Since $(CH_3)_3C^+(aq)$  achieves a steady-state concentration, the equations when directly added give the overall reaction whose rate constant is the product of the rate constants of individual reactions. Hence the rate law comes out to be as it is written.


Why do aqueous solutions always have to be electrically neutral?

Because the electrostatic force is much stronger than you can imagine. Yes, you can get charge-rich areas in a cell. However, the amount of charge held here will be negligible. A rather small amount of charge can overcome the field near the plates and prevent more charge from approaching. The amount is quite small; it's not something that you would notice (and it won't give you a static shock1

1. Everyday static shocks are stronger by orders of magnitude because they deal with a lot of charge accumulated on the surface of an insulator, while here it is a relatively small amount of charge dispersed through the volume of a conductor


When do lone pairs on nitrogen contribute to aromaticity?

Both nitrogens in your drawing are roughly $\mathrm{sp^2}$ hybridized, therefore they each have a p-orbital and 3 $\mathrm{sp^2}$ hybdridized orbitals. All of the p-orbitals, including those on the 3 carbon atoms in the double bonds, are perpendicular to the plane of the screen (e.g. they project above and below the plane of the screen). So there are 5 p-orbitals and each contains one electron except for the one p-orbital located on your "bottom" nitrogen (in green). This p-orbital contains the nitrogen lone pair, so there are two electrons in this p-orbital.

If we count the p-electrons we have $1+1+1+1+2 = 6$ electrons in $5$ p-orbitals that form a continuous loop. Therefore, the molecule is aromatic ($4n+2, n=1$).

Let's take a closer look at the "top" nitrogen in your drawing (in purple). Where is its lone pair? It resides in an $\mathrm{sp^2}$ orbital on this nitrogen. The $\mathrm{sp^2}$ orbital lies in the plane of the screen. This orbital is orthogonal (perpendicular) to the 5 p-orbitals; therefore these 2 lone pair electrons are separated from the electrons that make the molecule aromatic. They are not involved in the molecule's aromaticity.

Here's how to approach this situation in the future:

Identify the nitrogen(s)
Determine if the lone pair is in a p-orbital or an $\ce{sp^2}$ orbital.

If the nitrogen is in a double bond, then the lone pair is in one $\ce{sp^2}$ orbital, 1 electron is in the p-orbital and 1 electron is in each of the 2 remaining $\ce{sp^2}$ orbitals.
If the nitrogen is not in a double bond, then the lone pair is in a p-orbital and 1 electron is in each of the 3 $\ce{sp^2}$ orbitals.

Count the electrons in the cyclic p-orbitals to determine aromaticity


$\log K$ relates to the thermodynamic changes in entropy and enthalpy of a reaction.
$$\Delta H - T \Delta S = \Delta G = \Delta G^\circ - RT \ln Q$$

At equilibrium $Q = K$ and:

$$\Delta G = 0 = \Delta G^\circ - RT \ln K \Longrightarrow  \Delta G^\circ = RT\ln K$$

$$\Delta G^\circ = \Delta H^\circ -T\Delta S^\circ = RT \ln K$$

and:

$$\ln K = \frac{\Delta H^\circ -T\Delta S^\circ}{RT} = \frac{\Delta H^\circ}{RT} - \frac{\Delta S^\circ}{R}$$


How does an irreversible process maintain internal energy while dissipating excess heat?

The change in entropy of the system in both cases is the same, but, in the irreversible case, the entropy of the surroundings decreased less.  This is because less heat was transferred from the surroundings to the system.  In this sense, all the entropy generated within the system in the irreversible process is transferred to the surroundings.  So the interpretation would be that the "heat generated" within the system causes less heat to be transferred from the surrounding to the system and less work to be done.  In my judgment, this is a desperate attempt to provide some physical interpretation to what is happening.


How to calculate pH of basic salt solution mixed with strong base?

Exact Solution

I've fouled up this problem badly. Here is a more carefully thought out solution.

So let's start at the beginning with the ICE table. Making the normal assumptions:

Concentrations will be used instead of activities.
The $\ce{Zn(CH3COO)_2}$ dissociates completely in aqueous solution, thus the $\ce{Zn^{2+}}$ and $\ce{CH3COO-}$ ions can be considered separately.
The $\ce{NaOH}$ dissociates completely in aqueous solution, thus the $\ce{Na+}$ and $\ce{OH-}$ ions can be considered separately.
The $\ce{Na+}$ is a spectator ion and is left out of the table.

The problem only calls for the pH of the final solution, but a solution to the whole table will be calculated as a check.
Initial ICE Table
Let's let:

$x = $ the moles of $\ce{Zn(OH)2}$ formed. (The numerical value is the same as the change in molarity of $\ce{Zn^{2+}}$ since there is 1.0 liter of solution.)

$y = $ the molarity of the $\ce{CH3COOH}$.

\begin{array}{c | c c c c c}
               &\ce{[Zn^{2+}]}     & \ce{[CH3COO-]}        & \ce{[CH3COOH]}   & \ce{[OH-]}                         & \ce{Zn(OH)2(s)}  \\\hline
\text{I}  &\pu{0.050000 M}     & \pu{0.100000 M}           & \pu{0.000000 M}               & \pu{  0.025000 M}               & \pu{0 moles}  \\
\text{C} &  -x                           &  -y                               &   y                              & \pu{- 2x + y M}                  & \pu{x moles}  \\
\text{E} &\pu{0.050000 - x M} & \pu{0.100000 - y M}     &   y                              & \pu{0.025000 - 2x + y M}   & \pu{x moles}   \\
\end{array}
Applicable equilibriums
Wikipedia gives the $K_{sp}$ of zinc hydroxide is $K_{sp}=3.0\cdot10^{−17}$. So:
$K_{sp}=3.000\cdot10^{−17} = \ce{[Zn^{2+}][OH-]^2}$
Wikipedia gives the $\pu{pK_b}$ for sodium acetate as 9.25, which yields $K_b = 5.623\cdot10^{−10}$.  So:
$K_b = 5.623\cdot10^{−10} = \dfrac{\ce{[CH3COOH][OH-]}}{\ce{[CH3COO-]}}$
Exact Solution
Starting with the $K_{sp}$ equation for zinc hydroxide and making the appropriate substitutions for the final concentrations:
$K_{sp} = \ce{[Zn^{2+}][OH-]^2}$
$3.000\cdot10^{−17} = (0.050000 - x)\cdot(0.025000 - 2x + y)^2$
Starting with the $K_b$ equation for the acetate anion:
$K_b =  \dfrac{\ce{[CH3COOH][OH-]}}{\ce{[CH3COO-]}}$
$5.623\cdot10^{−10} = \dfrac{(y)\cdot(0.025000 - 2x + y)}{0.100000 - y} = \dfrac{0.025000y - 2xy + y^2}{0.100000 - y}$
Wolfram Mathworld's solution is:

x = 0.0134623
y = 0.00192466

$\ce{[Zn^{+2}]_{equil}} = 0.0500000 - 0.0134623 = 0.0365377$
$\ce{[OH-]_{equil}} = \sqrt{\dfrac{ K_{sp}}{\ce{[Zn^{2+}]_{equil}}}} = \sqrt{\dfrac{3.000\cdot10^{−17}}{0.0365377}} = 2.86543\cdot10^{-8}$
$\pu{pOH} = 7.542810$
$\pu{pH} = 14.000000 - 7.542810 =  6.457190 \ce{->[Round to 2 SF]} 6.46$
$\ce{[CH3COO-]_{equil}} = 0.100000 - 0.00192466 = 0.09807534$
There are two equations and two unknowns. Solving the equations yielded an exact solution. This is no problem with a computer but a mess even for a good calculator.


Which are stronger intermolecular forces: ionic bonds or hydrogen bonds?

Hydrogen bonds are the strongest intermolecular forces. Stronger than van der Waals' forces, namely (ordained by strength):

Permanent dipole - permanent dipole (Keesom's Force);
Permanent dipole - induced dipole (Debye force);
induced dipole - induced dipole (London's dispersion force);

Strong hydrogen bonds are created by the FONs (fluorine, Oxygen and Nitrogen), which are very electronegatives, what produces a strong attraction between one them and hydrogen (which is less electronegative). As example of this interaction we can cite the water, that's a polar molecule in a "V" shape, take a look at the figure below.

Then, the both hydrogens yields a $\delta^{+}$ that's attracted by the Oxygen (very strong electronegative, the $\delta^{-}$ side) of the another water molecule, yielding the hydrogen bonds, take a look at the figure below.


How can a nonspontaneous reaction occur?

When you add energy from somewhere, let's say a battery, you are creating a greater amount of entropy there (in the battery or whatever). That's all there is to it. The Second Law is not violated.
The electrolysis process is not non-spontaneous. Yes, it would not occur in the absence of the battery. But the fact that it does occur with the battery there must tell you that overall it is spontaneous. The fact that you are coupling a strongly spontaneous reaction (discharge of the batter) with a non-spontaneous reaction (the electrolysis) means that overall, the electrolytic process is spontaneous.
It's incorrect to treat the electrolysis as if it occurs without the battery.


Justification for Freezing Point Depression & Boiling Point Elevation in Solutions?

I think you are looking at the problem from slightly the wrong angle. The central quantity when dealing with colligative properties is entropy and not solute-solvent or solvent-solvent molecule interactions. Of course the interactions are important in the sense that they affect the entropy but since we are dealing with thermodynamics here the way you think about it at the moment clouds the main effect at work here imo.

So, when trying to picture how colligative properties work on a microscopic level think about the effect of the solute on the entropy. Boltzmann's entropy formula tells you that the entropy $S$ is proportional to the logarithm of the number of microstates available to the system. The solute is present in the solution in small quantities and the molecules are really dissolved and don't form a different phase. By their presence the solute molecules disturb the local order of the solvent molecules and introduce new available microstates into the system. As an extremely simplified example you can think of it like this: If you want to place 10 identical blue balls (solvent molecules) on 10 fixed spots there is only 1 way (microstate) to do this, since all the balls are alike. But if you replace one of the blue balls with a red ball (solute molecule) you suddenly have 10 possible distiguishable ways to arrange your balls on the 10 spots. So, when comparing the initial situations of a solution of pure solvent and one of solvent plus solute, the second solution will have a higher entropy to start with, i.e. $S_{\mathrm{l}}^{\mathrm{impure}} > S_{\mathrm{l}}^{\mathrm{pure}}$.

Now, what happens when you want to freeze both solutions? In order to freeze the Gibbs free energy of the frozen (solid) phase must be lower than or equal to the Gibbs free energy of the liquid phase, i.e. $G_{\mathrm{s}} \leq G_{\mathrm{l}}$ or put differently $\Delta_{\mathrm{m}} G \substack{{\scriptsize \mathrm{def}} \\ =} G_{\mathrm{l}} - G_{\mathrm{s}} \geq 0$. The Gibbs free energy is given by

\begin{equation}
 G = H - TS
\end{equation}

where $H$ is the enthalpy and $T$ is the temperature, so that

\begin{equation}
 \Delta_{\mathrm{m}} G = (\underbrace{H_{\mathrm{l}} - H_{\mathrm{s}}}_{\Delta_{\mathrm{m}} H}) - T (\underbrace{S_{\mathrm{l}} - S_{\mathrm{s}}}_{\Delta_{\mathrm{m}} S})
\end{equation}

Since there is only very little solute present in the second solution you can assume that $\Delta_{\mathrm{m}} H^{\mathrm{impure}} \approx \Delta_{\mathrm{m}} H^{\mathrm{pure}}$. So, the difference between $\Delta_{\mathrm{m}} G^{\mathrm{impure}}$ and $\Delta_{\mathrm{m}} G^{\mathrm{pure}}$ will mainly be caused by the entropical term. Let's have a look at that: the solid phase is assumed to be pure in both cases, i.e. consisting only of solvent molecules, and thus $S_{\mathrm{s}}$ must be the same in both cases. Furthermore, you generally find that the entropy of a solid phase is lower than the entropy of a liquid phase, i.e. $S_{\mathrm{s}} < S_{\mathrm{l}}$. Above I established that $S_{\mathrm{l}}^{\mathrm{impure}} > S_{\mathrm{l}}^{\mathrm{pure}}$. It follows that $\Delta_{\mathrm{m}} S^{\mathrm{impure}} > \Delta_{\mathrm{m}} S^{\mathrm{pure}}$. Putting all this together gives you the following picture: For $\Delta_{\mathrm{m}} G$ to become equal to zero the entropical term $T \Delta_{\mathrm{m}} S$ has to balance $\Delta_{\mathrm{m}} H$, which is assumed to be about the same for the pure-solvent and the solvent-plus-solute case. Since $\Delta_{\mathrm{m}} S^{\mathrm{impure}} > \Delta_{\mathrm{m}} S^{\mathrm{pure}}$ the temperature at which freezing becomes possible must be lower in the solvent-plus-solute case, i.e. adding the solute leads to a freezing point depression.

An analogous argument can be made for the Boiling point elevation. The focal point is again that the solute raises the entropy of the liquid phase, thus affecting the entropical term in the Gibbs free energy equation for the phase change. For the boiling process to happen the Gibbs free energy of the gaseous phase must be lower than or equal to the Gibbs free energy of the liquid phase, i.e. $G_{\mathrm{g}} \leq G_{\mathrm{l}}$ or $\Delta_{\mathrm{vap}} G \substack{{\scriptsize \mathrm{def}} \\ =} G_{\mathrm{g}} - G_{\mathrm{l}} \leq 0$.
Requiring the Gibbs free energy of vapourization

\begin{equation}
 \Delta_{\mathrm{vap}} G = (\underbrace{H_{\mathrm{g}} - H_{\mathrm{l}}}_{\Delta_{\mathrm{vap}} H}) - T (\underbrace{S_{\mathrm{g}} - S_{\mathrm{l}}}_{\Delta_{\mathrm{vap}} S})
\end{equation}

to be equal to zero gives

\begin{equation}
 \Delta_{\mathrm{vap}} H = T \Delta_{\mathrm{vap}} S
\end{equation}

Again, we can assume that $\Delta_{\mathrm{vap}} H^{\mathrm{impure}} \approx \Delta_{\mathrm{vap}} H^{\mathrm{pure}}$. Since the entropy of a gaseous phase is generally higher than the entropy of a liquid phase, i.e. $S_{\mathrm{g}} > S_{\mathrm{l}}$, and since $S_{\mathrm{l}}^{\mathrm{impure}} > S_{\mathrm{l}}^{\mathrm{pure}}$, it follows that $\Delta_{\mathrm{vap}} S^{\mathrm{impure}} < \Delta_{\mathrm{m}} S^{\mathrm{pure}}$. Thus the boiling temperature must be higher in the solvent-plus-solute case, i.e. adding the solute leads to a boiling point elevation.


Predict which hydrogen is more acidic in the following organic compounds?

In the second molecule, the charge on the carboxylate ion is not stabilised through resonance, so we can rule out any resonance stabilisation. Moreover, the fact that the benzene ring is in fact electron donating within the $\pi$ system doesn't help either.

More importantly, however, (as it was pointed out to me by @orthocresol and @hippalelectryon in chat) one can expect intramolecular hydrogen bonding.

Moreover, after the second deprotonation, the second molecule would have two negative charges in close proximity to each other, and this, again, is obviously destabilising.

A combination of these factors would make the second molecule a weaker acid.

In the first case, however, the inductive electron withdrawing effect would increase the acidity, (and the benzene ring is far enough to not have an effect).

Thus, I would say it is the first compound that is more acidic.

@Hippalectryon used a software to compute a pKa of 3.87 for the first one, and Wikipedia records 5.51 for the second one, which confirms my claim.


quantum number - question about the ms-value

As porphyrin clearly explained, the first three quantum numbers usually are enough to determine the orbital of a given electron.

You probably know this but I'd like to let it here for everyone else.

Each different value of the principal quantum number $n$ yields a different wave function and a different state of the electron. In atoms, $n$ defines the order of all states according to energy.

The second and third quantum numbers $l$ and $m_{l}$, arise from the quantization of rotation where $l$ gives the shape of an orbital and its angular distribution and $m_{l}$ determines the number of orbitals and their orientations in a subshell.

The spin quantum number of an electron is denoted $s$ and it has the value of $s=1/2$. The spin magnetic quantum number is denoted $m_{s}$ gives the projection of the spin angular momentum on the z-axis. Therefore, one electron can have only one orientation: either $m_{s}=+1/2$ which we commonly denote up-spin or $\alpha$ and $m_{s}=-1/2$ denoted as down-spin or $\beta$.

What I see in the comments are partial answers because to associate an electron (not an orbital) to these numbers you need to use $m_{s}$. Then, you have a spin-down or $\beta$ electron in a $3d$ orbital.

Answering your question in the comments. To have a $3d^{2}$ you'd need two pairs of four quantum numbers: $(3,2,2,+1/2)$ and $(3,2,2,-1/2)$. Here, two electrons occupy the same region in space but they have different orientations with respect to their spin angular momenta.


What causes methane to have 25 times higher global warming potential than CO2

First we have to look how the greenhouse effect works for all gases:

Light with a wavelength not absorbed by the atmosphere gets absorbed by the soil and heats up the earth. Because of black body radiation for 300K the earth starts to emit light itself with a maximum in the IR spectrum. Light with this frequency gets readily absorbed by vibrational excitation of molecules. But one has to keep in mind that intensity of absorption between different molecules and even between different vibration in one molecule differs a lot. This can not be explained with mass, or just vibration itself you actually have to look at https://en.wikipedia.org/wiki/Transition_dipole_moment or https://en.wikipedia.org/wiki/Selection_rule

But just believe me for a moment, methane is actually very bad at absorbing infrared light. Especially its by far worse than Water or $CO_2$.

But then there is another effect. You can not absorb more light than there actually is. So if you practically absorb 100% of the light for a given frequency then there is no effect in increasing the concentration of the molecule absorbing this frequency.

Both water and $CO_2$ absorb with nearly 100% in their frequency range (compare fig. 1 which shows the IR absorption of the atmosphere).
So in order to absorb more light you have to absorb in other frequency ranges.

This is where methane gets its role. If you look at figure 2 and keep in mind to change the wavenumber into wavelengths you will see that it absorbs in a frequency range that is covered neither by water nor by $CO_2$.


For a real gas, the internal energy varies with both temperature and molar volume.  The variation is described by the equation:
$$\mathrm dU=C_V\,\mathrm dT-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]\mathrm dV,$$ where $U$ is the internal energy per mole and $V$ is the molar volume.  This means that, unless the equation of state is such that pressure is directly proportional to temperature at constant molar volume (like an ideal gas), the internal energy will depend on molar volume.


Stability of transition and post-transition metal alkoxides

Alkali metal, main group, and transition metal (TM) alkoxides all have significantly different chemistry. Early- and Late-TM alkoxides also have different behaviour, so we should split this family into at least four categories. All of them are basic and will decompose via hydrolysis if dissolved in water; however, this type of chemistry is not usually carried out in presence of water.

In TM chemistry, alkoxide or phenoxide are mostly used as ancillary ligands, meaning they do not usually participate in the chemistry. Although they are liable to be hydrolyzed under protic conditions, alkoxides are actually known for being especially stable TM ligands because they do not undergo β-hydride elimination (BHE) (1), which is one of the most common decomposition pathways for TM alkyls. For instance; you often cannot make a stable t-butyl TM complex because the β-hydride will be transferred to the transition metal, liberating 2-methylpropene unless the metal is totally coordinatively saturated. An important example of a BHE-stable TM t-butoxide is the following Mo(IV) Schrock catalyst, which is stable:

$\hskip2.4in$

It should be noted, though, that the Schrock catalyst doesn't actually have any β-hydrides. Later, I'll show an example of a TM-alkoxide that does.

An important feature of alkoxides as ligands for TM chemistry is that they are electronically malleable. Since an alkoxide has 5 electrons it can behave either as an σ-donating X ligand or as a σ- and π-donating LX ligand (and formally as L2X). Presumably, this electronic malleability has something to do with their stability to BHE. An alkoxide's π-donor character might stop the metal center from liberating a coordination site to accept the β-hydride. For more information on BHE-stability, see (2), as cited by Schrock in (1).

Since early-TMs are more electropositive, the bonding character of their alkoxides is more ionic, although not to the same degree as that of alkali metal alkoxides. Alkoxides are very good bridging ligands, and if they are not sufficiently bulky, alkoxide complexes often polymerize. With a sufficiently bulky monomeric early-TM alkoxide, the main decomposition pathway is formation of the oxo complex. For example; the decomposition of a metal diolate shown in the next image was observed by Gable et al. (3) Their Hartree-Frock calculations implicated an empty metal MO at rhenium in the M=O bond formation, as shown in the MO picture in the next image.

$\hskip0.2in$

Late-TM alkoxides are expected to have weaker M-O bonds; and not as many have been isolated, but a new study (4) has examined these M-X (where X = O, N) bonds.

A really nice example illustrating a few important TM-alkoxide concepts is tantalum(V) ethoxide, Ta2OEt10. See wikipedia: https://en.wikipedia.org/wiki/Tantalum(V)_ethoxide

$\hskip1.5in$

Ta2OEt10 is a stable TM-alkoxide with true β-hydrides, bridging oxygens, and it's readily hydrolyzed when exposed to protons. It can be synthesized via salt metathesis with sodium ethoxide, a reaction that has equilibirium lying in the opposite direction of the alkali metal alkoxide formation you proposed.

In MPV reduction (and OPP oxidation), Al(O-i-Pr)3 is usually used as the catalyst. Aluminium is a metal, but it is a main group metal, and its chemistry is somewhere between that of boron, a semimetal, and that of early transition metals, except, of course, that it does not have a partially filled d shell. Al(O-i-Pr)3 is an electron deficient Lewis acid, comparable in chemistry to boronate esters, but much more reactive. The proposed mechanism for MPV/OPP [R]/[O] goes something like this:

$\hskip0.2in$

It is a true catalytic reaction, so the Al(O-i-Pr)3 is indeed regenerated, but I don't know if it's a routine part of the procedure to recover it. Al(OR)3's are not, however, soluble in IPA, but alcohols and ketones usually are, so you could probably precipitate Al(O-i-Pr)3 in IPA providing you have a way to seperate your ketone and your secondary alcohol afterwards.

Interestingly, Ti(O-i-Pr)4, is analogous to Al(OR)3, but its reactivity totally different, since titanium is a true TM and not just a Lewis acid main group metal like aluminium. Ti(O-i-Pr)4 is the catalyst for sharpless epoxidation, so this might be a good reaction to look at as an example of early-TM alkoxide reactivity.

To answer your questions in list form:

Pretty much all alkoxides are susceptible to acidic attacks, but, being fairly strong bases themselves, they are probably stable to weaker bases
Although they will react with acids, they will not necessarily react with electrophiles unless they are stronger electrophiles than the metal center; however, early-TMs are fairly strong electrophiles. I don't think people use TM alkoxides for ether synthesis though, and you always have to keep in mind d orbitals and the unique rules of TM chemistry when you're discussing TM complexes.
As mentioned above, and in my comment; "You can't have MX + M'(OR) -> M(OR) + M'X, where M=alkali metal and M'=TM, because the equilibrium would lie in the opposite direction, since the highly nucleophilic alkali metal alkoxide M(OR) will attack the electrophilic transition metal M'X (especially if it's an early-TM). As Schrock says in (1), "the relatively high electrophilicity of an early transition metal makes dissociation of alkoxide ion, at least in a neutral species, unlikely".
If you can find a suitable means of seperation like a solvent that will precipiate your metal alkoxide, you should be able to seperate it, although some catalysts will be deactivated after a certain number of cycles.

(1) R. R. Schrock, Polyhedron, Vol. 14., No. 22, pp. 3177-3195, 1995

(2) D. C. Bradley, R. C. Mehrotra and D. P. Gaur, Metal Alkoxides. Academic Press, New York, 1978

(3) K. P. Gable, J. J. J. Juliette. J. Am. Chem. soc, 1996, 118, 2625-2633

(4) J. R. Fulton, A. W. Holland, D. J. Fox, R. G. Bergman. Acc. Chem. Res., 2002, Jan; 35(1): 44-56


Spin spin coupling in a proton NMR of an ester?

In the most common case you can only observe 3J-couplings, any spins further away than three bonds do not cause a visible splitting in your spectra. The exception are couplings via double-bonds or aromatic systems, there you can often see small couplings over four bonds.

In this case the spins are too far away from each other to cause a visible splitting, you are correct that it would result in a singlet a quartet and a triplet.

In the case of OH and NH protons, chemical exchange can play a role. This is a bit more complicated, but the simple version is that protons that can exchange quickly with the solvent are often not observable in NMR (or just as a very broad signal) and due to that also don't couple. Though in the most common cases the OH proton is usually too far away from the other protons in the molecule to show a coupling anyway.


As noted in the comments, lye burns are very damaging to skin and especially mucous membranes. This damage can be drastically reduced if the lye is washed off as quickly as possible.
One reason that these solutions are considered particularly dangerous is that there is no discomfort immediately upon contact (in contrast to acids, which tend to produce burning sensations immediately). Instead, the affected surface simply feels a bit soapy.  As a result, the presence of lye on skin may not be noticed until it is too late to prevent damage by washing.
On a side note, the "it's sold in hardware stores" argument is not a good test for chemical toxicity. There are numerous highly toxic chemicals that can be purchased readily.


Fjords and bays

Fjord is a term originally developed by chemists studying polycyclic aromatic hydrocarbons, along with the term bay.

The descriptions aren't rigidly defined in the same way as stereochemical descriptors, but were designed to be 'self-explanatory', following the real world uses of the words1

Bay: "a broad inlet of the sea where the land curves inwards."
Fjord: "a long, narrow, deep inlet of the sea between high cliffs, as in Norway, typically formed by submergence of a glaciated valley."

In an attempt unify the usage of these terms, IUPAC issued a set of guidelines for their usage in 2014.2

Some historical context

The earliest publication of these terms was given in the references above by Harry Heaney, Keith Bartle, Denny Jones, and Peter Lees who were working at the now non-existent Institute of Technology (Bradford, UK). They were studying triphenylene, and trying to find descriptions for the various environments present.

  Circumnavigating triphenylene, the British crew sighted two essentially different kinds of H's; and Derry Jones spontaneously called one type "bay" and the other "peninsular." They published these descriptive "self-explanatory" names,40 but not without some apprehension. In fact, Professor Jones confessed to having dreamed that referees rejected their manuscript on account of its "bay" and "peninsular" nomencla ture

Reference:

1 Spectrochim. Acta 1966,22,pp941-951; Trans. Farad.  Soc. 1967,63,pp2868-287

2 J. Polycyclic Aromatic Hydrocarbons, 2015, 35, pp 161-176


What are disadvantages of using sublimation vs. recrystallization as a method of purification?

The most notable difference is that sublimation requires a volatile solid, whereas most solid can be recrystallized from a solvent system that is not too exotic.
Recrystallizations tend to recover more pure product.
In regards to green chemistry, sublimations are preferred as they generate no solvent waste.


Clear distinction has to be made between Arrhenius-equation and transition state theory. The formulas for TST can be derived, as opposed to Arrhenius equation, which by itself, is not based on a solid physical derivation. Basically any process can be modeled by an Arrhenius equation that has the formula:

$$ r \propto \exp{(-E_a/RT)} $$

Any molecular process can be reasonably approximated by this formula if it is a statistical transition between states with different energy. The reason for this is that the statistical probability of a state if proportional to the Boltzmann factor $g \exp{(-E/RT)} $, which is readily Arrhenius-like.

So what are some examples for Arrhenius-type behaviors? For example, if you look up the term Arrhenius-type equation on ScienceDirect you can see that research papers have been published on solid strain rates, membrane permeabilities, solubility rates and corrosion.

So, as you can see, an Arrhenius-type equation can describe a huge number of physical/chemical processes, so it is quite reasonable to use it for solid state reactions. The question is only how to find the energy of activation and the pre-exponent factor. For this, there is no general recipe. The best thing to do is to use them as empirical model parameters to be measured by experiments.


How exactly is activation energy defined?

You are right that the energy of the initial state is a function of temperature. But in practice for this equation the energy difference $E_\mathrm a$ usually is fixed for an arbitrary temperature (in the middle of the diapason of experimental measurements) and assumed temperature-independent.
From Wikipedia:

While this equation suggests that the activation energy is dependent
on temperature, in regimes in which the Arrhenius equation is valid
this is cancelled by the temperature dependence of $k$. Thus, $E_\mathrm a$ can be
evaluated from the reaction rate coefficient at any temperature
(within the validity of the Arrhenius equation).

As to the physical meaning of activation energy, citing Wikipedia:

Both the Arrhenius activation energy and the rate constant $k$ are
experimentally determined, and represent macroscopic reaction-specific
parameters that are not simply related to threshold energies and the
success of individual collisions at the molecular level.

In other words, activation energy $E_\mathrm a$ is a semi-empirical quantity and is not considered to have a strictly defined physical meaning (at least on molecular level). Note that the Arrhenius equation
$$k=A\exp\left(-\frac{E_\mathrm a}{RT}\right)$$
is itself considered to be a semi-empirical relationship.
In the IUPAC Gold Book (DOI: 10.1351/goldbook.A00102), activation energy is also defined to be an empirically determined quantity, in line with the above:

An empirical parameter characterizing the exponential temperature dependence of the rate coefficient $k$:
$$E_\mathrm a = RT^2 \frac{\mathrm d (\ln k)}{\mathrm dT}$$


Why is the solution of ammonium bifluoride more acidic than aqueous ammonium fluoride?

This question and the one referenced by Mithoron address NH$_4$HF$_2$ as an entity in itself which must be examined as a whole.
It is easier conceptually to rewrite the formula for the compound as NH$_4$F.HF. Consider adding NH$_4$F to water: you get a pH near 7. In a separate container, add HF to water (~0.1 M); you get a pH ~1 https://ehs.princeton.edu/book/export/html/197.
Combine the solutions and the pH will settle out somewhere between 1 and 7, because of the common ion effect: F$^-$ from the NH$_4$F inhibits the ionization of HF, but the solution is still acidic. The approximate pH can be calculated from the ionization constant of HF and the concentrations of NH$_4$F and HF.
The concept of the HF$_2^-$ moiety dominating the discussion for a solution is probably making the analysis murkier than it needs to be. It's just a hydrogen-bonded complex that happens to retain some semblance of existence in the solid. It is fairly stable: mp 126$^0$ C, bp 240$^0$ C (with decomposition). https://en.wikipedia.org/wiki/Ammonium_bifluoride


Retro Diels–Alder reaction in synthesis of corallopyronin A

Here is my assessment of the mechanism. You are correct about the Dess-Martin periodinane (DMP) oxidation of 1 to 2. The resulting ketone 2 can also exist as the enols (E)-3 and (Z)-4. Any one of the three can undergo a retro-Diels-Alder-like thermolysis, but I have chosen (Z)-4 for the convenience of cyclizing ketene 5, which has lost acetone in the thermolysis. Finally, the ketene 5 affords pyranone 6.
ADDENDUM: Citation for the "retro-Diels-Alder". A. R. Katritzky, Z. Wang, M. Wang, C. D. Hall and K. Suzuki, J. Org. Chem. 2005, 70, 4854. https://doi.org/10.1021/jo050307m


In the simplest terms and most convenient definitions, the lowering of vapor pressure is not a colligative property while the relative lowering of vapor pressure is indeed a colligative property.

A bit more explanation: A colligative property is dependent on the moles of solute only. The ratio in the definition you provided eliminates the temperature dependence of formulation (assuming ideal solutions) by making the measurement relative to another quantity, which in this case is the vapor pressure of the pure solution. Otherwise, the influences on vapor pressure are multifactorial. I.e. external pressure may affect it, as well as temperature.

Here is a somewhat hand-wavy out-of-the-pocket calculation to demonstrate. Assume the vapor pressure can be expressed as an ideal gas. Therefore pressure may be expressed using the ideal gas law, $P = \frac{nRT}{V}$. Also know Raoult's law for vapor pressure, $P_s = \chi_s P_o$ where $\chi_s$ is the mole fraction. Using your definitions, we have for absolute pressure difference
$$P_o - P_s = P_o - \chi_s P_o = \left(1-\chi_s\right)P_o = \left(1-\chi_s\right)\frac{n_oRT}{V}$$
This is therefore dependent upon temperature. Now do the same for the relative difference in pressures
$$1-\frac{P_s}{P_o} = 1 - \frac{\chi_sP_o}{P_o} = 1-\chi_s$$
The expression is now dependent upon the moles only, i.e. a colligative property. Of course, in reality this depends on the assumptions we made under ideal Raoult conditions.


How to determine the stereochemistry after elimination of anti-1-bromo-2-methylcyclohexane?

You diaxial intermediate is good and you can clearly see that there is no way to achieve a Zaytsev product in an $\mathrm{E2}$ elimination since that would require eliminating the methyl group. Carry on with the elimination and then you have a double bond in the top half of the molecule and the methyl group pointing downards — molecule E if inverted by a vertical plane of symmetry traversing the $\ce{C-Me}$ bond.

But the product molecules are drawn with the double bond on the bottom. Thus, you need to rotate the molecule by $180^\circ$. If you do this with a rotation axis in the paper plane (again one parallel to the $\ce{C-Me}$ bond), the double bond is in the correct position but the methyl group is pointing upwards.

You can verify this by determining the absolute configuration of the ipso-methyl carbon: it should be (R) according to your chair drawing and it is (R) in answer C.


Is it necessary to add prefixes of halogens (more than 1) to the iupac name alphabetically?

With the preferred IUPAC name (PIN), there is exactly one way that you expect to find a compound written.

Assume you want to buy the compound from the chemical supplier of your choice but don’t have a CAS number at hand. What do you do? You search for the compound in the catalogue. In the old days, these were always paper books that you could use as a weapon but nowadays electronic catalogues exist to. Still, if you’re looking for 1-bromo-3-chloropentane but it is entered into the catalogue as 3-chloro-1-bromopentane, you’re going to have a very hard time searching for it.

There are also other reasons why having one defined standard way of naming compounds is strongly preferred over random but clear.


When is the carbaldehyde suffix correct?

Both names, ‘benzenecarbaldehyde’ and ‘benzaldehyde’, are unambiguous and describe the same compound. However, the preferred IUPAC name (PIN) is ‘benzaldehyde’.
Concerning aldehydes, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-66.6.1 Systematic names of aldehydes
Aldehydes are systematically named in three ways:
(1) substitutively, using the suffixes ‘al’ for $\ce{-(C)HO}$ and ‘carbaldehyde’ for $\ce{-CHO}$;
(2) by changing the ‘ic acid’ or ‘oic acid’ endings of retained names of carboxylic acids into ‘aldehyde’; the nomenclatural properties of acids are transferred to aldehydes; thus, preferred names of aldehydes correspond to preferred names of acids, and carboxylic acids that are not substitutable generate nonsubstitutable aldehydes;
(3) by using the prefixes ‘oxo’, denoting $\ce{=O}$, or ‘formyl-’, denoting the substituent group $\ce{-CHO}$.

In particular,

P-66.6.1.1.3 The suffix ‘carbaldehyde’ is used when the $\ce{-CHO}$ group is attached to a carbon atom of a ring or ring system, or to a heteroatom.

For example, ‘cyclohexanecarbaldehyde’ is a PIN.

However, according to P-66.6.1.2.1, the PIN for the compound given in the question is ‘benzaldehyde’. (This name is analogous to the preferred name of the corresponding acid ‘benzoic acid’, see P-66.6.1 (2) above.)

Nevertheless, according to P-66.6.1.2.2, the PINs for phthalaldehyde and terephthalaldehyde are the systematic names ‘benzene-1,2-dicarbaldehyde’ and ‘benzene-1,4-dicarbaldehyde’, respectively. (These names are analogous to the preferred names of the corresponding acids ‘benzene-1,2-dicarboxylic acid’ [not ‘phthalic acid’] and ‘benzene-1,4-dicarboxylic acid’ [not ‘terephthalic acid’], respectively.)


The distance that CO2 molecules travel in the air before they collide with each other

You need to calculate the probability of one (CO2) molecule being the nearest neighbour of another.

Let $w(r)$ be the probability that the nearest neighbour occurs between distance  $r$ and $r+dr$ (on a spherical surface). This must be the probability than no molecules exist closer to the victim molecule,  which occurs at a distance $r$, and that this molecule exists in the shell $r$ to $r+dr$. Thus,

$$ w(r)=\left(  1-\int_0^r w(r)dr \right)4\pi r^2 n$$

where $n$ is the average number of molecules / unit volume. (1 mol/m3 $\equiv 10^3N_A/10^{27} = 0.6023 $ molecules / nm$^3$).

Solving this gives

$$ w(r)=4\pi r^2n\exp(-4\pi r^3n/3)$$

This function is zero at zero goes through a maximum and is  small at large. distances, see figure.

If you want a single number then the average distance between molecules is

$$\langle r \rangle =\int_0^\infty rw(r)dr =\Gamma (4/3)/(4\pi n/3)^{1/3} =0.55396n^{-1/3}$$

where $n$ is in number of molecules/volume then $n^{-1/3}$ is in units of length.

( $\Gamma(4/3)$ is the gamma function, sort of equivalent to a factorial but for a non-integer number)

The figure below shows $w(r)$ the probability of a near neighbour at distance $r$ for some concentrations.

If what you want is the total path taken accounting for all collisions then you would have to start with one the of the $w(r)$ distributions corresponding to your concentration and do a Monte Carlo computer calculation to work out the total path, not easy and v long calculation probably. There will be a statistical mechanics way of calculating this I'm sure but I'm not aware of it at present.


What is the temperature dependency of the van-der-Waals constants?

$a$ and $b$ are supposed to be constants. But, they in fact aren't, so your question is in principle valid.

vdW is not a very good model to describe $V_m(p,T)$ of any actual non-ideal gas over a larger pressure or temperature range, and if you take a really good model (those have 6 or 8 parameters), you can just mathematically derive $a(p,T)$ and $b(p,T)$.

However I doubt there is a general trend for a and b with increasing temperature or pressure. I'm sure you could find one for similar (e.g. mono- or diatomic, elementar) gases.

vdW is useful (just like virial coefficents and other models with only two or three parameters) to accurately model $p,V,T$ in a limited range without a large computational effort, e.g. for process control.

(The value of the vdW theory is imo an educational one: It gives a convincing reason why real gases are non-ideal, and correctly describes the trends.)


The pH = pKa of the equivalence point.
it is the best factor to decided during the titration .
because if the range of the indicator is less , then we can't find the correct volume used in titration.this factor should be considered but not most affective one.


Using Roman Numerals to indicate a compound

No, those are two distinctive species with different names:

$\ce{CuCl2}$: copper(II) chloride;
$\ce{[CuCl4]^2−}$: tetrachloridocuprate(II).

Oxidation number denoted with the Roman numerals in parentheses refers to the element it's placed after and doesn't carry any additional information besides the fact that copper exists as $\ce{Cu^2+}$ in both compounds.

Nomenclature rules were designed that way so that a name would unambiguously define a single compound only (I'd suggest the Red Book's Nomenclature by IUPAC [1] as a reference).

References

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


According to Wikipedia and my experience, glycine is soluble in water and is a polar molecule. The reference you saw, uses polar and non-polar to mean if there is another group other than the carboxylic acid and the amine group.

  Solubility in water   24.99 g/100 mL (25 °C)


What can I do if I don't get a phase separation between my organic and water phases?

One such cause, where separation does not take place is the nature of the organic solvent. Its better to check the miscibility of the solvent with water. There are several websites, where one can get this info. For example 1 and 2.

Even with hydrophobic solvent, if the problem persists, one can add NaCl and then check the separation.


Hellmann-Feynman Forces with Hartree-Fock

It is not at all surprising that the forces you get from application of the Hellmann-Feynman theorem do not match the true force. The reason is the appearance of Pulay forces: the atomic orbital basis set depends on the nuclear coordinates, and changing the position of the nuclei also changes the basis set. This yields the contribution you're missing in the plots.

cc-pVDZ and cc-pVTZ are still pretty small basis sets. I would expect that if you add diffuse functions, the result improves quite a bit, since the basis set covers more degrees of freedom. Furthermore, to get something meaningful you'd have to compare e.g. cc-pV5Z vs cc-pV6Z, which are closer to the complete basis set limit.

However, already in your plots you can see that the agreement is better from cc-pVDZ to cc-pVTZ when you look at the far-range behavior. A better way to plot these would be to plot the difference, i.e. the Pulay force.

I would also imagine that decontracting the basis sets helps, because if you imagine just moving the electron density but not the basis functions, then you will see that the description of the core orbitals is impoverished quite a bit as the contraction coefficients are wrong.

Actually, the result in cc-pV6Z is probably still not very close to the Hellman-Feynman limit. When you move the nucleus, 1s core orbitals start to gather p, f, etc character. Thus, when you move the nuclei but not the basis functions, your original basis set should have e.g. very tight p functions that are able to describe this effect.

If you want to toy around with this, you should look at calculations in larger and larger primitive basis sets. Even-tempered exponents should work fine. Start with a small set of s functions, and study how the result changes when you add steeper or more diffuse functions. Next, look at the p shell: which exponent changes the result the most? And then again start adding steeper and more diffuse functions. Repeat ad nauseam. (This is called completeness-optimization, and I've published an automatic procedure for this.)


Why is the gradient of the curve of a strong base titrated with strong acid small up until equivalence?

Let's calculate the $p$H by changing the volume of $\ce{HCl}$ added. I will use the unit millimole ($1 mmol = 0.001 mol$), which is more practical than the mole.
In the very beginning the solution contains $50 mL~ \ce{NaOH}~ 0.1 M$, or $5 mmol$ $\ce{NaOH}$. Let's take, for example, the point when you have added $30  mL~ \ce{HCl} ~0.1 M$ (or $3.0 mmol$$ ˙\ce{HCl}$). After this addition, the solution contains $5 - 3.0 = 2.0 mmol$ NaOH in a volume $50 mL + 30 mL = 80 mL$. The concentration of $\ce{NaOH}$ is $\frac{2.0 mmol}{80 mL} = 0.025 M$. The concentration $\ce{[H+] = 10^{-14}/0.025 = 4 10^{-13}M}$. So $p$H = $12.4$. You see that the pH has fallen from $13.0$ to $12.4$ for more than half the trip to the equivalencce point. A change from $13.0$ to $12.4$ is rather small. The point of the curve is still nearly at its initial value. The gradient of pH between the beginning and here is small.
If you take the same calculation with an addition of $45 mL~ \ce{ NaOH}~ 0.1 M$, or $4.5 mmol$, the solution contains still $ 5 mmol - 4.5 mmol = 0.5 mmol ~\ce{NaOH}$ in $95 mL$. The concentration in $\ce{NaOH}$ is $\frac{0.5 mmol}{95 mL} = 5.26·10^{-3} M$. $\ce{[H^+] =  1,9·10^{-12}}$, and $p$H = $11.7$ We are near the equivalence point but the pH has not decreased a lot. It is still $11.7$. As you see the gradient (or the slope) of the curve is not important in the part of the curve between $0 mL$ and $45 mL$.
And suddenly between $45 mL$ and $50 mL$, the $p$H drops from $11.7$ to $7$. This is a huge change in just $5 mL$.
If you want you can calculate one or two other points along this titration. Try for example $49 mL$. You will be surprised ! It still gives a rather high $p$H value.


What is actually the difference between valence bond theory and molecular orbital theory?

TL;DR

VB theory treats atomic orbitals (including hybridized orbitals) as providing a good mathematical/physical description of the true form of the molecular wavefunction. MO theory uses atomic orbitals (with Gaussian radial functions) as a tool of computational convenience in an effort to define a molecular wavefunction that in its final form often bears little resemblance to the original atomic orbitals, or to the VB theory description.

Long Version

The key difference between the VB and MO theories lies in their respective assumptions of the underlying mathematical/physical form of the electronic wavefunction.

VB theory holds tightly to the mathematical structure of atomic orbitals (AOs), assuming that the electronic structure of a molecular system is described adequately by retaining the atomic orbitals (possibly hybridized) from each constituent atom, and overlapping them to form the needed description of bonding.  VB calculations are computationally much more complex: the AO basis is generally non-orthogonal, and the linear algebra and other techniques required to cope with this non-orthogonality are much more involved.

More broadly, the need to invoke resonance (superposition of multiple valence-bond structures) and other concepts highlights that, in fact, often this adherence to atomic orbitals and their valence-bond overlaps is not adequate for a proper description, at least not without extensive use of methods analogous to the configuration interaction of MO theory.  VB theory does hold certain advantages, however—for example, I believe the AO basis does sometimes allow for a more direct linkage to fundamental chemical concepts such as hybridization and Lewis electon-pair theory. To note, Sason Shaik is one of the strongest modern proponents of the theory I know of—he describes some of these strengths and their applications in his recent retrospective article (doi:10.1016/j.comptc.2017.02.011).

MO theory, at its core, casts away most assumptions regarding the mathematical structure of the electronic wavefunction, as long as properties like indistinguishability and the Pauli exclusion principle remain satisfied.  Calculations within MO theory can use a wide variety of mathematical structures to try to represent the wavefunction, ranging from gridded numerical quadrature approaches (see here and here, and also orbital-free DFT) to Slater orbital computations to the Gaussian-orbital based calculations used most commonly today.

One potential point of confusion is that MO theory in its most common implementations makes direct use of atomic orbitals in composing the molecular wavefunction (see LCAO theory). The distinction to be made here is that instead of attributing significance to the atomic orbitals as an intrinsically good descriptor of electronic structure, they are used merely because they afford a convenient mathematical form for efficient computation, in trying to construct the MOs.  Particular advantages are obtained from the use of atomic orbitals constructed from Gaussian functions, since the product of two Gaussians is another Gaussian (even in multiple dimensions) and numerical integration of n-dimensional Gaussian functions is straightforward.

It should be noted that, for as much of an improved description it affords, MO theory still is an approximation. For a given system, the electronic wavefunction just IS the highly-abstract mathematical object that satisfies the appropriate Schrodinger equation.  MO theory is used extensively because it provides one of the most computationally tractable paths to getting in the neighborhood of the "real answer."

Practical Aspects of MO theory

The "simplest version" of MO theory for a time-independent system under the Born-Oppenheimer Approximation is the Hartree-Fock method, which provides a remarkably good description considering the significant simplification represented by its mean-field approximation of the electron-electron interactions.  Hartree-Fock is in general not good enough for computations at the level of chemical accuracy (~$1\mathrm{\ kcal/mol}$), however, and numerous methods have been developed to refine it: the most often used techniques are Moller-Plesset perturbation theory, configuration interaction, and coupled cluster.

In-Depth Reading on MO Theory

For a deep dive into the underlying mathematics of MO computations, it's hard to beat the classic Szabo & Ostlund. As a more readable, lighter-weight introduction to MO theory and its implementations, I would recommend Frank Jensen's Introduction to Computational Chemistry.


Yes, you definitely can, but it's complicated. Because you are a mathematician it might be easier to understand.
Introduction
On your picture is one-dimensional representation of the energy function of multiple variables. Variables are coordinates of all atoms in the molecular system. This function define the potential energy surface or PES. Sometimes you can simplified it and represent as a three-dimensional surface using a linear combination of coordinates. Your picture is the energy profile of the PES. Where minimums are equilibrium states of the system and saddle points (minimax) are transitional states.
The reaction progress on the picture is the intrinsic reaction coordinate (IRC) which is a one-dimensional representation of the a linear combination of vectors of your molecular system vibration.
Real energy profile
Consider, for example, the reaction $N_2O_2$ with $H_2$. Atoms in both molecules can oscillate in a specific well-defined way which gives you the molecular vibration. You need to solve quantum differential equations for the molecular system to get vibrational frequencies and vibrational vectors. To solve it analytically is incredibly difficult, but you can solve it numerically using quantum chemistry software. It will compute energy over intrinsic reaction coordinate (IRC) as a set of points with a certain displacement or step. Using simple spline function you can get a real energy profile.

Approximation
It is possible just to use a square function. According to the harmonic approximation it is precise at the vicinity of equilibrium states, but useless with transitional states. Sometimes to avoid this problem people use energy profile diagrams. Usually people are not interested in the form of function but the energy differences (activation energies, Gibbs energies and etc.)

If precision is not the case you just need to know energies of initial separated molecules, the transitional state energy (at the saddle point) and the energy of the product (which you can compute with quantum chemistry software) and join square functions at the found energies. You also need to remember that the energy profile is made for the elementary reaction.
Calculation
There are a lot of tutorials how to compute energies of initial systems, transitional states and products with quantum chemistry software. It should be mentioned that you will need to sum equilibrium energy of computed separately molecules to get your initial energy. The most difficult part is to compute the transitional state energy. You need to find a molecular geometry which would be very close to your potential transitional state. You need to calculate Hessian (second derivatives) and start minimax optimization (saddle point optimization).


Applying Huckel's rule to non-planar compounds

Number 2 is not aromatic because transannular $\ce{C-H}$ interactions cause the molecule to be non-planar although it satisfies $4n+2$. This problem was solved by the synthesis of number 4, which, while a little less than planar but rigid, displays aromatic properties (NMR, etc., Ref 1).
References:

E. Vogel, H. D. Roth, "The Cyclodecapentaene System," Angew. Chem. Int'l. Ed. 1964, 3(3), 228-229 (DOI:  https://doi.org/10.1002/anie.196402282).


What could possibly cause a "melting" reaction between vinyl and plastic?

It is probably a consequence of the plasticiser not the polymer

It is hard to be sure what has happened here without checking the details of which polymers are involved which the picture can't make clear. But we can make some intelligent guesses based on commonly used polymers.

First: the cable isn't made from natural rubber. Most likely it is made from plasticised PVC. This is a good and cheap cable material with good insulating properties and doesn't decay like rubber from interacting with atmospheric oxygen. This involves a lot of plasticiser, usually something like a phthalate. The cable might contain 10-40% of plasticiser by weight.

Depending on the exact polymer formulation and the specific plasticiser, the plasticiser can migrate into other materials if they are in contact for long periods of time. When I used to work in a lab, for example, the cheap PVC tuning we used in low pressure gas lines would sometimes accumulate a thick liquid in the lower parts of the tubing loops if used for years: this liquid was plasticiser that had migrated out of the PVC over years of use. This migration is worse when other volatiles are in contact with the polymer.

This may be what is happening in this case. In warm conditions the plasticiser in the PVC might migrate and damage other plastics. It isn't possible to tell from the picture what the breadboards are made from but ABS is most likely in a good quality breadboard, though acrylic, polystyrene or polycarbonate are all possible. All of these polymers can be affected by the migration of plasticiser from PVC as this study shows (though it depends on the specific plasticiser). Good choice of materials for the board and plasticiser can minimise this problem but cheap breadboards and cables are likely to have worse problems.

In short, this is the likely problem here.


Energy level of hybrid orbitals in the molecular orbital energy diagram

When we have an sp hybrid orbital, it is usually made of the s orbital and the p orbital that points in the bounding axis ($p_z$). It's energy will be the mean of the energy of the initial orbitals.

In the example of $\ce{N2}$, it is essential to bear in mind that each sp won't form an $\sigma$ and an $\sigma^*$ orbital. Only the one pair of the sp orbitals, the one that overlap the most, will do. The other pair will contribute to the $\ce{N2}$ lone pairs because there is nearly no overlapping. The 2 others bonds that form the triple bonds are made with the 2 left orbitals $p_x$ and $p_y$.

By the way, it is not necessary to use hybrid orbitals for the molecular orbital theory.

Credits to Philipp and Ben Norris.


Is there any food-safe way to break down capsaicin (chilli) so food is less spicy?

It is an interesting problem, from my experience the capsaicin behaves in a way which is consistant with it being a liphilic substance. I know that if you fry chilli pueppers with some cheese that the cheese helps to bring out the flavour of chilli peppers.

I know that anacdotal evidence suggests that a dairy product such as milk or yougurt helps to stop your mouth burning. I think that this is reasonable if we assume that the chilli pepper toxin dissolves in the fatty droplets in the milk or yogurt. This might make it a better mouth decontamination liquid than water.

I think that the way to test this would be to put known amounts of capsaicin on filter paers and then soak them in milk and water and then measure the capsaicin concentration in the liquids after soaking. It could be done with HPLC or if we were able to get C14 labled compound we could use liquid scintillation counting.

From my student days I recall that a vegatarian or sasuage chilli was hotter than one made with minced beef. So it is possible that the capsaicin adsorbs onto the surface of the beef. Maybe you should consider using the langmuir isotherm to model such a thing.

As a simple alternative just put less chilli in the food.


According to NIST Chemistry Webbook, for oxygen:

$$T_\mathrm{boil} = \pu{90.2\pm0.2 K}$$
$$T_\mathrm{fus} = \pu{54.8\pm0.2 K}$$
$$T_\mathrm{triple} = \pu{54.33\pm0.6 K}$$

The $T_\mathrm{boil}$ and $T_\mathrm{fus}$ are from Ref.1 while Brower and Thodos reported $T_\mathrm{triple}$ as $\pu{54.363 K}$ at pressure of $\pu{1.125 mmHg}$ (Ref.2). The uncertainties were given by Thermodynamics Research Center, NIST Boulder Laboratories, CO.

Accordingly, $T_\mathrm{boil} = \pu{-183 ^\circ C}$, $T_\mathrm{fus} = \pu{-218.4 ^\circ C}$, and $T_\mathrm{triple} = \pu{-218.8 ^\circ C}$. Therefore, as you correctly suspected, the values in the textbook is a misprint.

References:

Alex G. Streng, “Miscibility and compatibility of some liquefied and solidified gases at low temperatures,” J. Chem. Eng. Data 1971, 16(3), 357-359 (https://doi.org/10.1021/je60050a024).
Gerald T. Brower, George. Thodos, “Vapor pressures of liquid oxygen between the triple point and critical point,” J. Chem. Eng. Data 1968, 13(2), 262-264 (https://doi.org/10.1021/je60037a038).


While filling electrons, we follow Aufbau principle, but not while removing them. Why is this so?

Usually when adding electrons based on the Aufbau principle, you go from one element to the next highest one, e.g. from $\ce{Ti}: \ce{[Ar] 4s^2 3d^2}$ to $\ce{V: [Ar] 4s^2 3d^3}$. Thus you add not only an electron but also a proton to your atom.

When you remove electrons to get to a cation, you only remove electrons. Thus it is a different situation, with different interactions between nucleus and electrons, which affects the energy of the orbitals.

Also keep in mind, that the orbital picture is just a simplification. Strictly speaking (and from a purely theoretical point of view) orbitals do not exist, they are merely a mathematical crutch to solve the Schrödinger Equation. Although in practice this picture works surprisingly well, hence people use it a lot.

A proper quantum chemical calculation will give you the correct ground state, but you might have problems identifying it as something like $\ce{[Ar] 4s^2 3d^0}$. It will be a mixture (linear combination) of many electron configurations, although $\ce{[Ar] 4s^2 3d^0}$ might be the most important one for $\ce{V^{3+}}$.


Relationship between molar mass and density in solids

If your sample is solid and crystalline, knowledge of the unit cell's dimension and symmetry (e.g., fcc) and density allows you to determine the molecular mass of your compound.
In fact, in crystallography, you may determine the macroscopic density of your sample (e.g., find a liquid which i) wets your crystals and ii) lets your crystals float) in first place.  If you then know the dimensions of the unit cell by the diffraction experiment, knowledge of the chemical formula per unit cell may tell you the number of formula units per unit cell.  See e.g, this demonstration about Ne.
The example above about Ne may be extended to crystals consisting of molecules for that you may assign atomic volumes to each element that add up to a molecular volume. An early approximation for organic compounds[1] was to assign $\pu{18 Å^3}$ for any atom of $(\ce{C, N, O})$ and $\pu{2 Å^3}$ for any atom of $\ce{H}$.  Decades later,[2] backed by much more experimental data, it was possible to refine these assignments, e.g., to average volumes of $\pu{13.87 Å^3}$ ($\ce{C}$), $\pu{11.8 Å^3}$ ($\ce{N}$), $\pu{11.39 Å^3}$ ($\ce{O}$), and $\pu{5.08 Å^3}$ ($\ce{H}$), respectively.  Large deviations from these guesses from the then experimentally determined molecular volume may be caused by solvent-accessible voids in the crystal structure.[3]
References:
1 Kempster, C. J. E.; Lipson, H., A rapid method for assessing the number of molecules in the unit cell of an organic crystal in Acta. Cryst. B28, 1972, 3674; doi: 10.1107/S056774087200857X.
2 Hofmann, D. W. M., Fast estimation of crystal densities in Acta. Cryst. B58, 2002, 489-493; doi: 10.1107/S0108768101021814.
3 Florence, A. J., The use of supplementary information to solve crystal structures from powder diffraction in International Tables for Crystallography (2019). Vol. H. ch. 4.4, pp. 433-441; doi: 10.1107/97809553602060000959.


Partial Coverage During Zinc Plating Bath

It is hard to imagine this would be due to an electric field or a magnetic field.

I would say this has to be a different material or coating in the region where you are having bad coverage.

Would it be possible to sandpaper this piece to get rid of any potential coating? Even leftover glue from some sticky on top would cause an issue.

You could also try to electrochemically clean the surface immediately prior to the plating. Applying a highly cathodic potential should lead to some electrolysis that would evolve hydrogen gas which would knock any impurities off the surface.

Either method could roughen the surface somewhat, I don't know enough about motorcycles to say if this would be a dealbreaker.


How to find the temperature change of a solution?

The main problem here is a simple mistake in the algebra. You rearranged:

$\pu{n\timesΔH_{sol} = m\times C\timesΔT}$

to

$\pu{ΔT = \frac{m\times C}{n\timesΔH_{sol}}}$

rather than

$\pu{ΔT = \frac{n\timesΔH_{sol}}{m \times C}}$

Additionally, the molar enthalpy of solvation is given in the problem as units of kJ/g, not kJ/mol. Presumably, this is an error in the given problem. According to Parker, V . B ., Thermal Properties of Uni-Univalent Electrolytes, Natl .
Stand . Ref . Data Series — Natl . Bur. Stand .(U.S.), No .2, 1965, the molar enthalpy of solution for $\ce{NH4Cl}$ is $\pu{14.78 kJ/mol}$.

There was also a slight miscalculation in the moles of solute. Where you calculated $\pu{0.165 moles}$ of $\ce{NH4Cl}$, you should have gotten:

$\mathrm{8.5 g / 53.49\frac{g}{mol}=0.159~mol}$

These errors corrected, plugging in the values to solve for $\Delta \text{T}$ is trivial and gives:

$\pu{ΔT = \frac{\pu{0.159 mol}\times \pu{14.78 kJ mol-1}}{\pu{100g} \times \pu{4.186J~g-1~K-1}} = 5.6K}$


How should I measure out a solution given in nanorods per ml to make a solution of a set number of mols?

In order to avoid any possible misunderstanding, we should first clarify that “nanorods” (and thus also “nanorods per millilitre”) is not the name of a unit. A nanorod is a very small (so-called nanoscale) object.

The number of nanorods in one millilitre of the given solution is $N_\text{nanorods}=6.3\times10^{11}$, i.e.

$$\frac{N_\text{nanorods}}{V_\text{solution}}=\frac{6.3\times10^{11}}{1\ \mathrm{ml}}=6.3\times10^{14}\ \mathrm{l^{-1}}$$
where $V_\text{solution}$ is the volume of the solution.

Since

$$c=\frac nV$$

and

$$N_{\mathrm A}=\frac Nn=6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}$$

where $c$ is concentration, $n$ is amount of substance, and $N_{\mathrm A}$ is the Avogadro constant, the concentration of nanorods in the given solution is

$$\begin{align}c_\text{nanorods}&=\frac{n_\text{nanorods}}{V_\text{solution}}\\[6pt]&=\frac1{N_{\mathrm A}}\cdot\frac{N_\text{nanorods}}{V_\text{solution}}\\[6pt]&=\frac1{6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}}\cdot6.3\times10^{14}\ \mathrm{l^{-1}}\\[6pt]&=1.0\times10^{-9}\ \mathrm{mol\ l^{-1}}\\[6pt]&=1.0\times10^{-6}\ \mathrm{mmol\ l^{-1}}\end{align}$$

The given nanorods are approximately cylinders with a length of $l=40\ \mathrm{nm}=4.0\times10^{-8}\ \mathrm m$ and a diameter of $d=10\ \mathrm{nm}=1.0\times10^{-8}\ \mathrm m$, i.e. a radius of $r=5\times10^{-9}\ \mathrm m$.

Therefore, the volume of one nanorod is approximately

$$\begin{align}V_\text{nanorod}&=\pi\cdot r^2\cdot l\\[6pt]&=\pi\times\left(5\times10^{-9}\ \mathrm m\right)^2\times4.0\times10^{-8}\ \mathrm m\\[6pt]&=3.14\times10^{-24}\ \mathrm{m^3}\end{align}$$

The density of gold is $\rho=19.32\ \mathrm{g\ cm^{-3}}=1.932\times10^4\ \mathrm{kg\ m^{-3}}$. Thus, the mass of gold in one nanorod is approximately

$$\begin{align}m_{\ce{Au}}&=\rho\cdot V_\text{nanorod}\\[6pt]&=1.932\times10^4\ \mathrm{kg\ m^{-3}}\times3.14\times10^{-24}\ \mathrm{m^3}\\[6pt]&=6.07\times10^{-20}\ \mathrm{kg}\end{align}$$

The molar mass of gold is $M_{\ce{Au}}=196.97\ \mathrm{g\ mol^{-1}}=0.196\,97\ \mathrm{kg\ mol^{-1}}$. Thus, the amount of gold in one nanorod is approximately

$$\begin{align}n_{\ce{Au}}&=\frac{m_{\ce{Au}}}{M_{\ce{Au}}}\\[6pt]&=\frac{6.07\times10^{-20}\ \mathrm{kg}}{0.196\,97\ \mathrm{kg\ mol^{-1}}}\\[6pt]&=3.08\times10^{-19}\ \mathrm{mol}\end{align}$$

Note that this amount is equivalent to only about $190\,000$ gold atoms since

$$\begin{align}N_{\ce{Au}}&=N_\mathrm A\cdot n_{\ce{Au}}\\[6pt]&=6.022\,140\,76\times10^{23}\ \mathrm{mol^{-1}}\times3.08\times10^{-19}\ \mathrm{mol}\\[6pt]&=1.9\times10^5\end{align}$$

Therefore, the concentration of gold in the given solution can be estimated as follows:

$$\begin{align}c_{\ce{Au}}&=n_{\ce{Au}}\cdot\frac{N_\text{nanorods}}{V_\text{solution}}\\[6pt]&=3.08\times10^{-19}\ \mathrm{mol}\times6.3\times10^{14}\ \mathrm{l^{-1}}\\[6pt]&=1.9\times10^{-4}\ \mathrm{mol\ l^{-1}}\\[6pt]&=0.19\ \mathrm{mmol\ l^{-1}}\end{align}$$

or

$$\begin{align}c_{\ce{Au}}&=N_{\ce{Au}}\cdot c_\text{nanorods}\\[6pt]&=1.9\times10^5\times1.0\times10^{-9}\ \mathrm{mol\ l^{-1}}\\[6pt]&=1.9\times10^{-4}\ \mathrm{mol\ l^{-1}}\\[6pt]&=0.19\ \mathrm{mmol\ l^{-1}}\end{align}$$


Do hydrogen sulfide and oxygen produce pure sulfur or sulfur dioxide? If both, under which circumstances does each scenario occur?

Sulfur dioxide is not a problem ; in an aquarium it will make very dilute sulfurous acid which will likely neutralize with some carbonate mineral/gravel. H2S is lethal , extended exposure at 50 ppm for a day , or short exposure at 500 ppm. On the other hand , what you smell in aquarium sand is less then 1 ppm, the sense of smell is very sensitive to H2S. There are several answers on the internet but I expect most people can smell H2S at 0.1 ppm. An aerator will have no significant affect on H2S in an aquarium. When I clean my pond , which is well aerated ,the smell of H2s is strong from decomposing leaves, etc.
Regarding the first part of your question ; Hundreds of tons a day of elemental sulfur are made ,world wide, every day by burning H2S with limited air at high temperature. By recycling and control of gas compositions the only products are elemental sulfur and water,


Cyclopentyl to cyclohexyl carbocation rearrangement

Your first step is correct. The reactive exomethylene group attacks the proton leaving a secondary carbocation (remember Markovnikovs rule). Now the five-membered ring can rearrange into a more stable six-membered ring by releasing ring strain. This is a fast intramolecular step. Finally methanol attacks the rearranged carbocation releasing a proton which is used catalytically in the next reaction.

Remember though that you can often not be completely sure about a mechanism before it is verified in practice (i.e. in the lab through an experiment). Sometimes true mechanism can be quite different from what appears to be most reasonable on paper!


Are graphite and hexagonal boron nitride aromatic

Graphite is definitely aromatic and boron nitride is at least partially aromatic. For instance, in this paper, the authors calculate the percent resonance energy (%RE) of graphite as a comparison to other known aromatic systems. The %RE is a measure of the total resonance stabilization per carbon atom relative to some reference system.

They find the %RE of graphite to be 3.061 and that of benzene to be 3.528. This value for graphite is larger than that calculated for naphthalene (2.924), thus for all practical purposes, graphite is essentially as aromatic as benzene.

In the case of hexagonal boron nitride(h-BN), I haven't been able to find any reference which says explicitly that the system is aromatic, but there certainly is some delocalization of $\pi$ electrons as I've seen a few papers which note that aromatic systems adsorb to BN nanotubes (which ought to be electronically similar to h-BN) quite strongly and a large portion of this binding interaction comes from dispersion (i.e. $\pi$ interactions). See here for instance.

You mention "inorganic benzene" which is the molecule borazine. Borazine actually only exhibits about half the total resonance stabilization per atom as benzene. This is attributed to the polarity of $\ce{B-N}$ bonds which inhibits delocalization of electrons. See this paper on the aromaticity of borazine for instance.

However, the first reference I gave found that graphite is nearly as aromatic as benzene, so I would be willing to speculate that h-BN is about as aromatic as borazine.

  Can I explain this using Huckel's rules?

I would be hesitant to use Huckel's rules on a $\ce{B-N}$ system as Huckel's rules come from Huckel theory which is really only meant to describe fully conjugated hydrocarbons. Nonetheless, people will point out when other systems follow these rules just in case it turns out to be significant.

In contrast to what I just said, however, there are boron clusters which have been found to follow Huckel's rule, but this doesn't mean you can extend Huckel's rules to heteroatomic systems such as h-BN at will.


Graphene combusts at 620 K. See http://pubs.acs.org/doi/abs/10.1021/jp410044v.

It is postulated that it burns at this temperature "because of a large interlayer spacing of graphene sheets (i.e., 5.1 Å)".


After going through several literature, finally I am able to answer these questions. It might be important starting point for someone wanting to know carbohydrate chemistry, that's why I'm posting my answer.

Yes, I'm right. For α- and β-glucose 27702 configuration are possible. So, in total $27702 \times 2 = 55404$ possible structures for glucose [1].

  For xylose, 81 rotamers were optimized per puckering conformation for a total of 3078 conformations. For α-glucose, β-glucose, and β-mannose, 729 rotamers were generated for each pucker, resulting in 27 702 conformations for each of these six-carbon monosaccharides. For GlcNAc, 972 conformations per ring pucker were generated for a total of 36 936 geometries. Thus, the initial set of conformations considered for all molecules in this study comprised 123 201 geometries.

Actually these nomenclature is known as Cremer-Pople (Dr. Pople is the Noble prize winner chemist for his outstanding contribution to formulate DFT) nomenclature. They defined puckering parameters and generalized ring structure conformations for 3/4/5/6 or more carbon species. That's a real interesting world!
Courtesy: Shinya Fushinobu, University of Tokyo
It might not be clear from this picture. So I will try to explain them separately.

Chair: Two chair conformation possible. (O-C2-C3-C5 forms the reference plane). If C-4 is on top then $^4C_1$ and vice versa.

Boats: Boats forms when two plane intersects each other. The bottom two corner atoms are used to distinguish. If two bottom corners are upside down then they are used as superscript on the left and vice versa.

Skew-Boats: Same as chair. Reference plane is formed by (O-C2-C3-C5). If C5 is on the top and C1 is in bottom then it's named as $^5S_1$

Reference

Mayes, H. B.; Broadbelt, L. J.; Beckham, G. T. Journal of the American Chemical Society 2014, 136 (3), 1008–1022 DOI: 10.1021/ja410264d.


Dioxygen, $\ce{O2}$ is a very special molecule. A good majority of organisms on earth use dioxygen to breathe and survive. Oxygen is also the second most abundant element in the sky, as well as the most abundant elements in the earth's crust. What makes it so special? Its because of its abundance, which is obvious, and also because of it's strong oxidizing power.

If you are a student below 11th grade, you might be taught that oxidation is addition of oxygen or the removal of hydrogen. I will now, introduce a new definition of oxidation.

Oxidation: The loss of electrons from a substance is called oxidation.

The reagents that cause substances to become oxidized are called oxidizing agents. $\ce{O2}$ is an example of an oxidizing agent. A popular oxidizing agent in analytical chemistry is Potassium Permanganate ($\ce{KMnO4}$ ).

Many times, an actual loss of electron doesn't take place, rather a shift of electrons from one element to another within the same molecule occurs, which is also considered as oxidation. To explain such stuff, we introduce the concept of oxidation numbers. An easy way to find oxidation number of an element in a compound is:

Ox.num=(Num. of bonds formed with electronegative atoms)-(Num. of bonds with electropositive atoms)

For example, take carbon dioxide. It's structure is:

Note that bonds made by an atom with a similar atom are not considered (like $\ce{O-O}$ or $\ce{S-S}$).

Carbon makes four bonds with more electronegative atoms (oxygen). So it's oxidation number is +4.

If oxidation number of any atom of a molecule is increased through a reaction, we say that oxidation has taken place. Note that as this happens, the oxidation number of the oxidizing agent will decrease.

Now, finally (with a deep breath), lets come to your question.

Speciality of Oxygen:

Oxygen in nature, as dioxygen has a 0 oxidation number. Since it's so electronegative, it almost always forms bonds with electropositive atoms, and so its oxidation number in compounds is always negative. Oxygen itself has such a high tendency to rip off electrons from other molecules. This is what makes oxygen very special.

There are so many examples of oxidation reactions with oxygen:

Combustion: Burning stuff itself is an oxidation reaction. If you burn coal(carbon), its oxidation number  goes from 0 (in pure carbon) to +4 (in carbon dioxide).

Respiration: Even our own bodies have to do an oxidation reaction to keep us with energy to do our daily work, like asking and answering questions on StackExchange. Glucose, which has a oxidation number of 0 (equal number of bonds with oxygen and carbon in its formula of $\ce{C6H12O6}$ ) is oxidized by the oxygen we breathe to carbon dioxide.

Much of the interesting properties of oxygen is due  to its oxidizing power. While many oxidizers are known, like fluorine for example; oxygen is, and will always, be the most well known one!


Would silica have the same effect as activated charcoal?

The silica gel granules are granular silicon dioxide (SiO2), the same matrial from what is the normal sand.

The pure SiO2 has the same effect to health as when you eat sand.

But because the granules are larger, may cause appendix problems (im not an doc). Anyway, this is not chemical related problem.

It can cause dehydratation or other problems, what is caused by it absorbtion capabality.

The bigger problem is, than the standard silica gel granules aren't pure SiO2.  The common additivas are what causes color change when it is moistured,  like cobalt chloride or others and thise compounds can be toxic!

So, don't eat it!


Nuclear reactions involve naked nuclei only. True or false?

Nuclear reactions involve energies far greater than anything in chemistry. Whether or not the nuclei are stripped of their electrons, does not matter at all (except for electron capture, of course; to capture an electron, you must have one around in the first place).

So yes, a beta decay of a neutral atom does indeed result in a cation. As for alpha decay, here we have a massive particle leaving, so the remaining nucleus would be shoved in the opposite direction with considerable force, no doubt enough to break it free of any chemical bonds and strip it of quite a few outer electrons. If not for this factor, the process would leave behind an anion.


Does the carboxyl group count as a branched chain

In organic chemistry, the terminology of branched vs. unbranched is generally used to refer to the structure of the hydrocarbon skeleton of a molecule. Hence, heteroatomic functional groups are usually not relevant. The structure in your question is therefore unbranched. For an authoritative reference, see the relevant sections of the IUPAC blue book.


Why does the Wurtz reaction form such an unstable compound with torsional strain?

In the Wurtz coupling reaction, an alkyl halide first reacts with a metal in a one-electron transfer step to produce an alkyl radical which quickly reacts in a second electron transfer with the metal to form a carbanion.  In the final step of the reaction, the carbanion displaces a halogen from a carbon-halogen bond by a $\ce{S_{N}2}$ process to form a carbon-carbon bond.  In the case of 1-bromo-3-chlorocyclobutane two reaction pathways are possible, the carbanion could react intramolecularly to form bicyclobutane or intermolecularly to from a cyclobutane dimer.  Considering the equation $$\ce{\Delta G = \Delta H - T \Delta S}$$ we would say that even though the $\ce{\Delta H}$ term is less favorable for bicyclobutane formation due to the increased ring strain, the change in entropy required to bring two separate molecules together to form the cyclobutane dimer overwhelms the enthalpic term and drives the reaction to proceed by the entropically more favorable intramolecular process.

Look at the cyclobutyl carbanion drawn in the figure below.  The carbanion is ideally set up for an intramolecular $\ce{S_{N}2}$ reaction (backside displacement) on the carbon containing the chlorine diagonally situated across from it.  Entropically, this is extremely favorable.

In summary, because 1) the carbanion and the chloride to be displaced are in the same molecule, 2) because they are so well positioned for the $\ce{S_{N}2}$ displacement reaction to occur and 3) because entropy strongly favors the unimolecular process (bicyclobutane formation) over the bimolecular process (cyclobutane dimer formation), bicyclobutane formation readily occurs.


Refer to page 77 of Chemistry NCERT text book, class 12, part 1.

You were actually right about the $\mathrm{10^4\ S\ cm^2/mol =S\ m^2 /mol}$. Check if the $\mathrm{10^3}$ you are referring to has molarity included in its equation because molarity is given as moles per litre.

You can refer to the online NCERT book at its official website. Thank you, I hope this was useful and sorry if I got it wrong.


First of all, full arrow represents a movement of 2 electrons. Half arrows represent movement of 1 electron.

Diels-Alder is a 6π electron pericyclic reaction. The diene provides 4π electrons and the dienophile 2π electrons to the formation of the bonds. The Diels-Alder reaction belongs to [4+2] cycloadditions according to the Woodward-Hoffmann nomenclature.

A purely formal way of determining the participating electrons consists of simply counting the electron flow arrows that are used to illustrate the mechanism and multiplying by two. Each arrow moves two electrons.
You have 3 arrows representing a reaction, so 3*2=6 electrons

The number of electrons participating in the cycloaddition determines wether the reaction follows either a supra-supra (s+s) or a supra-antara (s+a) mechanism.

This is a nice source for Diels-Alder mechanisms


Why do charges result on atoms when coordinate bonds are formed?

The first thing to note about the charges that form during the formation of a coordinate bond: They are a formality.

The second thing to note about the charges that form during the formation of a coordinate bond: They are a mere formality.

There is little correlation, if any at all, between the formal charges assigned to two atoms that have formed a coordinative bond and the actual charges observed in the compound. Take $\ce{H3N-BH3}$: Nitrogen is so much more electronegative than boron, that the electrons are still closer to the nitrogen. Thus, nitrogen has $\delta ^-$ while boron has $\delta ^+$.

So where does the formalism come from? When drawing Lewis structures, after having filled in all bonds and lone pairs, check every single atom and mentally cleave all its bonds homolytically. Count the electrons, subtract the amount it should have as a neutral atom and the difference is its formal charge. Boron would have four electrons if all four bonds be cleaved homolytically, but it should only have three, so it gets a $1-$ charge.

This is only a formalism so that you can basically derive all structures the same way. It has no physical significance.

There are enough people who were unhappy about the ‘wrong’ charge distributions that they proposed different methods to draw these compounds: some used arrows from donor to acceptor, others just drew a simple bond, knowing that it would be different from a covalent bond. Each method has its advantages and disadvantages and all have survived in at least one niche.


Exothermic reactions at high temperature and the chemical equilibrium

There are two competing problems in the Haber Bosch process: the equilibrium position and the reaction rate. The reaction: $\ce{N_2 + 3H_2 -> 2NH3}\ (\Delta H = -92.4 \mathrm{\ kJ/mol})$ is exothermic and the equilibrium is actually favourable at room temperature, but the rate is very slow due to the large energy barrier to breaking the triple bond in $\ce{N2}$. To increase the reaction rate, a catalyst is used at high temperature. As you say, the high temperature pushes the equilibrium towards the reactants (by the Van 't Hoff equation) so there is a competition between yield and rate.

In practice, high pressure is also used to increase the yield (since the products side of the equation has fewer molecules), and like many industrial processes, it's run as a recirculating process. At manageable pressures and temperatures, they may only get 10 or 15% yield after one pass, but the ammonia is removed and the unreacted gases passed back into the reactor, giving quite high overall yield.


kinetic and thermodynamic parameters of biodiesel production from lard

I don't expect a significant energy difference per ester bond between glyceryl and methyl esters of a fatty acid. The way to drive such a reaction forward is by removing the desired product as the reaction proceeds. Then more product will form as the system seeks a new equilibrium.


What is the geometric formula of CaI2 and AlCl3

$\ce{CaI2}$ is mostly ionic in the solid and aqueous phases.  In the gas phases it is a linear molecule.  See Ionic or covalent? Can some first hints be derived from the solid state
structures of alkaline earth metal halide adducts?

$\ce{AlCl3}$ has an octaheral structure in the solid phase.

In the liquid phase:

Trigonal planar monomers also exist at high temperature.

Source: http://en.wikipedia.org/wiki/Aluminium_chloride


67 failures at extracting acetaminophen from Excedrin(R)

Did you notice any change of colour (e.g. to brown) and/or the formation of gunk upon adding the 3M NaOH?

I'd assume one would have to boil it, but is the following scenario conceivable under your conditions:

alkaline hydrolysis of the acetaminophen (paracetamol) to p-aminophenol
aerial oxidation to the corresponding quinone imine
partial polymerisation of the latter

You have confirmed that acetaminophen was initially there. A a change in the constitution of Excedrine since the publication of the article can thus be excluded. Isn't a reaction (hydrolysis or whatever) under your extraction not the most likely explanation for the vanished acetaminophen?


Does increasing one of the electrolytes in a galvanic cell increase the voltage?

As suggested in the comments, the key is Nernst equation. The fundation is the Gibbs free energy given by:
$\Delta G = \Delta G^0 + RT \ln Q_r$
where $\Delta G^0$ is the standard free energy for the process, while $Q_r$ is the reaction ratio. Remember that the reaction ratio is the analogous quantity of the equilibrium constant, but with the non equilibrium concentration. Remember also that the Gibbs free energy is proportional to the voltage. Then you need to increase the reaction ratio.


Isopropanol in contact lens solution plastic bottle

Based on your description, it looks like you've determined both plastics correctly. In general, they both would work fine for storing isopropanol; however, according to the Thermo Fisher's Labware Chemical Resistance Table, HDPE is preferred:

$\ce{i-PrOH}$ – PET

$\pu{20 ^\circ C}$: E Little or no damage after 30 days of constant exposure;
$\pu{50 ^\circ C}$: N Immediate damage may occur. Not recommended for continuous use.

$\ce{i-PrOH}$ – HDPE

$\pu{20 ^\circ C}$: E Little or no damage after 30 days of constant exposure;
$\pu{50 ^\circ C}$: E Little or no damage after 30 days of constant exposure.


Why is the standard enthalpy of formation of black phosphorus not zero?

Reproducibility is more important than stability. To obtain black phosphorus, you have to heat your sample under high pressure for quite a while, and even then it may still contain a significant number of random crystal defects, so its properties are not quite the same each time you make it. White phosphorus, on the other hand, is prepared by sublimation and hence considerably pure.

That's why white phosphorus, rather than black, is assumed standard and has zero enthalpy of formation.


How can I detect a specific molecule on a surface?

Simple UV spectroscopy is not an option.

You won't find anything at all above 254 nm - probably just a a weak band in the 220 nm range. Indolizidines are pretty dead in UV.

Derivatisation towards a fluorescent compound would require clevage of a $\ce{C-N}$ bond, formally a dealkylation of the tertiary bridgehead amine.

I have no idea if a reagent like 2-naphthyl chloroformate, which does form fluorescent carbamates upon heating with tertiary amines will work here and honestly, I wouldn't bother to try.

As pointed out in Martin's comment, GC/MS after washing/extraction probably is the way to go.


Why is 1-bromo-3-methylbut-2-ene named like that and not as 2-(bromomethyl)but-2-ene?

Here's 1-bromo-3-methylbut-2-ene's structure:

You have only two possible chains to choose from: a 3-carbon chain and a 4-carbon chain (doesn't matter which methyl group would be part of that chain, the other one will become a substituent on the chain).

You won't choose the 3-C chain for two reasons:

It is not the longest chain.
It does not contain the double bond in the molecule.

Thus, you are left with a 4-C chain, which necessarily includes the bromine atom.


When removing a reactant and equilibrium shifts left, doesn't this make an excess of the 2nd reactant, forcing equilibrium back to the right?

The thing to be careful about with Le Chatelier's principle is that it relates to equilibria :

  Le Châtelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to reestablish an equilibrium.

Le Châtelier's Principle

Now when you say in the above "But when the reaction shifts to the left, isn't a side effect  ..." note you can't use Le Chatelier's principle here again so fast because the system has not reached an equilibrium yet, it is just moving towards one.

It is likely that the next thing you will study is how to actually calculate the new equilibrium conditions. This gives usually a deeper understanding of how changes in various conditions make the reaction move forward/backward establishing new equilibrium conditions.


To what degree can the Rare Earth Elements be interchanged in its workaday applications?

It depends on the particular use of the "rare earth". Mischmetal, containing an assortment of mixed lanthanides, can be used as-is for lighter "flints" or fire-starters. On the other hand, ultra-high purity lanthanide elements are needed to get specific spectral lines for lasers, for example.

Other uses, such as in magnets and in high-temperature superconductors, may be somewhat less demanding of purity. Didymium glass, for example, used in glasswork to block yellow sodium light, is a mix containing primarily samarium, praseodymium and neodymium.

Another expense involved in lanthanide production is separation and disposal of the radioactive isotopes from the ore. At one time, for example, gas lantern mantles contained mildly radioactive $\ce{ThO2}$, but those are no longer available in many countries due to concern about the alpha emitting thorium and the alpha and beta emissions from daughters.


Identifying orbital given electron probability density plot

I think your inference is correct, though the plot looks very awkward. There should be an exponential decay at large values of $r$, but in the plot the curve plunges sharply onto the $x$-axis. Also the shapes of each peak should not be so symmetric, but instead skewed towards the left.


Preparation of Grignard reagent. Free radical rearrangement

Grignard reagent is an organomagnesium halide in which magnesium and halide bond is ionic. It is not like magnesium is always inserted in between carbon and halogen bond. Once formed, magnesium-carbon bond should be configurationally stable under the conditions of the reaction (Ref.1), and is polar covalent (at least partially) so that $\ce{Mg}$ has to be connected to its bonding partner $\ce{C}$ in the first place. In this polar covalent bond, carbon ends up being the negative end of the dipole, giving its nucleophilicity. The reaction undergoes in free radical mechanism (yet it does not involve a radical chain mechanism) to form Grignard reagent:
$$\ce{R-X + ^.Mg^. ->[\text{r.d.s.}] R^. + ^.Mg+X-} \tag1$$
$$\ce{R^. + ^.Mg+X- -> R-Mg+X-} \tag2$$
Magnesium metal has two valance electrons. The first step is the rate-determining step in this two-step mechanism, which happens on the surface of the metal. In this first step, $\ce{Mg}$ metal transfers one of its two valance electrons to the halogen radical, cleaving carbon-halogen bond homolytically. The process in step by step mode to explain the outcome:
$$\ce{R-X -> [R^. + X^.] ->[^.Mg^.] R^. + ^.Mg+X-} \tag3$$
This first step forms two radicals: $\ce{^.Mg+}$ and $\ce{R^.}$. These two radicals then get terminated by coupling each other to give the Grignard reagent in second step. The rearrangement of $\ce{R^.}$ (if there any) during the second step depends on the rate of rearrangement compared to the rate of the radical termination. However, according to the Ref.1, both chiral cyclopropyl-magnesium bond and vinyl-magnesium bond do in fact retain their configurations from parent alkyl halides. The authors carefully chose non-ionizable alkyl halide in solvent for this study so that any rearrangement should happen during the reaction (e.g., cyclopropyl tosylate solvolyzes at a rate of $\pu{5.7 \times 10^{-13} s-1}$ in actic acid at $\pu{25 ^\circ C}$, Ref.2). The excistence of $\ce{R^.}$ intermediate is proven by trapping it using $\ce{TEMPO^.}$ radical (Ref.3).
References:

H. M. Walborsky, "Mechanism of Grignard reagent formation. The surface nature of the reaction," Acc. Chem. Res. 1990, 23(9), 286–293 (https://doi.org/10.1021/ar00177a004).
P. von R. Schleyer, W. F. Sliwinski, G. W. Van Dine, U. Schöellkopf, J. Paust, K. Fellenberger, "Cyclopropyl solvolyses. III. Parent cyclopropyl derivatives and methyl-substituted cyclopropyl tosylates," J. Am. Chem. Soc. 1972, 94(1), 125–133 (https://doi.org/10.1021/ja00756a024).
Karen S. Root, Craig L. Hill, Lynette M. Lawrence, George M. Whitesides, "The mechanism of formation of Grignard reagents: trapping of free alkyl radical intermediates by reaction with tetramethylpiperidine-N-oxyl," J. Am. Chem. Soc. 1989, 111(14), 5405–5412 (https://doi.org/10.1021/ja00196a053).


How should we express concentrations and partial pressures in ΔG=-RTlnK?

What units of K should we use in ΔG = -RT lnK?

The full formula is:

$$\Delta_r G^\circ = - R T \ln K$$

The circle after the G refers to a standard state, where reactants and products are present at standard concentrations/pressures. The equilibrium constant has to be given with reference to the same standard state. The subscript r stands for reaction. The balanced chemical reaction has to match between the Gibbs energy of reaction and the equilibrium constant expression. Let's say our balanced reaction is:

$$\ce{ \nu_A A + \nu_B B + \nu_C C + ... <=> \nu_Y Y + \nu_Z Z + ...}$$

Here, the different $\nu$ are the stoichiometric coefficients.

The correct equilibrium expression for this reaction uses activities of species raised to their respective stoichiometric coefficient. Activities are dimensionless, and are equal to one at standard state. For dilute solutions, these can be estimated as the ratio of a concentration (or partial pressure) divided by the standard concentration (or partial pressure):

$$ a_\text{species} \approx \frac{c_\text{species}}{c^\circ_\text{species}} = \frac{p_\text{species}}{p^\circ_\text{species}}$$

As a consequence, the equilibrium constant expression will always be dimensionless, and will not depend on the concentration measure used, as long as the standard state concentration is expressed using the same measure.

  Since partial pressure can be in Pa, or bar/atm, with the difference between them will be roughly 100000 times.

No, because you are dividing by the standard state, using the same concentration measure (and for ease of cancelling units, in the same units). For example, let's say the standard state of a species in the gas phase is defined as one molar. At ambient conditions, this would correspond to a partial pressure of about 0.04 atm. If the equilibrium concentration is - say - 2 M, that would correspond to a partial pressure of about 0.08 atm. Either measure of concentration gives the same result: The species is present at twice the concentration (partial pressure) as the standard state. No matter what concentration measure I choose, I get the same value for the equilibrium constant.

The same goes for switching pressure units. As you are always dividing by the same standard pressure, you always get the same result (either by converting the units of the partial pressure, or the units of the standard partial pressure, and then cancelling).

Switching to a different standard state

In biology and biochemistry, a different standard state is used. This affects both the values of the standard Gibbs energy of reaction, and the values of equilibrium constants. However, these changes are self-consistent, i.e. you still can calculate K from the Gibbs energy and vice versa, and you can convert easily from one convention of the standard state to the other.


How to make p-chloroanisole from methanol and benzene?

Benzene is not basic enough to be protonated by methanol, which is precisely the first step you are proposing.

The use of free methanol to generate an aromatic methyl ether immediately inspired me to take a look at the Buchwald-Hartwig coupling. While the website only mentions the applicability to phenol couplings, the bible of organic chemistry, Strategic Applications of Named Reactions in Organic Synthesis also mentions the possibility of using alcoholates as a coupling partner.[1]

This leaves us the question of how to get to a compound suitable to be subjected to a Buchwald-Hartwig coupling. We would need an aryl halide into which the palladium catalyst may insert. Fortunately, we need halogenation anyway, so we might just make it a two-step procedure. Chlorination of benzene is well-known and well-taught, so I decided to check whether bromination of chlorobenzene is also known. Turns out that Ferguson et al. published $89~\%$ yield of the bromination of chlorobenzene using bromine and $\ce{AlBr3}$ in $\ce{CS2}$ — fully para-selectively.[2] This gives us the following sequence:

$$\ce{C6H6 ->[Cl2/AlCl3] Cl-C6H5 ->[Br2/AlBr3][CS2, $89~\%$] Cl-C6H4-Br ->[Pd(OAc)2, NaOMe][base] Cl-C6H4-OMe}$$

Wherein both the latter two are para-substituted.

The required sodium methanolate is easily synthesised from methanol using sodium hydride:

$$\ce{MeOH + NaH -> NaOMe + H2 ^}$$

References:

[1]: L. Kürti and B. Czako: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 70.

[2]: L. N. Ferguson, A. Y. Garner, J. L. Mack, J. Am. Chem. Soc. 1954, 76, 1250–1251. DOI: 10.1021/ja01634a013.


Role of catalyst in elementary reactions towards rate law

The catalyst uses a different reaction path with the zero net catalyst consumption, compared to not catalyzed reaction.
E.g. instead of $\ce{A + B <=> C}$, there is $\ce{A + M <=> AM}$, $\ce{AM + B <=> M + C}$.
If there is a homogenous catalyst, then its concentration in rate equations is processed by the same way as the reactants or products. But the different reaction path, compared to non catalyzed reaction, remains.
So it would not be $$\ce{O3 + M <=> O2 + O + M}$$
but rather e.g. :
$$\ce{O3 + M <=> O2 + MO}$$
and
$$\ce{MO <=> O + M}$$

OR:
$$\ce{O3 + M <=> O + MO2}$$
and
$$\ce{MO2 <=> O2 + M}$$
So you need to solve the kinetics of the system of linked reactions, considering catalyst form concentrations in steady states.

If it is a heterogenous contact catalysis, then the amount and activity of catalyst is implicitly involved in the catalyst-dependent kinetic constant:
$$\frac{\mathrm d[\ce{O2}]}{\mathrm{dt}} = k_{1,\mathrm{M}} [\ce{O3}] - k_{-1,\mathrm{M}}[\ce{[O2][O]}$$
where the ratio $\frac{k_{1,\mathrm{M}}}{k_{-1,\mathrm{M}}}$ remains constant, independent of the catalyst activity.