I've been reading up on this, and will take a shot at the answer. It is going to be an interesting answer, and probably not what you expect. There has been an interesting conversation going on in the Journal of Chemical Education in response to this letter advocating the retirement of the localized hybrid orbitals because they "aren't real" and cannot be used to predict "all properties" of molecules. Rebuttals included comments that delocalized molecular orbitals also fail to predict "all properties". However, the final word in the discussion came from this letter, which reminds us what orbitals are.

Orbitals are not real. The electrons that populate orbitals are real. Orbitals are mathematical constructs created to describe the curious wave-like probabilistic quantum mechanical properties of electrons. When we talk about atomic orbitals, we generally mean the 1-electron hydrogenic atomic orbitals (HAOs), which are mathematically represented as a wavefunction ($\Psi$) and graphically represented by the probability density function ($\Psi ^ * \Psi$) at either 90% or 95% level. Despite valiant and massive efforts, wavefunctions for multi-electron atoms cannot be solved exactly. They are approximated as a parameterized HAOs. When we talk about the atomic orbitals of carbon, we usually mean the HAOs that closely approximate what we think the AOs of carbon should be.

Molecules and their orbitals are even more troublesome. The LCAO approach creates a bunch of nice 1-electron wavefunctions out of the set of HAOs. These methods give rise to both the canonical delocalized molecules orbitals (CMOs), and the localized valence bond orbitals (VBOs). As the authors of that letter point out:

  The term canonical implies that CMOs are not the only possible MOs, and there are many orbitals that generate an identical “chemical reality”.

The approximate polyelectron wavefunction of a molecule (at least in HF theory) is the Slater determinant of a series of 1-electron CMOs $\psi _i$:
$$  \psi _1 \overline{\psi} _1 \psi _2 \overline{\psi} _2 \psi _3 \overline{\psi} _3 \psi _4 \overline{\psi} _4 ... $$

An interesting property of Slater determinants is that unitary operations (like rotations) on the set of CMOs return a different set of MOs, which produce the same expectation values for the polyelectron wave function as original set:

$$ \psi _1 \overline{\psi} _1 \psi _2 \overline{\psi} _2 \psi _3 \overline{\psi} _3 \psi _4 \overline{\psi} _4 ...~~ \underrightarrow{unitary~~transformation} ~~\psi _a \overline{\psi} _a \psi _b \overline{\psi} _b \psi _c \overline{\psi} _c \psi _d \overline{\psi} _d ...$$

Mathematically, it seems to me, that a linear combination of MOs that are themselves linear combinations of AOs would still be a linear combination of AOs, just a different one than you started with. If that is true, then all you have is a different set of MOs than you started with, and according to HF theory, that should give you the same expectation values.

Thus a set of "LCMOs" are no more or less valid than the original set of LCAOs. They likely do not produce any additional insights into the structure of the molecule. However, the approach might be computationally more efficient when trying to determine the MOs of very large molecules and polymers.


Boiling point of: Halogenoalkanes vs alkanes vs alcohols

The cause of this is the hydrogen bond that forms in the liquid state between the molecules, and debye bonds that form between polar molecules.

A hydrogen bond can only occur between an electronegative atom and a hydrogen atom, himself linked to an electronegative atom  (F, O or N).

In order to evaporate the liquid, you have to provide enough energy to break the hydrogen bond. So the more bonds there are, the higher the boiling temperature. Soft debye bonds can also occur between two polar molecules and have to be broken too.

The alkanes can not participate in any bonds, thus have the lowest boiling temperature.
The halogen alkanes may not form hydrogen bonds, but they are polar, thanks to the very electronegative halogens. They form debye links, that are not very strong but sufficient to raise the boiling temperature.
The alcohols have -OH extremities, very likely to associate themselves in hydrogen bonds, that are stronger than debye links, and thus alcohols have the highest boiling temperature.


(Star represents $\ce{^14C}$ atom)
This figure represents three out of the five possible resonance structures. Of these five structures, four have a negative charge on a $\ce{^12C}$ atom whereas one has negative charge on $\ce{^14C}$ atom.
As the first premise, let's consider stability of the two kinds of resonating structures. The more stable the resonating compound, the more likely it is to be protonated.

We can compare stability using the premise of electronegativity (or rather charge density). According to this property, the more heavy an atom, the larger is its volume. If volume increases, the charge density decreases making it less likely to be able to stabilize the negative charge (becomes less electronegative). Due to this we can say that the resonance structure with $\ce{^14C}$ having the negative charge is less stable and so is less likely to be protonated.
Now, moving onto the second premise, the number of resonating structures containing the negative charge on $\ce{^14C}$ and $\ce{^12C}$ respectively.
If you notice only one resonating structure has a $\ce{^14C}$ with the negative charge, whereas there are $4$ resonating structures containing $\ce{^12C}$ atom with the negative charge. This would mean that probabilistically (considering stability to be equal) The resonance structure having a $\ce{^12C}$ with the negative charge would be more likely to be attacked.
Now, if we combine the two premises, we see that both stability and frequency of the resonance structure containing $\ce{^12C}$ with negative charge is greater its $\ce{^14C}$ counterpart. Concluding, we can say that the resonating structure with $\ce{^12C}$ stabilizing the negative charge would be more likely to be protonated.
Further Reading on how stability of isotopic compounds can be determined :  Inductive effect of hydrogen isotopes


Explaining temperature change in an exothermic system.

In thermodynamics, we like to divide any situation into two parts, the system and the surroundings. In general the system is the part of the universe that is being studied while the surroundings is everything else that interacts with the system. Between the boundary of the system and the surroundings, energy and heat can be transferred (in an open system).

So in an neutralisation reaction, the system is the actual acid and base while the surroundings is the solvent (the water). Since it is an exothermic reaction, this means that energy is released from the system to the surroundings. Since the water gains more kinetic energy from the system as the reaction occurs, its temperature will increase. Also this means that the system loses energy and therefore become colder. So to answer your question, the solvent gets hotter while the system gets colder.


What possibilities are there for refining aluminium on the moon?

My theory was to use electromagnetic separation, aka the Calutrons used in the Manhattan Project.  Fine powder feedstock (check), into an arc discharge to vaporize and ionize (lots of good designs for high volume ion sources out there), electrostatic acceleration, then electromagnets to mass separate.  Power with a solar panel array.  No moving parts (except for the powder feed).  No vacuum systems needed.  And you separate all of the elements at once. Now, it only works during sunlight, but during the lunar night you could do maintenance and collect the separated elements.


Redox reaction with hydrogen peroxide with potassium iodide

With all redox reactions, we should first look at all oxidation states of our reactants and products.

$$\begin{array}{ccc}
\text{Element} & \text{Oxidation state in reactants} & \text{Oxidation state in products} \\
\hline
\ce{H} & +1 & +1 \\
\ce{O} & -1\text{ in }\ce{H2O2}; -2\text{ in }\ce{H2SO4} & -2 \\
\ce{K} & +1 & +1 \\
\ce{S} & +6 & +6 \\
\ce{I} & -1 & 0
\end{array}$$

Now, we have to look at what is being oxidized and reduced:

$$\ce{H_2O_2 + KI + H_2SO_4 -> K_2SO_4 + I_2 + H_2O}$$

Hydrogen has the same oxidation state of 1+.

Potassium has the same oxidation state of 1+.

Iodine is oxidized from -1 to 0.

Oxygen is reduced from -1 to -2.

For redox reactions, we have to write out our half-reactions.

$$\ce{H2O2 -> H_O}$$

$$\ce{KI + H2SO4 -> K2SO4 + I2}$$

For our two half-reactions, we first have to make sure everything is balanced other than hydrogen and oxygen.

For our second half-reaction, we need two moles of potassium iodide in order to balance out the potassium and iodines.

$$\ce{H2O2 -> H2O}$$

$$\ce{2KI + H2SO_4 -> K2SO_4 + I2}$$

In order to balance oxygens, we add molecules of water.

$$\ce{H2O2 -> 2H2O}$$

$$\ce{2KI + H2SO4 -> K2SO4 + I2}$$

Since we are in acidic solution, we add protons to balance hydrogen.

$$\ce{H2O2 +2H+ -> H2O}$$

$$\ce{2KI + H2SO4 -> K2SO4 + I2 + 2H+}$$

We add electrons to balance charge:

$$\ce{H2O2 +2H+ -> 2H2O + 2e-}$$

$$\ce{2KI + H2SO4 + 2e- -> K2SO4 + I2 + 2H+}$$

Now we add 'em(the half-reactions) both up and cancel out necessary terms:

$$\ce{2KI + H2SO4 + H2O2 -> K2SO4 + I2 + 2H2O}$$

This differs from your original equation since I think you forgot to balance it.

Hydrogen peroxide here is reduced to water in acidic solution.


Isothermal compression at critical temperature

https://youtu.be/RmaJVxafesU?t=242

carbon dioxide has a critical point at 31 degrees celsius.


Polypropylene is hydrocarbon and does not adsorb polar and ionic compounds. So, several treatments with mild hydrochloric acid followed with several treatments with clean water should work. However, I wouldn't store drinking water there without a proven test. For example, leave water there for several weeks and then perform analysis of the water for metal ions. If all concentrations are within safe levels, it should be alright.


How to determine the order of nucleophilicity for given chemical species?

Within a group of nucleophiles with the same atom, the nucleophilicity decrease with decreasing basicity of nucleophile.

Decreasing basicity means the decreasing affinity of electron pair for a proton. The decreasing order of nucleophilicity is shown in the figure.

But the relationship between nucleophilicity and basicity can be reversed by steric effect. Less basic but steric unhindered nucleophile therefore have higher nucleophilicity than strong basic but strically hindered nucleophile.


There is not one single most common oxidation state across all halogens. While fluorine is the most electronegative element of which compounds are known — meaning that it can only exist in the $\mathrm{-I}$ and $\pm0$ oxidation states — the electronegativity continually reduces down the group to iodine. Iodide in the oxidation state $\mathrm{-I}$ is easily oxidised to $\ce{I2}$ or $\ce{IO3-}$ with the oxidation states $\pm0$ and $\mathrm{+V}$, respectively.

What is true, however, is that all halogens except fluorine show oxidation states from $\mathrm{-I}$ to $\mathrm{+VII}$ as can be predicted from their electronic structure $n\mathrm{s^2}\,n\mathrm{p^5}$.


Why was the prefix 'bi' used in compounds, such as for bicarb of soda?

According to the Online Etymology Dictionary, the word bicarbonate is derived as follows:

  1814, bi-carbonate of potash, apparently coined by English chemist William Hyde Wollaston (1766-1828), from bi- + carbonate.

The question then becomes, what is the meaning of the prefix bi- here?  We know it means two, but it also can mean double or twice.  If that is the case, then sodium bicarbonate is sodium double carbonate.

Consider a given quantity of acid, let's say $\ce{HCl}$, that we are going to neutralize with sodium carbonate, $\ce{Na2CO3}$, and an equal quantity of acid that we are going to neutralize with sodium bicarbonate, $\ce{NaHCO3}$.  We add the sodium carbonate to the acid, and as the neutralization reaction goes forward, it produces some quantity of carbon dioxide.  The same process is repeated for sodium bicarbonate, and the carbon dioxide is again collected.

$\ce{Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O}$

$\ce{NaHCO3 + HCl -> NaCl + CO2 + H2O}$

In the first reaction, one unit of sodium carbonate is able to neutralize two units of acid.  In the second reaction, one unit of sodium bicarbonate can only neutralize one unit of acid.  Another way of looking at this is that double the quantity of carbon dioxide is produced from the sodium bicarbonate neutralization.  That is, the compound is a double carbonate.

Similar explanations make sense for bisulfate and bisulfite.  The acid-base reactions that a sulfate or sulfite ion might undergo are already partially completed in the bi- versions of the compound, so it takes twice as much sulfate or sulfite to do the same job as the non-hydrogen form.

The hydrogen phosphates don't fit this pattern as nicely.

Additionally, if the cation in these salts is a relatively light weight one, then the required mass of the bicarbonate or bisulfate is substantially more than that of the non-hydrogen form of the compound, though it's not quite double.  53g of $\ce{Na2CO3}$ could neutralize one mole of $\ce{HCl}$, but it takes 84g of $\ce{NaHCO3}$ to do the same job.  This is closer to accurate for the more massive sulfur and phosphorus based oxyanions.


It's hard to know exactly what “level” you want address, as the Wikipedia entry seems a decent introduction to me. But in any case, here's what I consider “special” about quasicrystals:

From the fundamental point of view:

Though they are regular (they are formed by symmetry operations that allow them to repeat throughout space), they are not periodic (the symmetry operations are not purely translations). Although there's nothing inherently “wrong” or “impossible” about that from theoretical physics and chemistry, it goes against the classical classification of solids: crystalline or amorphous.
Though hundreds of quasicrystals have been synthesized, there is only naturally occurring quasicrystal phase (icosahedrite)!
All quasicrystals are metal alloys (most of them being binary or ternary intermetallic alloys): given that there are other types of crystalline solids, why aren't there non-metallic quasicrystals?

Regarding applications:

Quasicrystals have found, so far, few practical and industrial applications. In these applications, they are used as other alloys would, and do not have any “magic quasicrystal property”.
One specific feature they have, though, is that quasicrystals tend to be rather strong alloys and that they age well, due to the way they are synthesized


Convention in electrochemical cell notation

Good question. You can remember the cell notation with the mnemonic "reduction on the right."
So, on the right hand side, you should reduction process and on the left hand side you should show the oxidation process.
For example, for a Daniel cell, one can write
$$\ce{Zn|Zn^{2+}, SO_4^{2−}}||\ce{SO_4^{2−},Cu^{2+}}|\ce{Cu}$$
Now coming to your cell, $$\ce{Pt,Cl_2}|\ce{Cl^-(aq)}||\ce{Ag^+(aq)}|\ce{Ag}$$ is indeed written incorrectly because the writer seems to have the misconception that even in the cell notation, half cells must be written as a reduction.


Why are silicon analogs of alkenes or alkynes so difficult to make?

The typical Si–Si single bond length in a silane is around 2.33 Å.  This is much longer than a typical C–C single bond (~1.53 Å) and helps explain why silicon-silicon single bonds are so much weaker than carbon-carbon single bonds (bond dissociation energy: ~53 kcal/mol for silicon-silicon vs. ~83 kcal/mol for carbon-carbon single bonds).  When we try to form a silicon-silicon double or triple bond, the large separation between the two silicon atoms results in even less effective p-orbital overlap and therefore even weaker π-bonds.  The result is that while disilenes and disilynes are known (see the linked SE Chem posts), they are extremely reactive.  Only when bulky substituents are placed around the double or triple bonds (to provide some "steric protection" against further reaction) can the molecules be isolated and characterized.


The term "monomer" only has meaning in the context of discussing polymerization.

A polymer is a molecular chain where the links of the chain are repeated identical or a least similar units.  Examples where the units are idential would be polypropylene, polyvinylchloride and polysiloxane.  Example where the units are similar but not necessarily identical would be DNA and proteins.

A monomer is a molecule that can react to form a single unit of a polymer.

For example, propylene is a monomer of poylpropylene and amino acids are monomers of a protein.


Why does Boron only need 6 valence electrons unlike the standard 8?

The principle of full octet is that most reactivity patterns are described well if there is a driving force / energetic gain for full octets in main-group atoms.

However, the only facet of an atom that dictates anything about the number of valence electrons is its nuclear charge.  Boron has a charge of 5.  This is balanced by 5 electrons.  Two of them are core electrons and the remaining 3 are valence electrons.  The valence electrons may participate in bonding through sharing with other atoms, to make three bonds.  Three bonds = six electrons.

That is the simple answer. The way this question is posed indicates you are very new to the study of chemistry, and this next bit goes well beyond an introductory chemistry course.

For a fourth bond to form, another atom would have to donate two electrons to the bond.  This happens in many cases, such as when an amine reacts.  In other cases, an atom already bonded may donate electrons, such as in boron trichloride.  Finally, borane gas itself can engage in 3-center two electron bonding which is really weird, but provides a full octet when two boron atoms share the bonding electrons in a boron-hydrogen bond.

Don't distress, boron has many ways to get is full octet.


Kinetics Vs Thermodynamics : Apparent contradiction in the definition of Enthalpy in the two disciplines

The reason might be that while drawing the reaction energy profile, we forget to mention what energy we are mentioning in the Y-axis. The following conventions are generally used:

If reaction conditions are constant NVT, energy in Y-axis should represent internal energy.
If reaction conditions are constant NPT, energy in Y-axis should represent enthalpy.
If reaction conditions are constant $\mu$VT, energy in Y-axis should represent Helmholtz free energy.
If reaction conditions are constant $\mu$PT, energy in Y-axis should represent Gibbs free energy.
If you are looking at single molecule, the energy will be the total energy of the molecule (kinetic energy + potential energy).

In case of reactions, the last three are generally used.
Hence, the "Energy" Y-axis changes based on reaction conditions. In the question, it seems you might have got mixed up somehow.

Summary of the terms used:

N: No of molecules
$\mu$: Chemical potential
V: Volume
P: Pressure
T: Temperature


How can batteries reach equilibrium if the reactions are irreversible?

Your confusion probably stems from the definition of a reversible path between two states: an infinite sequence of steps along a continuum of equilibrium states, such that at each step the equivalence condition of the second law of thermodynamics, $\mathrm dS_\text{total}=0$, is satisfied.

However, there is no requirement that a path between two states must be reversible. Many textbook problems are based on comparing the entropy changes associated with reversible and irreversible paths. For the reversible paths, $\Delta S_\text{total}=0$, for the (spontaneous) irreversible ones, $\Delta S_\text{total}\gt0$.

Your textbook explains that the discharging process of a Daniell cell is considered irreversible. The usage of the term irreversible in the book is rather similar to the common usage, which means "impossible to bring back to its original state, for practical reasons.". The answer that Karl has given you is correct. Basically it is not possible to bring the cell to its original state. The chemical and physical integrity of the recharged electrodes and electrolyte are altered compared to the original structure. There are entropic costs associated with cycling the cell.


Dehydrobromination of 3-bromo-3-cyclopentylhexane and number of possible alkene products

The structures (i) and (ii) have two isomers each,namely E and Z isomers.

For the structure (iii) there is a plane of symmetry (marked in the diagram) so it is not a geometrical isomer.

Thus Number of isomers=2 (of structure i) +2 (of structure        ii) + 1(of structure iii)= 5 isomers


If we consider **Ideal gas equation **
and ignore the type of interaction which are present in the real gases then ;
$PV=nRT $
wherevP= pressure ; v=vol. ; n= moles; R = universal gas constant ; T= temp.

$PV/RT=constant $
so $n=constant$

so if$ n$ is constant then if we multiply it with avogadro's constant $N_A $ it would give you number of molecules .
Again that's when ideal gas .


Chemical equilibrium and Le Chatelier's principle

1) Increasing Oxygen pressure will increase the reverse rate because to reestablish equilibrium some of the oxygen should be consumed.
2) Out of system due to precipitation.
3) Out of system due to precipitation.
4) This is endothermic reaction. So increasing temperature will increase the forward rate to consume the added heat.


What is Sand?

Scope of this Answer

In order to give a good answer, we first have to decide on the type of sand we want to talk about, because its composition is highly dependent on that. Is it beach sand? If so, where from? All these questions should have been addressed in the question. I will limit the scope of this answer to pure silicium dioxide $\ce{SiO2}$ sand.

Is Sand a Molecule?

Sand is, strictly speaking, not a molecule. It's a crystal. As such it has properties such as the lattice energy. Maybe have a look at that link, it might clarify it all for you already.

So now we have talked about what we want to talk about and clarified the basics, so let's dive right in, shall we?

Why does $\ce{SiO2}$ not Dissolve in Water?

Basically this is an energy calculation. On one hand you have the crystal, with its lattice energy, and on the other hand you have the solvated $\ce{SiO2}$ units. A tentative reaction equation would look like this:
$$ \ce{Si_{n} O_{2n} <<=>[E_\text{lat}] SiO2 <=>[E_\text{solv}] SiO2(aq)} $$

The energies I've indicated are the lattice energy $E_\text{lat} = 12535~\mathrm{kJ\,mol^{-1}}$ (Source) and the solvation energy $E_\text{solv}$ (see figure below, where I've also included sodium chloride as a substance to compare with).

Data from here (PDF) ($\ce{SiO2}$) and here ($\ce{NaCl}$). It is obvious that the solvation of $\ce{SiO2}$ monomers is highly unfavored (if I got all the sign conventions right, that is...)

Basically, for ease of understanding, imagine that the $\ce{SiO2}$ unit in the middle has to decide on which side it wants to go. Either it forms a crystal with other units (to the left) or it gets solvated by (surrounded by molecules of) water. Which way will it go? We already know that it will take the route to the left. This means that the whole system can release more energy this way. That is why $\ce{SiO2}$ crystals are stable in water.

Or somewhat. Some theoretical investigations point towards the possibility that the surface of quartz crystals undergo chemisorption of water, thereby generating $\ce{SiOH}$ terminal groups, which would then generate more favorable interactions between the crystal surface and the surrounding water.

Coming back to (a part of) your question:

  Is it because of the big structure that sand makes?

You are kind of on the right track here. It is because of the lower energetical state as a crystal that these big structures (crystals) are formed, vs. the units just being dissolved (solvated by water).


Measuring volume change from change in mass

You can't compute the volume of a solid from the mass alone.

You can measure the specific gravity as a function of the temperature of the liquid. You'd need to know how the density of the liquid changes with temperature to convert the specific gravity data at each temperature to actual density value which could then be used to calculate the effective volume of the solid at each temperate.


Is the Ebullioscopic constant temperature-dependent?

Short answer

As long as external pressure is constant, it does not make sense to discuss $K_b$'s temperature dependence. If external pressure is changed, boiling point of solvent changes and hence also the ebullioscopic constant.

Longer version

Condition for equilibrium (skip if familiar)

Boiling is a reversible equilibrium process. By the second law of thermodynamics,

$$\mathrm{d}S_\text{universe}=0.$$

Assuming constant pressure,

$$\mathrm{d}S_\text{universe}= \mathrm{d}S-\frac{\mathrm{d}H}{T}.$$

Provided additional constant temperature,

$$-T\mathrm{d}S_\text{universe}= \mathrm{d}H - T\mathrm{d}S = \mathrm{d}G_\text{P,T}.$$

We also know that $G=G(P,T,n_1,\ldots, n_k).$ Thus

$$\mathrm{dG}=V\mathrm{d}P-S\mathrm{d}T + \sum_{i=1}^{k}\mu_i\mathrm{d}n_i,$$

and therefore

$$\mathrm{d}G_\text{P,T} = \sum_{i=1}^{k}\mu_i\mathrm{d}n_i=0.$$
Currently, we have two phases: a gas and a liquid. Boiling is a dynamic equilibrium, the differentials are non-zero, $\mathrm{d}n_1 = -\mathrm{d}n_2$, and so

$$\mathrm{d}G_\text{P,T} = 0 \iff G_\text{liquid} = G_\text{gas} \iff \mu_\text{liquid} = \mu_\text{gas}.$$

Clausius$-$Clapeyron equation (skip if familiar)

Suppose a closed system with initial equilibrium, constant temperature and pressure. We saw that

$$G_\text{liquid} = G_\text{gas}.$$

Now change the temperature. To achieve a new equilibrium in a closed system, the pressure would also need to change. At this new equilibrium, $G'_\text{liquid} = G'_\text{gas}$, which is why

$$\mathrm{d}G_\text{liquid} = \mathrm{d}G_\text{gas},$$

or equivalently,

$$V_\text{liquid}\mathrm{d}P - S_\text{liquid}\mathrm{d}T = V_\text{gas}\mathrm{d}P - S_\text{gas}\mathrm{d}T.$$

Now

$$\frac{\mathrm{d}P}{\mathrm{d}T}=\frac{S_\text{gas}-S_\text{liquid}}{V_\text{gas}-V_\text{liquid}}.$$

Note that the RHS depends again only on the initial and final conditions of phase equilibrium. Thus we are allowed to use $\Delta S = \Delta H/T$ to arrive at

$$\frac{\mathrm{d}P}{\mathrm{d}T}=\frac{\Delta H}{T(V_\text{gas}-V_\text{liquid})}.$$

Because $V_\text{gas} >> V_\text{liquid}$, and due to the ideal gas law

$$\frac{1}{P}\frac{\mathrm{d}P}{\mathrm{d}T}\approx\frac{\Delta H}{RT^2}.$$

Assuming $\Delta H$ is not a function of temperature,

$$\ln \frac{P_2}{P_1} \approx -\frac{\Delta H}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right).\tag{Clausius–Clapeyron}$$

NOTE. There is a small but important detail to keep in mind here. The assumption that $\Delta H$ is independent of temperature will only make our equations more inaccurate but it will not affect our answer. See the definition further below.

Boiling point elevation

We have

$$\ln \frac{P^\text{s}}{P^\circ} \approx -\frac{\Delta H_{vap}(T^\circ_\text{b})}{R}\left(\frac{1}{T^\circ_\text{b}} - \frac{1}{T^\text{s}_\text{b}}\right)\tag1$$

where

$P^\text{s}$ – vapor pressure of solution at initial boiling temperature $T^\circ_\text{b}$,
$P^\circ$ – saturation pressure of solvent, by definition of boiling, $P^\circ = P_\text{ext}$,
$\Delta H_{vap}(T^\circ_\text{b})$ – entalphy of vaporisation at initial boiling temperature $T^\circ_\text{b}$,
$T^\text{s}_\text{b}$ – new boiling temperature at some external pressure $P_\text{ext}$,
$T^\circ_\text{b}$ – boiling temperature of solute at the same external pressure $P_\text{ext}$.

From Raoult's law

$$P^\text{s} \approx P^\circ X_\text{solvent} = P^\circ(1 - X_\text{solute}),$$

and thus from $(1)$

$$\ln (1 - X_\text{solute}) \approx -\frac{\Delta H_{vap}(T^\circ_\text{b})}{R}\left(\frac{1}{T^\circ_\text{b}} - \frac{1}{T^\text{s}_\text{b}}\right).$$

From first-order Maclaurin series, and since boiling point elevation is usually small for our equations to be a good approximation,

$$\Delta T_\text{b} \approx \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}X_\text{solute}.$$

Granted molar mass $M$ is in units $\mathrm{g/mol}$, and molality $c_m$ has units of $\mathrm{mol\ /\ 1\ kg\ solvent}$, then

$$\Delta T_\text{b} \approx \underbrace{\frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}}_{K_\text{b}}c_m \equiv K_\text{b}c_m$$

or

$$K_\text{b} \equiv \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}. \tag2$$

It does not matter whether our previous equations are mere approximations, this is now a definition of the ebullioscopic constant. It is evident that it is in no manner dependent on external temperature. It is only dependent on the boiling temperature of a pure solvent at some well-defined external pressure $P_\text{ext}$. To change this temperature, we would have to change the external pressure, or pick a different solvent altogether.

If the external pressure is transformed, $T^\circ_\text{b}$ will not remain the same. There is no reason to believe that $(T^\circ_\text{b})^2$ at this new pressure $P'_\text{ext}$ will cancel with the change in vaporisation entalphy. And thus, the ebullioscopic constant is, in fact, not constant.

Conclusion

If external pressure remains the same, it makes no sense to talk about temperature dependence. This is because the ebullioscopic constant is defined as $$K_\text{b} \equiv \frac{R(T^\circ_\text{b})^2}{\Delta H_{vap}(T^\circ_\text{b})}\frac{M_{\text{solvent}}}{1000}$$
which is dependent only on the original boiling temperature of the pure solvent at some well-defined external pressure.
If external pressure is changed, the pure solvent's boiling point $T^\circ_\text{b}$ will change to a new value. There is no cause to believe that the square of the boiling temperature will cancel with the altered entalphy of vaporisation. Hence, ebullioscopic constant will change, and is, in this sense, dependent on temperature. The exact function will vary with the solvent at hand but is most likely approximated by polynomials with empirical coefficients if need be.


Aluminum oxide coating can relatively easily react with either strong base or acid, dissolving in them, exposing naked elemental aluminum to reaction medium. You can read more  here.
To be concise, there are many ways to make aluminum react: you can use a base, you can use an acid, and you can scratch a sample of aluminum and coat it with mercury. Mercury would form an alloy with Al, but that wouldn't affect its reactivity towards the solution of anything, it would just make it harder for oxygen to oxidize the fresh aluminum.
Answering you question: as far as I know, the nitrate test involves strong base, which serves as a reactant for aluminum oxide, dissolving it and letting aluminum to react further.


The idea of a chemical equivalent relies on a compound's properties with respect to the usual reactions that occur with a few select species that tend to behave in very predictable ways.  In this case, the number of grams of sodium that would react with 8 grams of elemental oxygen, 23g Na, are one equivalent of sodium for oxidation-reduction purposes.

$\ce{Na_2O_2}$ is sodium peroxide.  It contains oxygen (normally a very well behaved element, chemically speaking) behaving in an unusual way, bonding not only with sodium but also with another oxygen atom.  The oxygen can be thought of as part of a peroxide ion, $\ce{O2^2-}$.

If you know about oxidation and reduction, then you'll remember that oxygen usually has an oxidation number of -2, but when it is in a peroxide it has an oxidation number of -1.  Because this particular compound involves the -1 state, it takes twice as much oxygen to react with a given amount of sodium (which in both compounds has an oxidation state of +1.

Finally, and perhaps more important in answering the root source of your trouble here, a scientific law is a thing that describes the behavior of the universe in specific conditions.  Outside of those conditions the law does not apply.  Even inside those conditions laws are often incorrect to some degree (though still useful approximations.)  It's not unexpected that we would find some exceptions to this law.


Selectivity in aldol condensation of 7-oxo(6,6-²H₂)octanal

There are two factors to take into consideration.

The acidity of the deprontonated $\alpha-\ce{H}$.
The electrophilicity of the carbonyl being attacked.

If 1 dominates, then the reaction is under kinetic control. While if 2 dominates then the reaction is under thermodynamic control. (Detailed arguments are omitted here.) Since addition to carbonyls is generally highly reversible, the reaction should generally be under theromodynamic control and yield 1-(2-hydroxycyclohexyl)ethan-1-one:

There are existing experimental results on aldol condensation of 7-oxo-octanal.
Ghobril et al. (Eur. J. Org. Chem., 2008: 4104-4108) used TBD as base and obtained the thermodynamic product, while Pidathala et al. (Angew. Chem. Int. Ed., 42: 2785-2788) used proline and obtained the kinetic product. Ghobril et al. also pointed out that

When unsymmetrical ketoaldehydes such as 3 are submitted to intramolecular aldol reactions, the ketone usually acts as the CH-acidic component, whereas the aldehyde plays the role of the carbonyl-active counterpart to afford regioisomer 4. This regiochemical outcome is also favoured when thermodynamic conditions are used.


Concentrations appearing in the Ka of buffer solutions

Because Na+ is a spectator ion in this situation.

$$ \ce{Na+ + CH3COOH <=> Na+ + CH3OO- + H+}$$

In other words, Na+ is on both sides of the chemical equation, so [Na+] would appear in both the numerator and denominator of the equilibrium constant if Na+ is included.

$[\ce{Na+}][\ce{CH3COO-}][\ce{H+}]/[\ce{Na+}][\ce{CH3COOH}] = [\ce{CH3COO-}][\ce{H+}]/[\ce{CH3COOH}]$


What's the mechanism for the oxidation of primary alcohol to carboxylic acid with HNO3?

http://connection.ebscohost.com/c/articles/25511593/oxidation-nitric-acid-aliphatic-alcohols-diols-carboxylic-acids

http://pages.towson.edu/ladon/orgrxs/reagent/oxidizer.htm

It is a dangerous thing to do for the easy detonation of nitrate esters.  Activated and stabilized alcohols, e.g., benzoin to benzil, are reasonable.


Why can syn-periplanar E2 elimination happen in molecules like vinyl bromides but seemingly not in other molecules?

Elimination from vinyl halides occurs faster when the halide and proton have a trans relationship.  When the halide and proton are located on the same side of the double bond (cis), the elimination to form an alkyne is much slower.  The faster elimination from the trans arrangement is consistent with a preferred anti-periplanar arrangement in the E2 reaction.

The fact that elimination still occurs, albeit at a slower rate, when the groups being eliminated are arranged cis to one another suggests that a different mechanism is involved in the "cis" case.  Two possible mechanistic alternatives are

cis elimination from a higher energy (higher energy, therefore less preferred, slower) syn-periplanar conformation, or
an E1CB mechanism where the proton is removed in a first step and a carbanion is generated, and this is followed by halide ejection in a second step.

Experiments in analogous molecules suggest that the E1CB pathway is most often followed in "cis" elimination from vinyl halides.


Difference in application between Standard Type Bulb Pipette Filler and Universal Type?

They both basically have same functions with same operations. However, universal type is longer than standard type with an adapter for pipette head (see the picture below: standard type is to your left).

Note: The bulbs shown are Deutsch-Neumann, but they are similar to HHH.


That depends how you define "chemist."

Today we think of chemists as working at the molecular level. However Lavoisier didn't have any concrete evidence of an atom's existence. He worked with Sulfates (SO4-2), metal oxides, and other molecular groups in an empirical an algebraic fashion.

Throughout much of the 1800s there was enormous debate about the very existence of atoms. Much of the problem arose from confusion over molecular weights and stoichiometry. Physicists were the most stubborn opponents of atoms, which we still can't fully explain even today. Eventually Advodagro's number and accepted stoichiometry combined to form irrefutable evidence of molecules. Many scientists were completely convinced that atoms existed long before they were proven though. Dalton was the first to introduce stoichiometry so he would be a good candidate as well.


Is a liquid vitamin E supplement possible without additives?

Vitamin E is not soluble in water. It is only soluble in oils. So it is no use swallowing pure Vitamin E, or this vitamin with water. It will poorly be absorbed by the intestine. You must absorb it in an oily solution.


How can I have iron powder without a fast reaction with oxygen?

All is matter of how fine the powder is ( surface per mass ratio ) and what is the level of prior surface reaction with oxygen.
If the powder is very fine ( I mean very ) and the surface is "naked iron", it would burn in a small firework of sparks.
If the powder is course or the surface is already passivated by oxygen because of the powder history of preparation, it would not provide noticeable reaction.
I remember some old book of chemical experiments, preparing pyrophoric powder by pyrolysis of ferrous oxalate,
but it was not iron powder, but pyrophoric ferrous oxide $\ce{FeO}$:

Preparation
$\ce{FeO}$ can be prepared by the thermal decomposition of iron(II) oxalate.
$$\ce{FeC2O4  -> FeO  +  CO2  +  CO}$$
$\ce{FeO}$ is thermodynamically unstable below $\pu{575 °C}$, tending to disproportionate to metal and $\ce{Fe3O4}$:
$$\ce{4 FeO(s) -> Fe3O4(s) + Fe(s)}$$

With the following oxidation by air;
$$\ce{3 Fe(s) + 2 O2(g)-> Fe3O4(s)}$$
respectively;
$$\ce{6 FeO(s) + O2(g)-> 2 Fe3O4(s)}$$


Ethane and ethene: which is easier to burn? Which one burns hotter? Why?

Why and how do enthalpy and $E_\mathrm a$ change when two single bonds and two H atoms are replaced with a double bond?

Olefins are less stable than their corresponding alkane analogues.  For example, when ethene is hydrogenated to ethane roughly 33 kcal/mol of heat is given off because ethene has a higher energy content than ethane.

Why is this, why is a molecule with a double bond higher in energy than a molecule without a double bond?

There are two equivalent ways to describe the orbitals in ethene.  One is the "double bond" description where a pi and sigma bond exist together side by side.  We know that a pi bond is not as strong as a sigma bond due to poorer overlap with the pi bond compared to the the sigma bond.  It only takes something on the order of 60 kcal/mol to break a pi bond and produce cis-trans isomerization about the double bond, whereas it takes something like 90 kcal/mol to break a carbon-carbon single bond. This poorer overlap with the pi bond causes ethene to be higher in energy than ethane.  Alternatively one can view ethene as a two-membered ring, no pi bond, just 2 sigma bonds between the 2 carbons forming a two-membered ring.  It is easy to see that such a ring system would contain a significant amount of strain.

Whether you use the pi bond or two-membered ring approach to describe olefinic double bonds, both lead to the conclusion that olefins are destabilized (higher in energy content) due to poor overlap or ring strain (which in itself is really a reflection of poor overlap).  Because of this destabilization, alkenes will generally have lower activation energies and more heat will be given off when they react compared to alkanes.

Example:
Let's consider the bromination of ethene and ethane
$$\begin{align}
\ce{C2H4 + Br2 &-> BrH2C-CH2Br}\\
\ce{C2H6 + Br2 &-> 2 CH3Br}
\end{align}$$

The heats of formation of ethene, bromine and 1,2-dibromoethane are 52.5, 0 and -37.8 kJ/mol respectively; the reaction is exothermic by 90.3 kJ/mol.  On the other hand, in the case of ethane, the heats of formation are -84.7, 0 and (2x) -37.4 kJ/mol; this reaction is endothermic by 10.3 kJ/mol.  The olefinic reaction is significantly favored for the reasons discussed above.


MO Theory: Why do hydrogen and lithium bond, but hydrogen and helium don't?

You've asked seemingly simple questions that get into quirky considerations. A molecule of HeH would very unstable:

$\ce{2HeH -> 2He + H2}$

As Oscar points out HeH probably can be made. In all of deep space where molecules could go days before colliding with another particle there must be at least one molecule of HeH. So it can be "made" and it does "exist." But how would you detect it?

Typically when we are talking about "making" something we are talking about making it in a earth bound laboratory. You make enough to at least observe it somehow. For instance a lot of radioisotopes have half-lives less than a second which really doesn't give an experimenter time to do any really "chemistry" with them. So in any practical sense the HeH molecule doesn't exist.


Generating mol files from a molecular structure image?

The MOL format specification doesn't specify an origin, so presumably any origin can be chosen, as long as the coordinate values are within the allowed range. (The format allows a range of roughly -10,000 to +100,000, but I don't know if all programs support that large a range.) For example, in one of the examples in the specification the atom closest to the origin is over 17.6 units away. Where the origin is actually set would depend on the software used to create the file.

The coordinates for each atom specify the nucleus of the atom. I think you have a misconception about atomic size. The size of an atom, and thus the space it takes up, can be defined in different ways. The representation of an atom as a sphere (with the nucleus at the center) is a convention, only approximating reality, and the size of the sphere is arbitrary (but usually based on measurable properties). The nucleus is the only point in the atom that has a well-defined location, and thus the only part that meaningful coordinates can be defined for.

In MOL files, the presence of hydrogen atoms can be implied from valence and number of other bonds. (They can also be specified explicitly if needed.) In the long set of zeros at the end of each atom line (just after the element), the fourth zero specifies to add however many single bonds to hydrogen atoms are needed for the atom's valence. The sixth zero indicates to use the element's normal valence. For example, the carbon of the $\mathrm{CH_3}$ group has only one explicit bond, so three bonds to hydrogen are implied by carbon's normal valence of 4. How implicit hydrogens are handled depends on the software (e.g., this molecule has four implicit hydrogens that, per the usual convention for skeletal formulas, are not shown in your image).

The specification for the MOL format is available from Accelrys. The this answer is based on the specification and my knowledge of chemistry. Also, it only describes the MOL V2000 format, which is what you sample file is (based on the header); there is a newer V3000 format.


Acidic strength comparison between sulphuric acid and nitric acid

The important thing about acidity is the stability of the conjugate base, which is enhanced by resonance effect and inductive effect. The more stable the conjugate base, the easier deprotonation becomes, and thus the stronger the acid.

In a $\ce{NO3^-}$ ion, the negative charge can delocalise among three O atoms, but in a $\ce{HSO4^-}$ ion, in addition to delocalisation among three O atoms, there is an additional OH group, which further stabilises the anion by inductive effect. So $\ce{HSO4^-}$ is the more stable conjugate base, and $\ce{H2SO4}$ is a stronger acid than $\ce{HNO3}$.


Why does radiocarbon dating only work in nonliving creatures?

There are plenty of good sources online explaining the principle behind radiocarbon dating. For instance, the wikipedia explains:

  During its life, a plant or animal is in equilibrium with its surroundings by exchanging carbon either with the atmosphere, or through its diet. It will therefore have the same proportion of $\ce{^14C}$ as the atmosphere, or in the case of marine animals or plants, with the ocean. Once it dies, it ceases to acquire $\ce{^14C}$, but the $\ce{^14C}$ within its biological material at that time will continue to decay, and so the ratio of $\ce{^14C}$ to $\ce{^12C}$ in its remains will gradually decrease. Because $\ce{^14C}$ decays at a known rate, the proportion of radiocarbon can be used to determine how long it has been since a given sample stopped exchanging carbon – the older the sample, the less $\ce{^14C}$ will be left. [16]

Your question is somewhat subtle: why doesn't decay begin immediately once the isotopes have been incorporated into an organism? Can't measurements of part of an organism tell us when that particular organ or body part was formed?

The key limitation is explained in a document from a company (Beta Analytic) that provides biobased content / renewable carbon measurements:

  Although carbon-14 is radioactively decaying away in the body, it is constantly being replaced by new photosynthesis or the ingestion of food, leaving the amount relatively constant.

But if a body part no longer cycles carbon (is in effect dead), then radiocarbon dating should be useful. One application, for instance, is in dendrochronology, the analysis of tree growth rings to assess age and location:

  Dendrochronology is useful for determining the precise age of samples, especially those that are too recent for radiocarbon dating, which always produces a range rather than an exact date, to be very accurate. However, for a precise date of the death of the tree a full sample to the edge is needed, which most trimmed timber will not provide. It also gives data on the timing of events and rates of change in the environment (most prominently climate) and also in wood found in archaeology or works of art and architecture, such as old panel paintings. It is also used as a check in radiocarbon dating to calibrate radiocarbon ages.1

Combining tree samples of known age (thanks to dendrochronology) and other sources with radiocarbon ratios provides a calibration method.


IS DMSO-o a σ-donor, a π-donor, or both?

The first thing to remember is that a σ-symmetric bond is always better than a π-symmetric one. Orbitals that can bond with each other are typically oriented in one direction or another (except for s-orbitals, but those can only participate in σ bonds, anyway). Approaching these orbitals from the direction ‘their lobes point towards’ and in the direction ‘your own lobe is pointing towards’ results in a much better overlap than if you attempted to create a π type interaction. Thus, typically any σ interaction will be preferable to any π interaction between a given set of two bonding partners.

To be perfectly honest, I am not aware of any ligand which does not initially bond as a σ donor.

The question remains what will happen if we add π effects into the picture. π effects can either be of the π-Lewis acidic type (e.g. $\ce{CO}$) or of the π-Lewis basic type (e.g. $\ce{F-}$). In the case of DMSO, the LUMO has much greater contributions on sulfur than on oxygen — in line with the general expectation that the LUMO is centred on electropositive partners while HOMOs centre more around the electronegative partners.

Also in line with what oxygen does in other ligands, I would expect DMSO-κO to be a π base, due to the probability of another populated orbital having significant contributions on oxygen. Contrarily, the picture of the LUMO shows how well sulfur in DMSO-κS can act as a Lewis acid.

For Lewis acidic bonding, i.e. π backbonding (also written as M$\leftrightarrows$L), it is beneficial if there is significant charge density on the central metal, i.e. if the metal has a somewhat Lewis basic character. Often, this will be enhanced by the presence of electrons in d-orbitals of the $\mathrm{t_{2g}}$ type.

For Lewis basic ligand bonding, i.e. π forward bonding (also written as L$\rightrightarrows$M), it is beneficial for the metal to be in a Lewis acidic, electron-poor state — often with empty d-orbitals.

In the DMSO-κS context, remember that sulfur carries a partial positive charge and is thus already electron deficient somewhat ($\pm 0$ oxidation state, but it almost has the same electronegativity as carbon, so it is not far away from $\mathrm{+II}$). Therefore, the σ forward bond will be weak per se. Only if sufficient stabilisation through backbonding can be supplied will a DMSO-κS complex be sufficiently stable.


Why does SiH4 have unusually large bond angles?

In hydrides that contain two or three hydrogens connected to a central atom such as $\ce{PH3}$ or $\ce{H2Se}$, the number of bonding partners does not exceed the number of available p orbitals on the central atom. In an extreme application of Bent’s rule, the most ideal bonding situation would be to have pure p-type orbitals form the required bonds and at least one lone pair reside entirely in an s-type orbital. This allows stabilisation of the s lone pair (as an s orbital’s energy is lower than a p orbital’s one) and better bonding (as p orbitals are directional and thus offer greater overlap with hydrogen; furthermore, they are higher in energy which better matches hydrogen’s orbital energy and again allows for greater bonding energy).

This ideal situation already falls apart in the second period where the bond angles of ammonia and water are unusually large; this is because the central atoms are too small to accomodate two or three bonding partners at $90^\circ$ angles so the bond angles are increased in a delicate balance until the lowest overall energy is obtained. The results are hybrid orbitals with a nonzero s component.

In $\ce{AH4}$ molecules, the situation is different and worse yet again. There are not enough p orbitals to accomodate four bonds so the s orbital has to help. Symmetrically, the optimal arrangement is a tetrahedron and this just happens to almost perfectly match an $\mathrm{sp^3}$ arrangement. As silane (and methane, germane, etc.) has four identical substituents, there is no energy to be gained by reducing symmetry so the ultimate geometry is that of an ideal tetrahedron whose bond angles are $109.5^\circ$.


Crystallisation of supersaturated sodium acetate is a good one, as seen in my answer here.

Another is the reaction of water with calcium oxide, which is commonly seen in self-heating coffee and soup cans. This reaction can get very hot, though.


Is purple paint more susceptible to UV damage than other colors?

I would also agree with you and the comment from ssavec. The only point that I would make is that it shouldn't be because of the colour of the pigment only the stability of the particular dye it contains (almost certainly organic). In fact the colour of the paint should make it less susceptible to fading as it is reflecting colours at the violet end of the spectrum (subject to your definition of purple) rather than absorbing them.


Can we synthesize alcohols by the photolysis of hydrogen peroxide?

I do not believe so.

The mechanism of decomposition of $\ce{H2O2}$ is:

$\ce{H2O2(g) -> 2HO.(g)}$

$\ce{H2O2(g) + .OH(g) -> H2O(g) + HOO.(g)}$

$\ce{HOO.(g) + .OH(g) -> H2O(g) + O2(g)}$

$\ce{.OH(g), HOO.(g)}$ are both radical compounds.

If you take a look at the first step in the mechanism, it is the rate limiting step and it is the same as your first step in the mechanism, the photolysis step.  Therefore if $\ce{H2O2}$ undergoes partial photolysis all that will happen is that the sample will undergo decomposition quicker.

Additionally, your mechanism of oxidation has been proposed before.  You can take a look at this article, in the section Mechanism of Oxidation by Peroxides: Radical or not Radical, That is the Question, in the end of subsection 4.1 and 4.2 they refer to the mechanism you proposed.  They end up proving that it proceeds through a concerted mechanism that requires the presence of a suitable metal catalyst.  It involves oxidative insertion or some type of $\ce{C-H}$ insertion by the metal followed by oxidation of the product of that reaction.

But say the generation of hydroxyl radicals does work out, the corresponding alcohol may not be the only product that forms.  Hydroperoxides have a tendency of forming as well.  This is because hydrogen peroxide tends to decompose into oxygen and water, the oxygen can interact with the alkyl radical to form hydroperoxides.  Along with the fact that the alcohol formed will tend to undergo oxidation in the presence of hydrogen peroxide and hydroxyl radicals.

Also if you are still going to try the reaction, the bond dissociation energy of the $\ce{O-O}$ single bond is around 146 kJ per mole. So the energy per molecule is $\ce{2.433\times10^{-19} J}$.  So the maximum wavelength of light to ionize this is $\ce{8.21\times10^{-7} m}$.  Your calculation is a little off, it is in the near infrared range (you may have forgot to convert from kilojoules to joules or something like that).  This is why hydrogen peroxide is stored in dark bottles and told to not be put in the presence of light.


At $\pu{20 °C}$ dynamic viscosity of liquid chlorine is about $\pu{0.35 mPa·s}$ (see Figure 4 from [1, p. 7]), whereas dynamic viscosity of water is about $\pu{1 mPa·s}$, e.g. water is nearly 3 times more viscous at room temperature than liquid chlorine, which should not appear that thick.

Figure 4. Viscosities of chlorine gas and liquid.

Note that your perception of viscosity and "thickness" may be distorted due to capillary effect if your chlorine sample is sealed in a tiny ampule.
References

Chlorine: Principles and Industrial Practice, 1st ed.; Schmittinger, P., Ed.; Wiley-VCH: Weinheim; New York, 2000. ISBN 978-3-527-29851-8.


pH level of fabric with regard to inhibiting fungal growth

Hypochlorite bleach is a potent biocide, so throughly washing clothes with it should kill all fungi. The pH part is true, bleach is alkaline, but so are all common detergents. But this should not even matter, if the clothes are properly rinsed with clean water, that should remove almost all of the bleach and restore the pH to close to neutral.
What I think happened, is that his clothes were not rinsed well enough, so traces of the bleach and the detergents stayed in the clothes, causing or exacerbating skin problems. Some people have skin especially sensitive to bleach and allergic reactions to detergents can happen to.
So I would recommend washing with bleach, but making sure the clothes are well rinsed. You may need to set your washing machine on an extra rinse cycle to get the desired effect. Using lemon juice should not be necessary.


What is the role of ammonium chloride in the workup of a Grignard reaction?

Ammonium chloride ($\ce{NH4Cl}$) is the work-up reagent that quenches the magnesium alkoxide product of the Grignard addition. It is the reagent of choice as it is a proton source without being acidic; acidic conditions could result in protonation of the tertiary alcohol product and elimination to the alkene. It also ensures that all inorganic salts of Mg will extract into the aqueous phase.


Is it true that "Voigt Circuit" would be able to be fitted to various electrochemical impedance spectra?

It is known that an electrochemical impedance spectra can be modelized with series-parallel combination of resistors and capacitors elements. What is more, for a given impedance spectra they are an infinity of different combinations of R-C elements which accurately model the same complex impedance.

Similar observations are commonly done in the field of imperfect dielectrics (non-Debye) for the complex permittivity spectra.

For examples: pages 10-11 , page 16 and pp.27-32 in the paper  https://fr.scribd.com/doc/71923015/The-Phasance-Concept . Each example of complex impedance is modeled with several different networks of R-C elements.

On theoretical viewpoint, all this is related to the energy efficiency of the phenomena in the experimental system, that is the presence of dissipative components (R) and storage components (C) distributed on various manner.  Even if the energy storage and energy dissipation are respectively on different physical nature in different experimental devices, possibly in some cases the complex impedance is likely to be similar due to a similar global energy efficiency behaviour. So, an infinity of different R-C models are likely to be derived.

This can be observed not only in experiments but also in purely theoretical studies for very simplified geometrical configuration : https://fr.scribd.com/doc/23155389/Theoretical-Impedance-of-Capacitive-Electrodes

In addition. To go beyond the question raised, the following quotation is extract from the first paper quoted above :

It has long been recognized that many impedance loci result in well-defined semi-circles with center depressed below the real axis. Linear loci are also commonly observed associated to rough electrode surface, porous electrode, diffusion, etc. "This features mean only that the system includes some part(s) transferring energy with constant efficiency in a range of frequencies. Any affirmation of the nature and mechanisms of the phenomena involved would be pure hypothesis if only based on impedance analysis."


Soda contains $\ce{CO2}$. Over time the gaseous $\ce{CO2}$ comes to the surface. The ice cream increases the rate of this  $\ce{CO2}$ from the soda causing increased fizzing. This increased rate is a result of the solvation of ice cream particles in the liquid. The particles serve as nucleation sites for the growth of gas bubbles. The bubbles get large quiet fast thus you observe fizzing.

After a bit of experimentation I found the following:

Effect of adding milk — There is a bit of extra bubbling and a curdy precipitate is formed. (Couldn't dare to drink it.)
Effect of adding salt (yes ice cream does contain salt) — Lots of bubbling around the salt. After the salt settles to the bottom, bubbles collect around it become big and then come to the surface.

Only thing left to try is sugar and vanilla essence but I ran out of soda.

So we come to know that the fizzing effect may be due to milk or salt or a combo of both.


It is not entirely inconceivable that a Galvanic cell could be created with an acidic anode and basic cathode.  Having both half-reactions occurring in the same beaker, however, is unlikely.  That has implications when determining the thermodynamics (cell potential) of the reaction but not in balancing the half reactions.  Given:
$$\begin{align}
\ce{2 H2O_{(l)} & -> O2 + 4 H+_{(aq)} + 4e^-} \tag{1}\\
\ce{ 2 H2O_{(l)} + 2 e- & -> H2_{(g)} + 2 OH-_{(aq)}} \tag{2}
\end{align}
$$

Multiply equation 2 by 2 and add:

$$\begin{align}
\ce{2 H2O_{(l)} & -> O2 + 4 H+_{(aq)} + 4e^-} \tag{3}\\
\ce{4 H2O_{(l)} + 4 e- & -> 2 H2_{(g)} + 4 OH-_{(aq)}} \tag{4} \\
\ce{6 H2O_{(l)} + 4 e- & -> O2 + 2 H2 + 4 H+ + 4 OH- + 4 e^-} \tag{5}
\end{align}
$$

The key here is that the  $\ce{4H+ + 4OH-}$ showing up on the same side equation 5 will yield $\ce{4H2O}$.  Combining acid and base (to make $\ce{4 H2O}$ on the product side) and canceling the $\ce{4e-}$ and $\ce{4 H2O}$ that occur on both sides of the chemical equation gives the desired equation.


When is the product a compound and when is it 2 ions in these reactions?

While FeCl3 is a possible product, the reaction in A is aqueous. Wikipedia shows that is readily dissolves in water (92 g/100 mL @ 20C). The products exist as ions, and are written as such.

I would suspect you're not supposed to have solubility constants memorized though, so in general if it is in water they will exist as ions.

B is a gaseous reaction and the product is a compound, not an ion set like A.


Why are pi bonds only formed when sigma bonds are formed?

A $\pi$ bond has a plane of symmetry along the bond axis. It cannot be formed by s-orbitals; it needs at least p-orbitals to be created. $90\,\%$ of all bonds described some time or another are somehow involving carbon, nitrogen or oxygen. (In fact, I probably underestimated). But these elements can only use p-orbitals to create $\pi$ bonds. To do that, one needs a p-orbital that is ortohogonal to the bond axis. So you run into the problem that you have an orbital pointing in one direction, but want to bond into another direction — hardly optimal, especially since there likely is already another orbital pointing in the direction you need to give a $\sigma$ bond.

Transition metals can use d-orbitals for $\pi$ bonding. They can actually point towards the atom they want to bond with so there is a greater chance of using them due to higher overlap. However, there will usually also be a different orbital pointing directly in the bonding direction which again will bond earlier and would give a $\sigma$ bond.


How to get a precise answer to a textbook precipitation problem?

Ok, let's do the "complete" and "exact" problem.

First, I'm just going to ignore the $\ce{Mn^{2+}}$. Using a first pass with the simplifying assumptions would be enough to show that MnS will not ppt.

Second the problem statement notes that:

  This solution is then saturated with H2S. The solubility of $\ce{H2S}$ is given as 0.1 mol/L and assumed independent of the presence of other substances in solution.

Chemically $\ce{H2S}$ would be bubbled through the solution to saturate it. While the $\ce{H2S}$ was bubbling $\ce{CuS}$ would be ppt'ing. So when the final concentration of $\ce{H2S}$ is 0.1 molar the ppt of $\ce{CuS}$ is complete.

Let:

The final $\ce{[H2S]_f = 0.1\bar0}$ molar
This means that the final concentration of all the sulfur species ($\ce{[H2S]_f, [HS-]_f,}$ and $\ce{[S^{2-}]_f}$) will be a bit more than $0.1\bar0$ molar.
The initial and final $\ce{[HClO4]} = 3.\bar 0\times10^{-3}$ molar
From $K_{sp}$ for CuS we know that:
$$K_{sp} = [Cu^{2+}]_f[S^{2-}]_f$$ or $$\ce{[S^{2-}]_f} = \dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}}$$
For $\ce{H2S}$ we can multiply $K_{a1}\times K_{a2}$ to get an overall ionization product $$K_t = K_{a1}\times K_{a2} = 1.2\bar0\times10^{-20} = \dfrac{\ce{[H+]_f[HS-]_f}}{\ce{[H2S]_f}}\times\dfrac{\ce{[H+]_f[S^{2-}]_f}}{\ce{[HS-]_f}} = \dfrac{\ce{[H+]_f^2[S^{2-}]_f}}{\ce{[H2S]_f}}$$
also that
$$\ce{[H+]_f} = \sqrt{\dfrac{K_t\ce{[H2S]_f}}{\ce{[S^{2-}]_f}}} = \sqrt{\dfrac{K_t\ce{[H2S]_f[Cu^{2+}]_f}}{K_{sp}}}$$
and
$$\ce{[HS-]_f} = \dfrac{K_{a1}\ce{[H2S]_f}}{\ce{[H+]_f}} = K_{a1}\ce{[H2S]_f}\sqrt{\dfrac{\ce{[S^{2-}]_f}}{K_t\ce{[H2S]_f}}}=K_{a1}\sqrt{\dfrac{\ce{[H2S]_f[S^{2-}]_f}}{K_t}}= K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}}$$
$\ce{[Cu^{2+}]_f}$ be the final concentration of copper.
The initial $\ce{[Cu^{2+}]_i}=2\times10^{-4}$ molar which we will assume to be $2.\bar0\times10^{-4}$ molar. Thus the amount of copper that is removed from solution is $2.\bar0\times10^{-4} - \ce{[Cu^{2+}]_f}$
For the final $\ce{[H+]}$ we know that
$$\ce{[H+]_f = [HClO4] + [HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - \ce{[Cu^{2+}]_f}) - (10^{-7} -[OH-])}$$ or $$ 0 = \ce{[HClO4] + [HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - [Cu^{2+}]_f) - (10^{-7} -\dfrac{K_w}{\ce{[H+]}}) - [H+]}$$
substituting for $\ce{[HClO4]}$
$$0 = 3.0000\times10^{-3} + \ce{[HS^-]_f + 2[S^{2-}]_f + 2(2.\bar0\times10^{-4} - [Cu^{2+}]_f) - (10^{-7} -\dfrac{K_w}{\ce{[H+]}}) - [H+]}$$
and collecting terms
$$0 = 3.3999\times10^{-3} + \ce{[HS^-]_f + 2[S^{2-}]_f - 2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$
substituting for $\ce{[S^{2-}]_f}$
$$0 = 3.3999\times10^{-3} + \ce{[HS^-]_f} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$
substituting for $\ce{[HS^-]_f}$
$$0 = 3.3999\times10^{-3} + K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +\dfrac{K_w}{\ce{[H+]}} - [H+]}$$
and finally substituting for $\ce{[H+]_f}$
$$0 = 3.3999\times10^{-3} + K_{a1}\sqrt{\dfrac{\ce{[H2S]_fK_{sp}}}{K_t\ce{[Cu^{2+}]_f}}} + 2\dfrac{K_{sp}}{\ce{[Cu^{2+}]_f}} - \ce{2[Cu^{2+}]_f +K_w\sqrt{\dfrac{K_{sp}}{K_t\ce{[H2S]_f[Cu^{2+}]_f}}} - \sqrt{\dfrac{K_t\ce{[H2S]_f[Cu^{2+}]_f}}{K_{sp}}}}$$

Since $\ce{[H2S]_f = 0.10}$ per problem statement, the only variable left is $\ce{[Cu^{2+}]_f}$. I'll leave it to the reader to solve this mess for $\ce{[Cu^{2+}]_f}$, but to 2 significant figures the answer should be $7.7\cdot10^{-21}$.

The whole point in using significant figures to simply the problem is not to get the "wrong" answer, but to avoid a computational mess that doesn't yield any better answer.


How to go from, Iron(II) Sulfate (anhydrous, FeSO4) to, Ferrous Sulfate Heptahydrate (FeSO4.7H2O)

According to this Wikipedia article:

  Iron(II) sulfate (British English: iron(II) sulphate) or ferrous
  sulfate denotes a range of salts with the formula $\ce{FeSO4·xH2O}$.
  These compounds exist most commonly as the heptahydrate (x = 7) but
  are known for several values of x.

While the anhydrous form appears as white crystals, the article above continues:

  Iron(II) sulfate can be found in various states of hydration, and
  several of these forms exist in nature.
  $\ce{FeSO4·H2O}$ (mineral: Szomolnokite, relatively rare)
  $\ce{FeSO4·4H2O}$ (mineral: Rozenite, white, relatively common, may be  dehydratation product of melanterite)
  $\ce{FeSO4·5H2O}$ (mineral: Siderotil, relatively rare)
  $\ce{FeSO4·6H2O}$ (mineral: Ferrohexahydrite, relatively rare)
  $\ce{FeSO4·7H2O}$ (mineral: Melanterite, blue-green, relatively common)

This all suggests that simply letting the anhydrous form sit in air overnight is not a reasonable procedure for the quantitative production of the heptahydrate form.

One qualitative hint that you could get regarding the degree of hydration of your old anhydrous material is the color.  While both the anhydrous and quadrahydrate are white, the heptahydrate is a bluish green color.

You could also measure the melting point of the old material; it should be $\mathrm{60-64^oC}$ if it has hydrated all the way to the heptahydrate.

If you really want to try to convert the old anhydrous material to the heptahydrate, you could expose it to a humidified environment overnight.  Then observer any color changes and measure the melting point. You could even weigh a sample of it then heat to ~$\mathrm{110^oC}$ for a few hours then re-weigh it and calculate the degree of hydration.

That's all a bit crude, but you could try it. I recommend just purchasing some of the heptahydrate compound if possible.


Speed at which methanol dissolve in water?

As already mentioned by @Zhe, the rate of the mixing of methanol in water (or the other way around) depends (among other parameters) on the rate of addition, the geometry of the vessel, if / how the solutions are mixed by a stirrer.  In addition, this process diffusion is dependent on temperature and gradients of concentration.  One approach to quantify diffusion is based on the differential Fick's laws by determination of the diffusion coefficient.

As difficult it may appear, the diffusion of methanol in water was determined experimentally, as here and here, for example.  If you do not have access to a university library, NIH / HHS made a more recent reference available here, and a more general value is given here


Where can I find reliable data for melting points of organic compounds?

You are absolutely right about suspecting the reliability of sources, based on their value spread. I think, I found what's going on with those internet (I assumed) sources. Yet, I know everybody agree with me that CRC handbook of chemistry and physics is The Bible for Physical Constants of Organic Compounds. Here is what I found on 2005 Internet Version of CRC handbook of chemistry and physics:

3-Methylpentane (CAS #: 96-14-0): physical form: liquid; Melting Point: $\pu{-162.90 °C}$; Boiling Point: $\pu{63.27 °C}$; and Density: $\pu{0.6598 g/cm^3}$ (pp. 3-398 of Ref.1)

Perfluoro-3-methylpentane (CAS #: 865-71-4): physical form: liquid; Melting Point: $\pu{-115 °C}$; Boiling Point: $\pu{58.4 °C}$; and Density: not listed (pp. 3-458 of Ref.1)

I think some reference sources may have mistaken with values of perfluoro-3-methylpentane as those of 3-methylpentane.

Nonetheless, evem PubChem listed two other values than $\pu{-162.90 °C}$ for melting point of 3-methylpentane (Ref.2): $\pu{-162.90 °C}$ from Lide, D.R., G.W.A. Milne (eds.). Handbook of Data on Organic Compounds. Volume I; 3rd ed.; CRC Press, Inc.: Boca Raton, FL. 1994., p. V4: 3865 (most trusted value since it is PEER REVIEWED - https://toxnet.nlm.nih.gov/cgi-bin/sis/search2/r?dbs+hsdb:@term+@rn+@rel+96-14-0); $\pu{-118 °C}$ from http://www.ilo.org/dyn/icsc/showcard.display?p_version=2&p_card_id=1263 (International Chemical Safety Cards (ICSC) are data sheets intended to provide essential safety and health information on chemicals in a clear and concise way); and third value of $\pu{-245 to -148 °F}$, basically, $\pu{-153.9 to -100 °C}$, from  OSHA Occupational Chemical DB (http://www.osha.gov/chemicaldata/chemResult.html?RecNo=764). I can only speculate that they all listed wrong values because they got hits for some other substituted 3-methylpentanes with search keyword of 3-methylpentane. For example, CRC Handbook of Chemistry and Physics listed $\pu{118.27 °C}$ for boiling point of 3-ethyl-3-methylpentane (CAS #: 1067-08-9).

Yet, only CRC Handbook of Chemistry and Physics consistently listed $\pu{-162.90 °C}$ as the melting point of 3-methylpentane since 1994. Thus, I'd take that value at any time.

References:

"Section 3: Physical Constants of Organic Compounds", in CRC Handbook of Chemistry and Physics, Internet Version 2005, David R. Lide, ed., http://www.hbcpnetbase.com, CRC Press, Boca Raton, FL, 2005 (pp. 3-398).
https://pubchem.ncbi.nlm.nih.gov/compound/3-methylpentane#section=Melting-Point


What is the exact nomenclature of ionization?

The reference for chemical terminology is the IUPAC Gold Book, or IUPAC Compendium of Chemical Terminology. The entry for ionization is as follows:

  The generation of one or more ions. It may occur, e.g. by loss of an electron from a neutral molecular entity, by the unimolecular heterolysis of such an entity into two or more ions, or by a heterolytic substitution reaction involving neutral molecules […] The loss of an electron from a singly, doubly, etc. charged cation is called second, third, etc. ionization. This terminology is used especially in mass spectroscopy.

It includes examples that you might find instructive. It confirms your intuition that the term has a broad understanding, covering a wide variety of reactions.


For the last question, I believe the latter would be true because if
  the anions produced at the negative electrode react with the cations
  that are attracted to the electrode, the cations that are attracted to
  the electrode would not be available to accept electrons and hence the
  current would become increasingly restricted until it stops all
  together. Could someone negate or confirm this reasoning?

You need to do more research to understand what an electrolytic and galvanic cell is and how it works. Especially salt bridges. A couple of links to check out would be.

Why the salts in a salt bridge?
Voltaic and electrolytic cells
Difficulties understanding how salt bridge works
Why is it important to use a salt bridge in a voltaic cell? can a wire be used?

Don’t be limited to those above, keep searching until you solve your dilemma.

Now for the Hard Part

To best explain this I will work you through a practice problem. It is not a perfect algorithm to write down, but this way you will get a general idea.

  Two electrodes are inserted into a solution of $\ce{NiF2}$ and a current of $2.2\ \mathrm{A}$ is run through them. A list of standard reduction potentials is as follows.

  $$\begin{alignat}{2}
\ce{O2_{(g)} + 4H+ + 4e- &-> H2O_{(l)}}\quad&&1.23\ \mathrm{V}\\
\ce{F2_{(g)} + 2e- &-> 2F-}\quad&&2.87\ \mathrm{V}\\
\ce{2H2O{(l)} + 2e- &-> H2_{(g)} + 2OH-}\quad&&{-}0.83\ \mathrm{V}\\
\ce{Ni^2+ + 2e- &-> Ni_{(s)}}\quad&&{-}0.25\ \mathrm{V}
\end{alignat}$$

So generally I would usually start of with determining with what is going on, you know that $\ce{Ni^2+}$ and $\ce{F-}$ are present in the solution ($\ce{H2O}$).

We can eliminate the third reaction since fluorine already exists reduced, and it does not like being oxidized. Then you have nickel and water, which of them is most likely to be reduced? Nickel.

Now we must decide the compound being oxidized. Fluorine and nickel are out of the question so that leaves us with Equation 1 and 3. As you can see even if we were to flip Eq. 3 it would make no sense since there is no $\ce{OH-}$ present in water. So that leaves us with Eq. 1. We must flip this equation for the reaction to make sense. That leaves us with:

Equation 1 (this one is flipped) and 4. $\ce{Ni^2+}$ is being reduced while $\ce{H2O}$ is being oxidized.

In general …

Eliminate which reaction cannot take place.
Look at what is left, and judge which reaction can best represent the reaction taking place in the cell.

With practice you will be able to solve these kinds of problems. If this is not what you are looking for, then hopefully you learned about these kinds of problems. :)


What is the product of magnesium reacting with carbon?

This asnswer is intended to extend my comment. Most of the information is taken from this site.

Both types of magnesium carbide($\ce{MgC2, Mg2C3}$) is formed when acetylene gas is passed over magnesium. At different temperatures, one type of carbide predominates over the other. At 570°C-610° C, $\ce{MgC2}$ converts to $\ce{Mg2C3}$:

$$\ce{2MgC2 -> Mg2C3 + C }$$

Above 610° C, $\ce{Mg2C3}$ decomposes:

$$\ce{Mg2C3 -> 2Mg + C}$$

Both types of magnesium carbide can be hydrolysed:

$$\ce{MgC2 + 2H2O -> Mg(OH)2 + C2H2}$$
$$\ce{Mg2C3 + 4H2O -> 2Mg(OH)2 + C3H4}$$

For more information on the structure of magnesium carbide, see this old paper from 1943. Also, how magnesium carbide behaves at high pressure is discussed in this new paper(2015). Another type of magnesium carbide, $\ce{Mg2C}$ exist at high pressure and it along with $\ce{MgC2}$ behaves as semiconductor while $\ce{Mg2C3}$ behave as insulator-metal transition. (OP is requested to see both papers if interested)


Thank you for attaching your input file.
The first place to start for improving these calculations is your choice of basis set. 6-31G is a rather small basis set and is likely a source of significant error here. If you're fond of using the Pople type basis then you would be better off using something like 6-31++G* so that you include polarization and diffuse functions. With your current basis I suspect that you are missing important polarization of the atoms upon distortion.

If you really wanted a very accurate answer I would recommend the following scheme. Optimize your geometry at the DFT level and then use that geometry as a starting place for a optimization and frequency calculation at a higher level of theory. For example use the cc-pvtz basis set (a correlation consistent basis set that's more efficient generally than the Pople type) and CCSD(T). This will provide a very accurate gas phase raman spectrum.

Edit:
The only additional Gaussian specific parameter you should need is the Geom=Checkpoint command for the CCSD(T) optimization and frequency run. This will use your existing checkpoint file as a geometry starting point.
The other keywords (basis and method) are exactly what they should be cc-pvdz and ccsd(t). For more information on basis keywords I recommend the users manual.


Why does free chlorine in the stratosphere lose its ozone-depleting potential after about 100,000 reactions?

Since the stratosphere is nowhere near a closed system, a chlorine atom will eventually leave it. Look at the phrasing again:

  It is estimated that one chlorine atom can destroy over 100,000 ozone molecules before it is removed from the stratosphere.

It is not stated, that it looses its potential, it just leaves the region, where there is sufficient ozone concentration.

In a closed system, the catalytic cycle would continue until the ozone concentration is small enough, that the chance of chlorine atoms recombining is bigger.

Thanks to Nicolau's comment I decided to further look into the issue. Also qwersjc gave some valuable insights. For further reading I suggest, the following sources:

http://www.nobelprize.org/nobel_prizes/chemistry/laureates/1995/ (overview)
Nobel Lecture by S. Rowland (Free Access)
Stratospheric Chlorine: a Possible Sink for Ozone. R. S. Stolarski, R. J. Cicerone; Canadian Journal of Chemistry, 1974, 52(8): 1610-1615. (Open access)
Original Paper by M. J. Molina and F. S. Rowland; Nature 249, 810 - 812 (28 June 1974) (Might not be free), reprint on ozone.unep.org

The Ozon Cycle

The ozone cycle is what protects us from quite dangerous UV-B radiation ($M$ is an inert Air molecule like $\ce{N2}$):
$$
\ce{O2 ->[h\nu] O + O} \\
\ce{O + O2 ->[<M>] O3}\\
\ce{O3 ->[h\nu] O + O2}
$$

The Troublemakers

The main reason for chlorine atoms in the stratosphere is man made. Chlorofluorocarbons were very common in many products. These molecules are colourless and water-insoluble. Diffusion taking its course the end up in the stratosphere, where they may undergo dissociation.
$$\ce{CCl3F ->[h\nu] Cl + CCl2F}$$
So the main problem is, that free chlorine radicals are produced in the stratosphere. Also these CFCs have a lifetime in the stratosphere of up to 150 years.

There are some natural sources (of close to insignificant importance) that also produces free Chloride Radicals, i.e. volcanic eruptions. Those may transport Hydrochloride into the stratosphere, where it can react with the Hydroxide radical:
$$\ce{HCl + OH -> Cl + H2O}$$

Removal of Chlorine

There are several steps involved, that lead to the removal of Chlorine from the stratosphere. Basically it will stay there for quite a while and cause significant damage. The stratosphere is very dry and water-soluble compounds, like HCl, will not be washed out quickly. In order to leave the stratosphere, the chlorine atoms or compounds therefore have to be near the troposphere.

Probably the most important reaction is with methane forming Hydrochloride, which may the be washed out by water (causing at least a part of acid rain nevertheless).
$$\ce{Cl + CH4 -> HCl + CH3}$$ The likeliness of this event is about 1000 times smaller than colliding with another ozone.

However eventually, after about two years (a number I picked up on the way), a chlorine will leave the stratosphere. In the troposphere is more water present, so Hydrochloric acid will be formed and washed out.

  In our estimates, termination of the Cl--ClO chain results from a downward diffusion of the longer lived species in the chain (HCl, ClO) and eventual removal by tropospheric process. From: Nature 249, 810-812 (28 June 1974).

There are also a number of reactions that decrease the potential of Chlorine radicals.
$$\ce{ClO + NO2 -> ClONO2}\\
\ce{ClO + O2H -> ClOH + O2}\\
...$$

However these also form reservoir molecules that might diffuse back to the stratosphere.

The calculations behind this are rather complex and use a wide variety of empiric data along with complex models.

I think there is still a lot of research going on in this area, considering more molecules like $\ce{CO2}$ and other greenhouse gases, as well as global climate change.

The Nobel lecture (1995) closes with the following remarks and we can only hope that this holds true:

  However, with the limitations just expressed, the threat of extensive further stratospheric ozone depletion during the 21st century appears to be under control.


Is water a possible fuel for jet engines?

It can't work because of the fundamental thermodynamics

What you are proposing is, basically, the plane carries water; the water is broken down into its components, hydrogen and oxygen; the components are recombined by burning them as fuel. Burning hydrogen and oxygen is a perfectly good way to create a lot of heat. But it doesn't much matter how you break the water apart into hydrogen and oxygen, the thermodynamics of the reaction won't work.

The problem is simple: you need to have a source of energy to split the water apart. In chemistry we know the energy levels of the reactants and the products and we can work out whether energy is released or stored in a reaction. Burning hydrogen and oxygen releases a lot of energy, but by the rule of thermodynamics, breaking water apart to its components requires the input of exactly the same amount of energy. You can't get round this. Worse, in the real world, there are losses at every conversion step so you can't even break even (ain't things unfair!)

In your plane you could, in principle, split water and burn it in the engine for propulsion. But you would need to have some other source of the vast amount of energy required to split the water. That implies both another fuel and another engine. In reality they would vastly outweigh any imagined savings in weight and cost.

Even if you could build some sort of SciFi engine that both splits water and then burns it again you would still be nowhere: the entire output of the burning would be required for the splitting with nothing left over to generate thrust (and in the real world there are losses so your plane would rapidly drop out of the sky).


Ways to determine the concentration of alkyl halides in organic solvents

You could form the Grignard and titrate it. The Grignard should be fairly straightforward to titrate (e.g. as described by Knochel). This also has the advantage of giving you precisely the concentration of your reactive specie. If you knew only the concentration of your alkenyl bromide, you would not know precisely the quantity of Grignard formed (as the reaction is not always perfect).


Difference in leaving group ability due to variation in nucleophiles

The data which you are using to explain a groups tendency to act as a good leaving group is via a nucleophilic substitution reaction.  Nucleophilic sub reactions are affected by 3 things mainly: the nucleophilicity of the nucleophile, the strength of the bond being broken and how easy is it for the leaving group to leave.

This means if you use a better nucleophile the substitution reaction will happen faster. so in the case of your data, you are comparing leaving groups using "relative rates". So the nucleophilicity of the nucleophile will end up impacting the rate. So by looking at two different nucleophiles you are looking at each nucleophiles nucleophilicity and the leaving group.


Ancient IUPAC literature

The IUPAC Interdivisional Committee on Nomenclature and Symbols states in recommendations from $1979$, Use of Abbreviations in the Chemical Literature,

  Since the style in which abbreviations are given (e.g., capital letters or small letters,
  with or without full stops) often depends upon well-established practices that vary widely
  from one journal to another, IUPAC does not recommend a specific style for general use.

  DOI: 10.1351/pac198052092229, archived version: here.

This means there is no one general guideline. You may compare this to units which are to be written, without exception, in upright Roman. (ACS Style guide does recommend Italic for molality $m = \ce{mol / kg solvent}$ but IUPAC disagrees.)

The IUPAC–IUB Combined Commission on Biochemical Nomenclature (CBN), in $1967$, set forth a document of revised tentative, i.e., provisional, rules. Abbreviations and Symbols for Chemical Names of Special Interest in Biological Chemistry does not list $\ce{DMSO}$ specifically, however does mention the similar $\ce{EDTA}$.

  7.$\ $Miscellaneous Compounds

  7.1 The following abbreviations are permitted; although they are fairly common, they should be defined in any paper if it is thought that readers might be unfamiliar with them. Some abbreviations are taken from the list published by Annual Review of Biochemistry.

  [---]

  $\ce{EDTA}$ – ethylenediaminetetraacetic acid (or acetate)

  DOI: 10.1111/j.1432-1033.1967.tb00070.x, archived version: here.

The above article mentions $\ce{EDTA}$ only as a separate compound, and is written once, exclusively in upper-case upright Roman.

Summary

Disclaimer: What follows is mostly personal opinion, and should be treated accordingly. It is a summary of my present understanding.

There really is not a recent IUPAC document discussing this topic.
IUPAC does not enforce a ubiquitous principle upon which rules of abbreviations are formed. Abbreviations should be used as rarely as possible, defined where possible, and the specifics depend on the field, journal, and purpose at hand.
That said, some rules exist. For instance, most new organic compound abbreviations should derive directly from acceptable full names set forth in IUPAC's $2013$ Blue Book.
If an abbreviation of a compound, such as $\ce{DMSO}$ or $\ce{EDTA}$, is applied in chemical formulae, they should be written in lower-case upright Roman. A non-exhaustive list of exceptions is listed in IUPAC's $2005$ Red Book, section IR-4.4.4, paragraph $\mathrm{(vi)}$.
The following cited quotation from the same paragraph is very confusing.

  Abbreviations for solvents that behave as ligands should also be in lower case letters [e.g. $\ce{dmso}$ for dimethyl sulfoxide {(methylsulfinyl)methane}, $\ce{thf}$ for tetrahydrofuran]; the practice of capitalizing the abbreviation of a solvent when it does not behave as a ligand is strongly discouraged as an unnecessary distinction.

I am fairly certain it means the following: If a collection of atoms attached to a central atom does not behave as a ligand, you should still use lower-case abbreviations when composing a chemical formula. Unless exception. This does not mean nor imply you ought to use lowercase when discussing a solvent on its own.
The formula $\ce{[CrCl2(dmso)4]Cl * dmso}$ seems correct to me. Not entirely sure if you could also use $\ce{[Cl2Cr(dmso)4]Cl * dmso}$.
I would include both options in the glossary but with a footnote. Mention that uppercase for $\ce{DMSO}$ is generally used the compound itself, lowercase is reserved for compiling a chemical formula. This might be overkill; if you prefer one, I would go with including only the upright version. But again, if space is not an issue, add both for maximum clarity.


Why are nanoliposome encapsulated antibiotics still so unavailable despite being so effective?

Liposomes are the first nano-drug delivery systems that have been successfully translated into real-time clinical applications. So far, all liposomal drugs are designed for intravenous applications with control release. Most of them are designed to certain targets such as cancer cells. That's why most of liposomal drugs are used in chemotherapy (Ref.1):

The main fact is those liposomal drugs are used mostly in hospitals with control of appropriate medical personals. For example, all seven liposomal drugs so far in chemotherapy are Doxil®, Depocyt®, DaunoXome®, Mepact®, Marqibo®, Myocet®, and Onivyde® as listed in Ref.1 used in cancer patients (in- or out-patients, but under the supervision of their medical doctors). Morphine sulfate extended-release liposome injection, DepoDur™, and non-opioid, bupivacaine liposome (trade name Exparel®) are analgesics used in pain management, specially after postsurgical procedures (e.g., bupivacaine liposome is a postsurgical analgesic used in the management of postsurgical pain and It can also numb an area of the body to relieve pain during surgery or medical procedures). In addition, VYXEOS® is recently FDA approved  (First approved August 3, 2017) liposomal drug for the treatment of adults with newly-diagnosed therapy-related acute myeloid leukemia (t-AML) or AML with myelodysplasia-related changes (AML-MRC). VYXEOS®, a liposomal daunorubicin/Cytarabine, is a fixed-dose combination of liposomal bound cytarabine and daunorubicin that delivers the two medications in a 5:1 molar ratio. Further, it is the ONLY treatment recommended by National Comprehensive Cancer Network® (NCCN®) for induction in sAML patients $\ge \pu{60 years}$. Also, in addition to its use in chemotherapy for leukemia, DaunoXome®, a liposomal daunorubicin, is also FDA approved to treat AIDS related Kaposi's sarcoma.
Based on these facts, therefore, it is safe to say that liposomal drugs are in use without aware of general public. Regardless to say that those treatments are very expensive as you have guessed.
I think main drawback to not available liposomal antibiotics is their high vulnerability to acidic conditions, thus no oral applications available (almost all liposome broke open in the stomach). So far, only literature available for acid stable liposome is Ref.2 and 3. However, we have to stop the development in that direction due to the lack of funding. Therefore, do not believe what you are hearing. Liposomal applications and their oral applications are coming pretty soon (someone else may follow our direction). Nonetheless, now we have a vaccine for COVID-19 within a year, don't we?
References:

Upendra Bulbake, Sindhu Doppalapudi, Nagavendra Kommineni, Wahid Khan, "Liposomal Formulations in Clinical Use: An Updated Review," Pharmaceutics 2017, 9(2), 12 (33 pages) (doi: 10.3390/pharmaceutics9020012)(PDF).
Adelphe M. Mfuh, Mathew P. D. Mahindaratne, Maritza V. Quintero, Frederick J. Lakner, Ande Bao, Beth A. Goins, William T. Phillips, George R. Negrete, "Novel asparagine-derived lipid enhances distearoylphosphatidylcholine bilayer resistance to acidic conditions," Langmuir 2011, 27(8), 4447-4455 (DOI: https://doi.org/10.1021/la105085k).
George R. Negrete, Mathew P. D. Mahindaratne, Adelph M. Mfuh, Maritza V. Quintero, "Compositions and Methods Related to Acid Stable Lipid Nanospheres," US20110268653A1 (https://patents.google.com/patent/US20110268653A1/en) (PDF).


Is it possible to manipulate some specific atoms of a liquid metal by dint of the magnetic field, electric field, etc?

While Field Structured Composites are a thing, they use nano- to micro-scale particles in a polymer matrix, with particle alignment controlled/influenced by applying electric or magnetic fields during curing.
Trying to do this at the atomic level, as you ask about above, runs into several problems. These include:

Overcoming the desire of the liquid to mix, so you need your desired element to have a fairly well tuned positive heat of mixing to just balance the entropic contributions to the free energy. Or, in other words, just barely soluble.

Overcoming diffusion in the liquid would require strong interactions of your field with the specific element (and only that element). This runs up against the limits below.

In a metallic liquid, applying an electric field isn't going to specifically target any one atom type since the free electrons will shield the bulk from the field. Ions in a dielectric liquid would be a better choice (much like the field structure composites above), but that is just the reinvention of ion chromatography.

The magnetic moments of atoms are pretty darned small, so you won't get the ferromagnetic boost that can be achieved with particles. Applied magnetic fields will have to be enormous if they would even work at all.


What is the connection between vapor pressure and intermolecular force explanations of boiling?

Consider boiling as the rapid escape of molecules.

In slow evaporation and in rapid boiling, only those molecules with sufficient kinetic energy are able to leave the liquid. As each "fast" molecule leaves, it subtracts its high KE from the average KE pool of the liquid, i.e. cooling it.

For evaporation to take place, there has to be a chance for molecules to escape and not to be replaced by those in the gas above the liquid. In fact, if the pressure is high (below the critical temperature), then the gas can condense to the liquid, faster than the liquid can evaporate to the gas.

For boiling to take place, there is the further restriction that localized temperature rise takes place faster than in the main body of the liquid, so that specific locations overheat, e.g. where heat flux is highest. Where pressure is a factor, then boiling may be prevented until the pressure is relieved, as in "black smokers". The water exiting the vent may be at 200 C or more, but does not boil unless the critical temperature is exceeded.


The H-C-H bond angle in ethene is ca. 117 degrees and the H-C-C angle is ca. 121.5 degrees.  There are two reasons that combine to explain this angular deformation in ethene.

First, from these bond angles and Coulson's Theorem (ref_1, ref_2) we can determine that the C-H sigma bonds are $\ce{sp^{2.2}}$ hybridized and the C-C sigma bond is $\ce{sp^{1.7}}$ hybridized.

From these hybridization indices (the index is the exponent "n" in the $\ce{sp^{n}}$ expression) we see that the C-C sigma bond has higher s-character content (1 part s to 1.7 parts p - 37% s) than the C-H bonds (1 part s to 2.2 parts p - 31% s). Since there is more s character in the C-C bond, it is lower in energy and the carbon sigma  electrons will tend to flow  towards this lower energy C-C bond.  Consequently, the C-C sigma bond will contain more electron density than the C-H bonds. Therefore, the electron repulsion between the C-C sigma bond and C-H sigma bonds will be greater than the electron repulsion between the two C-H bonds.  Hence the H-C-C bond angle will open up slightly from the $\ce{sp^{2}}$ ideal of 120 degrees and the H-C-H angle will close down slightly in order to minimize the bond-bond electrostatic repulsions.

Second, steric factors (which are also really just another way of describing electron-electron repulsion) may also come into play. To whatever extent the cis H-C-C-H hydrogen-hydrogen repulsion is more destabilizing than the geminal H-C-H hydrogen-hydrogen repulsion, it will also serve to increase the C-C-H bond angle and shrink the H-C-H bond angle.


Finding mass percentage of components of a solid mixture

We first start by writing the system of equations in this reaction, assuming that we don't know the ratios that the elements combine in.

Because we know the mass of the unknown mixture, we'll use $x$ to represent the mass of $\ce{K2CO3}$ and $y$ to represent the mass of $\ce{KHCO3}$, in grams. This is our first equation:
$$x + y = 2.184$$

Next, we know the final amount of moles of $\ce{HCl}$ that reacted. It was
$$\mathrm{0.03129\; \pu L \times \frac{0.742\; mol\; \ce{HCl}}{\pu L} = 0.0232\; mol\; \ce{HCl}}$$

Now, let's reconsider the balanced equations for the reaction. We have
$$\ce{2HCl_{(aq)} + K2CO3_{(aq)} -> 2KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$
$$\ce{HCl_{(aq)} + KHCO_{3(aq)} -> KCl_{(aq)} + CO2_{(g)} + H2O_{(l)}}$$

We can see that it takes $\ce{2 HCl}$ to neutralize one $\ce{CO3}$, but it only takes $\ce{1 HCl}$ to neutralize one $\ce{HCO3}$.

We also know the formula weights of the individual components. Now, we can write the second equation as an expression of the sum of the moles of the reactants. Our coefficients for $x$ and $y$ become:
$$
\ce{\frac{2\; mol\; HCl}{1\; mol\; K2CO3} \times \frac{1\; mol\; K2CO3}{138.2058\; g}} \times x
$$

$$
\ce{\frac{1\; mol\; HCl}{1\; mol KHCO3} \times \frac{1\; mol\; KHCO3}{100.1154\; g}} \times y
$$

Now we can cancel out units and write out the second equation:
$$\ce{\frac{2\; mol\; HCl}{138.2058\; g}}x + \ce{\frac{1\; mol\; HCl}{100.1154\; g}}y = \ce{0.0232\; mol\; HCl}$$

After that, this is straightforward linear algebra. This system of equations is simple enough to be solved by basic substitution and elimination. However, you can choose to use a matrix. In that case, we have

$$
\mathrm{
\begin{bmatrix}
    1 & 1 \\
    \frac{2}{138.2058} & \frac{1}{100.1154}
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
}
=
\begin{bmatrix}
2.184 \\
0.0232
\end{bmatrix}
$$

Rearranging gives
$$
\mathrm{
\begin{bmatrix}
    1 & 1 \\
    \frac{2}{138.2058} & \frac{1}{100.1154}
\end{bmatrix}^{-1}
\begin{bmatrix}
2.184 \\
0.0232
\end{bmatrix}
=
\begin{bmatrix}
x \\
y
\end{bmatrix}
}
$$

Using a calculator we find that
$$\mathrm{
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
0.309\\
1.875
\end{bmatrix}
}
$$
$$x = \ce{0.309\; g\; K2CO3}$$
$$y = \ce{1.875\; g\; KHCO3}$$

Finally, the mass percentages are determined to be:
$$\ce{\frac{0.309\; g\; K2CO3}{2.184\; g\; total} \times 100{\%} = 14.1{\%} K2CO3}$$
$$\ce{\frac{1.875\; g\; KHCO3}{2.184\; g\; total} \times 100{\%} = 85.9{\%} KHCO3}
$$


Reduction means that electrons are added, you are talking about oxidation. I haven't heared the term "fully reduced" in this context before, but I guess it means a full shell. In case of iron this would mean a oxidation state of -10, while as far as I know the highest we can get is -4.

And don't confuse oxidation states with charge of ions.

Well, now that I see the actual question I think there's nothing to really understand here. Just a wrong, made up answer.


Dissociation of zinc anode during electrolysis

Q: What is this process known as?

This is a particular case of dissolution. Call it electrolytic dissolution, if you want to be more specific. The word "dissociation" certainly does not apply here. It is used when talking about salts and other ionic compounds.

  Q: Do the $\ce{Zn^2+}$ ions remain "isolated"?

Well, $\ce{Zn^2+}$ just goes into the solution and gets solvated (hydrated), that is, surrounded by interacting water molecules, pretty much like any other ion. Some people in some contexts would probably call these ions "free", though they are surely not as free as in vacuum. The same applies to the $\ce{Na+}$ and $\ce{Cl-}$ ions out there, and no, they are not linked to each other in any way except for electrostatic interaction. Have you heard of dissociation? That's what it is about. A $\ce{Cl-}$ ion does not "know" that it "belongs" to $\ce{Na+}$ and not $\ce{Zn^2+}$. They all just coexist there together, probably with some other ions like $\ce{OH-}$. As the process goes on, I expect some $\ce{Zn(OH)2}$ precipitate to appear.


The answer should be D. Because of the principle of microscopic reversibility, the forward enthalpy is opposite to the reverse enthalpy. This also follows from the conservation of energy if you will.

In this StackExchange question, there is a nice picture given of a potential energy surface. There you can readily see that the forward reaction enthalpy is opposite to the backward reaction enthalpy.