text:
Taking volume contraction into account when mixing water with ethanol
To calculate the amount of purified water and the amount of ethanol (96% Vol.) which is needed for a desired/variable amount of a mixture - given a desired/variable percentage of ethanol - I use the following formula:
$$W = \left(\frac{T * P}{ 100}\right) \times \frac{1}{0.96}$$
and
$$E = T -  W$$
Where $T$ is the desired Total amount of product (in liters); $P$ is the desired Percentage of ethanol in the final product (in %Vol.); $W$ is the calculated amount of Water (in liters) and $E$ is the calculated amount of Ethanol (in liters containing 96 %Vol ethanol).
For example - using this formula - when I need $\pu{40 liters}$ product containing 50% vol. ethanol, the calculated amounts of water and ethanol for the mixture become $19.1667 : 20.833$.
The problem in the above calculation is that because of volume contraction the desired total amount of the final product will be less as given in the formula. To me, the problem became even more confusing after I learned that volume contraction depends on the mixing ratio of both components in a nonlinear way.
What changes should I made to the formula to take volume contraction into account?

How did you derive above equation? Or it is given?

@MathewMahindaratne  I tried to figure out this myself. I am not a chemistry guy. I use the industrial 96% ethanol as solvent for aroma's in a small lab setting. If measures scale up to 50-100 liters the volume contraction starts to play a significant role.

I am curious if there is a mathematical way to figure my problem out, so I can use it in a script (or spreadsheet).

This is almost a non-question. Separately adding the calculated volumes is actually fine as the total volume, even though it's contracted, does not preclude that they were in the volume ratio you wanted at the beginning. Analytical chemists will know all about this. To not mix to a known volume by adding one solvent to the other, but to measure them all separately and then mix together. If you need 40L and the calculated amount gives less than this, simply measure out the volume ratios and top up. Furthermore, any temperature change due to mixing is going to add more inaccuracy.

@Beerhunter It is a real question. I measure about 10% difference in calculated and measured values

@iep I should have said "non-problem"

Well since there is a volume change and you - I assume - do this at atmoshperic pressure there will be pressure volume work involved. This process comes toghether with other changes like temperature. I guess that the pressure volume work should be avoided or at least be sutracted when determining the mixing enthalpy.

@DavidJonsson Thank you for your comment. how would you calculate what you just wrote?

The adiabatic or isothermal (or some blend of the two) mixing enthalpy would need to subtract the pressure volume work of the contraction. For 100 proof ethanol. 50/50 by volume, of one cubic meter of initial volume the pressure volume work is 10^5 Pa * 0.04 m^3 = 4 kJ/m^3. What is this per mol? 783 kg/m^3 * 0.046 kg/mol =>  17 kmol/m^3 => 0.24 J/mol. This is practically negligible compared to the 140-800 J/mol mixing enthalpy.

Measuring out volumes can sometimes be easier than dealing with mass, but in this case, converting the calculations to mass-centric can make the process easier. (I'll use your symbols, but they will now be in mass units (kg). If you desire percentages of ethanol by volume, you still need to convert to mass units because of the volume contraction!))
First, calculate the total mass  in kg of the volume of solution needed; divide the volume by the density of the specific ethanol-water solution. You can look up this density in a handbook or Google it. One simple table is at https://www.engineeringtoolbox.com/ethanol-water-mixture-density-d_2162.html, although many tables are available, even to tenths of a percent.
Second, calculate the mass of 96% ethanol needed:  = (∗/100)×1/0.96. The ratio / will be higher than , the % desired, because the ethanol is only 96%. Convert this to volume in liters, if you desire, by dividing by 0.80138, the density of 96% ethanol.
Third, calculate the amount of pure water needed:  =  − .   / will be less than 100 -   because of the 4% water already in the ethanol. But the total mass will give you the volume required, and the % ethanol will be correct - BY MASS.
Commercial isopropanol solutions are frequently sold as 70% by volume; this calculates out to 64.7% by weight. Your procedure, if you desire to make variable % solutions BY VOLUME, will need to do a calculation before you even do the first step: convert the % ethanol by volume to % ethanol by mass.

The ethanol-water-mixture-density table is a great resource. It clearly shows the significance of my "problem". How did you find the the density of 96%?

I appreciate you took effort to demonstrate the calculations by mass. However, we have no tooling for precise measuring of masses > 50 kg. Measuring huge volumes is way easier in the current setting. Nevertheless, you made me thinking of "correcting" my formula with a "volume contraction ratio", probably derived from table values. Would such an approach work? I mean: say we have 96% ethanol at hand, and we want 100 liters of 70%, 60% or 35%, could we "just" multiply/divide with a value derived from the density table?

@iep unless you're using calibrated flow meters and if for example you were using a graduated vessel, your error in measuring the volume probably knocks out any concerns around the percentage volume loss due to contraction.

@Beerhunter No. The values do not knock out concerns.

@iep can't you use the ratios you derived above but use 11.5% more volume of each? You'll have more than 40L of 50-50 water-ethanol.  Use 40L (since you can measure accurately) and keep the rest for the next batch

Start with finding the volume required, but convert to mass, then calculate the amounts needed, then convert back to volume - all using the density table. For very common % solutions required, make a reference table for, say, 100 liters, then adjust as needed for the needed quantity. For density of 96% ethanol, I used https://handymath.com/cgi-bin/ethanolwater3.cgi?submit=Entry. (I didn't want to overwhelm you with all the decimals.) Don't over-calculate to more decimals than you can measure in your product.

@JamesGaidis I started making a reference tabel, as you suggested. The link you provided made me find another tool: https://planetcalc.com/1481/ This tool allows for temperature as well as calculating solutions by volume. From this discussion I learned that the use of tables is kind of mandatory when dealing with the nonlinear behaviour of volume contraction. As I noticed in the tables I created today, when plotting against a range of 5-25 degrees Celsius, the ratio ethanol/water needed for a desired %(/) draws a pretty straight line, though. Your suggestions are very much appreciated.

In addition to my previous comment, when plotting against a range of 1-25 degrees Celsius, the ratio ethanol 96% : pure water  calculated for a mixture containing 35 %vol. ethanol renders this figure:
https://i.stack.imgur.com/yNLmv.jpg

Based on the comment to an answer elsewhere by OP, I'd try to solve this problem, purely based on volumes. However, these calculations has ignored the volume contraction of 96% ethanol may have showed initially (that should be minimal since it is only ~4% of water by volume in there).
I'd say, your equation, regardless of how OP has derived it, is erroneous. I also like to change the variables OP has used to followings:

$V_{Tot}$ is the desired Total amount of desired ethanol-water solution (in liters);
$P_\%$ is the desired Percentage of ethanol in the final solution (in $\%(v/v)$;
$V_{W}$ is the calculated amount of Water (in liters); and
$V_{E}$ is the calculated amount of $96\%(v/v)$ ethanol (in liters).

Accordingly, the needed volume of pure ethanol $ = V_{Tot} \times \frac{P_\%}{100}=0.96V_{E}$. Thus,
 $$ V_{E} = \frac{V_{Tot} \times P_\%}{100 \times 0.96}= \frac{V_{Tot} \times P_\%}{96} \tag{1}$$
Note that this is actually OP's first equation, but it is for $V_{E}$ ($E$ in OP's notation) instead of $V_{W}$ ($W$ in OP's notation).
Now we can derive the equation for $V_{W}$. Actual $V_{W}$ is:
$$V_{W}=V_{Tot}\left(\frac{100-P_\%}{100}\right)$$
Yet, we cannot use $V_{W}=V_{Tot}-V_{E}$, because some water is coming from $V_{E}$. That amount of water is $ 0.04V_{E} = 0.04 \times \frac{V_{Tot} \times P_\%}{96}$. Thus, we can manipulate this equation as follows:
$$V_{W}=(V_{Tot}-V_{E})-0.04 \times \frac{V_{Tot} \times P_\%}{96}= V_{Tot}\left(\frac{100-P_\%}{100}\right)-0.04 \times \frac{V_{Tot} \times P_\%}{96}\\=\frac{96-P_\%}{96}\times V_{Tot} $$
$$\therefore \; V_{W}=\frac{96-P_\%}{96}\times V_{Tot} \tag{2}$$
Now we apply these two equation to OP's example of making 50% solution:
If $P_\% = 50$ and $V_{Tot} = \pu{50 L}$, from equation $(1)$ and $(2)$,
$$ V_{E} = \frac{V_{Tot} \times P_\%}{96}= \frac{50 \times 50}{96} = \pu{26.042 L}$$
$$V_{W}=\frac{96-P_\%}{96}\times V_{Tot} = \frac{96-50}{96}\times 50 = \pu{23.958 L}$$
Thus, theoretical total (disregarding contraction) is $\pu{50 L}$. Acutally, practical volume must be a little off but your percentage by volume is much close to 50% (only volume contraction did not account is initial 96% ethanol).
Late edition to fulfill OP's request:
If $P_\% = 35$ and $V_{Tot} = \pu{100 L}$, from equation $(1)$ and $(2)$,
$$ V_{E} = \frac{V_{Tot} \times P_\%}{96}= \frac{100 \times 35}{96} = \pu{36.458 L}$$
$$V_{W}=\frac{96-P_\%}{96}\times V_{Tot} = \frac{96-35}{96}\times 100 = \pu{63.542 L}$$

Do you mean notation rather than notification?

@airhuff : Yes, that's what I meant and corrected accordingly. I appreciate your careful reading. :-)


text:
Why don't molecular orbitals change when electrons are added or removed from a molecule?
My lecturer says that molecular orbitals don't change when electrons are added or removed from a molecule, but I don't understand why. As electrons are described by wave functions, surely more or fewer electrons means that the electrons interact differently and consequently the atomic orbital wave functions interact i.e. interfere differently to form different MOs.

They do change.

You are correct. While sometimes, for the sake of a simpler calculation, we might assume the orbitals remain the same with the addition/removal of an electron, they should always relax somewhat due to the new electronic configuration.

This ties into the difference of the concepts of vertical and adiabatic ionization.

The molecular orbitals do change when electrons are removed or added. But what does not change is the relative order of these orbitals with respect to other orbitals. If a given orbital has a higher energy than another one, it will remain so if you add one electron or it you remove one electron, whatever the absolute values of these energies.

"The molecular orbitals do change when electrons are removed or added. But what does not change is the relative order of these orbitals with respect to other orbitals" Proof or citation of proof, please - at the very least please explain how you are labelling the orbitals in varying geometries.

When the energy relative order of atomic orbitals changes by the electron filling,( like 4s vs 3d ), why it should not for molecular orbitals ?

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text:
Why does the nitrile carbon experience more shielding than a carbonyl carbon in CNMR?
In organic chemistry lecture, we were given a chart of approximate signal shift ranges for different carbon types in CNMR spectra. carbonyl carbons (C=O) have a range of about 155-210 ppm, but the nitrile group (C≡N) has a signal range of about 110-140 ppm. I understand that oxygen is a more electronegative atom, so it might pull electron density away from the bonded carbon (deshielding), but the C≡N is a more polar bond, which means the electron density spends a lot of time around nitrogen in this bond. Doesn't this imply the C≡N carbon would produce a signal more downfield than that of the C=O carbon?
From the given data ranges, it seems the opposite is true and the carbon of nitrile is more shielded despite the bond polarity.

TL;DR: It's complicated and likely has to do with paramagnetic shielding effects rather than the usual arguments based on electron density, ring currents, and so on.
Firstly, because of low-lying π* orbitals, π-bonded carbons typically have lower excitation energies than purely σ-bonded carbons and are thus more deshielded. However, due to the presence of a π-bond order term, alkynes experience less deshielding due to paramagnetic effects. So the general order of chemical shifts is: alkenes > alkynes > alkanes.
Carbonyls are more deshielded than alkenes, and nitriles more deshielded than alkynes (both due to the paramagnetic contribution), but this doesn't overturn the existing bias of alkenes > alkynes.

It is instructive to start by comparing alkenes (~120 ppm) vs alkynes (~70 ppm). There is already a large difference between those two. Carbonyls are essentially deshielded alkenes, and nitriles essentially deshielded alkynes. And so it is no longer too surprising that carbonyls come in at a higher chemical shift than nitriles.
The only question left is to ask why sp2 carbons are generally more deshielded than sp carbons. Although it's tempting to just look purely at the hybridisation and s-percentage, this doesn't explain why the pattern is generally sp2 > sp > sp3. I should first point out that the physical factors controlling 13C shifts are (usually) markedly different from those controlling 1H shifts, so any text that tries to "carry over" those ideas is likely incomplete or incorrect. The reason for that is described in further detail here, and you can see Hans Reich's excellent website for more information about the various terms that contribute to the chemical shift.
With that out of the way, I will briefly quote from Gunther's NMR Spectroscopy, 3rd ed. (p 410):

For the paramagnetic contribution, early theoretical considerations [...] led to the expression:
$$\sigma_\mathrm{p}^i \approx -\frac{1}{\Delta E} \left(\frac{1}{r_i^3}\right)_\mathrm{2p_z} \left(\color{red}{Q_{ii}} + \sum_{j \neq i} Q_{ij}\right)$$
where $\Delta E$ is a mean electronic excitation energy [...]

I have coloured $Q_{ii}$ in red because it is not in Gunther's text (I suspect this is an error). I have added it to bring it in line with an article by Pople (Mol. Phys. 1964, 7 (4), 301–306).
Note first that the paramagnetic shielding is negative, so a larger magnitude corresponds to a less shielded nucleus, or a higher chemical shift. Now, the "excitation energy" for an alkane is probably a $\sigma \to \sigma^*$ excitation, which is quite large. On the other hand, for unsaturated compounds (alkenes / alkynes), there are low-lying $\pi^*$ orbitals, so $\Delta E$ becomes smaller and $\sigma_\mathrm p$ more negative.* This accounts for why unsaturated groups have higher chemical shifts than alkanes.
It also explains why carbonyls/nitriles are more deshielded than alkenes/alkynes respectively: because a heteroatom (O/N) has a high-energy lone pair, the primary excitation becomes a $n \to \pi^*$ transition which has a smaller $\Delta E$.

[...] $Q_{ij}$ is a bond order term that originates from the presence of $\pi$-bonds. [...] The bond order term superimposes additional changes that are most pronounced for [...] the $\ce{^{13}C}$ resonances of alkynes. For [alkynes], theory yields $\sum_{j\neq i} Q_{ij} = 0$ as in alkanes, whereas for [the central carbon of] allene $\sum_{j\neq i} Q_{ij} = 0.8$.

Gunther doesn't say exactly what this term would be for a plain old alkene (not allene), but Pople writes that it is 0.44. According to Pople, the $Q_{ii}$ term (in red) is equal to 2, so the overall term in parentheses is equal to 2 for alkynes and 2.44 for alkenes. This means that an alkyne carbon experiences a smaller (less negative) $\sigma_\mathrm p$, and consequently is less deshielded than an alkene carbon.

* Note also that $\pi \to \pi^*$ excitations are symmetry-forbidden in the context of paramagnetic shielding, so the primary excitation is probably a $\sigma \to \pi^*$ or $\pi \to \sigma^*$.

The nitrile is sp hybridized while the carbonyl is sp2 hybridized, meaning that the orbitals around the nitrile carbon have more s character, leading to more shielding.

I'm afraid it's not as simple as that... if this argument was all it took, then sp3 compounds like alkanes should be even more deshielded than sp2 compounds like alkenes/carbonyls. Shifts in unsaturated groups generally have more to do with other effects that aren't related to electron density. This is especially true for 13C shifts. see e.g. https://www.chem.wisc.edu/areas/reich/nmr/06-cmr-02-shifts.htm

The reason for sp2 compounds being more deshielded is due to the planar node. It is an exception to the rule.

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text:
Synthesis of Aspirin
In my lab experiment for the synthesis of Aspirin we used hot water to quench the reaction. Why wouldn't we use cold water but hot water?

You used an excess of acetic anhydride, and you want to hydrolize the unreacted fraction. The hydrolysis reaction proceeds faster at higher temperatures.

In a commercial environment and larger lab scale, quenching of water sensitive reagents at too cold a temperature can lead to accumulation and then thermal runaway. Phosphorous oxychloride is notorious for this. It's sometimes safer to quench at warmer temperatures for the reason The_Vinz said, but it can be a balance.


text:
Why does a reaction move both forward and backward in dynamic equilibrium?
As a kid I never imagined equilibrium to be a state where both forward and backward reactions keep taking place. Although, by seeing one cannot tell. Why is it that both forward and backward reactions move at a constant rate once equilibrium is achieved(dynamic, of course).
For instance, why at 273K ice molecules turn to liquid water and liquid water molecules “adhere” to ice? Why can’t both be happy in their respective states?

Why ? Because nature of matter at molecular and atomic level is dynamic, not static. In classical mechanics, objects can be in long term mutual rest. Not in quantum mechanics and quantum chemistry.
Molecules have zero knowledge about the system being in equilibrium or not. If the process is supported by the thermodynamic and kinetic aspects, nothing is going to stop it. It may be only counteracted by the opposite process, leading to the zero macroscopic net effect.
The molecules at given temperature do not share the same mechanical energy, but their  energy follows particular statistical distribution. Even if we set all  these energies the same at some moment by a magic wand, they would get redistributed again in order of nanoseconds.
For water and ice, some of molecules of ice have therefore high enough energy to leave the ice structure, while some molecules of liquid water have low enough energy to get trapped by the ice lattice. At melting point, both processes are balanced, if their is zero external heat transfer.
The dynamic bidirectional state of the dynamic equilibrium applies not only to chemical equilibriums, but also to thermal, radiative and quantum mechanical equilibriums.
E.g. 2 objects of the same temperature simultaneously provide and receive thermal energy from and to each other, with molecules a/o atoms continually exchanging kinetic energy.
Similarly, molecules of air, that seems to be in rest for us, are moving by speed of a supersonic fighter plane ( and some of them even faster ), mutually colliding in average 10 billion times per second, all the time redistributing their kinetic energy.

@Karl The speed of sound is not equal to either of the 3 mean speeds ( most probable, linear, quadratic ). By easy thinking one must realize the sound speed must be slower, as molecules move in random directions, while the sound wave propagates linearly. In the increasing order: The air sound speed is sqrt{ 1.4 RT/M ) The most probable molecule speed is sqrt{ 2RT/M ). The linear mean molecule speed is sqrt{ 8RT/(pi.M) ). The quadratic molecule speed is sqrt{ 3RT/M ).

I take back the "exactly". ;) a factor 1.5 slower than the the quadratic mean.


text:
Why should we be interested in catalyst for electrolysis?
I've been reading papers on Phosphided metals (CoP, NiP, FeP) as a catalyst for electrolysis.  They increase the current density of the cell for a voltage.  If it still requires 39.4 kilowatt-hours per kilogram of hydrogen (https://en.wikipedia.org/wiki/Electrolysis_of_water), why does it matter how fast the reaction occurs?
For example, with no catalyst, an electrolyzer runs at 0.6A/cm2 at 2.3V or with a catalyst it runs at 1A/cm2 at 1.4V.  Both cases require about the same amount of power and should produce the same amount of hydrogen, correct?

The two examples you provide do not produce the same amount of hydrogen. Electrons are "stoichometric reagents/products" which can be indirectly measured in coloumbs ($\mathrm{1\ mol\ e^- =96485\ C}$). A higher current means more starting material is being consumed (or more product is being formed) for a given amount of time (recall that $\mathrm{1\ A=1\ C/s}$). This means that your second example would theoretically produce hydrogen $\mathrm{1/0.6=1.67}$ times faster.  The lower voltage also decreases the chance of electrochemical side-reactions which would also consume/release electrons.

Catalysts generally decrease the electrode overpotential, which leads to decreasing the needed voltage to perform electrolysis at given current densities toward its minimal theoretical value.
If higher current is possible at the same voltage due the catalyst, the lower voltage is needed for the same current, both leading to better efficiency.
As the energy needed to produce 1 mol of molecular hydrogen is
$$E=2 \cdot U \cdot F$$
where
$U$ is the used voltage
$F$ is the Faraday's constant cca 96490 C/mol
As the consequence, it has direct impact on efficiency of conversion of electricity to hydrogen for hydrogen based engines or just fuel.


text:
Hartree-Fock vs. DFT vs. Hohenberg-Kohn vs. Kohn-Sham
What is the difference between these four approaches?
My current understanding is that Hartree-Fock uses only 1 slater-determine (accounting for antisymmetry) but neglects correlation between electrons. HF orbitals are primarily determined self-consistently.
DFT is the idea that there is a functional (presently unknown) that maps energy in terms of density. Hohenberg and Kohn found a proof that it exists but did not actually find it.
$E[p] = T[p] + E(EE)[P] + E(EN)[P]$
I am rather confused about where kohn-sham theory fits into all of this. Is it's entire purpose simply to define kinetic energy? How does it do this?

The Kohn-Sham formalism reintroduced the orbital approximation into DFT with a little twist. I recommend reading in a textbook about it, as the topic itself is a little too broad for this site.

@Martin-マーチン I do not have a textbook that covers this material because it is not always covered in foundational physical chemistry courses/textbooks. I do not have the funds to purchase an entire textbook for only one unit of material for a course. Any help would be very appreciated.

I understand the predicament. I was a student once and did not always have the funds for all books. Especially highly specialised are expensive. Talk to your librarian about the books that are available, that sometimes also are looking for recommendations to buy new stock. You won't learn the specifics of computational chemistry from a physical chemistry book.

@Martin-マーチン I really do not need specifics. We covered the very basic in our phys chem course.  He wants us to understand the "ingredients" in each. We do not need the nitty gritty details.

My current understanding is that Hartree-Fock uses only 1 slater-determine (accounting for antisymmetry) but neglects correlation between electrons. | Accounting for antisymmetry actually does account for some electron correlation. Not for all of it, but for some of it.

The Kohn-Sham approach is one on top of the original DFT idea proposed by Hohenberg-Kohn.
They propose – to find the energy functional and its equations, which, as you said, are unknown – to start with what is known: the functional if there is no interactions. So that is what they do: approximate the kinetic energy and Slater determinant from the known, non-interacting system and add all the terms that are additionally known (i.e. Coulomb, kinetic energies and the electron/nucleus potential) until you are left with the Kohn-Sham equation, which has "the smallest unknown", the exchange & correlation potential.
I recommend to read more in a book or on Wikipedia in case you want a more detailed picture on any/all of the methods.


text:
Basic thermodynamics question about work
I am getting into thermodynamics, and I have a basic fundamental question about the definition of the change in internal energy $\mathrm{d}U$:
$\mathrm{d}U = T\,\mathrm{d}S - P\,\mathrm{d}V + \sum\mu_i\,\mathrm{d}n_i$
Firstly, I do not understand why it's not the case that $\mathrm{d}W$ is defined as $P\,\mathrm{d}V + V\,\mathrm{d}P$, but only as $P\,\mathrm{d}V$. I can imagine a situation where the volume is fixed, but number of particles increases, thus increasing the pressure. I would think that this increase in pressure would also increase the internal energy.
Secondly, in the same manner I also don't understand why $\mathrm{d}Q$ is defined as $T\,\mathrm{d}S$ and not as $T\,\mathrm{d}S + S\,\mathrm{d}T$. I would think that a change in temperature would surely change the internal energy.

p.V is not W, so it's differential p.dV + V.dp is not dW. At constant V, no mechanical work is done. Similarly, T.S is not Q, so TdS + SdT is not dQ. dS is defined as dQ/T.

@Poutnik dS is not $\frac{\text{đq}}{T} $, it is $\frac{\text{đq}{rev}}{T}$   Jacob: work is force x distance which, for pV work, gives $đw = - p{ext}dV$.  If you choose a reversible path, then $p_{ext} = p_{sys} = p$ => $đw_{rev} = - pdV$

@theorist Sure. But it is implied in the above formulas as well as in my note. I have just kept symbols.

@Poutnik  I'm all for shorthand, when it's appropriate.  But the distinction between $q$ and $q_{rev}$ is central to thermodynamics, and not understanding that distinction is a key source of confusion for new students.  Thus while you may understand that distinction,  using $q$ for $q_{rev}$ when connecting entropy to heat flow is not only a big mistake, it also spreads confusion.

@theorist I agree, OTOH it would then require explanation of q Vs q_rev difference , what does not fit the comment scope.

Well, the answer is tricky. But before the really tricky part, you may want to take a step back and consider a general statement of the 1st law,
$$dU = \sum_i \left(\frac{\partial U}{\partial q_i} \right)_j dq_i$$
where $q_i$ is an extensive coordinate of the system. Note this is consistent with our standard description of the relation between generalized forces and energy (the derivatives wrt coordinates are forces).
In the case of pV work, $q_i = V$ and
$$ \left(\frac{\partial U}{\partial V} \right)_j = -p $$
The partial derivative in question is evaluated holding all other extensive variables (including number of particles) constant. If there is a change in chemical composition of the system we can define a chemical force (although referred to as the chemical potential of substance i) as
$$ \left(\frac{\partial U}{\partial n_i} \right)_j   = \mu_i  $$
and an infinitesimally small amount of chemical work as
$$ \left(\frac{\partial U}{\partial n_i} \right)_j dn_i  = \mu_i dn_i $$
As noted in the comments, the formulation combining the first and second laws of thermodynamics presented in the OP is a particular representation of the first law for the case of reversible pV work, a more general formulation of the 1st law being
$$\begin{align} dU &= dq + dw\end{align}$$
Note that while $dU$ is exact, the differentials on the right-hand-side of the previous equation may be inexact. Distinguishing between pV and other types of work for a reversible process (for which the differentials $dw$ and $dq$ become exact) we can write
$$\begin{align} dU &= dq_{rev} + dw_{pV,rev} + dw_{nonpV,rev}   \\ &= dq_{rev} + dw_{pV,rev} + \sum_i \mu_i dn_i   \\ &= TdS + dw_{pV,rev} + \sum_i \mu_i dn_i \end{align}$$
The second equality follows from the first using the 2nd law. Then since $dw=-p_{ext}dV$ in the case of pV work  and since $p_{ext} = p_{sys}=p$ when the work is carried out reversibly, you arrive at the expression in the OP.

I can imagine a situation where the volume is fixed, but number of particles increases, thus increasing the pressure. I would think that this increase in pressure would also increase the internal energy.

You have to be careful how you define the process that alters the number of particles. Do you have an open system that allows the particles in from the surroundings?
Assume you transfer an ideal gas at constant T from the surroundings into the system, which consists of a vessel at lower pressure with a small orifice which we seal at some point. Since the gas is ideal there is no change in energy with pressure at constant T. There is no mechanical work done, since there is no volume change, but the change in the number of particles clearly amounts to a transfer of energy. This is accounted for in the term $\sum_i \mu_i dn_i $ of $dU$, provided the entropy is constant.  Changing the number of particles amounts to "chemical" work.

Secondly, in the same manner I also don't understand why dQ is defined as TdS and not as TdS+SdT. I would think that a change in temperature would surely change the internal energy.

A change in temperature can change the internal energy, but that does not require a change in the definition of heat or entropy. The equation $dQ=TdS$ holds only for a reversible transfer of heat, it is not general.
Now to the tricky part.
You can write
$$dU = -pdV-Vdp +SdT + TdS + \sum_i \mu_i dn_i + \sum_i n_i d\mu_i  $$
since
$$U = -PV + TS + \sum_i n_i \mu_i$$
The thing is you can show that
$$ 0 = -Vdp +SdT + \sum_i n_i d\mu_i  $$
This is known as the Gibbs-Duhem relation.

In general this is a good and thorough answer.  However, the equation: $ dU = dq + dw + \sum_i \mu_i dn_i $ is incorrect. It is valid only if the change of state is reversible (where $dw_{rev} = -PdV$ is PV work, and $dq_{rev} = TdW$).  See, e.g., Denbigh 1981, p 81.  Remember that work is defined as change in the surroundings.  It can only be equated with system state functions for a reversible change.  Calling $\sum_i \mu_i dn_i $ "chemical work" is misleading.

@ratsalad I need to give your comments further thought but I currently disagree. You can write $dU = dq+dw_{pV} + dw_{non-pV}$. The non-pV part includes chemical changes - changes in free energy that might be associated with chemical potential, and that can be used to do non-pV work. Whatever happens to the remainder is secondary provided the sum $dw + dq$ is conserved, since dU is exact. I might need to present this differently though.

Your new equation is indeed valid.  The problem is the identification of $dw_{non-PV}$ with $\sum_i \mu_i dn_i$, which can only be made for a reversible process.  It's similar to how $dq$ can only be identified as $TdS$ for a reversible path.  In general, $TdS \geq dq$, and $dw_{rev} = -PdV + \sum_i \mu_i dn_i  \leq dw$.  In many practical cases $dq = \sum_i \mu_i dn_i$, like when you short-circuit the terminals of a car battery, $dw = 0$ and the energy in $\sum_i \mu_i dn_i$ is liberated completely as heat.

@ratsalad Thanks for your comment. I see that I convoluted a couple of steps trying to make a point about work associated with a chemical process, when I should be talking more generally about non-pV work. Hopefully it's a bit more clear.

A note on notation: since $q$ and $w$ are inexact differentials, it is incorrect to write "$dq$" or "$dw$". Proper notation is $\delta q$ or $\delta w$.

@Andrew Thanks, I'm personally used to the cross-bar notation but didn't get around to looking up the mathjax equivalent. Instead I use subscripts to indicate reversibility and be explicit. Introducing $\delta$ requires also an additional definition, and although explicit once defined it requires keeping track of more notation.

@Andrew That convention is a subjective choice that is not generally followed in the thermodynamic literature.  For instance, Atkins and Denbigh use $dx$  for inexact differentials.  In fact the distinction is primarily a convenience, a hint to the reader, and is not mathematically necessary, since both inexact and exact differentials are, fundamentally, differential forms (i.e. infinitesimal changes).   An exact differential is a special case that can be expressed as the differential of a function.  Whether a given differential is exact or inexact has to be proven based on assumptions.

An alternate convention is to use $DW$ and $DQ$ (capitalized) to signify inexactness.


text:
Why aren't the Anton computers used more widely?
The Anton computers are purpose-built by DE Shaw research for molecular dynamics simulations. They achieve performance up to an order of magnitude faster than can be achieved with commodity hardware, but don't seem to be used all that much by the wider computational chemistry community. This is despite the fact that limitations on simulation timescales is a major impediment to using simulations to guide experiments, leading to all kinds of intricate theoretical methods to deduce longer-timescale properties from relatively short simulation trajectories.
Having read a bit about the Anton machines, which have been around for over a decade now, I'm perplexed as to why they haven't been more widely used. Is it because they are too inflexible, not available to enough researchers, not helpful in solving scientific problems, or is there some other reason I'm missing?

Perhaps 10 times as much commodity hardware is cheaper?

And does DE Shaw even make the purpose-built chips available for purchase?

@theorist No, not as far as I'm aware, but there is at least one machine available for researchers to request time on: https://www.psc.edu/anton-rfp

Do you have evidence that the computers are not widely used? Can you quote some numbers?

According to this 2016 article,  "As far as anyone is aware, there is only one Anton machine outside of the company’s New York-based research lab—at the Pittsburgh Supercomputer Center (PSC)" (https://www.nextplatform.com/2016/02/04/anton-sequel-makes-stronger-case-for-custom-supercomputing/).  I also gather from the article that DESRES has kept the technology needed to make it close to the vest, so that's one key barrier.  A second possible barrier is cost, and we can't assess cost if DESRES has never sold the tech to anyone else (it was donated to PSC).

@BuckThorn They aren't, as theorist mentioned. This is a special variety of supercomputer that's purpose-based for MD. Some of top 500 supercomps are also used for this, but are way more versatile.

Since this question has been viewed 43 times but still has no answer, perhaps it might be something that a dedicated computational chemical physics community would be interested in. There is a proposal for a Stack Exchange for Materials Modeling for example: https://area51.stackexchange.com/proposals/122958/materials-modeling?referrer=ODVjZmM2YzJmOTc0MzBkZDVkMTZmMTIzN2EyNmZmNTQyOWYyY2Y0YWFmNGYxNTQ2OTQzNmMwOWIyYjhhOTZjMhwblBomlTMtPu8EgXeAKtp-8dFsxghojkD0vBSe1wZl0. Perhaps you might consider supporting it if you think it would be useful for you!

Why this is a valid question, I would assume it is more of a business/ market research problem, than a scientific. Also, one order of magnitude is not a big deal. GPUs easily jumped that speed up in the last decade, and GPU based workstations / clusters are much more available and versatile.

I myself am a person using MD simulation for my research but one reason I am not so much interested in ANTON is not it's cost, limited resource etc. It's rather because one extremely long simulation wouldn't help much at least for my research.
Protein dynamics occur in multiple different time scales and it's thermodynamics as well as kinetics all depend on free energy landscape which is a priori unknown. To directly compute free energy from simulation, you generally need multiple order of magnitude faster machine even than ANTON. However, there has been many techniques such as Umbrella Sampling that helps resolve this issue on computers in our lab.
I would appreciate it if I can run my Umbrella Sampling in ANTON but it would be so much waste of time to wait on the line if I can run the same simulation for a week in my workstation.
Another reason I can guess why ANTON has not been widely used is because of its specificity. ANTON does not run typical MD simulation packages such as GROMACS, NAMD, AMBER etc. All packages use same algorithm to run the simulation but the commands, file structures etc, are all different. Typical MD user would prefer to use package that he/she is familiar with. I am not sure how much I would have to study to get used to ANTON MD package but why bother if I am not desperate?
So in summary, only those users who are interested in actual trajectories of protein and water in millisecond time scale would want to use ANTON but not many others. Other than such purpose, there are many techniques and algorithms that would work better than running a single extremely long simulation.

Thank you very much for this answer, it's exactly the type of information I was looking for. Is Free Energy an experimentally measured quantity? Also, if I understand correctly, Umbrella sampling does not involve solving the equations of motion, but using a clever sampling trick to make sure the whole coordinate space is adequately covered, is that right?

Well, is free energy an experimentally measured quantity? I am not an expert on free energy measurement so I would advise you to google about it for more details. About Umbrella Sampling, it is quite the opposite. US DOES solve equation of motions. In general, it does so with some constraints to sample restricted coordinate space. You run multiple US over space of interest to obtain good samples on the space and combine those multiple data set properly. I mean properly, as properly re-weighting. There are bunch of papers and articles about US, so maybe you want to look it up in the web.

Anton and its successor Anton2 are highly specialized pieces of machinery. In fact, the computer, the chip, and the only code that can run on the chip are all called Anton. There's a reason why the only Anton computer outside of D.E. Shaw is located in the Pittsburgh Supercomputing Center (Anton now deprecated there, replaced by Anton2): this highly specialized equipment can't be used for calculations other than biological MD simulations. Combine this with the fact that Anton is at the cutting edge (and therefore well beyond what most supercomputing centers would deem the threshold for diminishing returns), and you have your answer.
Note: I am in no way bashing Anton. I had the pleasure of running on both Anton and Anton2. While they made huge strides in user-friendliness with Anton2, the reasoning stays the same: Anton was made for MD simulations and nothing else.

Thanks for your answer, but my question was more about why more groups don't use the Anton machines for their MD work. I suspect that if there had been a long waiting list for access to the PSC machine, Shaw would have made more of them available, and potentially set up a commercial operation? Or am I incorrect in that assumption? Reading between the lines of your response, is Anton [perceived as] too difficult to work with compared to Linux-based supercomputers? Or perhaps the simulation timescale increases are not enough to justify the switching cost?

There is high demand for Anton. However, given Anton's prohibitive cost and difficulty to maintain and fine-tune (and perhaps D.E. Shaw's hesitation to make it a widespread computer), the majority of supercomputing centers wouldn't be able to justify the cost of owning a machine such as Anton, given that general purpose supercomputers are usually sufficient for MD users (and can serve the needs of other users of the center, unlike Anton). Without being on the board, that's my best guess why the individual group demands for Anton do not translate into mass implementation.

To use it you would need access. Outside of D.E Shaw Research there is only 1 available ANTON, and you simply can't go out and buy one as far as I can see.
Looking further into the business relationships of David Shaw, it's clear that he is part of Relay Therapeutics (https://relaytx.com/our-team/david-e-shaw-ph-d/) and I wouldn't be surprised if there is some exclusivity in access to the ANTON technology for Relay.


text:
Is chromate a suitable indicator for the titration of Ag⁺ with Cl⁻?
To titrate $\ce{Cl-}$ with $\ce{Ag+}$ we use chromate $\ce{CrO4^2-}$ as an indicator.
The titration reaction is:
$$\ce{Ag+ + Cl- <=> AgCl}\tag{R1}$$
$$K_1 = \frac{1}{K_\mathrm{sp}(\ce{AgCl})} = \frac{1}{1.8×10^{-10}} = 5.56×10^9\tag{1}$$
The theory says that after all $\ce{Ag+}$ are reacted with $\ce{Cl-}$ the end point of titration is detected when excess $\ce{Ag+}$ reacts with the indicator chromate to form silver chromate:
$$\ce{2 Ag+ + CrO4^2- <=> Ag2CrO4}\tag{R2}$$
$$K_2 = \frac{1}{K_\mathrm{sp}(\ce{Ag2CrO4})} = \frac{1}{1.1×10^{-12}} = 9.1×10^{11}\tag{2}$$
However, as you see, $K_1 < 100K_2,$ so when both $\ce{Cl-}$ and $\ce{CrO4^2-}$ are present, $\ce{Ag+}$ will react with $\ce{CrO4^2-}$ and not with $\ce{Cl-}$.
But our teacher and everywhere on Google they say $\ce{AgCl}$ precipitates before $\ce{AgCrO4}$. And that should be true since this method of titration (Mohr's method) has been used long ago.
But, how can that be true? I don't understand why. Where have I mistaken?

You are ignoring the fact that the conc. of chloride is way too high than chromate. See this reference https://books.google.com/books?id=imBzjH6SHi8C&pg=PA196&dq=Mohr+titration&hl=en&newbks=1&newbks_redir=0&sa=X&ved=2ahUKEwipzPmA3afmAhUCS6wKHQKMCDAQ6AEwAHoECAAQAg#v=onepage&q=Mohr%20titration&f=false

I edited you question, chromate is $not$ a catalyst, it is an indicator in Mohr's titration.

Be aware you compare Ksp of a binary and a ternary product. In such case, lower Ksp of the latter does not automatically mean it is less soluble. Do calculations for real analysis and you will see.

@M farooq oh yes i meant indicator what a horrible mistake! Thank you

@Poutnik aha yes indeed how did i forget that! thank you!!

You got the solubility part reversed.
The solubility of $\ce{AgCl}$ is lower than the solubility of $\ce{Ag2CrO4}:$
$$s(\ce{AgCl}) = \sqrt{K_\mathrm{sp}(\ce{AgCl})} = \sqrt{\pu{1.8E-10 mol2 L-2}} = \pu{1.34E-5 mol L-1}$$
$$s(\ce{Ag2CrO4}) = \sqrt[3]{\frac{K_\mathrm{sp}(\ce{Ag2CrO4})}{4}} = \sqrt[3]{\frac{\pu{1.1E-12 mol3 L-3}}{4}} = \pu{6.50E-5 mol L-1}$$
Therefore, if the $\ce{AgNO3}$ solution is gradually added to the solution containing the both $\ce{Cl-}$ and $\ce{CrO4^2-}$ ions, then initially the formation of a sparingly soluble $\ce{AgCl}$ salt occurs.
After the $\ce{Cl-}$ ions are almost completely isolated in the form of $\ce{AgCl},$ the $\ce{Ag2CrO4}$ precipitation starts to occur, signifying the equivalence point is reached.
The same reasoning can also be applied to the titration of even less soluble silver bromide $\ce{AgBr}$ with $K_\mathrm{sp}(\ce{AgBr}) = \pu{5.3E-13}$ (try it yourself).
Note, however, that while using Mohr's method it's imperative to titrate halide salts solutions with $\ce{AgNO3}$ and not vice versa.
Otherwise the precipitation condition
$$c(\ce{Ag+}) · c(\ce{Cl-}) > K_\mathrm{sp}(\ce{AgCl})$$
will be overridden by
$$c(\ce{Ag+})^2 · c(\ce{CrO4^2-}) > K_\mathrm{sp}(\ce{Ag2CrO4})$$
due to high concentration of silver ions in solution, favoring silver chromate precipitation (note squared term $c(\ce{Ag+})^2$) and thus shifting the equivalence point.

Very good answer. For the OP, another alternative to Mohr's method is Vollhardt's method.


text:
What products do 4-nitrobenzaldehyde and 3-nitrobenzaldehyde give in a crossed Cannizzaro reaction and why?

What are the products of  reaction of 4-nitrobenzaldehyde with 3-nitrobenzaldehyde in the presence of $\ce{NaOH}$ when they show Cannizzaro reaction?

To my understanding, 4-nitrobenzaldehyde has more electrophilic carbonyl because the nitro group is an electron withdrawing group and at para position it exerts a very strong mesomeric effect, so it should form the carboxylic acid. To my surprise, I was told that it's the other way around.
Can someone explain me why this is true?

The first thing to point out is that the products of the Cannizzaro reaction do not necessarily correspond to the most stable products, i.e. the reaction is not necessarily thermodynamically controlled. Instead, one should consider the transition state—a six-membered ring as taught generally in organic chemistry—and note that after the reaction the compound which formed the hydrate will end up as the carboxylic acid and the one which did not form a hydrate is reduced to the alcohol. The transition state is shown in figure 1 below.

Figure 1: transition state of the Cannizzaro reaction
Thus, the underlying question is which compound is more likely to form an aldehyde hydrate. You argued with the nitro group of 4-nitrobenzaldehyde exhibiting a strong $-M$ (mesomeric) effect. However, this mesomeric effect cannot, in fact, cross the $\ce{C-C(=O)}$ bond and thus cannot influence the carbonyl group as much as the simple $-M$ label implies.

Figure 2: some mesomeric structures of 4-nitrobenzaldehyde, demonstrating that the mesomeric effect cannot ‘cross the $\ce{C-C}$ bond’.
Now it is known that aldehydes with a more electron deficient carbon are more likely to form hydrates, hence why we are doing a mesomeric/inductive discussion in the first place. But since we actually do not have any $-M$ arguments that can help our case, we need to turn to $-I$ arguments. When considering inductive effects, distance matters much more so 3-nitrobenzaldehyde experiences a stronger $-I$ effect than 4-nitrobenzaldehyde. Therefore, this compound is hydrated more favourably and form 3-nitrobenzoic acid while the reduced product should be 4-nitrobenzyl alcohol.

@Jan has provided a well-reasoned explanation for the OP's question regarding the Cannizzaro reaction. However, the transition state provided in Figure 1 is not in accord with the experimental results. In 1956 Hauser, et al.1 demonstrated that benzaldehyde-d1 (1-d1) in the presence of aqueous KOH efficiently affords benzyl alcohol-d2 (3-d2). On the other hand, benzaldehyde-1-d0 in the presence of sodium deuteroxide in deuterated hydroxylic solvent failed to incorporate deuterium into the resultant benzyl alcohol (3-d0). I have presumed that the mono-alkoxy hydrate is operable and that the sodium counterion chelates with the carbonyl oxygen. [Equations 3 & 4 are from the Hauser paper, loc. cit..]
[Note: @Jan has subsequently corrected Figure 1.]

1) C. R. Hauser, P. J. Hamrick, Jr. and A. T. Stewart, J. Org. Chem., 1956, 21, 260.


text:
Is it possible to remove sugar from a sweetened drink but without boiling it?
Is it possible to remove sugar from a sweetened drink without boiling it? This might not be a good question but I'm wondering if you could separate sugar from a sweetened drink and replace the sugar for an artificial sweetener.

Not by an easy, fast nor cheap way, and you would not probably want to drink it afterwards. Focus rather on non sweetened drinks.

Any drink specifically? Many commercial beverages have added sugar.

Basically no. There is no simple way of doing this. Much much easier to make the drink with artificial sweetener from the beginning.


text:
Is there a "Comprehensive Solid State" book?
We have Comprehensive Coordination Chemistry, Comprehensive Organometallics, Comprehensive Organic Synthesis and probably few other similar works that aim to do a wide review of recent works at the time of writing.
Is there a similar review book, that reviews modeling, characterization, preparation, structures and properties of crystalline, non-molecular solids, maybe as a part of a wider topic or a series on smaller subjects?

A long time ago at a university not that far away we used "The Electronic Structure And Chemistry Of Solids" by Tony Cox for the undergraduate course. It was a good book in its time but I suspect it will be a bit dated now - especially on the modelling side, this was when something with a 50MHz clock was considered a fast computer!

I think the closest comprehensive analogical literature source for solid state materials to the aforementioned series like Elsevier's Comprehensive Coordination Chemistry II, Elsevier's Comprehensive Organic Synthesis or Wiliey's Inorganic Syntheses would be Springer's Springer Series in Solid-State Sciences currently consisting of 223 titles:

The Springer Series in Solid-State Sciences consists of fundamental scientific books prepared by leading researchers in the field. They strive to communicate, in a systematic and comprehensive way, the basic principles as well as new developments in theoretical and experimental solid-sate physics.

As for the universally recommended textbooks on solid-state chemistry in general, there are the following titles issued for the past 15 years:

West, A. R. Solid State Chemistry and Its Applications, 2nd ed.; John Wiley & Sons, Inc: Chichester, West Sussex, 2014.
Simon, S. H. The Oxford Solid State Basics, 1st ed.; Oxford University Press: Oxford, 2013.
Smart, L.; Moore, E. Solid State Chemistry: An Introduction, 4th ed.; CRC Press: Boca Raton, FL, 2012.
Tilley, R. J. D. Understanding Solids: The Science of Materials; J. Wiley: Chichester, West Sussex, England; Hoboken, NJ, USA, 2004.
Müller, U. Anorganische Strukturchemie, 6., aktualisierte Aufl., unveränd. Nachdr.; Studium; Vieweg + Teubner: Wiesbaden, 2009. (in German)
Dronskowski, R. Computational Chemistry of Solid State Materials: A Guide for Materials Scientists, Chemists, Physicists and Others; Wiley-VCH: Weinheim, 2005.

Might be worth extending Resources for learning Chemistry. I'm always surprised Müller's book never was translated. However, given that he had a "particular" appreciation for English, it is not that surprising any more.

@Martin-マーチン Good point, I added added these six titles to the solid state chemistry section.


text:
What prevents inner-sphere electron transfer from oxalato-iron(II) to chloro-platinum(II)?
In this book (p. 237), when talking about the reduction of $\ce{[PtCl4]^{2-}}$ by $[\ce{(C2O4)2Fe]^{2-}}$, the author states that

There is no easy route for imparting electrons onto the $\ce{PtCl4}^{2-}$ ion, but it may proceed more readily via the aqua-complexes $\ce{(H2O)PtCl3-}$

and

The pathway for the electron transfer from iron(II) to platinum(II) has not been elucidated, but in view of the labile nature of iron complexes with respect to ligand substitution it is unlikely to proceed via an 'inner sphere' mechanism, whereby the two coordination shells share a common bridging ligand.

What prevents the inner-sphere mechanism in this case?
In Taube's original example, the metal with the bridging ligand is low-spin before, and the other metal, which takes the bridging ligand, is low-spin after. Can inner-sphere transfer only occur in those kinds of cases? Here, both the Fe(III) and Fe(II) are high-spin.


text:
What happens when these 3 chemicals are mixed in air?
What happens when these 3 chemicals are mixed in air?

Pentanone/Di-isopropyl Ketone (2,4-Dimethyl-3-Pentanone)
Iodine, reagent grade
Sodium Iodide, reagent grade.

The 3 parts make up a liquid in a tilt switch.  The tilt switch can leak giving off a very unpleasant smell.  Also corrosion can occur on the part.  What is the smell?  How dangerous is the smell?  Using a fume cupboard etc, what exactly are we emitting into the atmosphere?

There's a chance you're getting iodination of the ketones. The resulting α-iodoketones could be acrid/lachrymatory.


text:
C−H bond length in organic compounds
I am reading Fundamental concepts of organic chemistry by Dr. Anju Srivastava (12/04/2007).
It shows that carbon-hydrogen bond in ethane is 1.1 Å long, in ethene it's 1.08 Å,  and in ethyne it's 1.06 Å. According to the book, it's clear that carbon-hydrogen bond decreases with increasing multiple bonds on the carbon atom onto which the bond is!
According to my understanding, the shorter the covalent bond, the stronger it is, and it's contrary to the fact that the hydrogen atoms in ethylene are more acidic than those of ethene and so is ethane. The more acidic the hydrogen atom, the weaker it is held in a bond.
Help please where am wrong, and also guide about the book concept.

Bond strength (bond dissociation energy, BDE) refers to homolytic cleavage of a bond ($\ce{C-H}$ in this case). Heterolytic cleavage is measured by $\mathrm{p}K_\mathrm{a}$. The two processes are the numerical inverse of one another. In the case of acetylene with $\mathrm{sp}$-hybridization (50% $\mathrm{s}$-character) of carbon, the electron density of the $\ce{C-H}$ bond lies closer to carbon when compared to ethylene (33% $\mathrm{s}$-character) and ethane (25% $\mathrm{s}$-character). Thus in acetylene, it is more difficult to break the $\ce{C-H}$ bond homolytically because the carbon is "unwilling" to give up an electron but is happy to keep both electrons and give up a proton. Thus, acetylene has a high BDE ($\pu{133 kcal/mol}$) and lower $\mathrm{p}K_\mathrm{a}$ ($=25$) than ethylene ($\pu{110 kcal/mol}$, $\mathrm{p}K_\mathrm{a} = 44$) and, ultimately, ethane ($\pu{101 kcal/mol}$; $\mathrm{p}K_\mathrm{a} \approx 50$).


text:
Is there a relationship between acidity of the water solution and oxidation/reduction reactions?
What I want to know is if there is a relationship between the acidity of the water solution and redox reactions and if there is a relationship what favors the oxidation and what favors the reduction? Or does it depends on the reaction?

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. || The pH definitely has an influence on a redox reaction, but I don't think it can be generalised. So, the answer to your last question would be: yes.

related https://chemistry.stackexchange.com/questions/73350/is-there-any-difference-between-balancing-redox-reactions-in-acidic-and-basic-me

https://chemistry.stackexchange.com/questions/32509/pourbaix-diagram-writing-reactions

By medium, do you mean water solutions, or solvents themselves ( what includes other solvents than water, e.g acetic acid or liquid ammonia ) ?

@Poutnik water solutions

Fenton chemistry based on iron is definitely more reactive at lower $\mathrm{pH}$. This is due, in part, to speciation issues and associated solubility. Source: 'Effect of $\mathrm{pH}$ on Fenton and Fenton-like oxidation', by
Jung YS, et al., at https://www.ncbi.nlm.nih.gov/pubmed/19278159.
However, the use of chelates (like EDTA) can expand the $\mathrm{pH}$ range of reactivity (see full paper at https://www.jstage.jst.go.jp/article/jwet/11/1/11_21/_pdf ).
There are other transition metals, like Copper, which appear to be active in a much wider $\mathrm{pH}$ range where the chemistry is referred to as Fenton-type. There is an old french patent on water purification (from microbes,...) based on copper ions ($\ce{CuSO4}$ in small amounts) and natural waters containing dissolved oxygen. Interestingly, the addition of EDTA is still recommended, but now as a means to TERMINATE the redox reaction at near-neutral $\mathrm{pH}$.
There is a similar property for Cobalt ions, however, its toxicity has generally precluded it from commercial applications (like remediation of polluted waters through advanced oxidation processes referred to as AOP). Here is a reference on non-iron fenton processes at https://www.sciencedirect.com/science/article/abs/pii/S0304389414003239 and a more general review here, click on pdf download feature at https://www.researchgate.net/publication/322020986_A_Review_on_Advanced_Oxidation_Processes_for_Effective_Water_Treatment/link/5a4b0cd5458515f6b05b5221/download).


text:
What reaction could rubidium undergo in a vacuumed chamber with Macor ceramic and glass?
In the center of the chamber on the photo below you can see the hexagonally-shaped Macor  glass-ceramic consisting of

Silicon dioxide $\ce{SiO2}~46\%$
Magnesium dioxide $\ce{MgO}~17\%$
Aluminum oxide $\ce{Al2O3}~16\%$
Potassium oxide $\ce{K2O}~10\%$
Boron oxide $\ce{B2O}~7\%$
Fluorine $\ce{F}~4\%$

The original color of Macor is white as you can see on the second picture on the side. I insert the rubidium into the chamber. It all became black at the temperature of approx. $\pu{150–200 °C}.$ What could rubidium possibly react with?
It was in the vacuum $\pu{E-8 Torr}.$ I have heard that the window material can also react with rubidium. Rubidium is in vapor state (according to optical spectrum at temperature $~\pu{100 °C}$ and $\pu{E-4 Torr}).$ Rubidium tends to condense on the cold part of the chamber, so it is possible to touch the ceramic.
I heat the chamber outside, so rubidium should be in vapor state inside of the chamber, and Macor is connected to the chamber, so it should be close but lower than the chamber temperature.

Please explain ! Is the chamber in a vacuum ? Is the rubidium in the metallic state ? Does it touch the ceramic ? What do you heat ? The rubidium ? The ceramic ? Both ?

Was it heated in air?

@maurice Yes, it is in vacuum

@oscarLanzi it was in the vacuum

This is a nice question and an impressive setup. I took a liberty to improve the "storytelling" a little by minor tweaking the wording, notations and I also re-grouped some text blocks and illustrations so it is subjectively easier to follow.

One candidate for reaction with the rubidium is silica, $\ce{SiO2}$.  Reference [1] reports that black powders are formed when alkali metals are contacted with silica.  Such powders are more stable in dry air than ordinary alkali metals.  The reference suggests that electrons are imparted from the alkali metals into the silica forming an electride complex.
Reference

Shatnawi, M.G., Paglia, Gianluca, Dye, James, Cram, Kevin, Lefenfeld, Michael, Billinge, Simon, "Structures of Alkali Metals in Silica Gel Nanopores: New Materials for Chemical Reductions and Hydrogen Production", Journal of the American Chemical Society 129, 1386-92 (2007).  https://doi.org/10.1021/ja067140e


text:
Reduction of an Allylic Enoate to afford allylic alcohol in presence of Benzyl Acetal Groups
I have a question that requires finding a reagent for the reduction of an enoate to the corresponding allylic alcohol. Preferably done in one step too with an acid workup at the end too.
What types of reducing agents would work?
Would Lithal be too strong and reduce the acetal groups or would it be okay?

What is the alkyl group on the ester?

@Waylander A Methyl group.

The actual structure of your cmpd. would be handy.

@user55119 compound is included now, was unsure of how to attach it.

Thanks for the structure. There may be a stereochemical issue in the lab but not on paper ;)

Your benzyl acetal groups should be stable to all varieties of hydride reducing agents.
For the enoate reduction I suggest Dibal-H or LiBH4/Et2O.
I don't have references for this as I no longer have Scifinder access, and Google is not finding me any references that are not behind paywalls.

okay thank you very much, LiBH4 was my original thought so I might stick with that. I have included the compound in the original post above if that helps your answer too.

If you are happy with the answer please accept it, and come back and tell us if the reaction worked.

Unfortunately I won't be able to carry out the reaction, it was an Organic synthesis lab regarding a 4 step process to produce the above compound before our Christmas break. We have been given further thought-experiments to carry out for our report, such as this one. I very much appreciate the help however.


text:
Why does phenolphthalein form in this reaction?
This reaction (http://orgsyn.org/demo.aspx?prep=CV1P0476) gives quinizarin as the major product in about 70% yield.
However, when the same reaction is conducted with phenol insteas of p-chlorophenol, the product we get is not 1-hydroxyantraquinone (as would be expected if we only knew the above reaction), but phenolphthalein.
Why does phenolphtalein form instead of 1-hydroxyantraquinone? Would 1-hydroxyanthraquinone form if we added only 1 mol of phenol instead of 2 in the phenolphthalein synthesis?

In the Org. Syn. link you provided phthalic anhydride 1 and p-chlorophenol 2 in 95% H2SO4 and boric acid at 200oC undergo condensation and cyclization. Phthalic anhydride 1 under acid catalysis reacts with p-chlorophenol 2 at the ortho position as shown below to form ketoacid 3. Chlorine is a weak o,p-director and boric acid, I believe, is complexing with the phenolic hydroxyl and the ketone to reduce bond rotation and aid in cyclization to form anthroquinone 4. There is an aqueous work-up at this point followed by treatment with hot aqueous KOH which forms quinizarin 5 by nucleophilic aromatic substitution which is aided by the adjacent keto group.
In the reaction of phthalic anhydride 1 with phenol 7 in the presence of sulfuric acid, (presumably no boric acid), cyclization is repressed because of a lack of electron density at the phenol meta-position in 6. In addition, protonation of the ketone in 6 gives extended conjugation with the phenolic hydroxyl. Preferrable addition of phenol to the ketone of 6 leads to 8 and on to phenolphthalein 9.

from 4 to 5 , Cl para to hydroxyl group is undergoing nucleophillic aromatic substitution. Aryl halides are known to undergo nucleophillic aromatic substitution if nitro group is at ortho or para position. While this us not the case a mechanistic apparoch would help.

There is a ketone ortho to the Cl, which is the reason why the nucleophilic substitution takes place, or am I wrong?

@FusRoDah; You are correct! The base conditions are fairly vigorous.

@Chakravarthy Kalyan: The mechanism of OH for Cl substitution is straightforward. I'll entertain a better suggestion.


text:
Methods to increase the speed of dissolving potassium sulfate in water apart from heating, grinding or shaking
I bought potassium sulfate to be used as a fertilizer. The fertilizer comes in the form of small pebbles 1-2mm in diameter. Wikipedia states that the solubility is around 111g/liter at 20°C.
I tested several concentrations
100g / liter
50g  / liter
25g  / liter
at room temperature (22° C). After 2 hours there are still solids remaining in the solution. The water here is hard (EC= 0.6ms / cm) which hinders the dissolution. What can I do to improve the solubility of this fertilizer in water apart from heating, using reverse osmosis, grinding the pebbles or vigorous shaking of the solution (which worked) but is very tedious.

Did you buy chemically pure (lab grade) potassium sulfate to use as a fertilizer, or did you buy it as fertilizer in the first place? A fertilizer often contains additives such as phosphates that may be less soluble in water. Also, have you tried to warm up the solution?

The label says potassium sulfate. (K2SO4) I'm using it to blend my fertilizer. It's definitely not lab grade. Warming up the solution will probably work but is impractical as I want to dissolve a large amount.

I would try to warm up a little portion to see whether the residue is soluble at all and it's not some stone dust. Also, it might help if you add a photo of the label: there might be some useful information regarding the composition of the fertilizer.

Shaking does not increase the solubility, it merely speeds up the process of dissolution.

Since you're really asking about speeding dissolution, not increasing solubility, use a mixer; either:

From above, e.g., a drill attachment for mixing paint, or
From below, e.g., a magnetic mixer/stirrer.

The links are just as examples; you could devise your own mixer.

An ultrasound probe would help too, but would be expensive

Actually, inexpensive ultrasonic humidifiers might help, except the beam is focused to a pinpoint. Would it be powerful enough to help if diffused?


text:
How is entropy change defined for a chemical reaction?
I know that:
ds = dq/T for reversible processes.
Can I write change in entropy of a system in which a chemical reaction is happening using this equation. Basically, my question is about whether a chemical reaction happening in an open vessel(isobarically and isothermally) be treated as a reversible process?

Google van’t Hopf equilibrium box.

You can't treat it as reversible. Any spontaneous chemical reaction at constant $T$ and $P$ is necessarily irreversible.

@ChetMiller van't Hoff

Can I write change in entropy of a system in which a chemical reaction is happening using this equation. Basically, my question is about whether a chemical reaction happening in an open vessel(isobarically and isothermally) be treated as a reversible process?

If the Gibbs energy of reaction is non-zero and you don't capture it as work, you can't (whenever the total entropy increases, it is non-reversible).
$dS = dq/T$ is most useful for calculating the entropy change of a waterbath that is heated or cooled reversibly.
There are other ways you could measure the entropy, though. Entropy is a state function, so if you can find a reversible path (electrochemistry perhaps?), you could measure the entropy using the given relation.
In most cases, it is easier to measure the enthalpy and Gibbs energy change, and determine the entropy change from those measurements (see answer by Maurice).

The simplest way is to calculate  $\Delta$H and $\Delta$G = $\pu{-RTln K}$ (or $\pu{\Delta G = -zEF}$) of the reaction, then calculate the difference $\Delta$H - $\Delta$G. And you divide this difference by the temperature T. The result is $\Delta$S of the reaction.


text:
Clarification required on whether to use $C_v$ or $C_p$ in this question on finding entropy change

$1 \text{ mol}$ of an ideal monoatomic gas at $300 \text{ K}$ is subjected to a reversible isentropic compression process until final temperature reaches $600 \text{ K}$. If initial pressure is $1 \text{ atm}$. Then find $\ln (\text{P}_2)$.

My Attempt
$$\begin{aligned}\Delta S &=nC\ln\left(\dfrac{T_2}{T_1}\right)+nR\ln\left(\dfrac{P_1}{P_2}\right)=0 \\ & \boxed{\ln(P_2)= \ln(P_1)+C\ln\left(\dfrac{T_2}{T_1}\right)}\end{aligned}$$

Now if I take $C=C_v$, I get the answer $\ln(P_2)=1.5\ln(2)$, but the answer is given to be $2.5\ln(2)$ which will be obtained when we take $C=C_p$. So how to decide which one to take?

The correct form can be identified as follows. Start from the combined 1st and 2nd laws:
$$\begin{aligned} d S & = \frac{dU}{T} +  \frac{p}{T}dV \end{aligned}$$
For an ideal gas
$$\begin{aligned} d U &= nC_V dT \\ \frac{p}{T} &= \frac{nR}{V} \end{aligned}$$
such that
$$\begin{aligned} d S & = nC_V \frac{dT}{T} +  nR \frac{dV}{V} \end{aligned}$$
Integrating
$$\begin{aligned} \Delta S & = nC_V \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{V_2}{V_1} \right) \end{aligned}$$
Substitute in the ideal gas law in the right-hand side (note $n$ is constant)
$$\begin{aligned} \Delta S & = nC_V \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{P_1 T_2}{P_2 T_1} \right) \\ & = n(C_V +R) \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{P_1 }{P_2 } \right) \\ & = nC_p \ln \left( \frac{T_2}{T_1} \right) +  nR \ln \left( \frac{P_1 }{P_2 } \right) \end{aligned}$$

The fundamental equation of thermodynamics ignoring all work terms except PV-work is

$$dU = T dS - p dV$$

For an ideal gas, we know that $dU = n C_v dT $, i.e. an ideal gas has an internal energy that is dependent only on temperature.
For an isentropic process, we also know that $dS \approx 0$.

Combining these three items gives us
$$n C_v dT = - p dV$$
Using the ideal gas law $V = \frac{n R T}{p}$ again, we can substitute for $dV$
by taking the differential using the quotient rule.
$$dV = d(\frac{n R T}{p}) = n R d(\frac{T}{p})$$
Thus, $dV = n R \frac{p dT - T dp}{p^2}$
We can now substitute for $dV$ in the $- p dV$ term:
$$- p dV = -p n R \frac{p dT - T dp}{p^2} = \left (-dT + T\frac{dp}{p}\right ) nR$$
Now, substituting this into the $n C_v dT = - p dV = - dT + T\frac{dp}{p}$ equation and rearranging gives
$(n C_v + nR) \frac{dT}{T} = n R \frac{dp}{p}$
This is finally a differential equation that can be solved, with the solution being:
$$\frac{C_v + R}{R}\ln{\frac{T_2}{T_1}} = \ln{\frac{p_2}{p_1}}$$
Up until now, we haven't used the fact that the gas is monatomic, only that it is ideal.  If we use the fact that monatomic ideal gases have $C_v = \frac{3}{2}R$, giving
$$\frac{5}{2}\ln\frac{T_2}{T_1}=\ln\frac{p_2}{p_1}$$
Since $p_1$ is 1 atm, the equation shows that $\ln p_2 = \frac{5}{2}\ln\frac{T_2}{T_1}$


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Greater ligand stabilization of the lower oxidation state increasing redox potential
According to Atkins, P.W., De Paula, J. and Keeler, J., 2006, Atkins' Physical Chemistry,
$$
\begin{align}
\ce{[Fe(CN)6]^3- + e- &<=> [Fe(CN)6]^4-} &\quad E^\circ &= \pu{+0.36 V} \label{rxn:1}\tag{R1} \\
\ce{Fe^3+ + e- &<=> Fe^2+} &\quad E^\circ &= \pu{+0.77 V} \label{rxn:2}\tag{R2}
\end{align}
$$
According to my lecture notes, if a ligand stabilizes a lower oxidation state more than the higher oxidation state, the redox potential increases i.e. becomes more positive. This makes sense as stabilizing $\ce{Fe^2+}$ in reaction \eqref{rxn:1} would have the position of equilibrium to the right, increasing electron charge transfer and hence the redox potential, meaning a more positive $E^\circ$ and hence more negative $ΔG$ as
$$ΔG = -nFE^\circ$$
Indeed, $\ce{Fe^2+}$ is more stabilised than $\ce{Fe^3+}$ by $\ce{CN-}$ as $\ce{Fe^2+}$ is a borderline soft/hard acid, $\ce{Fe^3+}$ is a hard acid and $\ce{CN^-}$ is a soft base.
I don't understand why $E^\circ$ becomes more positive going from reaction \eqref{rxn:1} to reaction \eqref{rxn:2}. Please help me to understand why!


text:
Calculating equilibrium concentrations of silver and halide ions

$\pu{500 ml}$ of $\pu{0.01 M}$ $\ce{AgNO3}$ is mixed with $\pu{500 ml}$ of solution containing $\pu{0.1 M}$ $\ce{NaCl}$ and $\pu{0.1 M}$ $\ce{NaBr}$. $K_\mathrm{sp}$ of $\ce{AgCl}$ is $10^{-10}$ and $K_\mathrm{sp}$ of $\ce{AgBr}$ is $5×10^{-13}$. Find $[\ce{Ag+}],$ $[\ce{Cl-}],$ $[\ce{Br-}]$ in equilibrium solution.

After adding all of these molarities will get halved. At equilibrium condition, both $\ce{AgCl}$ and $\ce{AgBr}$ would have precipitated out. So, using the $K_\mathrm{sp}$ expression, we can find maximum amount of $\ce{Ag+}$ required to precipitate both the salts, and this becomes the equilibrium concentration of $\ce{Ag+}$.
Plugging these values in $K_\mathrm{sp}$ of respective salts, we can get the equilibrium concentration of remaining anions.
The $[\ce{Ag+}]$ for precipitation of $\ce{AgCl}$ is $\pu{2E-9},$ and for precipitation of $\ce{AgBr}$ is $\pu{E-11}.$
So, $\ce{AgCl}$ gets precipitated at last.
$$K_\mathrm{sp}(\ce{AgCl}) = [\ce{Ag+}][\ce{Cl-}]$$
Therefore, $[\ce{Cl-}]$ at equilibrium is $\pu{5E-2}.$
$$K_\mathrm{sp}(\ce{AgBr}) = [\ce{Ag+}][\ce{Br-}]$$
Therefore, $[\ce{Br-}]$ at equilibrium is $\pu{2.5E-4}.$
$[\ce{Ag+}]$ at equilibrium is $\pu{2E-9},$ which is already calculated in first step.
However, this doesn't match the answer key. Where am I wrong?

Two points: a) The ions in solution should be net neutral. b) With the putative equilibrium concentrations, no silver chloride would precipitate because the product of concentrations is lower than the solubility by a factor of 10.

You could solve this with an ICE table and two variables x and y, see https://chemistry.stackexchange.com/a/111151

Both chloride and bromide ions are present in 10-fold excess over silver ions. That means that the chloride and bromide concentration in solution will not drop by much (they will remain major species).
The solubility product of AgCl is 200-times higher than that of AgBr. If both AgCl and AgBr precipitate, the chloride solution would be 200-times higher than that of bromide. That is impossible because so little silver ions are available, and bromide and chloride start at equal concentrations. Thus, no silver chloride will precipitate (if this explanation is not convincing, suspend your disbelief until we can check the equilibrium concentrations).
Most of the silver ions will precipitate. As a first approximation, let's say the silver ion drops from 5 mM to zero, so the bromide concentration will drop from 50 mM to 45 mM. Of course, this can't be the equilibrium, so we can dissolve some of the AgBr again.
$$[\ce{Ag+}] = \frac{K}{[\ce{Br-}]} \approx \frac{\pu{1e-11}}{\pu{45e-3}} = \pu{2.2e-10}$$
So the concentration of silver ions is roughly $\pu{2.2e-10 M}$, that of bromide ions roughly $\pu{45 mM}$ and that of chloride ions $\pu{50 mM}$, with no silver chloride precipitating.

At equilibrium condition, both AgCl and AgBr would have precipitated out.

Now that we have solved the exercise, you can verify that AgCl will not precipitate. The reason is that the silver ion concentration is extremely small due to the presence of 45 mM bromide ions. In fact, if NaCl and AgNO3 had been mixed first, AgCl would have precipitated. However, adding NaBr would have dissolved it again eventually, leading to the same final state.

However, this doesn't match the answer key. Where am I wrong?

As mentioned in the comments, the set of final concentrations does not make sense from a stoichiometry (charge balance) perspective, and also contradicts the claim that AgCl is saturated. If you assume that both AgBr and AgCl have formed and attempt to solve the problem by brute force, the solutions to the two equations with two unknowns will only yield solutions that are not attainable (negative concentrations, or concentrations higher possible with the available starting materials).

The problem could be solved with simultaneous equations, but the following are the wrong equations.
$$K_{\mathrm{sp,}\ \ce{AgCl}} = [\ce{Ag+}][\ce{Cl-}]$$
$$K_{\mathrm{sp,}\ \ce{AgBr}} = [\ce{Ag+}][\ce{Br-}]$$
The right equations to use would be:
$$K_{\mathrm{sp,}\ \ce{AgCl}} \ge [\ce{Ag+}][\ce{Cl-}]$$
$$K_{\mathrm{sp,}\ \ce{AgBr}} \ge [\ce{Ag+}][\ce{Br-}]$$
Rather than solving the problem using "exact mathematics" via simultaneous equations, a little chemistry knowledge, and a knowledge of significant figures makes the problem solution easy as Karsten has pointed out so well.


text:
Will the temperature of a faucet actually be lower than the countertop or will it only feel cooler due to specific heat?
Due to specific heat capacity of metals, they feel cooler because they don't retain  as well as other materials. But is their temperature actually lower if it could be measured?

Well, they are in contact with water inside, which is typically cooler than air.

Temperature is a funny thing. It isn't really a real thing... it is more of a sensation. $T=\left( \frac{\partial E}{\partial S} \right)$.

@Charlie Crown Temperature IS a real thing. But perception of temperature depends on multiple factors, the real temperature of objects is just one of them.

@Poutnik Could you please clarify your comment- it seems confusing.

@suse See the appendix in the answer.

The short answer:
A metal object as the faucet drains heat from fingers due its high thermal conductivity faster than the countertop.
The fingers get therefore colder than if touching the countertop.
The faucet perceived temperature is therefore colder, even if the faucet and countertop temperature are the same.
If it is really colder due cold water, the perceived coldness is even stronger.
The longer answer:
Metals have high thermal conductivity.
That causes higher thermal flow at the same temperature difference than for most non metal materials of the same temperature. ( Rare exception are diamonds, as their thermal conductivity is superior to all metals.).
That leads to perception like if metals were colder ( if being colder then body temperature ) or warmer ( if being warmer ) than they really are.
OTOH, metal (mass-)specific heat capacity  is mostly lower than for non metals, due heavy metal atoms, that in fact somewhat decreases the effect of high thermal conductivity.
For your specific case, it depends also on the real metal temperature, e.g if it is partially sharing temperature with cold or warm tap water.
Similarly, water feels like colder than air of the same temperature, as thermal flow from human body is higher. But at this time, the higher volume specific thermal capacity plays major role.

2 examples how our senses can be deceiving:
If you had not too much hot herbal tea in an aluminium cup, you could still be sipping the tea itself without burning your tongue, but the cup would feel like burning your lips. If you did the same with the stainless steel cup, it would not be feel like burning as much, but still hot. If we used thermometer sensors, we would realize the metal surface not in direct contact with tea is very slightly colder than tea. But in both cases you would have impression the metal is warmer than the tea itself.
This time our senses monitor real thing ( temperature ), but rather their own temperature, instead of the temperature  of the sensed object.

Let have 3 basins with cold water, lukewarm water and warm water. You measure their temperature with thermometers, so it is a real thing. Now you put your left hand to cold water and your right hand to warm water for a minute. Then you put both hands to lukewarm water. The left hand will give you perception it is warm water, while the right hand will give you perception it is cold water.
This time our senses process not the temperature itself, but rather the temperature spacial and temporal gradient.


text:
Why cobalt reacts favorably with nitrogen?
I used also iron, nickel, zinc, copper. However, cobalt gives the easiest and straight forward yield of nitrides (XPS and SAED verified).
It is so easy to get cobalt nitride that I don't even have to use high purity nitrogen as I need with $\ce{Fe},$ $\ce{Ni},$ $\ce{Zn},$ $\ce{Cu}.$
Is there any reason for that? Is it because of back-bonding?

I think this property has to do with the ease with which cobalt makes complexes with $\ce{NH3}$ and amines. The variety and the stability of the complexes containing $\ce{Co^3+}$, surrounded by $\ce{NH3},$ ethylenediamine, $\ce{H2O},$ and/or $\ce{Cl},$ is unparalleled. Iron $\ce{Fe}$ does not make complexes with $\ce{NH3}$ under usual conditions. Nickel, copper and zinc do make one or two complexes with $\ce{NH3},$ but they are easily destroyed in acidic conditions. The variety and the stability of the cobalt complexes is unequaled.
The stability of the cobalt complexes is related to its electronic structure. The cobalt atom is argon + 9 electrons. $\ce{Co^3+}$ is argon + 6 electrons. 12 electrons are missing in $\ce{Co^3+}$ and needed to build up the electronic configuration of the next noble gas, krypton. If the $\ce{Co^3+}$ ion is surrounded by 6 ligands like $\ce{NH3},$ $\ce{Cl-}$ and $\ce{H2O},$ it "achieves its goal" of looking like a noble gas.
Cobalt may react with nitrogen to produce $\ce{CoN}$ made of $\ce{Co^3+}$ and $\ce{N^3-}$. And maybe this compound $\ce{CoN}$ is surrounded by six $\ce{N2}$ in the same way as $\ce{Co^3+}$ is surrounded by six ligands in the $\ce{Co^3+}$ complexes. This should be checked by X-ray analysis.

According to the recent first principles ab initio DFT calculations done with VASP of structural, electronic and mechanical properties of 3d transition metals mononitrides $\ce{MN}$ carried out by Liu et al. [1], there is a local maximum of cohesion energy $E_\mathrm{coh}$ at cobalt for all possible/discovered structure types:

Ultra-soft Vanderbilt pseudo-potentials (US-PP) … as supplied by Kresse and Hafner …, with local density approximation (LDA) … and PW91 general gradient approximation (GGA) … were both used. It is known that LDA underestimates the lattice constant and overestimates the elastic constants and cohesive energy, while PW91 GGA does the opposite for the lattice constant and elastic constants by a smaller amount and yields more accurate values of cohesive energy.

Figure 1.
Cohesive energy per atom of the nitrides $(\ce{MN})$ versus their corresponding 3d transition metals $(\ce{M}).$
The three structures are zincblende (zb, black square), rocksalt (rs, red circle) and
cesium chloride (cc, blue triangle).
Experimental values for the rocksalt structure (green ×) … are also presented.

Since we are discussing synthesis from the elements, it's especially handy to use cohesion energy since for solids it signifies the energy required to disassemble the structure into an array of neutral free atoms.
As a rule of thumb, the higher the atomic cohesion energy is, the more stable the solid is.
Numerical values of $E_\mathrm{coh}^\mathrm{GGA}$ for cobalt nitrides are always exceeding the ones for iron, nickel, copper and zinc, and also $\ce{CoN}$ demonstrates mechanical stability across all structure types:

There is still a lot of controversy between the studies of preferred phase and magnetism [2]; I'm not sure though it is currently possible to interpret/rationalize or let alone predict this behavior by other, simplified means.
References

Liu, Z. T. Y.; Zhou, X.; Khare, S. V.; Gall, D. Structural, Mechanical and Electronic Properties of 3d Transition Metal Nitrides in Cubic Zincblende, Rocksalt and Cesium Chloride Structures: A First-Principles Investigation. J. Phys.: Condens. Matter 2013, 26 (2), 025404. DOI: 10.1088/0953-8984/26/2/025404.
Soni, H. R.; Mankad, V.; Gupta, S. K.; Jha, P. K. A First Principles Calculations of Structural, Electronic, Magnetic and Dynamical Properties of Mononitrides $\ce{FeN}$ and $\ce{CoN}.$ Journal of Alloys and Compounds 2012, 522, 106–113. DOI: 10.1016/j.jallcom.2012.01.100.

Very very good explanation. How can I adequately reference you?

@SSimon I don't think I'm worth referencing, it's the authors of these two papers who did all the work. And thank you for the bounty!

Metal nitrides can be produced from the amide ($\ce{NH2−}$) ion as the $\ce{N3−}$ source (Metal nitrido complex).
Also, heating water vapor to high temperature:
$$\ce{H2O ->[\Delta] .H + .OH}$$
$$\ce{.H + .H -> H2}$$
In the presence of atomic nitrogen from heated nitrogen:
$$\ce{.N + H2 -> .NH2}$$
Cobalt is an active transition metal which can assume a variety of oxidation states which facilitates the release of electrons:
$$\ce{Co(K) -> Co(K + 1) + e-}$$
Finally,
$$\ce{.NH2 + e- -> NH2-}$$
creating the favorable amide ion which is a reported source for $\ce{N3−}$.

thank you, but why other similar d blocks don't behave same?


text:
Positively charged gases
Theoretically, if you could could have a positively charged gas say O$^+$ enclosed (by a sufficiently inert material) around a conductor which is grounded, what would the reaction be at the conductor if the conductor was made of;
1) Tungsten, 2) Stainless Steel,  3) Copper.
Would the gas gain electrons from the conductor and form O$_2$, react with the conductor to form a coating, react with each other, react with the container or something else?

So the $\rm O^+$ is in an inert container, and this container is in grounded tungsten (steel, etc)?

The O$^+$ is in an inert container with a grounded metal rod protruding from the bottom of the container in contact with the O$^+$. There is no way the O$^+$ can escape from the container as it is essentially sealed from the outside environment. All it can do is interact with itself, the metal grounded rod or the container walls.

Colleagues, please refrain from close votes. This is a genuine question which naturally arises in some students' minds every now and then. Sending them away will not help them realize why this question should not be asked.

Could you please elaborate how you got the idea for this question? It`s very, well, unusual, to say the least. ;)

You totally can't have a gas (or any other state of matter) made of positive ions alone.
That's not what you asked, though. The question was: what if we theoretically could have it. Well, then in a pretty short instant you will stop having it. The said instant would be rather violent, too(*).
To put things into perspective, imagine an indestructible spherical vessel about as big as to contain 1 mole of gas at normal conditions, which implies the radius of 17.5 cm. Now let's magically put 1 mole of $\ce{O+}$ inside. That would give it an electric charge of about 96500 coulombs, which is quite a lot. The electric field outside the sphere would be $$|\vec E\,|={1\over4\pi\varepsilon_0}\cdot{q\over r^2}=2.836\cdot10^{16}\;\rm V/m$$
Do you know how big is this? Wrong! it is bigger than you think. Mere $3\cdot10^6\;\rm V/m$ would cause a breakdown in dry air. Good solid insulators can handle more, but I think we may safely assume that $10^{10}$ is enough to break anything that is made of atoms. Mind you, we have six orders of magnitude on top of that(†). The conductor will be instantly vaporized and thrown out with immense force, enough to start many nuclear reactions and create a bunch of weird elementary particles. What metal it was made of would be the least of your concerns.
So it goes.

(*) Reminds me of a certain what-if episode, but they used a bigger sphere.
(†) Thought we would break the Schwinger limit, but we didn't.

Thanks for the answer. If there was only a limited number of positive ions where the electric field strength wouldn't break down the whole system, what would happen?

Then they will simply rush to the grounded conductor, neutralize, and form a layer of metal oxide. Some might recombine into O2 and bounce away.

It is similar to a calculation I sometimes get undergrads to do, consider a tennis ball sized lump of NaCl with a 1 % excess of sodium ions. How much energy would you have in the tennis ball sized lump of "special salt" ? It is a lot of energy, if I recall rightly.

Okay, so what if you only have 1e-14 moles?

Then the numbers will decrease proportionally, and you won't perish in the explosion. But it is pretty hard to handle that little of anything.


text:
Anhydrous Chromium Alum production
I recently started a small project of making home-made chromium alum crystals, as can be found on many sites, videos and recipes all around the internet.
I tried calculating the amount of salt I needed from wikipedia's page for the chromium alum, and then went nowhere, since the solubility data stated there does not say if it is for the dodecahydrate or anhydrous form. I tried it for the dodecahydrate and failed since no seed crystals formed after two days.
Then I added about 100% more alum to the solution while heated to about 80°C (not measured) and it all dissolved, and then after some hours some nice seed crystals formed.
After the initial seed crystal creation, I was left with a small amount of saturated solution and some extra seeds that I wouldn't use.
Just for curiosity sake, I then went to try to heat this solution to boil of the water and try to make some anhydrous chromium alum, or at least the solid, crystalline hydrated form.
Then, first, the solution took a goopy, molten plastic-like appearance, and then, after several hours, it went from a dark purple (hydrated?) clumpy mass to a greyish-green colored powder that didn't change at all even at the maximum temperature of my (bad quality) hot plate that hovers around 250°C.
I believed to have made anhydrous chromium alum (wikipedia stated that it is a greyish-brown powder), and then proceeded to take a small sample of this powder and re-hydrate it.
To my surprise, it didn't dissolve in distilled water at all! Having read something somewhere about some salts being quite soluble in water but dissolving very slowly, I then waited several days, stirring it occasionally, and nothing changed.
What happened to my chromium alum?
Wikipedia states a melting point of 89°C that could be related to the plastic-like melt observed, but also states a boiling point of 400°C that while I was never able to observe since my hot plate never reaches anything close to this temperature, should have left me with molten chrome alum. I know some compounds increase  in viscosity with temperature increase, so I wouldn't be surprised to not get  a flowing liquid.
But instead I got a greyish-green powder at around 250°C that is not water soluble.
Wikipedia also does not state decomposition temperature for this compound.


text:
Is Pinacol rearrangement the source for the ring expansion?

I have no idea what's happening as in the product (the answer is (a)) there is a new ring being formed on the left benzene. But as I studied after $\ce{H2O}$ leaves, a carbocation is then formed on stabilized by the right ring.
Can someone please explain the rest of the steps of this reaction? The thing I don't get is how does the three-membered ring opening happens.
I attempted the question, but the answer that I came up with isn't the one matching in the book. Where did I go wrong with the mechanism?

Here is a hint. Form the most stable cation, migrate a C-C bond and do a Nazarov cyclization (Google it). To the rest of the audience, allow aditya to work on a solution before posting an answer.

Thanks I tried to attempt the question as per your suggestion but I didn't came to the right conclusion can you have a look on the mechanism as I have added  it in the question.

Your first and third structures are good but the arrow pushing in the second structure doesn't get the job done. As you have noted, diol 1 forms carbocation 2a the more stable of the two options. Now the lefthand cyclopropane bond with its pair of electrons migrates to the cationic center to form resonance-stabilized species 2b and on to ketone 3. This type of structure is a candidate for a Nazarov cyclization. The arrows in 4 may be drawn in the other direction as well. Rearomatization of 5 by loss of a proton and tautomerization leads to the product 6. [Note: Structures 2b and 4 are identical. Just a reprotonation of 3.]ADDENDUM: To answer your original question, yes it is a pinacol rearrangement. The bond that migrates (2a --> 2b) just happens to be attached via the cyclopropane ring to the site to which it is migrating. So, 1 --> 3 is a pinacol rearrangement.

Ok I understood most of it but I don't really get the breaking of cyclopropane and it's bond migration as I dont think I have ever studied the bond shifting or migration if a carb-cation is formed on  a 3 membered cyclic ring can u please suggest some study material or something like that for it

All I ever read was if there is a carb-cation on a carbon atom attached to it it undergoes dancing resonance and is highly stable but never read something about bond breaking if carb-cation is form on it

aditya: I split structure 2 into 2a and 2b.. One could have migrated the lefthand cyclopropane bond to give the tertiary carbocation where the gem-dimethyl group is located. This species also leads to 3. Structure 2b is the better choice IMO.

wouldn't the carbocation be more stable near An? Could you please explain the step which leads to 2b?

Yes, but now the cyclopropane ring has a chance to collapse and remove strain. The An cation is more stable than the Ph cation but that doesn't mean that An can't react. For 2b move the lefthand cyclopropane bond to the right--follow the red arrow-- and now you have a resonance stabilized double bond. 2b either deprotonates to form 3 or, since 2b = 4, undergoes Nazarov cyclization.


text:
Is the heat for constant pressure the same as for constant volume in the formation of nitric oxide?
In the reaction $\ce{N2 (g) + O2(g) -> 2NO (g)}$
my textbook[1]  says $q_v$ and $q_p$ are not the same, without any explanation.
I know that
\begin{align}
\Delta U &= q_p + w\\
\Delta U &= q_p - P\Delta V\\
\Delta U &= q_p - Δn_\mathrm{gases}RT
\end{align}
I also know that
$$\Delta U = q_v.$$
because under constant volume, work done ($-P\Delta V$) is zero, since there's no change in volume.
In the reaction mentioned,
$$n_\mathrm{gases} = 0,$$
so shouldn't
$$\Delta U = q_p$$
and therefore
$$q_p = q_v?$$
Here's a picture of question 143 in Chapter 7 of [1]. The answers say a) and d) are false:

Ralph H. Petrucci, F. Geoffrey Herring, Jeffry D. Madura, Carey Bissonnette: General Chemistry: Principles and Modern Applications, 11th edition. Published by Pearson (February 23rd 2016)

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup (more also here). I have also included a real reference to the book you are using, I hope it is correct. States of aggregation should not be subscripted, it is not wrong, but the recommendations (Sec. 2.1.) are different.

Kevin is right. In that reaction, the number of moles does not change. So carrying it at constant pressure or at constant volume does not make any difference. And $q_v$ = $q_p$

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".


text:
Is there a difference between (HO)2 and H2O2?
To me, it looks like the same thing. Is (HO)2 and H2O2 the same compound? When performing stoichiometric calculations, do they get treated the same?

There is no difference, except that nobody writes it as (HO)2. But that's just a matter of convenience.

Relevant

(HO)2 is identical to H2O2, except in the cases where ions are formed in water from this substance. For example, if you are using H2O2 as a proton donor, you may consider the following reaction :
$$\ce{H2O2->H+ + HO2-}$$
but this reaction could not be obtained and written from the formula (HO)2. So H2O2 is a better choice, and nobody used (HO)2.


text:
Why are alkali metal compounds like sodium hydride or sodium amide strong bases but weak nucleophiles?
I understand why compounds such as $\ce{NaH}$ or $\ce{NaNH2}$ are weak nucleophiles: as they aren't very soluble in organic solvents, they react only on the clusters' surface. But why are they strong bases?
I could understand why they are in water, as they are soluble in water, but in my organic chemistry class it was said that they accelerate E2 reactions. So how can they be strong bases in organic solvents if they aren't soluble in organic solvents?

Huh, you provide somewhat surprising reasoning. I'd be interested how valid it. Major problem is they aren't exactly "soluble" in water - while  they "dissolve" they also react and that's driving the process forward. Another thing is that they aren't completely insoluble in organic solvents, at least in the ones they are actually used.

H- and NH2- are both strong nucleophile and strong bases but we know that acid base reaction are very favourable and fast due to change in Gibbs free energy is highly negative.
So they prefer acid base reactions first as their conjugate acids are H2 and NH3 which are very weak bases.

You're right they're difficult to use in organic reactions because of the solubility problem you mention. However, by using crown ethers like dibenzo-18-crown-6 they can be captured and the crown ethers are subsequently dissolved so that the bases are "brought into solution" and react.


text:
Pyrolysis of calcium chloride dihydrate
Does the pyrolysis of $\ce{CaCl2 · 2 H2O}$ give $\ce{CaCl2 + 2 H2O}$ or $\ce{Ca(OH)2 + 2 HCl}?$

What temperatures are we talking about?

Ordinary heating

I did synthesis of uranium tungstates for quite some time and was imprinted to me that "ordinary heating" is supposed to be 900–1100 °C. An organic chemist would probably understand "ordinary heating" as some elevated temperature around the boiling point of water. So, again, what temperatures are we talking about?

About the boiling point of H2O

AFAIK nothing will happen. Dihydrate only starts to decompose (aka give away water of crystallization) at 175 °C.

@andselisk Actually, I think I would consider ordinary heating reflux for whatever the solvent is. Which can go down to room temperature for ether or DCM ;)

All calcium chloride hydrates give away water of crystallization upon heating without forming a hydroxide, and the anhydrous $\ce{CaCl2}$ melts without decomposition [1, p. 162]:

anhydrous salts melts at 772 °C, while the mono-, di-, tetra- and hexahydrates decompose at 260 °C, 175 °C, 45.5 °C and 30 °C, respectively; the anhydrous salt vaporizes at 1935 °C.

$$\ce{CaCl2 · x H2O ->[\pu{260 °C}] CaCl2 + x H2O}$$
Calcium chloride dihydrate would remain intact until its melting/decomposition point at 175 °C.
When heated quickly in the absence of desiccant, the powdered salt is dissolved in its own water of crystallization forming glass-alike solid upon cooling.
References

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


text:
Why are ions much more efficient than electrons at beam-induced deposition?
I have recently been reading about electron beam induced deposition and ion beam deposition. The idea is to send a high energy (5-30 kV) beam of electrons/ions focused down to a spot on a material and open a gas precursor source. The beam somehow breaks down the bonds of the precursor gas, and some of the components of the gas get deposited on the sample. (e.g. carbon, silver,  platinum,...)
My question is, why does this deposition process work more quickly with ions than electrons with the same kinetic energy? From what I've seen in the literature, ions like Gallium can be 10-1000 tines faster at deposition than electrons at the same energy.

The stopping power (energy loss) is much higher for ions. See, for example, https://www.nist.gov/pml/stopping-power-range-tables-electrons-protons-and-helium-ions - a 10keV electron in silicon has a stopping power of about 17 MeV cm$^{2}$/g while a 10keV alpha has an electronic stopping of about 212 and a nuclear stopping of over a 1000 MeV cm$^{2}$/g. So, the ion is much more likely to deposit energy at or near the surface where it counts for deposition.

@Jon Custer, is there some intuition for the difference in stopping power?