The reason behind the steep rise in pH in the acid base titration curve

I've decided to tackle this question in a somewhat different manner. Instead of giving the chemical intuition behind it, I wanted to check for myself if the mathematics actually work out. As far as I understand, this isn't done often, so that's why I wanted to try it, even though it may not make the clearest answer. It turns out to be a bit complicated, and I haven't done much math in a while, so I'm kinda rusty. Hopefully, everything is correct. I would love to have someone check my results.

My approach here is to explicitly find the equation of a general titration curve and figure out from that why the pH varies quickly near the equivalence point. For simplicity, I shall consider the titration to be between a monoprotic acid and base. Explicitly, we have the following equilibria in solution

$$\ce{HA <=> H^+ + A^-} \ \ \ → \ \ \ K_\text{a} = \ce{\frac{[H^+][A^-]}{[HA]}}$$
$$\ce{BOH <=> B^+ + OH^-} \ \ \ → \ \ \ K_\text{b} = \ce{\frac{[OH^-][B^+]}{[BOH]}}$$
$$\ce{H2O <=> H^+ + OH^-} \ \ \ → \ \ \ K_\text{w} = \ce{[H^+][OH^-]}$$

Let us imagine adding two solutions, one of the acid $\ce{HA}$ with volume $V_\text{A}$ and concentration $C_\text{A}$, and another of the base $\ce{BOH}$ with volume $V_\text{B}$ and concentration $C_\text{B}$. Notice that after mixing the solutions, the number of moles of species containing $\ce{A}$ ($\ce{HA}$ or $\ce{A^-}$) is simply $n_\text{A} = C_\text{A} V_\text{A}$, while the number of moles of species containing $\ce{B}$ ($\ce{BOH}$ or $\ce{B^+}$) is $n_\text{B} = C_\text{B} V_\text{B}$. Notice that at the equivalence point, $n_\text{A} = n_\text{B}$ and therefore $C_\text{A} V_\text{A} = C_\text{B} V_\text{B}$; this will be important later. We will assume that volumes are additive (total volume $V_\text{T} = V_\text{A} + V_\text{B}$), which is close to true for relatively dilute solutions.

In search of an equation

To solve the problem of finding the final equilibrium after adding the solutions, we write out the charge balance and matter balance equations:

Charge balance: $\ce{[H^+] + [B^+] = [A^-] + [OH^-]}$

Matter balance for $\ce{A}$: $\displaystyle \ce{[HA] + [A^-]} = \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}$

Matter balance for $\ce{B}$: $\displaystyle \ce{[BOH] + [B^+]} = \frac{C_\text{B} V_\text{B}} {V_\text{A} + V_\text{B}}$

A titration curve is given by the pH on the $y$-axis and the volume of added acid/base on the $x$-axis. So what we need is to find an equation where the only variables are $\ce{[H^+]}$ and $V_\text{A}$ or $V_\text{B}$. By manipulating the dissociation constant equations and the mass balance equations, we can find the following:

$$\ce{[HA]} = \frac{\ce{[H^+][A^-]}}{K_\text{a}}$$ $$\ce{[BOH]} = \frac{\ce{[B^+]}K_\text{w}}{K_\text{b}\ce{[H^+]}}$$ $$\ce{[A^-]} = \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}} \left(\frac{K_\text{a}}{K_\text{a} + \ce{[H^+]}}\right)$$ $$\ce{[B^+]} = \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} \left(\frac{K_\text{b}\ce{[H^+]}}{K_\text{b}\ce{[H^+]} + K_\text{w}}\right)$$

Replacing those identities in the charge balance equation, after a decent bit of algebra, yields:

$$\ce{[H^+]^4} + \left(K_\text{a} + \frac{K_\text{w}}{K_\text{b}} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]^3} + \left(\frac{K_\text{a}}{K_\text{b}}K_\text{w} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} K_\text{a} - \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}K_\text{a} - K_\text{w}\right) \ce{[H^+]^2} - \left(K_\text{a} K_\text{w} + \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}\frac{K_\text{a}}{K_\text{b}} K_\text{w} + \frac{K^2_\text{w}}{K_\text{b}}\right) \ce{[H^+]} - \frac{K_\text{a}}{K_\text{b}} K^2_\text{w} = 0$$

Now, this equation sure looks intimidating, but it is very interesting. For one, this single equation will exactly solve any equilibrium problem involving the mixture of any monoprotic acid and any monoprotic base, in any concentration (as long as they're not much higher than about $1~\mathrm{\small M}$) and any volume. Though it doesn't seem to be possible to separate the variables $\ce{[H^+]}$ and $V_\text{A}$ or $V_\text{B}$, the graph of this equation represents any titration curve (as long as it obeys the previous considerations). Though in its full form it is quite daunting, we can obtain some simpler versions. For example, consider that the mixture is of a weak acid and a strong base. This means that $K_\text{b} \gg 1$, and so every term containing $K_\text{b}$ in the denominator is approximately zero and gets cancelled out. The equation then becomes:

Weak acid and strong base:

$$\ce{[H^+]^3} + \left(K_\text{a} + \frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]^2} + \left(\frac{C_\text{B} V_\text{B}}{V_\ce{A} + V_\ce{B}} K_\ce{a} - \frac{C_\ce{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}}K_\ce{a} - K_\ce{w}\right) \ce{[H^+]} - K_\ce{a} K_\ce{w} = 0$$

For a strong acid and weak base ($K_\text{a} \gg 1$), you can divide both sides of the equation by $K_\text{a}$, and now all terms with $K_\text{a}$ in the denominator get cancelled out, leaving:

Strong acid and weak base:

$$\ce{[H^+]^3} + \left(\frac{K_\ce{w}}{K_\ce{b}}+\frac{C_\ce{B}V_\ce{B}}{V_\ce{A} + V_\ce{B}} - \frac{C_\ce{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}}\right) \ce{[H^+]^2} - \left(K_\ce{w} + \frac{C_\text{A} V_\ce{A}}{V_\ce{A} + V_\ce{B}} \frac{K_\ce{w}}{K_\ce{b}}\right) \ce{[H^+]} - \frac{K^2_\ce{w}}{K_\ce{b}} = 0$$

The simplest case happens when adding a strong acid to a strong base ($K_\ce{a} \gg 1$ and $K_\ce{b} \gg 1$), in which case all terms containing either in the denominator get cancelled out. The result is simply:

Strong acid and strong base:

$$\ce{[H^+]^2} + \left(\frac{C_\text{B} V_\text{B}}{V_\text{A} + V_\text{B}} - \frac{C_\text{A} V_\text{A}}{V_\text{A} + V_\text{B}}\right) \ce{[H^+]} - K_\ce{w} = 0$$

It would be enlightening to draw some example graphs for each equation, but Wolfram Alpha only seems to be able to handle the last one, as the others require more than the standard computation time to display. Still, considering the titration of $1~\text{L}$ of a $1~\ce{\small M}$ solution of a strong acid with a $1~\ce{\small M}$ solution of a strong base, you get this graph. The $x$-axis is the volume of base added, in litres, while the $y$-axis is the pH. Notice that the graph is exactly as what you'll find in a textbook!

Now what?

With the equations figured out, let's study how they work. We want to know why the pH changes quickly near the equivalence point, so a good idea is to analyze the derivative of the equation and figure out where they have a very positive or very negative value, indicating a region where $\ce{[H^+]}$ changes quickly with a slight addition of an acid/base.

Suppose we want to study the titration of an acid with a base. What we need then is the derivative $\displaystyle \frac{\ce{d[H^+]}}{\ce{d}V_\ce{B}}$. We will obtain this by implicit differentiation of both sides of the equations by $\displaystyle \frac{\ce{d}}{\ce{d}V_\ce{B}}$. Starting with the easiest case, the mixture of a strong acid and strong base, we obtain:

$$\frac{\ce{d[H^+]}}{\ce{d} V_\ce{B}}=  \frac{K_\ce{w} - C_\ce{B} \ce{[H^+] - [H^+]^2}}{2(V_\ce{A} + V_\ce{B}\left) \ce{[H^+]} + (C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}\right)}$$

Once again a complicated looking fraction, but with very interesting properties. The numerator is not too important, it's the denominator where the magic happens. Notice that we have a sum of two terms ($2(V_\ce{A} + V_\ce{B})\ce{[H^+]}$ and $(C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A})$). The lower this sum is, the higher $\displaystyle \frac{\mathrm{d}\ce{[H^+]}}{\mathrm{d} V_\ce{B}}$ is and the quicker the pH will change with a small addition of the base. Notice also that, if the solutions aren't very dilute, then the second term quickly dominates the denominator because while adding base, the value of $[H^+]$ will become quite small compared to $C_\ce{A}$ and $C_\ce{B}$. Now we have a very interesting situation; a fraction where the major component of the denominator has a subtraction. Here's an example of how this sort of function behaves. When the subtraction ends up giving a result close to zero, the function explodes. This means that the speed at which $\ce{[H^+]}$ changes becomes very sensitive to small variations of $V_\ce{B}$ near the critical region. And where does this critical region happen? Well, close to the region where $C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}$ is zero. If you remember the start of the answer, this is the equivalence point!. So there, this proves mathematically that the speed at which the pH changes is maximum at the equivalence point.

This was only the simplest case though. Let's try something a little harder. Taking the titration equation for a weak acid with strong base, and implicitly differentiating both sides by $\displaystyle \frac{\ce{d}}{\ce{d} V_\ce{B}}$ again, we get the significantly more fearsome:

$$\displaystyle \frac{\ce{d[H^+]}}{\ce{d}V_\ce{B}} = \frac{ -\frac{V_\ce{A}}{(V_\ce{A} + V_\ce{B})^2} \ce{[H^+]} (C_\ce{B}\ce{[H^+]} - C_\ce{B} K_\ce{a} + C_\ce{A} K_\ce{a})}{3\ce{[H^+]^2 + 2[H^+]}\left(K_\ce{a} + \frac{C_\ce{B} V_\ce{B}}{V_\ce{A} + V_\ce{B}}\right) + \frac{K_\ce{a}}{V_\ce{A} + V_\ce{B}} (C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}) -K_\ce{w}}$$

Once again, the term that dominates the behaviour of the complicated denominator is the part containing $C_\ce{B} V_\ce{B} - C_\ce{A} V_\ce{A}$, and once again the derivative explodes at the equivalence point.


Why are acid catalysed reactions of carbonyl compounds reversible?

This is a picture from wiki (Simple aldol reaction). Both Acid and base catalyzed reactions are reversible up to aldol intermediate. Dehydration is the only irreversible step. So this statement " I've noticed that all acid catalysed reactions are represented with reversible arrows, while base catalysed reactions are shown with regular arrows. " is incorrect.

Single arrow means that the step is irreversible. Reversible arrow means that reaction can go both forward and backward. Usually reversible steps are happening before the irreversible step and a happening faster than irreversible step.

This is easy to demonstrate with a pool table that nave a small obstacle near the hole. Reversible reactions are balls that move over the the surface of the table. They can get from any point to any other point of the table. Sometimes they get on that small obstacle near the hole and roll back to the table (they were not able to overcome activation barrier). Irreversible reactions are represented by balls that were able to overcome the barrier and now ended up in the hole. Now they have no chance to go back to the table on their own.

What makes reaction irreversible vs reversible. From thermodynamics perspective irreversible reactions are reactions with a large change in $\Delta$G. From organic chemists perspective reversible reactions are typically those that include protonation/deprotonation steps. Irreversible reactions are those where a strong bond like C-C or O-P forms.

Reversibility is a subject to conditions. 2H -> H2 (recombination of radicals) is effectively irreversible at room temperature, but is reversible at 3000K.

Why do we need to account for reversible reactions? (1) To understand the mechanism. (2) To know how to increase concentration of the reagents right before the first irreversible step to accelerate the overall reaction.

tl;dr: in reactions you are asking about formation of C-C bond is effectively irreversible step.


Does standard electrode potential depend upon temperature?

Yes, the standard reduction potential does depend on temperature. The definition of the standard reduction potential is stated in Ref.1 as:

A standard electrode potential $E^\circ$ is defines as the potential (in Volts, $\pu{V}$) of a half-reaction relative to a reference electrode at a specific temperature, all chemicals being at their standard states at unit activity. These states may be arbitrarily defined as pure crystalline solids, pure liquids, ideal gases at one atmosphere fugacity ($\pu{101325 \times 10^{5} Pa}$), and ideal solutes at unit molality. The most common temperature for the tabulation of standard electrode potentials, as for their thermodynamic data, is $\pu{25 ^\circ C}$ ($\pu{298.15 K}$).

Accordingly, tabulated data is chosen to be at $\pu{298.15 K}$ for the convenience. This is also beneficial that the most common reference electrode is the Standard Hydrogen Electrode (SHE) for the solvent water:
$$\ce{2H+_{(aq)} + 2e- <=> H2_{(g)}} \tag1$$
$E^\circ_{(\ce{H+/H2})}$ for the above half-reaction is been assigned as zero volts at all temperatures. Extended work over $\pu{298.15 K}$ by De Béthune and coworkers shown that the temperature dependence of $E^\circ$ is approximately linear between $\pu{273.15 K}$ and $\pu{373.15 K}$ according to the following equation (Ref.2-Ref.4):
$$E_T = E^\circ_{298} + (T - 298.15)\left(\frac{dE^\circ}{dT}\right)_{298} \tag2$$
References:

Steven G. Bratsch, “Standard Electrode Potentials and Temperature Coefficients in Water at $\pu{298.15 K}$,” Journal of Physical and Chemical Reference Data 1989, 18(1), 1-21 (https://doi.org/10.1063/1.555839).
A. J. De Béthune, T. S. Licht, N. Swendeman, “The Temperature Coefficients of Electrode Potentials: The Isothermal and Thermal Coefficients—The Standard Ionic Entropy of Electrochemical Transport of the Hydrogen Ion,” Journal of The Electrochemical Society 1959, 106(7), 616 (doi:10.1149/1.2427448).
G. R. Salvi, A. J. De Béthune, “The Temperature Coefficients of Electrode Potentials: II. The Second Isothermal Temperature Coefficient,” Journal of The Electrochemical Society 1961, 108(7), 672 (doi: 10.1149/1.2428187).
André Jacques De Béthune, Nancy A. Swendeman Loud, In Standard Aqueous Electrode Potentials and Temperature Coefficients at $\pu{25 ^\circ C}$; C. A. Hampel: Skokie, ILL, 1964.


Does the position of the functional group on the terminal carbon change the nomenclature?

They are not isomers, they are the same molecule, ethanol (the "1" is superfluous, there is no 2-ethanol), CH3CH2OH


What is the difference between "atomic hydrogen" and "nascent hydrogen"?

Is your book by chance very old? From the Wikipedia entry for "nascent hydrogen":

  Nascent hydrogen is purported to consist of a chemically reactive form
  of hydrogen that is freshly generated, hence nascent. Molecular
  hydrogen ($\ce{H2}$), which is the normal form of this element, is unreactive
  toward organic compounds, so a special state of hydrogen was once
  invoked to explain certain kinds of hydrogenations. Mechanistic
  understanding of such reactions is now available, and the concept of
  nascent hydrogen is discounted, even ridiculed.

Then, by example of how this concept came to be:

  Reductions of esters to give alcohols using a mixture of sodium and
  alcohols is called the Bouveault–Blanc reduction. It is an old
  reaction that has largely been superseded by alternative methods. At
  the time of popularity, the process caused much puzzlement because
  esters are unreactive toward hydrogen. It was also known that sodium
  reacts with alcohols to release $\ce{H2}$. it was concluded that some freshly
  generated ("nascent") hydrogen was responsible for this remarkable
  reaction. Subsequent studies have shown that this reaction proceeds
  via electron-transfer from metallic sodium to the ester substrate
  followed by protonation of the reduced intermediate. The evolution
  of hydrogen by the reaction of sodium and alcohol is purely a
  competitive reaction, the sole benefit being that in the presence of
  sufficient alkoxide, the sodium/alcohol reaction slows.

So in general, at least toward organic compounds, your relative reactivities of hydrogen species are correct if you take out "nascent" hydrogen so that:

$$\ce{H2} \lt \rm atomic~hydrogen$$


I think the mistake you're making is this conclusion: "Higher atomic charge. . .  -> higher electron density -> more pi-backbonding". With higher (ie more positive or less negative) atomic charge, there is less electron density donation towards the ligand, because the electron density is pulled to the metal center more strongly. Thus, $\pi$ back-bonding is weaker and C-O bond is stronger.

The more positively charged metal centers do not pull enough electron density from CO in the sigma interaction to be able to donate more back through pi interaction, which seems to be the way you're looking at it.


Failed esterification of 1-butanol and ethanoic acid

I don't think the brown color necessarily means that the reaction failed. It could be due to some contaminant in your reagents but in a small quantity that wouldn't cause the reaction to fail. The fruity smell could very well be due to your product see: n-butyl acetate has a sweet banana smell. I would take the distillate you collected (or skip the distillation altogether) and try an aqueous work-up: Add very carefully aqueous $\ce{Na2CO3}$ until basic $\ce{pH}$ and then extract this mixture with ether. This way any ester product will be in the the organic layer and sulfuric/acetic acid will stay in the aqueous together with most of the alcohol.

To finish the work-up dry the ether layers over $\ce{Na2SO4}$, filter and remove the ether (unless you have access to a Buchi evaporator, the safest way is to leave it in an open flask/beaker overnight). You can further purify the final product by distillation.

You might improve the yield if you cool the reaction mixture in an ice bath before adding sulfuric acid


Sound absorption in carbon dioxide with temperature

The acoustic wave is accompanied by a temperature oscillation. To a first approximation, the temperature oscillation is adiabatic and reversible in the bulk, but irreversible near solid boundaries. In many cases, boundary losses dominate (for example, in musical wind instruments), but as the frequency increases, and the surface-to-volume ratio decreases, bulk irreversibility may become important. Bulk losses in single-component polyatomic gasses are almost always due to loss of local thermodynamic equilibrium when the temperature oscillation describing internal degrees of freedom lags the temperature oscillation describing translational degrees of freedom.
A full discussion of these effects is given in the book "Ultrasonic Absorption" by A.B. Bhatia. His figure 5.1 shows that the attenuation for $\ce{CO2}$ peaks at about 30 kHz (STP conditions), and his figure 6.3 shows the velocity dispersion that occurs at about the same frequency. The dispersion is a large effect, with the velocity increasing by about 4% as the frequency goes from below to above the attenuation peak. The attenuation and dispersion occur when the acoustic frequency matches the rate at which the vibrational mode temperature relaxes to the translational temperature. The rotational modes relax much faster than the vibrational modes, so they do not play an important role under these conditions.
The relaxation time for the vibrational temperature will generally decrease with increasing temperature, so the attenuation peak will move to higher frequency. I am not sure what temperatures you are exploring, but if your frequency is above the attenuation peak, then you will see increasing attenuation as the temperature increases.


How closely should the energy components of ab initio SCF results at a stationary nuclear configuration preserve the virial theorem?

I hope I am not mistaken about this, but if I'm wrong then I'll learn so no big deal.

I believe this is a result of the fact that the electrons do not experience a true $r^{-1}$ potential due to electron-electron repulsion. Namely, we should expect that the electrons will repel each other which will cause the apparent attraction to the nuclei to be weaker than $r^{-1}$.

In other words, for a general potential,
$$
V(r)=r^{n}
$$
we have,
$$
\frac{dV(r)}{dr}=nr^{n-1}
$$

Because the Virial theorem can be stated as,
$$
\langle T\rangle=\frac12\langle\vec{r}\cdot\nabla V\rangle
$$
But for a potential which is less than coulombic, we have,
$$
\langle T\rangle=-\frac{n}{2}\langle V\rangle=-nE_{tot}
$$

That is to say, the electrons in water are feeling an effective potential of the form $V(r)=r^{-.99787}$ or so.

Or, in the language of the book you linked, the trial wavefunction of the form being tried for the hydrogen atom does not evaluate to a parameter $\zeta=1$.

Again, I am not completely confident that I nailed this answer, but for systems that have no obvious analytic form of the potential energy, there is no reason I see that we have should have $\langle T\rangle=-E_{tot}$ be exactly true. That is not to say the virial theorem is not satisfied, it certainly is, only that it is not as clean as we are used to.

I was a bit loose with some constants in my equations, please forgive me :)


Finding molarities of individual reagents given the chemical equation and total concentration in a container

The assumption you need to make is that at time zero, only $\ce{PH3}$ is present.  So relative to the moles at time zero, or $N_0$, you should find $\Delta N = N(t) - N_0$, the total new moles formed by the reaction since it began.

The reaction converts 4 moles of gas to 7.

So pressure increases mean the reaction is moving to the right.  Every mole of pressure increase must be $\frac{6}{7}$ due to $\ce{H_2}$ and $\frac{1}{7}$ due to $\ce{P4}$ formation.

So just find the total new moles of gas and multiply by $\frac{1}{7}$.


The given answer is wrong in many ways:

4 did not satisfy both criteria C and E, because it is not a chiral compound to begin with.
10 did not satisfy the criteria D, because it does give a precipitate with iodine in presence of $\ce{NaOH}$.

That gives only 5 in given answer to satisfy all criteria. Let's eliminate given compounds in systematic order:

That leaves only 5 and 6 to satisfy all criteria.


Calculate the number of moles of phosphorus in 15.95g of tetraphosphorus decaoxide

You can go about it more directly as follows.  You have $(15.95/283.889)=\pu{0.0562 moles}$ of $\ce{P4O10}$.  From the molecular formula you see that for every mole of $\ce{P4O10}$ you have 4 moles of phosphorous.  Therefor you have $4 \times 0.0562 = \pu{0.2247 moles}$ of phosphorous.

Your error involved converting moles of $\ce{P4}$ to moles of $\ce{P}$.  If you multiply your answer by 4 you get the correct answer.


Does mixing two different foaming soap make it ineffective?

The soap solutions were originally transparent. An interesting experiment, if you still have some of the original soaps, is to shine a light thru the solutions and look for a Tyndall effect:

The light source is on the far right, out of the picture. The beaker on the right shows light scattering from particles comparable in size to the wavelength of light due to the Tyndall effect; the beaker on the left shows no scattering, even tho the light beam is passing thru it. Its contents are unknown, but there seem to be no particles of sufficiently different index of refraction and size to scatter light.
The tiny particles which scatter the light (just like in a smoky room) are micelles, aggregates of several hundred or thousand soap molecules. In the original soap solutions, this scattering is small, due to selection of manufacturing conditions (exact formulation, concentration, pH, other ingredients, etc., etc.), so the solutions appear transparent. What is optimal for each soap separately may not be optimal for the combination: perhaps the mixing produced larger micelles which scattered more, so the mixture scattered more light and became cloudy to the eye. I'm going to suggest that both soap solutions show a Tyndall effect, and the mixed solution shows a much larger Tyndall effect. It may change with time.
Now the question becomes: is the combination as effective as either solution alone. Well, it is different (cloudy). It could be worse; it could be better! It might not be observably different.
The foaming test could cast some light on the effectiveness: If you can test the foaming (height of foam and time stability) of the original solutions and of the mixture, and if they are all the same (+/- ~10-20%), the combination is probably OK.
There is a case where this foaming test will indicate immediately that the combination is not good. The molecules in soaps and detergents have a hydrophilic end and a hydrophobic end. Soaps are salts of fatty acids, and have a negative (hydrophilic) carboxyl group at one end and a (hydrophobic) hydrocarbon at the other end, and a soluble alkali metal cation. There are nonionic detergents which have no charge; they are compatible with anionic or cationic soaps and detergents. Cationic detergents (usually amine salts) have a hydrophilic positively charged nitrogen at one end and a hydrophobic hydrocarbon at the other, and some anion, usually chloride. If you should mix solutions of a cationic detergent with an anionic soap, the positive and negative charges associate to produce a neutralized ion-pair with two hydrophobic hydrocarbon groups - together this looks like a big fatty molecule with a little polarity in the middle. Overall, this molecule is not very soluble in water and precipitates out (but it floats like scum). And since it is not soluble in water, it won't foam.


Cauchy-Schwarz integral screening different inequalites

As you have already stated in the question, (Ineq. 1) is the standard inequality used in Cauchy-Schwarz integral screening. (Ineq. 2) and (Ineq. 3) are not valid inequalities, which is clear if you select atomic orbitals such that a & b are spatially disjoint from c & d and the proposed upper bounds vanish. (Ineq. 4), (Ineq. 5), and (Ineq. 6) follow from a different use of the Cauchy-Schwarz inequality where the inner product is interpreted as being over the six spatial integrals and requires the Coulomb kernel to be pointwise non-negative so that you can take its element-wise square root,

$$ \begin{align} |g_{abcd}| &= \left| \int d\vec{r} d\vec{r}' \phi_a(\vec{r}) \phi_b(\vec{r}) V(\vec{r},\vec{r}') \phi_c(\vec{r}') \phi_d(\vec{r}') \right| \\
&= \left| \sqrt{V}\phi_a \phi_c \cdot \sqrt{V}\phi_b \phi_d \right| \\ & \le \| \sqrt{V}\phi_a \phi_c \| \| \sqrt{V}\phi_b \phi_d \| = \sqrt{g_{aacc}} \sqrt{g_{bbdd}} \end{align}. $$  (Ineq. 4) is not useful for integral screening because it is strictly weaker than (Ineq. 1) since $g_{abab} \le g_{aabb}$ (for real orbitals). Whereas $g_{abab}$ vanishes as orbitals a & b become spatially disjoint, $g_{aabb}$ only decays algebraically in the average distance between a & b. (Ineq. 5) and (Ineq. 6) are not strictly weaker than (Ineq. 1), but they have the same problem of being weakly decaying bounds that don't screen very much. For example, they are tight and thus tighter than (Ineq. 1) when a=b and c=d, but this is a class of integrals that will only be screened out for extremely large spatial separations between a & c.


Pure substances have only one kind of atom

Most of the issues can be resolved with precise definitions.

Let's start with purity. Purity is relative. In many cases (eg table salt) 99% is pure enough. What this means is that for all practical purposes we don't care about the remaining impurities. Chemists tend to describe something as "pure" when there is less of any contaminant than matters for what they are doing with the chemical. Sometimes purity of very high levels matters. Semiconductors behaviour depends on very precise doping with very small amounts of other substances. The wrong contaminant even at levels of 1 atom per million is a problem. So the industry have worked out good ways of making extremely pure silicon, for example. These techniques have been used to make some of the purest substances ever (the silicon sphere now being used to redefine the Kg is pure to parts per billion atoms.) Strictly speaking it isn't pure in the sense of having absolutely 100% of all its atoms as silicon, but that is as pure as it needs to be for its purpose. So, strictly, nothing is ever absolutely pure, but it doesn't matter.

The word "substance" is a bit vague. In chemistry it could be a compound an not just an element. Purity for a compound means that all the molecules in the substance are the same (to the level of purity that matters). Having more than one type of atom doesn't make it impure as long as all the molecules are the same.

Then there are some homogeneous mixtures that are hard to separate by any simple means, some metal alloys for example. We wouldn't normally describe these as "impure" but chemists might use special terminology to describe them (alloy rather than compound for example).

And as for "Pure substances contain only one kind of matter" it depends what you mean by "matter". Chemists would prefer the term "molecule" to "matter". Matter can be interpreted different ways (does it mean atoms or molecules: it's unclear). "Pure substances contain only one kind of "molecule" would be an unambiguous statement chemists would agree with.


Old patents of the 1930s, particularly from I.G. Farben often describe the use of hypochlorite/chlorine in the bleaching of bast.

However, from newer literature, such as this application manual for hydrogen peroxide bleaching from Evonik Industries, I would conclude that bleaching of bast can be performed without chlorine, supposed that the peroxide bleaching is performed

in alkaline medium
at elevated temperatures $(85-90\,°\mathrm{C})$


Relative acidities of alkanes, alkenes, and alkynes

You're right in that bond length, and therefore bond strength does affect acidity (see: $\ce{H2S}$, $\mathrm{p}K_\mathrm{a} = 7$ and $\ce{H2O}$, $\mathrm{p}K_\mathrm{a} = 15.7$). If we defined acidity with the following equation

$$\ce{HX -> H + X}$$

then the bond strength would indeed be the only deciding factor in the acidity of $\ce{HX}$, since the enthalpy change of that process is literally the bond dissociation energy.

However, acidity actually corresponds to the heterolytic dissociation of the $\ce{H-X}$ bond, with both electrons in the bond going to $\ce{X}$:

$$\ce{HX -> H+ + X-}$$

so bond strength isn't the only factor. In the case of ethane, ethene, and ethyne, the most important factor is the type of orbital that the electrons from the $\ce{C-H}$ bond end up in after deprotonation. For ethane, the two electrons end up in a carbon $\mathrm{sp^3}$ orbital; for ethene, an $\mathrm{sp^2}$ orbital; and for ethyne, an $\mathrm{sp}$ orbital.

Now, of these three, the $\mathrm{sp}$ lone pair is the most stable, since the $\mathrm{sp}$ hybrid orbital contains the most s character ($50\%$) and therefore has the lowest energy. So, of the three carbanions, the lone pair in the acetylide ion $\ce{HC2-}$ is most stabilised, making ethyne the most acidic molecule.

The acidity of terminal alkynes is in fact pretty useful from a synthetic viewpoint, since you can deprotonate them with a strong base like $\ce{NaNH2}$, forming a carbon-based nucleophile which can then be used to make new $\ce{C-C}$ bonds.

As for why the electronic stabilisation outweighs the bond strength, it's worth looking at some quantitative data. The $\ce{H-O}$ and $\ce{H-S}$ bonds have mean bond enthalpies of $463$ and $338~\mathrm{kJ~mol^{-1}}$ respectively (source: Physical Chemistry 9th ed., Atkins & de Paula, p 932), which is a difference of $\mathbf{125~kJ~mol^{-1}}$. This can, to some extent, explain the difference in acidities of $\ce{H2O}$ and $\ce{H2S}$.

On the other hand, the dissociation energies of the $\ce{H-C\mathrm{(sp)}}$ and $\ce{H-C\mathrm{(sp^2)}}$ bonds are given in J. Phys. Chem. 1987, 91, 17-19 as $132.6$ and $116~\mathrm{kcal~mol^{-1}}$ respectively. The difference is roughly $17~\mathrm{kcal~mol^{-1}}$, which is equivalent to $\mathbf{69.5~kJ~mol^{-1}}$ - a somewhat smaller value than before. You can see why bond length variations might therefore play a smaller role in the hydrocarbons.

An alternative

Another equivalent way of looking at it is that an $\mathrm{sp}$-hybridised carbon is more electronegative than an $\mathrm{sp^2}$-hybridised carbon, which is in turn more electronegative than an $\mathrm{sp^3}$-hybridised carbon.

As mentioned earlier, an $\mathrm{sp}$ orbital has the greatest s-character, and consequently electrons in a carbon $\mathrm{sp}$ orbital experience a greater effective nuclear charge and are more tightly bound to the nucleus than electrons in $\mathrm{sp^2}$ or $\mathrm{sp^3}$ orbitals.

Of course, the $\ce{C-H}$ bond also involves a contribution from the hydrogen 1s orbital. However, the above means that the electrons in $\ce{C(sp)-H}$ bonds are more strongly attracted to the carbon nucleus than electrons in $\ce{C(sp^2)-H}$ or $\ce{C(sp^3)-H}$ bonds. This means that the $\ce{C(sp)-H}$ bond is more polarised towards carbon, leaving less electron density on hydrogen, and a greater acidity.

The amount of electron density on the hydrogen can in fact be probed using $\ce{^1H}$ NMR spectroscopy. A lower electron density translates into a higher chemical shift. Even though there is a second, unrelated, factor - anisotropic shielding - that serves to decrease the chemical shift of acetylenic hydrogens, they still tend to show up at higher chemical shifts than hydrogens on saturated hydrocarbons (~2 ppm, compared to ~1 ppm). This is a clear indication that the local electron density on a proton attached to $\ce{C(sp)}$ is less than that of a proton attached to $\ce{(sp^3)}$.


When titrating a weak acid should only undissociated acid molecules be considered?

Acetic acid and acetate are connected by the dissociation equilibrium $(1)$:

$$\ce{CH3COOH + H2O <<=> H3O+ + CH3COO-}\tag{1}$$

So you basically have two acidic particles that can react with hydroxide: hydronium or acetic acid. Hydronium will actually react much faster on a molecular level even though its concentration is significantly lower as it is much more mobile in solution. However, any hydronium consumed is immediately reformed by acetic acid molecules reestablishing the equilibrium.

Therefore, on a very minute level hydroxide reacts with neither acetic acid nor acetate but with hydronium which is produced in the dissociation of acetic acid. In practice, that still means that hydroxide will be added until $n(\ce{CH3COOH})+n(\ce{CH3COO-})=n(\ce{OH-})$. From this point onwards, there will be more hydroxide than acidic particles causing the $\mathrm{pH}$ value to rise significantly, the indicator to change colour and the student to realise that they reached (and passed) the equivalence point.


Buffer formation in weak acid/strong base titration?

I think that the above diagram will explain a lot of your questions.

  it seems like the weak acid isn't initially a buffer

That is correct as to form a buffer you need a solution containing a weak acid and its conjugate weak base. So in the example above, initially the solution only contains acetic acid ($\ce{CH3COOH}$, but its conjugate base, acetate $\ce{CH3COO^{-}}$, is not present.

  But, as you begin to add strong base, conjugate base is created from the reaction between the weak acid and strong base

That is absolutely correct. For example, in the above diagram the reaction that is occurring when you titrate the acetic acid with sodium hydroxide is:
$$\ce{CH3COOH + NaOH -> CH3COO- + Na+ + H2O}$$

Now as you begin to titrate the acid, its conjugate base starts to form. This is supported by the fact that the pH doesn't really change when more $\ce{NaOH}$ is added to the solution.

To form an effective buffer solution you need roughly equal concentrations of the acid and conjugate base, hence that is why it says it is a buffer when half of the acid is titrated. However anywhere near that region can be considered as a buffer. Generally a solution is a buffer if the pH is within +/- 1 of its pKa. So for acetic acid, a buffer will have a pH between 3.76 and 5.76.

  Does this mean that a buffer is created from the titration?

Yes, a buffer is created. However is doesn't stay as a buffer for long. As I mentioned above, when the ration of the concentration of acid and the conjugate base becomes too high or too low, it won't be a buffer. That is evident as when more $\ce{NaOH}$ is added to the buffer, the concentration of the conjugate base becomes much greater than the concentration of the acid, resulting in the pH to increase dramatically.


First of all, the indicator does not change colour instantaneously at $\mathrm{pH = pK_a}$. The indicator is a different colour than it's conjugate base, and exists in an equilibrium with it in solution. For the colour change to be noticed, the concentration of the base has to be at least 10 times more than that of the conjugate acid, or vice versa. This is why there is a range of colour change(look at these pH charts), with the upper and lower bounds having a $\mathrm{pH}$ difference of $2$. As an example, here's a simplified reaction for Phenolphthalein ($\ce{Ph}$ is the phenolphthalein anion)
$$\ce{\underset{\text{colourless}}{H2Ph} <=> 2H+ + \underset{\text{violet}}{Ph^2-}}$$
If you take a simpler, weak monobasic acid indicator $\ce{HIn <=> H+ + In-}$, the $\mathrm{K_a}$ is
$$\mathrm{K_a} = \frac{\ce{[H+][In^-]}}{\ce{[HIn]}}$$
The relative concentration of the indicator and it's conjugate base is only determined by the concentration of the $\ce{H+}$ ion, as $\mathrm{K_a}$ is a constant.
Answering your second query, the author takes the condition of equilibrium, simply because the system is in equilibrium, that is $\mathrm{Q_a = K_a}$. It is assumed that the reaction is instantaneous or that titration is performed slowly enough so that the system is in equilibrium at every single instant.


What transition metal ions form without oxidant in water pH 7?

If you are counting lanthanides as "transition metals", then look at europium whose reactivity with water is similar to calcium.  Although the product listed in the reference is the trivalent hydroxide $\ce{Eu(OH)3}$, europium has a relativy stable +2 oxidation state and so the reaction msy proceed via a relatively soluble divalent hydroxide or cation ($\ce{Eu(OH)2, Eu^{2+}}$).  The combination of a highly electropositive element, possibly with formation of a relatively soluble, low oxidation state hydroxide/cation makes europium more reactive to water than most transition metals.


Thermal stability of interhalogen compounds

More electronegativity difference generally gives you stronger bonds,  therefore higher thermal stability (i.e. you need to put in more energy to break them). Also, there are 2 lone pairs on the central atom and 3 on each F. Thus, interhalogen compounds of this type with larger central atom would experience less electron pair repulsion. So, I believe that the correct order is $\ce{IF3>BrF3>ClF3}$ as you suggested.


Eggshell is 95% calcium carbonate. It seems more likely to be used as an additive in a composite plastic.  Testing composite plastics produced from adding ground eggshell to common plastics may make for an interesting science project.

References:

Dissertation on meat and bone meal derived plastics

Patricio Toroa et. al. Eggshell, a new bio-filler for polypropylene composites


Is there any non-toxic and non-solvable-in-air liquid?

You may look at silicon Si and its derivatives. For example, the ethoxytriethylsilicane, or triethylsilicol ethyl ether is boiling at 153 °C. Or the tetraethylsilicane is also boiling at 153 °C. This last substance is used as a reference in NMR spectroscopy.


In scenarios like the one you describe, the surroundings are typically modeled as an infinite reservoir, with the capacity to supply or absorb heat with negligible change in temperature.  If you treated it as a large finite reservoir instead, and evaluated its entropy change based on $\Delta S=mc\ln ((T_0+\Delta T)/T_0)$ and its heat absorbed by $Q=mc\Delta T$ (where m is the mass of the reservoir and c is its heat capacity), you would find that, in the mathematical limit of "mc" becoming infinite (at constant Q), the entropy change would approach $\Delta S=Q/T_0$.


Can citric acid protonation produce benzene and water

Here's a drawing of citric acid and benzene.  You would have to spend a lot of energy in order to break a large number of bonds (carbon-carbon bonds, carbon-oxygen bonds and carbon-hydrogen bonds) to rearrange citric acid into benzene.  This would be unlikely to occur in a natural process and would require many steps in the laboratory.


Derivation of the Temperature Dependence of the Boltzmann Distribution

What OP cited is the derivation of the Boltzmann distribution starting from the micro-canonical ensemble, characterized by a fixed energy and a fixed number of constituent particles. In mathematical terms, the problem statement is as follows:

  Maximize the function $\ln W(\{n_i\})$ under the constraints $\sum_i n_i \epsilon_i = E$ and $\sum_i n_i = N$.

The solution is a function of $E$ and $N$, so let's define

  $\ln \Omega(E, N)$ := (constrained maximum of $\ln W$ as a function of $E$ and $N$),

  $\tilde{n}_i (E, N)$ := (value of $n_i$ at the constrained maximum).

Notice that $\ln\Omega = \ln W(\{\tilde{n}_i\})$. Also, this quantity is what appears in Boltzmann's entropy formula $S=k\ln\Omega$ because $\Omega$ is the number of available states for given values of $E$ and $N$. This observation enables us to relate $\ln\Omega$ to the thermodynamic temperature:
\begin{equation}
\frac{\partial \ln \Omega}{\partial E} = \frac{1}{k}\frac{\partial S}{\partial E} = \frac{1}{kT}.
\end{equation}
Now, OP's question can be rephrased as follows:

  When deriving the Boltzmann distribution starting from the micro-canonical ensemble, $\beta$ is introduced as a Lagrange multiplier associated with the constraint $\sum_i n_i \epsilon_i = E$. How can we relate $\beta$ to the thermodynamic temperature, i.e., establish that $\beta = \frac{\partial \ln \Omega}{\partial E}$?

The underlying principle is really the chain rule in multivariate calculus:
\begin{equation}
\begin{split}
\frac{\partial \ln \Omega}{\partial E} &= \sum_i \frac{\partial \ln W}{\partial n_i}\bigg|_{n_i = \tilde{n}_i} \frac{\partial \tilde{n}_i}{\partial E}\\
&= \sum_i (\alpha + \beta\epsilon_i)\frac{\partial \tilde{n}_i}{\partial E}\\
&=  \alpha\frac{\partial}{\partial E}\Big(\sum_i \tilde{n}_{i}\Big) + \beta\frac{\partial}{\partial E}\Big(\sum_i \tilde{n}_{i}\epsilon_i\Big)\\
&= \alpha \frac{\partial N}{\partial E} + \beta \frac{\partial E}{\partial E}\\
&= \beta.
\end{split}
\end{equation}


How can molecular mass be calculated from vapour density?

According to Wikipedia, the vapour density of a molecule is "the density of a vapour in relation to that of hydrogen". The density of a gas, $\rho$, is proportional to its molecular mass, $M$:

$$\rho = \frac{m}{V} \propto \frac{m}{n} = M$$

where $m$ is the mass of the gas, $V$ is the volume, and $n$ the amount of gas. Therefore, if we denote your compound by $\ce{X}$:

$$\begin{align}
67.5 &= \frac{\rho(\ce{X})}{\rho(\ce{H2})} \\
&= \frac{M(\ce{X})}{M(\ce{H2)}} \\
M(\ce{X}) &= 67.5 \cdot M(\ce{H2}) \\
&= 67.5 \cdot \pu{2 g mol-1} \\
&= \pu{135 g mol-1}
\end{align}$$

Finally, you can use this piece of information to find that the molecular formula is $\ce{C8H9NO}$ as well.


You are starting with spin-paired starting materials (an even number of electrons, with half in the up state and half in the down state), and they react (exothermically) to create an oxygen molecule. Initially, spin is conserved so that the oxygen molecule also has to be spin paired (singlet), despite the fact that the triplet state of oxygen is thermodynamically more stable. It will then react with a substrate or eventually decay over minutes to hours (most likely thermally) to the triplet state.


Can you please elaborate on that part? I know that the E value of
  Mno4- is +1.51, and the one for BrO3- is +1.48. What does that
  suggest?? that Mno4- will more likely to be reduced by Fe+ than
  Bro3-?? please explain

The first and foremost hint, even before you calculate the electrode potentials is the solubility rule taught in general chemistry. Hydroxides of most metals are insoluble. Write the equations : Fe(II) + hydroxide ions = iron (II) hydroxide. This is why alkaline permanganate is not a good choice. However, this is not the final answer because you would like to further confirm whether C or D is the correct choice.

Next determine the E_cell= E(cathode)-E(anode) > 0 V or not.

Now you need to oxidize Fe(II) to Fe(III), so the iron half cell will be the anode. As a mnemonic, anode and oxidation both start with vowels; cathode and reduction don't. You would like to reduce the permanganate in an alkaline medium, this half cell is the cathode.

Repeat the same for bromate half cell in acidic medium and Fe(II). Both (C) and (D) would turn out to be positive E_cells. However, the (C) is not an answer because you don't want iron to precipitate out as iron(II) hydroxide.

In real life analysis, Fe(II) can be routinely titrated with potassium permanganate but in highly acidic medium. For practice, check the E_cell after writing a balanced reaction.


How can I tell whether or not a molecule is planar?

I think these general rules work:

If there is an $\ce{sp^3}$ hybridized carbon (or nitrogen), the molecular is NOT planar.

2) If there are no $\ce{sp^3}$ hybridized carbons (or nitrogens), and there is only one $\ce{sp^2}$ hybridized atom (carbon or nitrogen), it will be planar.

3) If there are no $\ce{sp^3}$ hybridized atoms, and there are 2 $\ce{sp^2}$ hybridized atoms (carbon or nitrogen) that are separated by an even number of double bonds and no single bonds, then the molecule will not be planar.

So a general simple rule is that:

  the molecule will not be planar if there is an $\ce{sp^3}$ hybridized carbon (or nitrogen) atom or two $\ce{sp^2}$ hybridized atoms of carbon/nitrogen which are separated by an even number of double bonds and no single bonds. Otherwise, its structure allows it to be planar.

Even though the molecule will have a structure that allows for it to exist in a planar conformation, there may be some/many that do not persist in a planar conformation due to steric effects, or complex three dimensional geometries.

In the problems you listed above, using this rule:

Not planar because there are no $\ce{sp^3}$ and the two $\ce{sp^2}$s are separated by an even number of double bonds.
Planar because there are two $\ce{sp^2}$s but they are separated by an odd number of double bonds (3) (and no single bonds)
Planar because there are no $\ce{sp^3}$s and only 1 $\ce{sp^2}$s that make 3 or more bonds (C or N).  The orbital geometry is NOT planar because the $\ce{sp^2}$ oxygen is separated from the $\ce{sp^2}$ carbon by an even number of double bonds.
Planar because 2 $\ce{sp^2}$s are separated by an odd number (1) of double bonds (and no single bonds)


No
Brønsted theory
In Brønsted theory $\ce{H+}$ isn't an acid at all. Acids lose protons, becoming conjugate bases, and $\ce{H+}$ is the proton itself.
Arrhenius theory
$\ce{H+}$ isn't an acid, because in this theory acids dissociate in water to form hydrogen ions.
Lewis theory
$\ce{H+}$ is an incredibly strong acid, but nuclei of other, heavier elements, for example alpha particles, are arguably stronger. I haven't found hard data for this and it may be rather difficult to get, these aren't your friendly neighbourhood Lewis acids ;)
Protonating agent
Bare $\ce{H+}$ might be the ultimate protonating agent. In proton transfer, with any Brønsted acid you could always try to find an acceptor weak enough that reaction constant would be lower then 1, proton would "prefer" to stay with acid then protonate base. That's not the case with bare proton, which is unbound. Therefore it may beat any Brønsted acid.
Why only "might"? Because whether bare proton can bind to a species depends on it's energy, which needs to be lower then proton affinity of a molecule to which it's supposed to bind. Otherwise proton may ionise the molecule instead, and even fuse with one of its nuclei, if energy is high enough.
Another thing is that acids like $\ce{H4O^2+}$, which are endothermic molecules, could beat a bare proton, because of their repulsive nature - they spontaneously lose protons and "throw them away" with positive charge of their conjugate base!


A student is provided with 500 mL of 600 ppm solutions of fructose. What volume of this solution in millilitres contains 0.15 fructose?

Assuming it's $0.15gm$ of fructose,$$ppm=\frac{Weight\;of\;solute_{(in\;gm)}}{Weight\;of\;solution_{(in\;gm)}}10^6$$
Since density of water is $1gm/ml$, $500ml$ of water would weigh $500gm$
$$\therefore600=\frac{W_b}{500}10^6$$
$$\therefore W_b=0.3gm$$

Thus, $500ml$ of water contains $0.3gm$ of fructose. You can find out what volume will contain $0.15gm$ of it.


Why is $\Delta G = -T \Delta S_{\mathrm{total}}$ valid only at constant pressure?

You have actually assumed in your equations that both pressure and temperature are constant. Firstly, we have that

$$\mathrm{d}S_\text{total} = \mathrm{d}S_\text{system} + \mathrm{d}S_\text{surroundings}.$$

By definition, differential of entropy is

$$\mathrm{d}S = \frac{\delta q_\text{rev}}{T}.$$

From the first law of thermodynamics, and the definition of entalphy, for a reversible process

$$\mathrm{d}H = \mathrm{d}U + P\mathrm{d}V + V\mathrm{d}P= \delta q_\text{rev} + \delta w + P\mathrm{d}V + V\mathrm{d}P.$$

When all work done is to expand, then $\delta w = -PdV$, and thus

$$\mathrm{d}H = \delta q_\text{rev} -P\mathrm{d}V+ P\mathrm{d}V + V\mathrm{d}P = \delta q_\text{rev} + V\mathrm{d}P.$$

Only if pressure is constant, do we get the relationship

$$\mathrm{d}H \overset{\mathrm{d}P=0}{=} \delta q_\text{rev} \to \mathrm{d}S = \frac{\mathrm{d} H}{T}.$$

Therefore,

$$\mathrm{d}S_\text{total} = \mathrm{d}S_\text{system} - \frac{\mathrm{d} H}{T}\tag{constant pres., exp. work}.$$

Additionally for constant temperature, or $\mathrm{d}T = 0$,

$$\mathrm{d}G = \mathrm{d}H - S_\text{system}\mathrm{d}T - T\mathrm{d}S_\text{system} \overset{\mathrm{d}T = 0}{=} \mathrm{d}H - T\mathrm{d}S_\text{system}.\tag{constant temp.}$$

Hence the result $$\mathrm{d}G = -T\mathrm{d}S_\text{total}$$ is true when pressure and temperature are constant (in addition to possible other constraints).

It is probably not the best practice on my part to write things like

$$\mathrm{d}H = \delta q_\text{rev} $$

which has the impression of sloppy (but probably not wrong) notation. Interpret this as that, given our assumptions, the number $1$ becomes an integrating factor for $\delta q$, equivalent to the inexact differential becoming exact under special conditions.

A better option might be to write this expression with definite integrals, i.e.,

$$\int_{\text{state 1}}^{\text{state 2}} \delta q_\text{rev} = \int_{H_1}^{H_2}\mathrm{d}H.$$


The short answer to your question is, usually, but not always.  The energy differences between these levels are very small, so sometimes other factors come into play that can cause exceptions.  Here are a few exceptions:

 Predicted configuration   Actual configuration

Cr    [Ar] 3d4 4s2  ...........  [Ar] 3d5 4s1

Cu    [Ar] 3d9 4s2  ...........  [Ar] 3d10 4s1

Ag    [Kr] 4d9 5s2  ...........  [Kr] 4d10 5s1

Au    [Xe] 4f14 5d9 6s2  .. [Xe] 4f14 5d10 6s1

There is a special stability associated with a full shell, in fact that's probably why you suspected that "transition metals first fill up their valence shell and then penultimate shell."  There also appears to be a stability advantage associated with a half-filled shell and this can explain many of the anomalies.  The reason that a half-filled shell may have some extra stability is that there are no paired electrons (Hund's rule) in such a configuration.


What is the explanation of the changes in stability going down a group for carbonates, bicarbonates, fluorides, and chlorides?

Carbonates

The quote from your text:

  Carbonates of alkaline earth
  metals are insoluble in water and can be
  precipitated by addition of a sodium or
  ammonium carbonate solution to a solution
  of a soluble salt of these metals. The solubility
  of carbonates in water decreases as the atomic
  number of the metal ion increases. All the
  carbonates decompose on heating to give
  carbon dioxide and the oxide. Beryllium
  carbonate is unstable and can be kept only in
  the atmosphere of CO2. The thermal stability
  increases with increasing cationic size.

So the stability that you are referring to is thermal stability. This is an important detail. So what is thermal stability? It's how resistant a molecule is to decomposition at higher temperatures.

What's happening to cause thermal instability?

So, lets look at the carbonate ion here:

This is just an illustration, and in reality the negative charge we see on the two $\ce{O}$ atoms is localized due to resonance.

Below the illustration shows where the negative charge is likely to be concentrated (colored in red).

So, when we create a carbonate complex like the example below, the negative charge will be attracted to the positive ion.

Because of this polarization, the carbon dioxide will become more stable and energetically favorable.

How does going down a group play into this?

Well as you go down the group, the charged ion becomes larger. The larger the ion, we see a lower charge density. Charge density is basically the amount of charge in a given volume. So, if a small ion has the same charge as a larger ion, the charge density will be greater for that small ion. Greater charge density, means a greater pull on that carbonate ion, and a greater pull causes the delocalized ions, and a more stable $\ce{CO2}$ molecule. So, the larger the ion, the lower the charge density, the less polarizing of an effect, and reduced stability of a $\ce{CO2}$ molecule, favoring the $\ce{CO3}$.

Chloride and fluoride stability

Stability of fluorides, chlorides, and other halogens, are likewise related to thier size. The halogens, specifically fluouride, is known for their electronegativity. Electronegativity, is the tendency to attract electrons to itself. As you move up the group, you see an increase in electronegtivity. This results in the creation of polar bonds.

Illustrated below, you see that as charge of the positive ions increase, polarizability increases (left), and as the halogen ion increases, polarizability and electronegativity decrease (right). When the ions electron cloud, is less polarized, the bond is less strong, leading to a less stable molecule.

Information and illustrations on carbonate ions were sourced from here.


Your definition of work is not quite correct.  A better definition is**

Work w, is the energy change that drives net shift in the distribution of mass in the system.

Thus energy here means total energy which is kinetic plus potential. Imagine going to the top of a building. You gain potential energy by virtue of being at the top (m.g.h, mass m, acceleration due to gravity g, and height h). If you are not sure that you have gained energy, think about what happens if you were to walk of the top of the building!

You can walk up the stairs and gain this potential energy but you could also run up, and if you could manage to run, (huff-puff-pant-groan) the amount of work you would have to do is far more than walking, but the potential energy gained is exactly the same. Therefore in general the work depends on the 'path' taken, not just the starting and ending conditions. The word 'path' used in thermodynamics generally means the sequence of all the changes made to go from start to end. Hope this helps to clarify.

** ref A. Cooksey Physical Chemistry


This question is a little bit broad, but thankfully, there are many resources available.

In answer to part of your question

  Which one is most suitable for packaging pastries, for example, biscuits?

The Natural Resources Defense Council webpage Food Storage Containers, they state that the safest plastics are PET, HDPE, LDPE, and PP.

For comparing the various types of plastic, the Plasticology 101 website compares the types of plastic according to:

Temperature tolerances
Clarity
Chemical, scratch and impact resistance
Rigidity
Food contact acceptability
Recyclable material

They suggest that PET, HDPE, PVC, LDPE, PP and PS are suitable for food, but temperature tolerances would be a major factor in what food they store.  Another chemical comparison that needs to be considered for the storage of food is information about the $\ce{O2}$, $\ce{N2}$, $\ce{CO2}$ gas and moisture transmission rates are included in Evergreen's document The barrier performance of common plastic film.

Specifically for biscuits, and any other baked goods, according to sevral websites, it depends on what the ingredients are - things to consider are the dairy content, whether the plastic will react with some of the ingredients etc.


As @MaxW has said in comments, the correct form is:

$\ce{SO_4^{2-}}$ and $\ce{Fe^{2+}}$

It is just as in math.
If you have $a+a$ it results in $2a$, not $a2$, and thus:

$\ce{SO_4^{--}=SO_4^{2-}}$ and $\ce{Fe^{++}=Fe^{2+}}$

It is not a stupid question.
IMHO it is a perfectly legitimate question.
I apologize for those who are downvoting your question.


Is the electrical conductivity only measurable in aqueous solutions?

Did you look into ionic liquids? Also in ionic complexes like poly (ethylene oxide)? For instance, look into these references:

P. V. Wright, Electrical conductivity in ionic complexes of poly(ethylene oxide), Polymer International, 1975, 7(5), 319-327 (https://doi.org/10.1002/pi.4980070505).
Q.-G. Zhang, S.-S. Sun, S. Pitula, Q.-S. Liu, U. Welz-Biermann, J.-J. Zhang, Electrical conductivity of solutions of ionic liquids with methanol, ethanol, acetonitrile, and propylene carbonate, J. Chem. Eng. Data, 2011, 56(12), 4659–4664 (https://pubs.acs.org/doi/abs/10.1021/je200616t).


Method to know the number of isomers of metal complexes

Do it many times over, so as not to get confused.

Alternatively, there is a thing called Polya's formula, but you won't be able to use it anyway. In trivial cases like this, the said formula is about 100 times more complicated than counting isomers by hand. It is not before polysubstituted fullerenes that its use in chemistry starts making any sense.


Is it possible to react two distinct carboxylic acids to form a dicarbonyl ester?

The standard term for such molecules is acid anhydride, as they can be viewed as the product of a condensation reaction between two carboxylic acids, with concomitant loss of $\ce{H2O}$. Asymmetric anhydrides (i.e., ones that decompose into two distinct carboxylic acids if hydrolyzed) can certainly be prepared. One possible route is to synthesize an acid chloride from carboxylic acid #1 (thionyl chloride, $\ce{SOCl2}$, being one common reagent used to make the acid chloride) and then react it with the carboxylate salt of carboxylic acid #2. In that reaction, the carboxylate functions as a nucleophile and attacks the carbonyl carbon of the acid chloride, with chloride anion expelled as the leaving group.

I don't think it's an efficient laboratory method to prepare asymmetric anhydrides by simple dehydration of stoichiometric mixtures of distinct carboxylic acids. I suspect that would probably yield a (roughly) statistical distribution of products, at least in the ideal case. In reality, various factors might skew the distribution one way or another.


Which is the correct definition for metamers (or metamerism)?

There is no IUPAC definition for metamers or metamerism in the gold book (not even one that makes it obsolete), and Wikipedia doesn't even have a real article about it, see Metamerism.

  In chemistry, the chemical property of having the same proportion of atomic components in different arrangements (obsolete, replaced with isomer). In organic chemistry, compounds having the same molecular formula but different number of carbon atoms (alkyl groups) on either side of functional group ( i.e., -O-,-S-, -NH-, -C(=O)-) are called metamers and the phenomenon is called metamerism.

I cannot find any authoritative source for this, but there is a question discussing it on our platform: What is metamerism? Also related: Are methyl n-propyl ether and methyl iso-propyl ether metamers?

Any of the two definitions cited in the question match the general definition of isomers.

  isomer (DOI: 10.1351/goldbook.I03289)
  One of several species (or molecular entities) that have the same atomic composition (molecular formula) but different line formulae or different stereochemical formulae and hence different physical and/or chemical properties.

More particularly, they match the definition of constitutional isomers.

  constitutional isomerism (DOI: 10.1351/goldbook.C01285)
  Isomerism between structures differing in constitution and described by different line formulae e.g. $\ce{CH3OCH3}$ and $\ce{CH3CH2OH}$.

Metamers/ metamerism is an archaic, ambiguous, and deprecated term in chemistry. Don't use it.


Selective transfer of least/most stable anion from organometallic reagents

It’s not really a dichotomy, it’s just two very different reaction mechanisms.

For the cuprate addition, which also applies to general Grignard-type additions, in principle the reaction is reversible (figure 1).

Figure 1: Reversibility of Michael additions.

However, the reverse direction is only really possible if the anion released is somewhat stable. In your example, thiophenolate may attack but may also be removed again in the course of the reaction. Once an alkyl chain has attacked, there will be no reverse reaction since the eliminiation of $\ce{R-}$ is too unfavoured.

For palladium-catalysed cross-coupling reactions, we need to consider the typical palladium catalysis cycle (figure 2).

Figure 2: typical catalytic cycle of palladium-catalysed ($\ce{M} = \ce{Pd}$) cross couplings. In a Stille coupling as in the question, $\ce{Y} = \ce{SnR3}$. Image taken from Wikipedia where a full list of authors is available.

The important step here is the transmetallation step: One of the ligands of metal $\ce{Y}$ ($\ce{Sn}$) must be transferred to the palladium ($\ce{M}$) centre. This will be the ligand that dissociates most easily — the best stabilised anion. So if the tin centre has three n-butyl ligands and one phenyl one, phenyl will be preferentially transferred since it dissociates more easily.

This is also part of the reason why $\ce{C}_\mathrm{sp^3}\ce{-C}_\mathrm{sp^3}$ cross coupling are so rare.


Basification of acid solution. pH calculation. How?

You've made some assumptions that we can deal with, and some that need adjustment.

1) 1 liter of water weighs 1000 grams. Correct. However, 1 liter of 5% acetic acid weighs 1005 grams. No big deal; 1 liter will still contain 50 grams of acetic acid.

2) As the commenter noted, neutralizing the acetic acid gives sodium acetate, a weak base with pH ~9. You can use phenolphthalein as an indicator because it changes color at this pH.

3) The liter of neutralized acid will be homogeneous, so the 50 grams of acid and 33.33 grams of sodium hydroxide will be distributed throughout; any sample of the solution will have the same pH.

(Although it is interesting to do an acid/base titration with phenolphthalein. As you approach the endpoint, a drop of base will splash red, then fade to colorless. The next drop will splash a larger red cloud, then fade; etc. Finally, you might get a huge red cloud and think you are done, but you are just shy of the last bit of neutralization and the solution turns colorless again. The next drop flashes totally red.  The reason I mention this is that you could say that for those few milliseconds when the solution is partially red and partially colorless, the pH is different in different locations, but this is not a steady state, so you mix it till you can get a repeatable measurement.)

4) What you measure when you stick a pH meter into the solution is the concentration of the hydrogen ion (hydronium ion) that results from the water molecules reacting with everything else all over the solution. The acetic acid is a reactant in the original solution that increases the hydrogen ion quantity, but the sodium hydroxide reduces it, so you have made a solution of sodium acetate, pH ~9.


Why is the Jacobsen's catalyst with tert-Butyl groups a better catalyst for epoxidation?

This type of oxygen transfer reaction consists of a two step
catalytic cycle, wherein, first an intermediate (salen)$\ce{Mn^V}$ oxo complex is generated, and then the activated oxygen is added to the olefinic double bond.

Electron withdrawing and electron donating group influence the strength of the $\ce{Mn=O}$ bond, with electron withdrawing substituents leading to a weaker $\ce{Mn=O}$, and electron donating substituents leadings to a stronger $\ce{Mn=O}$ bond.

An argument based on Hammond Postulate can be applied where in the case of electron withdrawing substituents (weak $\ce{Mn=O}$ bond), one expects an early transition state, and in the case of electron donating substituents (stronger $\ce{Mn=O}$) a late transition state is seen.

Enantioselectivity is governed by interactions of the salen ligand, which in the transition state geometry adapts a folded geometry forming a chiral pocket, and the incoming olefin.

A late transition state demands a certain proximity between the
olefin and the manganese oxo complex. This leads to a less  spatial separation between substrate and catalyst and consequently allows for better differentiation of stereoisomeric transition structures.

Thus the attenuation of the reactivity of the oxo-species by electro donating groups, leading to a comparatively late transition state is the reason for higher enantioselectivity.

References

This article looks at the electronic effects of 5,5' substituents using DFT calculations.
Cavallo, Luigi, and Heiko Jacobsen. "Electronic Effects in (salen)Mn-Based Epoxidation Catalysts." - The Journal of Organic Chemistry, 2008, 68(16), pp 6202–6207


How to recrystallize boric acid into flakes?

The ratio seems about right, though the solution could be a bit more saturated as the solubility of boric acid in $\pu{100 mL}$ of boiling water is $\pu{27.5 g}$, so for $\pu{400 mL}$ of water you can theoretically use $\pu{110 g}$ of boric acid.

From my experience it is crucial that initial nucleation occurs at the surface, then one gets flake-alike crystals. That's said, you need to start to cool down the solution very slowly, preferably using a warm water bath and a closed vessel with large cross-sectional area (wide beaker with a watch glass on top should do). Don't rush the cooling process, especially at the beginning – crystallization should begin on surface, not in the volume, otherwise you get a bunch of crystals of different morphologies.

P. S. When drying the crystals at the end after rinsing them with cold water on a filter make sure not to exceed $\pu{130 °C}$, otherwise boric acid will start to decompose and form metaboric acid $\ce{HBO2}$, then subsequently upon further heating, $\ce{H2B4O7}$, and, finally, $\ce{B2O3}$ [1, pp. 119–120].

Reference

Patnaik, P. Handbook of inorganic chemicals; McGraw-Hill handbooks; New York, 2003, ISBN 978-0-07-049439-8.


Why the ionization energy of Au is more than that of Cu?

Smaller size does not necessarily correlate with higher ionisation energy. The trends in 1st IE going across Period 2 offer a simple counterexample.

A better (but not 100% foolproof) guideline is to use the orbital energies (see: Koopmans' theorem). The higher the energy of the orbital from which the electron is removed, the lower the IE.

In the case of gold versus copper, the first ionisation corresponds to the removal of a 6s or 4s electron respectively. In gold, the 6s orbital is relativistically stabilised, leading to a larger than expected ionisation energy. See also Geoff Hutchison's answer here.


welcome to the chemistry StackExchange!
I think it would be helpful for you to review the wikipedia page on particle-wave duality. If you have access to a textbook on the origins of quantum mechanics, that would help clear things up,too.

The quantum mechanical model for particles states they have both particle-like and wave-like behavior, so it is incorrect to say they are never particles, which is what I assume you mean by "objects".

At the most fundamental level, quantum mechanics is just another tool with which to view and understand the world around us. It's an especially useful tool when we are looking at things in very fine detail (let's say, usually < $10^{-7}$ m). So, we can look at a given particle from the framework of classical mechanics or from the framework of quantum mechanics. This is an important point, because it is not quantum or classical mechanics that attributes mass to the particle. Rather, the particle had that mass to begin with.

So, to answer your question: A quantum mechanical treatment neither prescribes nor removes mass from a particle or an object. A photon is massless, a proton has mass, and that's that!


Reduction of nitro compound using protection group

For the first arrow you are correct that you need to protect the ketone before reducing the nitro group. For this I would use a dimethyl ketal. There are many sets of conditions for this; trimethyl orthoformate in refluxing MeOH with catalytic PTSA is a good bet. This will also give the methyl ester of the carboxylic acid but that is of no consequence as it will be reacted in the next step. Your next step is then the nitro group reduction. I see no reason why you should not use hydrogenation, just don't have any acid present as that may remove the ketal and you still need that for step 3. The aniline thus produced will cyclise easily, just heat it in neutral solvent if it does not spontaneously cyclise. Step 3 is reduction of the amide, LiAlH4 should do this. Finally you deprotect the ketone with a mild acid hydrolysis.

For the second arrow reduction of the nitro group by hydrogenation will give indole-2-carboxylic acid by spontaneous cyclisation as formation of the aromatic [5] is favoured over the non-aromatic [6]. This has been reduced to your target indoline-2-carboxylic acid by sodium/liquid ammonia  https://patents.google.com/patent/US4485241


Why does lattice enthalpy decrease with increasing ionic size?

Think about Coulomb's inverse-square law:
$$F = k\frac{q_1q_2}{r^2},$$
where $q$ is the magnitude of the charge, $r$ is the distance, and $k$ is the Coulomb’s constant.
Mathematically speaking, as $r$ increases, the magnitude of $F$ becomes smaller. As far as the charge density, as it increases, the magnitude of $F$ increases.
So both explanations are saying the same thing from different point of views. As ionic radii increases, charges are further apart so attraction decreases. As charge density increases, the attraction increases since the charges are increasing in magnitude.
Just as $\ce{MgO}$ is stronger then $\ce{NaCl}$ since $\ce{Mg}$ and $\ce{O}$ have smaller ionic radii and they both bare a $2+$ and $2-$ charge, respectively, making the lattice enthalpy greater.


Confusion about Ethinyl estradiol IUPAC name

For natural products and derivatives, three levels of nomenclature are recognized in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book): trivial names, systematic names, and semisystematic names. A new compound that can be isolated from a natural source is commonly given a trivial name. These trivial names are usually related to the biological origin of the material.

When the full structure of a natural product or derivative is known, a systematic name may be generated in accordance with the IUPAC recommendations for nomenclature of organic chemistry. However, the systematic name “(8R,9S,13S,14S,17R)-17-ethynyl-13-methyl-7,8,9,11,12,14,15,16-octahydro-6H-cyclopenta[a]phenanthrene-3,17-diol”, which is proposed in the question is not correct. This name implies that the carbon atoms 13 and 17 of the cyclopenta[a]phenanthrene parent structure remain sp2 hybridized; i.e. that a double bond remains between atom 13 and atom 17, which is not possible since both atoms have four bonds (to other skeletal atoms or to substituents). The correct systematic name is (8R,9S,13S,14S,17R)-17-ethynyl-13-methyl-7,8,9,11,12,13,14,15,16,17-decahydro-6H-cyclopenta[a]phenanthrene-3,17-diol.

Such systematic names may be too cumbersome. To overcome this difficulty and show the close similarity to related compounds, a semisystematic name may be formed. The semisytsematic name 17α-ethynylestra-1,3,5(10)-triene-3,17β-diol is based on the retained name estrane for the parent structure (8R,9R,10S,13S,14S)-13-methylhexadecahydro-1H-cyclopenta[a]phenanthrene. (Names of fundamental stereoparent structures are given in Table 10.1 of the 2013 IUPAC recommendations. The corresponding structures are shown in Appendix 3.) Addition of the three required double bonds to the estrane parent structure leads to the name estra-1,3,5(10)-triene, and addition of the missing substituents yields finally yields the name 17-ethynylestra-1,3,5(10)-triene-3,17-diol. To indicate the configuration at chiral centers, the ‘α/β’ system is used as described in IUPAC-IUBMB recommendations for the nomenclature of steroids.


Trying to understand why Sn2 is not possible in this reaction

Backside attack on the red bromide is not possible. The rest of the norbornyl skeleton completely blocks the trajectory required for $\mathrm{S}_{N}2$ reaction.


Neutralising sulfuric acid but not boric acid

You've got to know that sulfuric acid is a strong acid and boric acid is a very weak acid. So, sulfuric acid can neutralize most of the bases where boric acid cannot. So, you have to take a base that can be neutralized by only sulfuric acid and not boric acid. So, you should probably take group 2 hydroxides and try to react with the unreacted sulfuric acid because sulfuric acid reacts with group 2 hydroxides whereas boric acid can't.

$$\ce{M(OH)2 +  H2SO4 -> M2SO4 + 2H2O}$$

$$\ce{M(OH)2 +  H3BO3 ->nothing}$$

where $\ce{M=Be,Mg,Ca,Sr,Ba}$.

So, which group 2 hydroxide to use? Group 2 hydroxide react with the sulfuric acid to form group 2 sulfates. Now, solubility of group 2 sulfates decreases down the group. So, you should use calcium hydroxide so the calcium sulfate that forms will not dissolve in the solution due to its low solubility and form a precipitate. You can separate that precipitate by filtration.

Formation of calcium borate is highly unlikely and so only calcium sulfate will only form.

Also, barium hydroxide is the best chemical for this process because barium sulfate is very insoluble. But barium hydroxide is costly if you buy it or making it is difficult due to the cost of other reagents. Instead, calcium hydroxide is your best bet.


What effect on boiling and melting points does intramolecular hydrogen bonding have?

The effect of intramolecular forces versus intermolecular forces on boiling and melting points is sometimes seen when we compare the boiling points of ortho- and para-nitrophenol (or similar compounds).
Para-nitrophenol shows intermolecular hydrogen bonding, which causes an increase in boiling point as different molecules bond better with each other, as compared to ortho-nitrophenol, where there is intramolecular hydrogen bonding, which does not help in increasing the boiling point, hence p-nitrophenol has a higher boiling point than o-nitrophenol.
However, while comparing intramolecular bonding with no hydrogen bonding, no difference can be expected unless the intramolecular bonding affects the intermolecular bonding in some way, for instance by changing size and thus packing.


Your questions has two a two part answer.

(1) How would you get the $\ce{NaOH}$ out of a sealed bottle?

(2) A concentrated stock solution can be made for $\ce{NaOH}$. In the top of the bottle you use a drying tube stuffed with solid $\ce{NaOH}$ to remove $\ce{CO2}$ from air. On the bottom you have an outlet from which you pour the $\ce{NaOH}$. The gist is that $\ce{Na2CO3}$ is the least soluble carbonate. So most of the $\ce{CO3^2-}$ in the original solution participates. So if you let the solution sit for a couple of days it will be very stable.

The point with the stock solution as outlined is that $\ce{Na2CO3}$ "sticks" to inside of the container and doesn't flow out with solution. (You don't shake bottle up like you do with tomato juice that you drink to resuspend the solids.)

So for (1) it is high near impossible to make a $\ce{NaOH}$ solution without some $\ce{CO3^2-}$ in it. The $\ce{NaOH}$ itself has $\ce{CO3^2-}$ unless it is specially made, sealed in argon and so on.

So your stock solution is concentrated. I boil some distilled water to get rid of $\ce{CO2}$ and seal it till it cools. I then take $10~\mathrm{ml}$ of my stock $\ce{NaOH}$ solution and dilute to $250~\mathrm{ml}$ (or whatever ratios) to get the desired concentration of $\ce{NaOH}$. (That I can check with KHP titration.)


The SHE (Standard Hydrogen Electrode) is a standard equipment for measuring the relative electrode potential, as the process of measuring the actual potential is quite difficult. It consists of a platinum foil dipped in 1.0M $\ce{HCl}$ in the presence of $\ce{H2}$ gas at 1 bar of pressure.

Now when the SHE is connected to another half-cell, either oxidation or reduction may occur in the SHE.

Suppose we have to find the electrode potential of a half cell represented by "A" Now, on setting up the cell and connecting a voltmeter we get the standard potential(Estandard) of the cell i.e.Estandard = EA - EH2
Now as our our aim is to just find EA, we take EH2 to be equal to 0
Therefore,
Estandard = EA - 0
or,
Estandard = EA

The value of Estandard is taken from the voltmeter.
It should be noted that the electrode potential calculated using this method is relative(to the electrode potential of hydrogen). It isn't the absolute value.
Also, the potential for the SHE is taken to be 0 for the simplification of the process.


Passing chlorine gas through water yields a solution of $\ce{Cl2}$, hydrochloric acid ($\ce{HCl}$) and hypochlorous acid ($\ce{HOCl}$). The latter two are formed by reaction of chlorine with water under disproportionation.

$$\ce{Cl2 + H2O \rightleftharpoons HCl + HOCl}$$

Bleaches are chemicals which remove color by oxidation of dyes. $\ce{Cl2}$ and $\ce{HOCl}$ themselves are already strong oxidizers, but adding alkali metal hydroxides to an aqueous chlorine solution increases the solubility of $\ce{Cl2}$, and chlorine bleach is obtained, which contains alkali metal hypochlorite as active ingredient.

$$\ce{Cl2 + OH- \rightleftharpoons ClO- + Cl- + H2O}$$

Sodium hypochlorite is produced by this reaction and common household bleach is an aqueous 3-8 mass-% solution of the salt. The necessary chlorine gas is actually produced by electrolysis of $\ce{NaCl}$.

While gaseous $\ce{Cl2}$ itself can act as a bleaching agent by oxidation, the use of hypochlorite solutions has the benefit of avoiding handling of the poisonous gas. $\ce{H2}$ is a much better reducing agent, and therefore unlikely to be useful for bleaching.

A mixture of $\ce{H2}$ and $\ce{Cl2}$ will react to form gaseous hydrogen chloride when exposed to UV light. Hydrochloric acid is obtained when gaseous $\ce{HCl}$ is dissolved in water.

$$\ce{HCl + H2O \rightleftharpoons H3O+ + Cl-}$$


Are these chemicals suitable for disinfecting my water tank?

When you start asking questions about chemistry you are only left with more questions... in general terms of safety, using household bleach to create a concentration of 1 - 3 ppm sodium hypochlorite is relatively safe (especially compared with sewage). In specific terms, the answer is two-fold; you have to consider the amount of exposure and the component's toxicity (the difference between medicine and poison is the dose).

It is important to note that the SDS for household bleach only lists sodium hypochlorite in the hazardous ingredients section. This directly implies that the only chemical of significant hazard in the bottle is sodium hypochlorite, and the rest of the ingredients are not present in sufficient amounts to cause issues.

First let's look at what thise things are:

Water : you know
Sodium Hypochlorite : bleach- household bleach is made from chlorine, caustic soda, and water and DC current (electrolysis).
Sodium Chloride : table salt, leftover from production or evaporation of water and chlorine.
Sodium Carbonate : washing soda- kind of like baking soda but about 40% stronger. This may be considered a buffer or stabilizer or byproduct.
Sodium Chlorate : when bleach is cooked/boiled (and dried), the leftover crusty stuff is salt and sodium chlorate... so this would be a byproduct, probably only in a very tiny amount.
Sodium Hydroxide : lye (caustic soda, NaOH) - household bleach solutions are typically stabilized by lye as part of the manufacturing process. Lye is used for raising the pH of drinking water and for making many things including soap.
Sodium Polyacrylate : thickening agent - It may be used in food, toothpaste, and shampoo.

APPROXIMATE Household Bleach Amounts:

Water : 93%
Sodium Hypochlorite : 5.5%
Sodium Chloride : 0 - 2%
Sodium Carbonate : guestimating 0 - 0.2%
Sodium Chlorate : guestimating ~ 0 - 0.2%
Sodium Hydroxide : since the pH of bleach is about 11, the amount of sodium hydroxide must be less than 0.001 molarity, so less than 0.004%
Sodium Polyacrylate :  based on the viscosity of bleach and my personal experience with this thickening agent, about 0.5%.

Now lets look at the LD50 (a general measure of toxicity). The LD50 is the amount of material necessary to kill 50% of the test subject (usually rats). For these chemicals I listed acute (hard and heavy) oral dosage. Chronic exposure may be more relevant because you will not be drinking this, but chronic exposure data is also far more difficult to measure accurately, and with respect to all of these chemicals, chronic effects are generally nil, so I wouldn't have anything to say.

Water : greater than 90,000mg/kg, rat (yet drowning causes more deaths than all of the following chemicals combined)
Sodium Hypochlorite : 8910 mg/kg, rat
Sodium Chloride : 12357 mg/kg, human
Sodium Carbonate : 4090 mg/kg, rat
Sodium Chlorate : 1,200- 7,000 mg/kg, rat
Sodium Hydroxide : 100mg/kg, rat
Sodium Polyacrylate :  greater than 2000 mg/kg, rat. When something is above 2000 mg/kg the exact lethal dose is often not pursued.


What is the crystal structure of ammonium hydrogen sulfate? It's beautifully birefringent

Birefrigence is a result of structural anisotropy observed for crystals other than of the cubic Bravais class. Then, interaction of vectorial properties (e.g., light) with said material is anisotropic, too. It is a phenomenon seen both for inorganic, as well as organic materials (examples listed e.g., here). As seen e.g., by cooling films of molten polymers on a heat stage in polarization microscopy, local deviations from isotropy suffice to observe birefrigence, too (example). Thus, there is no «special functional group» which, if present, suddenly would switch on / switch off a sample to be birefrigent. It's observation and measurement by polarization microscopy allows to characterize the microstructure of e.g., minerals, and thus one of the  elder tools in geology (still used today, e.g., Stoiber's «Crystal Identification with the Polarizing Microscope» by 1994 lists some in its freely available back matter).  It may be used on organic samples, too (e.g., this tutorial).
The models to describe the crystal structure about ammonium sulfate, $\ce{(NH4)HSO4}$, depend on the temperature of the sample. This might increase or decrease birefrigence (I did not look out for a temperature-dependent recording of this about ammonium sulfate).  The explored variation of sample temperature in the work below (at ambient pressure) however do not change the unit cell symmetry this much that birefrigence suddenly would be cancelled out for symmetry alone.
At room temperature ($\pu{293 K}$), the best description of this material that there two hydrogensulfate groups per unit cell which are described independently from each other:

(credit 2012JChemPhysChemA223, Fig. 1a / $\pu{293 K}$)
In the illustration, these are marked by O (ordered) and D (disordered).  This is to say that the model describes the reality best assuming positions with $\ce{HSO^-4}$ with more than one orientation and local geometry.  (Similar to sitting either forward, or backward on a chair.)  This is, taking a different perspective on the sample (the representations omit hydrogen atoms):

(modified representation, based on 1972ActaCrystA445)
with ammonium ions, ordered and disordered $\ce{HSO^-4}$, the sites with the disordered would be either one of the following:

(credit 1972ActaCrystA445)
At lower temperatures, e.g. at $\pu{200 K}$ and at $\pu{125 K}$, it is energetically more favourable if both $\ce{NH^+4}$ and $\ce{HSO^-4}$ are in ordered position and orientation only, e.g.:

(credit 2012JChemPhysChemA223, Fig. 1b / $\pu{200 K}$)
This phase transitions and changes of unit cell symmetries (space group $P2_1/c$ (at $\pu{293 K}$) $\rightarrow$ $Pc$ (at $\pu{200 K}$) $\rightarrow$ $P1$ (at $\pu{125 K}$)) however will not cause a cancellation of  birefrigence because of symmetry reasons.

You do not state (this much) why knowing the the spatial arrangement of the atoms, and hence a model of the crystal structure is important for you.  However, because you mention to possess some command of chemistry and interest art, be aware among the crystallographers there are some equally with a sense for beauty.  One of the low-level entries is IUCr's Crystal growing competition for school children (note the their brochure with examples in the later section) considerably newer than their teaching pamphlet.
On the more artistically appealing side, though, selectively dyeing crystals may be of interest for you.   Work in this field was much popularized again by e.g. Bart Kahr:

(credit, includes protocol)

(credit 2001ChemRev893)
with protocols for chemicals easier to access (e.g., potassium dihydrogen phosphate [KDP] and amaranth dye, cf. here).  Beside beauty, the selective incorporation of dyes provides insight into the growth mechanism of crystals. You may find additional inspiration in other works by him, e.g., in 1996AdvMat941, 2001ChemRev893 including a 20 page long table of examples («reported
during the past 147 years»), or 2002CrystEngComm514 about «painting crystals» by an easy-to-follow protocol.


Is it true that a pH higher than the pKa of an acid make the H automatically disassociate?

I propose the following  answer:

Let's consider the following equilibrium:

$$ \ce{H}A+\ce{H_2O} \leftrightarrows A^-+\ce{H_3O^+}$$
The equilibrium constant $\ce{K}_A=\dfrac{[A^-][\ce{H_3O^+}]}{[\ce{H}A]}$. So if we take ''$-\log$'' of this equation, we can define the $\ce{pK}_A$ of an acid as: $\ce{pK}_A=\ce{pH}- \log\dfrac{[\ce{A^-}]}{[\ce{H}A]}$.

we rewrite this equation as follows:

$$
\ce{pH}=\ce{pK}_A+\log\dfrac{[A^-]}{[\ce{H}A]}
$$
From this equation, we can deduce that when $\ce{pH}>\ce{pK}_A$, $[A^-]>[\ce{H}A]$,  i.e. the concentration of the conjugated base is bigger than the concentration of the acid. But this doesn't mean that all the acid molecules are dissociated. The bigger the difference between the pH and the $\ce{pK}_A$, the bigger is the percent of dissociated acid molecules.


To answer this question you have to look at the reaction mechanism. Since the reaction is conducted in the dark, you don't have any radicals to worry about. Under these condition bromine reacts with an electron-rich alkene in a electrophilic addition reaction in the course of which a cyclic bromonium ion is formed. This bromonium ion is highly reactive and in the absence of other nucleophiles reacts with the leftover $\ce{Br-}$ ion in an $\text{S}_{\text{N}}2$ reaction which leads to the opening of the 3-ring and formation of the trans product.

Update:

As @BenNorris pointed out, the ring structure in the question are drawn wrongly. So, I will add the correct version: The dibrominated product will initially be formed in a diaxial conformation (this is always the case due to the ring geometry and the mechanism of the $\text{S}_{\text{N}}2$ reaction). This is usually a very unfavourable comformation so the ring will subsequently flip (if possible) to the diequatorial conformation.


How do polymer properties change when moving from bulk to nanosized fibers?

Nanofiber polymers in contrast with bulk polymers have high ratio of surface area to volume.
So, nanofiber polymers could be useful for any application that needs to maximize the ratio of surface area to volume.

Let's consider their application as scaffolds for tissue engineering: increasing the surface area of artificial-organ scaffolds gives cells more points at which to adhere.

Another application of nanofiber polymers is their use as electrodes  in fuel cells:  the greater the electrodes surface area, the more efficiently they catalyze the reactions that drive the cell.

Another interersting application of polymer nanofibers is in meshes that allow only nanoscale particles to pass through: ultrafine air, water filters, membranes for fuel cells.

Finally, as the capacitance is proportional to the surface area,  capacitors that store electricity, are one example of the possible applications of  polymer nanofibers.

For more details, please see this site and this article.


Why does methyl not shift in the mechanism of isoborneol to camphene?

This is a class of carbocation 1,2-rearrangement reactions, and historically named Wagner–Meerwein rearrangement after two chemists who discovered the reaction (see following diagram):

Although it can be explained considering as an rearrangement of a classical carbocation, there are a tremendous amount of work has been done considering 2-norbornene cation as non-classical carbocation (see comment of  orthocresol elsewhere). However, my interned task is to explain why it is not possible to have 1,2-methide shift instead of alkyl shift (Wagner–Meerwein rearrangement) considering classical carbocation.

First, why the reaction not go in favor of forming bornylene? Since the reaction is in acid medium, fast equlibrium of protonation (to give initial 2-norbonyl carbocation derivative) and deprotonation (to give bornylene derivative) could be expected until most thermodynamically stable product is formed. This can be expected from a $2^\circ$-carbocation, if there is a possibility of rearrangement to a more stable $3^\circ$-carbocation.

Now, the initial 2-norbonyl carbocation derivative is $2^\circ$-carbocation with a possibility of rearranging to a more stable $3^\circ$-carbocation by either 1,2-methide shift or alkyl shift (which is the one eventually happens). If you look at initial carbocation without the bridge-head, the bottom cyclohexyl part is in boat confirmation, which is locked in by the rigid bridge-head carbon. Now that restricts the 1,2-methide migration, which would made energetically unfavored triangularpyramidal $3^\circ$-carbocation. Thus, the readily available alkyl migration is energetically favored and fast. The rearanged crbocation would readily dehyrogenate to give camphene (2,2-dimethyl-3-methylidenebicyclo[2.2.1]heptane).

Also keep in mind that not only it is not favored by the steric resons, but also 1,2-hydride shift would generate a $3^\circ$-carbocation, which would have violate Bredt's rule.

Note: Nonetheless, a recent publication has discussed this scenario and 1,3-hydride migration to give 7,7-dimethyl-2-methylidenebicyclo[2.2.1]heptane as elimination product (Ref.1).

References:

Huidong  Zheng, Yi Lin, Mengchan Wang, Jie Liu, Dan Wu, Jingjing Chen, Guanghua Yin, S. Ted Oyama, Suying Zhao, “The influence of solvent polarity on the dehydrogenation of isoborneol over a $\ce{Cu/ZnO/Al2O3}$  catalyst,” Catalysis Today 2019, 323, 44-53 (https://doi.org/10.1016/j.cattod.2018.10.021).


Acid-base chemistry - determining stability

When we say that 'bigger the atom, more stable is the anion', we mean to say that the negative charge would be spread over a larger area. So, the interatomic repulsive forces would be spread over a large area, lowering the energy and stabilising the anion. Going down the group, the jump in size is large enough to compensate the decrease in electronegativity, so we use the 'size of atoms' argument to differenciate between different atoms in the same group.

Conversely, we use the 'size of orbital' argument to determine the basicity of the same atom but in different hybridisations. Hence, the size of the atom would be same and the smaller size of the sp orbital would mean that it would be better attracted by the nucleus making the anion more stable.


Why does STED microscopy cause photodamage?

Ballpark calculation: Wikipedia mentions the sun irradiation received at earth's sea level is up to about $\pu{300 W/m^2}$.  This is after the atmosphere filtered out already much.  You describe an experiment that where $\pu{0.5 W}$ set foot on about $\pu{0.2 µm^2}$ (or, equivalent to to this, $\pu{0.2 × 10^{-6} m^2}$).
This translates to a power density of
$$\frac{\pu{0.5 W}} {\pu{0.2 × 10^{-6} m^2}} = \pu{2.5 × 10^6 W/m^2}$$
which is more than 8000 times larger than above assumption.  This one of the reasons why samples are excited by pulsed laser radiation with repetition frequencies in the range of MHz.  And still degradation by photobleaching may occur, which one already tries to reduce, e.g., by excitation which do not hit the centre, but a tail of an absorption band.
(The mechanisms are different, but the results of exciting a small area with too much energy are similar to microscopic techniques based on second harmonic and complementary third harmonic generation (an example).)


How to calculate the equilibrium pressure when ammonia and hydrogen gasses react in a closed system?

Note: Not all the reactants will react to form the products, but rather exist in equilibrium. However since the question doesn't mention anything about the equilibrium constant, I will assume that the reaction goes to completion as otherwise it will be impossible to do this question.

If you are aware of the various gas laws, doing this question is actually really simple and doesn't involved many calculations.

Step 1: Finding Initial Pressure

Firstly, the question doesn't specify what the initial pressure of the container is. Therefore you can't just assume that it is $1\ \mathrm{atm}$. So the first step would be finding the initial pressure: $$p = \frac{nRT}{V} = \frac{6\ \mathrm{mol}\times 8.3125\ \mathrm{L\ kPa\ mol^{-1}\ K^{-1}}\times273\ \mathrm{K}}{8\ \mathrm{L}} = 1702.4\ \mathrm{kPa} = 16.8\ \mathrm{atm}$$

Step 2: Find Moles of Products

Looking at the stoichiometry of the chemical equation, $4$ moles of gas react to form $2$ moles of gas. Therefore the ratio is $2:1$. In this case, we originally had $6$ moles of gas. Therefore at the end of the reaction, there should be $3$ moles of gas.

Step 3: Find Final Pressure

Now we know that the total number of moles of gas has halved (from $6$ moles to $3$ moles). One of the gas laws state that the number of moles of gas is directly proportional to the pressure of the system. This is clearly obviously by looking at the ideal gas equation. Therefore, by halving the original amount of moles of gas should result in halving of the original pressure of the system.

Therefore the final pressure will be: $8.4\ \mathrm{atm}$


How can an ionic bond have partial covalent in character?

One way is Polarization. When a small cation approaches a large  anion it distorts the charge density towards itself thereby forming a type of bond which has directional characteristic i.e from ** (-) ----> (+) ** which is characteristic of a covalent bond.

This polarization has some rules given by **Fajan **

Smaller the cation better distortion/polarization.
More the amount of positive charge more is the polarization.
or more the charge : radius ratio of cation more is the polarization.

So in short every cation has the power of polarization and therefore no bond is 100% ionic.


What does the eta notation mean in the naming of a transition metal complex?

What do the $\eta^4$ and $\eta^6$ notations represent? Do they somehow specify the coordination "connectivity" of the Ru(0) and the COD and COT coordinating ligands?

They do.

The $\eta$-notation denotes the hapticity of the ligand, which is the number of atoms in the ligand bonded to the metal center.

In your image, shown above, cyclooctadiene (COD), has two alkene functional groups. A transition metal can coordinate to just one alkene, which would be $\eta^2$, or to both alkenes, which would be $\eta^4$.

Cyclooctatriene (COT) has three alkenes. A metal center can be coordinated at one ($\eta^2$), two ($\eta^4$), or all three ($\eta^6$) alkenes.


Preference for basic conditions in ester hydrolysis

Precedent - why change conditions that others have shown work well?
Control - it is much easier to react 1 eq of base/nucleophile under mild conditions, whereas acid hydroysis generally requires excess reagent and higher temperatures.
Ease of operation - base hydrolysis gives a carboxylic salt as product which frequently extracts into aq, giving an easy workup and purification. Acid hydrolysis requires neutralisation of excess reagent as a first step.


Scientific Reason for Salt Solution Gaining Volume

The vapour tension over the saturated table salt solution { the best  with the solid salt excess) is equivalent to the relative humidity about 75 %.

If the rel. humidity is lower than 75%, the solution evaporates and the salt crystallizes.

If the rel. humidity is higher than 75%, the solution absorbs the air moisture and the salt is dissolving.

Such systems may be informally called "relative humidity buffers". They are useful for hygrometer calibration or for making a constant humidity aerial baths for long term weathering of devices or materials.

See the articles with tables of relative humidity above saturated solutions of various salts:

Researchgate.net
PDF on NIST
PDF on Omega.com