Let's take a look at the first product you propose: 1-bromopropan2-one.

The mechanism includes deprotonation by hydroxide to form an enolate. The resulting enolate is better stabilized with more electronegative atoms (Br in this case) and will form more readily than another molecule of acetone turning into an enolate ion. Thus, as the reaction proceeds, the product distribution will be determined by the most stable enolate anion. This will be the tribromoacetyl intermediate enolate ion and will drive the reaction forward. So a whole mol of bromine will be converted completely to the tribromoacetyl intermediate with a 1/3 mol of acetone. Notice how only a 1/3 will react to give the monosubtituted intermediate, than 1/3 mol of bromine will react to give disubstituted product, and lastly the last 1/3 mol of bromine will react to form the trisubstitued product. The last step is displacement by hydroxide via the nucleophilic acyl substitution mechanism (not SN2!) to form acetate and release bromoform. This reaction isn't really great if bromine isn't in excess.


According to Schmitz [1], $\mathrm{p}K_\mathrm{a}(\ce{HIO2})\approx 6$ at $\pu{25 °C}$ (numerical references for the bibliographical entries have been updated):

Stanisavljev [2] has proposed a correlation between thermodynamic values of oxo-halogen ions and has obtained $ΔG_\mathrm{f}^\circ (\ce{IO2-}) = \pu{−51.8 kJ mol−1}$. He has also estimated the dissociation constant of $\ce{IO2H}$, $\mathrm{p}K_\mathrm{a}(\ce{IO2H}) = 4$, and obtained $ΔG_\mathrm{f}^\circ (\ce{IO2-}) = \pu{ −75 kJ mol−1}$. However, this value introduced in the above equation leads to $K_\mathrm{C}^\circ = \pu{2.8e12}$ and an impossibly large value of $k_\mathrm{+C}^\circ$. His correlation supports the idea that $k_\mathrm{+C}^\circ$ is very large but overestimates $ΔG_\mathrm{f}^\circ (\ce{IO2-})$ and probably also the dissociation constant of $\ce{IO2H}$. Figure 1 shows known $\mathrm{p}K_\mathrm{a}$ values of oxy-halogen acids, including the value of $\mathrm{p}K_\mathrm{a}(\ce{BrO2H})$ [3] not considered by Stanisavljev, and suggests that $\mathrm{p}K_\mathrm{a}(\ce{IO2H})$ is closer to $6$ than to $4$.

Figure 1. $\mathrm{p}K_\mathrm{a}$ values of oxy-halogen compounds $\ce{XOH}$ (•) and $\ce{XO2H}$ (◦).

References

Schmitz, G. Inorganic Reactions of Iodine(III) in Acidic Solutions and Free Energy of Iodous Acid Formation. International Journal of Chemical Kinetics 2008, 40 (10), 647–652. https://doi.org/10.1002/kin.20344.
Stanisavljev, D. Consideration of the Thermodynamic Stability of Iodine Species in the Bray-Liebhafsky Reaction. Berichte der Bunsengesellschaft für physikalische Chemie 1997, 101 (7), 1036–1039. https://doi.org/10.1002/bbpc.19971010708.
Faria, R. B.; Epstein, I. R.; Kustin, K. Kinetics of Disproportionation and $\mathrm{p}K_\mathrm{a}$ of Bromous Acid. J. Phys. Chem. 1994, 98 (4), 1363–1367. https://doi.org/10.1021/j100055a051.


Removing diamonds from a 14 karat gold ring with acid

I think that asking a jeweler is the best thing to do because gold is not something to throw away. However talking about the chemical concern, sulfuric acid can't oxidize gold, if the base metal (you should specify where and how it is) is iron theoretically you will be able to dissolve it with sulfuric acid:
$$\ce{Fe + H2SO4 ->FeSO4 + H2 ^}$$

Something similar can happen with copper and tin but I think that temperature of the acid is a determinant parameter for these reactions.

If the base metal is inside the alloy is quite unlikely that you can affect it because gold is not a porous material, so you can only interact with the superficial layer.


Tandem mass spectrometry: a1 and immonium ions

I don't know if you are still interested in this question. But yes, the a1 ion would have exact the same m/z as the immonium ion of all amino acids of the same type in the pepitde. Peptides typically fragment in low energy fragmentation methods such as CID by charg-directed mechanisms, which mainly produces y and b series ions. These ions can fragment further to lower b ions, a ions, or internal fragments and immonium ions. y and b ions are much easier to understand and we therefore mostly use these ions for interpretation of mass spectra.


What's the difference between "conversion" in condensation polymerization and in radical polymerization?

The reason they are defined differently is based on the differing mechanisms:

All of the monomer in step-growth (of which condensation is a subset) polymerization are reactive, and they are consumed very quickly, forming (mostly) dimers which then go on to react further.  At this point the average length of chains is 2-- not really a polymer, but also relatively monodisperse.  However this reaction is going to continue, and you'll get long chains after some time.  But the monomer itself is gone.

In addition polymerization (of which radical is a subset) there are very few reactive intermediates, and many monomers get added per chain. After 50% of the monomer is converted in addition polymerization, you have some really long chains, swimming around in the remaining monomer (half the original amount).  This contrasts the other situation.

Basically, you can't measure monomer concentration to reasonably monitor the progress of a step-growth polymerization, while you can do such for addition polymerization.  Conversely, you can generally monitor functional groups to monitor both kinds of polymerization.


What does $\mathrm{A}_r(\mathrm{H})$ represent when calculating the average natural weight of an elemental atom?

The relative atomic mass $A_r$ is 'the ratio of the average mass of atoms of an element ... to 1/12 of the mass of an atom of carbon-12' (from http://en.wikipedia.org/wiki/Relative_atomic_mass).

The atomic mass $m_a$ is the mass of a particular isotope of an atom, and if it is in 'u' (unified atomic mass units) it is relative to 1/12 of the mass of an atom of carbon-12.

So, to get $A_r$ you take a weighted average of $m_a$ in 'u' for each isotope.

Note also that the standard atomic weight is the relative atomic mass of an element using natural abundance on Earth (http://en.wikipedia.org/wiki/Relative_atomic_mass).


Inorganic chemistry for condensed matter physicists

Any text book suitable for an advanced undergraduate course in inorganic chemistry should cover these topics. Which text may be best for this purpose will be a matter of opinion, so I am not going to recommend a specific one. Links to specific books/editions will risk being out of date for future readers of this question and answer. However, some author combinations have been active in writing text books in this field for many years. I would look for a recent edition of the book by either:

Meissler and Tarr (as of this writing, the most recent edition is the 5th edition by Meissler, Fischer and Tarr, published by Pearson, 2013)
Shriver and Atkins (as of this writing, the most recent edition is the 6th edition by Shriver, Weller, Overton, Rourke, and Armstrong, published by WH Freeman, 2014)

Just be careful since the phrase "inorganic chemistry" in some uses refers to the introductory course sequence in chemistry, what is more commonly called "general chemistry".


These pi-type MOs are most commonly derived using Hückel MO theory. I've never been a fan of Wikipedia's technical articles, so for further reading I'd suggest using the QMUL resources online, which are very comprehensive. $\require{begingroup}\begingroup \newcommand{ket}[1]{\left|#1\right>} \newcommand{bra}[1]{\left<#1\right|} \newcommand{braket}[1]{\left< #1 \right>}$

Theory

Simple Hückel theory makes some key assumptions about the forms of the MOs as well as some of the integrals involved in a typical quantum chemical calculation:

pi-type MOs are linear combinations of p-orbitals, and no other orbitals contribute (so-called "sigma-pi separation")
the value of $\braket{a|b}$ (where $\ket{a},\ket{b}$ are p-type AOs on atoms $a$ and $b$) is equal to $\delta_{ab}$ ($1$ if $a = b$, $0$ otherwise)
the value of $\braket{a|H|b}$ is

$$H_{ab} = \braket{a|H|b} = \begin{cases}\alpha \text{ if } a = b \\ \beta \text{ if atom }a\text{ is bonded to atom } b \\ 0 \text{ otherwise}\end{cases}$$

With these simplifications the secular equations

$$\mathbf{H}\mathbf{c} = E\mathbf{S}\mathbf{c}$$

can be readily solved (the derivation of the secular equations will not be presented here). Here $\mathbf{H}$ refers to the Hamiltonian matrix and $\mathbf{S}$ refers to the overlap matrix. These are defined as the matrices with elements $\mathbf{H}_{ij} = \braket{i|H|j}$ and $\mathbf{S}_{ij} = \braket{i|j}$ respectively. In simple Hückel theory the matrix $\mathbf{S}$ can be neglected as it is simply equal to the identity matrix.

$\mathbf{c}$ is a column vector and one column vector represents one MO. The elements of $\mathbf{c}$ represent the contribution of each AO to the MO. This will be clearer with an example (to follow).

In general, for $n$ atoms there will be $n$ vectors $\mathbf{c}$ that will satisfy the secular equations, and $n$ corresponding values of the energy $E$. Thus, for the allyl cation (for example), there will be three column vectors $\mathbf{c}_1, \mathbf{c}_2, \mathbf{c}_3$ and four associated energies $E_1, E_2, E_3$.

The allyl cation

Using the atom numbering scheme as shown above, and the assumptions mentioned in the previous section, we find that the Hamiltonian matrix $\mathbf{H}$ is:

$$\mathbf{H} =
\begin{pmatrix}
\alpha & \beta & 0 \\
\beta & \alpha & \beta \\
0 & \beta & \alpha
\end{pmatrix}$$

For example the entry $\mathbf{H}_{12} = \beta$ because atoms 1 and 2 are bonded to each other, and the entry $\mathbf{H}_{13} = 0$ because atoms 1 and 3 are not bonded to each other.

The secular equations then reduce to the eigenvalue problem

$$\begin{pmatrix}
\alpha & \beta & 0 \\
\beta & \alpha & \beta \\
0 & \beta & \alpha
\end{pmatrix}
\begin{pmatrix}
c_1 \\
c_2 \\
c_3
\end{pmatrix} = E\begin{pmatrix}
c_1 \\
c_2 \\
c_3
\end{pmatrix}$$

Obtaining the eigenvalues and eigenvectors of a matrix is a process outlined in many maths books. The crux is that the "secular determinant" must equal zero:

$$\begin{vmatrix}
\alpha - E & \beta & 0 \\
\beta & \alpha - E & \beta \\
0 & \beta & \alpha - E
\end{vmatrix} = 0$$

from which one obtains the eigenvalues (here I chose an example where it's easy to do by hand, but in general you can use software like MATLAB). In increasing energy (note that $\beta < 0$), they are

$$\begin{align}E_1 &= \alpha + \sqrt{2}\beta & E_2 &= \alpha & E_3 &= \alpha - \sqrt{2}\beta \end{align}$$

and the corresponding eigenvectors

$$\begin{align}
\mathbf{c}_1 &= \begin{pmatrix} 1/2 \\ 1/\sqrt{2} \\ 1/2 \end{pmatrix}
&
\mathbf{c}_2 &= \begin{pmatrix} 1/\sqrt{2} \\ 0 \\ -1/\sqrt{2} \end{pmatrix}
&
\mathbf{c}_3 &= \begin{pmatrix} 1/2 \\ -1/\sqrt{2} \\ 1/2 \end{pmatrix}
\end{align}$$

This means that the lowest-energy pi-type MO, $\mathbf{c}_1$, is

$$\psi_1 = \frac{1}{2}\mathrm{p}_1 + \frac{1}{\sqrt{2}}\mathrm{p}_2 + \frac{1}{2}\mathrm{p}_3$$

where $\mathrm{p}_i$ is the 2p atomic orbital on carbon $i$. Note that the coefficients all have the same sign, indicating that the p orbitals are all in phase with each other. However, the coefficient for carbon 2 (the middle carbon) is larger than the others. Hence, a sketch of this MO should technically show that the middle p-orbital makes a larger contribution than the outer two p-orbitals. This is what TAR86 meant by their comment on your question:

  Hückel MO theory will also give you different absolute values for the linear combination coefficients, whereas your drawing suggests the same coefficients throughout.

Anyway, likewise for the LUMO we have

$$\psi_2 = \frac{1}{\sqrt{2}}\mathrm{p}_1 - \frac{1}{\sqrt{2}}\mathrm{p}_3$$

Note that 1) the p orbitals have different phases (different sign of coefficient), and 2) the p orbital on carbon 2 does not contribute to the MO - this leads to the node observed in the second MO. This is why the allyl cation is electrophilic on C-1 and C-3, but not on C-2: the LUMO has zero coefficient on C-2.

All in all, considering that the allyl cation only has two electrons in the pi system, this lets us construct the MO diagram


Is this molecule possible? And if so, what is it called? (Cyclic groups)

If you want to consider the diene as the main chain, you could call this 2-cyclohexyl-1,3-butadiene. This used to be the proper name for the compound, as double bonds used to take priority over any length of a carbon chain in IUPAC nomenclature. This compound exists, and you could theoretically purchase it at some place like this. In recent years, the proper priorities for IUPAC have changed. Now, the longer chain takes priority over double bonds, and thus this compound should be referred to using the cyclohexane as the main chain. This means that the compound should be called (buta-1,3-dien-2-yl)cyclohexane. For all practical purposes, either name should work, as they both describe the same molecule; however, the second name is consistent with IUPAC standards.


Dispersion of Lithium Ions in Nature (Ratio Similar to Isotopes?)

I believe I resolved my problem, the attributes I was looking at was spectra and combining the lines of both ions. The problem was I was comparing data given by two authors that used different units, so what I did, which I think solves my problem, was divide the sums of the given intensity by the emission probabilities for which I received the ratio of Li I to Li II as roughly 33 to 6 or 18% Lithium II. More than I or apparently anyone else thought so I may have still made a mistake, either way thanks to everyone who tried and helped.


$\psi$ does carry a great amount of physical meaning. The annoying part is, that it doesn't represent anything directly observable. The way (non-relativistic) quantum mechanics is constructed, makes $\psi(x_i, s_i, t)$ ($x_i$ - physical coordinates of $i$-th electron, $s_i$ - spin projection on $z$ of said electron) -  a direct analogue to trajectory, or, in case $\psi$ doesn't depend on $t$ (time),  orbit. *

Orbitals themselves, however, do not have direct physical meaning in that sense that they are nothing more than building blocks we use to construct collective $\psi$ of the electrons in the chemical system after we apply Born-Oppenheimer approximation (that we can ignore nucleus movement when we talk about electron movement). It's a good approximation, so it is almost always in effect in quantum chemistry.

In some special cases (hydrogen-like atoms) the electron wave-function represents exactly one electron of one atom and, ignoring spin contribution, orbital equals wave-function of that one-electron system.

Orbitals of hydrogen-like atoms can be represented ** as $\psi(x)=\psi(r,\theta,\phi)=\psi_r(r)\psi_{angle}(\theta,\phi)$. It is a useful representation, because even if $\psi_r$ changes from level to level, $\psi_{angle}$ doesn't, so you can see parallels between, say, $2p$ and $3p$ orbitals, which differ*** only by $\psi_r$. Again, neither part of the orbital is observable. What IS observable is "electron density" i.e. a probability to observe electron in this particular finite (small) volume. Those can be extracted from various electron or X-ray diffraction experiments. What, however, it is good at is to point out, how good outer electrons shield inner electrons and that chemical properties are mostly guarded by outermost reaches of outer shells.

$\psi_r$ on its own cannot be observed and has no simple interpretation. It has no inherent meaning on its own, and I'm unaware of obvious interpretation. It IS useful when one has to compare different orbitals between each other or to compare various approximations of the same orbital though.

$\psi^2_r$ is fairly useless, imho. It is observable for $s$ orbitals as electron density at given radius, but it isn't very useful. What is useful IMO is density integrated over spherical surface, which for $s$ orbitals is $ |rR|^2 = 4\pi r^2 |\psi_r|^2 $, which shows how much time electron spends at given distance from nucleus. It shows, how diffuse the orbital is, and when plotted against observed total electron density, it shows how good electron is shielded against outer influence.

I hope this helps.

* Note: Here is one of the non-obvious differences between quantum and classical systems lie: you CAN have a quantum system with large kinetic energy and $\psi$ independent on time.)

** Note: it is not the case for all mathematical functions, we are lucky that it is the case for atomic orbitals.

*** Note: orbitals on the same level CAN differ by $\psi_r$. The general rule is that orbitals on the same level have the same number of surfaces where $\psi$ is zero that is equal number of the level minus one. That means that we have only $s$ orbitals on the first level with zero such surfaces, but $2s$ and $2p$ orbitals. For $2p$ orbitals the "zero" surface is one of coordinate planes and for $2s$ orbital$ it is a sphere around the nucleus.


Does dissociation require energy to occur?

When the ions of an ionic compound disolve, do not stay isolated. They form new bonds with the solvents. This makes the overall process spontaneus when a salt is soluble.

Dissociation of course requires energy. But bond formation between ions and solvent molecules release energy.
This can result on a possitive (endothermic disolution, heat is absorbed in the process, like for example, $\ce{KI}$ in water. Bonds formed release less energy than bonds broken) or negative disolution enthalpy (exothermic disolution, like the one for $\ce{H_2SO_4}$ and water. Bonds formed -hydrogen bonds, stronger that ion-dipole interactions-, release more energy that is consumed by bonds broken).

Remember that if you want to know if a process is or not spontaneous, you do not have to take into acount just enthalpy, but Gibbs free energy, which is $\Delta G= \Delta H + T\Delta S$, which also takes into account entropic factors and the influence of temperature.


Is paper aging(oxidation) and paper burning(rapid oxidation) the same thing?

Rapid oxidation- burning - leads to two main products carbon dioxide and water. You cannot keep any records if paper is burned.
In case of slow oxidation some of paper components are oxidized which changes the colour of paper, but just that. Old manuscripts can be stored for hundred of years if no none burns them.


Relation between constant-pressure and constant-volume heat capacities: Cp - Cv = nR

Preliminaries

Consider $U = U(V,T, p)$. However, assuming that it is possible to write an equation of state of the form $p = f(V,T)$,  I don't  have to explicitly address  the $p$ dependence of $U$, and I can write the following differential:

$$\mathrm{d}U = \underbrace{\left ( \frac{\partial U}{\partial V} \right)_T}_{\pi_T} \mathrm{d}V  + \underbrace{\left ( \frac{\partial U}{\partial T} \right)_V}_{C_v} \mathrm{d}T \tag{1}$$

so one writes,

$$ \mathrm{d}U = \pi_T \mathrm{d}V + C_v\mathrm{d}T \tag{2}$$

also, for an ideal gas, the internal pressure ($\pi_T$) $= 0$

Additionally for changes in internal energy at constant pressure,

$$ \left (\frac{\partial U}{\partial T}\right)_p = \pi_T \left (\frac{\partial V}{\partial T}\right)_p +C_v \tag{3}$$

Here, I take the opportunity to define two new quantities, namely $\alpha$ and $\kappa_T$ (the expansion coefficient and the isothermal compressibility respectively)

$$\alpha = \frac{1}{V}\left (\frac{\partial V}{\partial T}\right)_p$$

and, $$\kappa_T = \frac{-1}{V}\left (\frac{\partial V}{\partial p}\right)_T$$

Rewriting (3) using $\alpha$

$$ \left (\frac{\partial U}{\partial T}\right)_p = \alpha \pi_T V +C_v \tag{4} $$

Now, although the constant volume heat-capacity is defined as $C_v = \left (\frac{\partial U}{\partial T} \right)_v$, if we let $\pi_T = 0$ in equation (4), we get $C_v = \left (\frac{\partial U}{\partial T} \right)_p$

This holds true for a perfect gas, and one can quickly obtain the desired relation at this stage.

Derivation: Difference between constant volume and constant pressure heat capacities for a perfect gas

Consider, $$C_p - C_v  = \overbrace{\left (\frac{\partial H}{\partial T} \right)_p}^{\text{definition of} \ C_p} - \overbrace{\left (\frac{\partial U}{\partial T} \right)_v}^{\text{definition of} \ C_v} \tag{5}$$

Introducing, $H = U+ pV = U+nRT$, and exploiting $C_v = \left (\frac{\partial U}{\partial T} \right)_v =  \left (\frac{\partial U}{\partial T} \right)_p$, equation (5) yields:

$$ C_p - C_v = \left (\frac{\partial (U+nRT)}{\partial T} \right )_p - \left (\frac{\partial U}{\partial T} \right )_p = nR$$

Derivation: Difference between constant volume and constant pressure heat capacities (general case)

However, as my contribution to this discussion I would like to derive a relation between heat capacities that is universally true for any substance, not just a perfect gas. So let's return to equation (5):

$$C_p - C_v  = \overbrace{\left (\frac{\partial H}{\partial T} \right)_p}^{\text{definition of} \ C_p} - \overbrace{\left (\frac{\partial U}{\partial T} \right)_v}^{\text{definition of} \ C_v} \tag{5}$$

Here, we substitute $H = U + pV $ and obtain,

$$ C_p -C_v = \overbrace{\left( \frac{\partial U}{\partial T} \right)_p}^{\text{evaluated in} \ (4)}+ \left( \frac{\partial (pV)}{\partial T} \right)_p - C_v \tag{6}$$

The first partial derivative was already taken care of in equation (4). For the second one, since the derivative is to be evaluated at constant pressure, we can do the following

$$\left( \frac{\partial (pV)}{\partial T} \right)_p = p \overbrace{\left( \frac{\partial V}{\partial T} \right)_p}^{\alpha V}$$

Putting all of this together, one obtains

$$C_p -C_v = \alpha \pi_T V + \alpha pV = \alpha(p+ \pi_T)V \tag{7}$$

At this stage, I will make use of the following relation (derived in additional comments) $$\pi_T = T \left (\frac{\partial p}{\partial T}\right )_v - p$$ After substituting this into (7) we get:

$$ C_p -C_v = \alpha T V \left( \frac{\partial p}{\partial T} \right)_V \tag{8}$$

I wish to transform the last remaining partial derivative, and to do so I consider $ V = V(T,p) $ which yields the following differential

$$ \mathrm{d}V = \left( \frac{\partial V}{\partial T} \right)_p \mathrm{d}T + \left( \frac{\partial V}{\partial p} \right)_T \mathrm{d}p $$

At constant volume, $\mathrm{d}V = 0$ so one gets,
$$\left( \frac{\partial V}{\partial T} \right)_p \mathrm{d}T  = - \left( \frac{\partial V}{\partial p} \right)_T \mathrm{d}p$$

$$\overbrace{ \left( \frac{\partial V}{\partial T} \right)_p}^{\alpha V} = \overbrace{-\left( \frac{\partial V}{\partial p} \right)_T}^{\kappa_T V} \left( \frac{\partial p}{\partial T} \right)_V$$

Note: One can avoid all of this work, and simply invoke the Euler Chain Rule

Rearranging,  $$\left( \frac{\partial p}{\partial T} \right)_V = \frac{\alpha}{\kappa_T}$$

We can finally substitute this into (8) to get

$$C_p -C_v = \frac{\alpha^2 TV}{\kappa_T} \tag{9}$$

This is true for any substance, not just a perfect gas. Now, for a perfect gas $ pV = nRT$ holds true, and thus $\alpha = \frac{1}{T}$ and $\kappa_T = \frac{1}{p}$. Making these substitutions into (9), we get our desired result

$$C_p -C_v = nR $$

Additional Comments

This might seem like an unnecessarily complex, and not to mention convoluted way to get to the desired result (especially, in light of a much simpler method presented by @orthocresol), however, I think the deriving the expression for a general case first, and then reducing it to the special case is illuminating. Moreover, in spirit and approach, it is not that far from what @orthocresol did.
Physical Significance of certain terms/quantities

$\pi_T$ is called the internal pressure (it has the dimensions of pressure) and is a consequence of the interactions between molecules. For an ideal gas it is necessarily zero.
$\alpha$ i.e the expansion coefficient is the fractional change in volume that accompanies a rise in temperature. A large volume of $\alpha$ implies that the sample responds very strongly to changes in temperature.
Similarly, $\kappa_T$ is a measures of the response to a change in pressure. The negative sign insures that $\kappa_T$ is a positive quantity, because a pressure increase causes a decrease in volume ($\mathrm{d}V$ is negative)
Since equation (9) holds true for any substance, for solids and liquids one might be tempted to say $C_p \approxeq C_V$ because $\alpha$ is small for solids and liquids. However, one must be careful because $\kappa_T$ can be small as well, which makes the fraction $\frac{\alpha^2}{\kappa_T}$ large. In other words, even though a little work has to be done to push back the atmosphere when a solid expands, a great deal of work will go into pulling the atoms apart.

Supplementary Derivation: For a system where, $N$ doesn't change the fundamental equation of thermodynamics is:

$$\mathrm{d}U = T\mathrm{d}S -p\mathrm{d}V$$

This seems to suggest that, $U = U(S,V) $. Thus, one can write the following differential and after comparing to the one above can equate $T$ and $-p$ (as indicated by annotations) with the partial derivatives given below:

$$\mathrm{d}U = \underbrace{\left ( \frac{\partial U}{\partial S}\right)_V}_{T} \mathrm{d}S + \underbrace{\left ( \frac{\partial U}{\partial V}\right)_S}_{-p}\mathrm{d}V$$

Moreover, dividing both sides of the fundamental equation by $\mathrm{d}V$ (yeah, I know), and imposing constraint of constant temperature) we can manipulate it into the following form:

$$\overbrace{\left( \frac{\partial U}{\partial V} \right)_T}^{\pi_T} = \overbrace{\left ( \frac{\partial U}{\partial S}\right)_V}^{T}\left ( \frac{\partial S}{\partial V}\right)_T - \overbrace{\left ( \frac{\partial U}{\partial V}\right)_S}^{-p}$$

Thus, we have $$ \pi_T = T\left ( \frac{\partial S}{\partial V}\right)_T - p$$

Invoking, the Maxwell Relation

$$ \left ( \frac{\partial S}{\partial V}\right)_T = \left ( \frac{\partial p}{\partial T}\right)_V$$

one finally gets, $$\pi_T = T \left (\frac{\partial p}{\partial T}\right )_v - p$$


Can I perform electrolysis without electrolyte? (with distilled water)

If I can, how high the voltage must be?

Not possible at home.

For obtaining pure hydrogen and oxygen it is still required to use an electrolyte.  However, there is some amount of them that are not electrolyzed, allowing to electrolyze only water. I'd try sodium/potassium hydroxide or sulfate.


Determination of initial excited state via luminescence

Sometimes formulae are more hindrance than help, and perhaps there are some red herrings in your problem.

  A one electron species initially in some excited state $n_i$ is
  irradiated with a light of wavelength 121 nm when the electron is
  promoted to a further higher orbit $n_f$. In returning back to the
  ground state, it gives an emission spectrum containing 15 bright
  lines. Out of these 15 lines, 9 lines were found to have wavelengths
  smaller than 121 nm while 5 lines have wavelengths greater than 121
  nm. What is the value of $n_i$?

There being fifteen bright lines suggest that $n_f = 6$, since $15 = 5 + 4 + 3 + 2 + 1$, there being $5$ transitions from the $n = 6$ state to $n = 5,\,4,\,3,\,2,\,1$; $4$ transitions from the $n = 5$ state, and so on. This is the basis for the formula you've employed.

Long wavelengths correspond to low energy; we see that the transition from $n_i$ to $n = 6$ is the sixth lowest in energy. $E \propto n_i^{-2}-n_f^{-2}$, so we pull out our handy-dandy calculator and work out: $$ 6 \to 5: 0.012,\,5 \to 4: 0.023,\,6 \to 4: 0.035,\, 4 \to 3: 0.049,\, 5 \to 3: 0.071,\, 6 \to 3: 0.083.$$ Thus $n_i = 3$. We did not need to use the exact wavelength of light used for irradiation and we did not need to know the identity of the species, because this problem is only concerned with the relative energy differences between energy levels.


Why does coke foam up when I add it to (cold) black coffee (no sugar, no milk)?

There are two likely contributing factors here. What you're seeing is rapid release of carbon dioxide from solution, indicating that its solubility has decreased and that the rate of re-equilibrating to that new lower solubility is quite fast. The solubility of carbon dioxide is a factor of

1) the partial pressure of CO2 in the air above the liquid (higher pressure forces more into solution),

2) the pH of the liquid (because higher pH leads to formation of more $\ce{HCO3-}$ and $\ce{CO3^2-}$ thus increasing the total CO2-related species in solution), and

3) temperature of the solution (like most gases, CO2 decreases in solubility as temp increases).

In your case, (1) and (3) are probably most important. [EDIT: I overlooked that the coffee was cold, so (3) is actually not likely to be relevant] The can before opening is pressurized with a high CO2 atmosphere, so it is overloaded with CO2 relative to just about any ambient condition. In order to equilibrate with ambient conditions, it needs to lose some CO2. On the other hand, you've diluted it into the coffee, so there is a slight offsetting effect there. The coffee is slightly acidic, but is not likely to be very well buffered, and probably is not much different than diluting with water.

In addition to these thermodynamic effects that change the equilibrium dissolved CO2 concentration, there is also the kinetic issue of how quickly the CO2 escapes in order to return to equilibrium. The CO2 can only leave solution at the boundary points where the liquid meets air. The stirring effect of pouring the liquid (even into an empty glass) creates more surface area of the liquid/air boundary and also keeps the CO2 well distributed throughout the solution so that the areas near the boundary are continuously resupplied with CO2 as it is lost to the air.

To determine if coffee specifically has any effect, I would try the same thing with water instead of coffee. I suspect the result will be very similar.

UPDATE based on new data:
I was wrong about the coffee itself not being important. The coffee mixture clearly foams more than the tea. There are two possibilities I can think of (not mutually exclusive):
1) the rate of CO2 evolution is greater in coffee. As the other answer postulates, that could be due to faster nucleation by particles in the coffee or, if the pH goes up very quickly, buffering by acids in the coffee once the pH reaches 5 or so (from the Coke starting point ~ pH 2).
2) the same amount of CO2 is released, but the coffee mixture foams more. Studies on sparkling wines such as the one described here indicate that proteins, especially ones with carbohydrates attached, contribute to foaming by acting as surfactants (that means like a detergent molecule, essentially making soap bubbles). It is likely that coffee contains similar molecules. I now think this is the likely explanation, though I also like the idea that there are more nucleation sites from the dissolved solids.

The way to test some of these hypotheses would be to filter the coffee with varying molecular weight cut-off filters, but I suspect you don't have access to that sort of equipment?


Yes, there are many examples. For instance

The common reducing agents LithAl, $\ce{LiAlH4}$, and sodium borohydride, $\ce{NaBH4}$ (last one you might argue about)
$\ce{[ReH9]^{2-}}$, see https://en.wikipedia.org/wiki/Potassium_nonahydridorhenate
Metal carbonyl hydrides, see https://en.wikipedia.org/wiki/Metal_carbonyl_hydride
More generally many organometallic compunds, e.g. $\ce{(C5H5)_2ReH}$

A quick google also shows up http://www.ilpi.com/organomet/hydride.html


Nomenclature of benzene derivatives with two functional groups

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the retained names “aniline” and “phenol” are preferred IUPAC names (PINs). Parent structures having retained names are called functional parent compounds.

For both retained names, substitution is permitted at any position. However, substitution of a functional parent compound is limited to substituent groups cited as prefixes in accordance with the seniority of the functional group that is explicitly expressed or implied in the functional parent compound name. (In systematic names, the presence of this characteristic group would normally be expressed as a suffix.)

  P-15.1.8.2.1 Substitution rules for Type 2a retained names

  Type 2a retained names include functional parent compounds whose name expresses or implies a characteristic group expressed as suffix in systematic names.

  P-15.1.8.2.1.1 Substitution by substituent groups, expressed as prefixes, having a lower seniority than that denoted by the suffix is allowed.

  (…)

  P-15.1.8.2.1.2 The senior characteristic group or functional class name must be
  expressed in the name.

Therefore, the correct name for the compound given in the question is “2-aminophenol” (not “2-hydroxyaniline”) since hydroxy compounds are senior to amines.


Why can't bicyclic structures undergo E2 elimination?

Elimination via a E2 mechanism cannot take place at the bridgehead carbon of a bicyclic compound due to the lack of the required trans geometry between the leaving group at the bridgehead carbon and any of the six β-hydrogens.

On the other hand, as you have correctly observed, a E1 mechanism has to be ruled out as well due to the strain at the bridgehead carbocation. In fact bicyclic structures prevent the tertiary bridgehead carbon becoming planar, which implies that is very high in energy because the nonplanar structure forces the cation to be an empty "sp3 orbital" instead of an empty "p orbital". This observation was made for the first time by Bredt, therefore is generally known as "Bredt's rule".


What the difference in data if calibrate pH-meter with 2 and 3 solutions?

You cannot "predict" any error by thinking that if we calibrate with two buffers the error will be x, and if we calibrate with three buffers the error will be y.
Ideally, in each case, the pH of the sample should be identical. Good quality pH meters tell you how close they are to the theoretical slope in terms of percentage. It should be close to 100%.
For example, at 25°C the theoretical slope is ~ -59 mV/pH. If your calibrate with two buffers and  is close to -58 mV/pH, the slope percentage is 98.3%. It is good to go. As electrode age, the percentages goes down.
As EdV said, three points might be better for the $12 pH meter.


Does precipitated copper carbonate contain impurities

Yes, the water contained in the freshly filtered precipitate will contain dissolved $\ce{Na2SO4}$, as the latter is the soluble by-product of your double displacement reaction. To get rid of it, you need to wash your precipitate thoroughly with distilled water (pH 7). If you really want to make sure that all $\ce{Na2SO4}$ has been removed, you can test samples of the wash water filtrate for $\ce{SO4^2-}$ by adding diluted hydrochloric acid and aqueous $\ce{BaCl2}$. When a white precipitate of $\ce{BaSO4}$ no longer forms, the aqueous sodium sulfate can be considered separated from the $\ce{CuCO3}$ precipitate.


In theory, any molecule that contains a H atom ("proton") can act as a Bronsted acid - it can donate that proton. How likely this is to happen is described by that proton's pKa value, the lower the value, the more likely this is to happen (the more stable the conjugate base that forms will be). You can look up pKa values for different types of proton on an inorganic or organic molecule by searching for pKa tables like this:

Different proton environments on the same molecule will have different pKa values:

The pKa describes the degree to which that proton will dissociate in a solvent (normally water, but other solvents are sometimes used) and high values mean that this essentially doesn't happen, or only to a tiny extent (in the figure above, these are labelled as "not acidic"). However, a very strong base could abstract these protons, so the molecule could still act as an acid.

To act as a base, a molecule normally needs to possess a lone pair. The Bronsted definition of a base only requires a molecule to accept a proton, but a lone pair is generally required in order for this to happen.

The example in the figure above shows ammonia acting as a Bronsted base, using its lone pair to bond to a proton. Any lone pair could (potentially) do this, but some functional groups (such as amines) have more propensity to do this. The pKb value for the functional group will quantify this propensity.

Molecules often contain both protons to donate and lone pairs, so can act as Bronsted acids and/or bases.

Lewis acids and bases are defined slightly differently. The definition here involves the electron pair, rather than the proton. So you can have Lewis acids (electron pair acceptors) that have no protons at all. These tend to be electron deficient species such as the $BF_3$ in this figure below:

Other examples of Lewis acids include small, highly charged cations that can accept coordinate bonds.

Some molecules can accept protons by using electrons from unsaturated moeities rather than lone pairs. An alkene is a classic example, which can use the pi electrons in the double bond to accept a proton, forming a carbocation.

So an alkene could be described as a base, but we don't normally categorize it in that way.

It is also worth looking at the distinction between acid/base and electrophile/nucleophile. There are some good posts on this site like this one


What is the meaning of signs in electrode potentials?

Chapter 14 of Harris' Quantitative Chemical Analysis has a helpful section entitled, an intuitive way to think about cell potentials.  Here's the corresponding graphic:

In this case, the author is describing a Galvanic cell with cadmium as the anode and silver as the cathode.  The potentials of the two half-cells are known and marked on the line-graph.  Electrons always flow to the more positive electrode, which in this figure is towards the right.  The difference between the two half-cells is the cell potential.

In your case, you know the separation (-0.07 V) and one of two points (+0.34 V).  If we ignore the sign for the moment, we can use the line-graph to narrow the possibilities down to C or D, since each of these differ from +0.34 V by 0.07.  The question now is, which direction are the electrons flowing?  When determining the cell potential using this line graph, we subtract the right-most number from the left-most number in order to obtain a positive potential.  Since the cell potential in your case is negative, we must be doing the opposite.  Therefore, the reference electrode potential is +0.41 V (D).


The missing resonance structure is

If you replace "SH" with "OH" in the 4 structures you've drawn (your 5th structure is the same as the one pictured on the exam) and the one I've drawn, you'll have all of the analogous resonance structures for phenol.

EDIT : response to OP's comment

In my view, it doesn't really matter how you order the structures, but your suggestion (between the 3rd and 4th structures from the left) seems logical.  I've redrawn the figure to show the electron movement.


Is the 'sweat' on the side of bottles distilled water?

Yes, here is the definition for distilled.

purify (a liquid) by vaporizing it, then condensing it by cooling
the vapor, and collecting the resulting liquid.

The sweat on the water bottle is the process of distillation, but only the condensation part of the process is seen on the water bottle.


Do both H (enthalpy) and ∆H, or ∆H alone have a physical significance?

Offcourse H and ∆H have physical significance. H represents the enthalpy of a system. Enthalpy refers to the sum of the internal energy of a system plus the product of the system's pressure and volume.

The ∆ symbol is used to represent change. Therefore, ∆H represents the change in enthalpy of a system in the reaction. So your book is completely correct.


How to find the change in internal energy (q+w) for an expanding monoatomc ideal gas?

The internal energy $U$ is a state function that can be expressed as a function of temperature $T$, pressure $p$, and amount of substance $n$:

$$U=U(T,p,n)$$

The change in internal energy $\Delta U$ is independent of the path by which the new state was reached. Hence, you do not necessarily have to calculate the exchanged heat $Q$ and the work $W$ done during a particular process.

The ideal gas gives simpler equations. According to Joule’s second law, the internal energy $U$ of an ideal gas depends only on the amount of gas and the temperature, and is independent of the volume or the pressure:

$$U=U(T,n)$$

For a monoatomic ideal gas (i.e. a model system of point-mass molecules) we find that the internal energy is proportional to the temperature $T$ and number of molecules $N$ or amount of substance $n$:

$$U=\tfrac32NkT=\tfrac32nRT$$

where $k$ is the Boltzmann constant and $R$ is the molar gas constant.

Since you know the temperatures $T_1$ and $T_2$, and since the amount of substance is constant $(n_1=n_2)$, you should be able to calculate the change in internal energy $\Delta U=U_2-U_1$.

Note, if you don’t know the number of molecules $N$ or amount of substance $n$, you may express the result as molar internal energy $U_\mathrm m$ (i.e. in $\mathrm{J/mol}$):

$$U_\mathrm m=\frac Un$$


When will the Lassaigne's method for detection of nitrogen and Kjeldahl's method for quantitative estimation of nitrogen fail?

You are right about the limitations of century old Lassaigne's method. Why are they still teaching it? Modern day CHNSO elemental analysis is done with gas chromatograpy systems. If you are quoting from a book, provide a reference.

This is an excerpt from a 1953 work Sulfur in Organic Compounds, by a Modified. Lassaigne's Method ACTA CHEMICA SCANDINAVICA 7 (1953) 1395

  Objections to Lassaigne's method are: a) low boiling substances evaporate
  before reaction with the sodium, b) diazo-, azo and polynitro compounds evolve nitrogen c) some heterocyclic compounds do not react.


Based on available theoretical considerations and a available literature, a tetrahedral geometry appears to be a good assumption for $\ce{Cr(NO)4}$.

The NO ligand can be rendered as a three-electron donor, if we take
it as uncharged.  It is actually similar to CO, really, in that it
interacts through its $\pi^*$ orbitals as well as its $\sigma$
orbital; but NO unlike CO has an electron to offer from its $\pi^*$.
That plus the $\sigma$ electrons makes three.
We then may identify the central metal atom as having 18 valence
electrons.  When this happens with multiple ligands we typically see
an arrangement with highest order symmetry, thus tetrahedral for
four-coordination.  This gives the best possible bonding to the
entire set of valence orbitals involved in the 18-electron
structure.  This is seen in the tetrahedral four-coordinationn of
$\ce{Ni(CO)4}$, and in the octahedral six-coordination of the series
$\ce{[V(CO)6]-, Cr(CO)6, [Mn(CO)6]+, [Fe(CO)6]^{2+}}$.  So, we should
expect $\ce{Cr(NO)4}$ to be tetrahedral.
Heberhold and
Razavi
found evidence for a tetrahedral structure in $\ce{Cr(NO)4}$ from
its IR spectrum.


Ambiguity in definition of primary valency in Werner's theory

George B. Kauffman has edited an excellent book, "Coordination chemistry: A century of progress." It was published as an ACS Symposium Series. In the chapter, he writes, "If history teaches us anything, it teaches us that the latest view is not always the best and that change is not always progress." I may add that there is no authority in science. I am glad that you challenged textbooks.
Basically, in Werner's time, the prevalent theory of the complexes was that the complexes are chained compounds, which was not consistent with the solution conductivity and optical isomerism of the complexes. Werner proposed the primary and secondary valence concept.
Contrary to most modern chemistry books, including that respectable Alan G. Sharpe's texts, Werner's primary valence matches the "oxidation" state of the metal in simple cases. The oxidation state matches Werner's primary valence when there is no anion in the coordination sphere. The textbook version of Werner's story fails for the type of compounds you just asked, such as $\ce{[Co(NH3)5Cl]Cl2}$.
Here the oxidation state is indeed 3. However, Werner's primary valency of this compound is 2.
Kauffman mentions

Werner postulated two types of valence—Hauptvalenz, primary or ionizable valence, and Nebenvalenz, secondary or nonionizable valence. According to Werner, every metal in a particular oxidation state, that is, with a particular primary valence, also has a definite coordination number, that is, a fixed number of secondary valences that must be satisfied. Now, whereas primary or ionizable valences can be satisfied only by anions, secondary or nonionizable valences can be satisfied not only by anions but also by neutral molecules containing donor atoms such as nitrogen, oxygen, sulfur, and phosphorus.

The keyword is that primary valence is the ionizable valence. The ionizable valence of the complex in question is 2. $\ce{[Co(NH3)5Cl]Cl2}$ would generate three ions in solution $\ce{[Co(NH3)5Cl]^{2+}}$ and two $\ce{Cl-}$. These experiments were easily doable in Werner's time by conductivity measurements. Secondly, as discussed in the comments, argentometric titrations would show only two chloride ions rather than three.


As @Amritansh Singhal said, $\ce{H2}$ would be produced at the cathode from the electrolysis of both the water and some from the acetic acid (main ingredient of vinegar after water) and $\ce{O2}$ would accordingly be produced at the anode.

The question in your comment "Would it create ethane?" is where some more interesting chemistry might take place.  The short answer is "very little at the most" since in the presence of water, the cell potential will be limited by the electrolysis of water as described above.

However, if you could obtain pure acetic acid and produce the proper cell potential, then you could see the Kolbe electrolysis reaction, which would produce ethane.  According to excerpts from this Wikipedia article":

  "The Kolbe reaction is formally a decarboxylative dimerisation of two
  carboxylic acids (or carboxylate ions)."

With respect to acetic acid specifically the above article states:

  "The reaction mechanism involves a two-stage radical process:
  electrochemical decarboxylation gives a radical intermediate, then two
  such intermediates combine to form a covalent bond. As an example,
  electrolysis of acetic acid yields ethane and carbon dioxide:"
  $$\ce{CH3COOH → CH3COO− → CH3COO· → CH3· + CO2}$$
  followed by:
  $$\ce{2CH3· → CH3CH3}$$

Where the latter product is ethane.


How to extract thymol from oregano leaves?

This sounds like a nice project and I'm tempted to answer in too much detail, but since this is homework […]. Here are some hints:

Find information about thymol. This is crucial!

What is the structure?
In which solvents is it soluble?
Is thymol soluble in water?

Is it soluble in acidic or alkaline solution?

What about the boiling point and the melting point?

Plant leaves have a cell structure in which your thymol is somehow bound and hidden. In order to break of the structure, you may want to

dry and powderize the leaves or
cut the fresh leaves into pieces and grind them in a mortar with sea sand

Oregano leaves are green. You know that they contain chlorophyll. There's probably a lot of other compounds too, not just thymol.
Supposed that you treat the leaves as described in 2. and start to extract using ethanol in a Soxhlet extractor, you'll end up with a nice cocktail. Do you need to separate thymol from the other extracted compounds? If so, how will you proceed? Knowing the properties of thymol will help you here!

Remember that the easiest work is the one that you can avoid ;-)

Think in analogies. Thymol is a monoterpene. A lot of other members from this family have a typical smell and are used as fragrances. There must be an established, mild procedure to obtain these compounds from plant material.

   Yes there is! Search for steam distillation :)

If there is an established prodecure that avoids column chromatography or other purification techniques, maybe somebody has already used it for thymol and wrote about it.

Back in the days, our professors used to say with a smile:

  A week in the lab saves you an hour in the library

Replace library with internet and have another search for a suitable and easy procedure already published.

Update

In a follow-up comment, you asked:

  Then, how much ethanol should I use?

My suggestion is: Avoid subsequent laborious workup and do not use any ethanol at all!

Read Introducing Organic Chemistry Students to Natural Product Isolation Using Steam Distillation and Liquid Phase Extraction of Thymol, Camphor, and Citral, Monoterpenes Sharing a Unified Biosynthetic Precursor by Katherine A. McLain, Kenneth A. Miller, and William R. Collins, J. Chem. Educ., 2015, 92, 1226-1228. Follow the given procedure, which was deliberately designed to be performed in a students' lab:

  One, 3 h laboratory period is required to complete the experiment. Dried plant material (10 g) or fresh plant material (20 g) is placed in a high-speed blender with distilled water (200 mL) and blended to a fine
  slurry. The slurry is transferred to a round-bottom flask (500 mL) with a distillation apparatus attached and a stir bar, and it
  is placed directly on a hot plate turned to high heat. A beaker
  (200 mL) is used as the receiving vessel to collect
  approximately 100 mL of distillate. The bioactive molecule is
  extracted from the distillate with an appropriate organic solvent
  (30 mL). The organic phase is dried with magnesium sulfate,
  and the organic solvent is removed on a hot plate at medium
  heat.

For further details, have a look at the Supporting Info.


Calculating the theoretical percent purity of a recrystallization

I can give you a hint, imagine you have a mixture containing one consitutant $A$ only in your solution with a solvant.

Doing the balanced equations about the mass of $A$ in each phases you have :

$$\begin{cases}
m_A^\mathrm{t}=m_A^\mathrm{c}+m_A^\mathrm{liq} \\
m_A^\mathrm{i}=x_A^\mathrm{c}\cdot m^\mathrm{c}+x_A^\mathrm{liq} \cdot m^\mathrm{liq}
\end{cases}$$

Where $m^\mathrm{t}$, $m^\mathrm{c}$, $m^\mathrm{liq}$ and $m^\mathrm{i}$ denote respectively the mass of $A$ total, in the cristal form, dissolute in the solvant (liquid), and at the begining (initial) and $x$ the corespondant mass-fractions.

Then you must obtain, $$m^\mathrm{c}=m^\mathrm{t}\left(\frac{\frac{m_A^\mathrm{i}}{m^\mathrm{t}}-x_A^\mathrm{c}}{x_A^\mathrm{liq}-x_A^\mathrm{c}}\right)$$

Notice that $\frac{m_A^\mathrm{i}}{m^\mathrm{t}}=x_{A,\mathrm{i}}^\mathrm{liq}$ which is the mass-fraction at the begining in the liquid and $x_A^\mathrm{liq}$ is the mass-fraction in the liquid at the end.

If you know how what you put at the begining, you can determine the yield. Now if you want to know the purity, you'll need to do the same with both of your component to see how much of them have cristalised at the end and then determine the purity of your marjoritory compound.

I hope it can help, have a good day ! :)

Be careful when you do the balanced equations if there is also a gas phase which can happend sometime.


Turning of a paddle wheel when exposed to light

Not the complete answer, but a short analysis of expectable forces:
The energy and momentum of light are related by equation $E=p \cdot c $.
Then force caused by light $F=\frac{\mathrm{d}p}{\mathrm{d}t}=\frac{\mathrm{d}E}{c \cdot \mathrm{d}t} = \frac Pc$
Therefore the light power 30 W (some cca 50-60W fluorescent bulb or cca 35-40 W LED bulb ) would cause direct  force cca  100 nN, equivalent of cca 0.01 mg.
Or twice as much if reflected.
Crookes radiometer shows the wheel is turning in the opposite direction than it should be, if caused by the light pressure.
The motion is caused by uneven heating and expansion of gas around the wheel, causing pressure differences in the opposite sides of paddles.
When pressure is too low, the gas there is not enough and there remains just light pressure, that is too weak to turn the wheel.

Quoting Effect observations from article.
The effect begins to be observed at partial vacuum pressures of several hundred pascals (or a few torr), reaches a peak at around $\pu{1 Pa}$ ($\pu{7.5e−3 torr}$) and has disappeared by the time the vacuum reaches $\pu{1e−4 Pa}$ ( $\pu{7.5e−7 torr}$) (see explanations note 1). At these very high vacuums the effect of photon radiation pressure on the vanes can be observed in very sensitive apparatus (see Nichols radiometer) but this is insufficient to cause rotation.

Nichols radiometer can register the true light pressure by mirrors on torsion balance.

For cathode rays = electrons
If there is
potential difference $U$
current $I$
elementary charge $e$
electron mass $m_e$
charge $Q$
Area $A$
pressure $P$ (avoiding confusion)
momentum of 1 Coulomb  $p$
electron momentum $p_e$
then for kinematics:
$$eU=\frac 12 m_e.v^2$$
$$v=\sqrt{\frac{2eU}{m_e}}$$
for momentum:
$$p_e=m_e.v=\sqrt{2m_eeU}$$
$$p=\frac 1e \cdot p_e=Q \sqrt{\frac{2m_eU}{e}}$$
for force and pressure
$$F=dp/dt= I \sqrt{\frac{2m_eU}{e}}$$
$$P=F/A= \frac IA \sqrt{\frac{2m_eU}{e}}$$
for $\pu{380 V}$ and $\pu{1 A}$ it gives force about $\pu{66e-6 N}$, what is equivalent to weight of the mass about $\pu{6.6 mg}$.
This is a model case. For a particular scenario, the geometry and electrostatic forces would play significant role.


Does the composition of a solution keep changing when being boiled?

It is not a messed up cycle. Let's consider what happens with an example.
If you heat up a mixture containing $10$% ethanol + $90$% water, the mixture boils at $92$°C, producing a vapor containing $50$% ethanol + $50$% water. With this operation the liquid loses more ethanol than water. So its concentration in  ethanol decreases, and it is necessary to heat it up to a higher temperature to maintain the process of evaporation. When about one tenth of the liquid has distilled, the residual liquid contains still $5$% ethanol, and it is necessary to heat it up to $95$°C to obtain and continue boiling. At $95$°C, the vapor emitted contains $35$% ethanol. At $98°$C, the liquid contains still $2$% ethanol and the vapor $20$% ethanol. When the last drops are vaporized, the temperature is $100$°C and the last drops are made of pure water.


Chemistry is an experimental science supported by computational methods. Qualitative explanations for reaction mechanisms follow as a result. More on that issue later.The mechanism of the Birch reduction has been investigated by Zimmerman and Wang.1 What are the experimental results? The Birch reduction is first order in substrate, electrons and alcohol. Therefore, the rate-limiting step is the protonation of the radical anion 2 of anisole 1. Based on the observation that the dianion of anthracene is more basic than its radical anion, 2 the authors designed an experiment utilizing 2% d1-t-BuOH as the alcohol source. The argument is that the radical anion, which is less basic than the anion, will be more discriminating,3 favoring protonation rather than deuteration (isotope effect). The more basic anion, 4 or 7, which is less discriminating, will give rise to a higher D/H ratio. Employing deuterium NMR, the meta position in 8 was found to have 7-times the amount of deuterium than the ortho-position. Thus, the ortho-position (2a --> 3) is protonated first; mechanism O (ortho) predominates over mechanism M (meta). --continued--

Calculations showed that the methoxy ortho-substituted cyclopentadienyl radical is more stable than the meta- and para-isomers (red box). But your request was for a qualitative rationale for preferring radical 3 over 6. Radical 3 has an advantage over radical 6 because radical 3 may be written with the radical on the ipso-carbon, the same radical that would be expected to be generated by auto-oxidation of 5-methoxycyclohexa-1,3-diene. Cross-conjugated radical 6 does not have this option.

1) H. E. Zimmerman and P. A. Wang, J. Am. Chem. Soc., 1993, 115, 2205.; H. E. Zimmerman, Acc. Chem. Res., 2012, 45, 164.
2) V. D. Parker, M. Tilset and O. Hammerich, J. Am. Chem. Soc., 1987, 109, 7905.
3) Consider by analogy that chlorine atoms are less selective than bromine atoms in the radical halogenation of alkanes.


Writing equation of reaction in stoichiometry of redox titrations

See the following steps:

We know what $\ce{KMnO4}$ will do by reading the question:

  Calculate the volume of […] $\ce{KMnO4}$ solution required to oxidize completely […] oxalic acid.

We need the structure of oxalic acid. (This is knowledge required to answer the question, no way around.)

Structure of oxalic acid. Image taken from Wikipedia, where a full list of authors is available.
Determine oxidation numbers of all the atoms involved. Shortcut: hydrogen has $\mathrm{+I}$, oxygen has $\mathrm{-II}$ and carbon has $\mathrm{+III}$.
Determine what can be oxidised. Only carbon really can.
Determine what it can be oxidised to. The final product must have $\ce{C^{IV}}$, which is most common as carbon dioxide or carbonate.
Solve the redox equation using either $\ce{CO2}$ or $\ce{CO3^2-}$. Both will work.


Hydrogen fuel cell - why do the H+ ions move through the electrolyte

In a galvanic cell, the anode is positively charged.  The cathode is negatively charged.  A good mnemonic I learned a while ago is that anions go to the anode, and cations go to the cathode.  This is because opposite electric charges attract.  So negatively charged anions are attracted to the positively charged anode.  And positive cations are attracted to the negative cathode.

The electron itself was originally discovered as the component particles of cathode rays from cathode ray tubes.  That electrons come out of the cathode is a hint about cathode charge: the buildup of electrons on the cathode eventually results in the emission of an electron beam.  If electrons are building up, the cathode must be negative.


Why is water ignored in the ionic equilibrium of HF?

Why are we not adding the contribution of water to the $\ce{H3O+}$ ions?

Write down the dissociation constant equation with and without $\ce{H2O}$.
What changes?
Can you find a formulation of the dissociation constant where it seems natural not to include $\ce{H2O}$?
Read up on "activity".

  Why are we assuming that the value $0.02x$ is dissociated from $\ce{HF}$ and not just $x$?

Write down and solve the equation with $0.02 x$ and with $x$ only. What is the difference?


What difference does a polar protic solvent make in the addition of X2 to alkenes?

In the presence of a polar protic solvent, the halogen will react with the solvent to form acids ($\ce{HX}$ and ${HOX}$).

$$\ce{Cl2 + H2O -> HCl + HOCl}$$

The high concentration of the cations formed (the hydrohalic acid and the hypohalous acid will exist as ions in a polar protic solvent) can act as electrophiles and initiate an electrophilic substitution reaction which results in the formation of a range of products.

$$\ce{CH2-CH=CH2 + X2 + H2O -> CH2-CH(Cl)-CH3 + CH2-CH(OH)-CH2(Cl) + CH2-CH(Cl)-CH3(Cl)}$$

This reaction is primarily initiated by the $\ce{H+}$ attacking the electron-rich double bond. This causes an electomeric shift of electrons to a carbon such that the carbocation formed is stable (Markonivkoff's rule). As carbocations are formed, rearrangements are possible. Not only are the mechanisms different, the products will be different too. This reaction is entirely different from a $SN_2$ reaction.


Decomposition products of sodium acetate

The fumes are likely to be acetic anhydride, as per this equation:
$$\ce{2CH3COONa -> Na2O + (CH3CO)2O}$$
This smells like vinegar, and reacts with moisture in the air to give acetic acid.


Why do phosphorus and sulfur have variable valences?

For a modern answer (with some hand-waving), we can simply say that valence electrons occupy more space as we go from first period to second period to third period. So hydrogen makes one bond, second period elements up to four, and third period elements up to six.
If you want to stick to Lewis structures and hybridization, you can explain $\ce{PCl5}$ with resonance:

Source: https://www.cusd80.com/cms/lib6/AZ01001175/Centricity/Domain/1108/003-center%204-electron%20bond.pdf
If you like molecular orbital diagrams, it is pretty straightforward to expand a $\ce{EX3}$ MO diagram to a $\ce{EX5}$ diagram without using d-orbitals, e.g. http://www.chembio.uoguelph.ca/educmat/chm364_preuss/5_5contd2%20MO%20Theory_ICl4%20PCl5.pdf. Like in the valence bond model, the bond order will decrease as you increase bound atoms.


Disproportionation of plutonium species - Latimer's diagram

In acidic solutions ($\mathrm{pH} = 0$), the values of the relevant reduction potentials of $\ce{Pu^3+/Pu^4+}$ ($\pu{+1.01 V}$), $\ce{Pu^3+/PuO2+}$ ($\pu{+1.03 V}$), $\ce{Pu^3+/PuO2^2+}$ ($\pu{+1.02 V}$), $\ce{Pu^4+/PuO2+}$ ($\pu{+1.04 V}$), $\ce{Pu^4+/PuO2^2+}$ ($\pu{+1.03 V}$), and $\ce{PuO2+/PuO2^2+}$ ($\pu{+1.02 V}$) are all very similar.
Therefore, plutonium has the remarkable feature that all four important oxidation states can be present simultaneously in aqueous solutions.

In aqueous solution, $\ce{Pu^4+}$ disproportionates into a mixture of oxidation states:
$$\ce{2Pu^4+ + 2H2O <=> Pu^3+ + PuO2+ + 4H+}$$
$$\ce{Pu^4+ + PuO2+ <=> Pu^3+ + PuO2^2+}$$
Thus, the net reaction of the disproportionation of $\ce{Pu^4+}$ is approximately:
$$\ce{3Pu^4+ + 2H2O <=> 2Pu^3+ + PuO2^2+ + 4H+}$$
Nevertheless, $\ce{Pu^4+}$ may still be the predominant oxidation state in the mixture.
Besides, the equations imply that the disproportionation of $\ce{Pu^4+}$ dcreseases as the acidity of the solution further increases.

Concerning pentavalent plutonium ($\ce{PuO2+}$), your expectations are correct. In typical acidic solutions, $\ce{PuO2+}$ indeed quickly disproportionates into a mixture of oxidation states:
$$\ce{2PuO2+ + 4H+ <=> Pu^4+ + PuO2^2+ + 2H2O}$$
$$\ce{Pu^4+ + PuO2+ <=> Pu^3+ + PuO2^2+}$$

However, the given values for the reduction potentials apply to pH = 0, and the various reduction potentials of plutonium depend on pH. Furthermore, the stability of individual oxidation states can be strongly influenced by hydrolysis, or the formation of complexes or compounds with low solubility products. Therefore, the predominant oxidation state can be changed by means of multiple factors. Furthermore, the valence of plutonium solutions may change as a consequence of radiolysis.


Why do orthoesters react with Grignard reagents, but acetals and ketals don't?

Grignard reactions are performed in ether or THF.
There you have only one -OR group on a carbon and, since it is the solvent, it is not reacting with it.
This shows that not always a -OR group is substituted in a Grignard reaction.

When you have three -OR's on one carbon, the formal positive charge ($\delta+$) on it is higher than when there are only two -OR's.
Therefore, the reaction proceeds on orthoesters and not on acetals or ketals.


Why is pyrrole more reactive than pyridine and benzene for an electrophilic substitution reaction?

From Peter Sykes:

  Pyridine (62), like benzene, has six at electrons (one being supplied bynitrogen) in delocalised it orbitals but, unlike benzene, the orbitals will be
  deformed by being attracted towards the nitrogen atom because of the latter's
  being more electronegative than carbon. This is reflected in the dipole of pyridine, which has the negative end on N and the positive end on the nucleus:

  Pyridine is thus referred to as a $\pi$-deficient heterocycle and, by analogywith a benzene ring that carries an electron-withdrawing substituent, e.g.$\ce{NO2}$ (p. 6.22), one would expect it to be deactivated towards electrophilic
  attack. Substitution takes place, with difficulty, at the 3-position because this leads to the most stable Wheland intermediate (63); the intermediates
  for 2- and 4-attack (64 and 65, respectively) each has a canonical state in which the charge is located on divalent N a highly unstable,i.e. high energy, state.

  There are certain formal analogies here to m- attack or
  p. 6.24), byt pyridine is very much more difficult to substi
  Thus nitration, chlorination, bromination and FriedelCraf
  really be made to take place usefully, and sulphonation on
  ing with oleum for 24 hours at 230°C, with an $\ce{Hg^2+}$ catalyst attack is due partly to the fact that pyridine has an available nitrogen, and can thus protonate (66), or interact with an electrophile (67):

  The positive charge will clearly further destabilise any of the $\sigma$-com
  plexes for electrophilic substitution, as did a substituent such as $\ce{-NR3^+}$, on the
  benzene nucleus (p. 6.24), but the destabilisation will be much more marked
  than with $\ce{-NR3^+}$; as the charge is now on an atom of the ring itself and not merely on a substituent.
  Pyrrole (68) also has 6 electrons in delocalised orbitals, but here the
  nitrogen atom has to contribute two electrons to make up the six (thus be
  coming essentially non-basic in the process, cf. p. 3.26), and the dipole of
  pyrrole is found to be in the opposite direction to that of pyridine, i.e. with
  the positive end on nitrogen and the negative end on the nucleus:

  Pyrrole is thus referred to as a π- excessive heterocycle and behaves rather like a reactive benzene derivative, e.g. aniline (p. 6.25), undergoing very ready electrophilic attack. This may be complicated by the fact that in
  strongly acid solution protonation (69) is forced even on the weakly basic
  pyrrole (it takes place on the 2-carbon atom rather than on N. cf. p. 3.26)

TLDR

Pyridine is a $\pi$-deficient heterocycle owing to nitrogen's more electronegative nature than carbon. Pyrrole on the other hand is $\pi$ excessive since nitrogen donates its lone pair to the ring.
The non-delocalized lone pair on pyridine tends to interact with the electrophile/$\ce{H+}$. This further destabilizes any of the $\sigma$ complexes formed in electrophilic aromatic substitution (EAS).


Is sign convention for delta H and enthapy change of a solution different?

The reaction is endothermic, meaning that it takes energy from the surrounding. The "surrounding" in your example is the solvent, so in the dissolution process, energy is taken from the solvent,  thereby decreasing its temperature.


How long burned tobacco/cigarette smoke remain toxic?

Toxicity of tobacco combustion products in different phases
For answering your question I think is better split it in two. First thing you need to understand that toxicity depends on the method of administation of the toxic agent. Tobacco smoke is a physical state of the tobacco combustion products. It is an aerosol originating from combustion where its particles are liquid, gas and solid (tar). Quoting Chemistry Central Journal 2012, 6:98

Cigarette smoke is a complex and dynamic aerosol consisting of at least 5,600 chemicals and toxicants found across two phases, the particulate (tar) and vapour phase.

(I would like to add the gas phase...) so when the product of the combustion of the tobacco are in this physical state you have to look at a different method of administration in this case vapour inhalation, Dust and Mist Inhalation and gas inhalation these are the most potentially dangerous method of administration. It seems that after about 40 minutes the liquid and solid part of the aerosol deposit (of course this depends on the condition, T, wind etc. etc.) so is probable that after this period of time the toxicity of tobacco combustion products decrease considerably. However this is not true for all the combustion products. For example: carbon monoxide is recognize tobacco combustion product and it is a gas so it wont deposit after this time and it will be "removed by reactions with OH radicals (85%), by soils (10%), and by diffusion into the stratosphere" very slowly.
Benzene is a good example of how the toxicity depends on the method of administration. LC50 inhalation value of benzene is 10000 ppm TFLo Dermal is 0.92 mL/kg so it is a big difference if you wait that the areosol deposit or if you inhale it.
If you are interested in rate of decomposition of benzene it seems that biodegradation of benzene can reach 0.95% at day according to Chiang.
Regarding 1-3 butadiene according to William A. McClenny and Donald Whitaker:

i.e., a 10-6 lifetime risk level for cancer due to inhalation exposure
of 0.03 $g/m^{3}$ (12.4 pptv at 20 °C and 1 atm pressure).1 This compound
is very reactive in the ambient atmosphere with a short
atmosphericlifetime, estimated to be 2–3 hr.

The half-life of acrolein is 14.4 hr.
You can also find a very complete EPA report EPA/600/P-98/001F
October 2002.
However determine the half-life of all the compounds is a very complex task because is related to how the ecosystem respond to these compounds so to the actual capability to decompose them through different mechanism is strictly related with thousands of factors.


Why is it vital that atoms in the same sample are all given the same charge? Mass spectrometry

This is a misconception that during ionization, each ion must get a +1 or +2 charge. It is desirable to have +1 so m/z corresponds to the molecular mass of the ion or its fragment. The state of ionization heavily depends on what you are using to ionize your target analyte. For example if you use electron impact, you will see fragments mostly with +1 charge. However in other techniques called matrix assisted laser desorption (MALDI), the state charge can be as high as +12 and so on. They are detected as such. In electrospray ionization, +2 is also common.

There is no way a mass spectrometer "knows" a priori the state of charge of an ion. All charge ratios are detected. However it can distinguish the sign. It depends on the user whether negative or positively charged ions reach the detector. Some ions are better detected in positive mode and some in negative mode.


Can you predict the resistance of conjugated polymers as a function of degree of polymerization (DP)?

Short answer: No.

First off, the degree of polymerization does not directly relate to the charge transport properties of conducting polymers (more below). You can find highly conductive oligomers.

Secondly, in my opinion, NEGF is not a suitable method to understand the charge transport in polymers. It works for nanoscale devices and molecular wires, but most polymer devices are on the ~100 nm to micron scale.

That's not to say that different oligomer and polymer lengths don't have different transport properties. Absolutely, different molecular weight polymers have different charge mobility, etc. The effects are usually due to a number of hard to characterize factors:

Concentration of defects and traps
Morphology of the film (e.g., higher MW might yield smoother or more ordered films)
Polydispersity (e.g., broader MW distribution is often problematic)
Effective conjugation length
Packing

etc.

The electronic structure of oligomers and conjugated polymers can be simulated fairly well using first-principals methods, including DFT. This can often predict components like the energy levels, band gaps, band dispersion, etc. for crystals and parameters for amorphous films such as the Marcus reorganization energy which is often related to the charge mobility in many of these materials.

What you usually find is that as a function of chain length, charges become more delocalized and the reorganization energy drops (increasing mobility). In reality, this trend should saturate, although most DFT methods incorrectly delocalize charges completely.

Also, most devices with conjugated polymers are amorphous or polycrystalline, making device performance predictions difficult. Transport is not usually band-like, but some form of variable-range hopping between chains or sites within a chain.


What are the products for the redox reaction HCl + SnCl₂ + KMnO₄?

When using $\ce{MnO4-}$ as an oxidant, pH is crucial.

In acidic medium: $\ce{Mn(VII) -> Mn(II)}$

$$\ce{8 H3O+ + MnO4^{-} + 5 e- -> Mn^{2+} + 12 H2O}$$

Under neutral conditions: $\ce{Mn(VII) -> Mn(IV)}$

$$\ce{2 H2O + MnO4^{−} + 3 e− -> MnO2 + 4 OH^{−}}$$

In strongly alkaline medium: $\ce{Mn(VII) -> Mn(VI)}$

$$\ce{MnO4^{−} + e− -> MnO4^{2−}}$$

As far as an excess of $\ce{HCl}$ is concerned, one might take into account that $\ce{Sn(II)}$ and $\ce{Sn(IV)}$ possibly exist as their anionic chloro complexes
$\ce{[Sn(Cl)4]^{2-}}$ and $\ce{[Sn(Cl)6]^{2-}}$, respectively.


You are assuming that all resonance structures contribute equally.  The resonance structures on the right and left are equivalent but have one aromatic ring and one non-aromatic ring.  This is in contrast to the center resonance structure, which has two aromatic rings and is more stable, thus giving the bonds between carbons 1-2; 3-4; 5-6; 7-8; and 9-10 more double bond character than the other bonds and making them shorter.


Predicting the product in a series of reactions

You seem very sure about your intermediate B; we’ll come back to that in a second.

For now, let’s concentrate on the reagents for your final step.

$$\ce{B ->[MnO2] C}$$

Manganese dioxide is the product you get from heating any manganese salts in the bunsen burner until red-hot; thus, one might call it the most stable or thermodynamic oxide. It seems very unreactive and in fact I only know of one type of reaction that it is able to perform on organic molecules: allylic and benzylic alcohols are oxidised to the corresponding carbonyl compounds while fully saturated (i.e. not allylic) alcohols are untouched.

Since you have that reagent, it might be helpful to go backwards; assuming that the final step is an oxidation of a benzylic or allylic alcohol to a corresponding carbonyl. Thus, $\ce{C}$ would either be a ketone or an aldehyde and $\ce{B}$ would have to be an alcohol. But your proposed structure $\ce{B}$ does not include an alcohol! Therefore, your proposed reaction of the second step is very likely wrong.

$$\ce{A ->[Ph-CHO][H2O] B}$$

Indeed, here you assumed that the anion $\ce{A}$ you proposed reacts with the phenyl ring of benzaldehyde in an electrophilic aromatic substitution. Take a good look at the two emboldened words and consider whether an anion can really be electrophilic. Then ask yourself what anions are in philicity terms. Once you have done that, attempt to locate the position of benzaldehyde with the corresponding electronic properties.

The solution is given in the spoiler below; please try to find it yourself, though; it is a much better exercise than just looking things up.

   Anions are negatively charged as are electrons. Thus, anions are practically never drawn towards negative charges — remember that like and like repulse each other. Instead, anions are drawn to positive charges: they are nucleophilic.

 Nucleophiles are drawn to the most positively charged bit of a molecule they are attacking. In the case of benzaldehyde, the carbon side of the $\ce{C=O}$ bond is most positively charged due to the resonance depiction we can draw below:

 $$\ce{C=O <-> \overset{+}{C}-\overset{-}{O}}$$

 Therefore, the acetylide anion will attack the carbonyl carbon of benzaldehyde nucleophilicly, creating a benzylic alcohol. It is now no secret that this is cleanly oxidised to the corresponding alkynophenone.

   $$\ce{Bu-C#C^- (A) ->[Ph-CHO][H2O] Ph-CHOH-C#C-Bu (B) ->[MnO2] Ph-CO-C#C-Bu (C)}$$


Surface Chemistry - Dipole Moment of Surfactant with Water and Phases of a Monomolecular Layer

Wow, that is a question!

Surfactants (aka tensioactive compound) have a dipolar moment at the air-water interface but that is not enough to characterize their behavior. There is a cohesive force between their lipophilic tails (due to Van der Walls force) and there is also a cohesive force between their polar head but only because water of metal ions can make a bridge between identically charged moieties.
I will address only your main question which seems to be "how to know the size of molecules in a monolayer"? Well, a surfactant floats at the surface of water. So you can make an apparatus filled with water and with a movable barrier which will move across the surface to slowly reduce the overall surface available for a tensioactive compound. We also know how to measure variations in the "tensioactive power" (whatever its name is) which depends on the number of molecules at the surface.

Using such a device, we can create a monolayer of a lipophilic compound, reduce the whole area (which we call a "compression") and detect the resulting effect (which correlates to the number of molecules at the surface). Whenever the available surface becomes too small for all the molecules to be nicely arranged at the surface (i.e. the "collapse").


Given $\ce{E + S <=>ES -> E + P}$

$\frac{d[\ce{ES}]}{dt}=-\frac{d[\ce{E}]}{dt}=  k_1[\ce{E}][\ce{S}]-k_2[\ce{ES}] -k_3[\ce{ES}]  $

  is the slope of $[ES]$ equal to $k_2$ ?

no

  is the slope of $[E]$ equal to $k_1 + k_3$

no


How much butane is typically in deodorant (percentage)

Here's a link to an MSDS for a deodorant; it claims the butane range is 30-60% (they don't state if this is by volume or weight). It might be cheaper to just go to a hardware store and buy a small butane cylinder, like the ones used for outdoor cooking, then you know exactly how much butane you are using.

  For a potato gun, is it wiser to use butane or propane?

My first though was butane because it gives off a bit more heat per mole (or per unit volume) when it burns, than does propane.

$$\ce{C3H8 + 5 O2 -> 3 CO2 + 4 H2O + \pu{2219 kJ/mol}}$$
$$\ce{C4H10 + $\mathrm{6.5}$~ O2 -> 4 CO2 + 5 H2O + \pu{2878 kJ/mol}}$$

However, according to "Common Fuels for Combustion Spud Guns",

  The key value for comparing two fuels based on their heats of combustion is not the actual heats of combustion. Instead, the "Heat per mole Oxygen" should be used since the amount of energy in the combustion chamber is limited by the amount of oxygen present in the chamber. Fuel is added to match that amount of oxygen. As you can see from the table, there is relatively little difference between the various fuels based on their "Heat per mole Oxygen" values.

Both propane and butane give off $\pu{105 kcal/mol}$ of oxygen, so they should perform equivalently in a spud gun. The article goes on to note that, of common fuels, only hydrogen ($\pu{119 kcal/mol}$ of oxygen) and acetylene ($\pu{120 kcal/mol}$ of oxygen) produce significantly more energy under these conditions than other fuels.

Now going back to your first question, if you know the volume of your chamber, you can back-calculate the correct amount of air (oxygen) and butane.


How actually do we decide acidity and basicity order?

I'm not sure about the first order, but I'll try to explain the reason for the second one. The acidic strength increases as $$\ce{HF<HCl<HBr}$$

One reason is the bond strength between the hydrogen and the halogen atom. Since the bond between $\ce F$ and $\ce H$ is very strong, the molecule does not easily lose $\ce H^+$. On the other hand, as the size of the halogen atom increases, the overlapping of the orbitals between halogen and hydrogen becomes less efficient. Thus, $\ce {Cl}$ and $\ce {Br}$ can more easily lose $\ce {H^+}$ than $\ce {F}$. Hence, the acidic strength increases as size of halogen atom increases.

Second reason is the stability of formation of conjugate base. When an acid loses $\ce {H^+}$ ion, it forms what is called as conjugate base. Similarly, when a base loses $\ce {OH^-}$ ion, it forms a conjugate acid. In case of $\ce {HF}$, we have $$\ce{HF <=> H^+ +F- }$$
Here, $\ce {F^-}$ is the conjugate base. Now, due to small size of $\ce {F}$, it is difficult for $\ce {F-}$ to hold negative charge. There are strong repulsions.

Now, for $\ce {HBr}$, we have $$\ce{HBr <=> H^+ +Br^-}$$

$\ce {Br}$ has a greater atomic radius. It is also less electronegative than $\ce {F}$. Hence, the repulsions are less in $\ce {Br-}$. Thus, $\ce {Br^-}$ is quite stable. The more stable is the conjugate base, the more readily the acid loses $\ce {H^+}$ ion, and so, more is its acidic strength.

Check this link. Maybe you'll find it useful.


why is delocalization less effective when there is a separation of charge?

It takes energy to separate the charges, even when the difference in electronegativity is in the same direction as the proposed charge separation.  The energy that must he put in becomes part of the energy in the charge-separated contributing structure, making that contribution less stable and less likely to contribute to the molecular ground state.


Does potassium nitrate solution degrade?

As mentioned in the comments, a likely culprit is microbial contamination.  In a $\ce{KNO3}$ solution, you are providing plenty of water, a nitrogen source (nitrate), and even a high-energy, respiratory means for growth (nitrate respiration).

Living things also need (i) food, (ii) trace amounts of sulfur, (iii) and even smaller trace amounts of phosphorus.  But if you are soaking hemp in this solution, enough leachable S and P-containing impurities are probably extracted into the solution to provide these.  (Or also low levels of S and P may already be present in your $\ce{KNO3}$, or water.)  The food could be organic compounds leached from the hemp.  Or (less likely) the microbial contaminants might be capable of photosynthesis, in which case their food is atmospheric $\ce{CO2}%edit$.


Will bleach combined with any of these chemicals create poisonous gas?

I need to know if a combination of bleach and the chemicals below will create poisonous gas.
Sodium laureth sulfate

Sodium laureth sulfate is $\ce{CH3(CH2)11(OCH2CH2)_{\bf{3}}OSO3Na}$, where the boldface 3 is an approximation and can actually vary between 1 and 5 or so.  The key point is it contains no nitrogen.

Sodium citrate

Sodium citrate can refer to mono-, di-, or tri-sodium citrate, which have formulae like $\ce{C6H_{7-\it{n}}O7Na_{\it{n}}}$, where $n$ is 1, 2, or 3, respectively.  Again there is no nitrogen.

C12-14 Pareth-3

According to Google, "C12-14 Pareth-3" is a "polyethylene glycol ether of a mixture of synthetic C12-14 alcohols with an average of 3 moles of ethylene oxide."  The most representative compound would presumably be the C13 alcohol with exactly 3 moles of ethylene oxide.  That would have the formula $\ce{CH3(CH2)11CH2O-(CH2CH2-O)3H}$, or, if I did my math right, $\ce{C19H40O4}$.  Again the key point is there is no nitrogen.

Will it create chloramine, and will there be any visible gas?

None of the compounds have nitrogen, so chloramine cannot be created from bleach (chemically known as sodium hypochlorite) and these chemicals.
I think the likely products of bleach treatment would be non-volatile compounds such as inorganic sulfate plus organic compounds like tridecanoic acid, tridecanol, ethylene glycol, as well as oxidized derivatives of all of those organics, possibly including some small amount of $\ce{CO2}$ gas.  But $\ce{CO2}$ gas is not poisonous and not visible.


Identifying nucleophiles and electrophiles in an acid-base reaction

In my opinion, your statements are both right, this is the IUPAC definition [1]:

Lewis acid: A molecular entity (and the corresponding chemical species) that is an electron-pair acceptor and therefore able to react with a Lewis base to form a Lewis adduct, by sharing the electron pair furnished by the Lewis base. For example:
$$\ce{\underset{Lewis acid}{Me3B} + \underset{Lewis base}{:\!NH_\phantom{3}} -> \underset{Lewis adduct}{Me3\overset{-}{B}-\overset{+}{N}H3}}$$

You can see that the author says molecular entity (atom, molecule, ion, ion pair, radical, radical ion, complex, conformer etc., identifiable as a separately distinguishable entity [1]) and the corresponding chemical species (a boarded concept, that in this case I interpret as the combination of different molecular entity), so I think that the exegesis of this definition is that both the acetic acid and the hydron can be called a Lewis acid. And in fact is what usually all the chemists use.
Reference

IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). Online version (2019-) created by S. J. Chalk. ISBN 0-9678550-9-8. https://doi.org/10.1351/goldbook.


Better than gallium, could the alloy Galistan be used? It’s a mixture of Gallium, indium and tin. Other liquids like Bromine, Cesium and NaK might have some potential but may just react explosively.


What gas law is involved when your ears "pop" when you are on a plane high in the air?

The temperature and the volume of the inner ear are constant. When your ears pop during descent, air from the cabin goes into the ear, increasing the pressure. The law is the following:
$$n / P = const$$
You can derive this from the ideal gas law. It has no special name.


At first you said that $x = \pu{2.0449\times10^−5 mol}$ and then later said $x=3$. It is clearly wrong even without going into the maths!

That said, I believe you are confusing the stoichiometry of the compound with the number of moles of the compound produced. The $x$ in $\ce{(A3B2)_x}$ is a part of the chemical formula for that compound, and not the moles of $\ce{(A3B2)}$ produced.

For example, $\ce{AlCl3}$ usually exists as a dimer in gas phase, so we write it as $\ce{Al2Cl6}$ or simply $\ce{(AlCl3)2}$ for clarity. You should hence realize that the $x$ in $\ce{(A3B2)_x}$ represents the number of $\ce{A3B2}$ molecules that have polymerized together to form a single molecule. And the question states that $\pu{1.231\times10^19}$ of such molecules was produced.

With the conceptual clarity attained, we'll now solve the numerical. Note that the question says that all the moles of elemental A were obtained in the compound. This implies:

$$
\begin{align}
\text{moles of A in the compound} &= \text{original moles of elemental A}\\
\implies 3x\times(\text{moles of compound }\ce{A3B2})&=1.84\times10^{-4}
\end{align}
$$

Solving this, you'll get approximately $x=3$, which is the expected answer.


Absorption bands are very small: smaller than the band which a typical monochromator is capable of isolating.

If a continuum light source was employed, the absorbed band would be much smaller than the band which would be isolated by the monochromator and which would be read by the detector. This implies a low SNR (signal-to-noise ratio), and a very poor resolution.

Using a single element lamp (or lamps with combined elements but with distinct emission frequencies) overrides the problem: a monochromator is well capable of isolating the different frequencies, and absorption, which occurs in a narrow band of frequencies, occurs in the same narrow band of light from the source: this implies a much higher SNR, and a much better resolution.

To make an easy to conceive example, let's say that the effect of absorption by the sample is the effect of a star shining in the sky: using a continuum light source implies watching it through the midday sky. A HCL lamp would be a midnight sky, in the metaphor.

Note that, aside from what textbooks say, continuum souce AAS (HR-CS AAS) does exist, but it requires, as expected, a sophisticated monochromator (high resolution monochromator, with a resolution of a few pm!)


What's the order of oxidising strength of the oxyacids of Chlorine?

For many acids, the proton $\ce{H+}$ is the oxidizing agent. For example, in the oxidation of zinc with $\ce{HCl}$, the oxidizing agent is $\ce{H+}$.

$$\ce{Zn + 2H+ -> Zn^{2+} + H2}$$

However, for some acids, the conjugate base (anion) is also an oxidant (and usually a more powerful oxidant than $\ce{H+}$. The oxyacids of chlorine are all oxidizing agents through their anions, as is nitric acid and some others:

$$\ce{Zn + 2H+ + 2OCl- -> Zn^{2+} + H2O + 2Cl-}$$

In the above example, hypochlorite is the oxidizing agent. For the oxyacids of chlorine and others, the strength of the oxidizing agent is independent of the strength of the acid.


There is no other Ba(OH)x so there is no need to note a difference between several forms.

This happens with any metal ion that only attains precisely one positive charge, whether it is +1 or +10.

Lithium ions have +1 charge. Carbonate ions have -2 charge. Therefore to balance the charge in the formula we need 2 lithium to 1 carbonate.

This happens with all formulae, and it is a simple mathematical check to decide what the correct ratios should be.


What is the product of reaction between 3,3-dimethyl butanal and SeO2 followed by conc. OH- ions?

In the first step, Riley Oxidation (selenium dioxide-mediated oxidation of methylene groups adjacent to carbonyls) is taking place. You will get A as:

A on reaction with conc. $\ce{OH-}$ (i.e. in strongly alkaline condition) undergoes redox reaction (famous Cannizzaro reaction). And terminal aldehyde will be oxidized to form the metal salt of acid, ketone will be reduced to form the alcohol, note that intramolecular reaction is taking place, to finally form the product B.


When an inert gas is added to a system of constant volume, equilibrium is not effected?

Addition of an inert gas at constant volume:
When an inert gas is added to the system in equilibrium of constant volume though the total pressure exerted by the gases increases there is no shift in the equilibrium.

There is no shift in the equilibrium since there is no change in the molar concentrations(kept at constant volume) of each of the reacting components.

BUT

If an Inert Gas is added to a mixture kept at constant pressure there is a change in volume of overall mixture and hence the molar concentrations of the reacting components also decreases.
Hence according to Le Chatelier’s Principle the reaction shift in direction of lower molar concentration.

Basically - ((Inert/Non Reacting) Gases)) do not affect the original concentrations and hence do not affect the equilibrium in case of constant volume situations


I'm not sure of any conventional ways of electroplating selenium, but you could use a technique such as magnetron sputtering to spray a thin film of selenium onto another metal. Heck, you could even use glass or wood if you wanted to do it that way. And if you're going for thin, this technique can get you some of the thinnest, smoothest layers of metal out there. A major downside is that your source metal sample has to be pretty pure selenium. But still, I think It's completely worth it.

However, If you don't have access to one of these devices, They can get expensive. Most major colleges with significant science majors will have one of these or something similar to coat samples for electron microscopy, or other reasons. Be sure to check there before going off to buy your own. Also, there are legitimate YouTube videos out there that show how to make your one device for magnetron sputtering. Check out this one if you're interested.


How does temperature and other factors affect sigmatropic rearrangements?

The increased temperature is necessary to overcome the activation energy of the reaction. Many [3,3] sigmatropic rearrangements have very high activation energies. The products are favored at equilibrium because they are thermodynamically more stable. For example, in the first case, you are exchanging monosubstituted alkenes for disubstituted alkenes.


Magnetic couplings using DFT in a spin frustrated system

DFT is fundamentally the wrong tool for a system like this. When you say that the spin 'resonates' between several different states what you are explicitly saying is that the wavefunction is strongly statically correlated. DFT is a single reference method, so any experimental agreement you get from it in a case like this is at worst wrong and at best right for the wrong reasons (and the range of functionals will likely cover that whole spectrum).

If we knew the exact density functional DFT would, of course, be exact. Unfortunately, real functionals are parametrized for mostly organic compounds and the occasional single metal, but certainly nothing as correlated as a spin frustrated Cu$_3$ cluster. If you want a result which is believable, you need to use a real multireference method. (It's not going to be a quick calculation, but you only need a single point so it's not too bad)


Finding equilibrium constant from standard free energy

Your problem seems to be one of calculation using large numbers, so look at it this way:
$$
\ln K = 1161.93596
$$
So
$$
\log_{10}K = \ln K \log_{10}(e) = 504.65\\
\Rightarrow K = 10^{504.65} = 10^{0.65}\times 10^{504} = 4.5 \times 10^{504}
$$
without your calculator exploding.


Why do fluorescence intensities decline with replicate number?

That looks like a bleaching to me. The notion that bleaching only occurs with excesive light intesity on small populations of fluorophores is news to me. Might also that in my field we just use the bleaching term loosely. Best way to test is to take a long exposure of your sample and monitor the fluorescence intensity over time.
With our chemistry we usually encounter intensity loss that can be fitted with exponential decay function which would explain the bottoming out that you are describing. We often take multiple such "bleaching curves" and fit them to correct for intesity loss in our experiments that can take multiple hours.
In case you are only exposing your sample for short periods i would just make sure you measure each sample the same number of times that should rule out any error between samples. Thus reducing to only one measurment will not make it more correct.
Bleaching is poorly understood thus far. There is not a way to predict what excatly happens with a specific molecule.

Also how big is the effect for example (Start value)/(3rd measurment value) ?


How are the hydrogens attached to the nitrogen atom in the ammonium ion?

The nitrogen atom, on both ammonium ion ($\ce{NH4+}$) and ammonia ($\ce{NH3}$), is $\ce{sp^3}$ hybridized. That's because we have 4 sigma bonds on ammonium ion and 3 sigma bonds + 1 electron pair in ammonia. The most stable orbital configuration for this set up is a tetrahedral shape.

Nitrogen can have a $\ce{sp^2}$ configuration if it has 3 sigma bonds and a pi bond, like in $\ce{H2C=NH2}$. The $\ce{sp^2}$ orbitals are "used" in the sigma bonds and the remainder $2p_z$ orbital is "used" in the pi bond. This gives a triangular plane shape to the molecule.

Finally, nitrogen can also have a $\ce{sp}$ configuration if it has 1 sigma bond and 2 pi bonds, in $\ce{N2}$ for example. In this case, the $\ce{sp}$ orbital is "used" in the sigma bond and the remainder $2p_y$ and $2p_z$ orbitals are "used" in the pi bonds. This gives $\ce{N2}$ a linear geometry.

Molecular orbital theory and hybridization are a little hard to process (at least for me it was and it's still being), so take your time to read some more about it, but at least with those examples you should have an idea of what to expect from some molecules geometries.


Why not an Avogadro's principle for solid matter?

Avogadro's law, which can be written as $V \propto n$, where $V$ is the volume of the gas and $n$ is the amount of substance of the gas (measured in moles), can be thought of as just another manifestation of the ideal gas law rewritten as follows,
$$
  V = (RT/p) n \, .
$$
Consequently, strictly speaking, Avogadro's law is applicable only for a hypothetical ideal gas, but since real gases under usual conditions (such as standard temperature and pressure) are not far away in their properties from an ideal gas, Avogadro's law usually works well enough for them also. Substances in condensed phase on the other hand are milage away from an ideal gas in their properties, and thus, Avogadro's law is inapplicable for such cases.

To clarify in which sense Avogadro's law is inapplicable for substances in condensed phases, note that for solids and liquids, as well as for gases, the volume is proportional to the amount of substance. But Avogadro's law states not just the proportionality. It is also insists that the coefficient of the proportionality is the same for each and every gas, and this is the thing which can not be generalised to substances in condensed phases. For real gases, the ideal gas description which depends on the following assumptions (too name just a few):

molecules are almost point-like (effectively zero-sized) particles;
molecules do not exhibit any intermolecular forcers;

usually works as a good approximation. But for liquids and solids such a description a priori is physically horrible: molecules here are relatively bigger and more tightly packed so that their size and intermolecular interaction between them couldn't be ignored anymore.