this last sentence differently, when we use a poor leaving group we raise the activation energy for the E1 mechanism.  Since the rate of the E1cb process is not affected by the leaving group, its rate remains unchanged.  Consequently, a poor leaving group will disfavor the E1 process making the E1cb process more competitive. If the leaving group is bad enough, we can disfavor the E1 process so much that we wind up pushing our reaction all the way over to the E1cb side.


What is more important, to have the parent chain be the cycle or the chain that has an hydroxyl group

The compound given in the question contains a characteristic group $(\ce{-OH})$. Since there is only one characteristic group, the seniority order of classes is not relevant in this case; thus, the $\ce{-OH}$ substituent corresponds to the principal characteristic group that is expressed as a suffix (‘ol’).

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the senior parent structure has the maximum number of substituents corresponding to the principal characteristic group.

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

Therefore, the correct name for the compound given in the question is ‘1-cyclohexylpropan-2-ol’ (not: ‘(2-hydroxypropyl)cyclohexane’). (Note that in ‘(2-hydroxypropyl)cyclohexane’, the locant ‘1’ of ‘1-(2-hydroxypropyl)cyclohexane’ is omitted since, according to Subsection P-14.3.4.2, the locant ‘1’ is omitted in monosubstituted homogeneous monocyclic rings.)

If there is still a choice (for example, if the ring and the chain each have a substituent corresponding to the principal characteristic group), the ring or the chain can be the senior parent structure; however, for the preferred IUPAC name, the ring is always selected as the senior parent structure.

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).


I think your suppositions and observations are basically correct. On the basis of general principles, it is certainly possible to imagine that such a mechanism might be at play. If one considers the structure of boric acid, it should be readily apparent that the boron atom could be quite electrophilic and hence act as a Lewis acid: it possesses a formally vacant $p$-orbital; it is bound to highly electronegative oxygen atoms, which withdraw electron density and create a partial positive $(\delta^+)$ charge on it (though this effect may be mitigated by the oxygen lone pairs, which can donate electron density into the aforementioned $p$-orbital); and finally, it has a planar geometry that makes it quite sterically accessible to approaching nucleophiles. Such a mechanism is also lent plausibility by the well-known propensity of various other boron compounds for forming Lewis adducts and being highly susceptible to hydrolysis (where applicable).

All of that having been said, I don't think it's fair to claim that such a mechanism is "obviously" correct for the reasons given above. Such arguments are strictly qualitative, and while they're very suggestive of the possibility, they are in no sense definitive. The only way to even approach something vaguely approximating certainty is to perform experiments (perhaps in concert with theoretical computations), and even there considerable difficulty exists in reaching firm conclusions. The confirmed presence of $\ce{B(OH)4-}$ under one particular set of conditions is not sufficient proof that the mechanism involving boron as a Lewis acid is always operative, nor does it rule out other pathways. (Note that I'm not necessarily disputing that mechanism, but only pointing out that extrapolating too widely from specific experimental results has certain liabilities.) As ron points out in his answer, the issue apparently isn't settled.

Edit: I would also add that there are arguments to be made against such a mechanism. There's likely some degree of $\pi$-bonding between the oxygen lone pairs and the boron, which would be disrupted by the pyramidalization of the molecule upon formation of the intermediate Lewis adduct resulting from nucleophilic attack by $\ce{H2O}$. The associated energy penalty might be relevant to the reaction kinetics under some conditions. My instinct is that this is probably not significant, but it may be a factor worth considering. Of course, this is all speculative on my part, though I think it further demonstrates that qualitative reasoning in chemistry is rarely clear-cut, and there's an awful risk of falling prey to fuzzy thinking. Hence, rigorous experimentation and calculation are vital.


Why the dilution of solvent doesn't affect the solid phase, in comparison to the liquid phase?

Instead of "making the jump" from liquid to gas, the water molecules are trying to stack together into an ice crystal. The salt molecules again get in the way.
Think of making a solid as like stacking shoeboxes, but the shoeboxes are vibrating slightly. The higher the temperature, the more they vibrate. At some temperature (the melting point), the vibration gets so powerful that the whole shoebox stack collapses. Now imagine that instead of a stack of just shoeboxes (pure ice), you start putting golf balls in between some of the shoeboxes. Those are salt molecules. They get in the way of the ordered stack, they mess everything up, they make the stack collapse really easily. There has to be a lot less vibration (lower temperature) to stack the boxes when there are golf balls getting up in everyone's business. So the freezing point is reduced.


As compared to the carboxylic acid, deprotonation of the amide is negligible.

I don't recall any data for THF or 1,4-dioxane as the solvent, but F.G. Bordwell has determined a lot of $pK_a$ values in DMSO.

In DMSO, the $pK_a$ of acetic acid is 12.6 [1], while the $pK_a$ of acetamide is 25.5 [2].


Is it valid to use Gibbs Free Energy for gases?

The Gibbs Free Energy has been found to be the most convenient thermodynamic function to work with when determining the equilibrium constant for a reaction or in establishing the chemical equilibrium state for a chemically reacting system involving multiple reactions.  The condition for equilibrium of these is that, at a given temperature and pressure, the Gibbs Free Energy is minimized with respect to changes in extent of reaction.  This applies to whatever temperature and pressure are currently present in the reactor, irrespective of whether they have been changing. For a mixture of ideal gases, it is possible to show that this Free Energy minimization is equivalent to requiring that the sum of the free energies of the pure reactants at pure gas pressures identical to their partial pressures in the reactor are equal to the sum of the free energies of the pure products at pure gas pressures identical to their partial pressures in the reactor.


Why the formation of PH4+ is difficult compared to NH4+?

The bond angle in $\ce{PH3}$ is about $92-93^\circ$, and by the percentage $s$ character formula i.e $\cos(\theta) = s/(s-1)$ you can calculate the $s$ character and that comes out to be 6%. And that where drago rule holds true , that is the orbitals involved in bonding are pure $p_x$ ,$p_y$, $p_z$ and as a character is high the lone pair is present in pure s orbital and highly penetrated towards phosphorus. Hence acts as a poor Lewis base.


Both of the statements have merit. When the orbitals are of the same type, the overlap is maximized. When the orbitals are nearly the same energy, the atoms are closer together (and give better overlap) than when the energy difference is larger. You don't really need to calculate the energies of the orbitals, just look at the size of the ions. The larger the ion, the higher the energy level of the bonding electrons.


Physical chemical property that determines redox potential?

As I understand it, the potential is related to the difference in energy between the oxidized and reduced state (via the Gibb's energy as Mithoron noted). While their individual electron orbital energies are probably similar given the similar nuclear charges, the oxidized and reduced states are differently occupied and so have different energies.

In zinc for example, you have the reduced state of 4s2, 3d10 and oxidized state of 3d10. Copper on the other hand goes from 4s1, 3d10 to 3d9. $\ce{Zn^2+}$ has a fully occupied d-shell while $\ce{Cu^2+}$ does not, meaning that the reduced state for zinc is lower energy and more stable. As a result, zinc has a stronger tendency to oxidize and consequently a lower reduction potential.


A saturated solution has the maximal amount of solute at equilibrium.

If equilibrium is not required, solutions can have more solute.  For example, when a bottle of soda is first opened, it is supersaturated in CO2, but it is not at equilibrium and CO2 effervesces.


Change in internal energy of a gas upon expansion against a constant external pressure

You did everything correctly with the exception of converting the units of your result for work.

First, calculate the work in joules:

$$w = -p\Delta V = (-2\,\mathrm{atm}) \times 9\,\mathrm{L} =
\left(-18\,\mathrm{L\cdot atm}\right)
\left({101.325\,\mathrm{J}\over\mathrm{L\cdot atm}}\right) = -1823.85\,\mathrm{J}$$

Second, add the heat absorbed (already in joules):

$$\Delta E = 200\,\mathrm{J} - 1823.85\,\mathrm{J} = -1623.85\,\mathrm{J}$$

Third, always remember to check your units!


What is the proper nomenclature for CH₂O₂?

The compound you provided can be written more clearly as $\ce{HCOOH}$

IUPAC name : Formic acid

Systematic name :Methanoic Acid

Other names:
Aminic acid, Formylic acid, Hydrogen carboxylic acid, Hydroxymethanone, Hydroxy(oxo)methane, Metacarbonoic acid, Oxocarbinic acid, Oxomethanol


Treatment of ambidentate ligands in geometrical and optical isomerism

To specify [Pt(NO2)4]2+ would mean the ligand donates electrons through the N-atom, and not the O-atom. If it were to donate through O, it would be written as (ONO) and not (NO2).

Thus, this compound would not show geometrical isomerism. Also note that, if NO2 ligands are replaced by ONO, then that is a form of structural isomerism in complexes, called linkage isomerism.


Confusion regarding calculating the pH of a salt of weak acid and weak base

Another approach, I believe, is this:

Definitions
$$\mathrm{pH} = \frac{1}{2}\left(\log K_\mathrm{b} - \log K_\mathrm{a} - \log K_\mathrm{w}\right) \\ 2\,\mathrm{pH}=\log(\frac{K_b}{K_a\,K_w})$$

It suggests equations' sum, like this:

$$\ce{HA <=> A- + H+} \tag{$\ce{K_a=\frac{[A-][H+]}{[HA]}}$}$$
$$\ce{BOH <=> B+ + HO-} \tag{$\ce{K_b=\frac{[HO-][B+]}{[BOH]}}$}$$
$$\ce{A- + B+ + H2O <=> AH + BOH}\tag{$\ce{K=\frac{[AH]\,[BOH]}{[A-][B+]}=\frac{K_w}{K_a\,K_b}}$}$$
Finding $\ce{[H+]}$ in relation to $\ce{K}$

If salt's concentration is $\ce{C_s}$, and the mass balance:

$$\ce{C_s = [AH] + [A-] = [BOH] + [B+] } \tag{Mass balance}$$

Dividing by $\ce{C_s}$ we define $\alpha$ and $\beta$ such that
$$\ce{\frac{C_s}{C_s} = \alpha{} + (1-\alpha) = \beta{} + (1-\beta)} \tag{Mass balance}$$

And if we admit that $\alpha$, $\beta <<1$ which also implies $\alpha\approx\beta$ (this is possibly a weakness of the answer) then $$\ce{\alpha^2\approx\frac{K_w}{K_a\,K_b}}$$

And $\ce{K_a=\frac{(1-\alpha)[H+]}{\alpha}\approx\frac{[H+]}{\alpha}}$

$$(\frac{[H^+]}{K_a})^{2}\approx \frac{K_w}{K_a\,K_b}$$
$$[H^+]^{2}\approx \frac{K_w\,K_a}{K_b}$$

And we get the equation.


Is steric inhibition of resonance or steric inhibition of protonation dominant in o-toluidine?

The reason for the decreased basicity is essentially the ortho-effect as outlined here: Ortho-effect in substituted aromatic acids and bases.

Your question on the other hand asks about a seeming contradiction, but I believe there is a small misconception on your part.

  We see that due the bulky methyl present on o-toluidine there is inhibition of resonance which increases the availability of nitrogen's lone pair. [...]

I don't see how resonance could be inhibited, as that essentially requires that the lone pair orbital of nitrogen would have to rotate significantly towards an in-plane arrangement. The methyl group is simply not big enough for that to happen, and if it were, the lone pair would more likely rotate toward the more shielded side, further reducing acidity.

The highest occupied molecular orbitals (HOMO) of the molecules show this quite nicely, as they are very, very similar.

On the DF-B97D3/def2-SVP level of theory the RMSD of the core structure (excluding the hydrogen and the methyl group) is negligible with only 0.03 Å.

From this I would conclude that there is no steric inhibition of resonance.

On the other hand, steric inhibition of protonation is given in 2-methyl aniline, where it is obviously missing in aniline.
Additionally there is a stabilising non-covalent interaction between the methyl and amine group, which can be analysed with NCIPLOT, and then displayed. The green blob is weakly attractive, red is strongly repulsive, blue would indicate strongly attractive interactions.

TL;DR: The dominant effect is steric inhibition of protonation, as steric inhibition of resonance is negligible (if possible at all).


What is formed when you leave iron(II) sulfate in plain air?

Usually, the original iron(II) sulfate is present as green $\ce{FeSO4.7H2O}$.

Oxygen can oxidize $\ce{Fe(II)}$ salts to $\ce{Fe(III)}$ salts; for example, at $\mathrm{pH}=0$:

$$\begin{alignat}{2}
\ce{[Fe(H2O)6]^3+ + e- \;&<=> [Fe(H2O)6]^2+}\quad &&E^\circ = +0.771\ \mathrm{V}\\
\ce{O2 + 4H+ + 4e- \;&<=> 2H2O}\quad &&E^\circ = +1.229\ \mathrm{V}
\end{alignat}$$

$\ce{Fe(II)}$ is even easier oxidized under alkaline conditions; for example, at $\mathrm{pH}=14$:

$$\begin{alignat}{2}
\ce{FeO(OH) + H2O + e- \;&<=> Fe(OH)2 + OH-}\quad &&E^\circ = -0.69\ \mathrm{V}\\
\ce{O2 + 2H2O + 4e- \;&<=> 4OH-}\quad &&E^\circ = +0.401\ \mathrm{V}
\end{alignat}$$

However, you cannot simply oxidize iron(II) sulfate $\left(\ce{FeSO4}\right)$ to iron(III) sulfate $\left(\ce{Fe2(SO4)3}\right)$ in dry air since you would need additional sulfate to balance the equation:

$$\ce{2FeSO4 + SO4^2- -> Fe2(SO4)3 + 2e-}$$

Nevertheless, $\ce{Fe(II)}$ can be oxidized to $\ce{Fe(III)}$.

The resulting $\ce{Fe(III)}$ is subject to hydrolysis. By way of comparison, the ion $\ce{[Fe(H2O)6]^3+}$ is only stable under strong acidic conditions. Already at $\mathrm{pH}=0{-}2$, it turns into yellow $\ce{[Fe(OH)(H2O)5]^2+}$ and begins to form $\ce{[Fe(OH)2(H2O)4]+}$:

$$\begin{align}
\ce{[Fe(H2O)6]^3+ \;&<=> [Fe(OH)(H2O)5]^2+ + H+}\\
\ce{[Fe(OH)(H2O)5]^2+ \;&<=> [Fe(OH)2(H2O)5]+ + H+}\\
\end{align}$$

Further addition of base causes precipitation of amorphous iron(III) hydroxide.

Accordingly, the likely product when iron(II) sulfate is oxidized is basic iron(III) sulfate, i.e. approximately:

$$\ce{4FeSO4 + O2 + 2H2O -> 4Fe(OH)SO4}$$

However, the real weathering and aging of iron(II) sulfate in dry air actually yields a mixture of various compounds, including iron(III) sulfate and iron(III) oxide-hydroxide.

You may observe this reaction in some iron fertilizers for lawns. The fresh product typically contains green iron(II) sulfate, which gradually becomes yellow.


What is the bonding type in hydrides of group 2 elements?

There is a quite extensive chapter about hydrogen and its compounds in Holleman and Wibereg's book "Inorganic Chemistry". As I only have the german version, I will briefly summarise what I read.

You are of course right, hydrogen forms salt-like structures with alkaline and earth alkaline elements, except for beryllium. Compounds like $\ce{LiH}$, $\ce{NaH}$, $\ce{KH}$, $\ce{RbH}$ and $\ce{CsH}$ crystallise in a sodium chloride structure. Every cation is coordinated by six hydride ions and vice versa. The compounds $\ce{CaH2}$, $\ce{SrH2}$ and $\ce{BaH2}$ crystallise in a fluorite structure. The hydride ions are tetrahedrally coordinated by four earth alkaline cations. The cations are cubically coordinated by eight hydride ions. This β-structure is persistent at higher temperatures. Below $780~^\circ\mathrm{C}$ ($\ce{CaH2}$), $855~^\circ\mathrm{C}$ ($\ce{SrH2}$), and $598~^\circ\mathrm{C}$ ($\ce{BaH2}$) it changes to the α-form, i.e. the lead dichloride structure. The magnesium compound exists in a rutile structure. Hydride ions are trigonally coordinated by magnesium ions. The cations are octahedrally coordinated by hydride ions.

However, according to the difference in the electronegativities, you would expect a predominant covalent character. The maximum is $1.3$ for $\ce{CaH2}$ and the minimum $1.0$ for $\ce{MgH2}$, which translates to a maximum $30\%$ and minimum $18\%$ ionic contribution. Maybe this is where the statement of the question originates from. Experimental findings rather support the more ionic nature of the compounds though, i.e. the found lattice energies were supported by calculations assuming a purely ionic structure.

The conclusion of this according to Holleman and Wiberg is that the elements are fused by ionic bonds with covalent contributions. The latter increases from $\ce{CsH}$ to $\ce{LiH}$ and from $\ce{BaH2}$ to $\ce{MgH2}$. The magnesium hydride is already borderline covalent.

The question about their reactivity towards water is mainly answered in the chapters for the elements itself. Magnesium reacts readily with water and releases hydrogen. Because of this it is sometimes used as a drying agent. $$\ce{MgH2 + 2H2O -> Mg(OH)2 + H2 ^ }$$
The other elements react similar. The reaction is highly exothermic, the reaction of calcium hydride releases $228~\mathrm{kJ/mol}$. It is often used as a drying agent for gasses because of this feature. Strontium and Barium also react exothermic, i.e. they release $177~\mathrm{kJ/mol}$ and $172~\mathrm{kJ/mol}$, respectively.


Do chemical compounds have spectral lines, or only elements?

Yes, but for "spectral lines" the gaseous state would generally be the implication. In solids and liquids photon interactions are generally over "bands" of energy not a very discrete wavelength as in a "spectral line."


Effect of reduction potential on particle size of chemically precipitated silver particles

Here is an example of a study of fish oil nanoemulsions which examines, to quote:

  The influence of surfactant-to-oil-ratio on particle size and physical stability was evaluated.

Here is yet another study to quote:

  Although the role of surfactant in the synthesis of particles in the nanometer range has frequently been documented...The surface coverage per surfactant molecule could be related to the surface tension of the latex, thus providing a relationship between particle size and stability.

And, another more expansive work to quote:

  It was reported that the formation process of these NPs could be divided into three stages: growth, incubation and Oatwald ripening stages. The higher boiling point of 300 °C of paraffin affords a broader range of reaction temperature and makes it possible to effectively control the size of silver NPs by varying the heating temperature alone without changing the solvent. Moreover, the size of the colloidal silver NPs could be regulated not only by changing the heating temperature, or the ripening time, but also by adjusting the ratio of oleylamine to the silver precursor.

While chemical reduction methods are connected to reduction potential, the multi-stage aspect of the creation process for NPs suggests only a partial correlation, so not a 'mathematical relationship', but perhaps a statistical one.


What happens when an electron in a metal is excited?

You seem to be misunderstanding what is a "sea of electrons". In fact, this is a metaphor upon a metaphor upon an abstraction. There is no sea. There is a huge bunch of orbitals. (Sure, the solid state people prefer to call them "states", but that's not really important.) The whole piece of metal is a giant molecule. It is not all that different from ordinary small molecules, except that it is very big, and many orbitals span the entire molecule (but then again, that's what they often do in normal molecules).

All these orbitals tend to have different energies. They are everywhere on the energy scale, very close to each other. You point your finger at any given energy, and you find an orbital with that energy. We can't really tell them apart. They kinda blend into a continuous spectrum. And that's what we metaphorically call the sea of electrons.

Electrons are not free, they are confined to some states. When a photon hits, any electron can get excited all right. It will move up to one of the empty states. In fact, it has an immense freedom compared to an electron in a small molecule. Instead of a few discrete states, it has an entire sea of them above. It can get excited to any of them, which in particular means that metal as a whole can absorb a photon of any energy. This, BTW, is also an explanation of metallic luster.

The excited electron is not always the valence electron (you may fire an X-ray photon at a substance and excite an electron from core orbitals deep down below), but that's irrelevant.


Sign of enthalpy for exothermic and endothermic reactions

Reason lies in definition of enthalpy of reaction. Enthalpy of reaction is heat exchanged between our system in which reaction happens and surroundings when reaction is carried at constant temperature and pressure. If reaction is exothermic, it releases heat and increases temperature of our system and so to keep it at the same temperature you need to give that heat to surroundings. If you do so, than since our system lost heat its enthalpy decreases (first law of thermodynamics) and because of that enthalpy of exothermic reaction is taken as negative. The opposite reasoning holds for endothermic reactions. You need to bring heat from surroundings into system to keep it at the same temperature and that heat increases enthalpy of the system.


Acid-catalysed isomerisation of phenyl epoxide to aldehyde

You are correct if this epoxide is a alkyl peroxide. However, this is a 2-phenyloxirane (styrene oxide or phenylethylene oxide),  which can stabilize a protonated intermediate by resonance, specifically with 2-hydroxyphenyl group. Thus, it can isomerize to phenylacetaldehyde through 1,2-hydride shift (similar to pinacol-pinacolone rearrangement). The following diagram illustrate how styrene oxide converted to phenylacetaldehyde (Ref.1):

Note that styrene oxide is not stabilized by extra hydroxy group here. That's probably the reason the authors have used slica-supported phosphotungstic heteropoly acid in cyclohexane (no other nucleophile there).

The identical isomerization has been used in the synthesis of 2-phenylnaphthalenes from styrene oxides (Ref.2).

References:

Vinícius V. Costa, Kelly A. da Silva Rocha, Ivan V. Kozhevnikov, Elena V. Gusevskaya, "Isomerization of styrene oxide to phenylacetaldehyde over supported phosphotungstic heteropoly acid," Applied Catalysis A: General 2010, 383(1–2), 217-220 (https://doi.org/10.1016/j.apcata.2010.06.005).
Kishor V. Wagh, Bhalchandra M. Bhanage, "Synthesis of 2-phenylnaphthalenes from styrene oxides using a recyclable Brønsted acidic $\ce{[HNMP]+HSO4−}$ ionic liquid," Green Chemistry 2015, 17(8), 4446-4451 (DOI: 10.1039/C5GC01170A).


Release of sulfur dioxide from odored fuel gases

According to this North American Energy standards board publication , the total concentration of sulfur in natural gas, both natural and from oderant additives, must not exceed $\pu{70 ppbv}$. Concentrations are actually more commonly about a tenth of this. For example, from the same source, the Union Gas system typically contains about $\pu{5 ppbv}$ sulfur.  Furthermore, the sulfur content of coal is many orders of magnitude higher than that of oderant-enriched natural gas at about $\pu{0.5 to 5\%}$ sulfur, and the worldwide combustion of coal is much greater than that of natural gas (although much of the sulfur from coal combustion is now scrubbed out at the source of combustion). Other fossil fuels like diesel also contain significant amounts of sulfur (regulatory sulfur content of diesel fuel in the US and EU was 10 ppm as of 2009).

The global atmospheric background concentration of $\ce{SO2}$, according to
Owen, Lewis A.; Pickering, Kevin T (1997). An Introduction to Global Environmental Issues. Taylor & Francis. pp. 33–. ISBN 978-0-203-97400-1, is around $\pu{1 ppbv}$, meaning the sulfur content of the Union Gas product, for example, is only five times that that found throughout the Earth's atmosphere.

Summary - TL/DR:
Based on the background $\ce{SO2}$ concentration and the overwhelming sulfur emision source from the combustion of coal and other fossil fuels, the sulfur originating from natural gas oderant additives represents a negligible contribution to the global atmospheric $\ce{SO2}$ concentration.


Increasing partition coefficient in an extraction

Generally, extraction of substances with organic solvents is influenced by various factors, among which there are the following:

nature of the extracted substance and the extractant (not a variable in your example);
temperature;
presence of electrolytes in aqueous solutions;
$\mathrm{pH}$ of the medium;
rate of agitation etc.

In your particular case probably only temperature and electrolytic presence are relevant.
Change in temperature often affects the distribution constant rather seriously, as not only the solubility of the extracted substances in each phase changes (unequally), but also the mutual solubility of the organic and aqueous phases changes too. When the temperature changes, the dissociation/association of the substance in the corresponding phase can change, influencing hydration (solvation) and therefore extractability. Ideally you might want to check solubility-temperature plots for your system, taking into account that this data often cannot be obtained by extrapolation.
Addition of highly soluble salts (electrolytes) to an aqueous solution of another substance can lower (salting-out effect) or increase (salting-in effect) its solubility in water. Increased high ionic strength also lowers activity of water in terms of depressing its competitive effects (first of all, solvation).
The salting-out effect of electrolytes depends on the nature and properties of the extracting substance and the concentration, charge and ionic radii of the salting-out ions. Smaller ions with larger charge have higher charge density than ions with a larger radius, therefore they are hydrated better.  $\ce{NaCl}$ and $\ce{Na2SO4}$ are commonly used for salting-out organic substances. However, this rule has a number of exceptions. Some non-electrolytes that are well soluble in water also have salting out-effect. For example, ethyl alcohol salts-out acetic acid from its aqueous solutions upon extraction with ethyl acetate.
The salting-in/out phenomenon is often explained based on the electrostatic model proposed by Debye and McAulay, reviewed in [1], and can be approximately described via equation:
$$c/c^\circ = 1 - K_D \cdot c_\mathrm{s} + A \cdot c_\mathrm{s}^{0.75},$$
where $c^\circ$ and $c$ are solubilities [$\pu{mol L-1}$] of non-electrolyte in pure solvent and solvent with salt, respectively; $K_D$ and $A$ - empirical coefficients; $c_\mathrm{s}$ - salt concentration.
Pretty much your explanation is on the same page with the original Debye theory:

This theory pictures the "salting-out" effect as a consequence of the aggregation of the water molecules of greater polarity around the ions owing to the field of the latter. The less polar non-electrolyte molecules tend to congregate in the portions of the solution most remote from the field of the ions, with the net result that an enhancement of the water molecules around the ions and of the non-electrolyte in the solution regions away from the ions occurs. A reduction of the solubility of the non-electrolyte, referred to the total water present, thus results, due to the increase of non-electrolyte-water ratio in those regions of the solution containing the non-electrolyte.
The criterion of polarity which Debye employs is the dielectric constant of the saturated aqueous solution of the non-electrolyte relative to water at the same temperature. On this basis if the saturated solution has a dielectric constant less than water, "salting out" occurs in the presence of the added salt. If the dielectric constant of the saturated non-electrolyte solution is above that of water, the non-electrolyte becomes more soluble in the salt solution or is "salted in". This means that the non-electrolyte tends to aggregate around the ions at the expense of the water, and so the nonelectrolyte-water ratio in regions removed from the ions is lowered and the total water present can hold more non-electrolyte in solution.

Also, in biochemistry so-called Hofmeister series are used to favor protein extraction by salting-out for given conditions.
References

Gross, P. M. The “Salting out” of Non-Electrolytes from Aqueous Solutions. Chemical Reviews 1933, 13 (1), 91–101. https://doi.org/10.1021/cr60044a007.


Yes, they're called multiple emulsions (and more specifically: double, triple emulsions, etc.). Usually they're made from alternating polarity phases—water inside oil inside water, for example. There are bulk ways of making them though there isn't great control over how many droplets end up in each encompassing droplet without using microfluidic techniques. David Weitz's lab at Harvard has some interesting work in building microfluidic devices to make multiple emulsions.

I don't know if there are any common natural occurrences but the pharmaceutical and cosmetics industries seem to be the the main parties interested in this stuff as they can be used for encapsulating drugs for time release and the like.


Can an EWG behave as an EDG and vice versa?

Of course not. That's why it is called EDG or EWG. It tends to destabilize it more in the absence of other effects like resonance and hyperconjugation.


Liquid and vapour can coexist at many combinations of temperature and pressure. But for most of them they are not in thermodynamic quilibrium.

For a given temperuture, there is only one (vapour partial ) pressure value with both phases in equilibrium.

For a given vapour partial  pressure, there is only one temperature value with both phases in equilibrium.

This temperature is also the liquid boiling point at the given external ( total ) pressure.

Standard boiling point is then temperature, when vapour partial pressure is equal external pressure 1 atm.

Note that boiling is not state of thermodynamic equilibrium, as there is net change of phase mass and thermal energy.

If partial pressure is higher than the equilibrium pressure for given temperature, the vapour condensates until the pressure reaches the liquid-vapour equilibrium curve of the phase diagram.

If partial pressure is lower than the equilibrium pressure for given temperature, the liquid evaporates until the pressure reaches the liquid-vapour equilibrium curve of the phase diagram.

If total pressure is lower than the vapour pressure for given temperature, the liquid boils until the total pressure reaches the saturated  vapour pressure.


All carbon atoms in fullerene are part of one "double bond", as you draw it. In quantum chemical reality, the structure of the fullerene is made up by fixed "single" bonds between those 60 sp2 hybridised carbon atoms, with the residual thirty pairs on pi electrons being somewhat delocalised.

In a sheet of graphene, the pi electrons are delocalised, and every bond lying between two six-rings gets the same share in the electron cloud.

In fullerene, they are as delocalised, but not equally distributed. The bonds between two six-rings are "more equal" (more aromatic) than those between five- and six-rings, and grab themselves a larger part of the electron density.

Problem of the five-rings is that the bond angles are not 120°, as they should be for a proper sp2 hybridised carbon atom. They become more sp2.4 or something, which let's the pi electrons move out of there a bit.


What is the greenish substance formed after the electrolysis of two graphite electrodes in a sodium sulphate solution?

I think that what possibly happened here is that the copper wire that connects the cable with the graphite electrodes is exposed in the solition (if you look carefully in Fig.1 there seem to be a lot of bubbles forming near the red cable insulation at the bottom of the tube which escape out of the tube). So copper is oxidising to give copper cations which give the coloured material in the solution.

This is just a theory though. To confirm, repeat the experiment first with copper electrodes then with your graphite electrodes but ensure that only the graphite is in the solution (so get rid of the tubes). If using the copper elecrodes gives the same result as above (blue green stuff) while using the graphite electrodes just gives O2 and H2 it means that the above theory is likely. Finaly you could isolate the greenish solids by filtration, dry them and do a flame test. If the flame becomes green it suggests presence of copper further confirming the above.

The only thing that doesnt fit with the above is that Cu2+ ions are blue and soluble in solution not green, so maybe the green stuff also (or only) result from some reaction with the indicator. Hence, it would be good to repeat the experiment with and without the indicator to see what happens (that makes 4 experiments in total).


Typesetting of the retardation factor (Rf)

According to Nomenclature for Chromatography (IUPAC Recommendations 1993) [1, pp. 843, 845] (also listed in IUPAC Gold Book), retardation factor is denoted as follows:

Column chromatography: $R$ (capital $R$ in italics):

  3.7.13 Retardation Factor ($R$)

  The fraction of the sample component in the mobile phase at equilibrium; it is related to the retention factor and other fundamental chromatography terms:

  $$R = 1/(k + 1)$$

Planar chromatography: $R_\mathrm{F}$ (capital $R$ in italics and upright capital $\mathrm{F}$ in subscript):

  3.8.04 Retardation Factor ($R_\mathrm{F}$)

  Ratio of the distance travelled by the center of the spot to the distance simultaneously travelled by the mobile phase. Using the symbols of Fig. 2:

  $$R_\mathrm{F} = b/a$$

  By definition the $R_\mathrm{F}$ values are always less than unity. They are usually given to two decimal places. In order to simplify this presentation the $hR_\mathrm{F}$ Values may be used: they correspond to the $R_\mathrm{F}$ values multiplied by 100.

  Ideally, the $R_\mathrm{F}$ values are identical to the $R$ values (see 3.7.13).

Reference

Ettre, L. S. Pure and Applied Chemistry 1993, 65 (4) DOI: 10.1351/pac199365040819.


Understanding elementary rate laws from a probability stand point

Let's divide the reaction volume into V little boxes and randomly throw ${\rm N}_A$ molecules of A into it and ${\rm N}_B$ molecules of B.  The probability $p_A$ that a given box has a molecule of A in it is ${\rm N}_A$/V, in the limit that $V >> {\rm N}_A$, i.e. an ideal dilute solution, and likewise the probability $p_B$ that it has a molecule of B in it is ${\rm N}_B/V$.  The probability it has two molecules of B in it is ${p_B}^2$, again in the limit that $V >> {\rm N}_B$ and ${\rm N}_B >> 1$.

Consequently, the probability all three atoms are close to one another is $({\rm N}_A/V)({\rm N}_B/V)^2$, which is proportional to the terms in your rate law.  (This isn't a derivation, by the way, but only a rationalization I hope is appealing to intuition.  I've swept all kinds of detail and qualification under the rug.)

Where your intuition may be leading you astray is in the assumption that squaring the molarity of B makes the RHS of your rate law bigger, and that doesn't match up with your intuition that it should be smaller if it involves a less likely event (a trimolecular collision).  But that is not the case.  Squaring numbers only makes them bigger if they are greater than 1, below 1 it makes them smaller.  And of course the prefactor will be different, too.  What the squaring really does is change the dependence of the rate law on changes in B concentration. It says the rate law is much more sensitive to changes in B concentration than A concentration.  And, given that you need two B molecules to collide with each A molecules, that should be intuitively compelling.


What is difference between aluminium mercury alloy and gallium aluminium alloy?

We can rationalize the differences in the reactivity as follows.  Gallium has a higher ionization energy compared to mercury.  During a redox reaction the metal is oxidized and it loses an electron, therefore how "easy" it is for a metal to lose its electon is iimportant. Since gallium is harder to inonze than mercurey, it implies that it is "easier" for mercury to lose its electron compared to Gallium, making mercury a better stronger reducing agent.

I may come back and add some quantitative information later, but i'm going to go eat now. I think looking in a table of standard reduction potentials would be reasonable.


Hygroscopic behaviour of DMSO - how bad is it?

Being curious about the hygroscopicity I have tried the experiment using pure, dried DMSO in our lab where the temperature was $21^\circ \text{C}$ and the relative humidity about $60\%$.

Have a look at the two images below. This is a DMSO droplet originally with $V=50\; \mu\text{L}$. You can clearly see that the volume of the droplet has increased substantially in 20 minutes from the different height.

Using a crude approximation, assuming that both shapes are caps of a sphere, I calculated that the change in height from 100 px, to 112 px is equivalent to a volume increase of $15\%$, which is much too big for an accurate flow experiment.


How to convert ethanol to 1,1-dichloroethane?

convert ethanol to 1,1-dichloroethane?

A standard route for producing gem-dichloro compounds involves treatment of a ketone or aldehyde with $\ce{PCl5}$ (see the second to last paragraph in the "Chlorination of organic compounds" section here and page 18 here).

(image source)

Therefore, you could carefully oxidize your ethanol to acetaldehyde and treat it with $\ce{PCl5}$.


The modern definition of aromaticity from deep theoreticians is that the π-system needs to support aromatic ring currents. Borazine can support it, so it is technically aromatic.

Aromatic systems that do not contain carbon are not really all that common, but they are known. Pentazole was detected, for example. The problem is, that for aromaticity to be a thing we need a π-system, and π-systems effectively restrict us to BCNOS elements, since other elements are not eager to form π-bonds. Aromatic systems are usually 5- or 6-membered rings. Add in the fact that rings with a lot of N/O atoms often happily go kaboom after a funny look, and it's not surprising we have very few actual examples here. Still, some people insist on doing unreasonable things with results from as far back as the 50s.


What is meant by excess energy in exothermic reaction?

The energy release is because of the total energy of valence electrons in the product bonds is lower  than in reagents.

In fact, the change of the mean total  potential energy of valence electrons is twice as much as the change of their total energy, as the half of that change is compensated by their increased mean total kinetic energy.

As a consequence, the energy released by forming chemical bonds of products is bigger than energy spent on breaking bonds of reagents.


How to find the plane of symmetry in biphenyls and substituted cyclohexanes?

Finding planes of symmetry is simple: try all planes, some (or maybe none) of them will fit.

As for your examples, the first image is just poorly drawn. The plane is supposed to be the one of the ring containing COOH and F, while the other ring is perpendicular to it (which, BTW, I don't believe is the native conformation of the molecule, but whatever).

The second example asserts right things, but for the wrong reasons. (This is not what you would normally expect of a textbook, to put it mildly.) True, each of these structures contains at least one plane of symmetry. Too bad the planes are drawn incorrectly in all structures except for the first two. You may take your time and fiddle around to find out where they really are.

In the third example, you would be right if the molecule were planar, which it is totally not. Steric repulsion between a and a' (also, between b and b') makes that impossible, even if all of the substituents a, a', b, b' are simply hydrogens.


Reaction between 1-methylcyclopentanol and potassium permanganate

There is no reaction.
t-BuOH may be used as a solvent for permanganate reactions. For example, see Abiko et al. [1].

References

Abiko, A.; Roberts, J. C.; Takemasa, T.; Masamune, S. $\ce{KMnO4}$ Revisited: Oxidation of Aldehydes to Carboxylic Acids in the Tert-Butyl Alcohol - Aqueous $\ce{NaH2PO4}$ System. Tetrahedron Letters 1986, 27 (38), 4537–4540. https://doi.org/10/c3t9xj.


Stereoisomers of metaldehyde and paraldehyde

Your analysis for metaldehyde is correct. There are four achiral stereoisomers. The example cited here is a bit different. To understand the use of R, S, r and s, see this discussion on the inositols.


Decomposition products of magnesium ethanoate

EDIT: Ok, here's your answer (Thermochimica Acta, 78 (1984) 17-27): It decomposes into the oxide, carbon dioxide and propanone:

$Mg(C_2H_3O_2)_{2(s)} \rightarrow MgO_{(s)} + CO_{2(g)} + CH_3COCH_{3(g)}$

PRE-EDIT:

The Russian Wikipedia favors the second reaction. This seems to support the idea.

This site mentions magnesium carbonate and propane-2-one as products, but this seems to be questionable, given the other sources.


That is a valid question, because acid rain has been known to damage paints, particularly high-gloss, i.e. clear coat. Best would be to test the effect of vinegar on a hidden area, and double-check under bright light from different angles. Your idea to test on glass is good, because it shows how long a soak is needed.

If it appears unaffected, rinse afterwards with a weak solution of baking soda or ammonia, and then rinse again with plain water to remove any remaining acid.
If damage is slight, can car wax effect a good repair?
If vinegar does damage in the test spot, you might look at various car polishes or even cutting compounds, and then wax (or reapply a gloss coat, which may mean refinishing a large section of bodywork to get it to match).


I think I can answer this in the affirmative. There are articles which find connections between abstract mathematics and chemistry, sometimes even bypassing physics altogether. Of course, these kinds of articles are considerably rarer, but they're sprinkled out there.
I can think of two articles which I'd love to discuss, but I literally do not have the necessary background, it just goes way over my head. The first one is Quantum Interference, Graphs, Walks, and Polynomials, Chem. Rev. 2018, 118, 10, 4887–4911. This is some rather pure graph theory which is not related to cheminformatics. In particular, I find it interesting that the connectivity in azulene is comparatively unusual, and this probably is deeply connected to its unusual photophysical properties. And then there's The Rouse Dynamic Properties of Dendritic Chains: A Graph Theoretical Method, Macromolecules 2017, 50, 10, 4007–4021, more graph theory with a little bit of physics, and whose content eludes me entirely.
Surely some amount of computer assistance was employed in deriving the results for these papers, because there's no reason to give up that tool. However, I don't think this counts as "extensive computations" in the sense you may have had in mind.


Testing for nitrate ions with NaNO3 in Devarda's test

Quoting from Svelha G., Vogel's Qualitative Inorganic Analysis,

  A solution of nitrate ion is boiled with zinc dust/aluminium powder
  and sodium hydroxide solution to get positive result. Excellent result
  is obtained by the use of Devarda's alloy (45% Al, 5% Zn, 50% Cu) but
  the ammonium ions must be removed by boiling the solution with sodium
  hydroxide solution or almost evaporating it to dryness before addition
  to metal.

  The overall reaction is:

  $$\ce{3NO3- + 8Al + 5OH- + 18H2O ->3NH3 ^ + 8[Al(OH)4]-}$$


Determination of the integrated rate equation for a reaction

After consulting the Lindemann mechanism, here is my attempt for an answer:

$$\frac{d[\ce{C}]}{dt}= k_1[\ce{A}]^2 - (k_2 +k_3 )[\ce{C}]^2 = 0$$
$$[\ce{C}]^2=\frac{k_1}{k_2 +k_3}[\ce{A}]^2$$
$$\frac{d[\ce{A}]}{dt}= -k_1[\ce{A}]^2 +k_2[\ce{C}]^2$$

Substituting $[\ce{C}]^2$,

$$\frac{d[\ce{A}]}{dt}= -k_1[\ce{A}]^2 +\frac{k_2k_1}{k_2 +k_3}[\ce{A}]^2$$
$$\frac{d[\ce{A}]}{dt}= \frac{-k_2k_1[\ce{A}]^2 + -k_3k_1[\ce{A}]^2 +k_2k_1[\ce{A}]^2}{k_2 +k_3}$$
$$\frac{d[\ce{A}]}{dt}= \frac{-k_3k_1[\ce{A}]^2}{k_2 +k_3}$$

Let $k'=\frac{-k_3k_1}{k_2+k_3}$, then

$$\frac{d[\ce{A}]}{dt}= k'[\ce{A}]^2$$
$$\frac{d[\ce{A}]}{[\ce{A}]^2}= k'dt$$
$$\int_{[\ce{A}]_o}^{[\ce{A}]} \frac{d[\ce{A}]}{[\ce{A}]^2}= \int_0^t k'dt$$
$$-\left(\frac{1}{[\ce{A}]}-\frac{1}{[\ce{A}]_o}\right) = k't$$
$$\frac{1}{[\ce{A}]}-\frac{1}{[\ce{A}]_o} = -k't$$
$$\frac{1}{[\ce{A}]}= -k't +\frac{1}{[\ce{A}]_o} $$

I wonder how is it correct when I also used the steady state assumption in my solution. How is it different from the answer posted by my professor? Which is more correct and corresponds to the principles of chemical kinetics? Why would the other one be wrong and the other correct? (That is if my current derivation is correct :) ).


Directing effect in Intramolecular Friedel Crafts Alkylation

Electron giving groups like $-OCH_3$ (by resonance) and $-CH_3$ (by hyperconjugation) are Ortho-Para directing.

But, in cases where there is no increased stability by H-bonding or other factors at ortho position, the para substituted product will give major yield.

This can simply be explained by stating that there is less steric hindernace on the para position compared to the ortho position. Which is further emphasized when you take this example:

So, the answer is para-substituted in both cases.

References

Aromatic reactivity


What are the similarities and differences of adiabatic and diabatic states?

A potential energy surface (PES) is generally the electronic energy written as a function of the nuclear coordinates. This is obtained by solving the Hamiltonian, considering explicit dependence on the electronic coordinates $r$, and parametric dependence on the nuclear coordinates $R$.

$\hat{H}\psi(r;R)=E\psi(r;R)$

Solving this, gives us the electronic energy as the function of the nuclear coordinates, $E(R)$. A plot of this function is the PES. This PES is obtained within the Born-Oppenheimer (BO) approximation, which allows us to assume that the kinetic energy of the nucleus does not matter. The physical interpretation is that the nucleus is very heavy compared to the electrons and, therefore, so much more slower that it can be considered almost stationary. This PES is generally written in the adiabatic basis, where the diagonal terms represent the electronic energy and the off-diagonal terms are the nuclear kinetic energy, $T_n$.

The off-diagonal terms can be neglected, as we already saw that $T_n$ is nearly zero. But in cases where two PESs come very close together (conical intersections and avoided crossings), there is coupling of the nuclear and electronic coordinates. This means that the nuclear kinetic energy can no longer be neglected, and there is a breakdown of the BO approximation. Further, the numbers become very high when the two states (PESs) cross as in a conical intersection and become too hard to handle. In such cases, a unitary transformation is done to change the basis from an adiabatic to a diabatic basis. By doing this, the diagonal term, $T_n$, gets converted to a potential energy term and is easy to handle.

Therefore, adiabatic and diabatic PESs differ by the basis they are written in. They can both be used to describe the same process, but there is no real need to use the diabatic representation when the BO approximations holds.

They can be referred to as adiabatic/diabatic PESs, or states, or surfaces, or representations. The term 'process' here is incorrect as it refers to thermodynamics terminology.


Finding all Chemical Reactions Given Products and Reactants

The purely combinatorial methods are a first step in modeling chemical reactions. They alone however cannot consider the possibility of the chemical reactions and the chemical stability of the reaction products.
$\ $

1.)

You can build all combinations of at least two formulas, all from the original set of given chemical formulas, and treat this as your set of given formulas. But you can also treat the whole original set of given chemical formulas.

I demonstrate the method with the first example given in the question:

$$\{ \ce{C, H2, N2, O2, CO, CO2, H2O, NH3, NO, NO2, NO3}\}$$

We set up one chemical equation with all chemical formulas that are in the given set. Let the stoichiometric factors in the chemical equation be denoted by $\nu$. Because we don't know which of the given substances will be educt and which of them will be product, we write the chemical equation without reaction arrow. Instead, a negative stoichiometric factor will later mark an educt, a positive a product:

$$\nu_1\ce{C}+\nu_2\ce{H2}+\nu_3\ce{N2}+\nu_4\ce{O2}+\nu_5\ce{CO}+\nu_6\ce{CO2}+\nu_7\ce{H2O}+\nu_8\ce{NH3}+\nu_9\ce{NO}+\nu_{10}\ce{NO2}+\nu_{11}\ce{NO3}$$

We have to obey one of the first fundamental laws of chemistry: the Law of multiple proportions. To fulfill this law, the total sum of each chemical element in our chemical equation has to be $0$. And if ions are among our given chemical formulas, the total sum of all electric charges has to be $0$ also. But no electric charges are involved in our given example.

We will now build a mathematical model of our chemical problem.

For each of the given chemical elements, we have to set up its balance equation from our chemical equation above:

$\ce{C}:\ \ \ \nu_1+\nu_5+\nu_6=0$
$\ce{H}:\ \ \ 2\nu_2+2\nu_7+3\nu_8=0$
$\ce{N}:\ \ \ 2\nu_3+\nu_8+\nu_9+\nu_{10}+\nu_{11}=0$
$\ce{O}:\ \ \ 2\nu_4+\nu_5+2\nu_6+\nu_7+\nu_9+2\nu_{10}+3\nu_{11}=0$

This is a linear equation system. In the general case, too, a linear equation system results. It consists of the coefficients (the numbers above) and the stoichiometric factors $\nu_i$, which are sought. In our example, the equation system has 4 equations (the number of different chemical elements and the electric charge) and 11 unknowns (the number of chemical formulas in the given set).

Linear Algebra says how a linear equation system can be handled. We use here the matrix presentation.

Each of the 11 given chemical formulas is presented by a column vector which contains the frequency of occurence of each chemical element in the chemical formula in a prescribed order. Each of the 4 given chemical elements is presented by a row vector which contains the frequency of occurence of the chemical element in the chemical formulas in a prescribed order.

All 11 column vectors or all 4 raw vectors of the equation system are combined to give the coefficient matrix $A$. Our 11 wanted stoichiometric coefficients build the solution vector $x$, which is sought. The matrix representation of our linear equation system is than:

$$A\cdot x=\emptyset,$$

wherein $\emptyset$ is the zero column vector with 4 rows. It is written out:

$$\left(
\begin{array}{}
1&0&0&0&1&1&0&0&0&0&0\\
0&2&0&0&0&0&2&3&0&0&0\\
0&0&2&0&0&0&0&1&1&1&1\\
0&0&0&2&1&2&1&0&1&2&3
\end{array}
\right)
\cdot
\left(
\begin{array}{}
\nu_1\\\nu_2\\\nu_3\\\nu_4\\\nu_5\\\nu_6\\\nu_7\\\nu_8\\\nu_9\\\nu_{10}\\\nu_{11}
\end{array}
\right)
=\left(\begin{array}{}0\\0\\0\\0\end{array}\right)$$

The solution vector $x$ can be found by solving the linear equation system by methods of Linear Algebra.

The solution vector of our example is:

$$x=\left(
\begin{array}{c}
2\nu_4+\nu_6+\nu_7+\nu_9+2\nu_{10}+3\nu_{11}\\-\nu_7+3\nu_3+\frac{3}{2}\nu_9+\frac{3}{2}\nu_{10}+\frac{3}{2}\nu_{11}\\\nu_3\\\nu_4\\-2\nu_4-2\nu_6-\nu_7-\nu_9-2\nu_{10}-3\nu_{11}\\\nu_6\\\nu_7\\-2\nu_3-\nu_9-\nu_{10}-\nu_{11}\\\nu_9\\\nu_{10}\\\nu_{11}
\end{array}
\right)$$

The different combinatorially possible chemical equations are obtained by choosing suitable values for the free variables $\nu_3,\nu_4,\nu_6,\nu_7,\nu_9,\nu_{10},\nu_{11}$.
$\ $

2.)

Further investigations could be made by searching the solutions that don't contain any other solution. That are the minimal solutions, that means the solutions that are linearly independent of the other solutions.

I found 663 chemical equations and 83 minimal solutions for the example above.

Generating only the minimal solutions in polynomial time needs particular algorithms:

Schay, G.; Pethö, Á.: Über die mathematischen Grundlagen der Stöchiometrie. Acta Chim. Acad. Sci. Hung. 32 (1962) 59-67

Pethö, Á.: Zur Theorie der Stöchiometrie Chemischer Reaktionssysteme. Wissenschaftl. Zeitschr. 6 (1964) 13-15

Pethö, Á.: Algebraic treatment of a class of chemical reactions in stoichiometry. Acta Chim. Acad. Sci. Hung. 54 (1967) 107-117

Pethö, Á.: On a class of solutions of algebraic homogeneous linear equations. Acta Math. Acad. Sci. Hung. 18 (1967) 19-23

Pethö, Á.: Kémiai reakciók egy osztályának algebrai elemzése. Magyar Kémiai Folyóirat 74 (1968) 488-491

Kumar, S.; Pethö, Á: Note on a combinatorial problem for the stoichiometry of chemical reactions. Int. Chem. Eng. 25 (1985) 767-769

Pethö, Á.: The linear relationship between stoichiometry and dimensional analysis. Chem. Eng. Technol. 13 (1990) 328-332

Szalkai, I.: Generating minimal reactions in stoichiometry using linear algebra. Hung. J. Ind. Chem. 19 (1991) 289-292
with a software code at the end of the article

Pethö, Á.: Mathematical discussion of the application of Hess‘s law. Hung. J. Ind. Chem. 21 (1993) 35-38

Pethö, Á.: Further remarks on the linear relationship between stoichiometry and dimensional analysis. Chem. Eng. Techn. Chem. Eng. Tech. 17 (1994) (1) 47-49

Pethö, Á.: Further remarks on the analogy between stoichiometry and dimensional analysis: The valuation operation. Hung. J. Ind. Chem. 23 (1995) 229-231

Laflamme, C.; Szalkai, I.: Counting simplexes in $R^n$. Hung. J. Ind. Chem. 23 (1995) 237-240

Laflamme, C.; and I. Szalkai, I.: Counting simplexes in $R^3$ Electron. J. Combin. 5 (1998) (1) #R40 11. Printed version in: J. Combin. 5 (1998) 597-607

Szalkai, I.: Handling multicomponent systems in $R^n$ I Theoretical results. J. Math. Chem. 25 (1999) 31-46

Szalkai, I.: On valuation operators in stoichiometry and in reaction syntheses. J. Math. Chem. 27 (2000) 377-386

Szalkai, I.: A new general algorithmic method in reaction syntheses using linear algebra. J. Math. Chem. 28 (2000) 1-34

Szalkai, B.; Szalkai, I.: Counting minimal reactions with specific conditions in $R^4$ J. Math. Chem. 49 (2011) 1071-1085

Szalkai, I.; Dósa, G; Tuza, Z.; Szalkai, B.: On minimal solutions of systems of linear equations with applications. Miskolc Mathematical Notes 13 (2012) (2) 529-541

Szalkai, B.; Szalkai, I.: Simplexes and their applications - A short survey. Miskolc Math. Notes 14 (2013) (1) 279-290

Szalkai, I.; Tuza, Z.: Minimum Number of Affine Simplexes of Given Dimension. Discr. Appl. Math. 180 (2015) 141-149

Szalkai, I.: Reakciómechanizmusok algoritmikus és matematikai vizsgálata (Algorithmic and Mathematical Examination of Reaction Mechanisms). PhD thesis, University of Pannonia, Veszprém, Hungary, 2014

Szalkai, I: An algorithmical and mathematical investigation of reactions. English extract of the author's PhD thesis, 2014
See the examples and references therein.

Tóth , J.: Reaction Kinetics: Exercises, Programs and Theorems. Springer, 2018
$ $

3.)

The answer above treats overall reactions. The underlying elementary reaction steps (elementary reactions) can be generated by working with the Ugi-Dugundji model: The chemical formulas are presented as graphs together with their graph matrices (Bond-Electron matrices (BE-matrices) and Reaction matrices (R-matrices)). The single elementary reaction steps will be generated successively by stepwise shifts of electrons or bonds in the set of given chemical formulas by stepwise shifting of single edges in the reaction matrix.

Ugi, I.; Gillespie, P.; Gillespie, C.: Chemistry, a finite metric topology - synthetic planning, an exercise in algebra. Trans. New York Acad. Sci. II 34 (1972) 416-432

Dugundji, J.; Ugi, I.: An algebraic model of constitutional chemistry as a basis for chemical computer programs. Top. Curr. Chem. 39 (1973) 19-64

Dugundji, J.; Gillespie, P. D.; Marquarding, D.; Ugi, I.; Ramirez, F.: Metric spaces and graphs representing the logical structure of chemistry. in: Chemical applications of graph theory. Balaban, A. T. (Ed.), Academic Press, London 1976, 107-174

Brandt, J.; Friedrich, J.; Gasteiger, J.; Jochum, C.; Wolfgang Schubert, W.; Ugi, I.: Computer programs for the deductive solution of chemical problems on the basis of a mathematical model of chemistry. in: Wipke, W. T.; Howe, W. J.: Computer-Assisted Organic Synthesis. American Chemical Society 1977, chapter 2, 33-59

Ugi, I.; Brandt, J.; Friedrich, J.; Gasteiger, J.; Jochum, C.; Lemmen, P.; Wolfgang Schubert, W.: The deductive solution of chemical problems by computer programs on the basis of a mathematical model of chemistry. Pure Appl. Chem. 50 (1978) 1303-1318

Ugi, I.; Brandt, J.; Friedrich, J.; Gasteiger, J.; Jochum, C.; Lemmen, P.; Schubert, W.: Computer Programs for the Deductive Solution of Chemical Problems on the Basis of a Mathematical Model of Chemistry. Pure Appl. Chem. 50 (1978) (11-12) 1303-1318

Behnke, C.; Bargon, J.: Computer-assisted topological analysis and completion of chemical reactions. J. Chem. Inf. Comput. Sci. 30 (1990) (3) 228-237

Ugi, I; Dengler, A.: The algebraic and graph theoretical completion of truncated reaction equations. J. Math. Chem. 9 (1992) (1) 1-10


Prediction/computation of mass spectral ionisation intensity

The absolute response coefficient (i.e. peak intensity in counts per mM or similar units) will be way too strong a function of instrument-specific parameters like solvent composition, pH, one or two or three temperature variables specific to the ion source, one or or two or three more voltage parameters particular to the ion source, whether a given emitter is dirty or fresh, how big it is, etc.
However, it may be possible to predict relative response coefficients across a series of compounds.  Most approaches I've seen rely on prediction of gas-phase basicity values, or alternately proton affinity of gas-phase neutral species (the assumption being that you are predicting the ionization efficiency of singly-charged ions in positive ionization mode arising from neutral molecules).
Folks have been working on "machine learning" approaches in this area.  Here is one example...


Why is UuX used as a symbol for unnamed elements on the periodic table?

In 1978, the IUPAC Commission on the Nomenclature of Inorganic Chemistry decided that it would be necessary to have a systematic naming for the elements with atomic number greater than 100, even for those which had not been discovered. The recommendations are as follows:

  The name is derived directly from the atomic number of the element using
  the following numerical roots:

  $$\require{enclose}
\small
\begin{array}{lcc}
\hline
\text{Number} & \text{Latin Word} & \text{Greek Word} & \text{IUPAC Nomenclature} \\
\hline
\text{0} & \enclose{circle}{\text{nihil}} & \text{mēdén} & \text{nil} \\
\text{1} & \enclose{circle}{\text{ūnus}} & \text{en} & \text{un} \\
\text{2} & \enclose{circle}{\text{duo}} & \enclose{circle}{\text{dúō}} & \text{bi } \\
\text{3} & \text{trēs} & \enclose{circle}{\text{tria}} & \text{tri} \\
\text{4} & \enclose{circle}{\text{quattuor}} & \text{téttara} & \text{quad} \\
\text{5} & \text{quīnque} & \enclose{circle}{\text{pénte}} & \text{pent} \\
\text{6} & \text{sex} & \enclose{circle}{\text{héx}} & \text{hex} \\
\text{7} & \enclose{circle}{\text{septem}} & \text{heptá} & \text{sept} \\
\text{8} & \enclose{circle}{\text{octō}} & \enclose{circle}{\text{oktṓ}} & \text{oct} \\
\text{9} & \text{novem} & \enclose{circle}{\text{ennéa}} & \text{enn} \\
\hline
\end{array}
$$

  The roots are put together in the order of the digits which make up the
  atomic number and terminated by 'ium' to spell out the name. The final
  'n' of 'enn' is elided when it occurs before 'nil', and the final 'i' of
  'bi' and of 'tn' when it occurs before 'ium'.
  The symbol of the element is composed of the initial letters of the
  numerical roots which make up the name.
  The root 'un' is pronounced with a long 'u', to rhyme with 'moon'.
  In the element names each root is to be pronounced separately.$^{[1]}$

Though not in the original text, I went ahead and added the words from which the IUPAC names were derived. Though not obvious, bi- comes directly from di-, which is a result of duo, found in both languages.

$^{[1]}$ Recommendations for the Naming of Elements of Atomic Numbers Greater than 100, Pure Appl. Chem., Vol. 51, No. 2, (1979) 381-384.


My proposed mechanism is a nucleophilic addition/elimination mechanism without a tautomerism step.

It is all concerted.

The first step, addition, features the oxygen in the (E)-propionaldoxime acting as a nucleophile to attack one of the carbonyl carbons in the acetic anhydride, which pushes electrons the carbonyl oxygen, making it negatively charged.

Then, the negatively charged oxygen strikes back, initiating a chain of concerted cyclic electron rearrangement which cleaves quite a few bonds, resulting in three molecules, completing the elimination step.


Why is Potassium in French and English not called Kalium?

(with thanks to @NilayGhosh, who provided the key source for much of this answer)

Short answer: Potassium was called Potassium (and Sodium Sodium) by Humphry Davy, who first isolated both; it was then renamed Kalium in Germanic countries because of a previous naming proposal by Klaproth, who was the first to show that potassium carbonate and sodium carbonate were different.

Until at least the Middle Ages no distinction was drawn between potash (potassium carbonate) and soda (sodium carbonate). Martin Heinrich Klaproth first distinguished them in 1797, and suggested the name 'kali' for the first and 'natron' for the second.

Metallic potassium and sodium were then isolated for the first time in 1807 by Sir Humphry Davy, by electrolysis of potassium hydroxide (KOH) and sodium hydroxide (NaOH) respectively. He named his discoveries Potassium and Sodium in the Bakerian lecture at the Royal Society on the 19th of November the same year:

  Potasium and Sodium are the names by which I have ventured to call the
  two new substances

(source)

(During the lecture he also apparently produced a nugget of metallic potassium and tossed it into a flask of water, "where the lump skittered around the surface of the water before exploding in lavender flames".)

This paper was translated into German by Ludwig Wilhelm Gilbert (supposedly in Annalen der Physik, although I can't find the original paper in any of the volumes from  1809). According to the article on vanderkrogt.net Gilbert replaced Potassium with Kalium and Sodium with Natronium "in agreement with German nomenclature" (i.e. following Klaproth's suggested names). From this point on, generally speaking, English- and French-speaking countries followed Davy's naming system, while the Germanic countries followed that of Klaproth/Gilbert, as can be seen from the sidebar lists here (potassium) and here (sodium).

Regarding the atomic symbols mismatch: the international system of atomic symbols was devised by the Swedish chemist Jacob Berzelius, a rival of Davy's, who followed the naming convention used by Germanic countries. Although actually, when first published (1813, in Thomas Thomson's Annals of Philosophy), Berzelius did follow Davy's names and used the symbols Po and So. Within a year he had switched to K/Na. So since the first publication of Berzelius' system actually followed Davy and used Po/So, the real surprise is that we ended up using K/Na.


If you want a reference to challenge the question, try page 248 of Advanced Study Guide Chemistry.  There is a table on that page that says the PbO is "ionic".


If fluorine has a lower electron affinity than chlorine, why does it have a higher ionization energy?

To quote chemguide:

  The first ionisation energy is the energy required to remove the most loosely held electron from one mole of gaseous atoms to produce 1 mole of gaseous ions each with a charge of 1+.

That means, the ionization energy of fluorine is the energy of the following reaction:

$$\ce{F -> F+ + e-}$$

The ionization energy of chlorine is the energy of the following reaction:

$$\ce{Cl -> Cl+ + e-}$$

Since the most loosely held electron in fluorine is closer to the nucleus than chlorine, it takes more energy to remove the electron from fluorine than in chlorine.

The ionization energy of chloride ion ($\ce{Cl-}$) is the energy of the following reaction:

$$\ce{Cl- -> Cl + e-}$$

The electron affinity of $\ce{Cl}$ is the energy of the following reaction:

$$\ce{Cl + e- -> Cl-}$$

As a result, their energies are equal in magnitude but opposite in sign.

Why does chlorine have a higher electron affinity than fluorine?

To quote this answer:

  Fluorine, though higher than chlorine in the periodic table, has a very small atomic size. This makes the fluoride anion so formed unstable (highly reactive) due to a very high charge/mass ratio. Also, fluorine has no d-orbitals, which limits its atomic size. As a result, fluorine has an electron affinity less than that of chlorine.

As can be seen in the second photo, this effect seems to be true for Period 2 elements.

(source: lardbucket.org)

(source: lardbucket.org)


Why doesn't water evaporate in a balloon filled with water?

The difference is the empty space in the latter case and lack of it in the former. By "empty" I mean "filled with air", but that's irrelevant; it could have been truly empty with the same result. Air pressure is also irrelevant; it could have been 10 atm, and that still wouldn't prevent water from evaporating.

See, when several gases occupy the same volume, they do not even know about each other. They do not care if another gas enters the same volume and starts to share it with them. But they care a great deal about the volume itself, and try to increase it with all their might; that's what we call pressure.

So it goes.


Voltage is indeed something that is measured at a particular point, but that doesn't mean its a property of only one point and in fact every point will have some voltage. In particular you can have a series of points that map out an "equipotential" where the potential is equal all along these points. A terminal of a battery is an equipotential so no matter where an electron "is" it will have the same potential energy U = e V.

Intuitively this is just like potential energy due to gravity. The battery is like a mountain with a flat top and level plane at the base. Everywhere along the top the altitude is equal and a ball will have the same potential energy. Similarly everywhere along the level base of the mountain the ball will have the same potential energy.


(Single) Amino acid rotational degrees of freedom

You should be concerned with bonds where the rotation is non-trivial. For example, rotation about a C-H or O-H bond doesn't do anything. But rotation around the serine side-chain C-O bound or the side-chain central carbon bond or the C-O carboxylate bond all matter.


First of all, you must understand that the reactivity of OR changes as you change the alkyl group (-R).

For example, consider, the methoxy group (-OCH3), as per proton NMR spectra of anisole and phenol, the reactivity is almost similar in both the cases. In fact the reactivity is so similar that undergraduate texts just mention the fact that anisole has reactivity similar to phenol. (Source 1)

On the other hand, if you consider more bulky alkyl groups, steric hindrance might be a contributing factor to reactivity, competing with the mesomeric effect. But to a large extent, it is sufficient to say that both hydroxyl and (-OR) groups have similar reactivity.


How can the number of molecules and atoms be the same in a gas?

The gas was one (or a mixture) of the noble gases, as every other gas is composed of more atoms than molecules.

Example

Helium: $1~\text{mol}~\ce{He}~\text{gas} = 1~\text{mol}~\ce{He}~\text{atoms}$
Nitrogen: $1~\text{mol}~\ce{N2}~\text{gas} = 2~\text{mol}~\ce{N}~\text{atoms}$


Why is "Spontaneous" a part of 2nd law wording?

In chemical thermodynamics we are interested in the amount of work that can be done by a reaction as it proceeds towards equilibrium and whether a reaction will spontaneously proceed towards products or, if not, how much work is needed to bring it about. Spontaneous is used because experimentally reactions and processes occur without us having to do anything to bring them about, and this needed an explanation.

The second law can be stated as$^1$  'Spontaneous changes are those which, if carried out under the proper conditions can be made to do work. If carried out reversibly they yield the maximum amount of work. In natural processes the maximum amount of work is never obtained.' This is also a general definition of equilibrium. A shorter definition of the second law is 'heat does not spontaneously flow from from a cold to a hot body'.

If there are differences in intensive properties then a spontaneous change is always that which eliminates the difference. Thus if there is a (perfect) piston with different pressures on either side, it will move to equalise the pressure. Similarly if there is a temperature difference between two bodies in contact with one another then 'heat flows' from the warmer to the colder. Similarly, a concentration difference in a solution will equalise. However, this cannot be used to predict the direction of a chemical reaction in which there may already be constant pressure, temperature or concentrations.

In a purely mechanical system, the direction of spontaneous change seems obvious and is one in which the potential energy is lowered to a minimum. It was thought at one time that the same would apply to chemical systems, i.e that the direction of spontaneous change was one that minimises the heat content of the system. But experiment shows that this is not the case, there are, for example, spontaneous endothermic reactions$^2$, so that another criteria is needed to predict the direction of a spontaneous reaction.  This means understanding why for example, a gas expands to fill a container even though no loss of energy occurs on doing this. The missing idea is the entropy$^3$.

It used to be thought, before the second law was properly understood, that the maxiumum work a chemical reaction could do was to be found by converting all the heat of reaction into work. This was not bourne out by experiment and it is the free energy $\Delta G$ that gives the maxmum amount of work. Gibbs was the first to understood that the maximum amount of work is given by the first law unless some heat is taken in or given out to the surroundings. Thus the work done may be greater than the heat of reaction.

The second law requires that there be an increase in entropy during any actual process. Thus $dS_{sys} + dS_{surr} \ge 0$ for the system and surroundings. The sum of the two entropy production processes is always a positive quantity, but which can be zero in the case of a reversible process.

(Your speculation about statistical mechanics is partly correct, it is possible for entropy to 'run backwards' if only a very small number of molecules is considered and then only for a very short time. In 'normal' lab chemistry the number of molecules involved is vast say, $10^{18}$ so deviations from the average are minute.)

$^1 $ E. B. Smith, 'Basic Chemical Thermodynamics'.

$^2$ e.g. $\ce{Ba(OH)2(8H2O)_{(s)} + NH4SCN_{(s)} +heat \rightarrow Ba(SCN)2 + 2NH3 + 10H2O}$

$^3 $ The second law is needed because the first law offers too much latitude compared to experimental observations; by the first law heat could spontaneously flow from a cold to a hot body or a broken glass could spontaneously reform itself.


Ring expansion from a given cyclic carbocation

I think your friend is thinking of the cyclobutyl carbocation which does ring contract to the cycylopropyl carbinyl carbocation (and also equilibrates with the  methallyl carbocation).

However, just as you thought, the cyclobutyl carbinyl carbocation does ring open to the cyclopentyl carbocation (ref_1, ref_2, ref_3).  This rearrangement is driven by carbocation stability (primary to secondary) and relief of the steric strain present in the 4-membered ring, again, just as you suggested.

Note: this question has been asked previously on SE Chem, however I believe the accepted answer is incorrect, as it seems to primarily address the cyclopropyl carbinyl case, and what is said about the cyclobutyl carbinyl carbocation (little to no ring expansion) is incorrect.


Triple Point of a compound and critical point

The essential condition is the compound thermal stability.
If it decomposes below its melting point, it does not have a triple point.
If it decomposes before properties of gaseous and liquid phases converge to each other,  it has just estimated, extrapolated triple point.
Beyond CP we talk about supercritical fluid, it kind of shares many properties of both liquid and gaseous worlds.
It is usually determined experimentally, or as an extrapolation.
See the picture of liquid/gas phase system below, at and above the critical point on Wikipedia: Critical_point


If the amount of extra C14 was unknown this might become a problem. If the change in atmospheric C14 was recorded and the records survive into the future, it would be easy to adjust the dates. At present we have good data on C14 concentration over time for various locations around the world. I don't know the spatial resolution though.


Your function $p$ is a function of the independent variables $V$ and $T$ i.e., $p(V,T)$. The other variables, $a$, and $R$ are constants.

Let me rewrite the differential as

$$\mathrm{d}p = A(V)\,\mathrm{d}T + B(T,V)\,\mathrm{d}V$$

where

$$A(V) = \frac{R}{V}$$

and

$$ B(T,V) = \left( \frac{2a}{V^3} - \frac{RT}{V^2} \right) $$

A differential is exact if

$$\left( \frac{\partial A}{\partial V} \right)_T = \left( \frac{\partial B}{\partial T} \right)_V $$

Note that subscripts denote that the variable is being held constant during the partial differentiation. The problem already told us what our $A$ and $B$ terms are, and I wrote them out explicitly above. Now, we just need to take the appropriate partial differentials and compare. If they are equal, then $\mathrm{d}p$ is exact!

Writing out the formulas took awhile, it seems a shame to stop now... A little bit of differentiating, remembering to hold the appropriate variables constant during differentiation leads to

$$\left( \frac{\partial A}{\partial V} \right)_T = -\frac{R}{V^2}$$

$$\left( \frac{\partial B}{\partial T} \right)_V = -\frac{R}{V^2}$$

If the two partial derivatives are the same, the differential is exact. I will let you be the judge.


Can I make a buffer from two different weak acids or bases?

Yes, having multiple weak acids and/or bases having different pKa values would give buffering capacity in more than one pH range, or over a greater pK range.

There is a term "universal buffer" which refers to a buffer having multiple pKas spaced thoughout the pH ~2-12 range to give some buffering capacity over the enter pH range.

As explained in A Neutral Buffered Standard for Hydrogen Ion Work and Accurate
Titrations Which Can be Prepared in One Minute J. Am. Chem. Soc., 1932, 54, pp 1911–1912, ammonium aceate has the property that if it is added to pure water it produces a buffered solution of pH 7.00, accurate to a hundreth of a pH unit.


Recommendations for textbooks with correct and up-to-date information

I disagree with the idea to forget about Lewis formulae because of their age. If you browse across the table of contents of scientific journals, however, you see Lewis formualae are widely used.  From the first December issue of Organic Letters, for example is taken the following screen photo:

If Lewis formulae were so off and wrong, chemists would use other forms to describe molecular structures in their publications.  As so often, however, it is a question about what you want to describe, and to know what limitations of the tools at disposition inheritated.
By convention (and training), for example, chemists do not read the rings in formula about nitronaphthalene (on the left) as an alternation of isolated single- and double bonds, but conceptionalize this as to depict aromatic systems.  Counting electrons, for example, this is easier this way as with the Thiele ring which accentuates the delocalization of the $\pi$-electrons.*)

(source)

So it is better to recognize where Lewis formulae may be misleading, with some instances listed e.g., here, and some you will encounter in study, e.g. diborane

(source)

It equally is helpful to know that dashes in formulae, depending on convention, may represent different aspects:

(source)

*) In 1899, Thiele's perspective was to emphasize that benzene's molecular structure is between the two mesomeric structures expressed as Kekulé-structure and thus used the (initially dashed) cricle in the structural formula.  Chemists' perspective of delocalization of $\pi$-electrons today obviously is influenced by quantum physics.


Why does cationic charge density affect metallic bonding?

A good way to conceptualize metallic bonding is that the valence electrons on metal ions become delocalized, leaving positive ions to be electrostatically attracted to the electrons.

Group II metals contribute 2 valence electrons to the delocalized "sea of electrons", and thus have a +2 charge on the remaining cation. These combined create a stronger attractive force.

It is true that the radii of Group II metal cations are smaller than those of Group I cations (within the same row). This also affects bonding, as the delocalized electrons are closer to the nucleus in the Group II case.

In summary: the increased coulombic force between a higher number of delocalized electrons and higher charges on metallic cations create a stronger metallic bond. Transition metals tend to have even stronger metallic bonding due to the participation of electrons from d orbitals.


"Shifting" an equilibrium

Unfortunately for learners of chemistry, statements like "the equilibrium shifts to the right" are quite common jargon.

What is meant by this, speaking in more accurate technical terms, is the following: A reaction had reached equilibrium (net reaction is zero), and then some change occurred so that it was no longer in equilibrium. Now, there is a net forward reaction in the direction of the products ("to the right" refers to reactants turning into products).

"Shifting to the left" describes a similar sequence of events, just in the reverse direction.

New equilibrium or reestablish equilibrium?

The OP is correct in saying that unless the temperature changed, the equilibrium constant does not change, and the reaction quotient Q of the first equilibrium state is equal to that of the second. So you could say you are back to the same equilibrium.

Typically, however, the mole fractions, concentrations or partial pressures will be distinct for the first and the second equilibrium. So you could also say the reaction reaches a new equilibrium.

If you say it reaches equilibrium again, you are sufficiently vague to be correct.

Position of equilibrium

The term "position of equilibrium" is ill-defined. To avoid any confusion, it is better to distinguish between reaction quotient (which is equal to the equilibrium constant for both the first and second equilibrium) and the set of concentrations (which is typically different).


First, a little bit of background. Transparency is not an absolute property of a material. Every substance is opaque, so long as light has to pass through enough of it, and opacity also changes according to ambient conditions.  Some substances, such as most metals, are opaque even in $100\ \mathrm{nm}$ thin films, while many gasses will let a small amount of light through mostly unperturbed even after several kilometres. For example, here is a measure of how far photons of different energies can travel on average in the pure elements before interacting once with their atoms (the graph is actually for high energy x-rays rather than visible wavelengths, but they're all photons nonetheless). It makes intuitive sense that a gas should let more light through, as the atoms in it are much more spaced than in a solid or liquid.

The two general processes that work to make materials non-transparent are photon absorption and photon scattering (reflection being included as an example of scattering). When we say a material is "transparent" colloquially, what we generally mean is that the material, at the thickness observed, does not scatter much light and thus allows a coherent image to pass through the material (basically, you can see through it to the other side without a lot of distortion, that is, the object is "see-through" or clear). Interestingly, this means that a clear material need not be colourless, as correctly pointed out by Geoff, because photons of a given colour may be absorbed while letting most other wavelengths pass through without scattering.

Gasses in general are highly transparent because they are both highly colourless (absorb little light) and very clear (scatter little light) unless you're looking through several kilometres of gas, as I mentioned previously. However, there are examples of highly clear but coloured gasses, as it is possible for them to absorb a significant amount of photons of a certain energy due to low energy electronic transitions, while allowing the rest of the visible spectrum to pass through unimpeded. While some gasses are coloured and others are not, I believe all pure gasses must be very clear (small amounts of gas will scatter almost no light), and they can only scatter a fair amount of light across a short distance in conditions where they are probably not best described as gasses anymore (plasmas, supercritical fluids, etc).

Perhaps the most classic example of a coloured gas is nitrogen dioxide, $\ce{NO2}$, a strongly brown-red noxious gas which is easily formed by decomposition of nitric acid, among several other ways. The unpaired electron in its structure is somewhat uncommon, and it can be easily excited to a higher electronic state upon absorption of photons in the blue region of the visible spectrum. Since blue light is strongly absorbed even in relatively thin samples (a few centimetres of $\ce{NO2}$ gas), the rest of the white light passes through with essentially no scattering, forming a perfect image of the other side but with a strong red tinge.

Even more interestingly, $\ce{NO2}$ reacts with itself at low temperatures or high pressures and dimerizes to form dinitrogen tetroxide, $\ce{N2O4}$, a colourless solid/liquid/gas (depending on conditions). In other words, the two substances are in a reversible equilibrium:

$$\ce{2 NO2 (g) <=> N2O4 (g) + energy}$$

Coincidentally, this equilibrium is characterized by an equilibrium constant which is close to 1 at ambient conditions, and this constant can easily be changed by varying temperature or pressure to higher or lower values. This means that under relatively easily achieved conditions, it is possible to study the reversible change of a coloured gas into a colourless one!

(Source)

Though I focused on $\ce{NO2}$ because it is a remarkable case, there are several other examples of clear but coloured gasses. As Geoff mentioned, the halogens all form coloured gasses, though bromine is a volatile liquid and iodine is a solid at room temperature, so a small amount of heating is necessary. Ozone, $\ce{O3}$, is somewhat blue (source), though its colour is best observed in the liquid phase. The same goes for oxygen gas, $\ce{O2}$, which is slightly blue, though it is not the cause for the blue sky. There are probably a few more examples out there.


Polysytrene factories and hypothetical health concerns

It is good to see the some authors want to build realism into their fiction: this clearly isn't common. And you have done extensive research already.

The problem is that industrial chemistry is very heavily regulated. So, while it is common for dangerous substances to be used, it is rare to find places where they can plausibly be misused. And the standards for harmfulness are very low. Even some things at the harmful end of the official scale are not that harmful.

The biggest problem you will have when it comes to realism isn't really to do with polystyrene manufacture at all. A factory that makes products with polystyrene won't usually make its own polystyrene but will buy in the plastic raw material as polystyrene beads which can be mixed with other ingredients (like colours) and melted in moulding machines which produce the product. So none of the chemicals used to manufacture the plastic will be used anywhere while making products from the plastic.

If you were to assume that the raw polystyrene was made locally (which isn't that realistic) then you might find plenty of chemicals that could be misused as you have already realised. CCl4 is one of the best candidates but i rarely used any more. Benzene is used in manufacture so is also plausible; styrene (the monomer that is used to make the plastic) is also nasty and possibly the best candidate you haven't considered.

However, if you want to stick with broad realism and a story that uses plausible things associated with a jewel case factory, then you need to consider a different approach. But it might be easier because you can make some things up that even experienced experts can't question. You could, for example, imagine that your factory adds some special ingredient to its products to create unusual colours or effects (maybe the cases glow in the dark, are unusually strong or contain inks that allow patterns to be printed deep into the plastic by lasers). The ingredients to create this effect could be anything you want (and you probably don't have to be too explicit about exactly what it is: give it an obscure acronym). This ingredient could plausibly have any bad properties you want without being unrealistic. Possibly the product will have been invented recently to give the products a special edge and nobody else has yet observed the side effects.


Why should one perform an extraction with multiple small portions of solvent as opposed to one large portion?

The important thing is the distribution ratio (for the extraction of a metal) and the partition coefficent for the extraction of a given chemical species. This can be used for a first approximation calculation.

The distribution ratio is the total metal concentration in the organic phase (grams of metal per litre) divided by the total metal concentration in the aqueous phase.

If we consider a batchwise extraction where a litre of water bearing x mg of a metal is extracted with an organic solvent. Assume that the distribution ratio is 2. We can do the calculation for the extraction.

The phase ratio (ratio of the volumes of the two phases) is 1, call this theta.

Theta D / (1 + (Theta D)) = the fraction extracted in each stage.

If D is 2, and theta is 1 then we will extract 66 % of the metal.

As an alternative, if we use one litre of water and extract this with 0.5 L of organic phase, then with D = 2 and Theta = 0.5 then we will extract 50 % of the metal.

If we repeat the extraction with another 0.5 L of solvent then we will recover 50 % of the remaining 50 % of the metal. Thus we will have recovered in total 75 % of the metal.

If instead we were to extract the 1 L of water four times with 0.25 L of water then in the first extraction we would leave 66 % of the metal in the lower phase. The easy way of working out the amount left in the aqueous later after n extraction stages with the same D and theta valve is to use

{1 - ( D Theta / {1 + (D Theta)})}^n = fraction left in raffinate. In solvent extraction raffinate is the term for the aqueous layer after an extraction has occured.

This gives me about 20 % left in the aqueous phase after four extractions with 0.25 L of solvent.

It is important to understand that the solubility of the substance is not what you should consider it is the distribution ratio or the partition coefficent which is needed.

If you were to take a dilute solution of DEHPA (150 mM) in aliphatic kerosene and to shake this with 0.1 nitric acid containing a small amount of Fe-55 with no stable iron as iron(III) then you might get a D value of 100 for arguments sake.

Now the organic phase can only extract 50 mM of iron per litre as to extract one iron atom three DEHPA molecules are needed. If we were to repeat the extraction experiment with a solution of 56 grams of iron per litre (roughly 1 moles per litre of iron) spiked with a little Fe-55 then we would measure a far lower distribution ratio.

This is due to loading of the organic phase.

Equally when you extract a large amount of an organic substance into an organic solvent the solvent properties of the organic layer at the end will be different to those of the pure solvent.

For example if I have 1 litre of water saturated with hexanol with 500 ml of water saturated hexanol floating on the top. Then if I was to add 500 ml of ether and then shake it. I would have a different Kd for hexanol to that which I would have if I was to shake a dilute solution of hexanol in water with ether.

I hold the view that the formula used above for the system with a D value of 2 will only hold perfectly true when the concentration of the metal is zero. This is why radioactive tracers are so useful for academic solvent extraction experiments. This is becuase they allow the measurement of distribution ratios with a zero metal concentration. The concentration in moles per litre of a 1 GBq per litre solution of Fe-55 will be very small.

The half life of Fe-55 is about 2.7 years, thus I have calculated that this very radioactive stock solution will contain 205 nanomoles of iron if the only iron atoms present are Fe-55. The best way to make the iron-55 would be to bombard managense with protons with a cyclotron and then to separate the iron from the manganese by solvent extraction or some other separation method.

I would in a lab expect to be able to run a solvent extraction experiment with only 700 Bq of Fe-55. If we assume that I was to shake 700 microlitres of aqueous phase with 700 microlitres of the organic phase. I would then be able to sample 200 microlitres of the two layers with a pipette.

If we assume that the highest distribution ratio I was to measure was 100 then I would expect that at the end of the shaking that the aqueous layer would contain 10 Bq per ml. If I was to sample 200 microlitres and to put this into a LSC vial and then to count it. Assuming that 100 % of radioactive decays will result in a count in the LSC machine then I would in one minute get 120 counts. If I was to count until I had 10000 counts (1 % error due to count number) I would have to wait about 83 minutes.

For a D value of 10 I would be able get 10000 counts in less than ten minutes for the least radioactive liquid phase. Thus it should be clear that with a radiotracer it is possible to measure the distribution ratio with less than 1 nanomole per litre of metal present.


Effect of the counterion on the colour emitted by a metalic salt

Recall the flame emission experiment is basically an atomic emission experiment. We don't say it is a compound emission experiment. The job of the flame/plsama or any high temperature medium is to decompose the compound into the constituent atoms or even ions for an experiment. Once a dilute metal salt solution is aspirated into the flame, the compound decomposes into constituents and then to the atoms. For example, you start with NaBr, NaF or NaI or NaNO3, in the flame you will get one and only emission at 589 nm from the sodium atoms. All of these compounds will make the flame bright yellow. So what happened to the non-metallic anion? Provided the flame temperature is high, one should "see" emission from non-metallic atoms as well. Our eyes cannot see the emission bc it in the deep UV region.

Sometimes the anion interferes with the atomization process such as phosphate for calcium. It forms very thermally stable compounds in the flame. You will see the signal of calcium go down quickly if phosphate ion is present.

Forget about copper as atomic emission, I guess you are thinking about the green flame of copper compounds. This only happens in low temperature flames. That color originates from molecular compounds of copper in the flame, and the anion may play a role. You need a very high temperature flame and copper emits in the UV.


Why Inductive effect operates only through single bonds?

If I'm not mistaken a good example of the inductive effect through multiple bonds is trifluoro-acetic-acid, or $\ce{F3CCOOH}$. The proton is so electron deficient that it is easily abstracted by anything that can act as a base. The not 'fluorinated' equivalent, acetic acid, is not nearly as strong an acid.

The reason for the difference in acidity, is the difference in electro negativity between hydrogen atoms and fluorine. Due to the very large electro-negativity of fluorine atoms the electrons in the acid protons are 'pulled' towards it, leading to a clear example of the inductive effect over multiple bonds.

A comment on one of the comments below the question: Electron displacement does occur in double bonds. A good example is the carbonyl functional group (C=O) in a molecule. The inductive effect results in a slightly more positive charge on the Carbon atom and slightly higher at the O-atom. This is also due to electro-negativity and results in the reactivity of the C=O functional group.

Hope this helped!


The flaw in the question is exactly what Gaurang Tandon said: $\ce{H3O^+}$ is formed by a water molecule and a proton ($\ce{H^+}$), not a hydrogen atom.

Compare the number of electrons in both sides of the following two equations:

$$\ce{H2O + H^+ -> H3O^+}$$

$$\ce{H2O + H^. -> H3O^+}$$

The last equation is wrong: there's an extra electron in the left hand side.