Does water ionically bond to chloride ion?

The most common mechanism for when $\ce{NaCl}$ dissolves is as folows:

As water is a polar molecule, as per the UC Davis ChemWiki page All About Water, shown below:

The caption reads:

  the positive and negative charges are not distributed uniformly. This is illustrated by the gradation in color in the schematic diagram here. The electronic (negative) charge is concentrated at the oxygen end of the molecule, owing partly to the nonbonding electrons (solid blue circles), and to oxygen's high nuclear charge which exerts stronger attractions on the electrons. This charge displacement constitutes an electric dipole, represented by the arrow at the bottom; you can think of this dipole as the electrical "image" of a water molecule.

So, when $\ce{NaCl}$ salt is dissolved in water, the sodium and chloride ions are pulled apart by the water molecules with the result explained in the image below:

Image source

You'll note, that there is no ionic (nor covalent) bonding present, rather an ion-dipole interaction between the water molecules and the sodium and chloride ions (University of Waterloo).

The UCLA page Illustrated Glossary of Organic Chemistry has a very nice definition of the ion-dipole interaction:

  A noncovalent attraction between one pole of a bond dipole and an oppositely-charged ion. The δ+ end of the bond dipole is attracted to an anion, and the δ- end of the bond dipole is attracted to a cation.

There is a neat YouTube clip showing the process here.

The ion $\ce{H2OCl+}$, does actually exist according to NIST, but I could not find any references that link it to the dissolution of $\ce{NaCl}$ salt.


Representing woody biomass concentration for kinetic reactions

First step is finding the molar masses of each compound. Since cellulose, hemicellulose and lignin are collections of similar molecules, I picked the following as a crude estimate:

cellulose: $\ce{C6H10O5}$ with $M_c = 162~\mathrm{g/mol}$
hemicellulose (L-Arabinose): $\ce{C5H10O5}$ with $M_h = 150~\mathrm{g/mol}$
lignin: $\ce{C31H34O11}$ with $M_l = 582~\mathrm{g/mol}$

Using a density of $700~\mathrm{g/dm^3}$ you will find the relative masses of the three compounds to be:

$0.48 \cdot 700~\mathrm{g/dm^3} = 336~\mathrm{g/dm^3}$ (cellulose)
$0.28 \cdot 700~\mathrm{g/dm^3} = 196~\mathrm{g/dm^3}$ (hemicellulose)
$0.24 \cdot 700~\mathrm{g/dm^3} = 168~\mathrm{g/dm^3}$ (lignin)

With $n = m/M$ we now find the number of moles of each compound in the sample ($1~\mathrm{dm^3}$):

$2.074~\mathrm{mol/dm^3}$ cellulose
$1.307~\mathrm{mol/dm^3}$ hemicellulose
$0.289~\mathrm{mol/dm^3}$ lignin

Keep in mind that this is a very rough estimate. You'll probably get a better result by taking the average of all species of hemicellulose, etc. in your sample and cross-checking it to make sure it fits the ratios of the elements (weight percent).


Why does density decrease across the period 4 and 5 transition metals?

Density is generally proportional to how far from a full octet (or in your case, an 18 electron set of outer-most orbitals) a material is, hence this effect, and its inverse - that earlier elements in a period are less dense. This does not apply only to transition metals, but all elements, as illustrated here. (Note: I am not sure what pressure is in that chart, so I suggest ignoring gas-phase elements, including O, F, and Cl.)

The reason for this is fairly simple. The further from a complete octet an element is, the more energetically favorable orbital overlap becomes. The more favorable orbital overlap becomes, the closer elements will pack at a given temperature. There is no energetic benefit to an element with a full octet (ideal gases) overlapping, so they typically do not, and thus form a gas. There is an intense energy benefit in elements with both half-full octets overlapping.


Colour of the transition metal complexes

Most colour in chemistry comes from selective absorption of light by electronic transitions of the appropriate energy

Colour is not limited to transition metal complexes. They have a particularly rich range of colours as the electronic transitions often match the energy range of visible light but also because they are easily altered by the nature of the ligands surrounding the metal (for example, Cobalt complexes can be almost any colour of the spectrum depending on the ligand).

But plenty of other compounds are coloured. Most dyes are purely organic and the colour comes from the absorption of light in transitions between molecular orbitals made from unsaturated chains in the molecules. Many elements are coloured because they, too, have some electronic transmission that matches the energy in some part of visible light. The mechanism for colour is the same in all these cases.

There are other ways to generate colour but they are less common. Some colours come from interference not selective absorption of light: some physical structures that contain features close to the wavelength of light  can reflect light but in a way that causes selective interference that cuts out some colours. This is why oil films on water produce rainbow colours but is also the source of unusual colours in some fruits, butterflies and birds.

Are things coloured if the light isn't present? That is a philosophical question like "does a tree falling the the forest make a sound if nobody hears it". From a chemical point of view the feature that creates colour is still present whether or not there is light. But in the absence of light nobody is going to notice it.


Would singlet oxygen in the engine cause a more efficient combustion?

This hypermileage scheme is one of the more interesting ones I've come across, so I think it's worth some discussion. I don't know much at all about the engineering aspects of internal combustion engines, so I'll stick to a theoretical analysis.

Let's start by describing the difference between singlet oxygen and triplet oxygen. Both of these are the same substance, molecular (di)oxygen, or $\ce{O2}$. However, they are in different energetic states. In its lowest energy state, a molecule of $\ce{O2}$ contains two unpaired electrons (which is a very interesting and important fact in itself), and this form is called triplet oxygen. This is the regular stuff you breathe all the time.

However, if some energy is given to triplet oxygen, it can be excited into a different state, and the two formerly unpaired electrons now pair up. Since there are now no unpaired electrons, we call this singlet oxygen (technically there is more than one singlet state possible, but for our purposes that can be ignored).

So in summary, singlet oxygen and triplet oxygen are the same molecule, except that singlet oxygen has inherently more energy. So why does this matter? If singlet oxygen has more energy, then one expects it to be more reactive. It is entirely conceivable that a fire supplied with singlet oxygen will burn hotter/more completely than the same fire supplied with triplet oxygen. So that part of the scheme is plausible.

That said, you may ask yourself, if all it takes is some energy to turn triplet oxygen into singlet oxygen, then shouldn't oxygen in any fire be continuously converted from the triplet state to the singlet state, simply by the heat of the roaring flame? This would mean preparing singlet oxygen separately is entirely superfluous! As it turns out, due to the quirks of quantum mechanics it is not easy to excite triplet oxygen into singlet oxygen. Technically speaking, this is called a spin-forbidden process.

However, due to special relativity (!), spin-forbidden processes turn out to not be completely forbidden, just very unlikely. A concept called spin-orbit coupling comes into play. The more spin-orbit coupling in a chemical system, the more likely it is for spin-forbidden process to nevertheless happen. Spin-orbit coupling becomes more and more relevant as you go further down the periodic table. This means heavy atoms such as platinum can favour the transformation of triplet oxygen into singlet oxygen. So again, the necessity of a metal honeycomb as pictured in the scheme is at least conceptually plausible for this singlet oxygen strategy to work.

(Of course, it is possible to produce singlet oxygen through other means, such as the decomposition of some oxygen-containing molecules like hydrogen peroxide and ozone, but doing so would defeat the whole purpose; you'd essentially have to buy a second kind of fuel!)

At this point it seems there is actually something going for this scheme, but here is where so many others like it stumble: the laws of thermodynamics. If it takes energy to transform triplet oxygen into singlet oxygen, then that energy has to come from somewhere. If the energy comes from the metal honeycomb itself (or more precisely, the energy imbued into the metal honeycomb during its manufacturing process), then yes, it is plausible to get increased mileage, but the mesh will lose energy until it is inactive and has to be replaced or reactivated. My bet is that this would cost much more than the meagre fuel savings.

In principle, the energy to make singlet oxygen can come from the extra energy from the more efficient fuel burn, leaving the mesh intact. However, now you're dipping into the very extra energy that was supposed to make the car get more mileage. The detailed energy balance is generally very hard to calculate, and I couldn't begin to tackle all of the real-world variables in play. That said, the answer to whether a hypermileage scheme is cost-effective almost always ranges from a flat-out "no" to "yes in principle, but no in practice". Processes that are cost-effective (precise spark timings, regulation of air-fuel ratio, etc) tend to make it into standard issue vehicles in the first place. I can't help but be reminded of the joke that alternative medicine which works is just called medicine.

Lastly, regardless of energy savings, singlet oxygen may not be great inside the engine. Not only is it more chemically reactive than triplet oxygen, but it can perform reactions which are completely different from triplet oxygen. It could cause more wear of the engine components, or could generate more of some unwanted sideproducts during combustion, such as nitrogen and sulfur oxides. Any singlet oxygen that manages to escape combustion and exit the tailpipe could also pose a significant biological and chemical hazard.


The simplest way to detect where a spin in a sample is is to place this in a magnetic field that changes in direct proportion to depth. Then the frequency at which spins respond to the r.f. radiation can be interpreted as a depth. Clearly it is going to be far more complex to get 3D images but in essence is the same; the detected signal has to be encoded with a position.


Can a organic compounds such as hydrocarbons contain an ionic bond?

Azulene has a 1.05 debye dipole moment,

A hydrocarbon as such bonded as an isolated cation-anion pair with no heteroatoms could happen in principle given sufficient steric hindrance.   Penta-tert-butylcyclopentadienide or pentakis(2,4,6-tri-tert-butylphenyl)cyclopentadienide with like fully substituted cyclpropenium or tropylium might work.

Similar substitution on fulvalenes' four inner carbons would dihedral angle rotate the rings by steric hindrance.  This prevents the double bond from forming, giving sort of a sigma covalent plus pi-dative bond.


Why there is no change in water level when salt is added?

This is an example of electrostriction - see, for example, this paper for a full explanation.

In short, the solvent molecules become more ordered in the vicinity of dissolved (charged) ions and less ordered as we go further away from the dissolved (charged) ions. The increase of solvent density near the ions is offset by the decrease of bulk density in the solvent as distance from the ions increases. This results in an overall net reduction in the observed volume of the solution.

From the Discussion section of the manuscript:

  The interaction of the electrostatic field of an
  ion with water tends to align the dipolar water
  molecules in the direction of the field. In this way
  the field tends to disrupt hydrogen bonded
  structures in liquid water, and to compress the
  water molecules surrounding an ion. These
  electrostatic effect give rise to a shrinkage of the
  water.


Its a qualitative answer, so you'll get a lot of odd things in these responses.

I find methanol to be like ethanol but shimmed by a plastic/wood-life taste.

More importantly, having a higher volatility, its taste will be more intense because there are more vapors to stimulate your odor receptors which frequently change or augment tastes.


What is the difference between ''1 C atom and 1 mole of C atoms?''

What is the difference between one egg and one dozen eggs?  A dozen is simply a certain count, in this case 12, that is rendered as one unit for our convenience.  Avogadro's number s just the same thing as a dozen, except we made the unit count larger than 12 in order to match it up with our conventionally used units.

Your assumptions about a "mole" are thus correct.


Force Constants for Angles From Gaussian To AMBER: which value should I put in the matrix equation?

I really don't know exact answer but I have an idea for you it may help: You can use some "force field developer toolkits". One of them is: VFFDT. It parametrizes (for amber forcefield *.frmcd files) bond and angle force constant from gaussian out/log file.


What compounds do Nor1 and Nor2 citalopram refer to?

Those names are referring to desmethylcitalopram and didesmethylcitalopram, which, as you correctly predicted, are the N-demethylated and N-didemethylated metabolites of citalopram. Per IUPAC conventions, the nor- prefix is used to refer to any structural derivative of some parent compound in which a single carbon and (possibly) its hydrogens have been replaced (typically by a hydrogen, but not exclusively). See the relevant Wikipedia page on the prefix for more detail.


What is the meaning of parentheses in a condensed molecular formula?

The parentheses just have a grouping function: $\ce{(CH3)3-C -}$ represents a carbon atom to which 3 methyl groups are attached, $\ce{-CH(CH3)}-$ is a carbon atom in the chain to which one hydrogen atom and a methyl group are attached.

Draw the molecule, find the longest chain for the base name, determine the position of the substituents and you're done.


How to remove Ethanethiol from household natural gas?

Just an idea: You can pass the gas through a bleach solution that can oxidize the thiol. I have no idea how effective this can be since the human smell is extremely sensitive to ethanethiol (to the scale of parts per billion). One possible set-up (very common in organic labs) is shown in this picture. The end Buchner flask should contain bleach and not HCl. The other conical is a safety measure in case you have suck back of the bleach solution.


Why do SN1 and SN2 reactions not occur at sp2 centres?

Sometimes, especially in introductory courses the instructor will try to keep things "focused" in order to promote learning.  Still, it's unfortunate that the instructor couldn't respond in a more positive and stimulating way to your question.

These reactions do occur at $\ce{sp^2}$ hybridized carbon atoms, they are often just energetically more costly, and therefore somewhat less common. Consider when a nucleophile reacts with a carbonyl compound, the nucleophile attacks the carbonyl carbon atom in an $\ce{S_{N}2}$ manner.  The electrons in the C-O $\pi$–bond can be considered as the leaving group and a tetrahedral intermediate is formed with a negative charge on oxygen. It is harder to do this with a carbon-carbon double bond (energetically more costly) because you would wind up with a negative charge on carbon (instead of oxygen), which is energetically less desirable (because of the relative electronegativities of carbon and oxygen).

If you look at the Michael addition reaction, the 1,4-addition of a nucleophile to the carbon-carbon double bond in an $\ce{\alpha-\beta}$ unsaturated carbonyl system, this could be viewed as an $\ce{S_{N}2}$ attack on a carbon-carbon double bond, but again, it is favored (lower in energy) because you create an intermediate with a negative charge on oxygen.

$\ce{S_{N}1}$ reactions at $\ce{sp^2}$ carbon are well documented.  Solvolysis of vinyl halides in very acidic media is an example.  The resultant vinylic carbocations are actually stable enough to be observed using nmr spectroscopy.  The picture below helps explain why this reaction is so much more difficult (energetically more costly) than the more common solvolysis of an alkyl halide.  In the solvolysis of the alkyl halide we produce a traditional carbocation with an empty p orbital.  In the solvolysis of the vinyl halide we produce a carbocation with the positive charge residing in an $\ce{sp^2}$ orbital. Placing positive charge in an $\ce{sp^2}$ orbital is a higher energy situation compared to placing it in a p orbital (electrons prefer to be in orbitals with higher s density, it stabilizes them because the more s character in an orbital the lower its energy; conversely, in the absence of electrons, an orbital prefers to have high p character and mix the remaining s character into other bonding orbitals that do contain electrons in order to lower their energy).


How are lowercase k and uppercase K used in general chemistry courses?

In addition to the original answer I also allowed myself to invite lowercase Greek letter kappa κ to the table as typographically it has a similar-looking appearance and add some more definitions:

Lowercase k

$k$ is one of the Miller indices.
$k$ is a symbol for thermal conductivity.

Uppercase K

$\ce{K}$ is a symbol for potassium.
$\pu{K}$ is a symbol for Kaiser, unit of energy $(\pu{1 K} = \pu{1 cm-1})$.
K is a symbol for aminoacid lysine.
K is used in spectrometry to denote spectral lines.
$K$ is sometimes used as a symbol for decadic absorption coefficient.

Lowercase Greek letter kappa κ

κ is used to denote denticity.
$κ$ is a symbol for [specific] conductivity.
$κ$ is a symbol for molar Napierian absorption coefficient.
$κ$ is a symbol for asymmetry parameter.
$κ$  is sometimes used as a symbol for bulk viscosity.
$κ$ is sometimes used as a symbol for magnetic susceptibility.

Uppercase Greek letter kappa Κ

Κ is a symbol for a kappa number characterizing bleachability.

Both lowercase k and lowercase Greek letter kappa κ

$k$ and $κ$ is a symbol for imaginary refractive index.

Both lowercase k and uppercase K

$k$ and $K$ is in general may refer to a coefficient (German "Koeffizient"); indices are often used to denote which one, e.g. mass transfer coefficient $k_\mathrm{d}$, octanol/water partition coefficient $K_\mathrm{ow}$.


Can you electrolyse a single H2O molecule?

In a traditional experimental setup this would never happen, as there are always several water molecules. But lets just think about it for the sake of it.

If you wanted to design an experiment where you only have one water molecule, you would have to get two electrodes extremely close together. Lets say you find a way to levitate the molecule between the two electrodes. You might be able to then apply a voltage and tear the molecule apart, forming one H$_2$(g) molecule at the cathode. The oxygen radical would then most likely immediately react with the anode to which it is attracted by the Lorentz force and form an oxide.

So the result of electrolysing several single molecules of water might result in gaseous hydrogen and no oxygen, but a buildup of oxide at the anode.

However, having two electrodes so close together would probably create a series of problems. Not only to hold the single water molecule in place (which could be achieved by optical levitation) but because quantum mechanical effects such as tunneling currents would occur. Probably there are several more complications that would by all means make it nearly impossible to actually observe this. The knowledge gain derived from such an experiment would likely not justify the effort and expenses.

EDIT

I just revisited this question and thought about it again:
it's not absolutely necessary that the two H atoms would form an H$_2$ molecule. Depending on the way the cathodes are formed, the electric field might not pull the charged particles together, but to a collision with something else in the vicinity; the cathode or any ambient gas.
Furthermore, my notion above that "H$_2$(g)" would form is a bit nonsensical, as a single molecule cannot really be attributed a state of matter (i.e., being gaseous in this case).


Carbon monoxide formation in broken boilers

The combustion of hydrocarbons works as follows:

$\ce{hydrocarbon + O2-> CO + H2O}$

The reaction can then continue to the complete combustion:

$\ce{CO + 1/2 O2-> CO2}$

An example of balanced equation is the combustion of methane:

$\ce{CH4 +  3/2O2 -> CO + 2H2O}$
$\ce{CO +  1/2O2 -> CO2}$

So, carbon monoxide is a product of the incomplete combustion of a hydrocarbon.

It is produced when there is too much fuel (eg: methane), but not enough oxidant (eg: oxygen from air).

I guess that a boiler which cannot mix properly air and fuel produces carbon monoxide.


Autodock Vina flexible metals and water molecules

For the water molecules, you can try hydrated ligand docking in which they are added in the ligand file.
The documentation here: http://autodock.scripps.edu/resources/autodock-hydrated-docking.
I am not sure if the same approach can be used for the metal atom.


Aminoxyl (trivial name: a nitroxyl radical or a nitroxide) denotes a radical functional group with general structure $\ce{R2N–O^•}$. It is well known that sterically unhindered aminoxyls baring $\alpha$‐hydrogens are unstable and undergo rapid disproportionation to nitrones and hydroxylamines (Ref.1: A new mechanism for the decomposition of $\alpha$‐hydrogen nitroxides is proposed). For example, decomposition of one of such compound is displayed below (The scheme is from Ref.1):

Sterically hindered aminoxyls without $\alpha$‐hydrogens such as TEMPO are stable radicals. However, it is not clear that steric hindrance at carbon $\alpha$ to nitrogen atom is important in the mechanism. Yet, the lone pair of electrons on nitrogen has played a significant role in the mechanism (Ref.1). Therefore, I speculate that 2,2'-dimethyldiphenylamine, when oxidized to relevant aminoxyl radical, may not be as effective as TEMPO, mainly due to the delocalization of the lone pair of electrons on nitrogen between two 2-methylphenyl groups.
Nonetheless, there are few available aminoxyl radicals, which are more efficient than TEMPO (Both following images are from Ref.2):

The research in Ref.2 has performed catalytic oxidation using few N-oxyl radicals, namely 2,2,6,6-tetramethylpiperidine-1-oxyl (TEMPO), 2-azaadamantane N-oxyl (AZADO), 1-methyl-AZADO (1-Me-AZADO), 9-azanoradamantane N-oxyl (Nor-AZADO), and 1,5-dimethyl-9-azanoradamantane N-oxyl (DMN-AZADO), in water at $\mathrm{pH} \  10$ on wood cellulose to determine the influence of different N-oxyl radical chemical structures on the reaction kinetics and structures of oxidized celluloses. The reaction time required for complete oxidation of wood cellulose using AZADO, 1-Me-AZADO, or Nor-AZADO was less than $\pu{20 min}$, while DMN-AZADO and TEMPO required $\pu{70 min}$ and $\pu{120 min}$, respectively.
There is also some comparative work has been done in Ref.3, which also showed the superiority of other N-oxyl radicals over TEMPO (Synthetic procedure to prepare Nor-AZADO has been given in the reference; see attached PDF).
References:

Aaron Nilsen,  Rebecca Braslau, "Nitroxide decomposition: Implications toward nitroxide design for applications in living free‐radical polymerization," J. Polymer Sci., Part A: Polymer Chem. 2006, 44(2), 697-717 (DOI:https://doi.org/10.1002/pola.21207).
Hiromasa Hondo, Tsuguyuki Saito, Akira Isogai, "Preparation of oxidized celluloses in a $\ce{NaBr/NaClO}$ system using 2-azaadamantane N-oxyl (AZADO) derivatives in water at $\mathrm{pH} \  10$," Cellulose 2019, 26, 1479–1487 (DOI:10.1007/s10570-018-2177-5).
Masaki Hayashi, Yusuke Sasano, Shota Nagasawa, Masatoshi Shibuya, Yoshiharu Iwabuchi, "9-Azanoradamantane N-oxyl (Nor-AZADO): a highly active organocatalyst for alcohol oxidation," Chem. Pharm. Bull. 2011, 59(12), 1570–1573 (DOI: https://doi.org/10.1248/cpb.59.1570)(PDF).


TL;DR; Addition in a chemical reaction does not mean mathematical addition, and translating this operation to a rate expression makes no physical sense.

Elaboration:
Your idea $\ce{C = A + B}$ changes the order of reaction, and has no
physical basis.

The product of two concentrations, $\ce{c_{a}}$ and $\ce{c_{b}}$ in the equation:
$${\partial c_\ce{AB} \over \partial t} = -k_\mathrm{off} c_\ce{AB} + k_\mathrm{on} c_\ce{A} c_\ce{B}$$
has the physical interpretation that $\ce{A}$ and $\ce{B}$ must collide for reaction to occur, and is hence a second order reaction.

Your proposed solution would change that to first order in a fictional component, $\ce{C}$, and leads you astray.

To help clarify, consider the following reaction:
$$\ce{Br^{\cdot} + Br^{\cdot} <=> Br2}$$

In this case A would be $\ce{Br^{\cdot}}$ and B would be $\ce{Br^{\cdot}}$, and thus your C would be $\ce{[Br^{\cdot}] + [Br^{\cdot}] = 2 [Br^{\cdot}]}$.  The rate expressions would therefore be first order in $\ce{Br^{\cdot}}$  as opposed to second order.

Other comments:

You also forgot to include the other reactions for the bimolecular case:
$${\partial c_\ce{A} \over \partial t} = {\partial c_\ce{B} \over \partial t} = k_\mathrm{off} c_\ce{AB} - k_\mathrm{on} c_\ce{A} c_\ce{B}$$Omitting these may be leading to some confusion.
Little k is a rate constant.  Equilibria are described with big K.  It is unclear what $\ce{k_{d}}$ means in your question.


Why do carbonates, oxides, and pure metals precipitate before metal sulfides?

Why Do Carbonates, Oxides, and Pure Metals Precipitate Before Metal Sulfides?

They don't. I cannot get to the particular paper you read, but I can provide both a logical and a quantitative argument that the metal sulfides precipitate first. The thermodynamics of this situation do not involve reduction potentials so much as simple solubilities. Metal sulfides are much less soluble in water than metal carbonates.

From your description, I imagine a system like the following image. There are two strata of sedimentation: an upper layer of metal oxides and a lower layer of metal sulfides.

There are three reasonable explanations why the sulfides are on the bottom.

The sulfides did not precipitate from the water; they were always there. The carbonates and oxides deposited on top of them.
The carbonates and oxides were deposited first, and the sulfides were deposited second, settling through the carbonates and oxides to form a deeper layer.
The sulfides were deposited first, and the carbonates and oxides were deposited second.

Explanation 1 is very reasonable, but counterproductive. We will ignore it.

Explanation 2 is the one supposed by the OP in the question, and perhaps it was also provided in the paper. Let us examine this idea a little. In order for this to happen, the metal sulfides would have to be denser than the metal carbonates and oxides to settle through. All it takes the counterexample of iron to see that this is not always the case. Iron (II) sulfide is denser than iron (II) carbonate but less dense than iron (II) oxide.
 - Iron (II) oxide, $\ce{FeO}$, 5.745 g/mL
 - Iron (II) carbonate, $\ce{FeCO3}$, 3.961 g/mL
 - Iron (II) sulfide, $\ce{FeS}$, 4.84 g/mL

Additionally, if the sulfides were settling through, the layers would not be so well stratified - the boundary between them would be indistinct.

  Metal sulfides are much less soluble in water than metal carbonates.

That leaves with Explanation 3, which is the correct one. Metal sulfides are almost universally less soluble that metal carbonates. We'll get to where the oxides come from later. Both metal carbonates and metal sulfides are nearly insoluble in water, so the minute amounts present are hard to quantify in the "grams of solute per liter of solution" kind of way. The saturated solution may only have few parts per billion of the compound. Instead, we talk in terms of the solubility equilibrium, in particular, the solubility product (constant), $k_{sp}$, the equilibrium constant associated with the combination dissolution and dissociation.

We can determine the concentration from the $k_{sp}$. Consider barium carbonate $k_{sp}=5.1\times10^{-9} \text{ M}^3$. The solubility equilibrium is:

$$\ce{BaCO3(s) <=>Ba^{2+}(aq) + CO3^{2-}(aq)}$$

And the equilibrium constant is (the concentration of solid barium hydroxide is constant and is thus part of $k_{sp}=k_{obs}$.
$$k_{sp} = [\ce{Ba^{2+}}][\ce{CO3^{2-}}]=5.1 \times 10^{-9} \text{ M}^3$$

Since $[\ce{CO3^{2-}}]=[\ce{Ba^{2+}}]$ (mass balance):
$$[\ce{Ba^{2+}}][\ce{CO3^{2-}}]=[\ce{Ba^{2+}}]^2$$
$$[\ce{Ba^{2+}}]^2=5.1\times 10^{-9} \text{ M}^3$$
$$[\ce{Ba^{2+}}]= 7.14\times 10^{-5} \text{ M}=>0.0141 \text{ grams per liter}$$

Now, let's look at carbonates and sulfides. All $k_{sp}$ came from here.

Cadmium:

Carbonate $\ce{CdCO3}$, $k_{sp}=5.2\times 10^{-12} \text{ M}^2$, $[\ce{Cd^{2+}}]=2.28\times 10^{-6} \text{ M}$
Sulfide $\ce{CdS}$, $k_{sp}=8\times 10^{-27} \text{ M}^2$, $[\ce{Cd^{2+}}]=8.94\times 10^{-14} \text{ M}$

Cadmium sulfide is less soluble than cadmium carbonate by a factor $10^8$ (100 million). The same behavior exists for cobalt, copper, iron, lead, manganese, nickel, silver, and zinc (all metals for which data for both the carbonate and sulfide are listed).

  Why might this paper suppose that the sulfides are deposited second when the sulfides are much less soluble than the carbonates?

The authors of the paper probably did not know that (and perhaps few others did either). That paper was published in 1901. Svante Arrhenius presented his thesis on the behavior of salts in water in 1884. It was not well thought of initially. He received a Nobel Prize for the work in 1903 (after the paper was published), but there is little reason to believe the authors (mining engineers) would have been taught controversial and cutting edge physical chemistry in whatever minimal introductory chemistry courses they had taken (if any). I may mention recent Nobel work in chemistry in my intro courses, but I certainly don't teach recent Nobel work in chemistry in into courses.

  Where do the metal oxides come from?

Metal oxides form either by decomposition of the carbonates or by oxidation of pure metals (which probably occur from the decomposition of the sulfides):

Decomposition of sulfides:

$$\ce{FeS(s) <=> Fe(s) + S(S)}$$

Oxidation of metal:

$$\ce{6Fe(s) + 3O2(aq) -> 3Fe2O3(s)}$$
$$\ce{2Fe(s) + 3H2O(l) -> Fe2O3(s) + 3H2} \text{ (acid is helpful here)}$$

Decomposition of the carbonate. Since the carbon dioxide would be carried away by the water, this equilibrium is continually pushed toward the oxide.
$$\ce{FeCO3(s) <=> FeO(s) + CO2(aq)}$$


What are the signs for work done by (on?) a gas during isothermal reversible compression?

The easiest way to remember the sign conventions is to think of the gas particles as "wanting" to get away from each other. To reduce the volume of the gas, you have to do work on the gas - the surroundings are doing the work. In chemistry we always take the point of view of the system. If we do work on the system, we add energy to it, and therefore, the sign of work is positive.

This means the opposite is true as well. When the volume of the system increases, the sign of the work term must be negative - the system is losing energy and doing work on the surroundings.

For your equations, all of this holds for the first one:

$$-nRT\ln  (Vf/Vi)$$

This is because a decrease in volume will give you a fraction in the $\rm{ln}$ term, which leads to a negative sign. The leading negative sign reverses that. The result is that any compression will have a positive work term, and any expansion will have a negative work term.

You can see how the correct sign is a result of the derivation in the wikipedia article.


Why would rinsing the flask with acid in this experiment affect its accuracy?

Probably because the use of hydrochloric acid (which is a reactant involved in this reaction!) could result in some residual aqueous HCl in the flask, which may add to the HCl added later.
Anything you use to clean the flask should be inert/not involved in that reaction. If you really want to use HCl, then you should later wash the flask with a lot of water, but that's not said in the answer.

Also note that the volumes involved here are not so large (only 5 ml of HCl are used), so a "small" residue might have a large effect on the result.

As for the stirring, the colour of the reaction is caused by the sulfur precipitate. But I think that, if you stir in each experiment, the times might probably decrease, but you would still observe a trend with temperatures.

That answer would be wrong if it involved stirring in some experiments and not in others.


Water-stable organoaluminum compounds in which aluminum is part of the aromatic ring

You're probably not going to get what you want putting aluminum into an aromatic ring.  Aluminum and carbon have relatively poor pi overlap, so your ring will not have the full delocalization necessary for string aromatic stabilization.  And even strong aromatic stabilization may not be all that strong compared with aluminum preferring to bond with oxygen or nitrogen rather than carbon.

Instead of trying to design a compound that somehow tames the reactivity of aluminum-carbon bonds, choose the right polar solvent.  Protic solvents are bad.  According to Wikipedia, organoaluminum compounds "readily form adducts with bases such as pyridine, THF and tertiary amines. These adducts are tetrahedral at Al."  The named examples are aprotic and coordinate through nitrogen or oxygen.


Let me opt for a paradoxical answer. They are not different phases, but you can't call them one phase either. There is no "them".

That is, you can't have a system in equilibrium which contains two different areas with the said compositions, either separated by a phase border or not. If you try, they will mix and form a single phase with averaged composition. It will be uniform all right.

Sure, you may have such compositions in different systems, and indeed many would call them "the same phase" then, based on the very close properties, similar structure, and the fact that they will mix if you let them. But this approach may quickly lead to nonsense. Say, water and acetone will mix in any ratio; are they one phase? Suppose yes. What about acetone and hexane? The same answer applies for the same reason. Now what about water and hexane? Er, well...

So let's not compare phases from different systems in terms of "sameness". Keep your apples and oranges apart.


What chemical compounds were used in CRT phosphors?

You should ask yourself: "For which application?". In fact there are different types of "phosphors" (most of them have no "Phosphorus" inside) for different applications and different compound were used during the evolution of CRT monitor. Some kind of phosphors may have advantages in terms of:

Efficiency of the luminescence induced.
Cheaper price.
Longer life.
Better color.
Shorter decay time.

Note however that in many application longer decay times are prefer because allow to the user to see transient signal one of this application is radar in this case is used the GB phosphors (it usually used the WTDS symbol to refer to a specific compound. In Phosphor Handbook
 by Shigeo Shionoya,William M. Yen,Hajime Yamamoto p. 607 you can find the relative compound in this case is  $(Zn,Cd)S:Cu + (Ca,Mg)SiO_3:Ti)$) that has a time decay between 100 ms and 1 s.

If you are  interested in RGB monitor and you want a list of the phosphors manly used.
I've copied the the following table from Practical Applications of Phosphors by William M. Yen,Shigeo Shionoya,Hajime Yamamoto p. 232 showing how phosphors compound for every RGB channel changed during time.

\begin{array}{c c c  c}\hline
&Blue  & Green  & Red \\ \hline
1               & ZnS:Ag^{2+}   &  Zn_2SiO_4:Mn^{2+}   & Zn_3(PO_4)_2:Mn^{2+}\\
2 & ZnS:Ag^{2+} &  (Zn,Cd)S:Ag    &  (Zn,Cd)S:Ag  \\
3& ZnS:Ag^{2+} & (Zn,Cd)S:Ag & YVO_4:Eu^{3+}\\
4 & ZnS:Ag^{2+} &   (Zn,Cd)S:Cu,Al & Y_2O_2S:Eu^{3+}or\space Y_2O_3:Eu^{3+} \\
Current & ZnS:Ag^{2+}       & ZnS:Au,Cu,Al  \space or \space ZnS:CU,Al    &Y_2O_2S:Eu^{3+}    \\ \hline
 \hline
\end{array}


How many resonance structures does the oxalate ion have?

Resonance structures are not real things. As also indicated by rbw, they have weight on the actual compound. Simpler example is ozone.

http://www.wou.edu/las/physci/poston/ch222/pdf/Ch08-s12-part2.pdf

From the first slide second image there;

you can see that the actual molecule is none of the "resonance" structures, and both bonds between leftmost and rightmost oxygen atoms with central oxygen atoms have equal lenght, and it is between 1 bond length and 2 bond length.

The general idea of drawing resonance structures is, after predicting the most appropriate ones, when you superimpose them and average the bond lengths you will be very near to see the real chemical bonds lengths.

This is straightforward with ozone, oxalate or carbonate etc. but things get complex with increasing atoms, though.

And wrapping up and also answering the question you posed, you should look atoms as if they are all unique, like oxygen 1 oxygen 2 oxygen 3 and oxygen 4. Flip the molecule and you will NOT get the other one, this is not like isomerism. It is just to predict the electron cloud density around atoms of a molecule.

Hope this also helps.


What does "a metallic oxide has 40 % oxygen" mean?

Your question is a good one, and no, you can't assume that the metal is $\ce{MO}$ necessarily.  However, you can find the equivalent weight of the metal without any assumptions about its formula.

The equivalent weight of a metal combining with oxygen is the mass that reacts with 8g of oxygen.  We know that the compound contains 40% oxygen, so it must contain 60% metal.  Suppose we had 100g of this compound; that would be 40g O and 60g M.

$40g\ \ce{O}:60g\ \ce{M}$ (Divide by 5)
$8 g\ \ce{O}:12g\ \ce{M}$

The equivalent weight of the metal must be 12g.

If you had been asked to identify the metal you would have needed the formula (or would have needed to test the likely formulas to see if they produced reasonable results.)

Given an equivalent weight for M of 12g:
$\ce{M2O-> 12g/mol, carbon}$ (not a metal)
$\ce{MO-> 24g/mol, magnesium}$
$\ce{M2O3-> 36g/mol, chlorine}$ (not a metal)
$\ce{MO2->48g/mol, titanium}$

Both magnesium and titanium fit the information that was given, $\ce{MgO}$ and $\ce{TiO2}$ are common oxides of those metals.


Solve for wavelength of photon given threshold frequency

Always work in energy not wavelength. You are given $\nu_0$ in Hz ($\mathrm{s}^{-1}$), multiply this by $h$ to convert it to Joules; ($h$ has units J s). You have the kinetic energy in eV, use the conversion  $1\mathrm{eV} = 1.602\cdot 10^{-19}$ J to get into joules. Now you have only to find energy $h\nu$ by subtraction. Finally convert energy to frequency using $E=h\nu$ and then to wavelength using $c=\lambda \nu$. Make sure $c$ is in m/s and as you answer will be in meters then convert to nm from metres.


Does solubility matter when measuring density with displacement method?

Volume is not a conserved extensive property of mixtures in the same way that other quantities like mass, net electric charge, number of particles, etc. are (within closed systems). That is to say, in reality, the volumes of components physically mixed together are not simply additive, with the total volume not necessarily equal to the sum of the individual volumes. I.e., $\sum{V_i} \not= V_{mix}$ generally. This occurs chiefly because intermolecular forces of attraction between particles will vary with the composition of the mixture. Mixtures of ethanol and water, for example, are always reduced in volume by comparison to the summed volumes of ethanol and water, respectively, in isolation.

In fact, the total volume can be calculated given the partial molar volumes of the individual components for a specific composition under given conditions. Specifically, $V_{mix} = \sum{n_i \bar{V_i}}$, where $\bar{V_i}$ and $n_i$ are the partial molar volumes and moles, respectively, of the separate constituents of the mixture.

A few good references I found for additional study: 1, 2. Any decent physical chemistry textbook should also cover partial molar properties.


Comparing relative negative inductive effect of substituents (dialkyl amine, amine, alkoxide, fluoride)

The inductive effect is the polarization of σ (sigma) bonds within a chemical species caused usually by differences in electronegativity between the neighboring atoms. The higher the polarization it causes, the higher the inductive effect. In this case, the decreasing order of electronegativity is:

$\ce{-F > -OR > -NH2}$

So, for practical purposes, the correct answer should be option 2.

However, substituents far from the atom you are considering can also affect the inductive effect and an amino ($\ce{-NR2}$) group can be more withdrawing than an alkoxide ($\ce{-OR}$) group.

For example: $\ce{-N(-CF3)2 > -OCH3}$

This amino substituent is probably more electron-withdrawing than the alkoxide in the right, because the first is bound to two highly withdrawing groups ($\ce{-CF3}$) while the second is bound to an electron-donating group ($\ce{-CH3}$)


Which has more metallic character: aluminium or magnesium?

Two types of metallic character
In fact, there are two type of metallic character if you look at the metal from the chemical point of view or if you look at the metal from the physical point of view:

So it really depends on how do you define metallic character

Chemical metallic character
Since metallic character in chemistry is defined as:

the tendency of an atom to lose electrons.

We can understand that it is directly related to electronegativity.

chemical property that describes the tendency of an atom to attract
electrons (or electron density) towards itself

Basically, it is the opposite (that would be Electropositivity...)! An so it increases from right to left. The picture below shows the electronegativity but can be used for the purpose.

I think the reasoning that the book wants you to do is not remembering all the subshell but just the basic chemistry principles.
So from the chemical point of view $\ce{Mg}$ has a greater metallic character compared to $\ce{Al}$.
Physical metallic character
The physical metallic character is defined taking in account the physical properties of the metal. A metal, from the physical point of view, has a high electrical conductivity, high thermal conductivity, and high density.
These properties are harder to predict:

In this case, however, $\ce{Al}$ has a more metallic character compared to $\ce{Mg}$.


I had some difficulties figuring out what the reaction was

The question mentions "oxidation of glucose", that's a very important hint that tells you that the reaction is going to be:

$$\ce{C6H12O6 + O2 -> ?}$$

Now, note that oxidation of any carbohydrate/hydrocarbon/organic compound only yields (in general) carbon dioxide and water.

Thus, your complete reaction should be:

$$\ce{C6H12O6 + 6O2 -> 6 CO2 + 6 H2O}$$

Notice that each of the products is just the "oxide" of the elements of the original compound $\ce{(C, H)}$


Why would hexacyanidocobaltate(II) be uncommon while pentacyanidoaquacobaltate(II) is common?

In Inorganic Syntheses (1946), a synthesis of $\ce{K3[Co(CN)6]}$ is described, which boils down to the following three equations:[1]

  $$\begin{align}\ce{CoCl2 (aq) + 2 KCN (aq) &-> Co(CN)2 v + 2 KCl (aq)}\tag{1}\\
\ce{Co(CN)2 + 4 KCN (aq) &-> K4[Co(CN)6] (aq)}\tag{2}\\
\ce{2 K4[Co(CN)6] (aq) + H2O &-> 2 K3[Co(CN)6] (aq) + 2 KOH (aq) + H2}\tag{3}\end{align}$$

The procedure requires precipitation of the cobalt(II) cyanide by carefully adding a $\ce{KCN}$ solution to a cobalt(II) solution. This precipitate is filtrated and washed, then redissolved in excess cyanide solution giving a deep-red solution. This in turn is heated to reflux for some 15 minutes, whereupon the colour changes to yellow indicating the formation of the hexacyanidocobaltate(III).[1]

I personally doubt equation $(3)$. Instead, I would assume that the oxidation of hexacyanidocobaltate(II) to the corresponding cobaltate(III) proceeds unter oxygen consumption, not hydrogen liberation. That would give equation $(\text{3'})$:

$$\ce{4 K4[Co(CN)6] + O2 + 2 H2O -> 4 K3[Co(CN)6] + 4 KOH}\tag{3'}$$

In any case, however, the presence of $\ce{[Co(CN)6]^4-}$ or hexacyanidocobaltate(II) as a stable intermediate in solution suggests to me that your premise is false and thus the book you cite from incorrect.

Reference:

J. H. Bigelow and J. C. Bailar (1946) Potassium Hexacyanocobaltate(III), in: Inorganic Syntheses, Volume 2 (ed W. C. Fernelius), John Wiley & Sons, Inc., Hoboken, NJ, USA. DOI: 10.1002/9780470132333.ch72.


Do all sigma bonds are stronger than pi bonds?

Every possible sigma bond is not stronger than every possible pi bond.

From this table at Wikipedia, we can see that the bond dissociation energy for the F-F bond in $\ce{F2}$ is 157 kj/mol. Meanwhile, the bond dissociation energy for the C-C bond in ethane $\ce{H3C-CH3}$ is on the order of 360 kJ/mol. The bond dissociation energy of the C=C bond in ethene $\ce{H2C=CH2}$ is approximately 710 kJ/mol. The note for this compounds suggests that the pi bond is around 270 kJ/mol (65 kcal/mol), leaving the sigma bond at 440 kJ/mol. Thus, the carbon-carbon pi bond is stronger than the fluorine-fluorine sigma bond.

A general statement of "sigma bonds are stronger than pi bonds" must thus be interpreted in the "all else equal" mode. The relative strength of the carbon-carbon pi bond is less than the relative strength of the carbon-carbon sigma bond. However, just as the strength of the carbon-carbon sigma bond has no bearing on the strength of the fluorine-fluorine sigma bond, the comparison of pi bonds to sigma bonds involving different elements is not helpful. The relative strength of the bonds depends as much or more on the atoms in the bond as it does the type of orbital overlap used to form the bond.


Is there any difference between Meprobamate and Equanil?

Pharmaceuticals with meprobamate as the active compound are marketed by different companies under different names.

Meprobamate is the International Nonproprietary Name (INN) for the drug.

Whether two pharmeceuticals with the same active components acutally have the same kinetics does depend on additional factors, such as the type of the filler.


Why formulate drugs as HCl salts when HCl is in stomach acid?

Another reason is that many drugs which are conjugated to hydrochloride salts is they are amines, and amines by themselves often stink. So people might be turned off from taking drugs that smell terrible (fishy).

Now, note that even though a label might say something such as "fexofenadine HCl" that doesn't mean there is HCl in the drug. Fexofenadine is an amine, and amines are somewhat basic - and more than basic enough for a strong acid such as HCl to completely protonate.

So if this is what you had in mind for the structure of fexofenadine, it isn't the most correct structure:

This would be a much better structure. The nitrogen in the ring has been protonated and it now bears a positive formal charge. Chloride ion is the counter ion.


One vague explanation I came up with is the following.

A prismatic arrangement of atoms (first figure) does not capture fully all symmetries of a hexagonal arrangement in a crystal lattice (which is the case of the second figure). However, the most overlap (in terms of the number of symmetries) between the two occurs at the $\Gamma$ point, since it is the point of the highest symmetry. Thus, the first figure resembles mostly the orbitals at the $\Gamma$ point in the first Brillouin zone of the crystal lattice.
Please, if I am thinking in a right direction, refine my answer to make it more strict and rigid.


Consider the equilibria in dilute aqueous solution:

$$\ce{H2A+ \rightleftharpoons HA + H+} \ \ \ \ \ \ \ K_{a_1}=\frac{[HA][H^+]}{[H_2A^+]}$$

$$\ce{HA \rightleftharpoons A^- + H+} \ \ \ \ \ \ \ K_{a_2}=\frac{[A^-][H^+]}{[HA]}$$

At the isoelectric point:

$$[H_2A^+]=[A^-]$$

Using the expressions for $K_{a_1}$ and $K_{a_2}$, we find:

$$\frac{{[HA]}[H^+]}{K_{a_1}}=\frac{K_{a_2}[HA]}{[H^+]}$$

$$[H^+]^2 = K_{a_1}K_{a_2}$$

$$pH_{isoelectric}=\frac{pK_{a_1}+pK_{a_2}}{2}$$

See also this section on Wikipedia.


Highstaker's answer is almost correct.  The antibonding orbitals that are occupied in NO are in fact pi symmetry, but when the dimer forms that is no longer relevant.  It is a sigma bond.

The enthalpy of the newly formed sigma bond in the dimer is weak because the net gain in bond is off set by the loss of a very odd set of single-electron resonance forms available for NO monomer.

Given $\Delta G= -17 \:\mathrm{kJ/mol}$ , and that dimerization is entropically disfavored, when the total free energy is considered there is no gain since entropic effects are on the order of $10-30 \:\mathrm{kJ/mol}$.  Thus any small gain in enthalphy is offset by the loss of entropy.

Always look at these things through Gibbs free energy and not just enthalpy.


Troubles with cinnemaldehyde extraction - A few general questions

One of the main problems is that, at the end, the receiving flask is full of water and has very few drops of oil whereas the video shows I should have a foggy mixture. [...] what's up with that?

It's tough to say without seeing your exact setup, but here are some possibilities for you to think about:

Too much water and not enough cinnamaldehyde is coming over in your distillation.  Try heating the still less rapidly.  You might also try using more cinnamon bark and/or less water.
You are using a degraded or very impure cinnamon bark that does not contain very much cinnamaldehyde.  Is there a way you can test a different batch?  What does the distillation smell like?  What does the condensate smell like?
Other possibilities I haven't thought of...

  Second, third and fourth: What is a good solvent to use other than DCM/ether?

The intended application of the solvent is extraction of cinnamaldehyde from an aqueous emulsion of the water-insoluble cinnamaldehyde.  Thus, the main criterion for the solvent is that it be insoluble in water.  If it isn't insoluble, adding it to the aqueous emulsion will just dilute the emulsion.  Nothing will be transferred from one phase to the other, because there will only be one phase.

A secondary feature of DCM in the video is that it is heavier than water.  This means that in the separatory funnel, the DCM/cinnamaldehyde layer will be lower than the aqueous layer.  In this particular extraction, the aqueous layer is repeatedly washed with fresh DCM, and the DCM layer is removed each time.  Doing this is a bit easier if the solvent layer is the bottom layer.  If it were the aqueous layer at the bottom of the funnel, each wash with the 60 mL of solvent would require draining the aqueous layer into one flask, then draining the solvent/cinnamaldehyde layer into another flask, then re-adding the water to the separatory funnel and washing it with fresh DCM, etc.  This requires more flasks and more manipulations than just being able to leave the aqueous phase in the funnel the whole time.

  I have isopropyl, methanol, acetone and naphtha [available].

Now that you know why DCM makes a good solvent, you can probably see why methanol wasn't a great choice.  It is miscible with water, so that explains why you didn't get multiple layers.  Of the solvents you have, only one, naptha, is immiscible with water.  The other two, isopropanol and acetone, are miscible with pure water, but are not miscible with very salty water.  So you might be able to extract the cinnamaldehyde by adding a lot of salt such as calcium chloride, magnesium sulfate, or sodium chloride to the distillate, and then trying to wash the now-salty emulsion of cinnamaldehyde to the solvent for washing and phase separation.  However this is likely to be more difficult than if you could access DCM.  (Getting DCM may be less difficult than you think, by the way.)

  I'm having difficulty understanding what the guy is doing after separation. I've watched the video several times but it still makes little sense to me.

I don't know what part is confusing or exactly what "separation" you mean when you say "after the separation".  But if you mean after the separatory funnel phase separations, then roughly the steps are:

"Drying" the DCM solvent.  Even though DCM and water don't completely mix (they form two phases, which, as explained above, is the reason why it works to extract cinnamalehyde from an aqueous emulsion), some water does leach into the DCM phase during the separation.  "Washing" the DCM with brine helps rid the DCM  of the small amounts of water dissolved in it after the phase separation.  Adding calcium chloride powder to the DCM is an even more effective drying agent that removes the last bit of water from the DCM.
DCM distillation.  After drying the DCM, all the cinnamaldehyde is still dissolved in a (now "dry" or water-free) DCM phase.  To get rid of the DCM and obtain the pure cinnamaldehyde oil, another simple distillation is performed.  Because DCM boils at around 40 °C and cinnamaldehyde doesn't boil until 248 °C, it is fairly easy to remove nearly all the DCM by simple distillation, leaving the oil.
NMR.  The very end shows a proton NMR of the oil.  This is something that requires big fancy equipment and won't be possible for home chemists.  Don't worry about it and enjoy your cinnamaldehyde!


For bond block the numbering can be done at random or it has to follow some order

Bonds don't need to be in any particular order, although usually bonds are created first for atom 1, then atom 2, etc. However, the first bond could be between atom 101 and atom 100 if you wish.

  Also in the atom block the order in which atoms occur can it be random or again it has to follow any order

Atoms can be in any order. Some programs use "canonical ordering" so that the same molecule will always have the same atomic order, but this is rarely used because it can be time-consuming.

  Is order of numbering done in bond block and atom block related

As indicated above, both the atom and bond blocks could be in random order. The relation, obviously, is that for a given molecular structure, there will be bonds between bonded atom pairs.

  Is there any free software which converts mol files to structure

There are many, but you don't describe whether you want 3D (presumably) or 2D structures.

Personally, I suggest Avogadro which can create 3D coordinates for files.

If you're generating a large batch of 3D structures, you will likely want to use Open Babel or RDKit or something similar. Both are open source and free.


Deriving kinetic equations for a complex reaction system

In many cases, complex systems of even simple differential equations do not have simple and obvious solutions. Therefore, it could be too difficult and not reasonable to find fully analytical solutions.

Usually, numerical solutions can be easily calculated using iterative algorithms. However, such computer codes can be susceptible to calculation artefacts. Furthermore, the validation and verification of the model and the code can be a difficult issue.

I made a compromise and used an analytical approach when solving systems of differential equations, but I allowed the code to evaluate subexpressions involving constants numerically. Thus, I made numerical calculations to combine all constant parameter values as soon as possible in order to limit the number of parameters in the equations (using excessive digits in order to avoid accumulation of round-off errors).

I found that this approach can be easily and quickly implemented using the commercial software Maple. However, other similar mathematical software could be suitable, too.

I successfully used this approach for the evaluation of radioactive decay of the inventory of a nuclear reactor core in dependence of time after shutdown. The calculation of the radioactive decay of the relevant radionuclides involved more than 100 individual differential equations, which mainly describe chains of simple first-order reactions, e.g.:

$$\frac{\mathrm da_{\text{I-131}}(t)}{\mathrm dt}=-\lambda_{\text{I-131}}\cdot a_{\text{I-131}}(t)+0.778\cdot\lambda_{\text{I-131}}\cdot a_{\text{Te-131m}}(t)+1\cdot\lambda_{\text{I-131}}\cdot a_{\text{Te-131}}(t)$$

> de_a_I131:=diff(a_I131(t),t)=-l_I131*a_I131(t)+0.778*l_I131*a_Te131m(t)+1*l_I131*a_Te131(t);

A large number of further differential equations could be added in order to model transport phenomena during postulated accidents (e.g. loss of coolant).

Differential equations that were dependent on each other and the corresponding initial conditions were grouped into systems of differential equations, e.g.:

> sys_131:={de_a_Xe131m,de_a_I131,de_a_Te131,de_a_Te131m,a_Xe131m(t0)=core_Xe131m,a_I131(t0)=core_I131,a_Te131(t0)=core_Te131,a_Te131m(t0)=core_Te131m};

The systems of differential equations were solved analytically and the subexpressions involving constant parameter values were evaluated numerically.

> sol_131:=evalf(value(dsolve(sys_131,[a_Xe131m(t),a_I131(t),a_Te131(t),a_Te131m(t)])));

The results could be easily rearranged into a suitable form, e.g.

$$\begin{aligned}a_\text{I-131}(t) ={} & -1.6030637832147524320\cdot10^{17}\cdot\mathrm e^{-6.4180294496291232354\cdot10^{-6}\cdot t}\\&-7.9769363886868464215\cdot10^{15}\cdot\mathrm e^{-4.6209812037329687295\cdot10^{-4}\cdot t}\\&+4.7213833147101620896\cdot10^{18}\cdot\mathrm e^{-9.9782796169607015473\cdot10^{-7}\cdot t}\end{aligned}$$

The results could also be directly used for further calculations within the same worksheet.


Part of the reason why p-tosyl chloride is seen so often is not only the ease of its preparation, yet that is a side product in the preparation of o-tosyl chloride towards saccharin.  Ease in storage and addition / use are beneficial.  While a p-nosyl chloride eventually may yield a better nucleofuge than the p-tosyl chloride, the very same nitrogroup may ease the hydrolysis of the chloride as well.


According to current nomenclature rules, $\ce{H3N}$ would be correct and acceptable.  However some chemical formulas, like $\ce{NH3}$ for ammonia, that were in use long before the rules came out, are still accepted today.


What is the origin of the "positive-to-negative" convention concerning dipole moments in chemistry?

Some sources point out the direction of the dipole moment is convention dependent. Since both conventions are mathematically correct it seems important when solving a problem to identify the convention.  Both conventions seem to be derived from the bones of dead theories.

The physics convention does seem to have Franklin's work identifying negative "resinous electricity" and positive "vitreous electricity" as one step in the progression to the modern theory.  In modern theory a positive to negative electric field will orient a dipole such that the field is in the same direction as a negative to positive dipole.

On the chemistry side, Peter Debye (who the unit for a dipole moment is named) developed the concept of the dipole moment in 1912 to explain anomalous dispersion. This dispersion was explained as being caused by a negative charge collecting on one side of a molecule leading to a dipole moment. This moment's magnitude explained anomalous dispersion, so direction did not seem to receive much attention.  Molecules during this time where described using the Thompson, or "plum pudding," model. In this model molecules were described as positively charged spheres.  (These spheres had no nucleus because the nucleus concept was in its infancy during this time period.) These spheres formed the basis of mathematical models used to describe the behavior of materials with a dipole moments.  The dipole moment in this sense went from the center of a molecule to the point of highest electron density.  With the center of the sphere as a pivot point, the rotation of molecules in an electric field could be explained. In this sense the positive to negative convention may be an artifact of the early spherical molecular models first used to describe dipole moments.

sources:

J. W. Williams   Dipole theory and the size of molecules.  Trans. Faraday Soc., 1934,30, 723-728

John W. Williams The Dielectric Constants of Binary Mixtures. VIII. The Electric Moment as a Vector Quantity, J. Am. Chem. Soc., 1928, 50 (9), pp 2350–2357  DOI: 10.1021/ja01396a005  Publication Date: September 1928

L. E. Sutton  Electric dipole moments and resonance in molecules, Trans. Faraday Soc., 1934,30, 789-801

https://www.quora.com/Why-does-a-dipole-have-a-direction-from-negative-to-positive-while-electric-field-lines-go-from-positive-to-negative

http://goldbook.iupac.org/E01929.html

http://www.ushistory.org/franklin/info/inventions.htm

http://www.wondersandmarvels.com/2011/12/accentuate-the-vitreous-eliminate-the-resinous.html


Well, yes, and this is not limited to bonds. The stablest anything has the lowest potential energy, because things just tend to find positions with lower potential energy and stay there. This is the very nature of our world.

On the other hand, in this case the energy required to get things out of this position (that is, to break the bonds, if we are talking about bonds) is the greatest.


Is the rate of electrolysis the same throughout as long as voltage is same

The rate of electrolysis (decomposition of water) is defined by the current that passes. If you apply a constant voltage, factors arise as suggested in the comment by MaxW.

At constant current, the rate of hydrogen (and oxygen) evolution would be somewhat different initially because on a microscopic level, hydrogen atoms have to cover the surface to some extent to form H2; H atoms may reduce some oxides on the electrode surface, and H2 has to agglomerate to form bubbles (and dissolve in water to saturate it with H2). And then the bubbles have to rise to the surface.


Is it possible to make liquid nitrogen with a bottle of CO2 gas and a freezer?

Short answer, no.

The method you are suggesting here is to use Joule-Thompson expansion to cool the air as it escapes and hope that it liquefies. This is a valid way to make liquid nitrogen and is used in both lab-scale and industrial air liquefaction. Unfortunately, you will be many orders of magnitude off from where you need to be. Plans I've seen for liquefying air on a lab scale require starting pressures of around 3000 psi. That air is sent through a long copper tube where it is rapidly expanded through a nozzle into an insulated vessel to atmospheric pressure (14.7 psi). Even then it does not liquefy. The cooled air is vented through another tube concentric with the first tube, where it chills the pressurized gas entering the vessel. My recollection is that if you start with a full-sized cylinder of nitrogen, after about 1/2 the tank is gone, you will start to condense liquid nitrogen and when the tank is empty, you'll be left with about a half-liter of liquid nitrogen. Those numbers are probably off, but the order of magnitude is correct. (Note: if you try this based on these "instructions", you will probably get yourself killed given the pressures involved).

Nitrogen liquefies at 77 K, or -196 C. Dry ice is only −78.5 C, and my freezer at home sits at about -5 C. I think a PET Coke bottle gets pressurized to 80 psi and a soda keg is rated to 120 psi. As you can see, the temperatures, pressures, and volumes you're likely to be able to achieve at home are off by one or more orders of magnitude.

Sorry.


Dependence of rates of neighbouring group participation on length of alkyl chain

Thanks to orthocresol's suggestion to look at Clayden's (2nd ed. p. 805-807), I now have a satisfactory answer which I can now present.

We can explain these trends using the Gibbs free energy equation. Except that now, we are now interested in the quantities: Gibbs free energy of activation, enthalpy of activation and the entropy of activation, since we are analysing rate.

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

Explaining the decreasing of $\ce {\Delta H^\ddagger}$ as the ring size increase

$\Delta H^\ddagger$ is associated with the ring strain of the resultant ring formed. The larger the ring strain, the larger the value of the enthalpy of activation. Clayden, Greeves, & Warren (2012) explains$\ce{^1}$:

  Firstly, small rings form slowly because forming them introduces ring strain. This ring strain is there even at the transition state, raising its energy and slowing down the reaction. The activation energy for forming a three-membered ring is very high, due to strain, but decreases as the ring gets larger.

The data (from pg. 108 of ref. 2) presented here of the strain energy determined for cycloalkanes of varying sizes also supports this trend.

\begin{array}{|c|c|c|c|} \hline
\text{Ring size} & \text{Strain energy}^\text{2,*}/ \text{kJ/mol} \\ \hline
\ce{3} & 115 \\ \hline
\ce{4} & 110\\ \hline
\ce{5} & 26 \\ \hline
\ce{6} & 0 \\ \hline
\end{array}

*Reported to the nearest whole number.

Explaining the increasing of $\Delta S^\ddagger$ as the ring size increases

Think of it this way: a long chain has a lot of disorder, and to get its ends to meet up and react means it has to give up a lot of freedom$\ce {^1}$. However, for shorter chains, it is far easier for their ends to meet since these chains have fewer degrees of freedom and thus, the loss of entropy upon cyclisation is less negative.

Consolidating the trends

Clayden, Greeves, & Warren (2012) reasons that based on the above trends for $\Delta H^\ddagger$ and $\Delta S^\ddagger$, the activation energy for the formation of the 4-membered ring is the largest. This is because the formation of the 4-membered ring likely has similar $\Delta H^\ddagger$ to that for the 3-membered ring while its $\Delta S^\ddagger$ is relatively more negative.

This seems to agree with the solvolysis rates I have presented:

For the ω-chloro alcohols, we see that the relative solvolysis rate is lowest for $\ce {Cl(CH2)3OH}$. For the ω-methoxyalkyl p-bromobenzenesulfonates, the formation of the 4-membered ring also has the 2nd lowest rate ($\ce {k_{rel} = 0.68}$), with the rate for the 3-membered ring being the lowest ($\ce {k_{rel} = 0.28}$).

Why does the solvolysis rate not peak at the formation of the 6-membered ring and instead, peak for the formation of the 5-membered ring? This is probably a coincidence. There are 2 opposing trends here so it may be the case that the combined effects of the two trends give different results under different circumstances.

ron has provided a graphical representation of the data for strain energy and the entropy of activation here. We do see that in general, there is a peak in the rate of formation of the 5-membered ring. In other words, from the formation of 5-membered ring to formation of 6-membered ring, $\Delta S^\ddagger$ becomes significantly more negative.

References

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.
Anslyn, Eric V., and Dennis A. Dougherty. (2006). Modern Physical Organic Chemistry. Sausalito, CA: University Science.


How can you use ICE tables to solve multiple coupled equilibria?

The way to use ICE tables in this case would be to combine the ICE tables, and use "$x$" for the changes due to one reaction and "$y$" for the changes due to the other. For each reaction, you have one unknown (how far it reacted, i.e. $x$ and $y$), so you need two pieces of information to solve it, in this case the two equilibrium constants.

Here is the combined ICE table:

$$
\begin{array}{|c|c|c|c|c|}
\hline
 &[\ce{Ca^2+}]  & [\ce{F-}]  & [\ce{H+}]&[\ce{HF}]   \\
\hline
I &  \pu{2.1e−4} & \pu{4.2e−4} & \text{N/A} & 0 \\
\hline
C &  +x  & +2x-y & \text{N/A} & +y  \\
\hline
E &  \pu{2.1e−4}+x & \pu{4.2e−4}+2x-y & 10^{-3.00} & +y \\
\hline
\end{array}
$$

Now, you can use the equilibrium constants for the two reactions to solve for $x$ and $y$, giving you the equilibrium concentrations.

What the combined table shows is that the fluoride concentration depends on both reactions, so you can't first deal with one equilibrium and then with the other but have to solve a system of two equations with two unknowns.

Once you combine the tables, you could also have initial conditions where nothing is dissolved yet, to get easier expressions (I'm using $p$ and $q$ because they will be different from $x$ and $y$):

$$
\begin{array}{|c|c|c|c|c|}
\hline
 &[\ce{Ca^2+}]  & [\ce{F-}]  & [\ce{H+}]&[\ce{HF}]   \\
\hline
I &  0 & 0 & \text{N/A} & 0 \\
\hline
C &  +p  & +2p-q & \text{N/A} & +q  \\
\hline
E &  +p & +2p-q & 10^{-3.00} & +q \\
\hline
\end{array}
$$


Can isopropyl alcohol vapor damage polymers?

I will use your example of the mouse pad to answer your more general question.

As you implied in your question, the mouse pad will be exposed to some isopropyl alcohol vapor if it is used to clean in the vicinity of the pad. As the vapor will quickly disperse to negligible levels, a simple solution could be to move the mouse pad (or other rubber-based or otherwise isopropyl alcohol sensitive item) to the other side of the room while cleaning the area around the computer. Once the area is dry and the smell is gone, there should be negligible isopropyl alcohol vapor remaining and you can put the mouse pad back. For the reasons discussed below, even this much effort is likely overkill and you can probably safely just clean around the pad and other items with isopropyl alcohol.

The key reason that isopropyl alcohol vapors are much less damaging to the rubber component of the mouse pad is simply how much more of the compound comes into contact with the rubber when in direct contact with the liquid phase vs secondary exposure via the vapor phase. Depending on a number of factors (contact time with the liquid, temperature of the surfaces that are cleaned with the isopropyl alcohol, proximity and quantity of liquid used in vicinity of mouse pad, etc.) the mouse pad could be exposed to a few orders of magnitude more isopropyl alcohol when in contact with the solution than when merely exposed to vapors from cleaning some area in the vicinity of the mouse pad. This means that the exposure of the sensitive items to the isopropyl alcohol vapor is very small and not likely to have any noticeable effects on the exposed items like the mouse pad.


On the stabler tautomer in amide-imidic acid tautomerism

The relative stability of the two tautomers (two isomers differing how protons are bound) may be assessed from the equilibrium constant you may formulate for the two.  Deriving from above illustration, it would be

to indicate that this is dynamic, with a back and forward reaction at microscopic level.  Once formulated as such, you probably recall $\Delta{}_RG = -RT \ln{}K$ to relate between the sample temperature $T$, the universal gas constant $R$, the free reaction enthalpy $\Delta{}_RG$ and the equilibrium constant $K$.
Depending on the substrate, you may use spectroscopic evidence (e.g., $\ce{^1H}$-NMR spectroscopy) to assess the concentration of the two forms.  Equally, based on prior work by others, there are databases to either list these values, or provide a prediction (i.e., an educated guess) about $K$.  Knowing these equilibria is important to anticipate the likelihood of chemical reactions. It equally offers insight regarding the pharmaceuticals and their uptake, anabolism, interaction with the body (metabolism), and catabolism and eventual elimination, too.
One of the databases available to the public is Tautobase (2020JChemInfModel1085). It is implemented in the freely available DataWarrior program allowing you a search-by-(sub)structure.  With some luck, one of the 1680 pairs matches your substrate closely enough, too.  In the simplest case one may imagine, however, the amide is strongly favoured:

The database backs this particular entry with the publications 1998JAmChemSoc10359 and 2001JAmChemSoc2681.


Does an axis of symmetry determine chiralty?

Chirality is left hand—right hand. The left hand is a mirror image of your right hand. If you take this mirror image and then rotate or shift it and you are able to superimpose it onto the original image, it is achiral. This is not possible with the hands, the thumb will always be wrong.

A plane of symmetry is sufficient but not mandatory to give an achiral compound. But it should be clear that mirroring an image that contains an intrinsic mirroring plane will give the same.

Another sufficient but not mandatory symmetry operation to generate a chiral compound is inversion. If some compound is point symmetric, it is never chiral.

And there are other elements of symmetry that are rare if not accompanied by planes or centres of symmetry but which are sufficient for generating achiral compounds: Improper rotation. In a nutshell, this is a symmetry element that requires a rotation with a subsequent mirroring perpendicular to the rotation axis. Take a tetrahedral compound such as methane: If you select the correct axis, rotate by $90^\circ$ and then mirror it perpendicular to that axis, you will receive the same product.

Strictly speaking, planes and centres of symmetry are special cases of improper rotation with a rotation of $360^\circ$ and $180^\circ$, respectively. So one could say that improper rotation is sufficient and mandatory to give an achiral compound.


What is meant by complete outer shell? Why do the noble gases have zero valency?

When we refer to "outer shells" we are talking about the highest n-level, but also limiting ourselves to the s- and p-orbitals. The reason for this has to do with how the effective nuclear charge "felt" by the valence electrons changes as you move through successively higher energy configurations.

In summary, $s$- and $p$- electrons are screened less effectively by inner shells, and so for a given n-value, the $s$ and $p$ orbitals of the next n fill before the current n (they have lower energy).

In your example using argon, this means that the 3d shell actually has no electrons - the condensed configuration is:

$$
\ce{Ar: \space [Ne]}3s^23p^6
$$

For the next noble gas, krypton, the condensed electron configuration is:

$$
\ce{Kr: \space [Ar]}4s^23d^{10}4p^6
$$

Note that it has $3d$ electrons at a higher energy than what argon had, but since the value of n roughly corresponds to the distance of the orbital from the nucleus, the $3d$ orbitals are physically closer to the nucleus than the $4s$ and $4p$ orbitals. In other words, for single atoms, the lowest energy electron configuration always fills the s- and p- orbitals first, and so when we are looking  for the valence or "outermost" electrons, that is where we need to look - we sort of ignore the $d$- and $f$-orbital electrons for these kinds of problems.


But I thought why not $\ce{SO2}$

While the reaction is stechiometrically correct, it is hard to perform.

In excess of $\ce{H2S}$ sulfur will be produced.
In excess of oxidant sulfate will be produced.

And exact balance is impossible to achieve.

That said, $\ce{SO2}$ reacts with  $\ce{H2S}$ finally producing sulfur. This reaction, just like many inorganic reactions, is an unholy mess of many different processes with many products obtained simulationaly and the only way to produce understandable and reproduceable results is to simplify the situation. One of the way to simplify the situation is to use a vast excess of one of the reagents.


How does High Frequency water electrolysis work?

First of all, have a look at the wikipedia page on electrolysis of water.

I also like this review: Zoulias et al.: A Review on Water Electrolysis, TCJST, 4 (2) (2004) 41-71
Specifically they list a number of actually existing installations (context: renewable energy) and their actually achieved efficiency.
Speed up does not necessarily have anything to do with higher efficiency. In electrolysis it is often the other way round: if you want to squeeze out the maximum free energy, you need to do the reaction infinitely slowly (despite thermodynamics having a dynamic name, it looks at infinitively slow processes).
Thus, speeding up usually means that you find a way to put more power through your system. The big issue is to find a way of doing this without loosing (too much) efficiency.
Pulsed/modulated DC: looking through a few papers I liked this one: Shimizu et al.: A novel method of hydrogen generation by water electrolysis using
an ultra-short-pulse power supply, Journal of Applied Electrochemistry (2006) 36:419–423, DOI 10.1007/s10800-005-9090-

They aim at avoiding the diffusion controlled situation by having the pulses short enough so that no depletion zone occurs.
Look at these diagrams:

So they report one setting where the pulsed electrolysis is actually more efficient than DC in their cell.

There's more research going on on this, however the papers I found reported increased efficiency compared to pure DC electrolysis, but the absolute efficiencies are around 10%.   However, compare their numbers with the 80% efficiency cited by the review for an industrial alkaline electrolysis. Note that one big difference is the voltage that is applied: for the DC it is around 1.85 - 2.05 V, so much less overvoltage. Note also that when they say that higher voltage speeds up ion transport, then this overvoltage is converted to heat (ions face friction in the medium when they travel) and thus basically lost.
Another line that looks real is that if you go to higher temperatures, a (small) part of the energy can be supplied by heat. As heat is cheap, this may help. One point one has to be aware though, that the efficiency calculations may be done with respect to the electric energy only (neglecting the heat input) and thus look artificially nice (like efficiencies of condensing boilers calculated against the lower heating value).
I found a bunch of nonsense claims in the internet, about the resonance frequency of water helping to split bonds.

The first thing to realize here is that there is no one resonance frequency of water. With suitable energy, you can excite rotational, vibrational and electronic states (I left out translation - there transition energies minute). At room temperature you can say as a rule of thumb that most molecules will be in some excited rotational state, but in the vibrational and electronic ground states. Excitation energies for rotation are in the far infrared or microwave energy/frequency region. Widely used e.g. in the microwave oven at 2.45 GHz ($\approx$ 12 cm). Actually, the whole region is full of bands where water absorbs. Note that microwave heating of water does not cause electrolysis. Vibrational transitions are around 2.9 μm = 105 THz = 3500 cm⁻¹ and 6μm = 50 THz = 1635 cm⁻¹ with lots of combinations and overtones throughout the near infrared region. Quite exceptionally, the visible region is basically free of water absorption. Electronic transitions (breaking of bonds) need energies in the UV, and here we meet bands that lead to photodissociation, e.g. at 166nm (taken from Wikipedia). That corresponds to 1.8 PHz = $1.8 \cdot 10^{15}$ Hz. Compare this to the kHz and MHz where your link claims dissociation.

This doesn't mean that the pulsed DC cannot help, nor that impedance spectoscopy won't give important information. But resonance frequencies in the kHz range are electrical LC-circuit resonances depending on cell and electrode geometries and electrical double layers etc. But neiter on vibrations nor breaking of the bonds of the water molecule.
To give the "method" you ask about some real world numbers,

at the very end of the Wiki page the energy efficiency for industrial water electrolysis is cited as usually between 50 and 80 %.
The paper then proposes to burn the gas in an internal combustion machine. As such a stationary process could be adjusted so that the engine is at its maximum efficiency, we may assume 1/3 or 35% efficiency here.
we then need a generator to convert the mechanical energy into electric energy. Fortunately, that step is rather efficient. Say, 95 %.
A fuel cell would be more efficient than the combustion - generator combination: ca. 40 - 60 % according to Wikipedia.
Unfortunately, also battery charging is not 100% efficient. Let's assume 80–90% (taken from Wikipedia on Li-ion batteries) For batteries that are charged with higher current (or current density) efficiency is less. Example would be lead-acid batteries as used in cars. Wiki quotes efficiencies between 50 and 80 %.

Taking these numbers together, I conclude that after going once through the cycle of the proposed "perpetuum mobile", 8 - 24 % of the energy are retained in a "useful state" while 76 - 92 % became heat. With fuel cell, we may be able to "boost" the energy efficiency to 43%.

Useful general knowledge (in addition to the law of energy conservation)

The US patent system is different from e.g. the German patent system in that here in Germany the patent application has to have commercial/industrial usability. This includes a technical argument why it works (according to the physical laws). A perpetuum mobile would be rejected on these grounds (of course the inventor could prove his case with a prototype). US patents to not have this technical check.
generators (mechanic -> electric conversion) are sources of current, while batteries (galvanic cells) are sources of voltage.


Do phosphoryl thiocholines, their uncharged analogs, and other organophosphates, interact differently with Acetylcholinesterase?

As a matter, of fact, yes, uncharged phosphoryl thiocholines do seem to interact with acetylcholinesterase differently compared to their charged analogs. For example, it has been reported that

Sundwall (1961) reported that P2S was considerably more effective as
an antagonist of the lethal action of 37 SN
+(isopropoxy-2-trimethylammonio-tiomethylphosphino oxide) in mice than it was of those of Sarin and 37 S-N
(isopropoxy-2-dimethylaminothiomethylphosphine oxide).

P2S refers to the methyl sulfonate salt of pralidoxime. This would seem to suggest that, at least in the case of 37-SN+, the adduct formed by the charged inhibitor is more susceptible to reactivation by oxime therapy as opposed to that formed by the neutral compound†.
Regarding the second half of the question, the answer is also yes. According to Boskovic, Acetylcholinesterase that has been inhibited by O-pinacolyl S-2-diethylmethylammonioethyl methyl phosphonothioate methyl sulfate undergoes aging two orders of magnitude more slowly compared to AChE that has been inhibited by Soman, and is also more susceptible to reactivation. The explanation proposed by the author is that the interaction between a phosphoryl-thiocholine and AChE results in the formation of an adduct in which the phosphoryl group is oriented differently compared to the adduct formed as a result of inhibition by an uncharged organophosphate. The author also points out that AChE which had been inhibited by the compound Ro 3-0411 is more easily reactivated compared to AChE inhibited by diisopropyl fluorophosphate.
In conclusion, it can be said that, for a set of organophosphorus compounds which only differ in whether the leaving group during AChE inhibition is a fluorine atom, a thiocholine, or dialkylaminoethylthiolate (And in which the other side chains are the same), each of these compounds would interact with AChE differently, which influences, probably among other things, their susceptibility to reactivation by oximes and the rate at which the inhibited enzyme undergoes aging.
†It should be noted that, under physiological conditions, some amount of these neutral compounds becomes protonated, although I do not know to exactly to what extent and how and if that affects their pharmacology.
References:
BOSKOVIC, B. (1981). The treatment of soman poisoning and its perspectives. Fundamental and Applied Toxicology, 1(2), 203–213. doi:10.1016/s0272-0590(81)80059-0
BošKović, B., Maksimović, M., & Minić, D. (1968). Ageing and reactivation of acetylcholinesterase inhibited with Soman and its thiocholine-like analogue. Biochemical Pharmacology, 17(8), 1738–1741. doi:10.1016/0006-2952(68)90237-2


Drug design methods - docking software, binding affinity

I mainly work with simulation (MD), so take these answers with a grain of salt.

  Is it the case that when designing drugs (i.e. these proteins) we attempt to optimize binding affinity (maximize?)

Computational docking usually employes two methods: A force field based estimation of the free energy of binding, and exploration of the conformational space for the ligand.

After a docking run you will end up with set of different ligands, a set of conformations (or poses) for each ligand, and a corresponding score value for each conformation. The score is usually based on the binding constant and Gibbs free energy (which is based on various numbers of physicochemical parameters, and may or may not include parameters like desolvation and entropic effects). Different software use different scoring functions and methods to estimate the free energy.

  How well does Autodock Vina compute this affinity? Is it opinionated? I've seen a few docking software, it looks like there are different methods to compute docking score? (can you recommend further reading for me in this regard).

As far as I can tell, it is hard or impossible to make general statements about which piece of software is the best, since it depends a lot on what you are trying to use it for. Some of the limitations of the AutoDock suite (implicit hydrogens, spherically symmetric H-bond potentials etc.) are given in this publication.

  Is there a linear correlation between binding affinity and drug efficacy? I.e. does it mean that higher binding affinity directly means we have a better drug, or is there some sweet point kinda thing?

Not necessarily. A docking run gives you an (rough) estimation of the ligand affinity for the receptor you chose. It might be that the ligand also binds strongly to a lot of other proteins, making it less effective in an in vivo setting.

  Why can't we use very complex molecules, more than 300 characters in SMILES length to combat viruses for example? I noticed everyone was trying to minimise ligands?

You can, but not necessarily with the AutoDock suite. Large compounds have too many degrees of freedom, and the software is not capable of properly exploring the conformational space. Usually these large compounds are split in to smaller fragments which are then docked individually.

  How can software like AutoDock tools prepare your proteins for docking, is there a standard method (they add hydrogens, add Gasteiger charges). Is that guaranteed to be accurate in every case?

Again, depends on the software. You listed some of the features of AutoDock, and you can find the details of the receptor preparation scheme from their website. It is definitely not accurate in every case. As far as I know, most docking runs are performed on protein structures obtained with X-ray methods, and there is no guarantee that the conformation of the protein in crystal is a good representation of the conformation in the solvent phase.


How photoluminescence (PL) spectroscopy works

The explanation that follows is valid for compound dissolved in liquid solution where PL is understood either as fluorescence or phosphorescence.

PL is measured with a fluorimeter. The main optical components of this instrument are: a lamp that provides the light used to excite the sample, two monochromators (M), one for the excitation (Mex) and the other one for the emission (Mem) and a detector, usually a photomultiplier.

To measure PL the Mex is fixed at the excitation wavelength (in your case 345 nm) and the Mem does the scan, i.e. it let pass the luminescence (a narrow band) in a fixed wavelength range. This is then measured with the photomultiplier. At the end you obtain a spectrum, i.e PL vs wavelength.

From the spectrum that you show this range goes from 200 nm to 1000 nm. Usually the range starts after the excitation, this is to avoid to record the excitation itself that is not desired. Because in the spectrum that you show there is not the excitation band i suppose they use a filter (cut-off) to cut it out.


Surface passivation creates different chemistry on the surface than the chemistry of the bulk solid. Imagine bulk Si with a clean surface (presumably in a vacuum and freshly made). It can be thought of as having "dangling bonds", though this is not necessarily the best way to think about it. Think more generally that the structure on the surface is different to the bulk structure. There may be different elements present and the bonds may be "satisfied". Or they may be strained compared to the bulk material.

Expose the Si to, for example, O2 and the surface layer (your dangling bonds, if you like) will react to give some SiO2 just on the top layer. This new layer may be much less reactive than bulk silicon in normal air, hence can be said to passivate the surface.

If the chemical and physical properties of the new layer are right, the layer will protect the bulk from further reaction: this is the definition of passivation. A similar situation occurs naturally for aluminium, for example, where the Al2O3 layer that forms is both strong and relatively impermeable to air preventing further reaction (aluminium with the layer removed will react so rapidly with oxygen it gets warm and disintegrates). Iron, on the other hand, forms rust which doesn't protect the surface as it is flaky and weaker than the bulk metal. To passivate iron you need to add other elements which encourage the formation of stronger passivating layers (this is how many steels are made not to rust).

Chemical manipulation of the surface can create specific types of passivation that serve a particular purpose rather than relying on the natural reactions of air and the surface. This is common in the semiconductor industry and can also be used to generate the coloured layers on aluminium and titanium products where the goal is aesthetic as much as protective.


The Gattermann reaction you mentioned in the question uses Copper powder and Hydrohalous acid to substitute diazonium group with a halide in an aromatic diazonium salt.
It uses $ArSN_1$  pathway, similar to Sandmeyer Reaction which uses Copper Halide and corresponding Hydrohalous acid instead of copper powder.
As diazonium group is a very good leaving group, it departs leading to a phenyl carbocation which is very unstable, this is in fact the slow step of the reaction.
The Nucleophile (Halide in this case) attacks the phenyl carbocation leading to the formation of Halobenzene(fast step).
Nitrogen gas and Hydrogen chloride are also formed.


How to calculate the concentration of the elements after dissolution of iron(III) chloride?

When answering such problems, I recommend writing everything know down, then use the appropriate formula to solve for the unknown. For the first part you already did that correctly.

  In the laboratory you dissolve $\pu{24.7 g}$ of iron(III) chloride in a volumetric flask in water to a total volume of $\pu{375 ml}$.

\begin{align}
  m(\ce{FeCl3 (s)}) &= \pu{24.7 g}\\
  M(\ce{FeCl3}) &= \pu{162.3 g//mol}\\
  V(\ce{FeCl3 (aq)}) &= \pu{375 mL}\\
\end{align}

  What is the molarity of the solution?

Molarity is another word for amount concentration, so you are looking for $c(\ce{FeCl3 (aq)})$.
\begin{align}
  c &= \frac{n}{V}\\
  n &= \frac{m}{M}\\
  c(\ce{FeCl3 (aq)}) &=
    \frac{m(\ce{FeCl3 (s)})}{M(\ce{FeCl3})\cdot V(\ce{FeCl3 (aq)})}\\
  c(\ce{FeCl3 (aq)}) &=
    \frac{\pu{24.7 g}}{\pu{162.3 g//mol}\cdot\pu{375E-3L}}\\
  c(\ce{FeCl3 (aq)}) &=
    \pu{0.406 mol//L}
\end{align}

  What is the concentration of iron(III) cation in M?
  And what is the concentration of chloride anion in M?

You are looking for the ion (amount) concentrations in your solution, i.e. $c(\ce{Fe^3+ (aq)})$ and $c(\ce{Cl- (aq)})$.
For these two questions you need to know how the substance dissolves. The wording here is already very explicit, as it says iron(III) cation, i.e. $\ce{Fe^3+ (aq)}$, and chloride anion, i.e. $\ce{Cl-}$. Otherwise, with some more experience you will know that this compound is a salt and it dissociates into ions when dissolved in polar solvents like water. Therefore the reaction equation is
$$\ce{FeCl3 (s) ->[H2O] Fe^3+ (aq) + 3 Cl- (aq)}.$$
From this you can derive the actual relationships between the ion concentrations.
You can clearly see that there is one iron(III) ion and three chloride ions per formula of $\ce{FeCl3}$; therefore
\begin{align}
  c(\ce{Fe^3+ (aq)})&= c(\ce{FeCl3 (aq)})
  &&=\pu{0.406 mol//L},\\
  c(\ce{Cl- (aq)}) &= 3\times c(\ce{FeCl3 (aq)})
  &&=\pu{1.22 mol//L}.
\end{align}

The other equation would probably be appropriate when you (thermally) decompose the substance (in absence of everything) as it is the reverse equation which can be used to prepare the molecule:
$$\ce{2 Fe (s) + 3 Cl2 (g) -> 2 FeCl3 (s)}.$$


Reference request on potassium cyanide production method

I did some literature research and I wasn't able to find any methods involving steam (as water in the gas phase) for the synthesis.
This in fact makes sense as alkali metal cyanides are prone to hydrolysis and tend to decompose to formate $\ce{K(HCOO)}$ over time or in hot water:

$$\ce{KCN(aq) + 2 H2O(l) -> K(HCOO)(aq) + NH3(g)}$$

which further decomposes to potash $\ce{K2CO3}$ upon further heating, so presence of water, especially at elevated temperatures, appears to be highly undesirable for the synthesis of cyanides.

I suspect the aforementioned article contains a typo.
Probably it should read "stream of nitrogen" — then it would make more sense as there are several industrial processes utilizing atmospheric nitrogen fixation for making cyanides with coal proposed in late XIX – beginning of XX centuries.

First, there is a German patent by Caro issued around 1900 (I couldn't find the full text or the exact date) mentioning synthesis of the cyanides of group 1 and 2 elements from carbonates using nitrogen and coal [1].

Second, there is Bucher process [2], which, I think, http://potassium.atomistry.com might refer to.
There is a comprehensive review of Bucher process [3]:

  The Bucher process, as known, consists in heating a mixture of sodium carbonate, charcoal, and iron powder (catalyst) in a stream of nitrogen at a temperature above 900°. The reaction is written as follows:

  $$\ce{Na2CO3 + 4 C + N2 -> 2 NaCN + 3 CO}$$

  […] A stream of nitrogen was passed  through the tube while a certain temperature between 900° and 1000° was maintained for several hours.
  The reaction  product was rapidly cooled by lifting the tube out of the furnace without interrupting the flow of nitrogen and was then examined microscopically.
  In all experiments sodium cyanide, together with some of the ingredients of the original mixture, proved to be present, which confirms the reaction given by Bucher.

There is also a brief overview of industrial methods used for manufacturing alkali metals cyanides throughout the past 170 years in Ullmann’s Encyclopedia of Industrial Chemistry [4]:

1850s–1900s: semi-industrial process (aka "the blood process") of fusion of potassium carbonate with potassium hexacyanoferrate(II) (obtained from animal blood and waste) initially proposed by F. and E. Rodgers [5].

1900s–1960s: Castner–Keller process; primarily employed at Degussa AG [6]:

$$\ce{Na + C + NH3 ->[\pu{600 °C}] NaCN + 3/2 H2}$$

1960s–present sodium and potassium cyanides are produced exclusively by the neutralization process, between a solution of sodium or potassium hydroxide with liquid or gaseous hydrogen cyanide:

$$\ce{MOH + HCN -> MCN + H2O}$$

  Gaseous or liquid $\ce{HCN}$ is first introduced continuously into concentrated solutions of $\ce{NaOH}$ or $\ce{KOH}$ at elevated temperature.
  The heat of neutralization is used for the subsequent vaporization of water. The use of special mixing devices and the addition of $0.2–3\%$ excess $\ce{NaOH}$ prevents polymerization.
  Simultaneously or in a second step, water is vaporized under reduced pressure and the alkali metal cyanide is precipitated below $\pu{100 °C}.$
  The shorter the residence time of the cyanide in solution, the cleaner the product and the lower the content of hydrolysis products and iron impurities.

  Figure 6. Flow chart of the main production steps for cyanide
  production ($\ce{NaCN}$ as example)

References

Caro, Nikodem. German pat. anmeld. C15812 (8-4-09). Verfahren zur Darstellung von Cyaniden aus Alkali oder Erdalkalioxyden, bezw. Carbonaten, Kohle und Stickstoff.
Bucher, J. E. The Fixation of Nitrogen. J. Ind. Eng. Chem. 1917, 9 (3), 233–253. DOI: 10.1021/ie50087a011.
Posnjak, E.; Merwin, H. E. Note on the Bucher Cyanide Process for the Fixation of Nitrogen. Journal of the Washington Academy of Sciences 1919, 9 (2), 28–30. https://www.jstor.org/stable/24521552
Gail, E.; Gos, S.; Kulzer, R.; Lorösch, J.; Rubo, A.; Sauer, M.; Kellens, R.; Reddy, J.; Steier, N.; Hasenpusch, W. Cyano Compounds, Inorganic. In Ullmann’s Encyclopedia of Industrial Chemistry; Wiley, 2011. DOI: 10.1002/14356007.a08_159.pub3.
F. and E. Rodgers; Ueber Darstellung und Anwendung des Cyankaliums. Ann. Chem. Pharm. 1842, 41 (3), 285–294. DOI: 10.1002/jlac.18420410303.
Degussa, DE 124 977, 1900; DE 126 241, 1900; DE 148 045, 1901; DE 148 046, 1901; DE 149 678, 1901; GB 265 639, 1926 (H. Freudenberg); DE 851
056, 1943 (W. Stephan).


Why are discharge tubes filled at reduced pressure?

Not only does it take a high starting voltage and a current-limiting device (e.g. inductor) to maintain a continuous discharge at higher pressures, but the quality of the light emitted is changed by atomic (and ionic) interactions, tending to broaden lines into a quasi-continuous spectrum.

A good example is the sodium-vapor lamp, often used for street lighting. In the example cited, compare the line spectrum of a low pressure lamp with that of a high pressure one (though there is also a contribution from other gases in the second image) .

The low-pressure hydrogen glow discharge corresponds well with calculated values and is a useful tool in showing how theoretical calculation of orbitals can be translated into an observable spectrum. Calculation of high-pressure phenomena is much more difficult.