Analytical gradient of the Hartree-Fock nuclear-electronic repulsion 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.


Is it possible for three carbon double bonds to form next to each other in a carbon chain?

There's two different ways to stick double bonds next to each other.

The first is to put the two double bonds right next to each other, the "cumulenes":

Image from Wikipedia Commons

In these compounds, the pi orbital of one double bond does NOT hybridize/delocalize with the pi orbital of the adjacent double bond. For example, in the structure on the far left, the three carbon propadiene, the two end carbons are hybridized $\ce{sp^2}$, and the center carbon is hybridized $\ce{sp}$. The remaining two $\ce{p}$ orbitals of the center carbon participate in the double bonds, one to each. But because the two $\ce{p}$ orbitals are perpendicular to each other, there's no communication between the two pi bonds - they're perpendicular to each other, too. One of them is in the plane of the screen, and the other is in the plane going in/out of the screen. (Or, it is in the simplest model of organic bonding that's taught. Once you get into things like molecular orbital theory things get more complicated, but the fact of two non-interacting orbitals remains.)

You can see this by looking at where the hydrogens are located in propadiene.

There's another way double bonds can be combined, and that with an interveneing single bond, the "polyenes":

Image from Wikipedia Commons

In a polyene, all the carbons are $\ce{sp^2}$ hybridized. Unlike a cumulene, the pi bonds of a polyene are all on the same "side" of the polyene. For example, in the polyene shown here, all the pi bonds are sticking in/out of the screen.

This means that you can draw resonance structures where the double bonds "travel" along the carbon chain, effectively swapping the double and single bonds. Like benzene, the distinction between double and single bond starts to break down, and it's more correct to think of a continous pi bond throughout the whole of the polyene chain.

But unlike benzene, the end effects for the bond-swap resonance form means it's less stable than the "normal" bond locations. This means it contributes less to the hybridized form. So while the polyene chain doesn't behave entirely like an alternating single/double bond form, it's not like it's really a "1.67" or "1.5" bond order, either.


Solute and Solvent classification in a Solution

By major/minor they mean whichever component has the greater/lesser presence in mass or moles.


Are there any descent symmetry tables for character tables of linear structures?

Here they are, finally I found them in some random presentation for a lecture:

$$\begin{array}{c|c} \hline
D_{\infty \mathrm h} & D_\mathrm{2h} \\ \hline
\mathrm{\Sigma_g^+} & \mathrm{A_g} \\
\mathrm{\Sigma_g^-} & \mathrm{B_{1g}} \\
\mathrm{\Pi_g} & \mathrm{B_{2g} + B_{3g}} \\
\mathrm{\Delta_g} & \mathrm{A_{g} + B_{1g}} \\
\mathrm{\Sigma_u^+} & \mathrm{B_{1u}} \\
\mathrm{\Sigma_u^-} & \mathrm{A_u} \\
\mathrm{\Pi_u} & \mathrm{B_{2u} + B_{3u}} \\
\mathrm{\Delta_u} & \mathrm{A_{u} + B_{1u}} \\ \hline
\end{array}$$

$$\begin{array}{c|c} \hline
C_{\infty \mathrm v} & C_\mathrm{2v} \\ \hline
\mathrm{A_1 = \Sigma^+} & \mathrm{A_1} \\
\mathrm{A_2 = \Sigma^-} & \mathrm{A_2} \\
\mathrm{E_1 = \Pi} & \mathrm{B_1 + B_2}\\
\mathrm{E_2 = \Delta} & \mathrm{A_1 + A_2} \\ \hline
\end{array}$$


Variable oxidation state of transition metal ions

Thermodynamics proposes, kinetics disposes.  The most stable form can have a high activation energy barrier surrounding it.  The medium matters.  Hard Lewis bases (water) stabilize high oxidation states (Cu(II), tungstate).  Soft Lewis bases (acetonitrile) stabilize low oxidation states (Cu(II); $\ce{W(CO)6}$ and http://www.google.com/patents/EP1995347A1?cl=en).

$\ce{Ti^{+4}}$ is the basement, especially with oxygen.  Now rationlalize titanium suboxide Magnéli phases.  They are spectacular as inert electrodes under extreme conditions, re Ebonex.

DOI:10.1016/j.electacta.2010.05.011
http://eprints.soton.ac.uk/cgi/request_doc?eprintid=152857&_action_null=Request+a+copy


How do you explain pKa to non-professional?

pKa and pH are related concepts but often confused. pKa is a property of a compound that tells us how acidic it is. The lower the pKa, the stronger the acid. pH is a property of a particular solution that depends on the concentrations and identities of the components.

For this discussion, I'm going to use the terms protonated and deprotonated to mean that a compound is associated or dissociated with a proton.

Based on the relationship between the pKa of a compound and the pH of a solution, we can predict whether a compound will be protonated or deprotonated. If the pH is lower than the pKa, then the compound will be protonated. If the pH is higher than the pKa, then the compound will be deprotonated.

A further consideration is the charge on the compound. Acids are neutral when protonated and negatively charged (ionized) when deprotonated. Bases are neutral when deprotonated and positively charged (ionized) when protonated.

Given the information you provided, if compound A (with pKa 7.9) is in a solution of pH 5.2, compound A will be in the protonated state.

Without knowing anything about the identities of A and B, the following is speculation. Most likely compound A is a base and compound B is an acid. The protonated version of A (its "conjugate acid") is more stable for long term storage than A itself.

Another possibility is that compound A is an acid and B is a stronger acid. If the deprotonated version of A is unstable, then compound B is added to ensure that if the mixture came in contact with a base, something more acidic than A would be present to react with the base.

EDIT BASED ON NEW INFORMATION:
As Mateus B said, lidocaine will be protonated by HCl at the amine nitrogen. This is likely done for two reasons. First, as its hydrochloride, it will be more water soluble. In solution, the compound will dissociate into protonated lidocaine cation and chloride anion. Second, as the nitrogen is protonated, it is less reactive to the environment. Over time, amines can react with oxygen to form the corresponding N-oxide and/or can absorb carbon dioxide.


What is the reaction equation for a mixture of trichloroethanoic acid and dimethylpropanoic acid?

TL;DR
Contrary to what the answers/comments have suggested, I would say that no reaction happens here. The acid–base reaction between the two given species is thermodynamically unfeasible, with an equilibrium constant $K \sim 10^{-9}$.

If you want an acid-base reaction to occur between A and B, it's no use comparing the $\mathrm pK_\mathrm a$ values of A and B. That tells you how acidic A and B are, but you're not interested in both of them acting as acids. You're interested in one of them being an acid, and the other being a base. So the relevant quantities are the $\mathrm pK_\mathrm a$ of A and the $\mathrm pK_\mathrm b$ of B!
In this context, let's keep trichloroacetic acid $\ce{Cl3CCOOH}$ as the acid; we already know that it has $\mathrm pK_\mathrm a = 0.65$. The question tells us that for the "base" $\ce{Me3CCO2H}$, its $\mathrm pK_\mathrm a$ is $5.05$. But we're not interested in that value; we're more interested in its $\mathrm pK_\mathrm b$.
To find that out, we need to look for data beyond the question text itself. The veritable Evans tables states that the $\mathrm pK_\mathrm a$ of $\ce{PhC(OH)+OH}$ is $-7.8$. That species is the conjugate acid of $\ce{PhCOOH}$, so the $\mathrm pK_\mathrm b$ of $\ce{PhCOOH}$ is $14 - (-7.8) = 21.8$. Recall that a larger $\mathrm pK_\mathrm b$ means that a species is a weaker base. This gigantic number should already tip you off to the fact that $\ce{PhCOOH}$ is a terrible base. The species in the question, $\ce{Me3CCOOH}$, will likely have a value that's broadly similar.
How do we use these values to find whether an acid-base reaction is plausible or not? Returning to the reaction
$$\ce{AH + B <=> A- + BH+,}$$
we have the equilibrium constant
$$\begin{align}
K &= \frac{[\ce{A-}][\ce{BH+}]}{[\ce{AH}][\ce{B}]} \\
&= \frac{[\ce{A-}][\ce{H3O+}]}{[\ce{AH}]} \cdot \frac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]} \cdot \frac{1}{[\ce{H3O+}][\ce{OH-}]} \\
&= \frac{K_\mathrm a(\ce{HA}) \cdot K_\mathrm b(\ce{B})}{K_\mathrm w}
\end{align}$$
For our molecules, we know that the relevant quantities are $\mathrm pK_\mathrm a(\ce{HA}) = 0.65$ and $\mathrm pK_\mathrm b(\ce{B}) \approx 21.8$. Putting in $K_\mathrm a = 10^{-0.65}$, $K_\mathrm b = 10^{-21.8}$, and $K_\mathrm w = 10^{-14}$, we get an equilibrium constant of
$$K = \frac{10^{-0.65} \cdot 10^{-21.8}}{10^{-14}} = 3.548 \times 10^{-9}$$
which is so small that it makes very little sense to claim that this reaction "happens".


Perturbation theory in Quantum/Computational Chemistry

Many people have already said this in the comments, but I will put it into an answer. Also, it would not be too hard to find a bunch of references in favor of how common any particular method is in quantum chemistry, so I won't bother to list a bunch of examples of perturbation theory being used.

That being said, if I had to guess, I would bet that second to DFT, second-order Moller-Plesset perturbation theory (MP2) is the most common method in quantum chemistry which includes electron correlation. Also, there are large communities in computational chemistry which virtually ignore the existence of DFT because they are willing to sacrifice computer-time (and possibly some accuracy) to be more certain that their answer is reliable. So, MP2 is a good example of perturbation theory being using in QM.

Also, you will often see people refer to coupled-cluster with singles and doubles excitations and a perturbative inclusion of the triples, CCSD(T), as the "gold-standard" in quantum chemistry. This is a bit more complicated application of perturbation theory, but I would say that CCSD(T) is used in virtually any problem where one can afford to do the calculation. There might be times where one could feel more confident about using DFT with a particular functional over MP2, but you would be hard-pressed to find a good argument to not do a CCSD(T) calculation if you can afford to do so. This is especially true if the alternative is to only use DFT because many functionals are parameterized against CCSD(T) calculations, so getting similar results from CCSD(T) helps validate the use of that functional for the class of problem.

There are also many other electronic structure methods which use perturbation theory such as CASPT2, one can also include perturbation excitations into truncated configuration interaction (CI) calculations, and probably lots of other but less-commonly-used methods. Also, one approach to including spin-orbit coupling in electronic structure is via perturbation theory.

One of the reasons there is such a vast array of methods which use perturbation theory is that they are diagrammatic theories, which means there is a rather simple way of understanding all of the terms physically and the connection between these terms and an algebraic representation of them is quite straightforward within the framework of second-quantization (basically using special creation and annihilation operators).

Also, I know you're mostly asking about electronic structure, but it should be noted that perturbation theory is often used to add anharmonicity onto a harmonic reference calculation of vibrational frequencies. Probably the most common method of doing this is via VPT2 which is basically the vibrational analogue of MP2.

In summary, perturbation theory is an extremely powerful tool which will almost certainly be applied to any problem where a simpler reference answer can be found. On this note, these days I don't think you could get away with doing only a Hartree-Fock calculation and expect people to believe your conclusions about the properties of the physical system. Also, there are such good, parallel codes for MP2 that if you can do a HF calculation, you can probably afford an MP2 calculation.

As a response to one of the comments, I would argue that CCSD by itself is not much more accurate than MP2 for the increased cost. Both methods can be understood as including double-excitations out of a reference determinant, and my experience is just doing CCSD does not improve much over MP2. Although MP2 and CCSD sometimes seem to be wrong in different ways, so they aren't strictly comparable. This is why people insist on CCSD(T) being the gold-standard and not just CCSD.


Probability of forming dihydrogen with molecular weight 3

Extending / reformulating Karsten Theis' answer: A probabilistic approach is to draw a tree about drawing two spheres where event $A$ (normal hydrogen, $\ce{^1H}$) dominates, but event $B$ (the next abundant isotope, $\ce{^2H}$) complements, $P(\ce{^1H} \cup{} \ce{^2H}) = 1 = \Omega{}$ and eventually to multiply the probabilities for each path (e.g., $P(\ce{^1H^2H})$ as shown on the left hand side).

Yet because in your case both paths $P(\ce{^1H^2H})$ and $P(\ce{^2H^1H})$ yield a dihydrogen molecule of mass 3, these individual probabilities need to be summed up.


Why doesn't a neutral solution at any temperature have pH 7?

As the temperature changes the value of $K_\mathrm{w}$ changes(as you wrote) and then the concentration of $\ce{H^+}$ and $\ce{OH^-}$ also changes and now the concentration of $\ce{H^+}$ will not remain $10^{-7}$ and hence the neutral point shifts. Therefore, at any temperature the $\ce{pH}$ of neutral point is not 7.


Why benzenediazonium bromide and iodide are not important?

There is indeed very little to be found in the literature.

A recent Russian paper (Ref.1) mentions benzenediazonium bromide as a reactant. This thesis from 1975 (Ref.2) contains a preparation of benzenediazonium bromide and an attempted preparation of benzenediazonium iodide; it notes that the iodide is unstable as does this reference here.  The discussion section of this Organic Syntheses' preparation of fluorobenzene (Ref.3) mentions an alternative preparation by formation of benzenediazonium fluoride and its decomposition referencing work from the 1900s.

My overall thought is that they are probably less stable than the well-known chlorides and undergo reaction in lower yields so they are of little interest.

References:

Yu. V. Ostapiuk, V. S. Matiichuk, N. I. Pidlypnyi, N. D. Obushak, "Convenient synthesis of $\alpha$-bromo ketones by the Meerwein reaction," Russian Journal of Organic Chemistry 2012, 48, 519–522 (https://doi.org/10.1134/S1070428012040094).
Geoffrey D. Picard, "Sulfonium Salts, Aliphatic and Aromatic Diazonium Salts, Formation and Isolation," Honors Theses 1975, 2119 (https://digitalworks.union.edu/theses/2119)(PDF).
D. T. Flood (Checked by W. W. Hartman and J. R. Byers), "Fluorobenzene," Org. Synth. 1933, 13, 46 (DOI: 10.15227/orgsyn.013.0046) and Organic Syntheses, Coll. Vol. 1943, 2, 295 (PDF).


There should be only one "/" symbol. When more are used, it should be unambiguous from the context what is the carrier, e.g. in your case it would be more appropriate to use a dash:

$$\ce{Mn-Fe/Al2O3}$$

From section 1.3 Composition, structure and texture of catalysts of 1975 IUPAC recommendations for heterogeneous catalysis [1, p. 79] (emphasis mine):

  Support. In multiphase catalysts, the active catalytic material is often present as the minor component dispersed upon a support sometimes called a carrier. The support
  may be catalytically inert but it may contribute to the overall catalytic activity. Certain bifunctional catalysts (§1.2.8) constitute an extreme example of this. In naming
  such a catalyst, the active component should be listed first, the support second and the two words or phrases should be separated by a solidus, for example, platinum/silica or
  platinum/silica-alumina. The solidus is sometimes replaced by the word "on", for example, platinum on alumina.

  Promoter. In some cases, a relatively small quantity of one or more substances, the promoter or promoters, when added to a catalyst improves the activity, the selectivity, or the useful lifetime of the catalyst. In general, a promoter may either augment a desired reaction or suppress an undesired one. There is no formal system of nomenclature
  for designating promoted catalysts. One may, however, for example, employ the phrase "iron promoted with alumina and potassium oxide".

References

Burwell, R. L. Manual of Symbols and Terminology for Physicochemical Quantities and Units. Appendix II. Heterogeneous Catalysis; Pure Appl. Chem. 46, 71, 1976. https://doi.org/10.1351/pac197646010071 (Open Access)


Lack of elaboration in IR spectra of compounds

In IR, wavenumbers below 1500/cm are called the "fingerprint" region. It's very tricky to assign anything there, because the transitions don't relate to specific bonds, but come from larger structures.
If you look deeper into the literature, you will surely find an explanation for that big hump at 800something inverse centimeters.
Generally, two substances with matching fingerprint region in the IR can safely be regarded as identical, but you cannot draw conclusions about the molecular structure. Hence the name.


Can the enthalpy change of a reaction be changed, and if so, will this affect the value of the equilibrium constant?

Short answer: No and yes.

Enthalpy is a state function. That means that it depends only in the current "state" - the identity of the substances, their concentrations, pressures, etc. That means that the absolute enthalpy of some substance is independent of how you produced it. However, that substance has different enthalpy in different states: gas versus liquid, pure versus in solution. If you change the state of the product(s), you with change their enthalpy. Same goes with any of the other thermodynamic potentials: entropy, internal energy, free energy.

And that of course will alter the equilibrium situation.


Vapour pressure and equilibrium constant

Equilibrium constants contain activities of all components involved. For gases, these are proportional to partial pressures, so we use those instead. For liquids and solids, they are constant, so we tend to disregard them altogether. But the activity of a liquid ceases to be constant as you start adding impurities. All of a sudden, you have to use $K=\dfrac{P_{gas}}{a_{liquid}}$.


Molar enthalpy vs Enthalpy of formation?

Molar enthalpy is the energy released when the substance undergoes combustion. It is essentially the energy required to fragment the substance into its component sub-compounds.

Heat of formation is the theoretical difference between the energy each atom typically has at STP and the total energy that the compound actually has. Essentially, it is the energy that goes into formulating the bonds between the atoms.

Molar enthalpy is generally used a lot more in experimental chemistry because it can be measured directly, and it gives useful values on the energy that can be extracted from a substance.

Heat of formation is preferred in theoretical chemistry, as it is the theoretical maximum amount of energy able to be extracted in perfect conditions.


Why is it easier to determine the radius of inert/noble gases?

There are several definitions of atomic radius. Guessing from context of your answer, I assume you are talking about  van der Waals atomic radius. By definition, it is a radius of atom in respect of equilibrium position with other atoms when only van der Waals forces are acting between the atoms. Since atoms of inert gases do not tend to make covalent bonds, the equilibrium distance between two in solid equals double atomic radius - plain and easy. Not so easy for other elements.

Since most metals and some non-metals form closely packed lattice in elemental form, there is no simple way to obtain their vdW radius. It is required to find a compound with the atom sterically unhindered and forming a crystal lattice containing close contact with said atom involved. Since most metals are electron-deficient and tend to interact with almost anything, no such experimental data are available for metals in a strict sense (though I can come with few examples when it may make sense). As such, vdW radii can be measured (and, actually, are meaningful) only for atoms of top-right corner of periodic table plus hydrogen. Furthermore, even for them vdW radius may fluctuate from compound to compound depending on the oxidation state and atomic neighborhood. Furthermore, the radius is pressure-dependent and, for chemically bound atom, may be unequal in different directions.

For covalent radii the problem is even larger, as the radius is dependent on near neighborhood and valence state in significant way (N-N bond legnth may vary from 1.10 Angstrom in dinitrogen to 1.45 in hydrazine)

As such, even if people talk about larger and smaller atoms, larger and smaller substituents, there is no strong definition for such relations.

As for atomic radius no related to vdW and/or covalent interaction, it depends on the definition of atomic border, which does not exist. A free atoms does not have a clear border, its electronic density fades exponentially with distance, but is non-zero for all finite distances.


Chemical equilibrium — why multiplication, rather than addition?

For the 8th grader, hmmm...

Let suppose this funny equilibrium equation:

$$\ce{2 S <=> H}$$

Where $\ce{S}$ are ice skaters on a giant skating ground and $\ce{H}$ are "collision heaps" of 2 skaters laying on the ice.

Now, what is the frequency of collisions leading to a heap, if there is "skater concentration" $\ce{ [S] }$ ?  It should be $k_1 \ce{[S]}$ as frequency should be proportional to $\ce{[S]}$, correct ?

But wait ! When there is more skaters, each skater would collide with  other skaters more frequently,  proportionally to $\ce{[S]}$ as well !

So collision frequency would be like $k_1 \ce{[S]}^2$

Now the heap "recovery":

It should be proportional to $\ce{[H]}$, as more heaps, more skaters would be  raising at the same time. So the heap recovery rate would be like $k_2 \ce{[H]}$.

Now imagine the equilibrium on the skating ground, where the rate of collisions is equal to the recovery rate
and number of collision heaps is constant.

Then $k_1 \ce{[S]}^2= k_2 \ce{ [H]}$

If we express the equilibrium constant $K=k_1/k_2$, then $$ K = \frac{\ce{[H]}}{\ce{[S]}^2}$$

Do you see ? No sum, but multiplication. Twice as many skaters leads to 4 times more heaps.

You can create similar system, where just for fun collide just boys and girls together.


How sure can we be about the structure of purine?

The article text (Ref.1) itself does go into details. The first thing to note: while aromaticity on the larger scale is retained in your structures 3 and 4, this is only true for the 10-π-electron system as a whole. In many of these types of 10 π systems made up of a six-membered and a five-membered ring, the two rings are not equal with the double bond in the five-membered ring often being much more susceptible to hydration. This prior thought should bias us against structures 3 and 4.
A pretty obvious first thought would be to take a look at the arrangement of individual molecules in the crystal structure. The paper does this in figure 3 which I have reproduced below. Wherever we expect the single proton, it should surely form a hydrogen bond to another purine molecule, meaning that these two nitrogens should be sufficiently close and in a favourable angle arrangement.

As the figure shows, the two nitrogens of the five-membered ring are in an arrangement that would suggest a hydrogen bond between them. The cut-off bits outside the unit cell are fragments of the next purine molecule not completed for brevity. This arrangement rules out 3 and 4 on the basis of no available hydrogen bonds and limits the discussion to 1 and 2.
As for the question whether the hydrogen prefers structure 1 or 2 this is admittedly more complex. As the authors note:

Since there is a possibility of tautomerism of the hydrogen between N(7) and N(9), particular attention was given to this problem. Two least-squares cycles, not quoted in Table 2, were computed in which a half-hydrogen was assigned to each of N(7) and N(9). Refinement of H(10), H(ll), H(12) and the two half-hydrogens resulted in the half-hydrogen attached to N(9) shifting to a position about 1 Å from N(7) of the neighbouring hydrogen-bonded molecule.

So basically, during refinement Watson attempted to put the hydrogen on both nitrogens equally. However, the refinement would show that the partial hydrogen on $\ce{N}$9 was moved sequentially along the $\ce{N\bond{...}H\bond{...}N}$ distance towards $\ce{N}$7. Taking the final $\ce{N-H\bond{...}N}$ distance of $\pu{285pm}$ and the $\ce{N-H}$ bond at $\pu{85pm}$, this means that there are approximately $\pu{115pm}$ between the two possible hydrogen positions on $\ce{N}$7 or $\ce{N}$9—and most of that was covered by the moving hydrogen. However, Watson also notes that the electron density corresponding to that $\ce{N-H}$ hydrogen was ‘much more smeared’ that those of the other hydrogens.
Sweet and Marsh arrive at the same conclusion using a different method that I did not fully understand.
All things considered, this gives us a pretty strong case that purine is best represented by structure 1 and that the principal tautomerisation structure would be 2 as that is where the hydrogen bond is directed.

Note: The authors state in the introduction to their paper:

This paper presents the results of two entirely independent investigations of the structure of purine (I).
The duplication of effort was discovered only when both three-dimensional studies were almost complete (Watson, 1964). Since different experimental techniques had been used, it was considered appropriate to publish both sets of results in the same paper and to make the necessary comparison between them. It has been a comfort to both parties that, except for systematic differences in the temperature factors, the two sets of results differ by less than their standard deviations.

This is why I focused so much on Watson and ‘ignored’ the other authors as the principal arguments are from Watson’s part of the paper.

Reference:

D. G. Watson, R. M. Sweet, R. E. Marsh, “The crystal and molecular
structure of purine,” Acta Crystallographica 1965, 19, 573-580
(https://doi.org/10.1107/S0365110X65003900).


Ignoring the difference of concentrations of reactants. Equilibrium

Let's look at your example and set up the ICE (intial-change-equilibrium) problem. Then I'll give you more inclusive of thumb.

Let's say reaction equation is

$\ce{A + B <=> 2C}$$

The law of mass action is:

$$K = \dfrac{[\ce{C}]^2}{[\ce{A}][\ce{B}]}=10^{-8}$$

Let's set up the ICE matrix. The change in concentration $x$, will be subtracted from $[\ce{A}]$ and $[\ce{B}]$ and added to $[\ce{C}]$.

$$\begin{array}{|c|c|c|c|}\hline
\ & [\ce{A}]\ (\mathrm{M}) & [\ce{B}]\ (\mathrm{M}) & [\ce{C}]\ (\mathrm{M}) \\ \hline
I & 10^{-20} & 10^{-20} & 0 \\
C & -x & -x & +2x \\
E & 10^{-20} - x & 10^{-20} -x & 2x \\ \hline
\end{array}$$

Now we pug in to the law of mass action:

$$10^{-8} = \dfrac{x^2}{\left(10^{-20} -x \right)^2}$$

Since all numbers have to be positive, let's take the square root of both sides:

$$10^{-4} = \dfrac{x}{10^{-20} -x }$$

Now we have a ratio that says that $x$ is four orders of magnitude smaller than $[\ce{a}]_i -x$. I could have easily used the initial concentration of $\ce{B}$ Put another way:

$$x \approx  K\cdot [\ce{A}]_i$$

If this value is much less than $[\ce{A}]_i$, then we can simplify.


Why the reduction potential of the reaction AgCl (aq) /Ag + Cl- is +0.22V

I don't know what your source is, but it should definitely be an $\ce{AgCl(s)}$ in your first equation. Wikipedia gets it right and if you find any textbook that doesn't explicitly state the phase of $\ce{AgCl}$, you can be pretty darn sure they meant solid because talking about aqueous $\ce{AgCl}$ makes little to no sense precisely because it is so insoluble in water.

So, essentially you've already answered your own question; precipitation of $\ce{Ag(I)}$ as $\ce{AgCl}$ stabilises it towards reduction and lowers the standard reduction potential relative to $\ce{Ag+/Ag}$.


Either reaction is valid for representing the reaction between zinc and sulfur. The important thing is that both equations stipulate the same stoichiometry (one atom of zinc for one atom of sulfur). The mathematics of chemical equations work just the same whether you add structural information about the reactants or not; the reaction equation is just shorthand for a system of mathematical equations involving the amounts and types of atoms.

In truth, neither equation is a correct description of the reaction at the microscopic scale (though I again restate that's not really the point of a reaction equation). Free zinc atoms do not interact with free sulfur atoms or $\ce{S8}$ rings. Rather, after a little bit of initial heating to get the reaction going *, liquid sulfur (a very complex mixture of sulfur chains ranging from $\ce{S6}$ to $\ce{S_{n}}$, where n is as large as 100,000) is in contact with a surface of zinc (or possibly the molten metal), and electrons are exchanged by some mechanism which probably hasn't even been elucidated, as electron transfer reactions are very quick and hard to monitor.

* You mention considering the reaction at room temperature. Even though thermodynamically, the reaction between zinc and sulfur is spontaneous at room temperature, it is kinetically very slow because both reactants are solid. Even if you mix them very very well, the reaction is diffusion-limited in the solid mixture; the atoms can't "reach" each other very well. Heating a little bit creates a small amount of liquid sulfur, which can then mix better with the zinc.


Usually, you shake chloroform with concentrated sulfuric acid, wash with water, dry over $\ce{CaCl2}$, and distill.

The density of chloroform is approximately $\rho = 1.48\ \mathrm{g\ cm^{-3}}$, whereas the density of concentrated sulfuric acid is approximately $\rho = 1.84\ \mathrm{g\ cm^{-3}}$.
Therefore, after you allow the layers to separate, the concentrated sulfuric acid is in the lower layer and the chloroform is in the upper layer.

After that, however, when you wash with water ($\rho = 1.00\ \mathrm{g\ cm^{-3}}$), the chloroform is in the lower layer and the water is in the upper layer.

If there is any doubt about the identity of a layer, add a drop of the layer in question to a small amount of water in a test tube. If the drop does not dissolve, the layer in question is the organic layer.

Caution: The mixture may become hot when the remaining sulfuric acid is removed with water, and chloroform boils at $61.2\ \mathrm{^\circ C}$.


How to identify the aromatic compound with the sum formula C7H6Cl2?

Hints to find the structure of (A):
(1) The given compound (A), is aromatic, hence, it is highly likely to have only one benzene ring (deduced from it's index of hydrogen deficiency/degree of unsaturation)
(2) Only those halides which can give a stable carbocation at the carbon to which the halogen atom is attached, can give precipitate of AgCl with AgNO3 solution. Therefore, one of the chlorine atoms is benzylic, as you mentioned.
(3) Now, oxidation of (A) yields a monochloro-benzoic acid, which on nitration, yields only one mono-nitro derivative. Out of the possible three options: ortho, meta, and para. now, chlorine atom on a benzene ring is ortho-para directing, while the COOH group is meta directing. Therefore, if the initial compound (A) would have -CH2Cl situated para to the Cl, it's obvious that, on oxidation, the COOH group would be formed from the CH2Cl group. Now, there is only two symmetric positions, at which both COOH and Cl will direct NO2+, hence, as both are symmetrical, they can be considered to be the same molecule. For that very reason, it only makes sense for (A) to be para substituted.


Nicco: Of course you can't "see it". If you were given the correct structure, the solution to your post would be straightforward. There should be a double bond in the lactam ring as shown in compound 1. The reaction is conducted in acetonitrile (bp. 82oC) at reflux. These conditions allow for a Claisen rearrangement to occur to form  ketolactam 2. Subsequent oxidative addition of palladium(0) [ligands ignored] to the C-Br bond creates the Pd(II) species 3 which adds to the double bond to afford cyclized species 4. Reductive elimination of Pd(0) regenerates the catalyst as alkene 5 is formed. The role of $\ce{Et3N}$ is to neutralize HBr that is formed. Addendum: Just found the original paper: https://doi.org/10.1016/0040-4039(94)88380-7


How important are those fields of chemistry in biotechnology and medicinal chemistry

Your question is broad and these chemistry which you have pointed out in your question is interlinked (Quantum ,Physical and Computational chemistry). The best way which I can explain this question is by giving links to main research carried out using these chemistry in biological science.

The study of quantum mechanics in photosynthesis is one such area and this nature paper(Nature 446, 782-786 (12 April 2007) is about that research

the highlights of the paper is the evidence for wavelike energy transfer through quantum coherence in photosynthetic systems

  two-dimensional electronic spectroscopy investigations of the FMO
  bacteriochlorophyll complex, and obtain direct evidence for remarkably
  long-lived electronic quantum coherence playing an important part in
  energy transfer processes within this system. The quantum coherence
  manifests itself in characteristic, directly observable quantum
  beating signals among the excitons within the Chlorobium tepidum FMO
  complex at 77 K. This wavelike characteristic of the energy transfer
  within the photosynthetic complex can explain its extreme efficiency,
  in that it allows the complexes to sample vast areas of phase space to
  find the most efficient path.

Other useful links

Nature 431, 256-257 (16 September 2004)

Nature 446, 740-741 (12 April 2007)


What are the units of your answer? You need a solubility, which is an amount per volume. If your amount is moles, the your solubility is in moles/liter: $\text{M}$.

That unit appears in the Henry's Law constant! $6.1\times 10^{-4} \frac{\text{M}}{\text{atm}}$. What can you do to get rid of the atmospheres in the denominator? You probably need the pressure and the mole fraction.


How does liquid gallium interact with a gold leaf

Broadly speaking, looking at the Au-Ga phase diagram one gets:

(Now, this is based on a paper from 2011, a more recent 2017 paper is open access).
Several things to notice (or get from the paper):

The solubility of gold in solid gallium can be considered to be zero.

The solubility of gallium in gold is not zero, but is on the order of a few atomic percent.

Slightly above the melting point of gallium, the solubility of gold in the liquid is also very small.

There are multiple intermetallic phases across the diagram, 4 of which are stable at (and below) room temperature.

Formation of an amalgam can be ruled out.

So, it is a pretty normal binary phase diagram. Now the phase diagram does not provide kinetics information, so the time needed to actually form any intermetallic compounds (and if those formed near room temperature would even be the equilibrium phases) is not known.
Now, if we look at how gold leaf is used, there is no glue used - the leaf is place, smoothed and hammered into place, and it just sticks. So try just sticking it.


Constructing a grenade (for a novel I'm writing.)

In short-no your technique will not work.  Reacting limestone with acid does not release enough energy to reach an appropriate temperature to ignite gunpowder and the acid would hinder combustion.

Fear not though, spontaneous ignition may be achieved by mixing glycerol and potassium permanganate which works well and is a classic example.  If you would like to be more out of the ordinary, zinc powder and ammonium nitrate with a minor amount of copper or ferric chloride and water will also create a spontaneous ignition for your gunpowder.


The first reaction is O-alkylation of p-cresol to give a 4-methylphenyl allyl ether derivative 3. The reagent in the first box should be 1-bromo-3-methylbut-2-ene (1; see the top box in the picture), which would undergo $\mathrm{S_N2}$ reaction with phenolic anion (2) in refluxing acetone. Note that potassium carbonate is a strong enough base to complete this reaction (this is surely a strong base than potassium bicarbonate).

The product 3 from alkylation would undergo Claisen rearrangement upon heating to give final alkylated cresol 4 in solvent-free condition. Remember, Claisen found his famous rearrangement first in solvent-free conditions, while he was trying to find the melting point of newly synthesized naphthyl allyl ether! (Picture is from https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3873100/ and modified accordingly)


According to Ionization constants and thermal stabilities of uracil and adenine under hydrothermal conditions as measured by in situ UV–visible spectroscopy Geochimica et Cosmochimica Acta Volume 93, 15 September 2012, Pages 182-204

  Adenine undergoes two ionization reactions. Density function calculations in the gas phase carried out by Russo et al. (1998) indicate that protonation occurs preferentially on the N1 nitrogen of the 6-member ring, in good agreement with the assignment given by Christensen et al. (1970b). The anionic species with a proton-deficient 5-member ring is the most stable.

Here, "Russo" is Protonation of thymine, cytosine, adenine and guanine DNA nucleic acid bases: theoretical investigation into the framework of density functional theory J. Comput. Chem., 19 (1998), pp. 989-1000 which says (emphasis added) :

  HOMO of adenine is formed essentially by the lone pairs of all nitrogens. The orbital of the amine group contributes with the higher coefficient, and that of N1 with the lower one. On the basis of this information, N1 protonation might be the less favorable process. On the contrary, the molecular electrostatic potential map (Fig. 5) indicates that N1, N3, and N7 are the better candidates for proton attachment. In particular, the N1 and N3 regions appear to be favored over N7. Experimentally, the preferred protonation sites of adenine are N1 and N3 with the former slightly favored.[reference 18] Our computation favors protonation at N1 over those at N3 and N7 by 1.12 and 7.38 kcal/mol, respectively. This trend is in agreement with previous theoretical computations.[reference 17] From bond orders analysis (see Fig. 6), and due to the small energy difference it is possible to view the N1 and N3 protonated forms in N1 the double bonds occur in C4—C5 and in the C2—N3 positions, whereas, in.the second, in the C4—C5 and N1—C2 positions as two resonance hybrids of a unique structure with a consistent charge delocalization. This observation allows better insight into the reason forw hich the protonation on the five-membered ring is unfavorable. In fact, as shown in Figure 6, the positive charge appears to be distributed only on the C8 and N9 atoms, because it is evident that no interactions exist between the two rings upon protonation.

"Christensen" of the first reference is the same as reference 18 of the second reference and is Thermodynamic pK, ΔH°, and ΔS° and ΔCp° values for proton dissociation from several purines and their nucleosides in aqueous solution Biochemistry, 9 (1970), pp. 4907-4913


Stereochemistry of product in Grignard-Gilman addition reaction

I have supplied a diagram that is more legible than the small one you have provided. The cis stereochemistry of the product is a function of axial, but not equatorial attack of the reagent. The stereochemistry is established after the addition occurs. Structures 1 and 2 illustrate the "chair-chair" interconversions of the cyclohexene ring. As the C-C bond is being formed by axial addition in 1, the carbon bearing the red p-orbital will rehybridize from sp2 to sp3 during the transition state (TS) for bond formation. This transition state (TS) will lead to enolate 3 through a low energy, chair-like pathway.
Alternatively, TS 2 would proceed through a higher energy, twist boat-like TS. Both routes occur through axial attack with 1 ---> 3 prevailing over 2 ---> 4. Enolate 3 is in equilibrium with enolate 5, which is unfavored by what is known as an allylic A1,3 interaction.1 The critical step is the kinetic protonation of enolate 3 on workup. Protonation of enolate 3 on carbon from the underside of the double bond (remote from the phenyl group) again occurs through a chair-like TS. Protonation on the top of the double bond would occur through a twist boat-like TS. This pathway is higher in energy and would have provided the trans isomer 9. Structures 6 and 7 are the two chair conformers of the cis-isomer. If one uses to vigorous a set of conditions upon workup, enolization of 6 and 7 can lead reversibly to enol 8 and eventually to the thermodynamic isomer 9.

1) Johnson, F., Malhotra, S. K., J. Am. Chem. Soc., 1965, 87, 5492.


You can’t really name it according to D/L.

This nomenclature requires you to draw a Fischer structure wherein the skeletal atom labelled ‘1’ by nomenclature rules is at the top; then, the orientation of the bottom-most functional group determines whether the compound is dextro, D (pointing to the right) or levo, L (pointing to the left. Below you can see Fischer depictions of all three tartaric acids.

Figure 1: D-tartaric acid, L-tartaric acid and meso-tartaric acid from left to right.

Now do one of the following: either rotate your PC screen/mobile device by $180^\circ$ or rotate your head by $180^\circ$.* You should notice that both D- and L-tartaric acid still have the bottommost hydroxy group pointing to the right or left, respectively. Both carboxylic acids are identical, so we would have to mark one carbon atom by using $\ce{^13C}$ to distinguish the top and bottom. The orientations lead to the same deduced stereochemistry.

This is not the case in the third molecule. Before rotation, both hydroxy groups are pointing to the right-hand side, afterwards they are pointing to the left-hand side. But, as I already stated, the two depictions are identical in the absence of isotope marking. Therefore, this compound eludes the D/L nomenclature and cannot be named unambiguously in accordance with it.

Thankfully, though, we have another descriptor that does not directly stem from D/L: it is the meso descriptor. It is used in case a molecule with asymmetric centres also features a plane of symmetry causing it to be identical with its mirror image and therefore achiral. While it is not part of the D/L terminology, it is the commonly accepted name for meso-tartaric acid.

*: Please only perform gymnastics if you can handle them. I decline all responsibility for injuries that may occur. Stack Exchange cannot be held responsible.


Why is the energy of an electron always negative?

First of all, regarding your comment, Rydberg's constant is not negative. If it were, then the equation you gave would have a positive energy all the time.

To be precise, Rydberg's constant is $$R_{\infty}=\frac{m_ee^4}{8\epsilon_0^2h^3c}$$ where all those constants have their usual meanings. Importantly, all of those constants are manifestly positive.

Whether the energy is positive or negative is entirely arbitrary. The energy of an electron can also be formulated as, $$E_n=R_h(1-\frac{1}{n^2}).$$

The only difference between the equation you wrote down and the one I just wrote down is that the above equation sets the zero of energy at the ground state of the hydrogen atom. You'll see that when $n=1$ the energy is zero and as $n\to\infty$ the energy approaches the Rydberg constant which is the ionization energy of the hydrogen atom.

When you say you wrote your exam I assume you mean you took your exam, not that you were writing the exam? Haven't heard that expressions before, but I suppose it makes sense. Maybe you were supposed to write that the negative energy represents the fact that the electron in a hydrogen atom is more stable (lower in energy) when associated with a proton than when isolated by itself.


Why is the IUPAC definition of adsorption so different from other definitions?

The IUPAC defintion isn't more limited than the ones you've read. Instead, it is more general.
Your confusion results from not being familiar with the term "condensed phase". You think a condensed phase is a 'more condensed liquid', and consequently think IUPAC is referring to a surface formed between a liquid and a 'more condensed liquid':  "where there is an interface of two liquids, one containing a solute, and one being more condensed".
No.  A condensed phase means a liquid or a solid.  Thus, when IUPAC says the following, what they mean is that adsorption can occur at liquid-liquid, liquid-solid, gas-liquid, and gas-solid interfaces.

An increase in the concentration of a dissolved substance at the
interface of a condensed and a liquid phase due to the operation of
surface forces. Adsorption can also occur at the interface of a
condensed and a gaseous phase.

Is the IUPAC definition completely general?  No.  It doesn't account for adsorption between supercritical fluids and solids. But it at least covers all the routine phase combinations.
As to the issue of whether IUPAC's defintion of solutions (which limits the solvent to be a condensed phase) contradicts its defintion of adsorption: While it could have been written more explicitly, a close read indicates there's not necessarily a contradiction.  Let's take each sentence in turn:
The first sentence concerns a liquid phase in contact with a solid or liquid surface.  Here the adsorbent is referred to as a "dissolved substance", which would make it the solute within a liquid solution:

An increase in the concentration of a dissolved substance at the
interface of a condensed and a liquid phase due to the operation of
surface forces.

Note that the solute can be (prior to its dissolution) a solid, liqud, or gas.
The second sentence concerns a gaseous phase in contact with a solid or liqid surface.  Here no explicit mention is made of a dissolved substance:

Adsorption can also occur at the interface of a condensed and a
gaseous phase.

Thus we could parse IUPAC's language to mean that adsorption involves either the solute within a liquid solution that is in contact with a liquid or solid surface, or a gas that is in contact with a solid or liquid surface (where that gas could be pure, or part of a mixture).


If there is an organic contaminant in the $\ce{Na2CO3}$, you could try to rinse the sodium carbonate with an organic solvent in which $\ce{Na2CO3}$ is insoluble.  Based on this paper, acetone and methanol would qualify.  Since those solvents, especially methanol, are toxic, I'm not recommending either one of those.  Based on the data for methanol and acetone, I'm guessing absolute ethanol would also be incapable of dissolving sodium carbonate, so it would probably the best (relatively) non-toxic solvent for extraction.  If you can't get absolute ethanol due to tax or beverage laws, you could probably used denatured ethyl alcohol instead.  It usually has small percentages of methanol, acetone, and/or isopropyl alcohol added to make it non-potable.

After you wash the sodium carbonate in this solvent, you should air dry it for several days, and then dry it in an explosion-proof drying oven at 100 - 200 °C.  Be very careful when combining flammable solvents with heated environments!

The problem could also be with the water you used, although you did say you used osmotically purified water.  The reason I think the water may be the culprit is that if the washing soda contained a brown contaminant, I would think it render the powder visibly brown.  If no brownness is evident until adding the soda to the water, it makes me think the high pH from the washing soda is precipitating minerals from the water, in particular iron.


Synthetic steps for the interconversion between aldehydes and ketones

You can easily convert aldehydes to ketones and vice-versa by using simple organic synthetic steps which are very much well known.

Converting aldehydes to ketones

For converting aldehydes to ketones there are numerous pathways. Here I am mentioning the simplest two ways, one of which is by using Grignard reagent and other by using thioacetal intermediates.

You can react aldehydes with Grignard reagents ($\ce{R^2 -MgBr}$) and perform acidic workup to generate secondary alcohols. Then you can oxidise the alcohol to get a ketone by commonly used oxidising agents like PCC (pyridinium chlorochromate).

The other way is to react the aldehyde with a propane-1,3-dithiol to generate a cyclic thioacetal . Now, interestingly, the proton attached to the carbon now can be removed by strong base like n-Butyllithium, and generate  Carbanion which can perform a $S_N2 $ reaction with an alkyl halide to generate cyclic thioketal. After that the thioketal can be hydrolysed using $\ce{HgCl2/CdCO3/H2O}$  to generate your desired ketone.

Converting ketones to aldehydes

The easiest step for this conversion is to perform a Bayer-Villiger Oxidation reaction on the ketone to get an ester. Then that ester can be reduced by Diisobutylaluminium Hydride (DIBAL-H) and an aqueous workup will generate desired aldehyde (along with an alcohol). (If you wish you can further oxidise that alcohol also in case you want aldehyde of that part).


Why does gas particle velocity affect rate of effusion?

Your point 1 is mistaken (or incomplete)
$$P=\frac{F}{A}=\frac1A\frac{\Delta p}{\Delta t}=\frac1A\times\frac{2nmv}{\Delta t}$$

This is assuming all particles hit perpendicularly (you can always modify this for all colissions by taking the average component of velocity in the perpendicular direction).

So, pressure is proportional to number of impacts, mass, and velocity. Pressure is not simply "number of impacts", rather it is a combination of number of impacts with other stuff. Since the energies are equal, we can say that $v\propto m^{-\frac12}$, and we get $P\propto n m^{\frac12}$. P is also the same, so $n\propto\frac1{\sqrt m}\propto v$.

Thus, when avg KE and pressure are the same, rate of diffusion is inversely proportional to square root of mass.


"The centre of positive and the centre of negative coincide" What does that mean?

Each molecule has negatively charged electrons, and positively charged nuclei.

If the centroid of the electrons' charges (electron probability density) is different than the centroid of the nuclear charges, then the molecule will have a permanent electric dipole moment.


How can I draw metallocenes in ChemDraw?

In the first step, separately draw the individual ring structures. ChemDraw comes with an template library of predrawn structures, including some suitable cyclopentadienyl rings.

You can place the ring structures anywhere in your document.

Select one ring structure. On the Structure menu, click Add Multi-Center Attachment. This adds a multi-center node as attachment point to the structure.

Repeat this process for the other ring structure. Asterisks indicate the created multi-center nodes.

You can use the multi-center attachment points to draw the coordination bonds. Note that bonds connecting to delocalized systems should be drawn as plain bonds; it is not acceptable to use dashed bonds. When one bond crosses another, if necessary, you can click Bring to Front or Send to Back on the Object menu to indicate which bond is in front. The bond which is behind will have a small section cut out.

Click to select the end (i.e. the terminal atom) of the coordination bond in one structure, and then shift+click to select the end of the coordination bond in the second structure. On the Object menu, click Join.

The two structures are joined so that they share an atom. Note that the second structure is moved accordingly.

Finally, you can add any substituents as required.

(Also note that for unsubstituted ferrocene in the gas phase, the cyclopentadienyl rings are actually eclipsed.)


Does a wavefunction have to be an eigenfunction of an operator in order to obtain an expectation value?

No, assuming that wavefunction is normalized, the expectation value $o$ of the operaror $\mathcal O $ is obtained as $\langle \Psi | \mathcal O | \Psi \rangle $.

EDIT

In response at your comment: I am not asking how to calculate the expectation value, I am asking why the wavefunction does not have to be an eigenfunction of the operator in order to calculate an expectation value

The fact that the expectation value is obtained with $\langle \Psi | \mathcal O | \Psi \rangle $ implies that there is no need that the wavefunction be an eigenfunction of the hermitian operator $\mathcal O$.

If it is an eigenfunction, then it is possible to get a simple expresion with:

$$ \langle \Psi | \mathcal O | \Psi \rangle = \langle \Psi | \mathcal O  \Psi \rangle =  \langle \Psi | \mathcal o  \Psi \rangle  = o  \langle \Psi | \mathcal   \Psi \rangle = o$$

If not the case, then you can not make the substiution  $\mathcal O  \Psi = o \Psi$. So, the application of the map $\mathcal O$ to $\Psi$ will be some function $\Omega$. Then, the expectation value will be  $\langle \Psi | \mathcal   \Omega \rangle$ that is an scalar that you can compute as
$$ \int d\mu \Psi^*\Omega$$


How do I calculate the proportion of respective conjugates of one base mixed with several acids (and/or conversely) at different temperatures?

If you want a general solution for complex mixtures, where simplified formulas do not apply, you may need to involve numerical mathematics.

As solving the equation set may lead to polynomial equation of higher order you may like.

Enumerate equations of the known.

First, charge balance.
Then, mass balances.
Finally, dissociation equilibriums.

Involve a function of [H+] with a minimum at the value being searched for.

About temperature, there is problem the  constants are rarely provided for other but the standard temperature $\pu{25 ^{\circ}C}$


Materials that resist adhesion with molten glass even when hot

Professional glass blowers use graphite paddles to "push" the glass around.  But to bend a heated glass tube you need to know that glass is a very poor conductor of heat.  Thus it is possible to grasp the heated tube six or so inches from the heated spot and just bend the glass.  This works well for small bore tubing.  Just be careful putting the bent tube down.  Putting it on a high heat conductivity surface will result in the glass breaking.  In the old days asbestos pads were used, but that's a big no-no now.


Can a polymer have different resonance structures?

There's no reason why the structural subunits of polymers could not exhibit resonance, provided there are conjugated $\pi$-bonds present. A polymer such as kevlar, for example, certainly benefits from resonance stabilization within each monomeric unit.

Whether there exist any polymers in which the resonance is truly distributed over the entire molecule (if that's what you're really asking), however, I'm unsure. From the perspective of valence bond theory, the electron delocalization required for resonance necessitates a planar geometry with unhybridized $p$ orbitals that align and overlap to form $\pi$-bonds perpendicular to the plane of the $\sigma$-bond skeleton of the molecule. In molecular orbital theory, the sort of extensive electron delocalization that resonance implies requires a bonding molecular orbital, distributed over all participating atoms, constructed from a set of symmetrical, in-phase $p$ orbitals, with all planes of symmetry present in the original molecule(s) conserved in the generation of new molecular orbitals.

Given the rigidity of those criteria, I sincerely doubt that polymers of any great length could exist in which the resonance is distributed over the entire polymer chain. However, that could be just a failure of imagination or a lack of knowledge on my part, and I hope somebody else weighs in with a more conclusive answer.


Nonmetals form covalent bonds with other nonmetals, not ionic bonds. This means you can't use ionic bonding rules to figure out what the structure will be. In fact, there is no way to predict the structure - all you could do is propose Lewis structures for various combinations to predict what it could be.

This is also why there is a special set of nomenclature rules for molecular compounds - it's not enough to just give the names of the elements, you also have to specify the number of each atom.

In your example, if all you are given is N and O, you would have no way of predicting the formula. It could be any one of maybe half a dozen compounds (maybe more). To end up with
$\ce{N2O5}$, you would need to know the name of the compound first.

For binary molecular compounds, the naming rules are pretty easy. First, you need to know the prefixes:

1: mono-
2: di-
3: tri-
4: tetra-
5: penta-
6: hexa-
7: hepta-
8: octa-
9: nona-
10: deca-

Then, for each element (starting with the least electronegative one - the one furthest to the left on the periodic table) you just use the prefix in front of the element name. $\ce{N2O5}$ would be called dinitrogen pentoxide.

There are two small details. The first is that we don't use "mono" when there is only one atom of the first element. This means that $\ce{CO}$ is carbon monoxide rather than monocarbon monoxide. The second is that when an element starts with a vowel and the prefix ends in a vowel, we drop the prefix's ending vowel. This means that we say pentoxide instead of pentaoxide.

To find the formula given the name, you just reverse this process.

$\ce{NO}$ - nitrogen monoxide

$\ce{N2O}$ - dinitrogen monoxide

... you get the idea.


What are the correct resonance structures of nitrous oxide?

Firstly, neither of the resonance structures that you drew for your test are possible because they both violate the octet rule. For the structure on the left, the leftmost $\ce{N}$ is in control of only 6 electrons, and will not exist in this form. The structure on the right is not possible because central $\ce{N}$ is participating in 5 bonds, which $\ce{N}$ cannot do. The maximum number of bonds that $\ce{N}$ can participate in is 3 (covalent bonding - like $\ce{NH3}$) or 4 (coordinate bonding - like $\ce{NH4+}$).

As long as you satisfy the octet rule (or at least as best as you can - here are some exceptions) and you wish to check the stability of a molecule that you have drawn, simply use formula for formal charge. This is given by the equation:

$$\mathrm{FC} = \mathrm{V} - (\mathrm{N_B} + \frac{\mathrm{B}}{2})$$

Wher $\mathrm{FC}$ is the formal charge, $\mathrm{V}$ is the number of valence electrons that the atom under consideration usually has, $\mathrm{N_B}$ is the number of non-bonded electrons, and $\mathrm{B}$ is the number of electrons shared in a covalent bond on the atom. If a molecule is stable, the sum of the formal charges of each substituent atom should be 0. If you were to test this on each of the resonance structures that the test provides as answers, you would see that this holds true.


Do all flames require the same minimum concentration of oxygen to sustain themsleves?

No, different substances will cease combustion at different concentrations of oxygen. The concentration needed can be described by the limiting oxygen index. This is the reason that the use of pure compressed oxygen can be so hazardous; not only will materials burn that would not have burned before, but now materials that do burn require a lower temperature/energy input to ignite.

You can understand this conceptually when you consider where the heat goes. On combustion in air some of the heat must be used to heat the $\pu{79\!\%}$ of the air that isn't oxygen and that heat is wasted in the inert gas and the heat is diluted. That means only ~$\pu{20\!\%}$ of the heat goes into activation energy to propogate the combustion. When the oxygen is more concentrated, less heat goes into inert gases and more goes into oxygen to propagate the combustion and vise-versa.


In practice, how dangerous is liquid oxygen as an oxidizing agent?

I'll have to acknowledge my question might be too vague for a straightforward answer, as can be seen by the various comments. The correct answer, as given by Paul J. Gans, probably is

[V]ery. I know of no way to quantify this.

However, using the comments as a starting point, I read some more around the topic, and I would like to expand on the comments to the question and to summarize what I found out.
Since there's no unit to measure "dangerous" with, we'll first have to get more specific. In the context above, we can offer the following dimensions:

ease with which LOX + X is set on fire or detonated, the easier, the more dangerous
predictability of ignition, the less predictable, the more dangerous
energy output of ignition, the more energy the more dangerous
speed of reaction, the faster the worse

The core problem with all of the above is that the LOX is not a fuel and will, on its own, do nothing at all. So before we can even speculate on possible reactions, we will have to narrow down the fuel options our spilled LOX might encounter. When we also want to account for oxygen-enriched atmosphere by boiled off LOX, a danger in itself, even without spilled, as matt_black correctly points out, the fuel can be almost anything burnable in the room.
That pretty much takes care of 2), predictability, which is low, so keeping the whole setup safe is difficult. -> Danger: high
Ease of ignition depends on the fuel, and the number we're after is the activation energy (Wikipedia). I found various Internet pages mentioning LOX spill accidents that turned tarmac into an explosive that can be set off by walking or driving over it, although I have not found an actual report of such an accident. So apparently, there are common materials that combine with LOX to a mixture requiring a very low activation energy. -> Danger: high.
Energy content and speed of reaction also depend on the fuel, but the fact that it's being used as rocket propellant and that LOX was used to build explosive charges in mining (so called Oxyliquits (Wikipedia), via permeakra's comment) means that we can expect plenty of both for a number of materials. The article also emphasizes the unpredictability, even of things that are meant to explode. -> Danger: high.
Finally, various safety guidelines inspire vivid imaginations of some scenarios:

The whole floor of an office is known to have caught fire when oxygen vapors contacted a lighted cigarette butt. -- Saftey Precautions for Oxygen Plants
If liquid oxygen spills on asphalt or other surfaces contaminated with combustibles, do not walk on or roll equipment over the area of the spill. Liquid Oxygen, Environmental Health and Safety


Part A: Why do you want to use the molality?! You have the concentration i.e the number of moles in one liter. So the mass of nitrogen in one liter is simply:

Number of moles in one liter $\times$ molar mass.

Part B: It's correct.


Does the oxidation for all the aliphatic monoaldehydes to carboxylic acids have the same value of electrode potential?

My answer to your main question is no.

Firstly, you are simplifying the process of oxidation of the monoaldehyde. The mechanisms for oxidation depending on the oxidising agent used, although they are generally the same (Refer to link). By considering only these 3 bond formation/cleavage events, you are not looking at this process in totality.

Secondly, the side chain does matter as it would affect the bond energies of the bonds that are broken and formed. Different side chains may give the aldehyde different steric strains etc. Depending on whether it is more electron-withdrawing or electron-donating, it can severely influence the bonds on the carboxyl group.

Due to the variations in oxidising agent, medium of reaction and side group of aldehyde, the oxidation of different monoaldehydes to carboxylic acids would have a different value of electrode potential as these factors would affect the feasibility and ease of oxidation.

Oxidation using chromic (VI) species:

http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch17/ch17-3-5-1.html


What gas was produced in this orbiting space-balloon from 33 pounds of benzoic acid and anthraquinone? Why these two chemicals?

There are three materials to investigate: the two solids in question have been described as having fast and slow rates of sublimation. The balloon was sent up with 4.5 kg of benzoic acid (bp 250C) and 10.6 kg of anthraquinone (bp 380C). https://en.wikipedia.org/wiki/Project_Echo
The third material is the film: 12.7 micron biaxially oriented polyethylene terephthalate metallized with 0.2 micron aluminum. The two sublimatible solids were used to rapidly inflate the balloon and then to keep it inflated in case of micrometeorite strikes.
Echo-2 was made with a thinner PET film (9 microns) sandwiched between two thicker aluminum films (4.5 microns each), and the balloon was inflated to a pressure that slightly deformed the aluminum sandwich and kept the whole structure more rigid, so continual sublimation was not required.
The PET films are so thin! By comparison, a human hair is typically 100 microns in diameter. Perhaps the choice of sublimatible materials was made for non-reactivity (with the film) rather than with any reaction to form a gas. But the rationale is still not clear - it certainly couldn't have anything to do with cost.


Why can't pure water conduct electricity since it can be reduced at cathode and oxidised at anode?

Indeed, water can conduct electricity by electrolysis. But this is hampered by the fact that water has very high resistance. From Electrolysis of water:

  Electrolysis of pure water requires excess energy in the form of
  overpotential to overcome various activation barriers. Without the
  excess energy the electrolysis of pure water occurs very slowly or not
  at all. This is in part due to the limited self-ionization of water.
  Pure water has an electrical conductivity about one millionth that of
  seawater.

And from Self-ionization of water we can see how small it is:

  Water molecules dissociate into equal amounts of $\ce{H3O+}$ and $\ce{OH-}$, so their concentrations
  are equal to 1.00×10−7 mol∙dm-3 at 25 °C


Draw a chain of three carbon atoms.
Make one of them $\ce{COOH}$. You have two carbon atoms to choose from.
Attach $\ce{OH}$ to one of the remaining carbon atoms to make it a primary alcohol. You have one carbon atom to choose from.


How do you solve a net ionic equation without a balanced equation?

This sort of question is really all about solubility. What you're really asking when you're trying to get a net ionic equation is What combinations of the ions are insoluble? Net ionic equations involve only ions that end up as precipitate on the right side of the equation. If you do enough of these problems, you get a feel for the general solubility trends of common ions, but there should be a table in your textbook or have a look here.

For this specific example, salts with alkali metal and ammonium cations are soluble. Nitrate and acetate anion salts are soluble. Therefore, all of those end up as spectator ions and we're left with just $\ce{Ba^2+}$ and $\ce{CrO_4^2-}$. Chromate salts other than those of alkali metal cations or ammonia are insoluble (rule 8 in the link), so the net ionic equation is:

$$\ce{Ba^2+_{(aq)} + CrO_4^{2-}_{(aq)} -> BaCrO_{4(s)}}$$

In this case, it's already balanced, but that may not necessarily be true.


What is the reaction mechanism for the hydrolysis of Amifostine?

You are expelling a thiol rather than an alcohol via alkaline phosphatase-catalyzed hydrolysis.  One can also view it as a free electron pair on "X" attacking the phosphorus to pop out its =O to an anion, then the charge poking back in to expel thiolate.


Why alkyl shift happens during hydroboration oxidation?

But why does this happen?

Well, if it was a plain old carbon in the middle, this migration wouldn't happen. It would sit there forever. Boron isn't carbon, though: it has fewer protons, is less electronegative, and doesn't like holding onto electron density all that much. That electron density gets pushed onto the adjacent carbons, which make them rather more nucleophilic, and happy to jump next door into the $\sigma_\ce{O-O}^*$ orbital.
You can see, after all, that there's something about that boron that's not quite perfect: it has a formal negative charge. While this shouldn't be taken as a sign of great instability, it does suggest that boron won't mind losing that formal negative charge too much.

Won't the boron atom become less stable as the empty p orbital is regenerated?

The energetics of this migration aren't just about the boron atom, it's about the system as a whole. You are also breaking a weak O–O bond, forming a stronger C–O bond in replacement, and so on. And indeed, even if this one step is energetically unfavourable, it might not even matter as long as the overall reaction is spontaneous.
Is the overall reaction energetically favourable? I'll leave it to you to think about it, based on the types of bonds that are broken and formed over the course of the entire reaction. You already saw some of the considerations in the linked question. (I don't get what you mean by the boron atom being bonded to oxygen before the reaction; it isn't, as it has three bonds to carbon.)

Won't this cause the $\ce{OH-}$ (strong nucleophile) to attack the boron (electrophile) again?

Well, yes, but not much happens if you go down that route. It's far more interesting if $\ce{OOH-}$ attacks the boron again, because that means the second alkyl group can migrate. That is precisely the mechanism that you have learnt.


The figures you see are isogyres and arise in substances that are optically active, which means that polarized light is transmitted through the material in different ways depending on the orientation of the sample. These figures are called interference figures. I highly recommend that you peruse this excellent write-up on the optical properties of minerals, particularly the sections pertaining to uniaxial and biaxial crystals, birefringence, and how a petrographic microscope works.

From the Wikipedia entry on conoscopic interference pattern:

  The figures are produced by optical interference when diverging light rays travel through an optically non-isotropic substance - that is, one in which the substance's refractive index varies in different directions within it.

Starch is uniaxial and the figures you see are characteristic of uniaxial materials when viewed using a petrographic microscope. See this reference (The Principles of Pharmacognocy. F. A. Flückiger and A. Tscirch. William Wood & Company, New York, 1887), pages 118-119.

As for the color seen in the image you reference: that arises from using a mica (or tint) plate which in turn indicates whether the substance is uniaxial positive or uniaxial negative, which refer to the propagation of light along differing optical axes. However, in your figure, the color of the isogyres and the background are both blue, which indicates to me that the blue color is simply a result of polarized light passing through the substance.


It's not necessary that an element shares some chemical property with all the others in its group. If it did share all chemical properties with the rest of the group, then we could just as well call the elements (something akin to) isotopes of each other. You'd have a pretty boring periodic table then.

What's happening here is that the most striking difference between elements in a group is at play. That difference being the number of electrons.

Our electron shells are broken into "orbitals". Every orbital can hold 2 electrons. $s$ orbitals come in sets of one, $p$ in sets of 3, $d$ in sets of 5, etc. Not every shell is made up of all orbitals--the first shell has only one $s$ orbital, the second shell has $s$ and $p$, and so on. Electrons fill into these orbitals by a triad of rules (which I won't go into here)-- the Aufbau principle, the Pauli exclusion principle (which is implicit in my statement that each orbital holds two electrons), and Hund's rule.

Now, an interesting thing is that bonding can only occur with orbitals with close energies. Thus, if the $4s$ orbital is involved in bonding, the $1s$ orbital will not be.

OK. Now, let's first look at Sulfur's outer electron configuration. There are 2 electrons in a $3s$ orbital, 4 in the $3p$ orbitals (so there is enough space for 2 more electrons here), and you have some "vacant space" in the $3d$ orbitals.

In the sulfate ion:

sulfur is "accepting" 6 electrons in this bond. There's some nifty mixing of orbitals that goes on, but basically, those electrons need some place in sulfur's outer shell to hole up--two electrons per orbital as usual.

Since we have 6 extra electrons, we need 3 orbitals. We have one "left over" $p$ orbital, and 5 $d$ orbitals, so we have enough space to accommodate those electrons and form a bond.

OK. Now, let's try the same thing for Oxygen. It has, in its outer shell, a full $2s$ orbital, 2 full $2p$ orbitals, leaving an "extra" $2p$ orbital empty. As mentioned before, the second shell only has $s$ and $p$ orbitals, so there obviously aren't any $d$ orbitals. You've probably noticed the issue here-- we have six electrons and only one extra orbital to stuff them in. This isn't possible, so oxygen never forms $\ce{OO4-}$. The other chalcogens($\ce{Se, Te}$, etc) all have $d$ orbitals in their outer shells, so they have no trouble forming such ions.

Note that I've grossly oversimplified the bonding process, deliberately leaving out a few crucial aspects (which would otherwise take extremely long to explain). If you want to know about the mechanism behind bonding, you may want to read up on orbitals first, then VSEPR theory, and then MO theory.


On the periodic table: Why are groups of elements organized by 'letter'

The letters are related to the electron orbitals, which were originally observed through spectroscopy. The lines shown in the spectroscope were named sharp, principal, diffuse and fine (or fundamental).

With a strong magnetic or electrostatic field, these separate into one, three, five or seven lines, or energy levels. There can be up to two electrons (with opposite spin) with the same energy level, according to the Pauli exclusion principle, so this implies there can be double the numbers above, or 2 s electrons, 6 p electrons, 10 d electrons and 14 f electrons. Thus, the Periodic Table is based on the number of electrons in each elements orbitals.


How to make sense of the crystal systems names

Monoclinic and triclinic

Ivan (in the comments) is spot on about these two. Among all the systems, these are the only ones with angles that are not fixed by symmetry, so you actually have to measure them.

Rhombohedral or Trigonal

These are two different settings of the same lattice. The unit cell can either be a prism (with angles of 90, 90 and 120 degrees, blue lines in the figure) or a rhombus (with all three angles equal, and all three sides equal, black lines in the figure).

In general, there are infinite choices for unit cells (you just need three translation vectors that are not linearly dependent, so pick an origin on a lattice point and three other lattice points and its a unit cell). The conventional choices usually have more of the symmetry elements along the axes, face diagonal or body diagonal, and keep the volume of the unit cell fairly small.

Orthorhombic

This used to be called pseudo-orthorhombic.

Again, you have a choice of unit cell, and the more uncommon one is a rhombohedral prism. When you switch from all right angles to a rhobohdral prism setting, face-centered turns into primitive and primitive turns into face-centered.

Here is a quote from Wikipedia:

  In the orthorhombic system there is a rarely used second choice of crystal axes that results in a unit cell with the shape of a right rhombic prism;2 it can be constructed because the rectangular two-dimensional base layer can also be described with rhombic axes. In this axis setting, the primitive and base-centered lattices swap in centering type, while the same thing happens with the body-centered and face-centered lattices.

Historically, the first data available to crystallographers (and minerologists) were the shapes of crystals. Orthorhombic crystals can have faces that are at 90 degree angles or not, depending on the kinetics of growth. Pictured below are three crystal forms with underlying orthorhombic symmetry, with the 2nd one showing is a shape with (+/- 1, +/- 1, +/-1) faces labeled orthorhombic pyramid:

Cubic, Hexagonal and Tetragonal

See OP's comments for those.

History

The German Wikipedia article on crystal systems has the terms Bravais used in 1886:

Assemblages terquaternaires (cubic)
Assemblages sénaires (hexagonal)
Assemblages quaternaires (tetragonal)
Assemblages ternaires (trigonal)
Assemblages terbinaires (orthorhombic)
Assemblages binaires (monoclinic)
Assemblages asymétriques (triclinic)


Why does latex tubing become white and brittle over time, especially where it's bent?

Natural rubber instantly crystallises (reversibly!) under mechanical deformation. That is actually very desireable, because it gets tougher under load! High quality tyres (for trucks, not regular car tyres) today still contain quite a large amount of the world's natural rubber production for that reason.

You can try to pull at a lenght of the rubber tube, it should become pale, and go back to brownish-opaque when you release the pull.

The drawback is this also makes the material a bit brittle, if deformed beyond a certain limit. I believe this is what you see with your rubber tubing. If you unbend the kinks, the white colour should fade, but the tube is probably damaged at this point.


Determining the boiling point of a substance given the triple point and critical point

What you need is the Clausius–Clapeyron equation, which is used to relate vapor pressure and temperature (or external pressure and boiling point).
$$\ln{\left(\dfrac{P_1}{P_2}\right)}=\dfrac{\Delta H_{\text{vap}}}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)$$
Step 1: Find the enthalpy of vaporizations $\Delta H_{\text{vap}}$ by plugging in the two sets of temperatures and pressures that you have.
Step 2: Use your $\Delta H_{\text{vap}}$ and the temperature and pressure of the triple point to determine the boiling temperature at 1 atm.


Independant variable for Asprin pill titration with NaOH solution

Independent variable means the variable which you can change easily. Direct titration aspirin is not a good idea as aspirin hydrolyzes into salicylic acid. There are a lot of non-sense high school/college level experiments doing direction titration and ignoring the side reaction of hydrolysis.

The proper procedure of aspirin analysis is by back-titration with NaOH and taking hydrolysis into account. Ask your teacher about it by showing him/her this procedure Back titration of aspirin

Coming back to independent variables...Just now you have another independent variable...comparison of two different methods of titration (direct vs. back). Think of other independent variables and share in your comments.


How to judge accuracy and precision of measurement devices?

Beakers, Erlenmeyer flasks, dropping funnels and the like are not volumetric instruments. They are not precisely calibrated, and their scales serve only as approximate guides. Therefore, the beaker that is mentioned in the question is right out.

Graduated cylinders, volumetric flasks, bulb pipettes, graduated pipettes, and burettes are volumetric instruments. They are calibrated; however, like any other measuring instrument, they only have a limited accuracy and precision.

The accuracy and precision do not only depend on the type of the instrument (e.g. graduated cylinder or pipette). They also depend on the dimensions, graduating divisions, and accuracy class of the instrument, which are usually internationally standardized.

In principle, volumetric instruments with a small diameter have a better precision because the same difference of the liquid level (i.e. the meniscus) corresponds to a smaller difference in volume. Therefore, typical long and thin pipettes are usually better than short and thick graduated cylinders. For example:

A standardized graduated pipette (Class A or AS) with a nominal capacity of 10 ml and an overall length of about 360 mm has a maximum permissible error of ±0.05 ml.
A standardized graduated measuring cylinder (Class A) with a nominal capacity of 10 ml and a maximum overall height of 140 mm (tall form) has a maximum permissible error of ±0.1 ml.
If you use an even larger graduated measuring cylinder (Class A) with a nominal capacity of 100 ml and fill it just to the 10 ml mark, the maximum permissible error is ±0.5 ml.

These errors represent the maximum permissible error at any point on the scale, and also the maximum permissible difference between the errors at any two points.

Since the same sources of error are, naturally, inherent both in calibration and use, a volumetric instrument should have a good precision in order to be able to be calibrated with a good accuracy.

However, note that graduated pipettes are usually calibrated ‘to deliver’ (TD, Ex), whereas graduated measuring cylinders are calibrated ‘to contain’ (TC, In). Hence, graduatted pipettes and graduated measuring cylinders may not be interchangeable at will.


How can there be decimal subscripts in a molecular formula?

First of all, in chemistry two types of formulas exist: structural and empirical. A structural formula shows the way atoms are connected. An empirical formula only summarizes atoms and their ratios. While often an empirical formula references a molecule, there are cases when the compound in question is not molecular, or it is unknown if it is molecular. In that case, since the total number of atoms in the molecule is undefined, the empirical formula may be scaled by any number, and as long as all indexes are scaled by the same number, it will denote exactly the same information.

So, formulations like $\ce{Ti_{0.9}O}$, and $\ce{Ti9O10}$ denote the same thing: a Ti+O phase with Ti/O ratio 0.9. Usually in such cases, if ratios are fixed, corresponding indexes are kept integral. So, it is  $\ce{Al2O3}$ because the Al/O ratio is fixed here, but $\ce{Li_{0.2}MoS2}$ because the Li/Mo ratio here is not.

Returning to the $\ce{Li_xFePO4}$ issue: in this phase $\ce{FePO4}$ part forms the host network similar to quartz and thus has a fixed composition. However, Li may and does migrate in and out of the host network, so its amount in the bulk is not fixed. This is possible because Fe ions have a lot of free orbitals and some of them have rather low energy, so Li electrons are passed to iron and the small Li ions may travel through the host network. Thus the composition of the electrode changes through the charge/discharge process.

In accumulator chemistry electrodes are rarely forced towards full transfer of traveling ions, because it may lead to irreversible degradation in the electrodes. For example, the $\ce{LiCoO2}$ phase would collapse if all or even a major part of Li left, so never is all or even a major share of the Li allowed to leave it. Consequently, Li-ion battery electrodes typically contain a significant amount of Li both in the fully discharged and fully charged states.

Similarly, it's usually technically impossible to achieve complete transfer even in more traditional batteries because the voltage and consequently the travel speed of ions between electrons falls down with discharge.


Correct flask for reacting and distilling on hotplate?

At under $100^\circ \mathrm{C}$ The best will be to use water bath and round-bottom flask. Most magnetic stirrers have enough range (magnetic forces do not stop working if vessel is in couple of centimeters from hot plate, though you should use non-magnetic dish for bath itself). Even if magnet in hot plate is weak I would suggest against flasks with flat bottom, because bottom may fall off under heating.


Which of the given compound has the highest rate of dehydration?

Notice, that the first compound (A)can show tautomerism, after drawing the enol form we find that the carbocation lies at the allylic position, which means that it is stabilised through resonance forms, hence the compound A has highest rate of reaction.
Hope it helps.


How to define a new solvent in Gaussian09 to run SCRF?

The correct input should be more like (replacing your Water2 with Generic)

#p b3lyp/6-31G(2df,p) int(grid=ultrafine) SCRF(Solvent=Generic,Read)

Comment

0 1
...

stoichiometry=H2O1
solventname=Water2
eps=78.3553
epsinf=1.77556

as it is stated in the Documentation under Additional Input for PCM Calculations.

Note - this only gets your calculation running. Your shown current input section might be insufficient to properly define sulphuric acid.

As Geoff noted, the Gaussian Documentary states an important fact short before showing some $\varepsilon$-values of common solvents that should be taken into account.

  We list the ε values here for convenience, but be aware it is only one of many internal parameters used to define solvents. Thus, simply changing the ε value will not define a new solvent properly.


Why other benzene derivatives do not undergo nitrosation

Anisole can be nitrosated in good yield reference here.  This, and other references, note that the PhOMe group is unstable to the reaction conditions being cleaved to give phenol.

m-Xylene can be nitrosated in good yield, o-xylene and toluene in modest yield in TFA or sulfuric/acetic acid mix under nitric oxide reference here

Nitrosonium tetrafluoroborate will also nitrosate a range of electron-rich aromatics, including anisole and xylenes reference here

This suggests to me that the authors of the textbook quoted are in error, as two of the references predate the publication date of the textbook.


Why do the unsaturated ketones have C-O bond weaker than saturated ketones? How it is related to resonance structures?

The statement can actually be said like this. In Unsaturated aldehydes/ketones the $\ce{C -O }$ bond has significant single bond character associated with double bond characters(i.e. partial double bond character), whereas, saturated aldehydes/ketones the $\ce{C -O}$ bond has almost pure double bond character.
 In unsaturated carbonyl compounds, as the no. of stable resonating structures increase,( the structure where the $\pi$ electron cloud of $C =C$ double bond delocalises is also very stable) the contribution of  that canonical form where the $C$ and $O$ has double bond between them also decreases . So, that's why, in the resonance hybrid of the molecule( which will give the IR spectra) the bond order of $\ce{C -O}$ bond becomes significantly less, and become close to the midway between double and single bond. But in saturated carbonyls, the other resonating form is relatively very less stable ( not complete Lewis Octet and very close  charge seperation)  than the canonical form of pure double bond. So, contribution of that other form becomes significantly less. So, the resonance hybrid becomes almost close to that consisting of pure double bond.So, in the IR spectra, the streching frequency of saturated carbonyls correspond to almost perfect $\ce{C=O}$ double bond.


Equilibrium constant and reaction Quotient

$K$ is just the specific value of $Q$ for which $\Delta_\mathrm r G = 0$.
In that case:
$$\begin{align}
\Delta_\mathrm r G &= \Delta_\mathrm r G^\circ + RT \ln Q \\
0 &= \Delta_\mathrm r G^\circ + RT \ln K \\
\Delta_\mathrm r G^\circ &= -RT \ln K
\end{align}$$

For any system not at the equilibrium it is possible that $Q < K$ or $Q > K$. What is the "forward direction" is just a convention, it depends on how you decide to write the equilibrium. There is nothing special in one direction or the other. The system will "move" toward the direction that leads to equilibrium in both cases. In the first case the reaction will consume reactants to generate products, while in the second case the opposite will happen.
You can express this idea also in terms of the ratio $\frac Q K$. If it is > 1 then the reaction will consume reactants. This is also often referred to as the "Le Chatelier principle".