What is the mechanism for this nucleophilic conjugate addition?

To figure out what's going on in an organic reaction, there are a few things you really should look out for:

what functional group(s) are present on the molecule(s);
what conditions the reaction is being run under;
which of the functional group(s) are most reactive under the given reaction conditions.

If you apply these steps to every stage of the reaction, then deducing the mechanism will be easier, and you also save yourself a lot of time by not having to memorise a bunch of mechanisms that you have been previously taught. If those mechanisms were taught by a competent lecturer, you should be able to apply the three steps above to see exactly why those reactions happen.

The solution (courtesy of @Jan)

So, without further ado:

  What functional groups are present on the molecule?

That's easy.

$\hspace{30 mm}$

  What are the conditions?

That's easy too. $\ce{EtO-}$ is strongly basic, but it can also act as a decent nucleophile.

  Which of the functional groups are most reactive in base?

Well, to begin with, the thiol is acidic, so it will be deprotonated easily. No question about that.

Now you have a couple of electrophiles - the two esters - as well as a couple of nucleophiles: the thiolate anion, as well as the ethoxide anion. A hard base such as ethoxide is not likely to react in a conjugate addition. Maybe it might react directly with the esters at the carbonyl carbon? But wait! Both esters are ethyl esters $\ce{RCO2Et}$. So, you can use ethoxide to attack those esters all you like, but you won't get anywhere doing it - you'll just get back the same product. In fact, that's exactly why we use ethoxide instead of a different base like methoxide, $\ce{MeO-}$: to prevent transesterification.

So, let's talk about the thiolate anion instead. It's pretty nucleophilic, and you also have an electrophilic α,β-unsaturated carbonyl, so these can react:

$\hspace{15 mm}$

Now, you have a nucleophilic enolate, as well as an electrophilic ester on the other side. They can immediately react with each other to form the product that you are given:

$\hspace{25 mm}$

This reaction has a name - it is called a Claisen condensation, but that's not important! The important part is that you realise that it is just the reaction of a nucleophile with an electrophile.

An alternative (my original answer)

There might actually be another possible product, though. The enolate that you formed above could pick up a proton from somewhere (either ethanol, or perhaps another thiol molecule) to give the conjugate addition product, which is exactly the same as the one you drew. However, your initial addition product had a positive charge, which is never a good idea in a basic medium. Under basic conditions, you should always deprotonate first and reprotonate at the end of any reaction.

If this were to be the case, you should probably go through the three steps again:

  Functional groups?

That's still pretty easy. You have a sulfide and two esters.

$\hspace{45 mm}$

  Conditions?

Still basic.

  Most reactive functional group?

Well, the sulfide is indeed nucleophilic and you could use it to attack either one of the electrophilic carbonyl groups. But there are a few things that don't really favour this:

Sulfur is a pretty soft nucleophile. It might love to react with $\ce{CH3I}$, but probably wouldn't like a carbonyl so much. In fact, this is the reason why the initial step consisted of conjugate addition instead of direct addition.
There's no more hydrogen on sulfur that can be taken off, so if sulfur does attack the carbonyl, it will be stuck with a positive charge. This makes the sulfur a pretty good leaving group, which means that any addition to the carbonyl group will be highly reversible.
If sulfur attacks the carbonyl group further away from it, it will form a four-membered ring. This is not particularly favourable. If it attacks the carbonyl group nearer to it, it will form a three-membered ring, which is even worse!

So, the sulfur is stuck where it is. You are better off thinking about the reactivity of the carbonyl group in base. You should know that they can be deprotonated to form enolates. Does it matter which carbonyl group is deprotonated? Of course it does. This molecule is not symmetric, so the two carbonyl groups are not equivalent. This obviously leads to the question of which carbonyl group should be deprotonated. Let's look at the possible anions formed from the deprotonation:

$\hspace{10 mm}$

The first of these two is more stable than the other, because the electronegative sulfur atom is capable of withdrawing electron density from the negatively charged carbon. (Actually, there are several theories as to why sulfur stabilises the carbanion, but I won't discuss that here. See this paper, for example.) So, it is that first conjugate base that will be preferentially formed.

At this stage, you have a good nucleophile (the anion) as well as a good electrophile (the carbonyl on the other side). So, what's left is to react them in the same Claisen condensation as above:

$\hspace{25 mm}$

There we go, side product.


Crystal structure in function of pressure

Seems like homework, so I'll give some hints:
What do you expect to happen to a crystal under pressure?
What does the crystal structure have to do with the response to pressure?
Why would the crystal structure want to change under pressure?
What crystal structures are common for metals? (there aren't many)
Since there aren't many, what distinguishes their responses to pressure that might make the metal want to change crystal structure (a phase change).

That should get you going.


Pressure exerted by the mixture of nonreactive gases

You did too many unnecessary steps and also didn't use unified units (e.g. liters instead of cubic meters, since you decided to use $R=\pu{8.314 J mol-1 K-1}$). I denoted both total pressures as $P_1$ and $P_2$ for simplicity since we don't need partial pressures here. From the ideal gas law:

$$P_2=\frac{n_2RT_2}{V_2}\tag{1}$$

Since both gases are not reacting at these conditions,

$$n_1=n_2=\frac{P_1V_1}{RT_1}\tag{2}$$

and (1) can be rewritten as following:

$$P_2=P_1\frac{V_1T_2}{V_2T_1}=\pu{1.00 atm}\cdot\frac{\pu{10.0 L}\cdot\pu{298.13 K}}{\pu{2.0 L}\cdot\pu{273.13 K}}\approx\pu{5.46 atm}\tag{3}$$


Coordination numbers of ionic and intermetallic compunds

NaCl

NaCl is not monoatomic. When discussing coordination number of cations in ionic compounds, the number refers to the anions the cations is surrounded by (or bonded to, or touching).

The blue Na+ cations are bonded to six other Cl- anions: top, bottom, left, right, forward, backwards (in this image). They are not bonded and do not touch the other Na+ cations in the diagonals.

CsCl

CsCl is not a "simple" simple cubic. Each of the different ions, Cs+ and Cl- by themselves are simple cubic, but they are interpenetrating to form something that looks like a body centered cubic:

A mono atomic simple cubic will have coordination number of six, but this is an ionic compound. Thus, each cation or anion that sits in the center of the cube is bonded to the eight anions or cations (respectively) that sit in the vertices of the cube.


Why can't hot water break disulfide bonds when the energy storage in hot water exceeds the bond energy?

Chemical reactions do not simply occur because there is enough energy around to make something happen. They also require a mechanism that enables something to occur. It isn't enough that the product will release energy on some reaction happening.

Some reactions will happen spontaneously with enough input of thermal energy to start them off (burning coal, for example). Others, like breaking disulphide bonds) will not. Even when there is a plausible mechanism for a reaction, you have to add enough energy to make the reaction happen which may be much more energy than you expect. Most reactions have an "activation energy" to get past the barriers specific mechanisms holding back the reaction.


Sensitivity vs. Limit of Detection of rapid antigen tests

In this situation, "sensitivity" refers to how often the test is able to correctly detect the presence of analyte. A test with high sensitivity means that there are few false negatives.
Limit of detection refers to how much of the analyte must be present before it is considered positive.
Imagine two different tests: test A uses an antibody with very high affinity for a specific spike protein sequence. Because of the high affinity, it can detect the spike protein even when there's not much around (good limit of detection). However, the antibody is less robust against other sequences, so it sometimes reports a false negative in the presence of spike protein variants (poor sensitivity).
Conversely, test B has an antibody with medium affinity for several different spike protein variants. It relies on high levels of the protein being present, but it works with many different variants and rarely gives false negatives. This test would have a high (aka poor) limit of detection and high (aka good) sensitivity.


Why doesn't graphite dissolve in organic solvent?

While you cannot directly make a suspension or dispersion of graphene nano-platelets from graphite, it is possible to perform very similar process, called exfoliation, on graphite oxide, as graphite oxide has functional groups decorating it that reduce the functional strength of the van der Waal's forces between layers.
The exfoliation process can create a dispersion of graphene nanoplatelets in an organic environment, however, outside energy is often (if not always) necessary to promote the exfoliation of the individual layers. Most often, this energy is provided by sonicating the solution.
A very quick search of suggests that this method has been shown to work using NMP, DMF, THF, and ethelyne glycol. (see Paredes, Villar-Rodil, Martínez-Alonso, and Tascón, 2008 for more info)


Brown stains on skin after working with Tollens' reagent

The stains are due to silver nitrate, the reactive portion of Tollens’ reagent. It is very prone to reduction to metallic silver.

If you got it on your hands (a clear signal you weren’t wearing gloves when you should have been) it will create small silver particles in your skin — this is why it won’t wash off. Since the particles are small, they don’t exhibit the classical metallic characteristics (looking silvery, reflecting etc).

Don’t worry, they will grow out within a week’s time or so, due to your skin renewing itself.
And wear gloves next time.


Do only outer electron shells take part in forming chemical bonds?

Great Question.
From this article, it seems that extreme conditions of pressure can lead to new type of molecules,bondings and how the electrons participate in the bonding.

  Under very high pressures, it appears, electrons in the atom's inner shells can also take part in chemical bonds.

  “It breaks our doctrine that the inner-shell electrons never react, never enter the chemistry domain,” says Mao-sheng Miao, a chemist at the University of California, Santa Barbara, and the Beijing Computational Science Research Center in China. Miao's calculations show that under extreme pressures cesium and fluorine atoms can form exotic molecules with inner-shell bonds.

  Miao identified two molecules that, at high pressure, would involve cesium's inner electrons as well. To form cesium trifluoride ($\ce{CsF3}$), a cesium atom would share its single valence electron and two inner-shell electrons with three fluorine atoms. Four inner electrons would go into making cesium pentafluoride ($\ce{CsF5}$).

  “That forms a very beautiful molecule, like a starfish,” Miao says. He reported his findings in Nature Chemistry. Both the shape of the resulting molecules and the possibility of their formation are “very surprising,” says Nobel Prize–winning chemist Ronald Hoffmann, a professor emeritus at Cornell University.

Found this in the comments about the stability of the exotic molecules at normal pressures

  "Since the compound is thermodynamically unstable it would decompose to the more stable form at lower pressures. Since this means that they are going to higher states of enthalpy I would expect that this decomposition would be a endothermic reaction which means that they would get cold during the process."


How effective is sulfur in detecting mercury vapors

The vapour pressure of mercury is appreciable, but the evaporation kinetics are slow.  In a well-ventilated lab there shouldn't be concern about mercury poisoning from broken thermometers, Schlenk lines &c.

That said, there is Alfred Stock's famous review (Angewandte Chemie, 1926) on mercury poisoning.  He suffered from a chronic case of mercury poisoning, which IIRC really took off after the war, when they had to save on ventilation.  In old laboratory buildings the quantities found under the wood flooring during renovation are at times staggering.  Our safety guys liked to show pictures of puddles.

To detect mercury there are commercially available detectors, usually build around the resistance of a vacuum-deposited gold film (a few 100 atoms thick).  A low-pressure mercury lamp might also work.


Ka relating to Dissociation of Weak acid and Water

Yes, water autodissociates according to the equation $\ce{2H2O <=> H3O+ + OH-}$.

But I should point out that the equilibrium constant is written as though the concentration of water itself is a constant, so the reactions becomes:

$\ce{H2O <=> H+ + OH-}$.

and the equilibrium is expressed as:

$K_\rm{w} = 1\times10^{-14} = \ce{[H+][OH-]}$

The pH is $\text{-log}\ce{([H+])}$ thus the pH of pure water is 7.

This also has to do with significant figures. Few pKa's are known to better than two significant figures. So:

For weak acid with pKa < 5 you can ignore water contribution.
For acid with pKa > 9 you can assume all the  $\ce{H+}$ is from water and that all of the acid is essentially present as the HA form.
For 9 > pKa > 5 you have to consider both ionization of the acid and water.


What does m± represent in the equation for mean activity of a salt solution?

The top $m$ is the mean ionic molality of the solution you have.  You must know the concentration since you need it to compute the activity coefficient.  The bottom $m_0$ is the standard molality which in general is $\pu{1 M}$.  It is there to make the activity dimensionless.

The required formula for $m_\pm$ is $$m_\pm^\nu = (m_+)^{\nu_+} (m_-)^{\nu_-}$$ where $m_-$ is the molality of the negative ions and $m_+$ is the molality of the positive ions. Also $\nu_-$ is the stoichiometric coefficient of the negative ions and $\nu_+$ is the same for the positive ions. Also $\nu = \nu_- + \nu_+$.

Thus in your specific case, where you have $\pu{0.0120 M}$ of $\ce{Na3PO4}$, we have

\begin{align}
\nu_+ &= 3 & \nu_- &= 1 & \nu &= 4 \\
m_+ &= \pu{0.0360 M} & m_- &= \pu{0.0120 M}
\end{align}

and hence

\begin{align}
m_\pm &= \left[(\pu{0.0360 M})^3(\pu{0.0120 M})^1\right]^{1/4} \\
&= \pu{0.0274 M}
\end{align}


Energy required to remove an electron from He

We can use Bohr's atomic model to calculate the ionization energy of $\ce{He+}$.
The ionization energy will be the amount of energy we must give to the electron of singly ionized helium to remove it apart from the nucleus to infinity. That gives $IE_2=E_{\infty}-E_1$ where $E_1$ is the energy in the first bohr-orbit (principal quantum number=$1$). Conventionally taking $E_{\infty}$ as $0$ we use the Bohr's equation to find the energy $E_1$:-
$$E_n=-\frac{Z^2me^4}{8n^2h^2\epsilon_0^2}$$
 where $Z$= atomic number, $e$= charge on electron, $n$= principal quantum number, $h$= Planck's constant, $m$= mass of electron, $\epsilon_0$= permittivity of free space.

Using $n=1$, $Z=2$ and the appropriate values for other constants we get $IE_2=-E_n=\pu{54.4eV}$.
Now the energy for removing both electron $IE_{net}=IE_1+IE_2$. Using $IE_{net}=\pu{79eV}$and $IE_2=\pu{54.4eV}$ we get, $$IE_1=\pu{24.6eV}$$
which corresponds to option (A), which I assume is typed wrong by you in the question.


"Hot" refers to the bead when just taken out of the flame of the Bunsen burner, while "cold" refers to the state when the bead cooled to room temperature. Provided the small size these beeds typically have (a toothpick like rod of magnesium oxide to carry them), this is just a matter of about a minute of exposure to ambient air. Duing this process, you may spot differences, too.  Conversely, intentionally cooling the hot glass-like beads under tap water often shatters them into pieces.

Side note #1:  Testing with a bead of Borax is a one quick pretest, to be confirmed by subsequent experiments.  And as a good habit: once you found the ions, compare your bead made with the analyte with a reference, a bead made with a sample of known composition.

Side note #2: Of course I miss one, an entry for cobalt ... regardless of the flame (oxidation or reduction), hot or cold: an intense dark blue. Almost Yale blue.


Maximum temperature attainable by thermite reaction

According to Wikipedia the maximum temperature is approximately 2500 °C, limited by the boiling point of aluminum, which is 2519 °C.  The article provides some discussion of how the temperature is limited, and how to calculate it using Hesse's law.


Are the units of mole of oxygen molecules the same with the units of mole of nitrogen molecules?

It is a good question which has already been somewhat addressed in the 1880s. This "field" was called quantity calculus. Calculus here is not the integration / differentiation, but rather the Latin calculus implying a method of calculation.
There is a very nice article "Quantity Calculus: Unambiguous Designation of Units in Graphs and Tables" by Mary Anne White in the Journal of Chemical Education. Please read this if you are seriously interested. Search on Google Scholar and it is free to download from there.
In quantity calculus Each physical quantity as the product of a numerical value and a unit:
physical quantity = numerical value × unit
This approach was introduced by British scientists and many leading physicists used it. Now there is there is nothing less or nothing more. Therefore your ambiguity arises from introducing another factor such as "oxygen" or "nitrogen". The unit mol does not know whether it belongs to oxygen or nitrogen.
As explained in the comments, suppose we write L symbolizing the height of a tree, then I can only write, L = 10 m. For mathematical purposes, I will not introduce "tree" anywhere in this equation. The tree is already incorporated in L (in your mind) but not in the mathematical equation. One can also write L/m =10. Now you have a pure number on both sides.
For fun: Think about taking the derivatives or log of quantities with units. What is the unit then?


How are these compounds strong bases, when they are sparingly soluble in water?

The typical definition of strong base that you will see at an introductory level is that there is complete dissociation to hydroxide ions, for example that NaOH completely dissociates into Na+ and OH-.

This is a terrible definition because reactions never go to completion.  The definition contradicts the concept of chemical equilibrium.

In reality:

$\frac{[Na+(aq)][OH-(aq)]}{[NaOH(aq)]} = K_B = ~4M$ (from table discussed below)

So even dissolved aqueous NaOH is not all dissociated.  It can exist as ion pairs or molecules in solution without being dissociated.

$K_B$ is quantification of how strong or weak the base is.  Strong base could be defined as $K_B$ exceeding a certain value. Let us say $K_B>1$ and therefore $pK_B<0$

So the strength of the base only refers to the equilibrium between the undissoicated dissolved species and the dissociated ions.  It is independent of how much or how little solid dissolves. Some hydroxides, like NaOH and KOH have high solubilities and some have low solubilities, but $K_B$ is a different equilibrium.

For quantitative data on the $K_B$ values of hydroxides, see:
DISSOCIATION CONSTANTS OF INORGANIC ACIDS AND BASES IN AQUEOUS SOLUTION
Look up the corresponding metal ion alphabetically, and a $pK_B$ value is given.

However, keep in mind that for each of $\ce{Mg(OH)2}$, $\ce{Ca(OH)2}$, $\ce{Sr(OH)2}$ there will be two $K_B$ values, since there are two hydroxides that can dissociate.

The $pK_B$ values for the second hydroxide dissociating are Mg 2.6, Ca 1.4, Sr 0.8, Ba 0.6 becoming stronger bases as you go down the column in the periodic table, increasing radius stablizing the charge of the ion.  The first $pK_B$ would be even lower, but I don't see any quantitative values.  It would be so low that it would be difficult to measure.  So, yes these are strong bases.

As far as how to tell, other than experimentally, how strong a base of a hydroxide is, there is a good lecture The Acid-Base Character of Oxides and Hydroxides in Aqueous Solution by Pilkington.  As you go from left to right across the periodic table (excluding the noble gases), the hydroxides progress from being strong bases to being acids.


How would you name this organic compound?

Yes, it's tetramethyl heptane, where are the 4 methyls? 2,4,4,6-tetramethylheptane.


Should we expect Jahn–Teller distortion in bis(ethylenediamine)nickel(II)?

There is a key difference between the $\mathrm d^8$ state and the $\mathrm d^9$ state which lies in exactly that extra electron. The result of the extra electron is that instead of nickel(II)’s $\mathrm{(t_{2g})^6\ (e_g^*)^2}$ state copper(II) has a $\mathrm{(t_{2g})^6\ (e_g^*)^3}$ state. There are two $\mathrm{e_g^*}$ orbitals, thus occupying each with a single electron with parallel spins leads to no degeneracy problems. An additional electron will mean that one of the two $\mathrm{e_g^*}$ orbitals is doubly occupied while the other is singly occupied—but this is a degenerate state as the energy is the same whether it is $\mathrm d_{z^2}$ or $\mathrm d_{x^2-y^2}$ that is doubly occupied.

This degeneracy of possible states in copper(II)—or any other $\mathrm d^9$ system—is the underlying reason for Jahn-Teller distortions. By lowering symmetry from $O_\mathrm h$ to $D_\mathrm{4h}$, these two orbitals are no longer degenerate and they can (and must) diverge in energy. Thus, you have a clear ‘winner’—the lower-energy orbital which receives an electron pair—and a ‘loser’—the higher-energy orbital which receives the single electron. Looking only at $\mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$ it doesn’t matter whether the symmetry reduction is performed by moving the $z$-ligands further away or closer but it does make a difference for the other three d orbitals ($\mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$) for which lengthening the $\ce{M-L}$ bonds along the $z$ axis proves favourable overall.

A $\mathrm d^8$ system at first glance has nothing to gain by this distortion. Rather than having a stabilisation that benefits two electrons at the expense of one, it can only stabilise one electron at the expense of one. This is energetically not favourable as there is nothing to be gained (but probably something to be lost) so a ‘simple’ Jahn-Teller distortion does not occur; nickel(II) prefers to remain octahedral in $\ce{[ML6]}$ systems.

Note however that there is one possibility when such a stabilisation is beneficial: if the ‘distortion’ is of such a large magnitude that the energy difference between $\mathrm d_{x^2-y^2}$ and the next-lowest orbital is larger than the spin-pairing energy. In this case, we could say that nickel(II) drops from a high-spin state to a low-spin state, although these terms are not usually used in this context. Geometrically, this ‘distortion’ is so significant that we term the resulting complex square planar rather than distorted octahedron; in general there is no longer any evidence of coordination bonding along the $z$ axis. A number of nickel(II) complexes with strong mid-field or strong-field ligands are indeed square planar for this reason such as $\ce{[Ni(dmg)2]}$, the dimethylglyoxime (or diacetyldioxime) complex.


Addition of acid to 3-methyl-2,3-dihydropyran

Apparently, you're using the Markovnikoff's rule here. Recall that it is only a basic rule intended for simple unsaturated hydrocarbons, and fails at more advanced examples involving heteroatoms or rings.

In this case, the first step is the addition of $\ce{H+}$ ion from the acid to the pi bond, which leads to the formation of a carbocation. The carbocation's positive charge is most stable next to the oxygen atom. Can you tell why?

In the second step, the $\ce{X-}$ ion will simply attach at the position where the positive charge was the most stable, leading to a product as you've shown in the second image.


Is the Henderson–Hasselbalch equation volume independent?

For a much shorter and clearer explanation of this problem, look here.
The Henderson-Hasselbalch (HH) equation is not to blame here. It is an approximate equation, with a certain region of validity. By its nature, it does not take into account the self-dissociation of water, which becomes increasingly important in dilute solutions. When concentrations reach somewhere around $\pu{10^-5 mol L-1}$ or lower, the true $\mathrm{pH}$ will deviate significantly from the value predicted by the HH equation. However, this is not the problem most of the time.
The reason the HH equation might produce poor predictions when calculating buffer $\mathrm{pH}$ is really because of an oft-made assumption which has nothing to do with the HH equation itself: the weak acid (or weak base) is assumed to be so weak that its ionization contributes almost no conjugate base (or conjugate acid) in comparison to the dissolution of the buffer salt. In other words, we assume the formal concentrations $C_X$ of species in the buffer are equal to their actual concentrations $[X]$; we replace
$$
\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{A^-}]}{[\ce{HA}]}
$$
for the worse approximation
$$
\mathrm{pH}
  = \mathrm{p}K_\mathrm{a} + \log \frac{C_{\ce{A^-}}}{C_{\ce{HA}}}.$$
This assumption is incorrect, but it is used to make the $\mathrm{pH}$ calculation much easier at the cost of accuracy. The resulting approximation breaks down in sufficiently dilute solution, and is already quite noticeable far before reaching water self-dissociation issues. It is also a noticeably poor approximation for weak acids/bases with relatively high ionization constants (say, $K > 10^{-2}$).
So how can we visualize the effect of diluting a buffer without approximations, and where do the approximations start to break down? Let us consider the simple case of solutions of a weak monoprotic acid $\ce{HA}$ with acid dissociation constant $K_\mathrm{a}$ where the concentration of acid and conjugate base are formally equal. This can be done by titrating half of the weak acid with a strong monoprotic base (e.g. $\ce{KOH}$). This problem can be solved exactly (assuming all activity coefficients are equal to $1$, which is generally a good approximation for solutions below about $\pu{1 mol L-1}$), and I have previously outlined the solution here. The equation relevant to this problem is the following:
\begin{multline}
0 =
  [\ce{H^+}]^3
    + \left(K_\mathrm{a}
      + \frac{C^\circ_\ce{B}V_\ce{B}}{V_\ce{A} + V_\ce{B}}
    \right)[\ce{H^+}]^2 \\
    + \left(
      \frac{C^\circ_\ce{B} V_\ce{B}}{V_\ce{A} + V_\ce{B}} K_\mathrm{a}
      - \frac{C^\circ_\ce{A} V_\ce{A}}{V_\ce{A} +  V_\ce{B}}
      K_\mathrm{a} - k_\mathrm{w}
    \right)[\ce{H^+}]
    - K_\mathrm{a}  k_\mathrm{w}
\end{multline}
$V_\ce{A}$ is the volume of the weak acid solution being titrated and $C^\circ_\ce{A}$ is its formal initial concentration (before mixing and reacting with the base), while $V_\ce{B}$ is a variable volume of strong base solution added with formal initial concentration $C^\circ_\ce{B}$ (before mixing and reacting with the acid). The presence of the self-dissociation constant of water $k_\mathrm{w}$ shows that it is being taken into consideration.
We can now substitute values as desired to obtain a polynomial in $[\ce{H^+}]$. I'll set up the volumes and concentrations arbitrarily so we can reach the acid and conjugate base formal concentrations of $\pu{0.3 mol L-1}$ as mentioned in the question. We can start off with $\pu{100 mL}$ of weak acid solution at a concentration of $\pu{0.9 mol L-1}$ with a dissociation constant $K_\mathrm{a} = 10^{-3}$ (this constant can be changed at will). We can reach the target concentrations of acid and conjugate base by adding $\pu{50 mL}$ of strong base at the same concentration $\pu{0.9 mol L-1}$ (neutralizing half the original acid). You can check the resulting concentrations after mixing the solutions, if you wish. Solving this polynomial, the resulting buffer solution containing formal concentrations $C_{\ce{HA}} = \pu{0.3 mol L-1}$ and $C_{\ce{A^-}} = \pu{0.3 mol L-1}$ has $[\ce{H^+}] = \pu{9.93399E-4 mol L-1}$. This results in $\mathrm{pH} = 3.00288$, which is very close to the value predicted by the approximations made at the start ($\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log 1 = \mathrm{p}K_\mathrm{a} = 3$). For now, they work.
After looking at this first case, we shall investigate the effects of dilution by factors of 10. This can be performed simply by diluting both the initial weak acid and strong base concentrations by 10 (i.e., $\pu{100 mL}$ of $\pu{0.09 mol L-1}$ $\ce{HA}$ plus $\pu{50 mL}$ of $\pu{0.09 mol L-1}$ $\ce{KOH}$ result in a buffer with acid/conjugate base concentrations of $\pu{0.03 mol L-1}$, and so on). Before going into the calculations, it's quite simple to see an issue with the approximation in the second paragraph. Independent of the level of dilution, so long as the concentration of acid and conjugate base remain equal, then according to the approximation, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} = 3$. However, consider the case when starting with a solution of $\ce{HA}$ with initial concentration equal to $\pu{9E-4 mol L-1}$. This is a weak monoprotic acid, so the initial solution must have $[\ce{H^+}] < \pu{9E-4 mol L-1}$ and thus $\mathrm{pH} > 3$. Clearly it's impossible to add any amount of strong base to this acid and get a buffer solution with $\mathrm{pH} = 3$! The approximations at the start have broken down at this point.
Now, for the calculations. To get the proton concentration, I replaced every value of $0.9$ in the previous Wolfram Alpha link with $0.09$, then $0.009$, and so on. With the proton concentration determined, I used equations shown in my other linked answer to calculate the actual concentrations $[\ce{HA}]$ and $[\ce{A^-}]$ for comparison with the formal concentrations $C_\ce{HA}$ and $C_{\ce{A^-}}$. Note that they are almost equal at high concentration, but the error increases quite suddenly at lower concentrations. While the ratio $\frac{C_\ce{A^-}}{C_\ce{HA}}$ is always equal to $1$, the ratio $\frac{[\ce{A^-}]}{[\ce{HA}]}$ rockets away.
$$ \scriptsize
\begin{array}{c|ccc|cc}
\hline
C_\ce{HA} = C_\ce{A^-} / (\pu{mol L^-1})  &
[\ce{HA}] / (\pu{mol L^-1}) &
[\ce{A^-}] / (\pu{mol L^-1}) &
\frac{[\ce{A^-}]}{[\ce{HA}]} &
[\ce{H^+}] / (\pu{mol L^-1}) &
\mathrm{pH} \\\hline
0.3 & 0.29901 & 0.30099 & 1.00664 & 9.93399\times 10^{-4} & 3.00288 \\
0.03 & 0.029061 & 0.030939 & 1.06464 & 9.39282 \times 10^{-4} & 3.02720 \\
0.003 & 0.00235421 & 0.0036458 & 1.54858 & 6.45751 \times 10^{-4} & 3.18993\\
3\times 10^{-4} & 1\times 10^{-4} & 5\times 10^{-4} & 5 & 2 \times 10^{-4} & 3.69897 \\
3\times 10^{-5} & 1.6539\times 10^{-6} & 5.83461\times 10^{-5} & 35.2778
 & 2.83464 \times 10^{-5} & 4.54750 \\
3\times 10^{-6} & 1.78596\times 10^{-8} & 5.98214\times 10^{-6} & 334.953
 & 2.98549×10^{-6} & 5.52498 \\
3\times 10^{-7} & 1.97992\times 10^{-10} & 5.99802\times 10^{-7} & 3029.42
 & 3.30096×10^{-7} & 6.48136 \\
3\times 10^{-8} & 6.96609\times 10^{-12} & 5.99939\times 10^{-8}  & 8612.15
 & 1.16115×10^{-7} & 6.93511 \\
3\times 10^{-9} & 6.09004\times 10^{-13} & 5.99939\times 10^{-9} & 9851.15
 & 1.01511×10^{-7} & 6.99349 \\
\hline
\end{array}
$$


What Methods Would I Use To Powder Cotton?

The content of cellulose in paper is not as high as that of cotton, however, it may be easier to use as a source if you do require a fine powder. Why do you require cellulose?


How does caesium’s density change with temperature?

Looking at the wikipedia page you can see that the density is $\rho=1.93$ kg/l at room temperature and $\rho=1.843 kg/l$ at it's melting point.

Additional information can be found in this paper, where they experimentally explore the density of liquid Cesium in the vicinity of its melting point. In the range of 302 to 375 K to be exact.

From their experimental data they deduce the following approximate equation for the density as a function of the temperature:
$$\rho= 1829.12- 0.61483 \left(T-T_{melt}\right) $$
where $T_{melt}=301.6$ K.

This shows that the density reduces with temperature. Plotting the information that we have from the paper and from wikipedia yields:

The jump at the melting point is from the mismatch between the paper and wikipedia. In the paper they specify specifically that the density is for the liquid case, on wikipedia it might (I'm not sure) be the density for the solid case at the melting point. If that is the case then the jump in density from solid to liquid is: $1843-1829=14 $kg/m$^3$


d-Menthol vs dl-menthol: Does an enantiomer and its racemic mixture have different melting points?

If solid menthol were just amorphous, then you would expect the racemate and each enantiomer to exhibit identical melting points. However, if you read further on the Wikipedia page you reference, you'll see that racemic vs. enantiopure menthol crystallizes differently:

  The two crystal forms for racemic menthol have melting points of 28 °C and 38 °C. Pure (−)-menthol has four crystal forms, of which the most stable is the α form, the familiar broad needles.

(This information is trustworthy and corroborated in the primary literature; here is one example)

The melting point of a crystalline solid has to do with how well the molecules can pack together. In other words, on a molecular level, there is an orientation dependence. The better the molecular packing, the stronger the intermolecular forces, and the higher the melting point - since it takes more energy to separate the molecules. This is why the enantiopure menthol has a higher melting point.


How is the formula of mean activity coefficient derived?

The problem is you were starting with an expression for the mean activity coefficient for salts where both the cation and anion were monovalent, and then attempting to derive, from this, a general expression for salts of all empirical formulas. To correct this, you need to start with the general expression for the mean activity coefficient of salts of any emprical formula.
If the empirical formula of the salt is of the form $\ce{A_pB_q}$, where A is the cation and B is the anion, then the general formula for the mean activity coefficient is:
$$\gamma_{\pm} = \sqrt[^{p+q}]{\gamma^p_+\gamma^q_-}$$
[Adapted from: https://en.wikipedia.org/wiki/Activity_coefficient, where I've taken the Wikipedia expression and substituted $\gamma_+$ for $\gamma_A$, and $\gamma_-$ for $\gamma_B$, to correspond to the nomenclature you are using.]
Charge balance dictates that  $|z_+| = q$ and $|z_-| = p$. [E.g., in $\ce{A_2B_3}$, the charge on A must be 3+, and that on B must be 2- (or some integer multiple of those).]
Substituting, we have:
$$\gamma_{\pm} = \sqrt[^{|z_-|+|z_+|}]{\gamma^{|z_-|}_+\gamma^{|z_+|}_-}$$
If you use that in place of eqn. (3-37) in your post, and work through your calculations again, you will get the result show in eqn. (3-38), which is a general result for any values of p and q.


What is the mechanism for the creation of P(CH2OH)4+ cation?

The acid serves as a catalyst to protonate the formaldehyde oxygen.  This makes the carbonyl carbon more electrophilic, a more reactive target for the weakly nucleophilic lone pair of electrons on phosphorous.  Here is a diagram of the basic reaction mechanism.

The proton transfer is thought to be intramolecular according to this link.


Okay, following up on my comment, I found the spectrum for the "correct" answer:

As you can see in this IR the absorption of the carbonyl peak is shifted down to 1718 as the problem statement claims.

Now, consider this IR spectrum for a conjugated aldehyde which is similar to a conjugated ketone:

In this spectrum, we see that the IR absorption for an aldehyde (normally around 1715) is shifted down below 1700.

Everything else in your answer seems right though. You always need to be careful to consider conjugation when looking at the peak of a carbonyl stretch.

If you want to look around at HNMR/CNMR/IR for basically any reasonable organic molecule, you should check out the SDBS Database. I actually found this to be a useful study tool and just kind of interesting to look around in.


What repulsion keeps non-hydrogen atoms at a distance: between inner shells or between nuclei?

I found a nice figure and the relevant statement in a paper by Frenking and Krapp (Unicorns in the world of chemical bonding models, 2006, https://doi.org/10.1002/jcc.20543):

The crucial term which is responsible for repulsive interactions in chemical bonds except in two‐electron systems such as H2 is the Pauli repulsion. The three terms (a) quasiclassical electrostatic interaction, (b) resonance, and (c) Pauli repulsion give a complete picture of the interatomic interactions which yield a chemical bond.

The figure shows how the distance-dependent repulsion for one scenario including the Pauli repulsion and two scenarios ignoring the Pauli repulsion; the latter do not show the drastic increase in repulsion that is observed.

Source: https://onlinelibrary.wiley.com/cms/attachment/2b2dc015-ca3c-4fd3-b51c-f768a6a924b5/mfig003.jpg
The entire essay is great, connecting the picture of bonding that bench chemists have with the developments of theory and computation in the 90 years since quantum mechanics was first applied to chemistry.


Light sensitive AgF decomposition reaction

As discussed in the comments, the asker only wanted to create some fluorine in situ for use in another flask, with no desire to store it. Chemistry with fluorine is difficult due to its dangerously high reactivity. Do not attempt anything without the proper safety procedures.

I actually still have doubts whether it is possible to obtain fluorine gas reliably from the exposure of silver fluoride to sufficiently high energy photons, and if there is any simple way to produce photons with the required energy. It is well known that the other silver halides easily suffer photolysis, especially for the heavier halides, but reactions which produce fluorine gas tend to be extremely endergonic, and therefore don't happen except in rather severe situations.

In case it is somehow possible to produce fluorine gas directly by photolysis of silver fluoride, then it is possible to transport the gas to another container where it can react, simply by connecting the flasks with a pipe. Few materials are capable of withstanding exposure to fluorine, however, so the containers must all be made of very well dried gas. Unless the asker has access to a very high power light source, I suspect that the rate at which silver halides decompose is quite slow, which may preclude their efficient usage as halogen generators. This is especially true for silver fluoride, which likely decomposes the slowest, and the fluorine formed in very low concentrations could easily be lost by reaction with impurities or residual water before even reaching the target. I do not know whether there are any catalysts which can be added to increase the photolysis speed of silver fluoride. Presumably some semicondutor with a high bandgap, such as titanium dioxide, could absorb the light and cause photoreduction of the silver/photooxidation of fluoride, but the catalyst would be quickly consumed by the fluorine formed.

If fluorine gas can be formed by silver halide photolysis fast enough, then by simply linking it to another flask with the desired reactants should cause the fluorine to diffuse and react, followed by a continuous equilibrium shift in the production of more fluorine in the flask containing the silver fluoride, and its consumption in the flask with the other reagents.


Reaction of chloroform, resorcinol and strong base

Well, turns out to be, the reaction proceeds to product a red colored species.

  Reaction with chloroform in the alkaline medium

  Dissolve 0.1 g in 1 ml of water R, add 1 ml of strong sodium hydroxide
  solution R and 0.1 ml of chloroform R, heat and allow to cool. An
  intense, deep-red colour develops which becomes pale yellow on the addition of a slight excess of hydrochloric acid R.


Prevent molten glass from cracking when cooling down

The surface cracks due to a mis-match in the temperatures inside and outside the ceramic. Things contract as they cool. So, the cooler outer surface is contracting against the hotter inner surface, which is still not contracting. The outer surface wants to be closer together than the inner material, but the inner material forces the outer material to stay apart. When the ceramic on the outside transitions to a solid (at what is called its "glass temperature"), it can no longer flow like a liquid. The only way it can continue to stay apart due to the forces from the inside is to "fall apart" ... i.e. to crack.

Your mother may be able to coat the pottery with a different glaze that forms a glass at a lower temperature than the one she has now. Also, the suggestions to cool VERY SLOWLY are directly on target. The cooling should happen in a place that has no air flow, ideally in a closed box. Absent an annealing oven, a different option may be to put the pots in an conventional NON-CONVECTION oven that is held at a temperature just below the glass temperature of the glaze being used. When the entire pot reaches that temperature, it can be removed to cool further in a box at room temperature, again with no air flow.


What's the difference between pipettes calibrated "TD", "TC", "Blow-out" and related terms?

Since you are explicitly asking for standards, you might want to start with ISO 4787 Laboratory glassware – Volumetric instruments – Methods for testing of capacity and for use. This international standard provides methods for testing as well as for use of various volumetric instruments made from glass. Further requirements for individual volumetric instruments can be found in

ISO 648 Laboratory glassware – Single-volume pipettes
ISO 835 Laboratory glassware – Graduated pipettes
ISO 385 Laboratory glassware – Burettes
ISO 1042 Laboratory glassware – One-mark volumetric flasks
ISO 4788:2005 Laboratory glassware – Graduated measuring cylinders

Most commercially available laboratory glassware is in accordance with these standards. However, some nonconforming volumetric instruments are also available, for example in non-standard sizes.
Due to retention of liquid on the inner surface of the volumetric instrument, the volume of liquid delivered is not identical with the volume of liquid contained by the volumetric instrument. Volumetric instruments are either adjusted ‘to contain’ (i.e. the capacity printed on the instrument corresponds to the contained quantity of liquid) or ‘to deliver’ (i.e. the capacity printed on the instrument corresponds to the delivered quantity of liquid; the wetting residue remaining in the instrument has already been taken into account in the calibration). According to international standards, the marking ‘Ex’ is used to indicate that the volumetric instrument has been adjusted to deliver; the marking ‘In’ is used to indicate that the volumetric instrument has been adjusted to contain. Most typical graduated pipettes, bulb pipettes, and burettes are usually adjusted to deliver, whereas typical graduated cylinders, volumetric flasks, and capillary pipettes (i.e. very small measuring pipettes with a capacity of up to 0.2 ml) are usually adjusted to contain.
ISO 648 and ISO 835 specify three classes of accuracy for pipettes:

Class A (higher grade, without waiting time)
Class AS (higher grade, with a specified waiting time)
Class B (lower grade, without waiting time)

‘Waiting time’ is the period to be observed in order to ensure complete delivery after apparent completion of the liquid delivery of the pipette (i.e. after the meniscus appeared to come to rest) and before the final reading of the delivered volume is taken (i.e. before the tip of the jet is removed from the inner surface of the receiving vessel).
Furthermore, four types of graduated measuring pipettes are standardized in ISO 835:

Type 1: Partial delivery (adjusted to deliver from zero line at the top to any graduation line; nominal capacity is represented by the lowest graduation line)
Type 2: Total delivery (adjusted to deliver from any graduation line down to the jet; nominal capacity is represented by the highest graduation line)
Type 3: Total and partial delivery (adjusted to deliver from zero line at the top to any graduation line; nominal capacity is obtained by delivery down to the jet)
Type 4: Blow-out (adjusted to deliver from zero line at the top to any graduation line; nominal capacity is obtained by delivery down to the jet where the last drop of liquid in the jet is expelled by blowing. Blow-out pipettes are only adjusted to accuracy Class B.)

Typical graduated measuring pipettes with a capacity from 0.5 ml to 25 ml are commercially available in accordance with ISO 835 as Type 1, 2, or 3 and in Class A, AS, or B. Bulb pipettes with a capacity from 0.5 ml to 100 ml are available in accordance with ISO 648 in Class A, AS, or B. They are all adjusted to deliver (‘Ex’); i.e. the residual liquid still left in the tip has already been taken into account during calibration and must not be expelled into the vessel by blowing out.
Very small measuring pipettes with a capacity of up to 0.2 ml (capillary pipettes) are designed as Type 4 (blow-out) in Class B; however, most available capillary pipettes are actually adjusted to contain (‘In’), which is strictly not conforming with ISO 835. Though, the formal difference between blow-out pipettes adjusted to deliver and pipettes adjusted to contain may be small for total delivery since the last drop of retained liquid in the jet can be expelled by blowing in both cases. Nevertheless, when the greatest possible accuracy is desired, it is not sufficient to simply empty pipettes that are adjusted to contain by blowing out the liquid with a pipetting aid; such pipettes should also be rinsed two to three times with the diluting medium.
The general procedure for the use of pipettes that are adjusted to deliver (‘Ex’) is described in ISO 4787 as follows. However, when the greatest possible accuracy is desired, pipettes should be used as closely as possible to the manner in which they have been calibrated (which is also specified in ISO 4787, ISO 648, and ISO 648).

10.5.1 Pipettes adjusted to deliver (see ISO 648 and ISO 835)
After rinsing with the liquid or reagent to be used, fill the pipette by suction to a few millimetres above the selected graduation line. Remove any liquid remaining on the outside of the jet. The final setting of the meniscus shall then be made by dispensing the surplus liquid through the jet. Remove any drops of liquid adhering to the jet by bringing an inclined ground glass vessel into contact with the tip of the jet. Delivery shall then be made with the tip of the jet in contact with the inner surface of the inclined receiving vessel.
If the setting after delivery is done at a lower graduation line, the liquid flow has to be nearly stopped a few millimetres above the graduation line. After observing a waiting time, if specified, complete the final setting quickly.
A waiting time, if specified, shall be observed before making the final setting for delivery of a given volume.

The corresponding procedure for the use of pipettes that are adjusted to contain (‘In’) is described in ISO 4787 as follows. Note that, when the greatest possible accuracy is desired, it is not sufficient to simply empty the pipette by blowing out the liquid with a pipetting aid; the pipette should also be rinsed two to three times with the diluting medium.

10.5.2 Pipettes adjusted to contain
Rinse the pipette with the reagent to be used to a few millimetres below the desired graduation line. Fill the pipette by suction to as close as possible above the selected graduation line. Remove any liquid remaining on the outside of the jet. Make the final setting of the meniscus to the line by withdrawing the surplus liquid by means of filter paper. For the discharge, rinse the pipette several times with the diluting medium.


What's the difference between povidone K29/32 and povidone K30?

Firstly, Povidone (polyvinylpyrrolidone, PVP)  is  a synthetic polymer used in the pharmaceutical industry as vehicle for dispersing and suspending drugs.

Polyvinylpyrrolidone

There are quite a number of commercial grades of povidone (and also its highly cross-linked counterpart called crospovidone). The K-value of different types of povidone is characterised by their viscosity in solution  and also the mean molecular weight:

The method of obtaining these values is stated in various pharmacopoeias e.g British Pharmacopoeia

These different chemical characteristics offer a variety of uses in pharmaceutical formulations, including as a binder for tablets and capsules, a film former for ophthalmic solutions, to aid in flavoring liquids and chewable tablets, and as an adhesive for transdermal systems.

These also affect the formulation processes during tablet manufacturing for example, the low viscosity and high adhesive properties of K29/32 make it highly efficient as a binder for wet granulation. On the other hand, K30 has the ability to complexate with the active ingredient providing taste masking and solubilization.

  The grade of povidone that is selected depends mainly on its molecular
  weight, as this dictates the viscosity, binding effect, the
  complexation capacity and how readily it is eliminated by the body.

Povidone is relatively inert and won't necessarily affect processes in the body instead, on the formulation providing different release characteristics of drugs for example modifying its release properties etc

References

British Pharmacopeia ed
Povidone K29/32
Kollidon® 30 – The Original from BASF


How do I make a 0.5 M citrate buffer at pH 4.5?

There are many ways to make simple pH buffers like this one. As @Poutnik noted, preparation via titration works fine and in very commonly used. To illustrate how this is done, several possibilities are presented. Assume the desired buffer characteristics are pH = 4.5, concentration = 0.5 M, final buffer volume = 1 L, and the following reagent grade chemicals are available: citric acid, sodium dihydrogen citrate, disodium hydrogen citrate and 2 M NaOH solution. Also assume a laboratory quality pH meter is available and that the pH meter has been properly calibrated with buffers of 4.00 and 7.00, bracketing the desired pH of 4.5.  Before proceeding, it is useful to examine citric acid’s $\alpha$ dissociation fractions.

Brief refresher on $\alpha$ dissociation fractions.

Citric acid, denoted $\ce{H_3Cit}$, is a triprotic weak acid. The successively less protonated citrate species are $\ce{H_2Cit^-}$, $\ce{HCit^{2-}}$, and $\ce{Cit^{3-}}$. Since all the citrate species are in the same volume of solution, molar concentrations (denoted by the usual square brackets, e.g., $[\ce{H_3Cit}]$) are proportional to numbers of moles. Define $D$ as the sum of $[\ce{H_3Cit}] + [\ce{H_2Cit^-}] + [\ce{HCit^{2-}}] + [\ce{Cit^{3-}}]$. Then $\alpha_0$ is $[\ce{H_3Cit}]/D$, $\alpha_1$ is $[\ce{H_2Cit^-}]/D$, $\alpha_2$ is $[\ce{HCit^{2-}}]/D$, and $\alpha_3$ is $[\ce{Cit^{3-}}]/D$. The sum of the $\alpha$ values equals unity, as expected: every citrate species must be somewhere in the solution.

At any given solution pH, all four citrate species are present (though possibly at negligible concentration), as determined by the pH, the three acid dissociation equilibria and their associated equilibrium constants: $\mathrm{p}K_\mathrm{a1}$, $\mathrm{p}K_\mathrm{a2}$, and $\mathrm{p}K_\mathrm{a3}$. For citric acid, the $\mathrm{p}K_\mathrm{a}$ values are 2.13, 4.77, and 6.40. At any given solution pH, the $\alpha$ fractions are simply the proportion of total citrate that is in the relevant ionic or molecular species.

The screenshot below shows the $\alpha$ fractions up to pH = 8 (it is boring above pH = 8):

For a desired citrate buffer at pH = 4.5, the figure shows that only $\ce{H_2Cit^-}$ (green $\alpha_1$ curve) and $\ce{HCit^{2-}}$ (blue $\alpha_2$ curve) are present in significant concentrations. Furthermore, $[\ce{H_2Cit^-}]/[\ce{HCit^{2-}}] > 1$.

Method 1

Weigh out an amount of citric acid equal to approximately 0.5 moles, i.e., about 91 g. Dissolve that, with stirring, in about 0.5 L of water in a large beaker. Be careful to not drop a magnetic stir bar into the beaker!  If 375 mL of 2 M NaOH solution were to be added, the pH would be about 4.76, which is a little too high. The reason is that 0.75 moles of $\ce{OH^-}$ would neutralize the first citric acid proton and lead to a roughly equimolar ratio between $\ce{H_2Cit^-}$ and $\ce{HCit^{2-}}$. This is too much added NaOH solution because $[\ce{H_2Cit^-}]/[\ce{HCit^{2-}}] = 1.82$, as is easily determined using the Henderson-Hasselbalch equation or the $\mathrm{p}K_\mathrm{a2}$ expression.

Add 250 mL of 2 M NaOH and then, with constant stirring and monitoring of the pH, slowly add more 2 M NaOH solution until the pH increases to 4.5. Finally, the last step is to add water until the final buffer volume is 1 L.

Method 2

Start by assuming $[\ce{H_2Cit^-}]/[\ce{HCit^{2-}}] = 2$, so the pH will be a little below the target value of 4.5, and weigh out about 71 g of $\ce{NaH_2Cit}$, i.e., about 1/3 mole. Likewise, weigh out 39 g of $\ce{Na_{2}HCit}$, i.e., about 1/6 mole. The total number of moles of citrate species is about 0.5. Dissolve the $\ce{NaH_2Cit} $ and $\ce{Na_{2}HCit} $, with stirring, in about 0.8 L of water in a large beaker. Then, with constant stirring and monitoring of the pH, slowly add 2 M NaOH solution until the pH increases to 4.5. The last step is to add water until the final buffer volume is 1 L.

What happens if a mistake is made and the pH is too high? Simply add a little more of the most acidic species, making sure it dissolves, and then titrate back up with NaOH solution. It can be helpful to have a lower concentration NaOH solution available, e.g., 0.1 M, either for this corrective procedure or for tweaking the pH.

Other possibilities include 1) citric acid plus trisodium citrate or 2) citric acid plus disodium hydrogen citrate or 3) sodium dihydrogen citrate plus trisodium citrate, but these are not considered further.


SN1 Polar Protic Solvent Stabilization of Substrate

I know that the standard textbook answer is that a polar protic
  solvent is able to better stabilize/solvate carbocations through
  hydrogen bonding.

In an $\ce{S_{N}1}$ reaction, a polar protic solvent can stabilize the leaving group through both hydrogen bonding and solvent dielectric effects.  While these same solvent dielectric effects can also stabilize the carbocation intermediate, I don't think hydrogen bonding plays a role in the stabilization of the carbocation itself.  First off, any protic molecule close enough to the carbocation to hydrogen bond would simply react with it instead of forming a hydrogen bond. Second, since the carbocation is positively charged it cannot form a hydrogen bond.  Again, solvent stabilization of the carbocation occurs through dielectric (dipolar) effects of the medium.


LogP of halogen derivatives of aromatics and aliphatics explanation

Here is a chart of the solubility of the species in water and the log of the partition coefficient you noted.

                           solubility       solubility
                  MW       g/100ml water    moles/liter   log(P)
  Benzene         78.11    0.179 (1)        0.0229         2.13
  Flororbenzene   96.103   0.155 (2)        0.0161         2.27
  Cholorbenzene  112.56    0.050 (2)        0.0044         2.84
  Bromobenzene   157.01    0.041 (2)        0.0026         2.99

(1) Wikipedia

(2) Cation Complexes of Compounds Containing Carbon-Carbon Double Bonds. VI. The Argentation of Substituted Benzenes
L. J. Andrews, R. M. Keefer
J. Am. Chem. Soc., 1950, 72 (7), pp 3113–3116
DOI: 10.1021/ja01163a085
weblink http://pubs.acs.org/doi/abs/10.1021/ja01163a085

There is a Wikipedia chart in article on Ionic radii.

 Ion     radii (picometers)
  F-      119
  Cl-     167
  Br-     182

The halide on a benzene ring is going to be relatively covalently bound. It definitely won't dissociate to form ions and only very moderately participate in hydrogen bonding.

So as the size of the halide increases the "volume" of the molecule increases. Thus the molecule would preferentially like to be in the non-polar phase.

Here is a scaled image of fluorobenzene molecule:

and a scaled image of the bromobenzene molecule.

Haloalkanes

The "anomalous" fluoroalkane is explained in a different manner. To get an n-alkane that has a boiling point above room temperature we need to go to n-pentane.

                   BP                 solubility      solubility
                   (deg C)   MW       g/100ml water   millimoles/liter
 n-Pentane         36        72.15    0.040 (1)         0.55
 1-Fluoropentane   63        90.14    1.270 (2)         14.1
 1-Chloropentane  108       106.6     0.197 (3)         1.85
 1-Bromopentane   130       151.05    0.127 (4)         0.838

(1) Wikipedia
(2) NIH
(3) Wikipedia
(4) Handbook of Aqueous Solubility Data, Second Edition
 By Samuel H. Yalkowsky, Yan He, Parijat Jain

There are two competing effects.

(1) The size of atom bound to the alkane.

    Br > Cl > F > H

    Scaled model of Fluoroethane

(2) The second effect is the electronegativity of the X group in the $\text{Alkane}\text{X}$ bond.

 F    >   Cl   >   Br   >   H
 4.0      3.2      3.0      2.2

The fluorine atom has a slightly larger radius in a fluoroalkane than the corresponding C-H group. But the halide groups allow more dipole forces. So by "volume" the overall effect is:

 fluoroalkane
 alkane
 choloroalkane
 bromoalkane


Let's start with the basics:

A representation is a set of matrices that fulfill the multiplication table for the point group.

So I might be able to use a set of 3x3 matrices, since of course I can define any of the symmetry operations over Cartesian (x,y,z) space. e.g.:

$$E = \left(
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array}
 \right) \;
i =
\left(
\begin{array}{ccc}
-1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1 \end{array}
 \right)
$$

But usually, I can fulfill the multiplication table with much simpler (smaller) matrices.

An irreducible representation is a set of matrices that are as small as possible.
A reducible representation is a set of matrices that can be decomposed into a linear combination of irreducible representations. This can occur by multiple techniques, including block-diagonalization.

The distinction exists, because for chemical point groups, there are usually a finite number of irreducible representations (given in character tables), but of course there are an infinite number of reducible representations.

The power of symmetry and group theory is that any function of a molecule, like orbitals, vibrations, etc. can be described using a representation. Thus, we usually label orbitals and vibrations by their corresponding irreducible representation.

I'll have to hunt down my copy of Ogden, but usually people use the Cartesian axes or $p_z$, $p_x$, and $p_y$ to illustrate the different irreducible representations. That is, a basis is some function (like the Cartesian axes or an atomic orbital) used to generate a representation. We look at how the basis is transformed under the different symmetry operations and write down the matrix or characters from those transformations.

(I try not to use that term because students find it confusing and different guides and texts use subtly different definitions for it.)

The reduction formula (or algorithm) is a way of decomposing the reducible representation (which is really a big matrix) into a combination of the different irreducible representations, but only needing to act on the characters of the representations.

That is, I don't need to actually use the reducible matrices themselves. Using the reduction formula, I can get the linear combination of irreducible representations only using multiplication of characters of the representations. It's much easier to do the reduction formula than to figure out how to transform a large reducible matrix.


The energy required to break the first $\ce{H-O}$ bond in water is $\pu{493 kJ/mol}$. For comparison, the energy required to break the first $\ce{H-C}$ bond in methane is only $\pu{435 kJ/mol}$. As a point of reference for just what these numbers mean in terms of bond strength, we can think about how long it takes for methane to be destroyed in the atmosphere. The first step of the destruction of atmospheric methane is to break that first $\ce{H-C}$ bond; because this is so difficult to do (again, requiring $\pu{435 kJ/mol}$) it takes an average of about 9 years for a molecule of methane to be destroyed! And the energy required to break the $\ce{H-O}$ bond in water to form hydroxide and hydronium ions is even greater.

I realize that my analogies involve very different processes, but the underlying issue in both cases (atmospheric destruction of methane and self ionization of water) is bond strength. I hope that this gives some insight as to why the self-ionization of water is as low as it is. Don't hesitate to ask for any clarifications in the comments.


Even for weak ligands, the electrons will eventually have to pair up, once the number becomes more than $5$. So, $\mathrm{d^7}$ is indeed possible with a weak field ligand. Consider $\ce{[CoCl6]^{4-}}$, for example.


Take a slightly more general problem of miscibility, all that in thermodynamic equilibrium. In your question you mix up kinetics and equilibrium, and additionally in a wrong way. Equilibrium is the most important part.

If you put two liquids together and shake them well, than in general, they do not mix. So you will observe a boundary between, say, diethyl ether (upper layer) and water (bottom layer).
If you separate them and analyze their composition, you will find, that the ether fraction contains little bit of water (1.5g / 100 mL) and that the water fraction contains little bit of ether (6g / 100mL). Knowing this, if you now take 100 mL of water and mix it with 5g of diethyl ether, you will observe, that no boundary is formed, and all ether has dissolved in water.

As another example, if you shake ethanol with water, they mix in any ratio.

That's observation, now the rationale. The particles of pure liquid interact with each other, in attractive way (otherwise you will have no liquid, but gas). Now you mix two pure liquids together (A + B) and look, if some of the particles of A can easily sneak into the B. If their behavior is reasonably similar, they can do it and interact with particles of B also in attractive way. The attraction should be comparable or greater, than within pure A. That is the enthalpy part.
But as we are dealing with free energy, there is also the entropy, which you have to take into account. On one hand, entropy favors mixing, as you add lot of accessible states. On the other hand, if the attraction of B to A is very strong, it starts to organize the molecules of A around itself, so they are not as free to move as they were in pure A, and limits the miscibility.

For the solubility of solids, it is technically similar, the A is solid and the interaction which hold sit together is the lattice energy.

As a side note, if you ask about solubility, miscibility and such, you should refrain from ideal approximations, as they do not predict such phenomena. Their calculation from pure substance properties is therefore terribly difficult, and physical chemistry tables are the only reasonable answer.


How can I know which molecule is more stable when bond order it's equal?

To recognize the more stable molecule you may extend the description with bond orders is the one considering how the mathematical concept to mix atomic orbitals by LCAO and visualizing the results in Molecular orbital diagrams and compute the overall energy of such a molecule.  Eventually, you compare the total energy of $\ce{O^-_2}$ with $\ce{C^+_2}$ using the same external reference.
As about the hydrogen molecule shown below:

Two important details about symmetric diatomic moleculs to retain are:

For $n$ atomic orbitals, LCAO yields $n$ molecular orbitals.
The split between the lowered bonding molecular orbital and the higher antibonding orbital is not symmetric.  The difference between the energy level of the initial atomic orbitals and the the antibinding molecular orbital is higher, than the difference between the initial atomic orbitals and the binding molecular orbital.

So you draw the diagram including the energetic levels, and populate the levels with electrons from the bottom up.  If there multiple molecular orbitals of same energy, you fill the orbitals first with one, and later with the second electron. You then compute the the overall energy of the molecule and later compare the results about the molecules in question against a reference in common, e.g. vacuum.
Note that the notation $\sigma$ and $\sigma{}^*$ is about the rotation symmetry around the bond between the molecule (contrasting to $\pi$ and $\pi{}^*$ as perpendicular to this), and not if the molecular orbital in question is conceptually formed from $s$, or $p$ atomic orbitals.  Equally, the energetic consecution of levels of the molecular orbitals may change, e.g. between nitrogen and oxygen:

(using material from here)


Atomization enthalpies of transition elements

The sorts of generalities always sweep aside other considerations.

$\ce{^26Fe}$ has a body-centered cubic crystal structure.

$\ce{^27Co}$ has a hexagonal close-packed crystal structure.

$\ce{^28Ni}$ has a face-centered cubic crystal structure.

Here is a nice table showing the overall trends.


Why do we need incubation time during the glucose test by chronoamperometry?

Well, Although I'm not a chemical student, I think I could share my understand of the Chronoamperometry by the parameters that could affect the result in my project.

According to my test result, the incubation time will affect the magnitude of the current response at the chemical cell(also for the trigger time). This is related to the property of the glucose test strip(Truetrack in my design) and the circuit design. And I assume that this is because of the distribution of the glucose in the test strip, specially the amount of the glucose close to the surface of the electrode.

I hope if there is a chemical guy that can explain the detailed principle of Chronoamperometry, very detailed stuff. Because I have finished my project and thus the theoretical staff used for the lecture is not enough to attract my attention.
Also, if someone has interest in this, I can share my experiment result and discuss with him personally.
Here is my email: matthew97829782@gmail.com


Mechanism for the synthesis of 2-pyridone from pyridine N-oxide

Pyridine N-oxide itself is generally fairly resistant to nucleophilic attack as the ring is already 'electron rich' and doesn't want to accept additional electron pairs.

When treated with acetic anhydride, however, the carbons at the 2 position become electrophilic (the acetate is electron withdrawing making the C=N bond more like an iminium), hence it is possible to have acetate attack, with subsequent elimination to re-aromatise the ring, leaving us with a 2-substituted pyridine.

If the acetate is hydrolysed (water, or mild aqueous base), the 2-hydroxy pyridine is formed, which may tautomerise to the more stable 2-pyridnone.

It's worth noting that this mechanism differs significantly from the Boekelheide reaction (the initial pyridine has an additional methyl group at the 2-position), whereby the acetate bound to nitrogen is migrated by means of a [3,3] sigmatropic rearrangement. This rearrangement isn't possible/favourable in this case, as the [3,3] requires a 6 membered transition state, whereas without the methyl group only a 5 membered TS exists.


Opening a shaken up bottle quickly = no fizz?

The trick is simply waiting for the CO2 to be released ahead of time.

For a full and cold soda this trick will not work, the pressure in the head-space will be slightly lower due to the lower temperature and with less total volume of CO2 -which makes the "pop" part of the trick fail- and the CO2 initialization points within the solution will be distributed -which causes the fizzing and overflow of liquid.

With a half filled soda that's been sitting out in a comfortably warm room, the soda is warmer. This means that you are working with a (even more) super saturated solution of CO2 in water and acid. This additional instability combined with a greater head-space for pressurization allows a greater build-up of CO2 in the bottle allowing for the "pop".  If you didn't wait for the bubbles to go away you would still have fizz formation (and even flatter soda). If you did wait for the bubbles to disappear but you opened the bottle slowly, the soda would still be just as flat as with the pop-trick but with a slight hissing sound and limited additional bubble formation.


Why is aniline less basic than methylamine?

N in aniline is forced into an $\ce{sp^2}$ configuration because only then can its p orbital interact to form aromatic MO’s. By contrast, methylamine has its N in the usual $\ce{sp^3}$ configuration. An $\ce{sp^2}$ lone pair has a great contribution from an s AO than an $\ce{sp^3}$ lone pair (33% vs 25%), and because an s AO is lower in energy than a p AO (due to a region of electronic density closer to the nucleus), it is also lower in energy. Therefore, the HOMO of aniline has a poorer energy match with the H 1s LUMO than the HOMO of methylamine, and the resonance energy stabilizing the protonated adduct is smaller. Hence, aniline is less basic.


Research-backed articles on cast iron seasoning

There is this research on Chinese equivalent of seasoning process called "Kitchen God's Blessing" [1].
They argue that the non-stick property of seasoning comes by iron oxide $(\ce{Fe3O4})$ nanoballs.
If they are correct then the oil or its polymers is not exactly what is causing the non-stick property but the those nanoballs on the surface. Their seasoning method is somewhat different however. It uses temperatures up to 450 °C.
Reference

Gao, C.; Yang, N.; Li, C.; Wang, X.; Yu, X.; Zhang, L.; Wei, Z. Seasoning Chinese Cooking Pans: The Nanoscience behind the Kitchen God’s Blessing. Nano Materials Science 2020. DOI: 10.1016/j.nanoms.2020.06.001.


How to run crystal simulation on GAMESS?

As you said, there are many packages capable of handling calculations of crystals, slabs, etc. with periodic boundary conditions. VASP is only one of dozens. (That's probably a separate question in its own right.)

GAMESS does not include treatment of periodic boundary conditions, so you would need to run a calculation on a "cluster."

Since you mention wanting to treat a surface reaction, you'd use a builder (e.g., Avogadro) to create a large cluster (e.g., Build->Supercell). So rather than one primitive unit cell, you create maybe a 2x2x2 or 3x3x3 slab. If needed, you'll terminate dangling bonds (e.g., adding hydrogens to cap oxygens in $\ce{TiO2}$ for example.. notice the terminal "uncapped" atoms on the back and right of the unit cell).

Finally, after running calculations in GAMESS, you'd need to check for convergence. So you'd want to run calculations with increasing system size and see if the properties converge.

Now, this isn't as strange as it sounds. In some cases ($\ce{MgO}$ and $\ce{TiO2}$ are good examples) the properties require more advanced calculation methods like multi-configuration QM, and this may be easier to do in a quantum chemical package with a cluster, rather than a solid-state crystal code.


Why do cyclic hydrocarbons have higher boiling points than their acyclic isomers?

Your intuition is indeed correct! Several sources provide the same answer (1, 2, 3). Perhaps the simplest and most direct evidence comes from comparing the densities of the liquid unbranched alkanes and cycloalkanes:

(Source)

The cycloalkanes have slightly lower molecular mass than their parent unbranched alkanes with the same number of carbon atoms, yet the cycloalkanes are consistently more dense in the liquid phase by a factor of ~20%. The only way this can happen is if each individual cycloalkane molecule effectively occupies less volume in the liquid phase than the related alkane. This means cycloalkane molecules are closer on average in the liquid, and as van der Waals forces are attractive (at intermolecular separations) and their strength varies inversely with distance, the intermolecular attractions are stronger in cycloalkanes, and so the boiling point of the cycloalkanes should be higher.

This can be explained by considering that the unbranched linear chains in alkanes have many possible conformations, and thus are in more geometrically irregular contact in the liquid, leaving more space on average between the molecules. Cycloalkanes do not have as many conformations available, and so their molecules will approach each other more orderly and effectively, leaving less intermolecular space.

Presumably this effect can be extended to all substances (including the ethers mentioned in the question you linked), but it need not always be the dominant factor in determining boiling points.


I have access to several programs or similar solutions for the work with so-called steam tables. I usually prefer to use the NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9. It is a program that can calculate the thermodynamic and transport properties of water and steam as well as some other important fluids and their mixtures. It was developed by the National Institute of Standards and Technology (NIST) and is commercially available here.
Although I usually don’t do this myself, REFPROP can be used with Microsoft Excel. This is described in the corresponding Help file as follows:

Excel Spreadsheets
The REFPROP software includes dynamic link library ('.dll') support that allows a broad variety of applications to take advantage of REFPROP's capabilities.  REFPROP is distributed with front-end software that links Microsoft Excel applications to fluid property functions.
The REFPROP Excel front-end is supplied as a Microsoft Visual Basic module.  Excel automatically incorporates Visual Basic as a macrofunction execution, scripting and interface utility.  When the REFPROP front-end module is imported into an Excel spreadsheet, REFPROP functions become accessible to the user.  The proceedure for doing this is described below.
The information given below demonstrates the front end usage; however, the user is not bound to this format; rather, the preferred method would be to write one's own visual basic code that calls the DLL directly.  A dedicated module will increase the speed of the application and give the ability to use the REFPROP functions to their fullest.
The following supplementary instructions are intended for users who wish to create and modify their own spreadsheets that utilize the REFPROP functions based on the front-end module.  The Excel application included with the software is called REFPROP.xls and is located in the directory where the REFPROP program was installed.  Sample FORTRAN and Visual Basic files are also included.
Properties are calculated by calling various functions.  For example, the following line would be placed in a cell to calculate the density of argon at 300 K and 10 MPa:
=Density("argon","TP","SI",300,10)
Other properties can be calculated by using any of the following functions:
(…)


What is the correct way to calculate the rate constant for the reaction of NO and O2?

Either definition is acceptable, as long as you clearly identify relationship between the calculated rate value $R$ and the rates-of-change for each of the species in the system.

In case (1), you've effectively calculated the rate constant defined as:

$$
R_1 = k_1\ce{[NO]^2[O2]}
    = -{\mathrm d \ce{[O2]}\over \mathrm dt}
    = -{1\over 2}{\mathrm d \ce{[NO]}\over \mathrm dt}
    = {1\over 2}{\mathrm d \ce{[NO2]}\over \mathrm dt}
$$

In case (2), you've instead calculated the rate constant defined as:

$$
R_2 = k_2 \ce{[NO^2][O2]}
    = -2{\mathrm d \ce{[O2]}\over \mathrm dt}
    = -{\mathrm d \ce{[NO]}\over \mathrm dt}
    = {\mathrm d \ce{[NO2]}\over \mathrm dt}
$$

Both cases accurately describe the kinetics, but you have to be careful to identify which definition of the rate you've used in calculating $k$.


E2 Elimination; Anti-periplanar beta hydrogen

A reasonable chemist would be on the lookout for two products, especially given that E2 elimination from 1-bromo-1-methylcyclohexane produces some methylenecyclohexane

In the transition state for the E2 elimination the pi bond is already partially developed.  This means that the relative thermodynamic stabilities of the products will play a role in determining the product distribution.  Since more highly substituted olefins are thermodynamically more stable than less substituted olefins, one would predict that the product with the double bond internal to the ring would predominate over the analogue with the external double bond.

Reaction temperature, base size, steric constraints around the various hydrogens that might be involved and the numbers of the various hydrogens all factor into the final product ratio, but these effects are usually secondary unless you change circumstances so much that the reaction mechanism changes and shifts away from E2.


How were silver and gold separated using the salt cementation process?

This is an old process

M. Berthelot gives it from Papyrus V from the Leyden Papyrus X, in his Introduction à l'étude de le Chimie, des anciens et du moyen âge (Introduction to the study of Chemistry, of forefathers and the middle age).
The papyrus is at least from the III° century, the process very probably older. From page 14 (my translation):

  "Take piquant vinegar, thicken, take some [lacuna], 8 drachmas of common salt, 2 drachmas of lamellar alum (schist), 4 drachmas of litharge, crush with vinegar for 3 days, separate by decantation and employ. Then add to the vinegar 1 drachma of copperas, half an obole(1) of [lacuna], 3 oboles of chalcite(2), one and a half oboles of sory(3), one silique(4) of common salt, two siliques of Cappadoce salt(5). Make a lamella having two quarters (of an obole?) Submit it to the acion of fire... until the lamella breaks, then take the pieces and view them as refined gold.

  1 drachma = 6 oboles
  copper ore, such as pyrite
  product of alteration of pyrite, which can include both copper sulfate and basic iron sulfate.
  silique = third of an obole, a measure of weight
  a variety of rock salt"

What about the chemistry?

Berthelot cites the process of départ par cémentation (departure by cemetation) from Macquer's Dictionnaire de la Chimie (1778). From the latter, page 312 (still my translation):

  "First prepare a cement composed of 4 parts crushed bricks passed on a sieve, one part green vitriol calcined to red, and one part common salt: mixe the whole very-exactly, and make a firm paste, moistening it with a little water or urine. This cement is called cément royal, because it is used to purify gold, which chemists view as the king of metals.

  On the other hand, reduce the gold one wishes to prepare, to lamellas about as thin as billon pieces: at the bottom of a crucible or cementation pot, place a layer of the cement of thickness the width of a finger: stratify the gold lamellas on this layer: place on top a new layer of cement: thus fill the pot, always placing the gold between two layers of cement; and cover it with a lid sealed with sand and clay. Place the pot in a furnace or oven ; heat progressively, until the pot is a dull red ; maintain this level of heat for about twenty four hours: it is very-essential that the heat not be able to melt the gold. After that, let the pot cool down, and open it to remove the gold, which must be separated thouroughly from the surrounding cement: it also has to be boiled in large quantities pure water several times. One can try this gold on a touchstone or otherwise ; and if it is found insufficiently pure, submit it a second time to the process.

  The vitriolic acid of the brick and of the calcined vitriol, releases the acid from the common salt during this cementation ; and this last one dissolves the silver alloyed to the gold, and seperates it by this mean.

Berthelot has this to say about the chemistry occuring here, (page 16):

  By proceeding thus, the silver and other metals dissolve in the sodium chloride, with the help of the oxidative -- and then chlorinating -- action exerted by the iron oxide derived from the vitriol ; while the gold remains unchanged.

And elsewhere (page 15):

  [...] At a stretch, it could be that sodium chloride, in the presence of basic salts of [ferric oxide] [...], releases chlorine [...].

So what I think happens is: the silver reacts with $\ce{Cl2}$ and maybe $\ce{FeCl2}$ (speculation?, b.p.= 1023°C, vapor pressure at 800°C = 0.1bar) to give $\ce{AgCl}$. Then the $\ce{AgCl}$ evaporates (b.p.= 1547°C, vapor pressure at 800°C = 0.1mbar), condensates on the cement, and dissolves in it. It is unclear whether the silver stays in oxidized form. I would bet yes, but I am no inorganic chemist.

Recovering the silver

Macquer also says (page 313):

  We can see it is essential to wash thouroughly the gold, after the process is completed, to remove all the disolved silver parts, which would otherwise stay mixed with it.

  Concerning this silver, it can subsequently be separated from the cement, by heating it with a sufficient quantity of lead and litharge, and taking the resulting silver containing lead slag and cupelating it to remove the lead."

So apparently you can redissolve the silver in lead, and then remove the lead by normal cupellation.

Food for (even more) thought

Macquer (page 313) also mentions that the process works just as well if you replace common salt by saltpeter. Which is weird because $\ce{AgNO3}$ decomposes at 440°C.

He then adds that you can add both common salt and saltpeter, and not only does the silver still get dissolved, the gold doesn't. Which is also weird because you have all the components of aqua regia, but it doesn't attack the gold. Macquer does acknowledge his low confidence in his sources for this last process.


Why can alkali metal hydrides usually not act as nucleophiles towards carbonyl groups?

The problem is that neither of these nucleophile are hard. Vanilla hydrides such as potassium hydride or sodium hydride release very soft hydride ions. Hydrogen only has one proton. Giving it two electrons to stabilize as in the case of the hydride anion is a tall order. Note the proton to electron ratio: 1 to 2. Even carbanions which are unstable species don't have such a ratio.

Therefore, the hydride ion has a massive ionic radius and is easily polarized, making it an extremely soft base. Electrophilic carbon atoms in carbonyl groups are usually hard(er).

Plus the 1s orbital of hydrogen is too small to effectively overlap with anything other than another hydrogen. Is your textbook written by Clayden? I distinctly remember Clayden's remark about hydrogen's 1s orbital with regard to its anion's nucleophilicity.

Usual reagents used for hydride delivery/reduction of carbonyl compounds (and related) include LAH, NaBH4, and NaBH3CN, in order of decreasing strength. As far as I know, the lithium ion in LAH plays an integral role in its mechanism of reduction by bonding with the carbonyl oxygen of carboxylic acids.

Also, these aforementioned reducers are in order of decreasing strength because they are listed in order of decreasing partial negative charge on their hydride ions. For example, the nitrile group in sodium cyanoborohydride withdraws electron density from the hydrogens. Hence the reason it is the weakest of the three aforementioned reducers. NaBH3CN will only act on strongly polarized iminium ions rather than plain imines or carbonyl compounds.


Is d3-acetic acid more acidic than acetic acid?

The following articles suggest that the data reported from Solomons, Fryhle and Snyder Organic Chemistry is likely incorrect.

Maya Paabo, Roger G. Bates and R. A. Robinson ("Dissociation of Acetic Acid-d3 in Aqueous Solution and Related Isotope Effects from 0 to 50°", J. Phys. Chem. 1966, 70, 2, 540-543) report that the $\mathrm{p}K_\text{a}$ of the deuterioacid is greater (less acidic) than that of the protio analogue.
See Table 1 on p. 134 of C. L. Perrin's chapter ("Advances in Physical Organic Chemistry", Volume 44, edited by John P. Richard) where it is also reported that the $\mathrm{p}K_\text{a}$ of the deuterioacid is greater (less acidic) than that of the protio analogue.
Rose McConnell, Walter Godwin, Brant Stanley, and M.Shane Green ("Acidity Studies of Deuterated Acids and Bases Commonly Used as Buffers in NMR Studies", Journal of the Arkansas Academy of Science, Volume 51 Article 21) also report (see Table 1 on p. 140) similar findings.

These findings indicate that acetic acid-d3 is less acidic than acetic acid-h3.  This would be consistent with the operation of a simple inductive effect where protium is more electronegative than deuterium (as expected).  Hence, the more electron withdrawing $\ce{CH3}$ group is better able to stabilize the electron rich carboxylate anion in the the ionized acid as compared to the less electron withdrawing $\ce{CD3}$ group.


Why does nitric acid not turn yellow everytime?

Pure nitric acid is colourless, the brown colour is caused by dissolved nitrogen dioxide, $\ce{NO2}$.

Nitrogen dioxide is formed by the slow decomposition of nitric acid, according to $$\ce{4 HNO3 -> 4 NO2 + 2 H2O + O2}$$

Nitrogen dioxide is a good oxidant and it does so - supposed that it finds a suitable partner.

This would be chloride. So, the more chloride in solution the more redox chemistry furnishing colourless products.


Is Hydrogen Bonding a Type of Dipole Dipole Interaction?

Well, it turns out that this is a very active area of research. I will only summarize what I understand to be true about the covalent nature of the hydrogen bond, so I'm sure the explanation could be more detailed and potentially more accurate in some places (I hope someone gives a more detailed answer), but here's what I've got.

As you said, it has been found that the hydrogen bond (specifically in water) is about 90% electrostatic and 10% covalent. See this link for a sort-of-review/overview of hydrogen bonding and some relevant research.

Upon following the paper trail, I found the following research article: E.D. Isaacs, A. Shukla, P.M. Platzman, D.R. Hamann, B. Barbiellini, C.A. Tulk, J. Phys. Chem. Solids, 2000, 61, 403-406. (Mirror)

Essentially, they use some method which I don't really understand and find quite conclusive evidence of the (word of the day) anisotropy of hydrogen bonds. That means that the bond is indeed directionally dependent. Bond direction is one defining characteristic of covalent bonds.

That raises the question of what is going on with the orbitals that allow for hydrogen to participate in more than one covalent bond at a time.

A simple explanation of the orbital interactions for a water dimer is to view the oxygen 2p,2s hybridized orbital as having superposition with the hydrogen 1s orbital and then interacting covalently with that same 2p,2s hybridized orbital on the other oxygen in the dimer. This superposition which includes the hydrogen then explains the way that hydrogen is participating in one covalent bond and another partially covalent interaction.

Furthermore, this explanation of what the orbitals are doing fits well with what we would expect from a hydrogen participating in two different covalent interactions, which is that the energy curve has two minima where one is much deeper than the other.

Further evidence for the covalent nature of the bond is the fact that the O-H bond elongation and O---H hydrogen bond length always change together. That is, as the O-H bond lengthens, the O---H interaction length shortens. This implies a shift in electron density towards hydrogen which is most easily explained by a delocalization of electrons participating in a covalent interaction.

So, I hope that's a good start toward understanding the covalent nature of the hydrogen bond--it was for me!

But to answer the simple part of the question which you clearly already know is that yes dipole-dipole is a significant part of hydrogen bonding, and it certainly should be, but as is always true of chemistry, there's a lot more going on than what one might initially think.

EDIT (06/01/2016)

Since I posted this answer, I think I have developed a better understanding of the situation, and feel obliged to update my answer. First, let me split hydrogen-bonding into two obvious parts, electrostatic and covalent characteristics. This is an obvious split, but it is not completely clear yet what we mean by covalent, so let us try to define that by looking at some examples. Furthermore, we must be careful when talking in about hydrogen-bonding because in every hydrogen bond there is a donor and acceptor, and these two roles are often played best by different chemical species.

Hydrogen-Bonding in thiols

Here we go breaking yet another rule. They told us that sulfur couldn't hydrogen bond. Well, that's not exactly true. Okay, but that hydrogen-bonding has got to be basically negligible right? Not exactly.

The first thing we need to do here is get some information. First, sulfur has an electronegativity of 2.58 on the Pauling scale. This is roughly the same as the electronegativity of carbon which we do not consider to hydrogen-bond well. So, if there are hydrogen-bonds involving sulfur, we should expect that they will be primarily covalent as any dipole interactions will be pretty small.

As it turns out, sulfur is a much better hydrogen-bond acceptor than it is donor, and at times is a comparable acceptor to oxygen. Here is a table comparing certain dimers from a great source you can reference at the bottom of the page$^3$.

As you can see, when the sulfur in these dimers (which are all named in the caption of the table) acts a hydrogen bond acceptor, the shift in vibrational frequency of the $\ce{O-H}$ and the binding energy are comparable to when an oxygen is the acceptor atom.

Looking at other research$^4$ involving sulfur hydrogen bonds reveals that approximately 70% of the interaction energy between sulfur and oxygen when sulfur is the acceptor comes from dispersion. That is the key physical insight in all this. Despite the loss of electronegativity, sulfur gains something in its greater dispersion interactions which makes it a good hydrogen bond acceptor. It should be noted, however, that the electronegative component of these hydrogen-bonds maintains their directionality.

Also, to compare, according to that wikipedia article I linked, the interaction energy of a water dimer is only 24% dispersion. In water electrostatics dominate and induction and exchange play meaningful roles.

Conclusions

I have more I could say, but it's perhaps a longer explanation than is worth going into. To conclude, what is meant by covalent interactions in hydrogen-bonding varies based on the species you are considering. In something like water, we can think about this in the usual way by saying that there must be some orbital overlap and hence electronic exchange which will be stabilizing. With other species, we can see that what we mean by covalent interactions might be different. That is, there is still some orbital overlap, but we must also consider non-electrostatic interactions like dispersion.

3 Noncovalent Forces-Chapter 2

4
Wennmohs, F., Staemmler, V., & Schindler, M. (2003). Theoretical investigation of weak hydrogen bonds to sulfur. The Journal of chemical physics, 119(6), 3208-3218.


Think of oxidation in a classical sense for a minute or so. Oxidation is the addition of oxygen to a molecule or it can also mean the removal of hydrogen. If you carefully look at the structure of ascorbic acid, you will see that the two -$\ce{OH}$ groups are now converted into carbonyl groups. Those two hydrogens can be lost as a proton, $\ce{H+}$.


Why is manganese(II) coloured although the transition should be spin-forbidden?

Selection rules

The intensity of the transition from a state $\mathrm{i}$ to a state $\mathrm{f}$ is governed by the transition dipole moment $\mu_{\mathrm{fi}}$ (strictly, it is proportional to $|\mu_{\mathrm{fi}}|^2$):

$$\iint \Psi_\mathrm{f}^*\hat{\mu}\Psi_\mathrm{i}\,\mathrm{d}\tau \,\mathrm{d}\omega \tag{1}$$

where $\mathrm{d}\tau$ is the usual volume element (representing integration over spatial coordinates) and $\omega$ is an additional spin coordinate (since the states $\Psi$ also specify the spin). For an electronic transition, $\hat{\mu}$ is the electric dipole moment operator, and only depends on the spatial coordinates $x$, $y$, and $z$. (I've also ignored the vibrational component i.e. the Franck-Condon factor since it's not really relevant for the purposes of this answer.)

Normally in determining selection rules, we are more interested in whether the integral in $(1)$ is zero or not. If the integral is zero, the transition is said to be forbidden. The usual way to determine whether it is zero is by using symmetry: if the integrand is antisymmetric under any symmetry operation, then the integral is zero. (It is exactly analogous to the integral of an odd function $f(x)$ over a symmetric region; for example, $\int_{-a}^a \sin x \,\mathrm{d}x = 0$.)

If we assume that both the initial and final states can be separated into spin and spatial components, then we get:

$$\iint (\psi_\mathrm{f}^\mathrm{spin}\psi_\mathrm{f}^\mathrm{space})\hat{\mu}(\psi_\mathrm{i}^\mathrm{spin}\psi_\mathrm{i}^\mathrm{space})\,\mathrm{d}\tau\,\mathrm{d}\omega$$

(I also dropped the complex conjugates - they aren't important.) Since $\hat{\mu}$ does not operate on the spin coordinate, we can separate this multiple integral:

$$\left(\int \psi_\mathrm{f}^\mathrm{spin} \psi_\mathrm{i}^\mathrm{spin} \,\mathrm{d}\omega\right) \left(\int \psi_\mathrm{f}^\mathrm{space} \hat{\mu} \psi_\mathrm{i}^\mathrm{space} \,\mathrm{d}\tau\right)$$

The first integral is nonzero if $\Delta S = 0$. The quantum number $S$ is related to the eigenvalue of a hermitian operator (the actual eigenvalue is $S(S+1)\hbar^2$), and eigenfunctions of a hermitian operator corresponding to different eigenvalues are necessarily orthogonal, making the integral identically zero if $\Delta S \neq 0$. This is the spin selection rule.

The second integral is nonzero if $\psi_\mathrm{f}^\mathrm{space}$ and $\psi_\mathrm{i}^\mathrm{space}$ have a different character under the inversion operation, i.e. one is gerade and another is ungerade. This is because $\hat{\mu}$ itself is ungerade, so for the integrand to be gerade, we need $\psi_\mathrm{f}^\mathrm{space}\psi_\mathrm{i}^\mathrm{space}$ to be ungerade ($\mathrm{u \otimes u = g}$). In turn this means that $\psi_\mathrm{f}^\mathrm{space}$ and $\psi_\mathrm{i}^\mathrm{space}$ must have different symmetry. This is the Laporte selection rule. It can be relaxed when the geometry is not perfectly octahedral, or tetrahedral for that matter, since inversion is no longer a symmetry operation. However, I digress.

Spin-orbit coupling

The problems arise because the total states are not separable into spin and spatial components. This occurs because both electron spin, as well as orbital angular momentum, are both angular momenta. They therefore interact with each other, and this is the spin-orbit coupling spoken of. Mathematically, it is treated as a perturbation to the Hamiltonian; the result is that the transition dipole moment cannot be separated into two integrals and this leads to the invalidation of the selection rules.

When considering spin-orbit coupling, the term symbols such as $^3\!F$, $^1\!D$, which imply a simultaneous eigenstate of the total spin operator $\hat{S}$ and the total orbital angular momentum operator $\hat{L}$ are no longer meaningful since both angular momenta are coupled to each other (by $\vec{j} = \vec{l} + \vec{s}$) and behave as one "combined" source of angular momentum. The quantities $S$ and $L$ are called "nearly good" quantum numbers: they are no longer strictly conserved, but the "contamination" of different quantum numbers is relatively small. Mathematically we could express this as an admixture of different spin states into the original wavefunction:

$$^3\!F_J = c(^3\!F) + c_1(^1\!F) + \cdots$$

where $^3\!F_J$ denotes a state with spin-orbit coupling (not an eigenstate of $\hat{S}$ and $\hat{L}$), and $(^3\!F)$ denotes the original $^3\!F$ state before spin-orbit coupling is considered (which is a true eigenstate of $\hat{S}$ and $\hat{L}$). If $c_i \neq 0$, then one obtains a non-zero transition dipole moment between "states" of nominally different $S$.

The fact that the selection rules are still a thing is solely because spin-orbit coupling is a small effect ($c_i$ close to 0), so the compounds behave "almost" as if they obey the selection rules. More properly, the transition dipole moment between states of different $S$ is small, and therefore the d-d bands in $\mathrm{d^5}$ spectra are extremely weak. The magnitude of spin-orbit coupling, however, does vary with $Z^4$ ($Z$ being the atomic number). Therefore, while spin-orbit coupling is rather small in the 3d series, it becomes an extremely important mechanism by which the spin selection rule in the the 4d and 5d congeners, as well as the f-block elements, is relaxed.