How do I show that a transition is magnetic dipole allowed with group theory/symmetry?

For a state-to-state transition coupled by an operator to be allowed, the direct product of the irreducible representations (irreps) of all three components must contain the completely symmetric irrep of the point group you're working in. That is, we are determining whether the following integral is forced to be zero by symmetry:

$$
\int \Psi_{i}(\tau) \times \hat{\mathbf{A}}(\tau) \times \Psi_{f}(\tau) \, \mathrm{d}\tau \overset{?}{=} 0
$$

where $\Psi_{i}$ and $\Psi_{f}$ are the initial and final states, $\hat{A}$ is the operator (electric dipole, magnetic dipole, one-electron spin-orbit, there are many more), and $\tau$ is the integration coordinate, usually over electron position. Why bother with this if the integral can be evaluated exactly?

It's no more complicated than this pencil-and-paper calculation to gain qualitative understanding about whether or not a system with some symmetry will have a certain kind of spectroscopic response.
Why bother calculating the integral if it's going to be zero anyway? It's just extra work, though it usually isn't very expensive.

For $O_{\mathrm{h}}$, the completely symmetric irrep is $\mathrm{A_{1g}}$. It's always listed as the first irrep in a character table.

$$
\small\begin{array}{c|cccccccccc|cc}\hline
O_\mathrm{h} & E & 8C_3 & 6C_2 & 6C_4 & \begin{aligned}3C_2 \\ \scriptsize=C_4^2\end{aligned} & i & 6S_4 & 8S_6 & 3\sigma_\mathrm{h} & 6\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & 1  & & x^2+y^2+z^2 \\
\mathrm{A_{2g}} & 1  & 1  & -1 & -1 & 1  & 1  & -1 & 1  & 1  & -1 & & \\
\mathrm{E_g}    & 2  & -1 & 0  & 0  & 2  & 2  & 0  & -1 & 2  & 0  & & \begin{aligned}(2z^2-x^2-y^2,\\ x^2-y^2)\,\,\,\,\,\, \end{aligned} \\
\mathrm{T_{1g}} & 3  & 0  & -1 & 1  & -1 & 3  & 1  & 0  & -1 & -1 & (R_x,R_y,R_z) & \\
\mathrm{T_{2g}} & 3  & 0  & 1  & -1 & -1 & 3  & -1 & 0  & -1 & 1  & & (xy,xz,yz) \\
\mathrm{A_{1u}} & 1  & 1  & 1  & 1  & 1  & -1 & -1 & -1 & -1 & -1  & & \\
\mathrm{A_{2u}} & 1  & 1  & -1 & -1 & 1  & -1 & 1  & -1 & -1 & 1  & & \\
\mathrm{E_u}    & 2  & -1 & 0  & 0  & 2  & -2 & 0  & 1  & -2 & 0  & & \\
\mathrm{T_{1u}} & 3  & 0  & -1 & 1  & -1 & -3 & -1 & 0  & 1  & 1  & (x,y,z) & \\
\mathrm{T_{2u}} & 3  & 0  & 1  & -1 & -1 & -3 & 1  & 0  & 1  & -1 & & \\ \hline
\end{array}
$$

The magnetic dipole operator is the angular momentum operator with some constants in front:

$$
\hat{\mathbf{m}} = \gamma_{\mathrm{e}} \hat{\mathbf{l}} = -\frac{e}{2m_{\mathrm{e}}} \hat{\mathbf{l}} \approx \left(\hat{l}_{x}, \hat{l}_{y}, \hat{l}_{z}\right)
$$

The angular momentum operator components have rotational symmetry, so $\hat{l}_{y}$ would use the irrep corresponding to $R_{y}$. Constant prefactors don't affect the symmetry of operators, so all three components of the magnetic dipole operator have $\mathrm{T_{1g}}$ symmetry.

Now we must show that

$$
\left< \Psi_{i} | \hat{\mathbf{m}} | \Psi_{f} \right> \equiv \int \Psi_{i}(\tau) \times \hat{\mathbf{m}}(\tau) \times \Psi_{f}(\tau) \, \mathrm{d}\tau \rightarrow \mathrm{T_{2g}} \times \mathrm{T_{1g}} \times \mathrm{E_{g}} \overset{!}{\ni} \mathrm{A_{1g}},
$$

and to do this we need a product table.

$$\begin{array}{c|ccccc} \hline
 & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\ \hline
\mathrm{A_1} & \mathrm{A_1} & \mathrm{A_2} & \mathrm{E} & \mathrm{T_1} & \mathrm{T_2} \\
\mathrm{A_2} & & \mathrm{A_1} & \mathrm{E} & \mathrm{T_2} & \mathrm{T_1} \\
\mathrm{E} & & & \mathrm{A_1 + [A_2] + E} & \mathrm{T_1 + T_2} & \mathrm{T_1 + T_2} \\
\mathrm{T_1} & & & & \mathrm{A_1 + E + [T_1] + T_2} & \mathrm{A_2 + E + T_1 + T_2} \\
\mathrm{T_2} & & & & & \mathrm{A_1 + E + [T_1] + T_2} \\ \hline
\end{array}$$

Product tables are symmetric, so only the lower or upper triangle needs to be shown, and inversion symmetry is left out here as long as you consider the rule

$$
\begin{align}
\mathrm{g} \times \mathrm{g} &= \mathrm{g} \\
\mathrm{g} \times \mathrm{u} &= \mathrm{u} \\
\mathrm{u} \times \mathrm{u} &= \mathrm{g},
\end{align}
$$

just like multiplying even and odd integers. Some product tables show all of the redundant information.

Working from left to right, find the product of the symmetry of the initial state with the symmetry of the operator:

$$
\mathrm{T_{2g}} \times \mathrm{T_{1g}} = \mathrm{A_{2g}} + \mathrm{E_g} + \mathrm{T_{1g}} + \mathrm{T_{2g}}.
$$

The product of degenerate irreps with each other leads to multiple irreps, hence the sum. You might think that the next step is

$$
\left(\mathrm{A_{2g}} + \mathrm{E_g} + \mathrm{T_{1g}} + \mathrm{T_{2g}}\right) \times \mathrm{E_g} = \dots,
$$

however we take advantage of two nice rules:

For the transition to be allowed, the final product only needs to contain the completely symmetric irrep; irreps other than $\mathrm{A_{1g}}$ are allowed to be present as long as $\mathrm{A_{1g}}$ is there.
The product of any irrep with itself always gives the completely symmetric irrep; see the diagonal elements in any product table.

From the last equation, we see that $\mathrm{E_g}$ is in the sum, so the other terms can safely be ignored because $\mathrm{E_g} \times \mathrm{E_g} = \mathrm{A_{1g}}$. Additionally, because all three components of the magnetic dipole operator transform as the same irrep, this electronic transition will be allowed if you apply a magnetic field along any of the three Cartesian axes in the molecular frame.


Magnetic non-equivalence of acetal protons

This link on SE explains why the methylene hydrogens of the ethyl groups are different or diastereotopic. For your specific case, if a deuterium is substituted for Ha on the left side of the structure, it leads to an R-configuration while substitution of deuterium at Hb leads to an S-configuration. Both of these hydrogen sites "see" the same handedness at the methine carbon. Thus, the two hydrogens, Ha and Hb, are in different environments, they have different chemical shifts and they are coupled to each other and the vicinal methyl group. The same argument applies to the ethyl group on the right side of the structure. This difference in chemical shift is more apparent at high field (the QE300 MHz spectrum) rather than at low field (60 MHz).
There is no restricted rotation and temperature will not make a difference. Don't feel bad, these arguments were made for a monoalkyl derivative of diethyl malonate by a chemistry Nobelist!
ADDENDUM: Here is an additional link to this answer. Here is a related spectrum: Intermediate Level #12 at this link.


Chemical shift value and proton NMR spectroscopy

Firstly, the ppm scale is a dimensionless scale that describes the energy level of a signal relative to a reference (usually TMS). Significantly, it is magnetic field independent, and so allows comparison of signals recorded on different systems (not just a 300MHz vs a 600MHz spectrometer, but between all different 600MHz spectrometers, because no two will have been set up at identical Bo fields).

The chemical shift scale is defined as:

$$\delta =(v-v_r )*10^6/v_r$$ , where $v_r$ is the spectrometer reference frequency (of TMS, say)

Now, for a 600MHz NMR spectrum, each ppm represents 600Hz. So, if the resonance frequency of a signal at 0ppm is 600,000,000Hz, the signal at 1ppm represents an energy of 600,000,600Hz. At 2.5ppm, the frequency is 600,001,500Hz.

The chemical shift scale is often confusing in its labelling, but the thing to remember is that as you go to higher chemical shift values, you go to higher frequencies.

Higher ppm == downfield == Deshielded == higher frequency

Lower ppm == upfield == Shielded == lower frequency


Is selenocysteine C3H6NO2Se or C3H7NO2Se?

Selenocysteine can be either of the following: $\ce{C3H8NO2Se+, C3H7NO2Se, C3H6NO2Se-}$ or $\ce{C3H5NO2Se^2-}$. This is because the neutral form of selenocysteine is both a base and a diprotic acid: the amino group can be protonated and both the carboxylic group and the selenol can be deprotonated.[1]

The above graph taken from the chemicalize.com property calculation site[2] shows the distribution of the different forms of selenocysteine in aqueous solution. The underlying data has been calculated by chemicalize.com and is not experimental data. For the cell, neutral pH is most important. We can see that the most common form by far at neutral pH is $\ce{C3H7NO2Se}$, but a small quantity is present as $\ce{C3H6NO2Se-}$. Once selenocysteine has been incorporated into a protein, only the selenol function remains potentially acidic in aqueous solution; however, it’s $\mathrm pK_\mathrm a$ value is higher than neutral pH by three logarithmic units. Thus, we can assume that it will not be deprotonated notably.

The actual $\mathrm pK_\mathrm a$ values have been measured by Huber and Criddle.[3] These show that chemicalize.com’s predictions are rather bad. They determine the $\mathrm pK_\mathrm a$ values to be $2.01,5.24$ and $9.96$ for the carboxylic acid, the selenol and the amino group, respectively. Using these values, we expect selenocysteine to be present in the monoanionic form $\ce{C3H6NO2Se-}$ at cellular pH. A distribution graph based on these values (provided by Canadianer) can be seen below.

PubChem’s suggestion of $\ce{C3H6NO2Se}$ is wrong. If you add up valencies, you will notice that it is only possible as a radical structure. It would be acceptable if a negative charge had been included (which I don’t see on the site).

On the other hand while the chemical formula on Wikipedia or ChemSpider is correct and gives a neutral compound with all spins paired, this formula does not correspond to the species predominantly present in strongly buffered aqueous solutions.

Notes:

[1]: This way of writing it assumes the uncharged neutral form. In aqueous solution, the molecule is expected to adopt a zwitterionic neutral form with the carboxylate being deprotonated and the amino group being protonated. From this form, the carboxylate is the base while the amino group is one of the two acidic functions.

[2]: Unfortunately, I am not aware of a way to link directly to the calculation. However, this calculation is free if you register, since the molecule contains less than twelve heavy atoms.

[3]: R.E.Huber, R.S.Criddle, Arch. Biochem. Biophys. 1967, 122, 164–173. DOI: 10.1016/0003-9861(67)90136-1.


Any program for conformational searching for sugars?

Use RDKit

It sounds like you're only considering free or open source conformer generation tools. In that case, I highly recommend using RDKit, since the distance-geometry (and in turn ETKDG) methods will intrinsically generate all possible ring geometries. That includes not just the two chair forms you indicate, but all possible ring-puckering forms.

There are other free tools which do this, but RDKit has been repeatedly shown to be the best or among the best free conformer generation tools. (Based on current benchmarks.)

"High-Quality Dataset of Protein-Bound Ligand Conformations and Its Application to Benchmarking Conformer Ensemble Generators." J Chem Inf Model 2017 57, 529-539.
"Freely Available Conformer Generation Methods: How Good Are They?" J Chem Inf Model 2012, 52 (5), 1146–1158.

I'd advocate the newer ETKDG method with RDKit, since it seems to produce somewhat better geometries.


Why do we have an alkyl shift rather than epoxidation in a pinacol reaction?

Excellent question.

When the first Pinacol rearrangements were being studied, there was a suggestion that the product of the reaction was indeed the epoxide (migrations/shifts weren't well known reactions at this point in time, and methods of characterisaton were limited). The reaction that was being attempted is shown below (source: Wikipedia), however at this point there was even dispute over the structure of acetone, further adding to the complications.

  One of the first reports of a Pinacol rearrangement

Today, we of course know that Pinacol and semi-Pinacol rearrangements occur when vicinal diols are treated with protic acid (or Lewis acids) to give the carbonyl product as a result of a [1,2]- alkyl shift.

  Source: Strategic Applications Of Named Reactions. Wiley.

Looking at the mechanism (below), we first have a rate-determining loss of water via protonation-elimination. This generates the all important cationic intermediate.

The simple answer to your question is that rearrangement occurs faster than epoxidation (which is never observed to any significant extent), this is intuitive if we consider that:

Under the acidic conditions, the alcohol required to form the epoxide is fully protonated, and hence not a very good nucleophile (in agreement with R-OH generally being a poor nucleophile except in basic conditions where we get the anionic alkoxide RO(-)
The migration generates a more stabilised carbocation by the influence of the oxygen LP donating into the carbocation to form the resulting ketone.

    Source: Strategic Applications Of Named Reactions. Wiley.

As an aside, if you'd kept reading through Clayden by another page or so, you would have discovered that epoxides also undergo Pinacol-type rearrangements, so realistically even if the epoxide did form, it would simply re-open under the acidic/Lewis-acidic conditions before undergoing a rearrangement reaction.

  Source: Organic Chemistry, 1 ed. Oxford University Press (Warren & Clayden)


Why does the energy difference of a reversible process not equal that of an irreversible process in an adiabatic expansion?

I think the premise of the question itself is false: if you start with a given state (P1, V1) and undergo a reversible, adiabatic (hence isentropic) expansion, you can get to state P2, V2. But if you don't have the requirement of reversibility, I don't think it is possible to gain the same second state - there will be either internal heating/cooling, causing the expected final volume (or pressure, depending on the mechanism of expansion) to not be P2 or V2.


The answer should be A.

Since 3 is vinylic hydrogen, therefore it is least acidic.
Both 1 and 2 are allylic and are at the alpha position with respect to the carbonyl group.
Nitrogen is quite electronegative, therefore hydrogen 1 is more acidic than hydrogen 2. Hence the result.


Can someone explain my professor's answers for these alkene additions?

In example A, you are right that the cyclohexane ring is more stable than the cycloheptane ring. There are two choices for rearrangement in this case:

1) Ring expansion to form the cycloheptane ring

2) 1,2-hydride shift to yield the tertiary alkyl bromide

The second product will be more stable and thus be the major product

In example B, you are right, a ring expansion is possible, so there should be the ring expansion product too. You will have a mixture of products in this case (1,2-alkyl shift, ring expansion, and no carbocation rearrangement)

In examples C, E and F, it is possible for the H-A to add across the double bond either way, so again, a mixture of products would be obtained

A lot of times, only one product is shown for addition reactions such as hydrohalogenation; however, it is important to remember that there are often multiple products possible.

Chemists very rarely utilize reactions that involve a carbocation rearrangement because a mixture of products is obtained, which results in very poor yields


Why does the cathode ray tube only start glowing at low pressures?

At 1 atm. it is difficult create stable plasma, because breakdown voltage is very high.
Your electrodes are located at least 10-15 cm I think, and it is simply you do not have enough electrical potential between them to create initial plasma channel.

Then pressure goes down, voltage is fixed. When you reach breakdown voltage you get your glow (the Townsend discharge happens just as glow discharge becomes visible)-normal glow discharge. What happens? You ionized your gas, you have free electrons and ions, which are floating in different directions: ions are floating to negatively charged cathode and electrons flow to positively charged anode.

Because charge carriers are moving in space. They are rubbing against each other, part of the energy is dissipated and goes into heat and other is emitted and you see your glow. Ions are moving relatively slow, electrons, on contrary,  move very fast comparing to ions, because they are light weighted. So this fast free electrons, accelerated by a sufficiently strong electric field, give rise to electrical conduction through a gas by avalanche multiplication caused by the ionization of molecules by ion impact.

Then you continue to reduce your pressure more and more, less and less gas molecules can be ionized by free electron flow, less and les collisions are happening (less energy is dissipated), so the intensity of glow is going down, and finally it stops, because the breakdown voltage increased and overcome your applied voltage. (see Paschen Curve, lower is a pressure higher is the breakdown voltage.)

This is generally how it works.


As each of the groups surrounding the double bond are all identical there is no optical isomers of this molecule, as no matter which way you rotate or swap groups around you will always have the same molecule. therefore as it has no optical isomers it is neither cis nor trans


Electrolysis/Galvanic cell is basically redox reaction split in half-reactions, running on different electrodes. For example, hydrogen fuel cell has two electrodes. On one hydrogen donates electrons to electrode:

$H_2 = 2H^+ + 2e^-$

on another, oxygen drains electrons from electrode

$O_2 + 4 e^- + 4 H^+ = 2 H_2O $

The electrons travel from one electron to another via external electricity consumer. If electrons are forced to travel by external force, the reaction may be reversed. This ideas are utilized in electric chemical batteries, accumulators, fuel cells, industrial electrolyzers and so on.

This way, half-reactions are separated between different electrode. Running half-reaction by itself without other half elsewhere is problematic at best, because of electroneitrality violation. You'll need extreme voltage source and even then it will stop quickly.


How to solvethis problem related to redox titration?

Chromium(VI) is reduced to chromium(III) while oxidizing iodide to iodine. So one equivalent of chromate corresponds to 3 equivalents of iodine ($\ce{I}$, not $\ce{I2}$).

$$\ce{BaCl2 \cdot 2 H2O} \equiv \ce{BaCrO4} ~\mathrm{(chrome(VI))} \equiv \frac{3}{2}~\ce{I2} \equiv 3~\ce{S2O3^{2-}}$$

If you take this into account you should probably end up with 87.5% purity.


How did Thomson calculate the momentum of electron when he didn't know the mass of electron?

Thomson described the mechanical effect produced by cathodic rays on the first edition of its book Conduction of Electricity Through Gases (1903, pp. 501–502) and again later on the second edition (1906, pp. 629-630).
As you mentioned it refers to the Crookes experiment with the paddle wheel in which the electron beam collides with the mill coausing it to move (fig.below).

The electron beam carries a current ($I$) that can be measured and is equal to the number of electrons hitting the paddle every second (N) multiplied by the charge of the electron (e).
$$ I =N\cdot  e$$
Considering that the electrons are much smaller than the paddle you can consider that they bounce back with the same velocity (like when you hit a wall with a ball) and therfore the momentum transferred to the paddle per unit time ($p_{tot}$) is twice the momentum of the electron beam.
The momentum of the electron beam per unit time can be calculated with the well known formula $m\cdot v\cdot N$ with N to account for the number of collisions per unit time.
Therefore the total momentum per unit time transferred to the mill is:
$$ p_{tot} = 2m\cdot v\cdot N$$
This is equivalent to (multiplied and divided by the same quantity $e$):
$$ p_{tot} = 2\cdot \frac{m}{e}\cdot v\cdot (N\cdot e)= 2\cdot (\frac{m}{e})\cdot v\cdot I$$
Thompson considered an hypothetical best case scenario with a very high current of $\pu{10^{-5}A}$ carried by the electrons and a very high velocity of $\pu{10^8 m/s}$.
These are the highest reasonable values he could think could be applied in the Crookes' experiments (he even mentioned that the actual current measured was only $\pu{10^{-7}A}$.
The current can be easily measured while the velocity can be estimated from various methods using magnetic and electric fields (i.e. see section Charge-to-mass ratio of the Wikipedia where v is derived). And $\frac{m}{e}$ was already determined (around $\pu{6e-12Kg/C}$.
With those data, you can estimate that in the best case scenario (very high velocity and many particles hitting the paddle per unit time) the momentum transferred to the paddle per unit time is:
$$p_{tot} = 2\cdot (\frac{m}{e})\cdot v\cdot I = 2\cdot 6\cdot 10^{-12}\cdot 10^8\cdot 10^{-5}= \pu{1.2e-8 N} $$
This corresponds to $\pu{1.2e-3 Dyne}$ in the cgs units (units used by Thompson).
If you consider a paddle of $\pu{1cm2}$ then the pressure on the paddle is $\pu{1.2e-4 Pa}$, that is extremely small (an atmosphere is around $\pu{101 325 Pa}$!). Because of this it is reasonable to  attribute the movement to heating transfer.
As you see here, Thompson couldn't calculate the momentum defined as $m\cdot v$ but instead he calculated the  momentum per unit time  (that is a force)! He also considered the momentum carried by many electrons so that he could use the current in the formula of $p_{tot}$.


Factors controlling enzymatic phosphorylation of D-ribulose 5-phosphate

After looking at the website I found the reaction is based off of an enzyme called: phosphoribulokinase. In biological systems many reactions occur under the auspice of enzymes or proteins. These are incredibly large molecules made of an extremely long chain of amino acids which are a set of 26 or so small molecules that can 'hook' into the chain structure of a protein. Each amino acid has different properties and thus experiences different forces pulling it towards and away from other amino acids on the chain. What this means is the whole chain folds up into an extremely complex structure. For a protein to allow a reaction to occur it has a small region that the reacting molecule can hook into. There are then sets of atoms and molecules that act as scaffolds that allow atoms in the reacting molecule to be moved without the need for nearly as much energy as would be required without the scaffold. Now this is how it changes structure. Now whilst the protein can just as easily allow the molecule produced to be converted back to the original, this may be less energetically favourable which brings us to equilibrium chemistry involving dynamic forward and backward reactions. What makes this more complicated is that proteins often have separate 'hooks' for molecules to disable them. That is if a lot of substance A is around then the protein will be disabled and no longer convert B into A until more A is consumed and thus stops inhibiting the protein.


Mathematical equivalence of ionic-bond resonance structures and d-orbital participation in hypervalent molecules

You are essentially correct, but the distinction is the degree of bonding.

As you described, in a polarized or partially ionic bond, the atomic orbital of the more electronegative atom is lower in energy. Thus, the bonding molecular orbital is closer in energy to that atomic orbital, and the antibonding molecular orbital is closer in energy to the atomic orbital of the less electronegative atom. Because of this, the relative contribution of the lower energy AO to the bonding orbital is greater, and the contribution from the higher energy AO correspondingly lesser.

The distinction between "covalent" and "ionic" is therefore blurry. There's a continuum of degree of ionic bonding. In the extreme case of a completely ionic bond, there is no stabilization of the "bonding" MO relative to the lower energy AO. Furthermore, the higher energy AO no longer contributes at all. At that point, we no longer have bonding and antibonding MO's, but rather both are nonbonding orbitals comprised entirely of a single AO each.

But there are plenty of cases where there is just a little bit of stabilization, and we still describe the bond as "ionic". That's partly because it's something of a judgement call at what point we conclude that there is no bonding interaction. We can't solve the Schrodinger equation exactly, so we use computational approximations. Our computational methods aren't precise enough to tell the difference between a tiny bit and none at all. For the $d$ orbitals in the compounds in question, most methods predict at least a small participation of the $d$ orbitals in most compounds. But how much above 0 does it have to be before it's no longer correct to say that $d$ orbitals are not involved? We can say with certainty that the participation of the $d$ orbitals is nowhere near the $sp^3d^2$ view that used to be taught.

In your question, you described the compounds as "better represented" by ionic bonding "rather than utilization of $d$ orbitals."

You're correct that this distinction isn't either/or, but the key words are "better represented". Even if the $d$ orbitals contributed as much as 5% to the bonding MO, isn't the ionic representation "better"?


I can explain some of this, and actually some of it is explained on the Wikipedia page you referenced.

Most water models represent the oxygen using a Lennard-Jones potential, which encapsulates the repulsive interaction that stops the molecules getting too close (this is physically because the electron shells don't want to overlap each other) and a long-range attraction due to Van der Waals forces. The Lennard-Jones potential can be parameterized in several ways, two of which are:
$$
v_{\text{LJ}}(r)
= 4\epsilon \left[ \left(\frac{\sigma}{r}\right)^{12} -
\left(\frac{\sigma}{r}\right)^{6} \right]
=
\frac{A}{r^{12}} - \frac{B}{r^6}
$$
The first of these appears in your force field specification file.
The $\epsilon$ parameter is actually given in Kelvin, which is $\epsilon/k_{\text{B}}$, where $k_{\text{B}}$ is Boltzmann's constant.
Hence the note "this is K".
This may seem an odd choice, but there are some historical reasons.
However, you'll see on the same line,
in a comment for the user,
that this number is translated into SI units and also into old-time cgs units of kcal/mol.
The $\sigma$ parameter is also given in your file, in old-style Angstroms.
Just to be clear, this is the interaction between oxygen atoms,
hence it is described as the O-O Lennard-Jones term.

The second form appears on the Wikipedia page. The values of $A$ and $B$ appear in the table, in old-time units. If you do the necessary multiplications
you can check that (to within a few significant figures, anyway),
$4\epsilon\sigma^{12}$ and $4\epsilon\sigma^{6}$ do indeed agree with the
tabulated $A$ and $B$, taking $\epsilon$ in kcal/mol and $\sigma$ in Angstroms,
for the SPC/E model.

There are some redundant entries in your file: BEND and BOND. For a molecule like this,
one could allow the OH bonds to vibrate, and the HOH bond angle to vary,
and these terms would need some force constants and equilibrium bond lengths and
angles.
But SPC/E is a rigid model: the bond lengths and angles are fixed,
so it looks like the force constants ($K$) have been set to zero.

One might imagine that the H atoms deserve also to be represented as Lennard-Jones atoms,
but they are quite small, and in this model this aspect is neglected.
Hence we see that $\epsilon$ is zero for the H-O and H-H interactions.

So then, all that is needed to complete the model is the charge distribution,
and it looks like you've identified the three charges,
one on each atom,
characterizing SPC/E.

This isn't the most sophisticated model,
as you can see if you scroll down that page.

For more details concerning the input file,
I would hope that your software package and/or exercise instructions,
give enough information.
Hope this helps at least to understand the correspondence between the
numbers in your file,
and the numbers on the Wikipedia page.


Portable library to render 2D structural formulas as vector graphics from SMILES or InChI

You might want to have a look at Open Babel.

It is licensed under GNU GPLv2
It has bindings for Python
It has bindings for PHP
It has bindings for Ruby
It can read (and write) SMILES
It can read (and write) InChI
It   can write SVG
It can write a list of painter commands


Role of amalgamated zinc in Clemmensen reduction

As the illustration below would show, you can use either amalgamated zinc  (in a Clemmensen reduction) or hydrazine (in a Wolf-Kishner reduction). This reaction as drawn hides some of the secondary products which should be zinc peroxide and mercury oxide*

Metallic zinc is “activated” by mercury by “dissolving” it into individual atoms of zinc that are distributed evenly in the alloy. You're basically creating a zinc solution.

The reaction of zinc with HCl releases hydrogen gas.
However, when this reaction takes place with zinc amalgam, the individual hydrogen molecules that are formed by the reaction of zinc with HCl remain in a reactive state (sometimes called “nascent hydrogen”) and reacts with the ketone to initiate the reduction reaction.

So the mercury in zinc amalgam serves to “trap” the active hydrogen as it is formed, and allows it to attack the carbonyl compound rather than be released as a gas.

*Disclaimer: I have never actually performed this reaction and I could be dead wrong.


The test of metal hydroxides and metal oxides

Litmus paper does not test for either metal oxides or metal hydroxides; indeed, it will not change color at all. Try it: put dry $\ce{CaO}$ on dry litmus paper. Litmus does change color with pH+, the negative log10 ratio of $\ce{H+}$ ions to $\ce{OH-}$ ions (the ions may actually be more complex, such as $\ce{H3O+}$, but the idea is the same). This means the substance is in aqueous solution, i.e. in water.

Litmus also changes color with non-metals in solution, such as $\ce{NH3}$, which forms $\ce{NH4OH}$ in water, or with acetic acid, having an excess of $\ce{H+}$ ions.


Probably the easier way to explain it is to rely on the fact that if Li indeed had a ground state with $S = 3/2$, this would necessarily comprise four levels with $M_S = +3/2, +1/2, -1/2, -3/2$. This sort of splitting can be observed under a magnetic field, for example (the Zeeman effect).

Note that it's not possible for the Pauli principle to forbid $M_S = +3/2$ and $-3/2$ while simultaneously allowing $M_S = +1/2$ and $-1/2$. The Pauli principle doesn't mean that "this state can exist, but the electrons can't reach that state". It means that "this state simply cannot exist": because any state that violates the Pauli principle is equal to zero, which isn't a valid state. So it's not as if the $M_S = \pm 3/2$ states are there, and that the electrons simply choose to go into the $M_S = \pm 1/2$ states. The $\pm 3/2$ states are nonexistent. And consequently, because a $S = 3/2$ state necessitates the existence of $M_S = \pm 3/2$ states, it cannot exist either.

However, since that line of argument doesn't seem to fully convince you, you may want to go through the "exercise" of proving it. Unless I'm terribly wrong, this should be a completely foolproof (albeit terribly tedious) method. You can construct the full wavefunction for the Li atom as a Slater determinant (without loss of generality we choose $\alpha$ for the third spin):

$$\Psi = \frac{1}{\sqrt{6}} \begin{vmatrix}
1s(1)\alpha(1) & 1s(1)\beta(1) & 2s(1)\alpha(1) \\
1s(2)\alpha(2) & 1s(2)\beta(2) & 2s(2)\alpha(2) \\
1s(3)\alpha(3) & 1s(2)\beta(3) & 2s(2)\alpha(3) \\
\end{vmatrix}$$

so "all" we need to do is to apply $\hat{S}^2$ to this wavefunction and see what eigenvalue it returns. We know that

$$\hat{S}^2|\Psi\rangle = S(S+1)|\Psi\rangle$$

and so if $S = 1/2$ then the eigenvalue will be $3/4$; and if $S = 3/2$ the eigenvalue will be $15/4$. The $\hat{S}^2$ operator can be expanded as

$$\begin{align}
\hat{S}^2 &= (\hat{S}_1 + \hat{S}_2 + \hat{S}_3)^2 \\
&= \hat{S}_1^2 + \hat{S}_2^2 + \hat{S}_3^2 + 2(\hat{S}_1 \cdot \hat{S}_2) + 2(\hat{S}_1 \cdot \hat{S}_3) + 2(\hat{S}_2 \cdot \hat{S}_3) \\
&= \hat{S}_1^2 + \hat{S}_2^2 + \hat{S}_3^2 + 2(\hat{S}_{1x} \hat{S}_{2x} + \hat{S}_{1y}\hat{S}_{2y} + \hat{S}_{1z}\hat{S}_{2z}) \\
&\quad + 2(\hat{S}_{1x} \hat{S}_{3x} + \hat{S}_{1y}\hat{S}_{3y} + \hat{S}_{1z}\hat{S}_{3z}) + 2(\hat{S}_{2x} \hat{S}_{3x} + \hat{S}_{2y}\hat{S}_{3y} + \hat{S}_{2z}\hat{S}_{3z})
\end{align}$$

We already know what the individual operators $\hat{S}_i^2$, $\hat{S}_{ix}$, $\hat{S}_{iy}$, and $\hat{S}_{iz}$ do to the basis states $|\alpha(i)\rangle$ and $|\beta(i)\rangle$. Bearing in mind that operators don't act on different degrees of freedom, you can work out what this monster operator does to the Li wavefunction.

If you look online, there are examples of how to perform this calculation for a two-spin system (e.g. "prove that the singlet and triplet wavefunctions are eigenstates of $\hat{S}^2$"). The extension to a three-spin system is not difficult, but is extremely tedious, which is why I gave up on writing it out in full. But I am fairly certain that at the end of the day you will find the eigenvalue of $3/4$.


Will a film of oil on water retard evaporation (of water) and prevent air and atmospheric impurities from diffusing into it?

I don't know about oil (fun fact, putting oil on water was a traditional approach to calming waves, thus the phrase "pour oil on troubled waters").  But cetyl alcohol has been used to decrease evaporation from reservoirs, which might be what you heard about.  For instance, here is a report on this, with lots of references.  They used a mix of steryl and cetyl alcohols (long-chain alcohols that will form a monolayer on the water).


How do chemists verify certain phenomena like 1,2-hydride shifts, 1,2-methyl shifts and ring expansions?

I wanted to put this as a comment but I don't have enough rep :p

It happens the other way round, first the products are obtained, then the mechanism is proposed.

The point is, according to me that once it has been verified which product is major and which minor, chemists may try to come up with a suitable explanation for the product obtained, on paper. They may try some things, come up with possible mechanisms and sometimes even phenomena like hydride shift, methyl shift etc. Once they come up with such an explanation, they try to apply it to other related problems and upon reaching the right solution every time, they can say this is what might have happened to a high degree of certainty.

So you see, chemistry is also sometimes glorified problem solving :p


As much as I understand, there is nothing wrong with the question, if your translation is correct. Let's see the reaction:

$$\ce{2 C2H4 (g) + O2 (g) -> 2 C2H4O (g)}$$

The flow reactor was fed with constant flow rates of $\ce{C2H4}$ at $\pu{200 kmol\:h-1}$ and $\ce{O2}$ at $\pu{90 kmol\:h-1}$. Only $6\%$ of the ethene reacts in the reactor. That means the reaction rate of the reactor is $\pu{12 kmol\:h-1}$ per $\ce{C2H4}$. Therefore, assuming a perfect world, $\pu{6 kmol}$ of $\ce{O2}$ would be consumed in each hour to produce $\pu{12 kmol}$ of $\ce{C2H4O}$ as well (see the chemical equation).

a) Thus, the gas outlet of the reactor would release 98% of unreacted $\ce{C2H4}$ ($0.98 \times \pu{188 kmol}$) and $\ce{O2}$ ($\pu{84 kmol}$), and $\ce{C2H4O}$ ($\pu{12 kmol}$) in each hour as a mixture. The sum ($188 \times 0.98+84+12 \approx \pu{280.2 kmol\:h-1}$) is the rate of gas flow out of filtrator of the flow reactor. Of course, this is assuming filtrator does not filtered off any of these materials (it wasn't clear on the translation). If filtrator separates the product and oxygen from the mixture (e.g., by gas partition by difference in melting points; melting points of oxirane and ethene are ~$\pu{161 K}$ and ~$\pu{104 K}$, respectively, and melting point of oxygen is ~$\pu{54.4 K}$), the outlet rate of $\ce{C2H4}$ out of the filtrator would be $\pu{188 \times 0.98 kmol\:h-1} \approx \pu{184.2 kmol\:h-1}$ (with no recycling of $\ce{C2H4}$). If  the flow of gas out of the filtrator means only the gases pass through the filtrator, that should be only oxygen and oxirane. Thus, the outlet rate of gases out of the filtrator would be $\pu{(84 + 12) kmol\:h-1} = \pu{96 kmol\:h-1}$.

b) I believe, the recycling of $\ce{C2H4}$ means unreacted $\ce{C2H4}$ is adding to original rate of $\pu{200 kmol\:h-1}$. Thus, rate of $\ce{C2H4}$ fed into reactor after first hour is: $\pu{(200+184.2) kmol\:h-1} = \pu{384.2 kmol\:h-1}$. Only 6% of it ($\pu{23.05 kmol\:h-1}$) reacts with $\pu{11.53 kmol\:h-1}$ of oxygen to give $\pu{23.05 kmol\:h-1}$ of oxirane. Therefore, outlet rate is:

$$\left((384.2 - 23.05) + (90 - 11.53) + 23.05\right) \ \pu{kmol \: h-1} = \pu{462.67 kmol \: h-1}$$

This would be the outlet rate without filtration. If you have to consider only 98% of $\ce{C2H4}$, then the rate would be:

$$\left((384.2 - 23.05)\times 0.98 + (90 - 11.53) + 23.05\right) \ \pu{kmol \: h-1} = \pu{455.45 kmol \: h-1}.$$


Comparing ionic character of group 1 elements

Well, the second source is certainly more wrong. There is a major difference between ionic character of a bond and the stability (lattice energy) of an ionic lattice. $\ce{LiH}$ is certainly more covalent than $\ce{CsH}$ (meaning $\ce{CsH}$ is more ionic), which can be justified simply by comparing charge densities on $\ce{Li}$, which is very high and draws electron density from the $\ce{H}$ atom, forming a more covalent bond.


Difference between sodium ion and a transition metal ion dissolving in water?

There is a slight difference between solvated ions and aqua complexes: The nature of the bond.

When solvating a sodium ion, the electron pairs of the oxygen in water never really interact with the sodium in a way that creates a bond of some sort: It's all just electrostatic interactions.

However, when a transition metal ion gets surrounded with these water molecules, the lone pair electrons of the oxygen form a bond with the ion. This is because they interact with the $d$ orbitals of the transition metal (which usually also contains some electrons), thus changing the energetic state of the ion itself. First and second group elements don't have that energy-stabilizing effect because they don't have electrons in the orbitals that get split up by the formation of the complex.

It is not unimaginable that there then exist further electrostatic interactions between the transition metal aqua complex and the surrounding water molecules, thus leading to the solvation of said complex.


Why is the melting point of TiCl4 lower than TiCl3

-The high melting point of  $\ce{TiCl3}$
is mainly due to the short bond length $\ce{Ti-Ti}$ , which infer
 a strong metallic character to these bonds between titanium ions in  $\ce{TiCl3}$.

-While in $\ce{TiCl4}$, the main intermolecular forces are of van der Waals type.


why do we denature proteins before mass spectroscopy?

Trypsin is an enzyme, and has to bind to a target protein in a specific way in order to be active. Without first denaturing your protein, most of the protein will be conformationally/sterically inaccessible to trypsin, and you will only get partial digestion (or maybe no digestion at all).


How to interpret TLC data regarding the purity of one’s sample

TLC is a great and poor method at the same time. It is great because it is quick and accurate, and it is poor because it can only show substances that can be coloured in some way and are sufficiently non-volatile to not evaporate away.

You can be rather sure of the following points:

Your impure sample does not contain any more free acid since a spot of the same $R_f$ value is missing.
Your impure sample contains a probably very high amount of aspirin, since the spot has the same $R_f$ value.

However, you cannot be sure about:

Impurities that evaporated away during spotting, development or colouring of the TLC plate
Impurities that simply don’t show up in your method of colouring (e.g. due to the lack of a benzene ring for UV fluorescence).
Solvent impurities that will never show up due to both points above.

To prove purity you would need to resort to different analytic methods such as melting point (historically), HPLC-MS or GC-MS, NMR, etc.


$\ce{N5P3}$ is more commonly written as $\ce{P3N5}$, and known as triphosphorus pentanitride. It's a crystalline solid at ambient conditions and not a molecular compound. From the first publication that reported the pure compound and its structure [1]:

In the solid a three-dimensional cross-linked network structure of corner sharing $\ce{PN4}$ tetrahedra has been identified with 2/5 of the nitrogen atoms bonded to three P atoms and 3/5 of the nitrogen atoms bonded to two P atoms.

Only the combination of spectroscopical and diffraction methods, especially electron microscopy for structural analysis, enables a detailed structural and crystallographic characterization of $\ce{P3N5}$ as no conventional single-crystal data for structure determination have been available for this compound.
A representative for an unique binary structure type was found for the first time which combines a tetrahedral structure with the very rare molar ratio of 3:5 for the constituent elements.

References

Schnick, W.; Lücke, J.; Krumeich, F. Chem. Mater. 1996, 8 (1), 281–286 DOI: 10.1021/cm950385y.


What are the ways to make radioactive water drinkable again?

The water itself isn't radioactive, rather particles dissolved in it, because water is a quite good solvent. Therefore, filtration (e.g. with activated carbon) or precipitation can remove such particles. Also a ion exchanger can remove radioactive species. But this can take quite long, especially if the needed amount is large.


Is hydrogen bonding separate from permanent dipole interactions?

Consider the three hydroxyphenols catechol (ortho-hydroxyphenol), resorcinol (meta-), and hydroquinone (para-).  All three engage in hydrogen bonding with their hydroxyl groups.  But they differ in polarity.  Hydroquinone, with its hydroxyl groups symmetrically placed on opposite corners of the benzene ring, has a limited (average) dipole moment between two conformations, one polar and one nonpolar, whereas the other two isomers have an additional dipole from their hydroxyl groups being nonsymmettically located and have larger overall dipole moments.


Predicting the generation of singlet oxygen

Molecular orbitals are actually, to the best of my knowledge, not really helpful in explaining this observation. Instead, a much more fundamental concept does a much better job: the conversion of (electronic) spin.

The reacion you mentioned can be analysed in terms of spin states. Nobody will disagree that hydrogen peroxide, sodium hypochlorite, sodium chloride and water are singlet compounds. Therefore, using generic Greek letters as placeholders for the different compound to illustrate the point that their true doesn’t matter much, we can write the equation as:

$$\ce{^1\unicode{x391}\ + ^1\unicode{x392}\ -> ^?\Gamma\ + ^1\Delta\ + ^1\unicode{x395}}\tag{1}$$

We see that we are reacting two singlet compounds two three compounds and two of the products are also singlets. This means, every spin on the left-hand side of the equation is paired and most of the spins on the right-hand side, too. To generate a triplet product, we would need to flip a few spins but that is a slow process on the timescale of a chemical reaction. (Note that the conversion from singlet oxygen back to triplet can be observed as a faint red glow demonstrating that it is not instant but delayed.) Thus, in the moment of the particle collision, generating all-singlet products from all-singlet reactants (if only a single product potentially can have an unpaired spin) is the only option. Thus:

$$\ce{^1\unicode{x391}\ + ^1\unicode{x392}\ -> ^1\Gamma\ + ^1\Delta\ + ^1\unicode{x395}}\tag{2}$$


For IUPAC nomenclature, does the number of multiple bonds take precedence over the chain length?

According to the new recommendations of IUPAC nomenclature rules, when choosing the senior parent structure, you have to first consider the senior parent structure, which has the maximum number of substituents corresponding to the principal characteristic group or senior parent hydride in accord with the seniority of classes. Then, if there is a choice, choose the principal chain, which has the greater number of skeletal atoms. Next in order is the chain, which has the greater number of multiple bonds, and then has the greater number of double bonds (This is according to Blue Book in 2013, which was displayed in 5 steps you have mentioned from Wikipedia).
Note that the new IUPAC recommendations (Blue Book in 2013) on order of chain length and number of multiple bonds have been changed (You can find more details here and here):

In acyclic parent structures the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed. Thus, the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second.

For example consider following structures:

In all first three structures, the longest chain has chosen over the multiple bonds (the old names are in orange). In the typical case, the principle group is $\ce{-CHO}$ and hence hence to hexanal is the principal structure which has $\ce{-OH}$ substituent and longest chain. Thus, the name is 2-ethynyl-5-hydroxyhexanal.


What's the usual agent added to discourage drinking of hazardous substances

It is denatonium benzoate. This comes from wikipedia page of denatonium:

Denatonium, usually available as denatonium benzoate (under trade
names such as Denatrol, BITTERANT-b, BITTER+PLUS, Bitrex or Aversion)
and as denatonium saccharide (BITTERANT-s), is the most bitter
chemical compound known, with bitterness thresholds of 0.05 ppm for
the benzoate and 0.01 ppm for the saccharide. It was discovered in
1958 during research on local anesthetics by MacFarlan Smith of
Edinburgh, Scotland, and registered under the trademark Bitrex.

With respect to discouraging drinking of hazardous substances it again comes from wiki:

The bitterness of the compound guides most applications of denatonium.
Denatonium benzoate is used to denature ethanol so that it is not
treated as an alcoholic beverage with respect to taxation and sales
restrictions.


Comparing acidity of anilinium ion and para-fluoro anilium ion

The best way to compare acidities of organic compounds is by drawing out the conjugate bases and figuring out which one is more stable. Molecules with more stable conjugate bases are more acidic, so doing the same here:

The conjugate bases are aniline and para-fluoro aniline respectively. The $\ce{-NH2}$ group shows +M (mesomeric effect), so on drawing out the resonance structures we find that the second resonance structure is more stable as the resulting negative charge is inductively withdrawn by fluorine group.

For halogens the inductive effect dominates over their positive mesomeric effect, especially in the case of fluorine (because it is small and the lone pairs experience a high effective nuclear charge,in case of large atoms like chlorine, bromine and iodine, the outer electrons are held weakly)


Why would sulfuric acid turn gray/black?

The only reason I can think of that would cause the solutions would blacken over time is if the solutions reacted with organic substances. Sulfuric acid is an extremely strong oxidizing agent, and dehydrates many organic compounds to produce carbon in the form of graphite, and water. For example, the reaction of sulfuric acid with paper (cellulose) is:

$$(\ce{C6H10O5})_{n}\ce{(s)} \xrightarrow{\ce{H2SO4}} {6n~\ce{C(s)} + 5n~\ce{H2O(l)}}$$

and can be seen in the following photo:

If a considerable quantity of organic material was reacted in your solutions, the $[\ce{H+}]$ in your solutions would actually decrease, so I would recommend titrating one of your solutions to see if this has changed since you bought them.


Do certain cumulated dienes present geometric isomerism?

Properly speaking, these sorts of allenes/cumulated dienes do not exhibit geometric isomerism in the way that alkenes can. Rather, when each of the two respective carbons bear two different substituents, the molecule is chiral and will exhibit enantiomerism. More specifically, these asymmetric allenes exhibit axial chirality. This is consequent to the two consecutive $\pi$-bonds being mutually perpendicular, which results in the molecule lacking any mirror plane of symmetry.

Generally, an even number of successive double bonds in a cumulene produces a potential axis of chirality (again, assuming each carbon at the end of the double-bond system bears two different substituents), while an odd number results in the substituents at the ends of the double-bond system being coplanar, which permits geometric isomerism akin to that seen in alkenes.


How can I perform structure alignment on an inorganic species?

Inorganic structures – that could be proper solid or molecular metalloorgnic. For the later one you can use xp (program written G.M.Shedrick - comes with Bruker diffractometers). If you have running license of Cambridge Crystal Data Base there is program called IsoStar, and I think that it should do the job. I haven't used it my self but apparently Olex2 can make aligning of fragments (this program is free so that's the place were you can start). I'm not quite sure but I think that old GROMOS code could also make a fit of orthogonal coordinates.


Due to their low mass, protons can tunnel through barriers, they don't always have to go over the barrier in an activated process.  Proton tunneling becomes particularly important at low temperatures where the available thermal energy may be inadequate for an activated process (e.g. over the barrier) to occur.  A nice, brief and not overly-mathematical introduction to tunneling can be found here.

dashed line - thermally activated process

solid line - tunneling process

A second reason why proton transfer is so facile relates to hydrogen bonding.  If hydrogen bonding is present, then the proton only needs to elongate one bond (the initial bond) while simultaneously shortening the hydrogen bond in order to effect transfer from one molecule to the next.  This energetically compensating process reduces the activation energy required for proton transfer.  In solvents with hydrogen bonded clusters or chains (water would be one such example), this process can be repeated within the cluster thus allowing the proton to cover large distances (from one side of the cluster to the other) on a very short timescale.  In other words, only minor mass movement is required; by simply shortening some bonds a bit, and slightly elongating others, a proton at the left side of a group of molecules (top row in the following figure) can "reappear" on the right side of the chain (bottom row) extremely rapidly. This process is referred to as the Grotthuss mechanism after its discoverer.  Some of the finer points of the mechanism still constitute an active areas of research.


Basic chemistry of the Shinoda test for flavonoids?

I found one source indicating that the reaction involves the conversion of the flavonoid to the corresponding anthocyanidin. The structure of that resulting anthocyanidin contains an extended system of conjugated $\pi$-electrons, which constitute a chromophore. As it happens, the electrons of molecules containing conjugated $\pi$ systems have energy level transitions corresponding to absorbance in the visible (and also ultraviolet) region of the electromagnetic spectrum. This accounts for their perceived color.

As for the precise mechanism involved, I was not able to find any sources whatsoever. It's obvious that the flavonoid is reduced, with the magnesium metal being the reducing agent. It could be the process occurs by single electron transfer (SET) with radical and anionic intermediates, or via a cyclic transition state and carbenoid intermediates with coordination of the magnesium, or something else entirely. In any case, the purpose of the acid is likely to activate the carbonyl of the flavonoid to reduction by protonating it, and providing a source of $\ce{H+}$ ions for the subsequent formation of water.

(I find it unlikely that a definitive mechanism for this reduction has been determined, if only because the mechanisms of many similar reductions involving metals and mineral acids widely used in synthetic organic chemistry have yet to be entirely resolved.)


What is the reaction mechanism of nitrogen leaving from diazo compounds?

The reaction is $\ce{S_N1}$

  Among the evidence for the $\ce{S_N1}$ mechanism with aryl cations as
  intermediates, is the following:

  The reaction rate is first order in diazonium salt and independent of the concentration of Y.
  When high concentrations of halide salts are added, the product is an aryl halide but the rate is independent of the concentration of the
  added salts.
  The effects of ring substituents on the rate are consistent with a unimolecular rate-determining cleavage.
  When reactions were run with substrate deuterated in the ortho position, isotope effects of $\sim 1.22$ were obtained. It is
  difficult to account for such high secondary isotope effects in any
  other way except that an incipient phenyl cation is stabilized by
  hyperconjugation, which is reduced when hydrogen is replaced by
  deuterium.

  That the first step is reversible cleavage was demonstrated by the observation that when  $\ce{Ar^{15}N+ #N}$ was the reaction species,
  recovered starting material contained not only $\ce{Ar^{15}N+ #N}$ but
  also $\ce{ArN^{+} #N}$This could arise only If the nitrogen breaks
  away from the ring and then returns. Additional evidence was obtained
  by treating $\ce{PhN+ #N^{15}}$ with unlabeled $\ce{N2}$ at various
  pressures. At $\pu{300 atm}$, the recovered product had lost $3 \%$ of
  the labeled nitrogen, indicating that $\ce{PhN2+}$ was exchanging with
  atmospheric $\ce{N2}$

Source: March's Advanced Organic Chemistry: Aromatic Substitution: Nucleophilic and Organometallic


Why is the conjugate of a strong oxidizing agent a weak reducing agent?

It is not the magnitude of the standard cell potential that indicates whether or not the cell reaction will be spontaneous, but the sign. In the same way a reaction is only spontaneous if $\Delta G < 0$, a substance only oxidizes/reduces if the corresponding cell potential is positive.


Using baking soda paste (alkaline) to "neutralize" honeybee venom (acidic). Is this scientifically sound?

Honeybee venom is acidic. It contains formic acid, also known as $\ce{HCOOH}$. However,
Rubbing baking soda probably isn't really reducing the pain because

The formic acid isn't causing the pain
The venom is injected under tissue
The ratio of venom to baking soda is way out of proportion

For more information, visit this website.


Which single organic molecule has strong absorbance throughout the visible range?

Quite a few dyestuffs will have strong absorption across the visible range of wavelengths, though few will have an entirely even absorbance. This won't matter much is they are concentrated enough and absorb most of the light at each wavelength.

Acid Black 1, for example, shows strong absorbance from the UV to just under 700 nm, which is most of the visible. Absorbance is stronger in the red but overall absorbance is large enough that it is sometimes used in black inks. See this paper for a sample absorbance spectrum:

There are quite a few other dyes with strong absorbance at most visible wavelengths and whether they appear black depends on the concentration.


Why are all 3 of propane, butane and isobutane used in deodorant sprays?

Bu/iBu or Bu/iBu/Pr mixtures are easier/cheaper to get instead of pure components as Ivan says, as there is extra effort for their separation, not the extra effort for their mixing.
Butane(Bu)/isobutane(iBu)/propane(Pr) boiling points are $\pu{0 ^\circ C}$, $\pu{-11  ^\circ C}$ , $\pu{-33  ^\circ C}$  respectively. Real boiling points would be higher than boiling points of pure hydrocarbon mixture, as deodorants are solution of active components in volatile liquid hydrocarbons, what increases the boiling point.
Propane (and isobutane too to a lesser extent) shifts usability toward lower temperatures, similar as Pr/Bu mixture of liquefied gas fuel cartridges for outdoor usage. Butane itself does not work well around $\pu{0 ^\circ C}$ either as propelent either as fuel, as it has not enough pressure to overcome the external one.
The goal is to provide the vapour pressure within the given pressure range for the given operational temperature range. Pure compounds have usually too low vapour pressure at the minimal temperature or too high vapour pressure at the maximal temperature.
Another reason is the control of evaporation rate of the solvent when it is being sprayed out. Different compositions of deodorant matrix may prefer different rates of hydrocarbon evaporation and different related cooling effect.


It is a semantics question with an open ended discussion. If you recall the old problem "How many angels can dance on a pinhead? Medieval problem, this issue here is similar.

Basically in chemistry, all the terminology is controlled or endorsed by the International Union of Pure and Applied Chemistry (IUPAC). They have a very open ended definition of a salt: "A chemical compound consisting of an assembly of cations and anions."

If we follow some really old chemistry texts (>1850s), yes an acid was defined as a salt of hydrogen ion, i.e., an acid is a salt whose electropositive element is hydrogen Definition of an acid.

Did this definition by a few chemists become popular? The simple answer is no. Not in a general sense.

The modern interpretation of an acid, again from IUPAC is, "A molecular entity or chemical species capable of donating a hydron (proton) or capable of forming a covalent bond with an electron pair."
If you have heard of ionic liquids, they consist of a large organic cation and an anion, this is a salt as well.

Then you have so-called acid salts, such as sodium hydrogen sulfate e.g.,  $\ce{NaHSO4}$. Their pH is very acidic and these "salts" will neutralize bases like a typical acid.

Thus one may loosely say, that all acids may be considered as salts but not all salts are acids.


Why is HSO4- shown as an example of a weak acid instead of HSO4?

The bisulfate molecule doesn't exist in solution. I can only find references to the hydrogen sulfate molecule as part of salts, which implies that hydrogen sulfate is usually found as an ion.

The bisulfate (hydrogen sulfate) ion is indeed an acid. It is a relatively strong weak acid too, with a $\ce{K_a}$ value of $1.2 * 10^{-2}$.

Part of the reason for its acidity has to do with its electronegative oxygens isolationg electron density away from the hydrogen atom bonded to the oxygen. This makes the hydrogen more partially positive and thus more reactive (more likely to be taken by a base).


How to calculate concentrations and what is the concept of 'mole'?

Concentration of a liquid can be given by the formula C=N/V, where V is the volume of the solution and N is the number of moles present in it.

To help you understand exactly what a 'mole' is, think of it like this: when you ad a certain amount of powder to water, say sugar or a fruit drink, how many spoons/ grams of powder you add is the number of moles. (1 mole is actually 6.02214129 * 10^23 particles of the substance)

The concentration can be determined by how much water you add to the drink: If you add more water, the solution is said to be diluted, since the ratio of solute:solvent is lower. So in a drink with 30% concentration, the ratio would be 3:7, 3 moles of powder for 7 moles of water. The concentration can ONLY BE CHANGED if the amount of water changes. The ratio remains same throughout, so concentration would be 30% if you take 20ml, 10ml, or any measurement of volume.

Hope this train of thought helped your mind to get the reasoning :)


Low-lying, empty antibonding orbitals will suffice.  The reaction of carbon dioxide and water to form carbonic acid serves as a parallel example.  Water donates one of the lone pairs on oxygen towards bond formation, water is the Lewis base in this case.  Carbon dioxide has empty, low-lying $\ce{\pi}$* orbitals that serve to accept these electrons; carbon dioxide is a Lewis acid in this example.


How is sulphur trioxide formed and why is it stable?

Synthesis

Synthesis of $\ce{SO3}$ can be done by heating $\ce{CaSO4}$ with $\ce{SiO2}$, which produces $\ce{CaSiO3}$ and $\ce{SO3}$. Another way of synthesis is the heating of $\ce{NaHSO4}$ to $\ce{Na2SO4}$, $\ce{SO3}$, and water. (source)

As to why $\ce{SO3}$ is stable...

Sulfur forms an expanded octet. That means that it doesn't really obey the octet rule, allowing it to take on extra electrons. Sulfur is a 3rd-period element; hence it can use its 3d orbitals to make more than 4 bonds. Wikipedia has a page on hypervalency, and from an answer here:

  Regarding the octet concept, it's only a guideline at best, and nothing more. I wouldn't put too much stock into the concept. If you insist on using it, then only apply it to the following elements: C,N,O, and F.

EDIT:

Users such as @Mithoron have pointed out that this analysis is not correct. Papers seem to show that there is no d-orbital utilization. According to this comment:

  ...the doubly bonded picture is a very crude oversimplification, though. -@Martin

This enlightening image

shows that sulfur trioxide is actually a resonance of several different structures, and should be more accurately expressed as several single bonds with a dashed line to show that the bond order is not actually two. This also solves the question as to why it is stable with apparently six bonds, as in reality it only has four bonds.


Would electrolysis work with a conductive (but waterproof) divider between anode and cathode?

A cell with two electrodes separated from each other and an aluminum divider plate in the middle is really two cells: one has an aluminum plate as a cathode, and the other cell has the same aluminum plate as the anode.

In this situation, the solution does not directly connect the external anode and cathode; the aluminum divider separates the solution into two separate cells.

When you pass current (electrons) thru this arrangement, ions travel. Protons (or other cations) travel toward the negative electrode. In the cell attached to the external positive source, protons move away, and discharge at the aluminum. Simultaneously, protons in the adjacent cell move away from the aluminum and discharge at the stainless steel cathode. Hydroxyls (or other anions) do the reverse.

You will need twice the cell voltage to get a high current, and hydrogen will be evolved from the external cathode (-) and the farther side of the aluminum divider plate. Oxygen will be evolved from the other side of this aluminum plate and from the external anode (+).

Now make a small change: let the aluminum divider go from the top of your case to below the water line, say, halfway down. Then ions can travel thru the solution. This is now just one cell. Hydrogen will be evolved at the (one) cathode and oxygen will be evolved at the (one) anode. Your separation of the gas phase by the aluminum divider allows hydrogen to be collected separately from the oxygen. BTW, you might as well use stainless steel for this divider, since it is used for the electrodes.

When collecting the gasses, hydrogen will be evolved at twice the rate of oxygen. If you build up pressure, the cell will respond by pushing liquid from the cathode compartment into the anode compartment to equalize the pressure. You may have to use a check valve or some kind of flow equalizer to keep liquid from spurting  out of one of the gas outlets.


Finding chiral carbons in aromatic or cyclic compounds

If an atom is part of a double or a triple bond it is never a stereogenic center because then this atom and the atoms bound to it form something planar, so that you can always find a symmetry plane.
But an atom that is part of a double bond can be prochiral, if it can be converted from to a stereogenic center in a single step.


How should I think about potential energy surfaces for reactions of complex systems?

In order to understand what the PES really is, it is useful to see where this concept comes from.

A system of $M$ nuclei and $N$ electrons should be described by a multidimensional wavefunction $\Psi(\vec{r}_1,\dots,\vec{r}_N,\vec{R}_1,\dots,\vec{R}_M)$, where $\vec{r}$ denotes electronic coordinates and $\vec{R}$ denotes nuclear coordinates (we neglect spin for simplicity). Within the Born-Oppenheimer approximation, the total wavefunction is split in an electronic part $\phi$ and a nuclear part $\chi$:
$$
\Psi(\vec{r}_1,\dots,\vec{r}_N,\vec{R}_1,\dots,\vec{R}_M) = \phi(\vec{r}_1,\dots,\vec{r}_N;\vec{R}_1,\dots,\vec{R}_M)\chi(\vec{R}_1,\dots,\vec{R}_M)
$$
where the electronic wavefunction depends parametrically on nuclear positions (as it will be clear in a moment).

If you inject the Born-Oppenheimer ansatz in Schrödinger equation for a system of interacting nuclei and electrons and you integrate out electronic degrees of freedom you get an equation for the nuclear wavefunction only
$$
\left[ -\sum_I \frac{1}{2M_I} \nabla^2 + \sum_{I<J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|} + E(\vec{R}_1,\dots,\vec{R}_M) \right]\chi = E_\text{tot} \chi
$$
where $E(\vec{R}_1,\dots,\vec{R}_M)$ is the solution of the electronic problem
$$
\left[ -\sum_i \frac{1}{2}\nabla_{i}^2 +   \sum_{i<j} \frac{1}{|\vec{r}_i-\vec{r}_j|} -\sum_{i,I} \frac{Z_I}{|\vec{r}_i-\vec{R}_I|} \right]\phi = E(\vec{R}_1,\dots,\vec{R}_M)\phi
$$
As you can easily see the solution of the electronic problem (both the energy and the wavefunctions) depend parametrically on nuclear coordinates. The potential energy surface is the potential energy appearing in the equation of the nuclear wavefunction, i.e.
$$
U(\vec{R}_1,\dots,\vec{R}_M)=E(\vec{R}_1,\dots,\vec{R}_M) + \sum_{I<J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|}.
$$
It is clear that finding the PES is a formidable task, since you have to solve the electronic problem for each possible set of nuclear coordinates. So, yes the PES is a multi-dimensional function of nuclear coordinates. If you have $M$ nuclei, your PES will be a function of at least $3M-6$ coordinates (once you remove translation and rotation of the whole system).

Since a multi-dimensional function is impossible to visualize, usually we study the PES by considering few interesting degrees of freedom (as the O-C distance and O-C-I angle in your example). Therefore, as you say, you can think as if you are exploring the PES locally and in a reduced space.

ADDITIONAL INFORMATIONS

Since the nuclear wavefunction is sharply picked, (heavy) nucley can usually be considered as classical particles. Thus instead of the nuclear quantum Hamiltonian
$$
H_n = -\sum_I \frac{1}{2M_I} \nabla^2 + \sum_{I<J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|} + E(\vec{R}_1,\dots,\vec{R}_M)
$$
you can consider its classical counterpart
$$
H_n^\text{cl} = \sum_I \frac{\vec{P}_I^2}{2M_I} + \sum_{I<J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|} + E(\vec{R}_1,\dots,\vec{R}_M) = \sum_I \frac{\vec{P}_I^2}{2M_I} + U(\vec{R}_1,\dots,\vec{R}_M)
$$
where $\vec{P}_I$ is the momentum of the nucleus $I$. From the laws of classical mechanics you can easily verify that the force acting on nucleus I is simply given by
$$
\vec{F}_I = -\nabla_{\vec{R}_I} U(\vec{R}_1,\dots,\vec{R}_M).
$$
So even if you consider the nuclei as classical particles, the PES has the same meaning, i.e. is the potential felt by the nuclei. Here you can easily apply a classical picture. A transition state is simply a state along your reaction coordinate where you have a local maxima of the PES for the reduced coordinates you are considering(as the C-O bond in you example).

I never looked to this into details but I think computational programs climb the PES uphill from the initial and final state of your system (i.e. they follows -$\vec{F}_I$) in order to find the TS.


What is the oxidation state of oxygen in Na2O2?

Oxygen can take multiple oxidation states. This particular compound is sodium peroxide.

You're right that usually oxygen has a charge of -2, but in this case, there's no way that each $\ce{Na}$ can have an oxidation state of +2.

So you then work backwards, deciding if it's $\ce{Na+}$ then you have +2 from the sodium, and oxygen must have an average oxidation number of -1 per oxygen atom.

Besides peroxides ($\ce{O2^{2-}}$) there are also superoxides ($\ce{O2^{-}}$).

It's not surprising that these compounds are usually very reactive.


Clarifications about the mechanism of the Wittig reaction

1) Why is only triphenyl phosphine used here. Why not some stronger
  nucleophile like say trimethyl phosphine?

Trialkyl phosphines can particpate in the Wittig reaction just fine, but they create some other problems.  Let's say you were going to create the phosphonium salt from ethyl iodide and trimethyl phosphine.  It would form in high yield, but in the next step where you deprotonate it and form the ylide you would have a problem.  You could form the ylide from deprotonation of the desired ethyl group, or you could form an ylide by deprotonating any of the 3 methyl groups.  This would give rise to a mixture of ylides and consequently a mixture of products.  On the other hand, there are no protons on triphenyl phosphine that can be removed once the phosphonium salt is formed.

The following additional reasons also favor usage of triphenyl phosphine:

it is cheap
it is very cheap
it is a solid and easy to handle and weigh out, many of the smaller trialkyl phosphines are liquids
triphenyl phosphine is less toxic and doesn't smell as bad as many of the smaller trialkyl phosphines

  2) Why would the lone pair on P attack the C of the alkyl halide in the
  first place? Isn't the lone pair in resonance with the benzene ring?

Triphenyl phosphine has the aromatic rings arranged in a non-planar, propeller-like geometry which makes delocalization less likely.  Also $\ce{2p-3p}$ overlap would be involved which is not as effective as $\ce{2p-2p}$ overlap.

  3) Why is butyllithium used particularly. Can't any other reagent be
  used to abstract H?

In the phosphonium salt, the protons alpha to phosphorous are not very acidic (pKa ~ 20-35), a strong base is required to remove them.  Other strong bases could be used, but BuLi is commercially available in high purity at a reasonable cost.  When BuLi abstracts the proton butane is formed and escapes as a gas - one less side product to worry about separating out.

  4) Why is the four membered ring formed in step 2?

The 4-membered ring may not be as strained as you imagine.  Remember, phosphorous is in the third row of the Periodic Table.  It has longer bond lengths and can tolerate smaller bond angles.

  5) Why does the ring collapse in step 3?

The bond between phosphorous and oxygen is an extremely strong bond, in other words its formation is very exothermic.  Therefore, this last step is being driven by thermodynamics.  It is the exothermicity of this step that drives the reaction to completion.


Hair is largely (~90%) composed of a protein called keratin, which originates in the hair follicle.

Now, keratin is composed of a variety of amino acids, including the sulfur containing amino acid, cysteine. All these amino acids are joined to each other by chemical bonds called peptide bonds to form these long chains that we call polypeptide chains. In the case of human hair, the polypeptide that we're talking about is keratin. The polypeptide chains are intertwined around each other in a helix shape.

The average composition of normal hair is 45.2 % carbon, 27.9% oxygen, 6.6% hydrogen, 15.1% nitrogen and 5.2% sulfur.

(I got that diagram off of Google Images)

Now, there are a whole bunch of chemical interactions that maintain the secondary and tertiary structures of proteins, such as van der Waals forces, hydrophobic interactions, polypeptide linkages, ionic bonds, etc. But there is, however, one additional chemical interaction in proteins that contain the amino acids cysteine and methionine (both of which contain sulfur) called disulfide linkages. You can see that in the diagram above (it's been marked in yellow, which is fortunately, a very intuitive color when you're dealing with sulfur).

When you burn hair (or skin or nails... anything that has keratin in it for that matter) these disulfide linkages are broken. The sulfur atoms are now free to chemically combine with other elements present in the protein and air, such as oxygen and hydrogen. The volatile sulfur compounds formed as a result is what's responsible for the fetid odor of burning hair.

Quite a few of the "bad smells" we come across everyday are due to some sulfur containing compound or the other. A great example would be the smell of rotten eggs, which can be attributed to a volatile sulfur compound called hydrogen sulfide. Yet another example (as @VonBeche points out in the comments) would be that of tert-butylthiol, which is the odorant that is used to impart the characteristic smell of Liquefied Petroleum Gas (LPG).


Aluminium iodide ($\ce{AlI3}$) is first introduced in 1984 as an easily accessible, a highly regioselective versatile ether-cleaving reagent with novel cleavage pattern (Ref.1). Application of $\ce{AlI3}$ in organic synthesis has been reviewed (Ref.2) and it seems like acetonitrile is the best solvent to be used in catechol ether demethylation (e.g., 2-methoxyphenol or 1,2-dimethoxybenzene).

For example, recent article discussed an one‐Pot cleavage of aryl alkyl ethers by aluminum and iodine in Acetonitrile (Ref.3). Their proposed mechanism for the cleavage is depicted in following diagram:

Accordingly, since it needs acid work-up for final product, I'd suggest you may use acetonitrile as the solvent of reaction, and use your solvent of choice to extract the final product from aqueous work-out.

If you are not happy with that, you may use either carbon disulfide ($\ce{CS2}$), which is also toxic, or toluene (their are references have been in literature using benzene as a solvent with mixed results) (Ref.2). Use of $\ce{CS2}$ is applied for specific reactions such as substrates having allyl functionality. Yet, it should work with simple compound as yours.

I'm little curious of your intend to use non-toxic solvent in the reaction since DMSO is also toxic in certain concentrations. In that case, you may use solvent-free conditions (you need to use enough DMSO for convenience, which is not critical according to the mechanism)(Ref.4).

Note: Cyclohexane is not a suitable reagent for this reaction, although it was involved in some publications (Ref.3). For example, when demethylation of eugenol reaction has bee done in acetonitrile at $\pu{80 ^\circ C}$ for $\pu{18 h}$, hydroxychavicol was achieved in 78%. However, when the reaction was run in cyclohexane under identical condition, no hydroxychavicol was achieved.

References:

M. Vivekananda Bhatt, J. Ramesh Babu, "New reagents 3: aluminium iodide - a highly regioselective ether-cleaving reagent with novel cleavage pattern," Tetrahedron Letters 1984, 25(32), 3497-3500 (https://doi.org/10.1016/S0040-4039(01)91058-5).
Juan Tian, Dayong Sang, "Application of aluminum triiodide in organic synthesis," ARKIVOC 2015, (vi), 446-493 (http://dx.doi.org/10.3998/ark.5550190.p009.309).
Juan Tian, Huaxin Yue, Pengtao Yang, Dayong Sang, "One‐Pot Cleavage of Aryl Alkyl Ethers by Aluminum and Iodine in Acetonitrile," Chemistry SELECT 2019, 4(1), 38-41 (https://doi.org/10.1002/slct.201803469).
Mehran Ghiaci, Jila Asghari, "Dealkylation of Alkyl and Aryl Ethers with $\ce{AlCl3 - NaI}$ in the Absence of Solvent," Synthetic Communications 1999, 29(6), 973-979 (https://doi.org/10.1080/00397919908086060).


Maybe these links would help:
US3046204A - Method for making diamond tools - Google Patents
https://patents.google.com › patent

The Basics: Electroplating & Vacuum Brazing - Diamond ...
https://www.dpcanada.com › articles › pl...

US4079552A - Diamond bonding process - Google Patents
https://www.google.com › patents

https://www.jstage.jst.go.jp/article/matertrans/43/10/43_10_2604/_article/-char/ja/

https://www.sciencedirect.com/science/article/pii/S0263436808001649

https://patents.google.com/patent/US6286498B1/en


Why isn't HCl called hydrogen monochloride?

Why not call it monohydrogen monochloride? If it's unambiguous you leave away anything that could describe "mono-".


Why is $^3P_1$ an atomic term symbol for carbon?

with $J=L+S$

Nope. In general, it can be shown that the permitted values of the $J$ quantum number for the total angular momentum that arise from two sources characterized by quantum numbers $J_1$ and $J_2$ are given by
$$
    J = J_1 + J_2, J_1 + J_2 - 1, \dotsc, |J_1 - J_2| \, .
$$
In this particular case, $J_1 = L$ and $J_2 = S$, so $J$ ranges from $L + S$ to $|L - S|$ in integral steps and not necessarily equal to $L + S$.


A first-order reaction at infinite concentration

When you go to the limit of infinite concentration (obviously you can never get there), a reaction cannot stay first order, because you will always run into some other rate-limiting mechanism thus making the reaction zeroth order in the highly concentrated species.

To give an example, for the binding of 2 molecules if both molecular species are 'infinitely' concentrated, the rate will not be limited by molecules 'finding' each other (as is normally the case for first order reactions), but it will be limited by the finite time binding process itself takes.


$\ce{^{99m}Tc}$ is a nuclear isomer of $\ce{^{99}Tc}$, which is excited by 142.6 keV. Therefore, when one atom decays to $\ce{^{99}Tc}$, it releases 142.6 keV, mostly as gamma radiation, but also some internal conversion occurs. Therefore, for 1 kg of $\ce{^{99m}Tc}$ (10.1 mol):

$$
(142.6\ \mathrm{keV})(10.1\ \mathrm{mol})N_A = 8.673\times 10^{29}\ \mathrm{eV} = 139\ \mathrm{GJ} = 38.6\ \mathrm{MWh}
$$

In general, the total energy released by a nuclear reaction in a given mass is simply the decay energy (commonly known as the Q value) multiplied by the number of atoms. The Q value is simply the difference in mass between the starting material and final product converted to energy:

$$Q = (m_\mathrm{i}-m_\mathrm{f})c^2$$

Q values are easy to just look up, though. (Wolfram|Alpha is a good place)

Of course, depending on the application, all this energy is not accessible or useful. If one were trying to make an RTG, an enormous amount of shielding would be needed to turn the gamma rays into heat, since they penetrate matter very well, where things like Pu-238 work better because they release mostly alpha particles, which are easily absorbed, delivering their kinetic energy as heat.


Mechanism for the chlorination of Fluorene with NCS?

According to this paper here

it has been described recently that at very low pH ($\mathrm{pH <  3}$)  $\ce{Cl2}$  functions  as  an  active  chlorinating  agent

In the authors' view the system works by $\ce{NCS}$ oxidising chloride to chlorine which then provides $\ce{Cl+}$ as the chlorinating agent.


Electroless Plating of Zinc onto Copper in NaOH Solution

An explanation is provided in this reference. The idea is that the electrode reactions are

$\ce{Zn(s) + 4OH^- -> [Zn(OH)_4]^{2-} + 2e-}$
and

$\ce{[Zn(OH)_4]^{2-} + 2e- -> Zn(alloy) + 4OH^-}$

So the zinc powder remaining will go into solution and the zincate is reduced on the copper surface to form a form of zinc rich brass. Brass is an alloy and the atoms are held together via metallic bonding.


Why does water volume decrease when salt is added?

What you are experiencing is the phenomenon of partial molal properties - components of a mixture contribute unequally to the properties of the mixture.

Sometimes the components of a mixture do not contribute linearly to the properties of a mixture. A well-known example is the water-ethanol system, which has negative excess volume. When 5 mL of water and 95 mL of ethanol are mixed, the final volume is less than 100 mL. See the following graph constructed by Wikipedia user Wilfried Cordes:

Similar behavior is occurring in your sodium chloride and water system. Now, as for why these phenomena occur, a lot of it has to do with intermolecular forces. For example, you could consider that the hydrogen-bonding and dipole-dipole interactions between water and ethanol are perhaps stronger than between water and water or ethanol and ethanol. Additionally, water molecules are smaller than ethanol molecules. There is also a thermodynamic consideration. This state of affairs must represent minimum free energy (maximum entropy).

How can you relate the interactions of sodium and chloride ions with water to the observation of negative excess volume?


Why are there exactly 5 types of two-dimensional lattices, and what distinguishes them?

First, about the black lines. I think they are helpful for the hexagonal and the rhombic lattice. For the hexagonal system, it shows you the hexagons. One hexagon contains exactly the same space as one conventional unit cell. You just have move (translate by a lattice edge) certain areas as shown below.

For the rhombic lattice, it is nice to have the black lines when you are looking at the green cell. Without it, you would not understand why you would call it rhombic.

So why does a rhombic lattice have 2 types of unit cells (the blue and the green) whereas a hexagonal lattice only has 1 (and why do the solid black lines for these two lattices are different?).

The green cell is a centered cell, which makes the math easier because it has right angles. On the other hand, you suddenly have a translation by $½a + ½b$ on top of the lattice translations that you have the keep track of. In fact, the rhombic type is usually described as a centered rectangular lattice,  e.g. https://en.wikipedia.org/wiki/Bravais_lattice#In_2_dimensions.

The same question also applies to rectangular lattice. Isn't it just a special case of oblique lattice where γ = 90°?

Compared to the oblique lattice, all the other lattices are special cases. They aren't special cases because of some rare coincident, but because of the presence of point group symmetries, i.e. mirror planes and twofold, threefold, fourfold and sixfold rotations. A mirror plane will always allow you to choose a unit cell with a right angle. A fourfold axis will always allow you to choose a unit cell that is square, and a threefold or sixfold axis will always allow you to choose a unit cell that is of the hexagonal type. As you can see, the conventional unit cell is not a hexagon but a rhombus with a 60 degree angle. This makes the math easier.
I don't know why the illustrator chose to offset the black lines and the unit cell by half a unit cell length in both $a$ and $b$ for all lattice types (except the hexagonal one). Maybe it is for better visibility. In general, any point in the unit cell will have symmetry mates in other unit cells related by lattice translations.
If you want to see some examples of lattices that contain objects, check out this link: https://www.iucr.org/education/pamphlets/21/supplement
Here is the illustration for a lattice that contains a hexagonal axis. In this case, you have six identical objects in the unit cell, related by point group symmetry. They do no fit directly into the conventional unit cell, but it has the same volume as six objects (you can try to draw the black lines showing hexogons in your mind). The object is called the asymmetric unit, and it can have various shapes as long as it maps onto the unit cell filling the space without overlap.


Defining and testing custom made DFT functionals?

Depending on what you're after, you might want to look for programs using libxc, so you can completely define your own functionals.

From the libxc section of the Octopus wiki, these include:

  Abinit – plane-wave code
  APE – an atomic code
  Atomistix ToolKit – numerical orbitals code
  AtomPAW – projector augmented wave functions generator
  BigDFT – wavelet code
  CP2K -  A program to perform atomistic and molecular simulations of solid state, liquid, molecular, and biological systems.
  DP – Dielectric Properties, a linear response TDDFT code
  Elk – FP-LAPW code
  ERKALE – a DFT/HF molecular electronic structure code based on Gaussian orbitals
  exciting – FP-LAPW code
  GPAW – grid-based projector-augmented wave method
  JDFTx – plane-wave code designed for Joint Density Functional Theory
  MOLGW - a small, but accurate MBPT code for molecules
  octopus – real-space (TD)DFT code
  Yambo – solid state and molecular physics many-body calculations code

On a more personal note, and this is just my opinion, if you're just arbitrarily mixing functionals, you should really make sure you have a scientific justification for doing this, otherwise you're just generating numbers without basis.


The reasons that you cite are both important for the solubility of an alcohol. Generally speaking, if the molecule is branched its (hydrophobic) surface exposed to water is less than in a linear molecule. This effect is more pronounced when the number of atoms is bigger. On the other hand the effect of the degree of substitution on the solubility is less pronounced when the molecular size rise.

This would partially justify your second question, but the problem here is that your second question is wrong, 2-methyl-2-propanol is soluble with water in every ratio, while n-butanol is not.


what is the white fuzz left behind on basement floor after puddle evaporated

Three pieces of information from the comments:

It fizzed and disappeared in vinegar.
It had a yellow-orange color in blue flame.
It's salty (note that it's usually a bad idea to taste random stuff you find on your floor).

The only white crystalline material that is a plausible candidate for being there and fizzes in acetic acid (ie vinegar) is calcite - $\ce{CaCO3}$. It's also a mineral that has 'retrograde solubility' meaning it dissolves more readily in cold water. This is probably how it got to the water in the first place. This is also why it precipitates on your electric kettle. It is the same stuff.

The salty stuff that burns yellow-orange is obviously $\ce{NaCl}$, aka table salt. It also probably dissolved with the vinegar as well. Why is the salty taste not quite like table salt? Well, first of all it is mixed with calcite. But, calcite has no taste. It's possible that there is some $\ce{KCl}$ or $\ce{MgCl2}$ in there as well. They are slightly more bitter than $\ce{NaCl}$. This is the stuff they use in low sodium salt.

There's another possibility - it could be any of the epsom salt family, $\ce{MgSO4.$n$\,H2O}$.

As to would this be dissolved in the first place, you can only speculate. You say that it came from a table that was outside. Did people eat there? Could people had spilled salt on it? Is the airborne dust in your area contain a lot of calcite? What about your basement? Could someone have spilled some salts on the floor, maybe even years ago?


Do electrons go to even lower ground states after exothermic reactions?

Yes, electron energy levels are like that, especially for single atoms.

Yes, forming molecular bonds creates different electron energy levels, with exothermic reactions leading to chemical bonds with lower ground levels then the reagents had.

Note that by an electron ground energy level is meant the minimal energy of  energy levels unocuppied by other electrons.

As an electron is forbidden by quantum rules to be in the same quantum state as any other electron within an atom or molecule.

Thermal, photon or collision excitations may keep part of electrons in excited state.

Much more info you can get reading about

molecular orbitals
molecular orbitals diagrams
molecular orbital theory

on Wikipedia.


When you are applying 31.9V across I am assuming cm-ish distances, you are definitely causing a lot of electrochemistry to happen in the solution.

You are oxidizing one of your aluminum electrodes to $\ce {Al^3+}$, therefore precipitating $\ce{Al(OH)_3}$ and probably reducing water on the other electrode to evolve $\ce{H_2}$.Along with these, you are probably doing a lot of other reactions with any and every impurity around, but these two should dominate.

Measuring ionic conductivity properly requires doing an AC measurement since you cannot have a sustained DC current without doing electrochemistry of some sort. The electrochemistry on the electrode surface ends up the main factor influencing your measurement in any DC setup.


Both "monooxide" and "monoxide" are used in the literature, yet "monoxide" is being  used more often (Google Books Ngram Viewer).
Although this is an accepted elision, it is not the preferred one, and must not set a precedent for other cases when multiplicative prefix ends with the same vowel as the root word begins with, such as "diiodine".
According to the current version of Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 [1, p. 31]:

IR-2.7 ELISIONS
In general, in compositional and additive nomenclature no elisions are made when using
multiplicative preﬁxes.
Example:

tetraaqua (not tetraqua)
monooxygen (not monoxygen)
tetraarsenic hexaoxide

However, monoxide, rather than monooxide, is an allowed exception through general use.

Further, from section IR-5.2 Stoichiometric names of elements and binary compounds [1, p. 69]:

The multiplicative preﬁxes precede the names they multiply, and are joined directly to them without spaces or hyphens. The ﬁnal vowels of multiplicative prefixes should not be elided (although ‘monoxide’, rather than ‘monooxide’, is an allowed exception because of general usage).
[…]
Examples:
[…]

$\ce{NO}$ nitrogen oxide, or nitrogen monooxide, or nitrogen monoxide

References

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2. IUPAC website


Why are some organic reactions run at elevated temperatures?

Though they can produce a similar effect (faster reaction rate), heat doesn't act as a catalyst.

Reaction rate is dependent on how often reagents can overcome the activation energy of a reaction. Therefore, a lower activation energy, $E_\mathrm{a}$, means a faster reaction.

A catalyst lowers the activation energy of a reaction by producing lower energy transition states. This is how catalysts speed up reactions. A generic version of this is shown in the image below. Transition states occur at the local maxima of the graph in this image.

Heat, on the other hand, increases the average kinetic energy of the molecules within a system (i.e. most molecules are moving more quickly). By doing this, the chances of two particles colliding and having enough energy to react are higher because the particles will move faster. Therefore, they will collide more often, and when they do, they will have more "momentum" (so to speak) to get over this energy hill. By making it so more of the reactant particles can overcome this activation energy, heat also speeds up reactions.


Mathematical basis of why enthalpy of mixing is 0 for ideal gas

A strategy that often works is to look at the initial and the final states and then compare them. A good place to start is usually the Gibbs free energy of the system (with the intention to use $\Delta_{\mathrm{mix}}H = \Delta_{\mathrm{mix}}G + T \Delta_{\mathrm{mix}}S$ later): for isothermal and isobaric conditions it is

\begin{equation}
  G = \sum_{i} n_{i} \mu_{i} + \mathrm{constant} = n_{\mathrm{tot}} \sum_{i} \underbrace{x_{i}}_{= \, \frac{n_{i}}{n_{\mathrm{tot}}}} \mu_{i} +  \mathrm{constant} \hspace{4em} (1)
\end{equation}

with $n_{i}$, $\mu_{i}$, and $x_{i} = \frac{n_{i}}{n_{\mathrm{tot}}}$ being the amount, the chemical potential, and the mole fraction of the $i$th component in the system, respectively, and $n_{\mathrm{tot}} = \sum_{i} n_{i}$ being the total amount of gas in the system.
So, now we need an expression for the chemical potential of an ideal gas.
It can be shown that this is given by

\begin{equation}
  \mu_{i} = \mu_{i}^{0} + R T \ln\Bigl(\frac{p_{i}}{p^{0}}\Bigr) \hspace{18em} (2)
\end{equation}

where $\mu_{i}^{0}$ is the chemical potential of the $i$th component in a given standard state, $R$ is the universal gas constant, $T$ is the temperatur, $p_{i}$ is the partial pressure of the $i$th component, and $p^{0}$ is the standard pressure (i.e. the pressure connected to the standard state).
For ideal gases we can apply Raoult's law stating that

\begin{equation}
  p_{i} = x_{i} p_{\mathrm{tot}}
\end{equation}

where $p_{\mathrm{tot}}$ is the total pressure of the system.
Using this with equation (2) we get

\begin{equation}
  \mu_{i} = \mu_{i}^{0} + R T \ln\Bigl(\frac{x_{i} p_{\mathrm{tot}}}{p^{0}}\Bigr) = \mu_{i}^{0} + R T \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \hspace{6em} (3)
\end{equation}

Furthermore, it is known that the partial pressures in a system must add up to the total pressure, i.e.

\begin{equation}
  p_{\mathrm{tot}} = \sum_{i} p_{i}
\end{equation}

Ok, now that the stage is set we simply have to apply equations (1) and (3) to the initial and the final state of the mixing process.
At the initial state the chambers are seperated (whereby chamber $i$ contains only one sort of gas labeled by $i$) and so these subsystems are completely independent from each other.
Thus, the Gibbs free energy of the total system of chambers is simply the sum of the Gibbs free energies of each chamber, whereby the gas in each of the chambers feels the ambient/total pressure $p_{\mathrm{tot}}$ in the system, since $x_{i} = 1$ (each chamber/independent subsystem holds just one kind of gas).
According to equation (1) the Gibbs free energy of the $i$th chamber is then given by

\begin{align}
  G_{i} &= n_{i} \mu_{i} = n_{i} \biggl( \mu_{i}^{0} + R T \underbrace{\ln 1}_{= \, 0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr) \\
        &= n_{i} \biggl( \mu_{i}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)
\end{align}

and the Gibbs free energy of the total system at the initial state is

\begin{align}
  G_{\mathrm{ini}} &= \sum_{i} G_{i} \\
                 &= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( \sum_{i} x_{i} \mu_{i}^{0} + \sum_{i} x_{i} \underbrace{R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr)}_{= \, \mathrm{constant}} \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \underbrace{\sum_{i} x_{i}}_{= \, 1} \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)
\end{align}

At the final state the chambers are connected and the gases mix, so the chambers are not independent subsystems anymore and the Gibbs free energy cannot be broken down into independent parts like before.
So now, the complete system of connected chambers must be treated via equation (1)

\begin{align}
  G_{\mathrm{final}} &= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( \sum_{i} x_{i} \mu_{i}^{0} + \sum_{i} R T x_{i} \ln x_{i} + \sum_{i} x_{i} \underbrace{R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr)}_{= \, \mathrm{constant}} \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \underbrace{\sum_{i} x_{i}}_{= \, 1} \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)
\end{align}

Now, that we have the Gibbs free energies of the initial and final state we simply need to subtract the former from the latter one in order to obtain the change in Gibbs free energy during the mixing process:

\begin{align}
  \Delta_{\mathrm{mix}}G &= G_{\mathrm{final}} - G_{\mathrm{ini}} \\
                       &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)  - n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \ln\Bigl(\frac{p_{\mathrm{tot}}}{p^{0}}\Bigr) \biggr)  \\
                       &= n_{\mathrm{tot}} R T \sum_{i} x_{i} \ln x_{i} \hspace{22em} (4)
\end{align}

Differentiating equation (4) with respect to the temperature we get the entropy of mixing since $\frac{\partial G}{\partial T} \big|_{p, \{ n_{i} \}} = - S$

\begin{equation}
  \Delta_{\mathrm{mix}} S = - \frac{\partial \Delta_{\mathrm{mix}}G}{\partial T} \bigg|_{p, \{ n_{i} \}} = - n_{\mathrm{tot}} R \sum_{i} x_{i} \ln x_{i}
\end{equation}

Finally, you get the desired result by using the famous equation $G = H - TS$

\begin{align}
  \Delta_{\mathrm{mix}}H &= \Delta_{\mathrm{mix}}G + T \Delta_{\mathrm{mix}}S \\
                        &= n_{\mathrm{tot}} R T \sum_{i} x_{i} \ln x_{i} - T n_{\mathrm{tot}} R \sum_{i} x_{i} \ln x_{i} \\
                        &= 0
\end{align}


Regioselectivity in Electrophilic Aromatic Substitution of Pyrrole and Indole

The traditional explanation for the reactivity of indole at C–3 is that attack at C–3 does not disrupt the aromaticity of the benzene ring in the cationic intermediate (see any organic chemistry book; for heterocycles a good book is Joule and Mills as has been suggested in the comments):

You have drawn many of the resonance structures for the intermediate, including the two shown above. So I should mention that the major resonance contributor should be the one where the positive charge is borne by the nitrogen. In this case, the positively charged nitrogen has four bonds and a complete octet; whereas if the positive charge is on carbon, that's formally a six-electron, electron-deficient carbocation. Many thanks to Buttonwood for pointing this out.
To a first approximation, one could consider the reactivity of indole to be as if it were a benzene ring plus an enamine, and enamines are very, very happy to react β to nitrogen and not α. This is of course not an enamine, but the idea is the same.
For pyrrole, there is no "other benzene ring" to worry about, so this consideration does not apply.