text:
Analytical gradient of the Hartree-Fock nuclear-electronic repulsion term
The analytical gradient of the Hartree-Fock energy is given by (Szabo & Ostlund eq. C.16):
$$ \frac{\partial E}{\partial X_{A}} = \sum_{ij}{P_{ji} \frac{\partial H_{ij}}{\partial X_{A}}} + \frac{1}{2} \sum_{ijkl}{P_{ji}P_{kl} \frac{\partial (ij|kl)}{\partial X_{A}}} - \sum_{ij}{Q_{ji} \frac{\partial S_{ij}}{\partial X_{A}}} + \frac{\partial V_{NN}}{\partial X_{A}} $$
Of course, the Born-Oppenheimer approximation means that the nuclear repulsion term is easy to handle, and the derivative is given as:
$$ \frac{\partial V_{NN}}{\partial X_{A}} = Z_{A} \sum_{B} \frac{Z_{B} (X_{B} - X_{A})}{R_{AB}^{3}}$$
The remaining terms require computing the derivatives of the Gaussian functions (depending upon the molecular integral evaluation scheme used). However, in Szabo & Ostlund the nuclear-electronic attraction term is provided as:
$$ \frac{\partial V_{Ne}}{\partial X_{A}} = -Z_{A} \sum_{i} \frac{X_{i} - X_{A}}{r_{iA}^{3}}$$
In the Hartree-Fock equations, the molecular Hamiltonian is formed as $H = T + V_{Ne}$. Presuming that the gradient of the molecular Hamiltonian is formed the same way, how can I treat this formulation of the analytical gradient with only a scalar value for the nuclear attraction term, when I must take into account the density matrix? Is this solution to the nuclear attraction gradient of no use in calculating the gradient of the energy?

[1] A. Szabo and N. S. Ostlund, Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory, Dover Publications Inc., New York, pp. 441-442.

@TAR86 edited- only the first equation has a number in the text but I have included a page reference in the post.

I would note that the definition of $\partial X_A$ changes on page 442. I think that the latter equations are of operators, not energies. Further, I believe equations up to C.12 to be correct, because I once correctly implemented a toy gradient (using numerical derivatives for the integrals because my integral code is a joke) based on this chapter.

@TAR86 the definition of $\partial X_{A}$ doesn't actually change, it just refers to an arbitrary nuclear coordinate at first and then to the specific $x$ coordinate of nucleus $A$ on page 442: this is done as the derivative of the Gaussian function is given in terms of the $x$ coordinate and therefore the $l$ angular momentum term.

There are a few points to discuss:

Since there are $3N$ possible $\{X_{A}\}$, each term where an $X_{A}$ appears will result in $3N$ matrix elements. In your first equation, that will be $S$, $T$, $V$, and the 2-electron contribution, which should really be rewritten into a Fock matrix term $F$ where the ket is precontracted with the density.
There is nothing wrong with this solution, though it is a little strange because it's in the MO basis, and we normally prefer to work in the AO basis to avoid any unnecessary transformations. I'm going to drop the $Ne$ from here on out since $V$ is understood in this context to only be the one-electron nuclear attraction term.
I think you're asking what happens to the density matrix. The derivative of $V$ expands into multiple parts, as we will see below. I will reference a bunch of equations that use $h$, where $\left( h = H = H^{\text{core}} \right) \equiv T_{e} + V_{Ne}$, and differentiating into a sum should hopefully be clear. What follows should work for any real-valued one-electron operator. I will use the book by Yamaguchi with liberties taken. In many places I use $p,q$ rather than $i,j$, as these indices run over all MOs, not just occupied ones.

Expanding $\frac{\partial V_{ij}}{\partial X_{A}}$ using the product rule gives
$$
\begin{align*}
\frac{\partial V_{ij}}{\partial X_{A}} &= \frac{\partial}{\partial X_{A}} \left( \sum_{\mu\nu}^{\text{AO}} C_{\mu i} C_{\nu j} V_{\mu\nu} \right) \tag{3.80} \\
&= \sum_{\mu\nu}^{\text{AO}} \left( \frac{\partial C_{\mu i}}{\partial X_{A}} C_{\nu j} V_{\mu\nu} + C_{\mu i} \frac{\partial C_{\nu j}}{\partial X_{A}} V_{\mu\nu} + C_{\mu i} C_{\nu j} \frac{\partial V_{\mu\nu}}{\partial X_{A}} \right), \label{3.81}\tag{3.81}
\end{align*}
$$
where the third/last term is the true AO integral derivative, and the first two terms, the MO coefficient derivatives, come from differentiating the density matrix.

The simplest term to write out, but the most difficult to implement, is the AO integral derivative. I will use $\mu,\nu$ rather than $\chi_{\mu},\chi_{\nu}$, so they refer to both AO basis functions and their matrix indices.
$$
\begin{align*}
\frac{\partial V_{\mu\nu}}{\partial X_{A}} &= \frac{\partial}{\partial X_{A}} \left< \mu | \hat{V} | \nu \right> \tag{3.24} \\
&= \left< \frac{\partial \mu}{\partial X_{A}} | \hat{V} | \nu \right> + \left< \mu | \frac{\partial \hat{V}}{\partial X_{A}} | \nu \right> + \left< \mu | \hat{V} | \frac{\partial \nu}{\partial X_{A}} \right> \label{3.25}\tag{3.25}
\end{align*}
$$
What you have presented,
$$
\frac{\partial \hat{V}}{\partial X_{A}} = -Z_{A} \sum_{i} \frac{X_{i} - X_{A}}{r_{iA}^{3}},
$$
is only the derivative of the operator. The basis function derivatives are also required, as the AOs in this case depend on the nuclear positions. Contrast this with an electric field perturbation, where only the derivative of the operator is required, or magnetic field derivatives, where basis functions may be perturbation-dependent if gauge-including atomic orbitals (GIAOs) are used. Also important is that the index $i$ here refers to an electron, not an occupied MO. This is an electric field integral.

Now for the derivative of the MO coefficients/density matrix. Clearly they don't appear in the final HF gradient expression. They disappear through the magic of Wigner's $2n + 1$ rule. From page 25:

When the wavefunction is determined up to the $n$th order, the expectation value (electronic energy) of the the system is resolved, according to the results of perturbation theory, up to the $(2n+1)$st order. This principle is called Wigner's $2n+1$ theorem [29-31].

More explicitly, we have the zeroth-order wavefunction, so we must be able to calculate the first-order correction to the energy. Worded differently, any first derivative of the energy must be easily calculated without differentiating MO coefficients, which is only required for second derivatives, such as the molecular Hessian or the dipole polarizability.
First, rewrite the MO coefficient derivatives
$$
\frac{\partial C_{\mu i}}{\partial X_{A}} = \sum_{m}^{\text{MO}} U_{mi}^{X_{A}} C_{\mu m}, \tag{3.7}
$$
where the index $m$ runs over all occupied and unoccupied/virtual MOs. The same goes for $p,q$. I've replaced $a$ from the text with $X_{A}$, but this holds for any general perturbation (such as $\lambda$, pick your favorite unused index). The key insight is that we can write the effect of a perturbation on the MO coefficients as the contraction of the unmodified MO coeffcients with a unitary matrix describing single-particle excitations from occupied to virtual MOs, as well as deexcitations from virtual to occupied MOs. In matrix form, this is
$$
\mathbf{C}^{(X_{A})} = \mathbf{C}^{(0)} \left( \mathbf{U}^{(X_{A})} \right)^{T},
$$
where it is usually easiest to have the dimension of $\mathbf{U}$ be $[N_{\text{orb}}, N_{\text{orb}}]$, with only the occ-virt and virt-occ blocks being non-zero. The problem now becomes solving for $\mathbf{U}$ and eliminating it from the final gradient expression. I will walk through parts of section 4.3 to show how this is done.
Given the energy expression for an RHF wavefunction,
$$
E = 2 \sum_{i}^{\text{d.o.}} h_{ii} + \sum_{ij}^{\text{d.o.}} \left[ 2(ii|jj) - (ij|ij) \right], \tag{4.1}
$$
the first derivative with respect to $X_{A}$ is
$$
\frac{\partial E_{\text{elec}}}{\partial X_{A}} = 2 \sum_{i}^{\text{d.o.}} h_{ii}^{X_{A}} + \sum_{ij}^{\text{d.o.}} \left[ 2(ii|jj)^{X_{A}} - (ij|ij)^{X_{A}} \right] + 4 \sum_{m}^{\text{all}} \sum_{i}^{\text{d.o.}} U_{mi}^{X_{A}} F_{im}, \label{4.16}\tag{4.16}
$$
where the Fock matrix is defined as
$$
\begin{align*}
F_{pq} &= h_{pq} + \sum_{k}^{\text{d.o.}} \left[ 2(pq|kk) - (pk|qk) \right] \tag{4.6} \\
&= h_{pq} + 2J_{pq} - K_{pq},
\end{align*}
$$
and I've introduced the Coulomb and exchange matrices $\mathbf{J}$ and $\mathbf{K}$ as well. I skipped all the steps in expanding the first derivative, and $\eqref{4.16}$ is what results upon collecting terms with $\mathbf{U}$.
Using the RHF variational conditions, the Fock matrix from a converged calculation is diagonal in the MO basis, corresponding to the MO energies
$$
F_{pq} = \delta_{pq} \epsilon_{pq}, \tag{4.7}
$$
so $\eqref{4.16}$ simplifies to
$$
\frac{\partial E_{\text{elec}}}{\partial X_{A}} = 2 \sum_{i}^{\text{d.o.}} h_{ii}^{X_{A}} + \sum_{ij}^{\text{d.o.}} \left[ 2(ii|jj)^{X_{A}} - (ij|ij)^{X_{A}} \right] + 4 \sum_{m}^{\text{all}} \sum_{i}^{\text{d.o.}} U_{mi}^{X_{A}} \epsilon_{im}, \tag{4.17}
$$
which can be further simplified as
$$
\frac{\partial E_{\text{elec}}}{\partial X_{A}} = 2 \sum_{i}^{\text{d.o.}} h_{ii}^{X_{A}} + \sum_{ij}^{\text{d.o.}} \left[ 2(ii|jj)^{X_{A}} - (ij|ij)^{X_{A}} \right] + 4 \sum_{i}^{\text{d.o.}} U_{ii}^{X_{A}} \epsilon_{ii}. \tag{actual 4.17}
$$
Now we use one of the most important tricks in quantum chemistry. Given the orthonormality of the MOs,
$$
S_{pq} = \delta_{pq}, \tag{3.44}
$$
we must have
$$
\frac{\partial S_{pq}}{\partial X_{A}} \overset{!}{=} 0. \tag{3.45}
$$
This is where not using general notation in the above is a bit confusing because it seems to conflict with your original expression, but remember that similar to $\eqref{3.81}/\eqref{3.25}$, this is in fact multiple terms: two for the basis functions, and two for the MO coefficients, giving
$$
\begin{align*}
\frac{\partial S_{pq}}{\partial X_{A}} &= \sum_{\mu\nu}^{\text{AO}} C_{\mu p} C_{\mu q} \frac{\partial S_{\mu\nu}}{\partial X_{A}} + \sum_{m}^{\text{all}} \left( U_{mp}^{X_{A}} S_{mq} + U_{mq}^{X_{A}} S_{pm} \right) \tag{3.40 + 3.43} \\
                                       &= S_{pq}^{X_{A}} + \sum_{m}^{\text{all}} \left( U_{mp}^{X_{A}} S_{mq} + U_{mq}^{X_{A}} S_{pm} \right). \tag{3.43}
\end{align*}
$$
The sum over all MOs can be eliminated by reusing the orthonormality condition, so in the first term $m \overset{!}{=} q$ and for the second term $m \overset{!}{=} p$, and the overlap matrix in the MO basis is unity for those terms, giving
$$
\frac{\partial S_{pq}}{\partial X_{A}} = S_{pq}^{X_{A}} + U_{qp}^{X_{A}} + U_{pq}^{X_{A}} \overset{!}{=} 0. \tag{3.46}
$$
Recognizing that we only need diagonal terms, this can be rewritten as
$$
U_{pp}^{X_{A}} = -\frac{1}{2} S_{pp}^{X_{A}}, \tag{4.20}
$$
which is then plugged back into the first derivative expression to give
$$
\begin{align*}
\frac{\partial E_{\text{elec}}}{\partial X_{A}} &= 2 \sum_{i}^{\text{d.o.}} h_{ii}^{X_{A}} + \sum_{ij}^{\text{d.o.}} \left[ 2(ii|jj)^{X_{A}} - (ij|ij)^{X_{A}} \right] + 4 \sum_{i}^{\text{d.o.}} \left( -\frac{1}{2} S_{ii}^{X_{A}} \right) \epsilon_{ii} \\
                                                &= 2 \sum_{i}^{\text{d.o.}} h_{ii}^{X_{A}} + \sum_{ij}^{\text{d.o.}} \left[ 2(ii|jj)^{X_{A}} - (ij|ij)^{X_{A}} \right] - 2 \sum_{i}^{\text{d.o.}} S_{ii}^{X_{A}} \epsilon_{ii}. \tag{4.21}
\end{align*}
$$
Rewrite the last term
$$
\begin{align*}
\sum_{i}^{\text{d.o.}} S_{ii}^{X_{A}} \epsilon_{ii} &= \sum_{i}^{\text{d.o.}} \sum_{\mu\nu}^{\text{AO}} C_{\mu i} C_{\mu i} \frac{\partial S_{\mu\nu}}{\partial X_{A}} \epsilon_{ii} \\
                                                    &= \sum_{i}^{\text{d.o.}} \sum_{\mu\nu}^{\text{AO}} C_{\mu i} C_{\mu i} \epsilon_{ii} \frac{\partial S_{\mu\nu}}{\partial X_{A}} \\
                                                    &= \sum_{\mu\nu}^{\text{AO}} W_{\mu\nu} \frac{\partial S_{\mu\nu}}{\partial X_{A}} \tag{4.24}
\end{align*}
$$
to use the energy-weighted density matrix $\mathbf{W}$, which in your expression is called $\mathbf{Q}$.
Again, the elimination of the $\mathbf{U}$ matrix is one of the most important results in quantum chemistry, as it means the coupled-perturbed SCF equations do not need to be solved for first derivatives of SCF wavefunctions. This is why you do not see any density or MO coefficient derivatives in the gradient expression.
References

Yamaguchi, Yukio; Goddard, John D.; Osamura, Yoshihiro; Schaefer III, Henry F. A New Dimension to Quantum Chemistry: Analytic Derivative Methods in Ab Initio Molecular Electronic Structure Theory (International Series of Monographs on Chemistry); Oxford University Press: 1994.
Epstein, Saul T. General Remainder Theorem. J. Chem. Phys. 1968, 48, 4725.
Epstein, Saul T. Constraints and the $V^{2n+1}$ theorem. Chem. Phys. Lett. 1980, 70, 311.

If you don't have access to the Yamaguchi book, the seminal article on HF derivatives contains a condensed version of this derivation, but be aware of typos.

Pople, John A.; Raghavachari, Krishnan; Schlegel, H. Bernhard (Berny); Binkley, J. Stephen. Derivative Studies in Hartree-Fock and Møller-Plesset Theories. Int. J. Quantum Chem. 1979, 16 (num. S13), 225-241.

The book doesn't explicitly list all the terms you need to evaluate $\frac{\partial H_{\mu\nu}^{core}}{\partial X_A}$ (which I'll abbreviate as $\partial H_{\mu\nu}^{core})$.
$H_{\mu\nu}^{core} = \langle \mu |\hat{T} + \hat{V}_{eN}| \nu \rangle$ so
$\partial H_{\mu\nu}^{core} = \langle \partial \mu |\hat{T} + \hat{V}_{eN}| \nu \rangle + \langle \mu | \partial \hat{V}_{eN}| \nu \rangle + \langle \mu |\hat{T} + \hat{V}_{eN}| \partial \nu \rangle$
where $\partial \hat{V}_{eN} = - Z_A \sum_i \frac{X_i-X_A}{r^3_{iA}}$


text:
Why is mer-AlQ3 more stable than fac-Alq3?
Several research papers and books say that the meridional isomer of tris-(8-hydroxyquinoline) aluminium is thermodynamically more stable than the facial isomer.
What is the reason behind this stability?

The stability of meridional isomer over facial isomer can be deduced by considering the interactions between atoms of the ligand 8-hydroxyquinoline.
 In considering the interactions if we take the repulsion between the electron rich atoms dominated in 90 degrees and at 180 degrees these interactions become negligible, we will find that in meridional isomer, there are dominating
i) 2 N-N interactions
ii) 2 O-O interactions
iii) 8 N-O interactions.
 But if we look into the facial isomer of tris(8-hydroxyquinolinato)aluminium we will find there are dominating
i) 3 N-N interactions
ii) 3 O-O interactions
iii) 6 N-O interactions.
 So in facial isomer two N-O interactions of mer isomer has turned into 1 N-N interaction and one O-O interaction. Due to high charge density on small sized O atoms O-O repulsions are much much greater than N-O interactions and thus the electronic repulsions are much more greater in fac isomer which decreases its stability than meridional isomer

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text:
Synthesis of Hydrogen Fluoride
At almost 300°C, $\ce{CaF2}$ reacts with sulfuric acid and this reaction produces $\ce{HF}$. Is there any way to make a relatively dilute (10-20%) hydrofluoric acid with $\ce{CaF2}$ as the source of fluoride ions without that much amount of heat?

CaF2 is already source of F- anions.

But the synthesis described above is too dangerous to be safe to do in my lab. Of course I know that $CaF_{2}$ is a source of $F^{-}$ ions.

Well, if you really need acid, then better buy it.

But I do not need even 100ml of a 5% solution of it and I do not want to store that much of such a nasty chemical which I won't need anymore

Hmm, then you might borrow a bit ;) Well what you mention is rather industrial method, but it shouldn't be hard to find lab version.

Well If I find or experiment a safer method for directly producing the dilute acid I will post it here as an answer :)

HF is pretty nasty stuff... can you substitute something such as ammonium acid fluoride, NH4HF2 (which is almost as bad, admittedly)?

In desperation, for some unknown reason, I need 15% aqueous $\ce{HF}$. What do I really need? A weak solution of $\ce{H+}$, say about pH 2. And a solution of fluoride ions to wreak havoc on some unsuspecting metal or glass(Ooops, I may have given away the secret).
My subconscious suggests that I could get them from two different sources, like $\ce{NaF}$ and $\ce{HCl}$. Yeah, that'll work.
Maybe. Maybe even with $\ce{CaF2}$(The use of $\ce{H2SO4}$ with the calcium salt suggests the reason was to precipitate the $\ce{CaSO4}$ and flash off the $\ce{HF}$, thereby separating, not just generating, the $\ce{HF}$).

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text:
What does "sine" mean?
I see the suffix "sine" (seen/sin) a lot, adenosine, cytosine, lysine, tyrosine, etc. Most of where I hear it is in amino acid R groups, but it's usually only the prefix that is recognized as significant.

Actual suffix is just -ine and it's standard. "s" comes from various (typically greek) words, like tyros - cheese.

Comparing adenine/adenosine and cytosine/cytidine couldn't the argument be made that the suffix was -osine for these two?

According to Textbook of metabolism and metabolic disorders (1964):

The ribosides derived from purines have the suffix -osine; those from pyrimidines, the suffix -idine. The corresponding deoxyribosides, with the exception of thymidine, do not have any such simple designation. Occasionally, instead of, for example, guanine-deoxyriboside, the term deoxy-guanosine is used.

Beyond the more common adenosine and guanosine, there are also:
inosine  and
xanthosine

That's a pretty old reference, but what about lysine? Or, what is lys?

@JohnJoe I picked an old reference intentionally, because I am interested in when the convention started.  A 2016 book says (page 453)  "For purine bases, the suffix -osine is used (adenosine, guanosine)" https://books.google.com/books?id=HRCdBQAAQBAJ&pg=PA453&dq=%22suffix+-osine%22&hl=en&sa=X&ved=0ahUKEwitm7HmlIfZAhVku1kKHZlEA-sQ6AEIKTAA#v=onepage&q=%22suffix%20-osine%22&f=false "Lysine" has an unrelated word origin http://bioetymology.blogspot.com/2012/03/lysine.html

@JohnJoe see also: https://en.wikipedia.org/wiki/-ine


text:
How does atomization allow droplets to evaporate below their boiling point, cooling the surrounding vapor to an even lower temperature?
I understand why the vapor is cooled in such a scenario, because of the heat required to change states, but I don't fully understand why the state change even occurs if below the BP. Why does it seem that in many cases, atomization almost forces evaporation, minimizing the role of temperature?
I understand that air can naturally hold a certain amount of water, but I've never learned how this "humidity" concept applies to other solvents. Is there some kind of formula I can use to estimate how much of a particular solvent a particular volume of air can hold at room temperature, without the solvent being at boiling point? Or really any sort of math model to understand how atomization assists in evaporation below boiling point?

Atomisation just increases surface, it's enough.

@Mithoron - it also introduces a (large-ish) surface energy term through the curvature. So, evaporation is favored if the radius is below the nucleation size.

Every gaseous phase that is in contact with a liquid phase contains some vapor from the liquid phase.
If the liquid phase is pure and liquid and gaseous phase are in equilibrium, the partial pressure of the liquid is called vapor pressure.
As it turns out the vapor pressure not only depends on temperature but also on the shape of the surface. The Kelvin equation shows that the vapor pressure for small droplets is greater than for larger droplets or plain surfaces.
So atomized liquids have higher vapor pressure, and more liquid can evaporate. It also shows that larger droplets grow on the cost of smaller ones.
The drop in temperature is due to the required latent heat of vaporization, i.e. the energy required by the transition liquid to vapor.


text:
Hydrolysis of transition metals' halides?
So, it's a relatively common known solubility rule that any hydroxide with a cation not in the first two groups is basically insoluble. So supposing we have a transition or post-transition metal $M$ and say it forms a charge of $n+$ (so it's cation is $\ce{M^{n+}}$) why would $\ce{MX_n}$ (where $\ce{X}$ is a halide) remain in solution? Why doesn't the following reaction occur?
$$
\ce{MX_n(aq)} + n\ \ce{H_2O(l)} \rightarrow \ce{M(OH)_n(s)} + n\ \ce{HX(aq)}
$$
My thought was that perhaps this reaction does occur, but just very slowly and not fast enough for the $\ce{M(OH)_n}$ to precipitate in noticeable amounts. However, I don't really know and this is just a guess. Could someone explain to me why this doesn't happen?

$\ce{MCl_n(aq)} + n\ \ce{H2O(l)} \rightarrow \ce{M(OH)n(s)} + n\ \ce{HCl(aq)}$ does tend to go to the right. "Basically insoluble" is not quantitative enough to generalize. If the hydroxide is very insoluble, the acid produced by hydrolysis may not keep it in solution. But if the hydroxide is even only slightly soluble, it may not precipitate out. The acidity of the solution is a subtle indicator that the reaction has shifted a little bit to the right. The reaction isn't slower, just not always so extreme.
If you put $\ce{SiCl4}$ in water, it gives $\ce{HCl}$ + silica gel because $\ce{SiO2}$ (or "$\ce{Si(OH)4}$") is so insoluble.
If you put $\ce{MgCl2}$ in water, no precipitate appears because $\ce{Mg(OH)2}$ is slightly soluble (~2 mg/L), and the $\ce{HCl}$ produced by the hydrolysis reaction is acidic enough so that if a significant amount of $\ce{Mg(OH)2}$ were produced, it would just redissolve. $\ce{Mg(OH)2}$ gives a pH of about 10 in water, so at pH's lower than 10, $\ce{Mg(OH)2}$ doesn't precipitate out. BTW, $\ce{MgCl2}$ is slightly acidic because of the hydrolysis reaction.
$\ce{AlCl3}$ is complicated because of ionic complexes, but in water it hydrolyzes to give a very acidic solution and some aluminum hydroxide complexes or gels which are of varying solubility. $\ce{AlCl3}$ has been used as a water treatment because it produces flocculent precipitates that carry down impurities.
$\ce{NiCl2}$ does not precipitate in water because $\ce{Ni(OH)2}$ is even more soluble than $\ce{Mg(OH)2}$.
$\ce{FeCl3}$ is quite acidic in water. You might expect the commercial 40% solution to be a cloudy product because of the production of acid, but it is a clear, deep red color. Oh, but when it is diluted and the pH rises because the acid is diluted, it precipitates in flocs like $\ce{AlCl3}$. It is used to clarify water by precipitating dirt and dust.

How is nickel hydroxide more soluble than magnesium hydroxide?  Ksp values show that at the same pH magnesium ions would be more soluble by about a factor of $10^4$.

I was surprised too. The CRC handbook gives 0.0009 g/100mL for Mg(OH)2; 0.013 g/100 mL for Ni(OH)2. I did not look up Ksp. I wonder if there is some other factor/difference in the measurements that makes sense.

If the numbers are right, nickel hydroxide in solution must be a weak base, while the limited amount of magnesium hydroxide that dissolves is much stronger.


text:
Choice of Lewis Acid for EAS reaction
Why is AlCl3 or FeCl3 might be preferred for chlorination reactions and AlBr3 or FeBr3 for bromination reactions? Why not use AlCl3 for bromination reactions?
Source: https://www.masterorganicchemistry.com/2011/07/22/reagent-friday-aluminum-chloride-alcl3/

Can you elaborate on your answer?

This does not explain why FeBr3 would be a better catalyst for bromination reactions over FeCl3.

yes, but by your logic, FeCl3 would be a better catalyst than FeBr3 for a bromination reaction, or any type of EAS reaction for that matter, which is not the case.

In Electrophilic Aromatic Substitution reaction (EAS), the choice of catalyst may influence the yield of the desired product.
First of all, note the following data:
Bond dissociation energy (in kJ/mol) -
Al–Br : 429.2
Al–Cl : 502
As the bond dissociation energy for Al–Cl bond is greater, it must be stronger.
Now, consider using aluminium bromide as catalyst for an EAS chlorination.
In the first step, i.e. interaction of chlorine molecule with Lewis acid aluminium bromide, one can't possibly neglect the formation of bromonium ion which may act as electrophile - decreasing yield of chlorobenzene and giving unnecessary side product of bromobenzene. This possibility is supported by thermodynamics, as shown by the data mentioned above.
To make sure that yield of product is maximum and no unnecessary side reactions occur - aluminium chloride catalyst is employed for chlorination and aluminium bromide for bromination.
Hope this answers your question.
P.S.
Correction to @AvnishKabaj's answer:
If I had to give a reason it would be that when the AlCl3-Br2 complex is made there might be a teeny tiny chance that instead of the bromine breaking away the chlorine breaks away.
As per the bond energy data I have, from this exhaustive table ;it is easy to conclude that the bond strength of Al–Cl bond is greater than Al–Br bond. Hence, problems would arise when chlorination is done in presence of aluminium bromide catalyst. In the AlBr3-Cl2 complex, there'd be a teeny tiny chance that instead of the chlorine breaking away, the bromine breaks away. (based on thermodynamic favourability)

Oh! Yes! Also I just wanted to point out that aluminium bromide for chlorination is a bigger problem than using aluminium chloride for bromination; as I've mentioned in my edit in the postscript.

Why are you comparing using Al-Cl bond strength? Mind you, the chlorine atom that is abstracted from the complex is the terminal chlorine atom, which is attached to the central chlorine atom in the complex.

Honestly speaking I think there won't be a difference as in clayden they have used $\ce{AlCl3}$ for bromination

If I had to give a reason it would be that when the $\ce{ AlCl3-Br2}$ complex is made there might be a teeny tiny chance that instead of bromine breaking away the chlorine  breaks away.

My basis of saying that the probability of chlorine breaking away is low is because there's a +tive charge on $\ce{Br}$ in the complex.Chlorine on the other hand is free from any such effects. There are no factors supporting the breakage of the $\ce{Al-Cl}$ bond. So the chances of $\ce{Br-Br}$ bond breaking are higher than the $\ce{Al-Cl}$ bond.
The same reasoning can be applied for $\ce{AlBr3}$ with $\ce{Cl2}$
That's why in Master Organic they've written preferred not necessary.

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text:
Pressure-Volume Formula Question
I have a chemistry homework problem that states the following:

$6.2$ L of $N_2$ at $0.74$ bar is mixed with a $15.2$ L sample of $O_2$ at $0.35$ bar. The gaseous mixture is placed in a container of volume $12.0$ L at the same temperature. What is the pressure of the system, in bar?

I answered the question and I believe I have the right answer, but I'm unsure as to whether it's a fluke or not. Here's my approach:
Use the formula $P_1 * V_1 = P_2 * V_2$.
$6.2 * 74 = 21.4 * P_2$, $P_2 = 21.4$ kPa.
Repeating the process for $O_2$, I get $P_2 = 46.26$ kPa.
I add the volumes together since the gases are now mixed, and I also add the pressures. Now that the mixture is placed into a $12.0 L$ container, we use the formula again:
$46.26 * 21.4 = P_2 * 12$. $P_2 = 83$ kPa. This is equal to 0.83 bar.
I'm still not 100% sure of my reasoning for this question, wondering if someone could verify my logic and provide perhaps a better way of doing this question that I might've overlooked.
Thank you!

I arrived at the same conclusion, but I am not sure your approach is optimal, for efficiency and conceptual reasons. The exercise states nothing about where the two samples are mixed, and I think it is reasonable to simply mix them in the final container:
$$
p_i \cdot V_i / V_{\text{final}} = p_{i,\text{final}},
$$
where $i$ runs over $\ce{O2}, \ce{N2}$. Add the $p_{i,\text{final}}$ to obtain total final pressure $p_\text{final}$.
The actual calculation (I dropped the units) is thus:
$$
( 0.74 \cdot 6.2 + 0.35 \cdot 15.2) / 12 \approx 0.83,
$$
confirming your result.


text:
Preparation of di-phenyl ester from phenol
How to create di-phenyl ester from phenol in laboratory. What would be the steps. I am asking this question as it can not be done by Williamson ether synthesis.

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

Does this  help

Do you mean diphenyl ether?

No sorry I want the preparation of di-phenyl ester (ph-o-ph)

PhOPh is Diphenyl Ether!

Diphenyl ethers have been prepared from condensation in melt phase of potassium phenoxide and aryl bromides in the presence of Copper and Copper(I) Chloride - J. Chem. Eng. Data, 1978, 23 (2), 185–186
US patent 4564712 A also describes a similar process


text:
Obtaining Benzene from toluene
What will be the reaction for producing benzene from toluene?
My attempt:
Toluene→Benzoic acid→Sodium benzoate→Benzene(decarboxylation) ...
Is there any other ways, like a direct reaction?

There is also a disproportionation reaction that converts toluene to a mixture of benzene and xylene (Bawa et al), also, see wikipedia.  It has commercial applications (because the benzene/xylene mixture can often be sold for more than toluene) and can be licensed.  However, it is more of an industrial process than one you would use in the lab, as it is really driven by an exothermic dealkylation in vapor phase over zeolite catalyst.
Interestingly, as zeolites act as a Lewis acid, this is strangely similar to Friedel-Crafts alkylation, which uses Lewis acid catalysis in liquid phase.  The paper by Bawa et al discusses experiments in which the trans alkylation was done with Frieden-Crafts catalysts in liquid phase.

And how does that work? (a good answer should explain, not just provide links to, the material). We really don't like link-only answers.


text:
Calculate the cathode electrode potential in this redox reaction
In this problem I'm asked to calculate the electrode potential of the magnesium electrode ($E_{Mg^{2+}/Mg}$) given the red-ox reaction:$$\text{Mg(s)}+2\text{Ag}^+(10^{-2}\text{M})\leftrightarrow 2\text{Ag(s)}+\text{Mg}^{2+}(10^{-2}\text{M})$$
The information I'm given is:

Temperature equal to $278\,K$ (the standard).
The molar concentration of the $\text{Mg}^{2+}$ ions is of $0.01\,\text{M}$.
The standard reduction potential of the $Mg$ electrode is: $E^{º}_{Mg^{2+}/Mg}=-2.34\,V$.

I know I have to use the Nernst Equation, but my solution and the book's answer are very different.
My solution:
$$E_{Mg^{2+}/Mg}=E^{º}_{Mg^{2+}/Mg}-\dfrac{0.059}{n}\,\color{#ff3300}{\mathbf{\log\frac{\left[\text{Mg}^{2+}\right]}{1}}}=\ldots=\boxed{-2.28\,V}$$
My book's solution:
$$E_{Mg^{2+}/Mg}=E^{º}_{Mg^{2+}/Mg}-\dfrac{0.059}{n}\,\color{#00ff00}{\mathbf{\log\frac{1}{\left[\text{Mg}^{2+}\right]}}}=\ldots=\boxed{-2.40\,V}$$
($n=2$ because 2 moles of electrons where exchanged per 1 mole of reaction)

I've colored the part where I go wrong (basically the logarithm). I don't know exactly why the book put the concentration in the denominator, since $\text{Mg}^{2+}$ is a product (not a reactant), therefore following what I studied, its concentration should be in the numerator.
Any help? Sorry for asking such an elementary question, but I'm still learning the basic stuff in Chemistry!
Note
I attach an image of the solution provided by the book

Zhe's answer is the best. Though, I will try to offer a different (mathematical) approach, which will hopefully enable you to see the answer yourself, or rather, verify that your book's solution is indeed correct.
According to you, the standard oxidation potential of magnesium electrode would be $E_{Mg/Mg^{2+}}=E^{º}_{Mg/Mg^{2+}}-\dfrac{0.059}{n}\,\mathbf{\log\frac{\left[\text{Mg}^{2+}\right]}{1}}$. Agreed?
Also, note that $E_{Mg/Mg^{2+}}=-E_{Mg^{2+}/Mg}$ always holds. Agreed?
Then, $E_{Mg^{2+}/Mg}=-E_{Mg/Mg^{2+}}$ $$=-(E^{º}_{Mg/Mg^{2+}}-\dfrac{0.059}{n}\,\mathbf{\log\frac{\left[\text{Mg}^{2+}\right]}{1}})$$ $$=(-E^{º}_{Mg/Mg^{2+}})-\dfrac{0.059}{n}\,\cdot(-1)\cdot\mathbf{\log\frac{\left[\text{Mg}^{2+}\right]}{1}}$$.
$$=E^{º}_{Mg^{2+}/Mg}-\dfrac{0.059}{n}\,\mathbf{\log\frac{1}{\left[\text{Mg}^{2+}\right]}}$$.
Which is exactly the answer given by your book.
I hope it helps!

The equation that corresponds to the standard reduction potential is:
$$\ce{Mg^{2+} + 2e- -> Mg}$$
Notice that this is not the actual half reaction in your total equation, where magnesium is oxidized. But the book is determining the reduction potential, so the magnesium ion is a reactant not a product. In the reaction quotient, the concentration/activity of magnesium ion should be in the denominator.
I'm not sure I agree that we should choose the standard reduction potential as the electrode potential for the electrode where oxidation occurs, but at least the book is consistent.

According to my book, the equation that corresponds to the standard reduction potential is the same as you wrote, but with the arrow pointing to the other side. I'm a bit lost, could you clarify this in your answer please? I am beginning with the Chemistry of reactions

That would be incorrect. The reaction I have written is the reduction. If you reverse the reaction, that would be an oxidation. @JoseLopezGarcia

But the $Mg$ is the anode in the galvanic cell. It's being oxidized by the $Ag$. Is this correct?

(In my first comment I said "reduction" but I meant "oxidation", apologies)

Yes, that's my whole point. I don't understand why the book is talking about the half reaction as if it were a reduction. It doesn't make sense to me. But if you are going to talk about it as a reduction, then the magnesium ion is a reactant not a product, which means it is in the denominator of the reaction quotient.

I don't know either, but I still can't seem to understand the problem. I've uplodaded a screenshot of the solution, would you have time to check it. It's in Spanish but the equations and solution flow I think are easy to follow.

In the image I uploaded, note that both logarithms are taken the same way (as if $Ag^+$ and $Mg^{2+}$ were BOTH products, which is not possible). One of those logarithms need to be wrong in my opinion.

Maybe this is a language issue, but you're repeating what I'm telling you is confusing. The solution is treating representing the potential of the magnesium electrode as a reduction. It's not ideal, but potentials are generally given in tables of standard reductions. So while not ideal, not insane. Your opinion is not correct here. As written, both potentials are reduction potentials. Yes, the magnesium is the anode, but it still has a reduction potential.

You can't just use the wrong logarithm. That's coupled to the standard potential $E^{\circ}$.


text:
Several carbonyl groups, which one does ethylene glycol protect?
We know that ethylene glycol is used to protect carbonyl functional groups. But, consider this:

Here, we have two carbonyl groups, but only one equivalent of ethylene glycol. Obviously, only one carbonyl group will get protected in this case. But will it be the $\ce{C}^1$ or $\ce{C}^2$ carbonyl? And why?

Several factors influence the overall rate of a reaction under various conditions. Among the crucial factors are:

structural features of the carbonyl compound;
the role of protons or other Lewis acids in activating the carbonyl group toward nucleophilic attack;
the reactivity of the nucleophilic species and its influence on subsequent steps; and
the stability of the tetrahedral intermediate and the extent to which it proceeds to product rather than reverting to starting material. (Ref.1)

Limiting ourselves to current molecule, aldehydes  are more reactive than ketones. Electron-withdrawing substituents render the carbonyl group more electrophilic.

Image credit: Organic Chemistry, Structure and Function Seventh Ed by Peter Vollhardt, University of California at Berkeley, Neil Schore, University of California at Davis
Because aldehydes form acetals more readily than ketones, we can protect an aldehyde selectively in the presence of a ketone. The following example shows the reduction of a ketone in the presence of a more reactive aldehyde (Ref.2):

With reference to your question, C-2 is more reactive then C-1 carbonyl group for the above reasons.
References

Advanced Organic Chemistry, Fifth Edition, Part A: Structure and Mechanisms by Francis A. Carey and Richard J. Sundberg, University of Virginia
Organic Chemistry, Eighth Edition, by L. G. Wade Jr., Whitman College
Reactivity of Aldehydes & Ketones - Chem LibreTexts


text:
Numerical related to crystallisation of water below its freezing point
Question:

There is $16\ \mathrm g$ of pure water in a container at temperature $-20\ \mathrm{^\circ C}$. A small piece of ice is added to start crystallization. Amount of water in container when temperature reaches to $0\ \mathrm{^\circ C}$ is: (assume specific heat of water below $0\ \mathrm{^\circ C}$ is $1\ \mathrm{cal\ g^{-1}\ ^\circ C^{-1}}$.)

My attempt:
I have read this: Liquid water below freezing temperature. So, I understand that the small piece of ice is only added to provide a nucleation site to promote the freezing of water.
But, all physical processes must follow the Law of Calorimetry, right? So, for water below $0$ degrees to freeze, it must first accept heat from somewhere, to reach zero degrees. But, the container is at an even lower temperature, so it won't provide water with any heat (negative temperature gradient). We have also not been given the mass of ice added, so we can't calculate the amount of heat the water will gain from the ice (the latter is at a high temperature than water). The system is isolated, so there is no other possible heat source...
I am apparently at a dead end in this problem. I hope I have put everything in detail and correctly. What is the logical mistake in my thinking? And what is the correct way to approach this problem? Thank you!

sigh You start with water -20 end up with ice in higher temperature. Supercooled water is inherently metastable and no more activation energy then the nucleation site provides is needed, as this is kinda "chain reaction".

Another answer explains more generally how to interpret the problem. This completes that spoiler:

 If one can assume that the enthalpy of fusion ($\pu{334 J/g},~$2) is independent of temperature between the normal melting point ($\pu{0^\circ C}$) down to $\pu{-20 ^\circ C}$, then the heat released during freezing equals that absorbed during heating of the liquid to $\pu{0^\circ C}$ so that $$\begin{align} \pu{334 J/g} \times m &= \pu{20 K \times  16 g \times 4.18 J/g*K} \\ \rightarrow m&= \pu{4.0 g} \end{align}$$ Note the important condition that the heat capacity of water remains constant and equal for liquid and solid over that temperature span.

Borrow some energy to warm up all the liquid. Then repay - interest-free - by making some ice. https://memegen.link/custom/state_functions.../are_path--independent.jpg?alt=https://iili.io/H0DOYb.jpg

In the question Crystallization begins after you add a small piece of ice
.
 So it has to begin that's been given in the question.
I don't know thermodynamically whether it's plausible or not, but it's a given in the question a logical alternative way to approach the question (according to me) is

I formatted it as a spoiler just in case you want to think a bit more or anybody else does

 The temperature has increased because latent heat of fusion is being released i.e ice being formed from water is  releasing heat.

PS: I may be wrong

Water around the ice was initially entirely at $-20^\circ$ celsius. How would you account for the heat required to take water from $-20^\circ$ celsius to $0^\circ$ celsius? Will water start freezing at $-20^\circ$ celsius itself?

Latent heat released by ice formed = increase in temperature of the system

This equation is dimensionally incorrect...

You did get what I'm trying to say. If not then multiply $\ce{[\delta]T}$ into the specific heat of the ice formed and water remaining

Will water start freezing at $−20^∘$ celsius itself?

The freezing point of water is 0° Celsius. It won't matter when at what temperature  though because all you need is the energy for two states.

Ignoring any differences in specific heat between solid and liquid...

@JonCuster Enthalpy is a state function. You can choose a path that is easy to calculate, i.e. warm up all the liquid water first, and then turn some of it into ice. Along this path, you need the specific heat of the liquid, only. Going along the actual path, any warming up of ice should cancel out because the answer has to be the same.

So, for water below 0 degrees to freeze, it must first accept heat from somewhere, to reach zero degrees.

No, water can freeze at lower temperatures, see e.g. https://www.youtube.com/watch?v=Fot3m7kyLn4.
As it freezes, the exothermic process warms up the ice and the surrounding water. The difference in enthalpy between liquid and solid water is larger than the difference between liquid water at the two temperatures, so not all the water freezes. Instead, you end up with liquid and solid water at equilibrium once the water and the ice reach the normal freezing point (not because they are warming up from the environment, but as ice continuously forms).
Instead of adding ice to start the process, you can also shake the container like in the video.

The crystallization is an operation that delivers heat, and a good deal of heat. This is sufficient to heat the metastable water form -20°C to 0°C.

The previous answers are good ones. I just want to add a (longwinded) metaphorical picture to help understanding the freezing of water at -20C.
Imagine a tower of bricks (2 x 4 x 8 inches each) about 20 high. It is metastable, but the tiniest push will topple it because gravity pulls each brick to the lowest level. This is a metaphor for the freezing of water at -20C, right? Yes, but there is a great dissimilarity, because the bricks fall down to a lower height, but the water temperature rises as it freezes. We all know that in reality, we can't reduce the temperature of water to -20C and have it be as stable as a pile of bricks 20 high. Then, too, freezing is a process that 99.9% of the time requires a reduction of temperature.
Now let's try another metaphor: take 20 bricks of low-density foamed polystyrene and push them down into a large tank of water. You may be able to hold them down perfectly straight, or even balance a heavy lead brick on top of them to hold them down while they are semi-floating, but if the slightest imbalance occurs (like the ice-seed), the polystyrene bricks will all float up to the surface (~ ~ supercooled water will freeze) and the heat released will equal the energy required to push the polystyrene into the tank. The heat released as the polystyrene bricks rise is really from the water settling in a gravity field (~ ~ the heat of fusion that was added to the water to keep it from freezing while it was being cooled to -20C.).
Now how can you cool water to -20C without removing the heat of fusion? In your imagination, separate the specific heat and the heat of fusion into different accounts. Take the heat out of the one and leave the other, but don't let the water know. This allows you to mentally reduce the temperature without allowing the solidification process to occur - it's like applying a "negative catalyst". Then when you remove the "negative catalyst" (i.e., add the ice-seed), some of the water freezes, releasing bottled-up heat of fusion to flow into some of the -20C water - the temperature rises. The result will be some ice and some water, and when the system comes to equilibrium, the temperature will have risen to 0C.
And the process can be stated much more succinctly by an equation, but sometimes equations are too succinct!


text:
What are the key differences between Raoult’s Law and Henry’s Law
Raoult’s Law ($p_A = \chi_A \cdot p_A^\circ$) becomes a special case of Henry’s Law ($p_A = K_H^{px} \cdot \chi_A $) when $K_H^{px} = p^\circ_A$, but where do these laws differ from each other?
There’s a statement in my text book which I’m having trouble understanding.
The statement:

As a real solution approaches the limit of infinite dilution its components behave more ideal. The solvent obeys Raoult's law whereas solute (minor component) obeys Henry's law for dilute solutions.

To my understanding is Raoult's law used to describe the component with the higher concentration in a mixture. Henry's law, on the other hand, will describe the compound with the lower concentration better.

@GaurangTandon No, this a photo\scan from Cengage Physical Chemistry: Part 2, 2nd ed., p. 2.16 by K. S. Verma.

This is discussed here: https://www.quora.com/What-is-the-relationship-between-Henrys-law-and-Raoults-law and i think it is reasonably clear.

Raoults law is for finding the vapour pressure of different component of the solution dissolved in it whereas the henry law is for finding the amount of solute that could be dissolved in the solution or to find the solubility of the solute.

Raoult's law can be seen to be valid for ideal liquids. The assumption behind Raoult's law is that both the liquid phase and the vapour phase behave ideally. This means that the interaction between any two molecules in the liquid and the vapour have no interaction.
The Raolt's law just states the mole fraction of a component in the liquid phase is proportional to the mole fraction of the same component in the vapour phase. This is valid only for mixtures of ideal liquids.
When you add a very small amount of impurity in a solvent (equivalent to infinite dilution) the interaction between the impurity and the solvent molecule is minuscule or technically infinitesimal. Therefore, the liquid still behaves ideally. Hence Raoult's law is applicable for the solvent.
Whereas Henry's law is also defined for a ideal mixture only this time the assumption is that the mole fraction of the component is tending to zero. this means that when very few molecules a present in a vast space, the interaction between them is almost zero. This makes the system ideal. therefore, an infinitesimally small quantity of impurity does not feel any interaction with another molecule of the same impurity in the solvent. Therefore, Henry's law is used for the solute. That is why Henry's law constant have huge values. Huge values makes sure that even small amount of the solute is accounted for when calculating the mixture properties.
For further clarity, please refer to chapter 6 of 'Molecular Thermodynamics of Fluid phase equilibria' by John M.Prausnitz et. al.

I would argue against the use of the term infinitesimal in any context of physical chemistry except maybe "infinitesimal" fluctuations around the equilibrium, which is explained by any serious textbook as a simpler notation for time derivatives. As far I as I know, there is no record of other uses in the thermodynamic literature and mathematicians surely wouldn't approve it since infinitesimals are only well defined in the extended reals.

Raoult's and Henry's laws are limiting laws, generally applicable when the solute concentration goes to zero. In this limit the vapor pressure of any component in the solution depends linearly on its mole fraction, implying the absence of solute-solute interactions.
Raoult's law describes the dependence of the vapour pressure of a solvent as a function of its mole fraction $\chi_1$:
$$\lim_{\chi_1\rightarrow1}\left( \frac{p}{\chi_1}\right) =p^\ast$$
where $p^\ast$ is the vapour pressure of the pure solvent.
Henry's law describes the dependence of the vapour pressure of a solute as a function of its concentration. In terms of mole fraction $\chi_2$:
$$\lim_{\chi_2\rightarrow0}\left( \frac{p}{\chi_2}\right) =K$$
For a binary mixture of pure substances it can be shown that the laws are complementary: if one law holds for one component then the other law holds for the other component. It can also be shown that the laws imply that the solution satisfies other criteria for ideality, including zero enthalpy and volume of mixing.

Could you explain how the laws imply zero enthalpy of a solution?

For either solute or solvent, we can write that in the mixture the component has molar enthalpy $H_{m,i}=-T^2\left(\frac{\partial( \mu _i/T)}{\partial T}\right)_p=-T^2\left(\frac{\partial( \mu _i^\circ/T)}{\partial T}\right)_p=\textrm{constant}$ where I substituted the definition of μ according to the ideal laws. Since the enthalpy of either component in the mixture is unchanged from the enthalpy in pure form, it follows that the enthalpy of mixing is 0.

Compare and contrast
As stated in the caption of the figure posted by the OP, Henry's law applies to the solute and Raoult's law applies to the solvent. If a solute is non-volatile (e.g. sucrose or sodium chloride), Raoult's law still works, but invoking Henry's law, while possible technically (by setting Henry's constant to zero), would be kind of pointless.
As you can see for the example shown in the picture posted by the OP, Henry's law describes the partial pressure of a component at very low concentration. On the other hand, Raoult's law describes the pressure of a component at a concentration close to pure liquid.
Raoult's law is independent of the nature of the solute; the only parameter is the mole fraction of the solvent (or solute, if written as $\Delta p = \chi_\mathrm{solute} \cdot p^\circ$). Henry's law, on the other hand, depends on the nature of both solute and solvent (i.e. the Henry's constant is different for different solvents).
Continuous transition from Henry's to Raoult's law
For a binary system where both components are volatile, are liquid in the pure state and are miscible across the entire range, you can discuss the equivalence or non-equivalence of the two laws. The situation for the ethanol: water mixture is shown below:

The figure was taken from R K Thomas's course page, which has a great explanation of the entire topic. Note that the vapor pressure of either component increases linearly near 0% and 100% composition, but not with the same slope. In this particular case, both water and ethanol show positive deviation from Raoult's law at low concentration. There are also cases of negative deviation such as the chloroform: acetone system.
The case described by the OP, where components conform to Raoult's law for any composition, is realized for mixtures of the same molecule with different isotopic composition, e.g. ethanol with carbon-13 on the first or second carbon. To a good level of approximation, the boiling points of the two species are identical, and the interactions between identical or isotopically different molecules are the same, so Raoult's and Henry's law would give the same values. I don't know of a less trivial system (i.e. where the boiling points of the pure components differ) where that is the case.

[quoted in OP's question] As a real solution approaches the limit of infinite dilution its components behave more ideal. The solvent obeys Raoult's law whereas solute (minor component) obeys Henry's law for dilute solutions.

I don't think this statement is true. Henry's law does not require ideal behavior (the solute: solvent interaction can be stronger or weaker than the solvent: solvent interactions), and the distinct Henry's law constants depending on solvent reflect that. For example, more oxygen will dissolve in water than in cyclohexane at the same partial pressure of oxygen. The range where Henry's law applies is called ideally diluted solution. Roult's law will apply for any solute, so it also does not require an ideal solution.
If the solution is ideal (no difference between solvent: solute and solvent: solvent interactions), the condition $K_H^{px} = p^\circ_A$ mentioned in the OP's question becomes true.

The solute in the Henry's Law region behaves ideally in the sense that it only "sees" the surrounding solvent, not other molecules of its own species. Its chemical potential is defined by its concentration and these interactions, not by interactions with itself, as would be seen in the Raoult's Law region.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How come ∆G ≤ 0 is used for spontaneity in electrochemistry, not ∆G ≤ W(other)?
The spontaneity condition for a system at constant temperature and pressure in which the only type of work accomplished is of the $pV$ type can be expressed as:
$$\Delta G\le0\tag1$$
In case of other types of work (such as electrical, friction, etc.) there is an additional term and the previous equation becomes:
$$\Delta G\le W_\text{other}\tag2$$
In case of an electrochemical process the maximum work is the electrical work and can be expressed as:
$$W_\text{other}=-nFE\tag3$$

Question: Why it is still said that the electrochemical reaction is spontaneous if the change in Gibbs free energy is lower than zero, thus using Equation $(1)$, even though the proper equation to be used is number $(2)$?

Most likely the source is wrong, or it has purposfully omitted the correct equation to pseudosimplify a problem. It might be that an electrochemical reaction was under discussion, not a cell. A further possibility is pursued in  the appendix. I recommend adding a source where you came upon the quoted exchange to avoid a straw man argument; even though in this case  I have observed (and possessed) the misconception first hand.
One definition of spontaneity
I define the true spontaneity condition as such when there is a tendency toward net 'something different' (chemical reaction, expansion etc) to establish state 2 instead of some original state 1. An example could be a chemical reaction where there is a higher reaction extent than at equilibrium, so a net reverse reaction takes place.
You are correct that the true spontaneity equation (from Clausius's inequality) in the thermodynamic limit is, for net-constant pressure and temperature in the system,
$$\operatorname{d} G-\delta w_\pu{other}\leq 0.\tag1\label1$$
When electrochemical work is the only component besides expansion work, this implies (due to $|\operatorname{d} n_\pu{e}|=|\nu_\pu{e}|\operatorname{d}\xi$)
$$\Delta_\pu{r} G_{T,P} +|\nu_\pu{e}|FE\leq0\tag2\label2$$
where $\Delta_\pu{r} = \partial/\partial \xi$; here $\xi$ is extent of the reaction. The Greek $|\nu_\pu{e}|$ signifies an absolute value of the stoichiometric coefficient of an electron in some half-reaction. Equation $\eqref2$ assumes that only one net reaction occurs. The term $|\nu_\pu{e}|FE$
should be a good indicator of electrical work.
It might also be that they are discussing an electrochemical reaction, not the cell itself. We can have the process
$$\ce{Zn(sln) + Cu^2+(sln) -> Cu(sln) + Zn^2+(sln)}\tag3\label3$$
without harnessing its electrical work via an external circuit. So $\operatorname{d} G\leq 0$ would hold for spontaneity.
Appendix: An alternative
A different definition of spontaneity might be in play. Namely, the term spontaneity is also used to mean a large enough standard equilibrium constant (especially in biochemistry).
$$RT\ln \frac1K = \Delta_\pu{r} G^\circ_{T,P} \tag4\label4$$
Equation $\eqref4$ is technically a definition of the standard equilibrium constant. When non-negative absolute temperatures are quaranteed, the LHS of equation $\eqref4$ will become non-positive for all $K$ big enough, i.e., $K\ge1$ . That implies
$$\Delta_\pu{r} G^\circ_{T,P} \leq 0. \tag5\label5$$
Note, however, that $K\ge1$ doesn't necessarily imply much about the reaction balance itself because the equilibrium constant comprises of activities (not concentrations), and because the stoichiometric coefficients in the denominator may drown out the coefficients in the numerator. (Equation $\eqref6$ assumes that surrounding fugacity is equal to an agreed fugacity in the standard state, denoted here and elsewhere by '$^\circ$'.)
$$K(\pu{in solution, solvent \ce{A}}) = \left[a(\ce{A})_\pu{eq}\right]^{\nu_\ce{A}}\prod_i \left[a(\ce{B_$i$})_\pu{eq}\right]^{\nu_i}. \tag6\label6$$
Reactants–products (other than solvent) are designated by $\ce{B_$i$}$. But still, for $K$ big enough (2nd definition of spontaneity), equation $\eqref5$ will hold by definition. Also keep in mind $\Delta_\pu{r} G^\circ_{T,P} \neq \Delta G$ (even their dimensions are different!). So the source is presumably still at fault for poor notation.

Some details (such as derivations and motivations for definitions) have been omitted for brevity. If something is unclear, please point it out in the comments, and I will do my best to clarify (edit) as soon as possible.

Seems a matter of definitions.  "The" fundamental equation that I'm familiar with is $\Delta G = -S dT + (work terms)$, and if you have electric fields, magnetic fields, surface tension, gravity, etc. etc., then that is part of the definition of $dG$.  So with that definition, $dG \le 0$ is still a valid spontaneity criterion.  If your definition of $dG$ only includes $PV$ work, then YMMV.

@CurtF. Thank you for your comment. I took some time before this reply to think, and think some more. The conclusion: I stand my ground (for now). As long as the definition of Gibbs energy remains $G = U + PV - TS$ (is this challenged?), then $\operatorname{d} G \leq 0$ is a necessary and sufficient condition for spontaneity iff (1) a thermodynamic limit is applied, (2) net-constant pressure, (3) net-constant temperature, (4) only non-expansion work is done. The effect of external electromagnetic fields, surface tension, gravity, friction and so on have to be included in $w_\pu{other}$.

see also: the appendix (b) of my answer here

I think people commonly use your definition of $G$, but also not uncommonly use other definitions too.  For magnetocaloric materials, people use $dG = -SdT - M dB$, for example.  https://chemistry.stackexchange.com/questions/33329/fundamental-equation-of-thermodynamics-for-magnetocaloric-materials

And to emphasize, I'm not saying your answer is wrong at all, in fact +1 from me.  Just depends on your definition of $G$.

@Linear Christmas I totally agree with your answer and is coherent with what I was thinking of, so maybe my misunderstanding is related to the boundary of the system.
For example, please refer to https://chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Basics_of_Electrochemistry. It is stated that the redox reaction is spontaneous if the Gibbs energy change is lower equal than zero so that the potential difference between the two electrodes is positive. But this seems to me incoherent in view of what said in these answers. What am I missing?

@CurtF. No offence taken; I am strongly invested in getting this answer as correct as possible. Wrt the linked question, (as I see it) that includes a differential of Gibbs energy (not an absolute quantity), assuming that $G = G(T;P;M, N_i)$ and is perfectly valid in certain situations. But regardless spontaneity is not guaranteed by $\operatorname{d} G \leq 0$ in cases where $\delta w_\pu{other} \neq 0$.

@CurtF. see doi: 10.1021/ed400453s equation (9b) and the assumption for (10b), and the following-preceding discussion

@horowitz Say we have an electrochemical cell with only one reaction taking place. Assume temperature and pressure are net-constant (they may change or even be undefined during the process, as long as initial and final values are equal). Then the condition that is both necessary and sufficient for spontaneity of the entire cell (for Clausius's inequality to hold) is: $\partial G/ \partial \xi + |\nu_\pu{e}|FE \leq 0$, this includes equilibrium (two-way spontaneity) when it's equal to $0$. The spontaneity for simply the reaction itself is still $\operatorname{d} G \leq 0$.

@horowitz (2/2) Here 'the reaction' denotes the reaction when the external circuit is effectively not present. I recommend to stick around in case of updates for a few days.

@LinearChristmas So normally it is used DeltaG lower equal than zero to refer to the reaction itself and the complete equation is used when the cell is taken into account?

@horowitz What did you mean by 'DeltaG' here? Did you have in mind $\operatorname{d} G$, $\Delta_\pu{r} G$, or $\Delta G$?

By spontaneous we normally mean that the reaction is thermodynamically allowed WITHOUT forcing it (that is, without doing non-PV work from the outside). This is described by your eq. (1).
A non-spontaneous reaction we can make thermodynamically allowed by adding enough work, this is your eq. (2). A typical chemistry lab example of the last is electrochemical splitting of water into oxygen and hydrogen gases by means of two electrodes and an external power supply.
Eq. (2) also tells the thermodynamic limit (i.e. maximum) electrical work one can possibly extract from the given electrochemical reaction, if it is spontaneous and fulfills eq. (1).


text:
Basicities of nitrogen atoms in purine

Arrange the sites in purine in order of basicity.

I feel that 9 is the least basic as nitrogen's lone pair is delocalised. However, I am unable to rank 1, 3, and 7 in a particular order. Is there a particular theory to decide basicity of sites?

Why don't you try to draw out all the resonating structures?

@AvatarShiny What should I look for in those ?

Which site has the most number of times  a negative charge in the Canonical structures ?

Possible duplicate of Most basic nitrogen in Adenine

I'm not convinced that this is a duplicate, but on a conceptual level they are clearly related.

@Abcd If anything, this should be closed not the other way round...

I don't think that question is trivial and except for what you have already deduced, I would not have an idea how to solve that without employing computational chemistry.

The problems already start that I suspect that the proton from position 9 readily exchanges to position 7 in any kind of polar solvent (and in absence why not with itself).
Well from that point on forward, I'd say you need at least a lot of acid to protonate 9, as the π system is aromatic and will most likely not react first.
After that you could probably deduce from the approximate bond angles which of the nitrogen would have the largest s- and p- character, then rank them from highest to lowest p contribution. The one with the most will probably be the HOMO and therefore react first. To be honest though, this is just guessing.
I have absolutely no clue what the lesson with this exercise it, except for: "They're pretty much the same."
Let's have a closer look anyway. I have computed the neutral molecule at the DF-B97D3/def2-TZVPP level of theory, and used NBO6.0 for the partial charges:

As expected, there is not much of a difference; I'd even go as far and say they are the same. Position 9 is a little bit more negative, because it has a hydrogen to draw from.
Next up, let's have a look at the highest molecular orbital, as we would assume to protonate there:

Again, there is only little to no difference. Or in numbers (contributions > 3%):
Alpha occ 31 OE=-0.217 is
N1-p=0.3635 N1-s=0.0382
N3-p=0.2551
N7-p=0.0667
C5-p=0.0563
C4-p=0.0473
C2-p=0.0452
C6-p=0.0382

From this we would conclude that since N1 has the largest contribution, the proton is most likely to go there.
Therefore I also calculated all the protonated species, and their relative energies are:
\begin{array}{lr}\hline
\text{Position} & \Delta G(\pu{298.15 K}, \pu{1 atm})/(\pu{kJ/mol})\\\hline
1  &     0.0  \\
3  &    41.2  \\
7  &    26.1  \\
9  &   196.8  \\\hline
\end{array}
From this we conclude that the most likely position to be protonated is N1, followed closely by N7, and also N3. Off the charts is, as expected, N9 as it breaks the aromaticity.
TL;DR: Order of protonation $1 > 7 > 3 > 9$, calculated at DF-B97D3/def2-TZVPP.

tl;dr: Do you mean to say this question cannot be solved theoretically on pen and paper and needs special software?

@GaurangTandon Yes. And even then it is only a very educated guess.

@Martin Why "educated guess"? Aren't computational softwares accurate enough to accurately mimic real molecules?

@ApoorvPotnis While computational chemistry can certainly do a lot, and explain a lot, the level of theory I used is also an approximation. Without proper calibration and cross-reference with more accurate levels it remains a very good educated guess. Given the small energy differences a quantitative statement is not possible. At least the calculations are good enough to gain the understanding, that there is not much of a difference in this case.

@Martin-マーチン If the calculations are very accurate, then can we theoretically predict anything about any compound?

@KingTut that's really more of a philosophical kind of question. I believe that with a better understanding (which may include yet unknown mathematics, physics, or chemistry) we would be able to predict anything about any compound. With the current state of the art, that may already be in reach for simple compounds. However, for complicated molecules, there will be changes which we can't get tackle. In any case, the situation here is by far too complicated to be treated without (rudimentary) quantum chemical models.


text:
Distinguishing propan-2-ol from ethanol using IR spectroscopy
A question requires me to distinguish between propan-2-ol and ethanol from IR spectra. To me, they both look nearly identical, even though I know both of them contain the same functional group and that IR spectra are only capable of identifying functional groups.
My question is, how do you distinguish the two, without using a mass spectrometer?

The isopropyl group shows characteristic coupled peaks around ~1350 cm-1

Here is the spectrum for ethanol:

Here is the spectrum for propan-2-ol:

Here are the tabulated peak positions ($\pu{cm^{-1}}$) and percent transmittance (%T) for the two. I've tried to match the appropriate peaks.
$$
\begin{array}{cccc}%
\hline%
\text{ethanol position} & \text{ethanol %T} & \text{propan-2-ol position} & \text{propan-2-ol %T} \\
\hline%
3358 & 12 & 3346 & 9  \\
     &    & 3334 & 9  \\
2974 & 13 & 2972 & 5  \\
2927 & 38 & 2933 & 19 \\
     &    & 2907 & 23 \\
2887 & 35 & 2884 & 20 \\
     &    & 2722 & 60 \\
     &    & 2669 & 60 \\
     &    & 2521 & 74 \\
     &    & 2408 & 79 \\
     &    & 2387 & 81 \\
     &    & 2198 & 84 \\
     &    & 1903 & 85 \\
     &    & 1766 & 84 \\
1455 & 59 & 1467 & 24 \\
     &    & 1409 & 29 \\
1381 & 54 & 1379 & 16 \\
     &    & 1368 & 17 \\
1330 & 68 & 1341 & 28 \\
1274 & 74 & 1309 & 23 \\
     &    & 1162 & 10 \\
1090 & 22 & 1130 & 9  \\
1050 & 7  & 1110 & 15 \\
     &    & \color{red}{954}  & 4  \\
881  & 37 & 818  & 13 \\
669  & 67 & 660  & 37 \\
     &    & 654  & 37 \\
     &    & 490  & 66 \\
     &    & 436  & 68 \\
\hline%
\end{array}%
$$
One way of differentiating between the two is that I don't believe the peak at 954 wavenumbers in propan-2-ol appears in ethanol. Another better way that doesn't require looking at the fingerprint region is to see that the hydrogen-bonding peaks (the broadest ones to the left) are red-shifted in propan-2-ol compared to ethanol.


text:
Can't nobelium form compounds like other lanthanides/actinides do?
Over the past week I've been browsing through WebElements just to find out some interesting facts/properties about certain elements in the Periodic Table.
I've just come across the uncommon element nobelium and it looks like it doesn't form compounds with anything despite being an actinide. For example, actinium can form Actinium Trichloride and promethium can form Promethium(III) Fluoride.
Is there a reason why nobelium is so 'special'?

Maybe because we have not been able to do the right experiments yet?  When you have to create elements artificially and they decay rapidly, chemistry can get hard to study.

Beyond number 100, there is practically no chemistry to speak about and no reason to ask why is that so.

Wikipedia reports that nobelium does form compounds, but unlike other later actinides it could adopt a +2 oxidation state as well as +3, which complicates identifying specific compounds (e.g. nobelium reacts with chlorine but the specific product is not known).  The +2 state is stabilized by the full shell configuration $[\ce{Rn}]5f^{14}$.  Experiments in aqueous solution seem to show a preference for the +2 oxidation state, and $\ce{No}^{2+}$ has been coprecipitated into solids.  But we have yet to definitively identify a neat nobelium compound in either specific oxidation state.


text:
Transfer of odor molecules from oil to water
I brew beer, I'm not a chemist.  I want to add natural peanut aroma to a beer.  The most common way this is being done by brewers is the addition of peanut powder, which is essentially the product left over after pressing most of the oil out of roasted peanuts.  This product is preferable in brewing to whole peanuts because the introduction of significant amounts of oil into beer will affect desirable characteristics like the "head retention".
I don't like the idea of using this powder for various reasons. First, odor molecules like to be in oil, so pressing out the oil means pressing out a lot of the aroma.  Second, by adding all that powder to the beer, the beer ends up with far too much body and the powder absorbs some of the beer which lowers the yield.
If you've ever tasted roasted peanut oil, you'd notice that it is intensely flavorful of roasted peanuts.  What I'd like to do is extract odor molecules from the peanut oil and ultimately get them dissolved into a water based solution (a.k.a. beer).  This is where I was looking for help/ideas.
I've mixed aromatic fats with ethanol/water solutions (like everclear) and I've found there to be a reasonable transfer of aroma into the ethanol/water. I don't see this working in this case though.  The amount of odor molecules transferred to the solution would be insufficient when used to flavor beer.  The alcohol content of the beer would end up too high by the time enough of the solution was added to get the desired aroma intensity.

In chemistry, the word "aromatic" refers to a property of the electronic structure - it is unrelated to whether a molecule has an odor. It seems like you used the word "aromatic" to refer to odor molecules. If that's really the case, you might want to clarify.

Thanks, I updated the question.  My experience is with beer, so beer with me.

According to this publication https://www.ncbi.nlm.nih.gov/pubmed/20866052, the molecules responsible for roasted peanut odor are not particularly hydrophobic. Maybe they are easily extracted to an aqueous environment?

I suppose I could just combine the roasted peanut oil and water in a stir plate and let it go for a while and do some tastings of the resulting water, resulting oil, and the original oil for comparison.


text:
Why is cyclopropylmethyl radical opening so fast?
Ring opening of the cyclopropylmethyl radical is reliable and fast, and so has been used as a radical trap and a radical clock.
Could someone please explain what exactly makes this reaction so fast, besides the ring strain? Why is the anionic ring opening slower?

Cyclopropylmethyl radical undergoes a very rapid ring opening rearrangement that relieves the ring strain and is enthalpically favorable. The rate-constant for this reaction is 8.6×107 s−1 at 298 K.

@ChakravarthyKalyan Thanks! But why is the anionic process much slower?

https://chemistry.stackexchange.com/questions/19090/what-is-the-reason-for-the-exceptional-stability-of-the-cyclopropylmethyl-carboc


text:
Molecular formula and decimal number
Today I did some stoichiometry exercise about finding the ratio between the elements. What confused me was that while the ratio was right, but it was decimal.
Ex.

Reducing $\pu{0.800 g}$ of lead oxide with an excess amount of hydrogen, $\pu{0.725 g}$ of elementary Pb formed. What formula of the oxide?

In the chapters I've learnt so far, no oxidation number or stuff was mentioned. So I did as the examples states. I did this:
$$m(\ce{PbO})=\pu{0.800 g},$$
$$m(\ce{Pb})= \pu{0.725 g},$$
$$m(\ce O)= 0.800-0.725=0.072(\pu g).$$
The ratio is therefore $x/y$ where $x =0.725/207.2
$ and $y = 0.072/15.9994$
which is equal to $11.59/14.91 =0.7$, so the formula should be $\ce{Pb_{0.7}O}$ and if I multiply it by 2 I get $\ce{Pb_{3\text/2}O2}$; Multiply it by 2 again I get $\ce{Pb3O4}$, the answer.
Why should I have to do this? Is there anyway to go from decimal to fraction with the correct answer?
Now, I didn't find the answer until later in another problem, where the same thing happens. I was stuck on the decimal ratio. It was:

$\pu{10.0 g}$ of elementary phosphorus reacts completely with $\pu{7.75 g}$ of sulfur. What is the molecular formula?

I did the same thing: number of moles of $\ce{P}$ is 0.32, number of moles of $\ce{S}$ is 0.24. I calculated the ratio and it was 1.33:1, so the formula should be $\ce{P_{1.33}S1}$, but if I multiply by 3, I get the answer neatly. Why is that? Am I doing this wrong?

What is supposed to be the question here? Why proportions of elements aren't written as fractions?

well 0.800-0.725 = 0.075 not 0.72. The ratio also shouldn't be rounded to 0.7. You got lucky that you got the right answer.

You must find integers as for your are requested to find a molecular formula, or, in case, a minimal or a  formal formula for ionic compounds.
Obviously each atom must enter in that formula as an integer, because a molecule is composed of atoms and not by parts of them.
If you were requested to find the ratio between elements in a sample, then does not really matter if the answer is given as 1.5:2 or as 3:4.
Cases exists in which the ratio is not that of two integers, these compounds are called non stoichiometric and results from vacancies in ionic lattices.
Doping is another case in which that happens.
But you will never be asked to find the molecular formula of something that has none.
For an unknown sample, you would find what case applies. A ratio of integers would indicate a possible molecular nature of one compound.

By convention, chemists have decided to use whole numbers for molecular compounds and most simple solid-state chemistry. (Exceptions are doped compounds, where the exact composition is not fully fixed and the fractions involved are tiny.)
So, no, I do not think that the work shown in the question is wrong. By convention, you multiply by the lowest number that will give you integer numbers until you get an indication (usually by a higher molecular mass, think mass spectrometry) that the molecule is even larger than that. Consider the example of ethane ($\ce{C2H6}$), wherein a ratio calculation like the one carried out in the question would yield $\ce{CH3}$ -- clearly a wrong formula for that compound.

To TAR86 and most importantly to @Federico. " by convention" is an unhappy choice or words.  It is at the basis of modern atomism. There is 1 atom of H. Half of it doesn't exist.

@Alchimista I see your point for molecules, for FeO/Fe2O3 systems, not so much.

Doping is different story as well as alloys and ionic crystals. Obviously I refer to molecular formula. This is what I want to highlight especially to the OP as dramatically important for a beginnert. I had no doubt you know what you mean, but in general the incipit of your answer is potentially misleading. Better to point out the existence of non stoichiometric compounds, for which no molecular formula can be attained. Else the question make little sense, as does not relly matter if not just integeres can be obtained.

For the iron system you mentioned, indeed the only way to use whole numbers is in FeO/Fe2O3 as you did.


text:
How does baseline correction affect the absorbance in UV-vis spectrophotometry?
I did experiments with UV-Vis spectrophotometer to obtain a set of wavelengths vs absorbance for various concentrations. I found that I missed one trial of different concentration and I needed to obtain the data for the trial I missed. I remade a baseline solution and obtain the data I want. I need to graph the data from the trial I missed with the rest of data that I obtained before. I wonder if baseline correction from the previous data has effects on my another trial that did not run at the same time. Although I use the same concentration of solution for baseline solution, it will not lead to exactly the same data with the other data I did before.
I wonder if the baseline correction affects the absorbance of my new data and can I compare the new data with the previous data that has the same concentration baseline but not run at the same time?

If I understand ..... obviously you have to assume that your baseline solution (what is? Usually just solvent but it could have been a matrix) was correctly prepared in both cases.


text:
What are azeotropic mixtures?
I’m not able to get the qualitative idea of azeotropic mixture.
What does it mean in simple language?

Try google eg https://en.wikipedia.org/wiki/Azeotrope

If you have two components in a mixture, say water and ethanol, and you heat the liquid to the boiling point of that mixture (at the given pressure), you may find that the composition of the resulting vapor and the remaining liquid are different. In that case, you have a non-azeotropic mixture and the composition of vapor and liquid will change as you continue boiling.
For the case of water/ethanol, at some point, the composition of vapor and liquid will match (and no longer change, no matter how long you boil, until all liquid has evaporated). That composition/mixture is called an azeotrope.
The liquid mixture of oxygen/nitrogen is an example for a system that does not show an azeotrope.

After the composition of vapour and liquid matches why is there no change in composition no matter how long we boil?                              This is the point that I’m unable to grasp.

@RitwikDas My explanation to that takes us into the realm of statistics.

would love to hear it.


text:
Atropisomerism of naphthyl alcohol

In March's Advanced Organic Chemistry (6th ed.), it's stated that the above two are atropisomers. I don't get why is it so. How is the topmost $\ce{C}$ locked in its place? Surely, there's steric hindrance due to presence of two methyls but how does that prevent rotation? It can certainly rotate with the methyls being "rigid" in their positions. Also, are they chiral? I believe they aren't because they are superimposable mirror images.

Think of the di methyl hydroxy group as the fins of a fan and planar hydrogen of naphthalene jutting out. Rotation will be restricted.

The hindered rotation is due to the hydrogens at the naphthyl moiety. The following shows the rotation calculated at the DF-B97D3/def2-SVP level of theory. To better visualise this, I have chosen a mode with large atoms (not actually using the van-der-Waals radii, because that looked very strange; click here for a ball and stick version):

We can observe three minima and three transition states corresponding to that rotation, the depiction you have posted is slightly inaccurate as it shows one local minimum and a transition state.
From the three states that should be observable, the global minimum is achiral (1), while the other two are enantiomers (2, 3). The latter are connected via a low lying transition state (TS-2-3); probably because of the hydrogen bonding interaction guiding it. The global minimum on the other hand is separated via a very high energy barrier. These transition states (TS-1-2, TS-3-1) are again enantiomeric to each other.
I have calculated the associated energy barriers in $\pu{kJ/mol}$ at the DF-B97D3/def2-TZVPP level of theory.
\begin{array}{c*{6}{r}}\hline
\text{State} &
\Delta E_\mathrm{o} &
\Delta G^{\pu{50 K}}&
\Delta G^{\pu{75 K}}&
\Delta G^{\pu{100 K}}&
\Delta G^{\pu{298.15 K}}&
\Delta G^{\pu{500 K}}\\\hline
%
\mathbf{(1)}     &
    0.0 &  0.0 &  0.0 &  0.0   &  0.0  &  0.0\\
\mathbf{\text{(TS-1-2)}}     &
   32.9 & 32.9 & 32.9 & 33.1   & 35.7  & 39.7\\
\mathbf{(2)}     &
    0.7 &  0.7 &  0.7 &  0.8   &  1.4  &  2.3\\
\mathbf{\text{(TS-2-3)}}     &
   12.0 & 12.1 & 12.1 & 12.3   & 14.9  & 18.9\\
\mathbf{(3)}     &
    0.8 &  0.8 &  0.8 &  0.8   &  1.5  &  2.3\\
\mathbf{\text{(TS-3-1)}}     &
   32.9 & 32.9 & 33.0 & 33.1   & 35.7  & 39.8\\\hline
\end{array}

From these values we can estimate rate constants for the energy barrier with the Eyring equation, which employs the transition state theory:
$$
k = \kappa \frac{k_\mathrm{B}T}{h}\exp\left\{\frac{-\Delta G}{RT}\right\}$$
I use $\kappa = 1$, i.e. I am ignoring it, to get a guesstimate at which temperature we will be able to distinguish between all, or some of the states.
\begin{array}{c*{5}{l}}\hline
k/\pu{s-1};\ T= &
\pu{50 K}&
\pu{75 K}&
\pu{100 K}&
\pu{298.15 K}&
\pu{500 K}\\\hline
%
\mathbf{(1) \to (2)}     &
   \pu{4.9E-23} & \pu{1.8E-11} & \pu{1.1E-5} & \pu{3.5E6}  & \pu{7.5E8}\\
\mathbf{(2) \to (3)}     &
   \pu{2.6E-1}  & \pu{5.4E3}   & \pu{7.8E5}  & \pu{1.5E10} & \pu{1.1E11}\\
\mathbf{(3) \to (1)}     &
   \pu{2.8E-22} & \pu{5.9E-11} & \pu{2.8E-5} & \pu{6.1E6}  & \pu{1.3E9}\\
\hline
\end{array}
From this we can see, that at room temperature the rotation is pretty much free with a couple million transitions per second. At around $\pu{100 K}$ the is hindered; from the global minimum (1) to the adjacent minima the methyl groups simply cannot pass. Without haven calculated the values, I'd guess that this rotation starts to become hindered at around $\pu{150 K}$.
The other barrier still is easily manageable. That essentially means, that the enantiomers still readily interconvert. You will spectroscopically observe a superposition, which gives rise to the schematic structures that were used in the book. Even at $\pu{75 K}$ this barrier is taken a couple of thousand times per second
Only around $\pu{50 K}$, the interconversion between the enantiomers will also be hindered, and you should be able to distinguish all three states.
TL;DR:
You will start seeing atropisomerism at about $\pu{150 K}$, but at that temperature you'll only see the superposition of 2 and 3 (probably what is depicted in the book, i.e. left in your question) and the achiral, global minimum (on the right).

Side note:
It should be fairly obvious, that non covalent interactions also play an important role in this molecule. The barrier between the enantiomers 2 and 3 is probably significantly lower because of the hydrogen bond. However, one shall not forget, that this interaction is also present in the local minima to a not insignificant amount. The following pictures show the non covalent interactions analysed with NCIPLOT-3.0 and plotted with VMD; green areas are weakly attractive, blue areas are stronly attractive (hydrogen bonds), red areas are repulsive.


text:
Do catalysts themselves need an input of energy to work?
Do catalysts themselves need an input of energy to work?
My book says "catalysts do not per se require an energy source". What does it mean by per se? If they don't use energy how do they activate?

Come to think of it, do the reactants need energy?

@IvanNeretin Well then what is the activation energy? I always thought of it as the minimum energy needed to be supplied to the reactants ...

The word "supplied" may be misleading. We don't have to supply anything, since thermal energy is already there. If it isn't, then of course we have zero K and no reaction.


text:
Base used in Claisen Condensation
Here's what I read about which base to use in Claisen Condensation reaction (two esters or ester + carbonyl compound forming a beta keto-ester or beta-diketone):

The base used must not interfere with the reaction by
  undergoing nucleophilic substitution or addition with a carbonyl
  carbon. For this reason, the conjugate sodium alkoxide base of the
  alcohol formed (e.g. sodium ethoxide if ethanol is formed) is often
  used, since the alkoxide is regenerated. In mixed Claisen
  condensations, a non-nucleophilic base such as lithium
  diisopropylamide, or LDA, may be used, since only one compound is
  enolizable. LDA is not commonly used in the classic Claisen
  or Dieckmann condensations due to enolization of
  the electrophilic ester.

Here's what I didn't understand, particularly -

Why is the conjugate sodium alkoxide base of alcohol generated, used? The paragraph talks about regeneration of alkoxide ion; but I didn't really understand that part. (Seems like something related to chemical equilibrium?)
They said that Lithium Diisopropylamide is NOT generally used due to enolisation of electrophilic ester. What do they mean by it? Why can't we use LDA? Which bases can we use instead?

Could someone please answer the above two questions, with proper explanation? Thanks a lot!

Consider that there may be other reactions also going on, not just acid-base chemistry. So, for instance in the Claisen reaction of ethyl acetate, if methoxide was used instead of ethoxide then you may get methyl acetate formed as well.

Are there other advantages? I thought that formation of the alcohol and its successive ionisation (negligible, but non zero) in aqueous solution maybe a way of regenerating the catalyst so that the reaction is thereafter self sustaining. Is it true?

This is NOT done in aqueous soln. If an ester group is part of the reaction then the leaving group is the alkoxide of the ester, this is how the alkoxide is regenerated.

The Claisen condensation of ethyl acetate (pKa=~25) is usually conducted in concentrated EtONa/EtOH (pKa=~16) or EtONa in an inert solvent such as toluene. The product, ethyl acetoacetate (AAE; pKa=~11) , is deprotonated by the EtONa generated in the condensation. Unlike the aldol condensation, which is catalytic in base, the Claisen condensation must be conducted with stoichiometric base.


text:
When boiling a liquid, how vapour pressure and ambient pressure get equal?
When boiling a liquid, how vapour pressure and ambient pressure get equal?
As vapour pressure will increase, it will lead to increase in ambient pressure also.
let, Initially there would be some ambient pressure acting on the liquid, in open vessel.
Now, when we start heating the liquid, temperature of liquid will rise, thus vaporisation of liquid will take place. The vapour will also exert pressure on liquid along with atmosphere. Thus, ambient pressure will become higher and so the boiling point will increase and liquid will not boil. This will go on and on, and liquid will never boil. Obviously, liquid will boil at some point of time.
Pls. tell where I am getting wrong.

In a tiny closed vessel, you are kind of right. But if your container is in a room, then the amount of liquid initially boiled off will increase the ambient pressure only very, very slightly. The vapour pressure will catch up easily.

@TAR86 In that condition, pressure applied by vapour will also be small.

Pressure applied by vapour will be zero. See, the atmosphere of Earth is pretty huge. It presses at you with 1 atm, and you won't change that when you boil your liquid in an open vessel.

Related: https://chemistry.stackexchange.com/questions/38653/vapor-pressure-and-boiling-point-for-solutions-in-open-containers/38654#38654

@IvanNeretin Thanks, I did't found that question earlier.

Thanks to @IvanNeretin {https://chemistry.stackexchange.com/a/38654},

Boiling is not about the upper surface at all. Boiling means that bubbles are forming at the bottom and probably within the bulk of the liquid. Bubbles have pure water vapour. That's where the equilibrium between vapour and water actually comes into play.
'Vapour pressure of this vapour in bubbles' and 'ambient pressure' should be equal. This is because, the bubbles must withstand the external pressure, otherwise they would not be able to form in the first place (or if they will, they will be instantly squeezed into nothing). That's why they actually appear only when vapour pressure becomes equal to ambient pressure.

With the complication that this is in a small pot. Otherwise hydrostatic P add on top.


text:
IUPAC way of writing prefix of alkyl isomers of an ether
Consider this compound:

I am able to identify the base chain as propane. But, I am having trouble naming the other side of the chain. Will it be named as:

2-(isopropoxy)propane,
2-[(propan-2-yl)oxy]propane, or,
2-(2-propoxy)propane

What is the preferred IUPAC way for such a naming of isomeric alkyl groups?

why is option 2 preferred over "1-methylethyloxy"?

@EashaanGodbole I think "1-methylethyloxy" is wrong because there is no principal chain (i.e. no "ethane"/"propane"/etc.)

@GaurangTandon I asked because we used "1-methylethyl" and "1,1-dimethylethyl" for isopropyl and tert-butyl respectively as the point of connection of the side chain is often considered the "1'" carbon.

@EashaanGodbole As far as I can guess this will lead you to the common name of this compound, diisopropyl ether

According to Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book) (P-63.2.2.2), the preferred prefix for the substituent $\ce{(CH3)2CH-O\bond{-}}$ is 'propan-2-yloxy'. Both 'isopropoxy' and '1-methylethoxy' remain permissible in general nomenclature.
Correspondingly, the PIN of the symmetrical ether $\ce{[(CH3)2CH]2O}$ is 2-[(propan-2-yl)oxy]propane.

I'm quite sure the double parentheses are required based on other similar examples given in the Blue Book, but am not 100% sure of the rationale for it and would be grateful if somebody could confirm that. I'm thinking one set comes from P-16.5.1.1 as 'propan-2-yloxy' is a compound substituent, and one more comes from P-16.5.1.2 to separate the locant '2' in 'propan-2-yl' from the remainder of the name, but don't quote me on that.

You may have found another bug, e.g. 2-chloro-4-(5-chloropyridin-2-yloxy)pyridine (P-15.3.2.4.2) was fixed to 2-chloro-4-[(5-chloropyridin-2-yl)oxy]pyridine in the errata.

@mykhal Mmhmm, there is a similar erratum on pg704. I’m not sure whether anything else is actually wrong, though. Maybe the prefix should be named ‘(propan-2-yl)oxy’ instead of ‘propan-2-yloxy’. However, the entire section on alkoxy prefixes doesn’t have that set of parentheses (P-63.2.2.1.1), so I’m inclined to believe they’re right.

Maybe not preferred officially in this case, very nice IUPAC names are the symmetry-based ones (multiplicative name or ether functional class names):

2,2′-oxydipropane
di(propan-2-yl) ether
diisopropyl ether

(One might be tempted to replace "di" with "bis", but it should not be the case for this simple substituent.)
2-[(propan-2-yl)oxy]propane PIN name in orthocresol's answer is correct. For symmetrically characteristic-group-disubstituted derivatives, above mentioned name type is used, e.g. $\ce{HO-C(CH3)2-O-C(CH3)2-OH}$ 2,2′-oxydi(propan-2-ol).

The backbone has 3 carbons so it is -propane. On the middle carbon (carbon number 2), there is a branch consisting of an oxygen atom and further branching off to three methyl groups. Three is prop- and oxygen is -oxy, so we have propoxy-. So your second option is discarded.
So currently we have $\text{O-CH}_3\text{-CH}_3$ with the $\text{CH}_3$ next to the $\text{O}$ branching off to a methyl group. In other words, this $\text{CH}_3$ is in the middle, with only one branch, so we call it iso-.
As said previously, the carbon number 2 of the backbone has a branch, so the IUPAC name for the compound is 2-isopropoxypropane (your first option) and is more commonly known as diisopropyl ether.

Note that, according to current (2013) IUPAC recommendations, the prefix ‘isopropoxy’ is retained for use only in general nomenclature.


text:
Converting phenol to phenetole
Disclaimer: high school question
I was hoping to make phenol do a nucleophilic attack on $\ce{C2H5Cl}$. But then, I noted that the $\ce{O}$ atom in phenol is NOT as nucleophilic as the regular $\ce{R-OH}$ (since it delocalises with the benzene ring). So, I am afraid phenol's $\ce{O}$ atom might not be a good enough nucleophile to achieve the conversion by this method.
Is there any other method to achieve this conversion?

Actually it is a better nucleophile than standard -OH. Your idea of Phenol plus ethyl chloride under basic conditions is perfectly sound. I would suggest using ethyl iodide though as it is more reactive and ethyl chloride has a boiling pt of 12C making it difficult to keep in the reaction mixture.

@Waylander oh, I didn't have basic conditions. You're right, I missed the basic concepts. Closing as a duplicate. Thanks a lot for your help! :D


text:
Which is the most acidic hydrogen in vitamin C
I thought it would be the lower first $\ce{-OH}$ because it's close to the double bond with oxygen, so I thought there would be more electron withdrawal from the oxygen atom due to its electronegativity.But the correct answer is actually the second lower $\ce{-OH}$, which I really don't understand

My attempt to the question:

The first lower implies to the one closer to carbon double bond oxygen group

And the second lower is to the right of first lower

Try drawing both of the structures. You'll notice something.(add them to the question as well)

Compare the $\ce{OH}$ bonds in Vitamin C (ascorbic acid) and decide which one is the most acidic. The  most acidic proton  in ascorbic acid is the one whose  conjugate base is most resonance stabilized.

Removal of either of these $\ce{H}$'s  at hydroxyl group A or B does not give a resonance stabilized anion:

$\bf{Scheme \ 1}$

The proton  at D is less acidic, since its conjugate base is less
resonance stabilized with two resonance structures:

$\bf{Scheme \ 2}$

The proton at C is the most acidic proton in ascorbic acid since the conjugate base is most resonance stabilized:

$\bf{Scheme \ 3}$

Good answer except for one detail, the molecule in the answer is not oriented the same wah as in the question, so it is not clear that the answer was the lower right hydroxyl group in the original structure.  How hard is it to get consistent orientation?  Failing that, add text saying that C was the lower right hydroxyl group in the original structure.

@Oscar Lanzi I will reorient the post as you have suggested right I in the morning.Thankyou

The correct anion is stabilized by the mighty resonance.

That's about it.
Hydrogen bonding is irrelevant, since it affects both variants the same way.

As shown, if the hydrogen on the bottom-right position is removed, the negative charge is stabilized by more number of resonance structure (more conjugation) as compared to when the hydrogen on the bottom-left position is removed.
So the hydrogen on the bottom-right position is more acidic.


text:
experimental evidence of charge separation in zwitterions
Zwitterions vs ampholytes
The former having charge separation, the latter not. See the abstract from J.E.A. Comer's reference here.

In particular where he says 'Both molecules have two pKa values. In clioquinol, the base pKa is lower than the acid pKa, so at pH values between the pKa values, the molecule is neutral. However, the acid pKa for ampicillin is lower than the base pKa, and the molecule is charged across the entire pH scale.'

Now, the part that I don't get is how one decides that the 'acid' pKa is this value and the 'base' pKa is that value, and therefore that there is in fact charge separation in the molecule or not (depending on whether intramolecular deprotonation from a hypothetical non-charged species is thermodynamically favoured or not).

Please correct me if I'm wrong. If you have a real substance in front of you and you want to know the pKa's, maybe you dissolve it in water and titrate it with a strong base, starting from very acidic pH, following the pH curve vs the equivalents of base added, and you should get the classic 'jumps' at the points corresponding to the pKa values.
Suppose that in the 0-14 pH range you find 2 pKa's, let's call them pKa1 and pKa2, with pKa1 < pKa2. At that point all you know is that this molecule is initially protonated N times; at pKa1 it will be half N-protonated and half (N-1)-protonated; at pKa2 it will be half (N-1)-protonated and half (N-2)-protonated.
In what was does that tell you where protonation takes place?
In the example from the reference, ampicillin and clioquinol are mentioned.
OK, I agree that from knowledge of chemistry we can say that the pKa of phenols is about 10 and the pKa of pyridinium is about 5, so I don't expect pyridine to deprotonate phenol, or clioquinol to be zwitterionic. But in fact this is an assumption - what do I know of what happens in a molecule where both groups are present, maybe in some particular arrangement that may favour reactions that wouldn't happen intermolecularly?
And for ampicillin, OK, benzylamine is usually sufficiently basic to be protonated by an average carboxylic acid. But how do we know for sure that the ampicillin molecule exists (in solution? in the solid as well?) with a deprotonated carboxyl oxygen and a protonated amino group?
Based on what experimental evidence does one decide that a molecule is zwitterionic or ampholytic, in the sense defined in the above cited abstract?
And are there 'borderline' cases, meaning: can we find in one solution molecules (of the same substance, of course) with zero total charge, but some with charge separation and some not?

I will add this comment in case other people have the same question. I can't post an answer because the original post was closed down (...). It turns out that a molecule with overall charge z may indeed exist in all its allowed tautomeric forms, so the answer to the last question in the post is: yes. We currently have access to some software that allows to calculate the % of each species according to its charge and the % within a single charge class.


text:
Question about resonance structures
I stumbled upon an exercise asking to draw the resonance structure of $\ce{SO4^{2-}}$.
We can choose between the resonance structures by calculating the formal charges and by checking which one gives out the least value, and which one has the negative charges on the more electronegative atom.
My Chemistry Textbook "Chemistry 10th edition" by Chang states that:

The general rule for elements in the third period and beyond is that a resonance structure that obeys the octet rule is preferred over one that involves an expanded octet but bears fewer formal charges.

As I've searched more about this example , I've found out that most exercises that treated this molecule have labeled the one that has the least formal charge as the more stable.
This got me confused, is the resonance structure which obeys the octet rule the more plausible or is it the one which has the least formal charge?

Your textbook is correct.

@Mithoron is right.  The "expanded octet" model was built to fit molecules like $\ce{SF_6}$ to pairwise bonds.  Today we use molecular orbitals instead of pairwise bonds, and the molecular orbitals can be (and are) constructed without expanded octets -- including the case of $\ce{SF_6}$.

https://chemistry.stackexchange.com/questions/47086/why-does-calculating-formal-charges-help-us-draw-correct-lewis-structures