What is the chemical structure / hybridization of sulfuric acid?

It is much easier to explain it on simpler example – $\ce{O3}$ molecule. It has structure of resonance hybrid of $\ce{O=O+-O-}$ and its mirror. And of course, central atom has hybridization state $sp^2$. One bond here is normal covalent bond and another bond is dative: an electron pair is donated onto vacant orbital of $\ce{O}$ atom with all electrons paired. in $\ce{H2SO4}$ molecule to bonds are simple covalent ($\ce{S-OH}$ ones) and two are dative ($\ce{S-O}$ ones). A common concept of electron unpairing to my knowledge is proved brocken by quantum chemistry calculations and spectral experiments for hypervalent compounds of $\ce{P}$ and $\ce{S}$.

A little more interesting example is $\ce{XeF2}$ molecule, where three-atom four-electron bond $\ce{F-Xe-F}$ are formed, that can be think of as reasonance hybrid of structure $\ce{F-Xe+\ F-}$ and its mirror $\ce{F- \ Xe+-F}$

Of course, this scheme is still far from perfect, as reality is much more complicated, but if you do not wish to take course of quantum chemistry, it should be enough. However, I'll recommend to search for "MO LCAO model": it is quite simple and very useful. It is also often used in advanced chemistry books and articles.


If phosphorus trioxide is present in the atmosphere of Venus, could it not react with water to form (sufficient) phosphorous acid?

The issue seems to be not the reaction between $\ce{P4O6}$ and water, but whether $\ce{P(III)}$ can form at all.  The OP's first reference says no whereas the second reference says yes.  Basically the two references used different thermodynamic assumptions.  The first one assumes an atmosphere in equilibrium while the second assumes only a partial equilibrium; in the latter case $\ce{CO}$ is allowed to form from nonequilibrium (radiation-driven) dissociation of $\ce{CO2}$, and minor species are then equilibrated with the $\ce{CO/CO2}$ couple.  In the case of phosphorous oxides this favors $\ce{P4O6}$ over $\ce{P4O_{10}}$, although the former could be oxidized by other species such as sulfuric acid.
The issue has become important because the recent discovery of $\ce{PH3}$ among the clouds of Venus has led to a hypothesis that this compound is generated by Venusian microbes, but if $\ce{P(III)}$ is available for a disproportionation reaction it could provide an alternative abiotic (non-life) source.  As noted above, the references disagree on this, and the question of whether this alternative to a biological source of $\ce{PH3}$ is viable awaits resolution.
Relevant to the question of whether $\ce{PH3}$ is a true biosignature is this question about organic compounds on Venus, which apparently spawned the current one.


Can hydrogen and oxygen gasses be turned back into water?

When hydrogen burns, pure water is created; the reaction is exactly the reverse of the electrolysis and shown in the equation below.

$$\ce{2 H2 + O2 -> 2 H2O}$$

If you just have hydrogen, then it will burn with a flame as you might know from a methane gas flame. However, once you have a certain mixture of hydrogen and oxygen gas in your container, the reaction becomes explosive. If you have the exact stoichiometric mixture of $2:1$, the reaction is a very strong (and loud) explosion; a tiny amount created a louder bang in Munich’s chemistry lecture hall than all other explosions the experimentors showed on that day.

The fact that hydrogen burns to give water and only pure water (vapour) was one of the main points used to try and sell hydrogen-powered cars, since water vapour is environmentally friendly and does not cause global heating like $\ce{CO2}$ does.

The electrolysis/explosion difference shows how different conditions for forwards and backwards reaction can be.

It bears repeating: do not attempt the combustion at home unless you know exactly what you are doing!


Degradation mechanism of hydroxymethylfurfural to 2,5-dioxo-3-hexenal and 4-oxopentanoic acid

Here is a link to a detailed reference on HMF.  In section 3.3 (page 32) they show a mechanism for the acid catalyzed degradation of HMF.  As you noted, they do propose two pathways for decomposition, one leads to polymeric materials, the other leads to 2,5-dioxo-3-hexenal, which then can react further to produce 4-oxopentanoic acid. The work they describe was done by Horvat (Horvat, J.; Klaič, B.; Metelko, B.; Sunjic, V. Tetrahedron Lett. 1985, 26, 2111).

I don't want to reproduce their drawing, but follow the link and look for Scheme 17.  Most of the steps involve protonation of either

a double bond to form a resonance stabilized carbocation followed by addition of water and elimination of a proton to yield an alcohol
a hydroxyl group followed by elimination of water to yield a resonance stabilized carbocation followed by elimination of a proton to yield a double bond
or protonation of the furan oxygen followed by ring opening to generate a protonated, resonance stabilized carbonyl

If you have any specific questions on the mechanism, just post back here.


How to calculate the required concentration for sodium sulfite to remove oxygen as an oxygen scavenger from a solution?

From the equation
\[\ce{2SO3^{2-} + O2 -> 2SO4^{2-} } \]
you could conclude that you need (at least) 2 moles of sodium sulfite $(\ce{Na2SO3})$ for each mole of oxygen in solution.


I can suggest only two possibilities:

Your acetone or sodium are not pure. Acetone contains water, or something.
Sodium might react like magnesium does, though it required different conditions. That's possible, but maybe you just discovered a new reaction or had some weird impurities that served as a catalyst.


Do catalysts shift equilibrium constant towards 1?

TL;DR Your Maxwell–Boltzmann diagram up there is not sufficient to describe the variation of rate with $E_\mathrm{a}$. Simply evaluating the  shaded area alone does not reproduce the exponential part of the rate constant correctly, and therefore the shaded area should not be taken as a quantitative measure of the rate (only a qualitative one).

There is a subtle issue with the way you've presented your drawing. However, we'll come to that slightly later. First, let's establish that the "proportion of molecules with sufficient energy to react" is given by

$$P(\varepsilon) = \exp \left(-\frac{\varepsilon}{kT}\right) \tag{1}$$

Therefore, for a reaction $\ce{X <=> Y}$ with uncatalysed forward activation energy $E_\mathrm{f}$ and uncatalysed backward activation energy $E_\mathrm{b}$, the rates are given by

$$k_\mathrm{f,uncat} = A_\mathrm{f} \exp \left(-\frac{E_\mathrm{f}}{kT}\right) \tag{2} $$

$$k_\mathrm{b,uncat} = A_\mathrm{b} \exp \left(-\frac{E_\mathrm{b}}{kT}\right) \tag{3} $$

The equilibrium constant of this reaction is given by

$$K_\mathrm{uncat} = \frac{k_\mathrm{f,uncat}}{k_\mathrm{b,uncat}} = \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \tag{4}$$

As you have noted, the change in activation energy due to the catalyst is the same. I would be a bit careful with using "$\mathrm{d}E$" as the notation for this, since $\mathrm{d}$ implies an infinitesimal change, and if the change is infinitesimal, your catalyst isn't much of a catalyst. So, I'm going to use $\Delta E$. We then have

$$k_\mathrm{f,cat} = A_\mathrm{f} \exp \left(-\frac{E_\mathrm{f} - \Delta E}{kT}\right) \tag{5} $$

$$k_\mathrm{b,cat} = A_\mathrm{b} \exp \left(-\frac{E_\mathrm{b} - \Delta E}{kT}\right) \tag{6} $$

and the new equilibrium constant is

$$\begin{align}
K_\mathrm{cat} = \frac{k_\mathrm{f,cat}}{k_\mathrm{b,cat}} &= \frac{A_\mathrm{f}\exp[-(E_\mathrm{f} - \Delta E)/kT]}{A_\mathrm{b}\exp[-(E_\mathrm{b} - \Delta E)/kT]}  \tag{7} \\[0.2cm]
&= \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \frac{\exp(\Delta E/kT)}{\exp(\Delta E/kT)} \tag{8} \\[0.2cm]
&= \frac{A_\mathrm{f}\exp(-E_\mathrm{f}/kT)}{A_\mathrm{b}\exp(-E_\mathrm{b}/kT)} \tag{9}
\end{align}$$

Equations $(9)$ and $(4)$ are the same, so there is no change in the equilibrium constant.

The question then arises as to how eq. $(1)$ is obtained. The simplest way is to invoke a Boltzmann distribution, which almost by definition gives the desired form. However, since you have a Maxwell–Boltzmann curve, I guess I should talk about it a bit more. The fraction of molecules with energy $E_\mathrm{a}$ or greater is simply the shaded area under the curve, i.e. one can obtain it by integrating the curve over the desired range.

$$P(\varepsilon) = \int_{E_\mathrm{a}}^\infty f(\varepsilon)\,\mathrm{d}\varepsilon \tag{10}$$

where the Maxwell–Boltzmann distribution of energies is given by (see Wikipedia)

$$f(\varepsilon) = \frac{2}{\sqrt{\pi}}\left(\frac{1}{kT}\right)^{3/2} \sqrt{\varepsilon} \exp\left(-\frac{\varepsilon}{kT}\right) \tag{11}$$

At first glance, we would expect this to be directly proportional to the exponential part of the rate constant, i.e. $\exp(-E_\mathrm{a}/kT)$. Alas, it is not that simple. If you try to work out the integral

$$\int_{E_\mathrm{a}}^{\infty} \frac{2}{\sqrt{\pi}}\left(\frac{1}{kT}\right)^{3/2} \sqrt{\varepsilon} \exp\left(-\frac{\varepsilon}{kT}\right) \,\mathrm{d}\varepsilon \tag{12}$$

you don't get anything close to the form of $\exp(-E_\mathrm{a}/kT)$. Instead, you get some "error function" rubbish, and some nasty square roots and exponentials. (You can use WolframAlpha to verify this.)

Why is this so? Well, it turns out that there are other terms that also depend on $\varepsilon$ and therefore need to go inside that integral (they aren't constants and can't be taken out).

The simplest example is that faster molecules tend to collide more often, so even though the right-hand tail of the diagram seems to contribute very little to the "proportion of molecules with sufficient energy", it actually contributes more significantly to the overall rate because these molecules collide more often. In collision theory this is described using the "relative velocity" of the particles $v_\mathrm{rel}$.

There is also another complication, in that the Maxwell–Boltzmann distribution, the direction of the particles is not accounted for. (For more insight please refer to Levine Physical Chemistry 6th ed., p 467.) Therefore, there has to be yet another term that takes into account the direction of movement of the particles. The idea is that a head-on collision between two molecules is more likely to overcome the activation barrier than is a $90^\circ$ collision. The term that compensates for this is the "collision cross-section" $\sigma$.

If you go through the maths (and I don't really intend to type it out here, it's rather long, but I will give some references) then you will find that at the end you will recover the form $\exp(-\varepsilon/kT)$. Once you have arrived at this, it's very straightforward to see that the increases in rate of both the forward and backward reaction cancel each other out.

Now, as for the promised references, Pilling and Seakins's Reaction Kinetics pp 61-2 have a short outline of the proof. Atkins's Physical Chemistry 10th ed. has a slightly longer proof on pp 883-4.


Is it possible to make broad spectrum light absorbing paint?

Carbon nanotube black is highly absorbent in and outside the visible spectrum, see http://www.extremetech.com/extreme/186229-its-like-staring-into-a-black-hole-worlds-darkest-material-will-be-used-to-make-very-stealthy-aircraft-better-telescopes.

Of course, that does not make it ideal for energy conversion, as it would also need at least the following qualities to be practical.

A good conductor of heat to the heat-exchanger surface below
An enduring finish, resistant to abrasion, heat and environmental degradation
Economical, both in material cost and application
Safety in use and in manufacture


Having recently had an opportunity to play around with some gallium of my own, I have a few thoughts.

It's possible you have a different metal than what you thought you had, but frankly, I don't think that's very likely.

It takes a surprising amount of absorbed heat to melt gallium. Putting a sizable chunk of gallium in $35~^\circ\mathrm C$ water doesn't cause it to melt immediately. Even boiling water doesn't melt it instantly. Because gallium is a metal, it conducts heat fairly well, so when heated gently, it won't melt until the temperature of the whole piece gets close to the melting point.

As mentioned by user5134, a shiny metal like gallium isn't going to pick up heat very quickly from a radiative source like a heat lamp. If you wait long enough, it'll definitely melt. If you want to melt the gallium relatively quickly, then you'll want to put it into physical contact with something warm. The standard method for melting solid gallium in a container is to just immerse the closed container in hot water (not boiling water, especially if the container is glass).


Why hydrogen peroxide sizzles in the bathroom sink?

Hydrogen peroxide contains an peroxide linkage and it decompses rapidly to give oxygen.

But when substances like detergent come in its contact in a bathroom (as in the question) it not only  catalyses the reaction but if it was a foam making detergent it would make the decomposition process more evident and visible as oxygen bubble would produce effervescence in foam.


Can acidified or neutral KMnO4 oxidise toluene to benzoic acid?

Here are the three equations describing the reduction of manganese (and concurrent oxidation of whatever substrate may be present) under basic, neutral and acidic conditions respectively.

$$\ce{Mn^{+7}O4- +e- ->~ Mn^{+6}O4^2-~~~~[basic]}$$
$$\ce{2H2O + Mn^{+7}O4- + 3e- ->~ Mn^{+4}O2 + 4OH-~~~~[neutral]}$$
$$\ce{8H+ + Mn^{+7}O4^{-} + 5e- ->~ Mn^{+2} + 4H2O~~~~[acidic]}$$

Acidic conditions are the most economical as 5 electrons are transferred per mole of manganese, but oxidation can be achieved under all 3 conditions. A problem with neutral conditions is the precipitation of insoluble $\ce{MnO2}$ and the need for its subsequent separation. A problem with all 3 conditions is that oxidation will only occur if the material to be oxidized is somewhat soluble in the reaction medium.

The milder basic conditions are generally preferred for the oxidation of aromatic alkyl side chains because with the milder basic conditions an organic co-solvent that is soluble in water can be employed to facilitate solubility of the organic substrate.


It's somewhat unclear what you mean by "using $\Delta G$", but here's the full working. The information you are given is:

  $$\begin{align}
\ce{MnO4- + e-} &\rightarrow \ce{MnO4^2-} & E &= +0.564~\mathrm{V} \tag{1}  \\
\ce{MnO4^2- + 2e- + 4H+} &\rightarrow \ce{MnO2 + 2H2O} & E &= +2.261~\mathrm{V} \tag{2}
\end{align}$$

Converting these to Gibbs free energy changes is fundamentally the correct step. We have $\Delta G = -nFE$. (Standard state symbols omitted for clarity, they are implied throughout.) For half-equation (1), $n = 1$, and for half-equation (2), $n = 2$. So:

$$\begin{align}
\Delta G_1 &= -(1)(96485~\mathrm{C~mol^{-1}})(0.564~\mathrm{V}) = -54417.7~\mathrm{J~mol^{-1}} \\
\Delta G_2 &= -(2)(96485~\mathrm{C~mol^{-1}})(2.261~\mathrm{V}) = -436306.7~\mathrm{J~mol^{-1}} \\
\end{align}$$

You need to now sum up these two equations, which also means summing up the changes in Gibbs free energy. Let's define a new reaction $(3) = (1) + (2)$:

  $$\begin{align}
\ce{MnO4- + 3e- + 4H+} &\rightarrow \ce{MnO2 + 2H2O} & E &= E_3 \tag{3}
\end{align}$$

As you can see, all I have done is to add the two half-equations together and cancel out any terms that appear on both sides. We have $n = 3$ for this half-reaction.

$$\begin{align}
\Delta G_3 &= \Delta G_1 + \Delta G_2 \\
&= -490724.4~\mathrm{J~mol^{-1}} \\
E_3 &= -\frac{\Delta G_3}{nF} \\
&= 1.695~\mathrm{V}
\end{align}$$

Last thing, please always quote your units in your answer! You cannot have an electrode potential that is unitless.


Racemization of ketones in presence of catalytic acid

You are right that the racemization occurs through the enol form.

The chiral center is on the carbon attached to the chlorine atom. In B, that carbon participates in the enol form, but in A not. Therefore, in A the enol is formed but there is no racemization.


What is the origin of rotational energy barrier in ethane?

As you say, this issue appears to be unsettled. I find the argument in favor of hyperconjugation compelling, but there is theoretical evidence that steric strain contributes as well.
This debate is about 20 years old. Here are two papers summarizing the opposing views:

Pophristic, V.; Goodman, L. (2001). "Hyperconjugation not steric repulsion leads to the staggered structure of ethane". Nature. 411 (6837): 565–8. doi:10.1038/35079036.
Bickelhaupt, F.M.; Baerends (2003). "The case for steric repulsion causing the staggered conformation of ethane". Angew. Chem. Int. Ed. 42 (35): 4183–4188. doi:10.1002/anie.200350947.

A more recent paper that has an interesting take:

Baranac-Stojanović, M. (2015). "Theoretical analysis of the rotational barrier in ethane: cause and consequences". Structural Chemistry. 26 (4): 989–996. doi:10.1007/s11224-014-0557-5.

I cannot read the entire article, but the abstract offers the conclusion that hyperconjugation is responsible for the initial energy rise during rotation, but that steric factors are needed to fully explain the high energy of the eclipsed conformation.


To my knowledge, Friedel-Crafts alkylation of pyrrole with aluminum chloride ($\ce{AlCl3}$) is unattainable. However, recent literature suggests F-C alkylation of pyrrole with other catalysts is possible. For example, an asymmetric Friedel–Crafts alkylation of pyrrole with nitroalkenes catalyzed by a copper complex of a bisphenol A-derived Schiff base is reported (Ref.1):

  Abstract: The asymmetric Friedel–Crafts (FC) alkylation of pyrrole with nitroalkenes was mediated by $\ce{CuBr2}$ and a novel bisphenol A-derived chiral catalyst at room temperature. The catalyst was found to be applicable for the asymmetric FC alkylation of pyrrole with a wide range of nitroalkenes, affording optically active alkylated pyrroles with enantioselectivities up to 94%. Furthermore, enantiomerically pure 3-nitro-2-arylpropanamides were prepared by the oxidative cleavage of the pyrrole rings in the FC products with $\ce{NaIO4/RuCl3}$ to demonstrate the synthetic application of the products.

For other example, see Ref.2 & 3. For a review, see Ref.4.

References:

Halide Sedef Özdemir, Engin Şahin, Murat Çakıcı, Hamdullah Kılıç, “Asymmetric Friedel–Crafts alkylation of pyrrole with nitroalkenes catalyzed by a copper complex of a bisphenol A-derived Schiff base,” Tetrahedron 2015, 71(19), 2882-2890 (https://doi.org/10.1016/j.tet.2015.03.059).
Yuan-Zhao Hua, Xing-Wang Han, Xiao-Chao Yang, Xixi Song, Min-Can Wang, Jun-Biao Chang, “Enantioselective Friedel–Crafts Alkylation of Pyrrole with Chalcones Catalyzed by a Dinuclear Zinc Catalyst,” J. Org. Chem. 2014, 79(23), 11690-11699 (https://doi.org/10.1021/jo5023712).
Hugh J. Anderson, C. W. Huang, “Pyrrole chemistry. X. Friedel–Crafts alkylation of some pyrrole and furan derivatives,” Canadian Journal of Chemistry 1970, 48(10), 1550-1553 (https://doi.org/10.1139/v70-252).
Hai-Hua Lu, Fen Tan, Wen-Jing Xiao, “Enantioselective Organocatalytic Friedel-Crafts Alkylations,” Current Organic Chemistry 2011, 15(24), 4022-4045 (https://doi.org/10.2174/138527211798109196).


What are the constraints on the two electron integral matrix?

General Answer:

The basic premise of semi-empirical methods is to replace the computationally expensive integrals with values fit to experimental data or, in some cases, to high level ab initio calculations of the relevant quantity [1]. You are completely correct to point out that making this substitution will break the theoretical validity of the variational principle.

The short answer to your main question then is that the constraints are given by the experimental values to which one fits. That is to say, if you fit the values of your integrals to experimental data and all of a sudden you get out energies which are unbounded, discontinuous, or have some other undesirable and un-physical feature, then the problem is either in the fitting or in the corollary between the experiment and integral you are fitting. So, as in many physical contexts, the answer is circular. Sure, we could have any value for the integral, but the answers which experiments give live in a very small region of all possible answers to the question.

Approximations:

First of all, there are tons of semi-empirical methods which differ slightly in how they do the fitting, what approximations are made, etc. One important thing to note is that no semi-empirical method is really claiming to get the physics of these interactions correct for the right reason. They all rely on cancellation of errors, but this is built into the method on purpose, so this cancellation tends to be rather robust. One can achieve this because the behavior of the integrals arising in electronic structure is quite systematic, so if you are careful to know what you neglected in one part, you can try to neglect something of the opposite sign in another part. Reading ref. [1] is a good place to start to see how this is done in practice.

DFTB:

I should also point out that density function tight binding (DFTB) is often described as a semi-empirical method, but DFTB does obey a variational principle[2, 3]. That is, one can expand the total energy from regular density functional theory with respect to variations of the charge density (the central object in DFT). To zeroth order, one gets a non-iterative theory which depends only on properties of the individual atoms in the molecules and then corrects for interatomic repulsions. This is trivially variational.

If one includes terms up to second-order in the density variations, then you get self-consistent-charge DFTB (SCC-DFTB). This still defines an approximate Hamiltonian, so one can variationally solve for the energy. So, you might ask why this is considered semi-empirical. I'll just quote from ref. [2] as they give a nice summary.

  Comparing SCC-DFTB with MNDO-type methods, the
  underlying philosophy and the derivation of the applied approximations seem quite different at first sight. If one focuses
  on the actual working equations and their implementation, there
  are, however, many similarities: All these methods are valenceelectron SCF-MO treatments with a minimal valence AO basis
  set, only one-center and two-center terms are included, the twocenter two-electron integrals are represented by damped Coulomb interactions with the correct limit at large and small
  distances, and there are repulsive atom-pair terms that correct
  for deficiencies in the formalism. The methods are semiempirical
  in the sense that they employ empirical parameters which are
  adjusted to reproduce selected reference data, i.e., DFT energies
  and potential curves in the case of SCC-DFTB, and experimental
  heats of formation and other experimental data in the case of
  the MNDO-type methods.

Hopefully I've addressed your main concerns in the original question and left you with some resources so you better approach further questions.

References:

[1]: Thiel, W. (2014). Semiempirical quantum–chemical methods. Wiley Interdisciplinary Reviews: Computational Molecular Science, 4(2), 145-157.

[2]: Otte, N., Scholten, M., & Thiel, W. (2007). Looking at self-consistent-charge density functional tight binding from a semiempirical perspective. The Journal of Physical Chemistry A, 111(26), 5751-5755.

[3]: Seifert, G. (2007). Tight-binding density functional theory: an approximate Kohn− Sham DFT scheme. The Journal of Physical Chemistry A, 111(26), 5609-5613.


Why is ionic strength (mu) calculated differently for these 2 examples?

The third pKa of phosphoric acid (going from disodium to trisodium phosphate) is 12.32.

When trisodium phosphate is added to water it is a relatively strong base, beginning to titrate H20 to OH-, to a significant degree.  In other words, to be above the pKa of 12.32, the concentration of OH- needs to be at least 0.02M.

In the other examples, since the other two pKas of phosphoric acid are 2.15 and 7.2, the pH in those examples will be about midway between 2.15 and 7.2 for the monosodium salt and about midway between 7.2 and 12.32 for the disodium salt, so the concentrations of H+ and OH- are insignificant compare to the other ions.


Mixing of molecular orbitals in polyatomic molecules

The standard way is to solve for the eigenvectors of your Fock matrix. Since you have a single electron, that's your Hamiltonian. Any number of programming languages have algorithms for this implemented. Each of your coefficients would correspond to an entry in the eigenvector.


How to rationalise the difference in halogen bonding?

A halogen bond can, quite analogous to a hydrogen bond, be considered a mostly electrostatic interaction with some covalent interaction. Hence the halogen bond acceptor is the electron donating molecule, i.e. a Lewis base. The halogen (bond) takes the role as the Lewis acid.

In (b) you compare $\ce{H3C-I}$ and $\ce{F3C-I}$. Fluorine is more electronegative than hydrogen, and the statement $$\angle(\ce{H-C-H})>\angle(\ce{F-C-F})$$ holds. The carbon directs mor $\ce{p}$ orbital character into the carbon fluorine bonds than in the carbon hydrogen bonds. As a consequence the $\ce{s}$ orbital character in the carbon iodine bond (from the carbon) is higher in $\ce{F3C-I}$ compared to $\ce{H3C-I}$. That causes the bond to be slightly shorter and having more electron density at the carbon. This is also making the iodine a little more electron deficient and a better electron acceptor, i.e. a stronger acid (see also Bent's Rule). The electronic interaction will therefore also be slightly stronger.
This observed effect is also often called inductive effect, which like you correctly stated is larger in $\ce{F3C-I}$.

In (c) you are comparing the Lewis bases $\ce{NMe3}$ and $\ce{OMe2}$ acting towards $\ce{F3C-I}$ (or $\ce{HC3-I}$, the reasoning would remain the same). Oxygen is more electronegative than nitrogen, hence the lone pair orbitals of oxygen are more compact, i.e. the maximum electron density is closer to the oxygen nucleus. This causes dimethylether a weaker electron donor. Hence the the oxygen iodine interaction in $\ce{F3C-I\bond{~}OMe2}$ can be expected to be weaker than the corresponding nitrogen iodine interaction in $\ce{F3C-I\bond{~}NMe3}$. (Just like it is stated in your sheet.)
Additionally, since methyl groups are usually slightly electron donating, and trimethylamine has one more methyl group than dimethylether, it can be expected to be an even better electron donor, stronger base.

Resonance does not contribute to these bonding situations in a significant way. (Looking at this from a valence bond perspective would certainly exceed this format and it is not necessary for a good model explanation.)

Last a word of caution. Oxygen (in neutral molecules) can in most cases not be described as $\ce{sp^{3}}$ hybridised. Symmetry considerations and it's electronegativity are usually the reason for this. It is better described as $\ce{sp}$ hybridised in terminal position (like in ketones), or $\ce{sp^{2}}$ hybridised in bridging positions (like in ethers).


What drives a molecule/atom towards stability?

It boils down to thermodynamics. No, molecules do not possess intelligence. But they respect the laws of thermodynamics, which in a general look state:

Energy of a closed system is constant.
The entropy of the universe is always increasing.
You can't get to absolute zero temperature! (Ok, this one is less important for this analysis)

The second law is the key: in order to augment the entropy of the universe a certain transformation can do one of two things: increase its own entropy or the entropy of its surroundings. The latter can be achieved by releasing heat, which is something exothermic reactions do. The sum of these effects is described by the system's Gibbs free energy:

$$G = H - TS$$

For a chemical reaction occurring at a constant temperature and pressure, $G$ has to decrease and becomes minimized at equilibrium. We then say the system is (thermodynamically) "stable". So when you hear someone say "chemical reactions minimize their energy", it would be better spoken as "... minimize their free energy".

Now you might wonder why the second law is true. Why does entropy has to increase? Why does the free energy has to be minimized? It certainly doesn't want to.

The truth lies in the molecular interpretation of entropy: it measures the amount of micro-states at which system can be found. Put together a bunch of blue balls with another bunch of red balls in a box, shake it all and look at how they end up. They surely won't be found in separate sides of the box, they will mix! The scenario where they are separated corresponds to a very low-entropy state: you can think of the red ones to one side or vice-versa: not that many possibilities of arrangements. The mixed state though has a much higher entropy: think of how many arrangements are possible by exchanging the positions of red and blue balls!

Since molecules in a closed system are very very much like tiny balls being shaken inside a box - not by a mechanical source but by their own thermal energy -, their behavior with respect to entropy is the same. Again, it's not that they want to increase the entropy of universe, there just is a higher probability that they do rather than don't. Given that we are usually observing systems of sextillion molecules or more, statistics kicks in and tell us probabilities mean more or less certainty: we are pretty much certain that chemical reactions will keep on trying to maximize the entropy of the universe.


Why there exist no void with co-ordination number 5

Trigonal bipyramidal complexes are abundant.  However, if you want a periodic crystal, you must tesselate volume with translation - the unit cell of the crystal lattice.  A five-fold symmetric figure cannot do this, for the angles do not add up without gaps.  2-, 3-, and 6-fold symmetries are allowed, offering 230 unique 3-D space groups overall.

If you allow two kinds of unit cell, you can get approximate periodicity, as in quasicrystals and gas-water clathrates.  Both can have 5- or 10-fold symmetry overall and give sharp x-ray or neutron diffraction spots (the IUC definition of a crystal).


Boron is a covalent solid with high melting point, like diamond (though not quite), and hence its crystals are hard to make. Unlike diamond crystals, they are not nice and probably wouldn't make a great display.

The table on http://periodictable.com/Properties/A/MolarVolume.v.log.html seems to corroborate your findings about boron molar volume being the smallest among all elements. Pity it's wrong, and so are you. (Or rather, it is technically right, but in a way that conveys a wrong impression.) Some elements just tend to have multiple polymorphs (called allotropes in this case), and carbon is one of them. All data in the standard tables are for the standard polymorph, which is graphite. But diamond at $3.5\;\ce{g/cm^3}$ is much denser, and decisively beats boron in the contest for the smallest molar volume.

Sometimes it takes walking around the world to realize that the aim of your quest has been in your pocket all along. The picture of the "smallest mole" is the one you brought here.

So it goes.


Cell potential for redox titration of iron(II) with cerium(IV)

Let's start at the beginning, with a solution where $\ce{[Fe^{2+}}] = 0.01$ M. In order to apply Nernst's law, we also need $\ce{[Fe^{3+}}]$, which is not known, but must be very very low, but cannot be zero. We will see that its exact value is without importance. Let's suppose it is  $\ce{[Fe^{3+}}] = 10^{-6}$ M. The redox potential of the solution is : $$\ce{E(Fe^{3+}/Fe^{2+}) = E°(Fe^{3+}/Fe^{2+}) + 0.059 log\frac{[Fe^{3+}]}{[Fe^{2+}]} = 0.77 + 0.059 log(10^{-4}) = 0.53 V}$$ Now this potential cannot be directly measured. The only thing that can be measured is the difference of potentiel between this electrode and a reference, which is here SCE, at - $0.24$ V. So if you dip a platinum electrode and a SCE electrode in this solution, you will measure a voltage between them to be : $$\ce{E_{cell} = 0.53 V -(- 0.24) V = 0.77 V}$$ This not a reliable value, as it supposes that we know the amount of $\ce{Fe^{3+}}$ in the solution. Anyway, let's go on.
First addition of Cerium(IV).
Now you add $5$ mL $\ce{Ce^{4+} 0.001 M}$, that is $50~\mu$moles. This cerium ion will react with $\ce{Fe^{2+}}$ ions according to the equation $$\ce{Ce^{4+} + Fe^{2+} ->  Ce^{3+} + Fe^{3+}}$$ The initial solution contained $250~ \mu $moles $\ce{Fe^{2+}}$. Adding $~50~\mu$mole $\ce{Ce^{4+}}$ consumes $~50~\mu$mole $\ce{Fe^{2+}}$ and produces $~50~\mu$mole $\ce{Ce^{3+}}$ and $~50~\mu$mole $\ce{Fe^{^3+}}$. The final amount of $\ce{Fe^{2+}}$ is $250 - 50 = 200~ \mu$mol. The potential of this solution is, according to Nernst's law : $$\ce{E(Fe^{3+}/Fe^{2+}) = E°(Fe^{3+}/Fe^{2+}) + 0.059 log\frac{[Fe^{3+}]}{[Fe^{2+}]} = 0.77V + 0.059 log\frac{50}{200} = 0.77V - 0.036 V = 0.734 V} $$ The potential of the whole cell (solution + SCE) is : $$\ce{E_{cell} = 0.41 V - (-0.734 V) = 1.144 V}$$ It is worth the trouble mentioning that here we could also have got the same result using the Cerium potential and Nernst's law. In principle. But, to apply Nernst's law to the Cerium couple we would need to know the residual concentration in $\ce{Ce^{4+}}$ which must be very small but not zero. So we have to use the couple $\ce{Fe^{3+}/Fe^{2+}}$.
Second addition of Cerium(IV).
Here you add $25$ mL $\ce{Ce^{4+} 0.001 M}$, that is $250~\mu$moles. The same reaction happens. But here, the amount of Cerium added is equal to the amount of original Iron in the solution. So at the end, there is in solution  $250~\mu$moles of $\ce{Ce^{3+}}$ and the same amount of $\ce{Fe^{3+}}$. The residual amount of $\ce{Ce^{4+}}$ and $\ce{Fe^{2+}}$ are very small, but unknown.
The potential of the solution must be the average of $\ce{E°(Fe^{3+}/Fe^{2+})}$ and $\ce{E°(Ce^{4+}/Ce^{3+})}$, and this is $(0.77$ V + $1.61$ V)/$2$ = $1.19$ V. The potential of the cell after this $2$nd addition is then $$\ce{E_{cell} = 1.19 V - (-0.24V) = 1.43 V}$$
Third edition of Cerium(IV).
Here there is a huge excess of Cerium(IV). There is exactly twice the amount of cerium(IV) necessary to oxidize all Iron(II). The final solution contains the same amount of both Cerium ions. The potential of the solution cannot be calculated with the Iron couple, as the amount of remaining $\ce{Fe^{2+}}$ is unknown and very low. So the potential of the cell must be calculated by Nernst's law applied to the Cerium couple. As the final amount of both ions $\ce{Ce^{3+}}$ and $\ce{Ce^{4+}}$ are equal, the potential of the solution is equal to the standard redox potential, which is $1.61$ V. And the measured difference of potential between the solution and the SCE is : $$\ce{E_{cell} = 1.61 V - (-0.24 V) = 1.85 V}$$
To summarize all these calculations, the measured potential in the cell (solution + SCE) increases steadily from about $0.77$ V, to $1.14$ V, then to $1.43$ V, then to $1.85$ V.


Trend in the boiling point of the hydrogen halides

There are formulae you can use to calculate the strength of interactions between two molecules, which are derived from first principles. You can find an overview here on Chemistry LibreTexts. The two relevant ones are:

$$U_{\mathrm{dipole}} = -\frac{2}{3kT} \frac{p^4}{(4\pi\varepsilon_0)^2r^6}$$

$$U_{\mathrm{dispersion}} = -\frac{3}{4} \frac{\alpha^2 I}{(4\pi\varepsilon_0)^2r^6}$$

$p$ is the molecular dipole moment
$r$ is the average separation between neighbouring molecules
$\alpha$ is the polarisability of the molecule, which in a nutshell, measures how easy it is to create an induced dipole in the molecule
$I$ is the first ionisation energy of the molecule

If you plug all the numbers in for the hydrogen halides $\ce{HX}$, you will find that the magnitude of the dispersion forces is much larger than the magnitude of the dipole-dipole attractions, i.e. $\left|U_{\mathrm{dispersion}}\right| \gg \left|U_{\mathrm{dipole}}\right|$.

Unfortunately, I don't have all the necessary data on hand, but I can quote some results that were given in first-year lecture notes at Oxford (which should be legitimate). For $\ce{HCl}$, dispersion forces contribute $86\%$ to the intermolecular attractions, and for $\ce{HI}$, they contribute $99\%$.

Considering this fact, it is not surprising that variations in the magnitude of dispersion forces affect the boiling point much more than variations in the magnitude of the dipole-dipole attractions. If you had the all the data required and evaluated the interactions using the two formulae above, you would reach the same conclusion quantitatively.

In fact, this trend is hardly unique to the hydrogen halides. The hydrides of the Group 15 and Group 16 elements behave exactly the same way:

$$\begin{array}{|cc|cc|}
\hline
\text{Species} & \text{Boiling point / }\mathrm{^\circ C} & \text{Species} & \text{Boiling point / }\mathrm{^\circ C} \\
\hline
\ce{PH3} & -87.5 & \ce{H2S} & -60.3 \\
\ce{AsH3} & -62.4 & \ce{H2Se} & -41.3 \\
\ce{SbH3} & -18.4 & \ce{H2Te} & -4 \\
\hline
\end{array}$$
(source: Greenwood & Earnshaw, Chemistry of the Elements 2nd ed., pp 557, 767)

Of course, the first-row hydrides are left out of the discussion because of hydrogen bonding, which makes their boiling points anomalously high.


I can think of two reasons:
1) Steric repulsion
2) Electrostatic interaction

Steric repulsion: cis means in same side. So, sterically it is suitable to be on opposite side. It is a fundamental nature of both human and chemical entity to define a personal boundary (My personal opinion :) )
In the image you can see the personal boundary of $\ce{-CH_3}$ is the furthest in the trans configuration and that's wh the trans configuration is more stable than the cis configuration.

Electrostatic interaction: Due to electrostatic interaction, molecules or functional group try to distance themselve as far as possible. This also results in lower energy in trans configuration.


Metallic lead is very low risk.

Lead compounds are fairly poisonous: they slowly build up in the body and cause many harmful effects. But lead metal is very inert and you would need to do something fairly risky with it to create much likleihood of generating dangerous lead compounds.

Don't eat it, for example. Don't put it in contact with food (especially acidic foods which can solubilise lead). But normal handling should not be a problem.

Lead is so inert that it has been used to pipe clean water supplies to people for most of history (or at least the parts of history where somebody had access to clean water). This is safe unless the water supply is particularly acidic and leaches lead from the pipes.

In short: don't worry about lumps of lead.


How is the selective titration testing equation simplified?

I finally found the answer to the question I have made a huge mistake in the testing equation
In order for a titration to happen the Formation constant $\ce{K}$ should be greater than or equal to $\ce{10^7}$
What I have written in my note book is

$\ce{K''_{MY^{n-4}}} = \ce{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}-α_{Y^{'}}}$

Which is incorrect it,
Should be like this

$\ce{K''_{MY^{n-4}}} = \ce{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}α_{Y^{'}}}$

The conditional formation constant of metal $\ce{M}$ should be multiplied by $\ce{α_{Y^{'}}}$ not subtracted
Substituting the equation for $\ce{α_{Y^{'}}}$ will give the simplified equation
$$\ce{α_{Y^{′}}=  \frac{1}{(1+K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}])}}$$
$$\ce{K''_{MY^{n-4}} =  \frac{K_{MY^{n-4}}α_{M^{n+}}α_{Y^{4-}}}{(1+K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}])}} $$
But since
$$\ce{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] >> 1 }$$
$$\ce{K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] + 1 \approx K_{NY^{m-4}}α_{N^{m+}}α_{Y^{4-}}[N_{total}] }$$
Which gives the following final simpified experssion
$$\ce{K''_{MY^{n-4}} =  \frac{K_{MY^{n-4}}α_{M^{n+}}}{K_{NY^{m-4}}α_{N^{m+}}[N_{total}]}} \geq 10^7 $$
Finally the testing equation is sloved for selective titrations


How to create black precipitate in clear liquid

I can think of iron(II) sulfide($\ce{FeS}$). There are many ways to make iron(II) sulfide but one reaction procedure will suit your purpose. It uses iron(II) sulfate and lime sulfur(calcium polysulfide) to form a clear black precipitate of iron(II) sulfide with a clear supernatent. See this e-book excerpt.

You can use other sulfides as well. See this picture (full picture here):

From left to right: silver sulfide ($\ce{Ag2S}$, black), lead sulfide ($\ce{PbS}$, black), nickel sulfide ($\ce{NiS}$, black-brown), copper sulfide ($\ce{CuS}$, dark indigo). These are formed by reacting $\ce{Na2S}$ with corresponding metal ions. If you want to fully make your solution black, then your options would be to form silver sulfide and nickel sulfide. You could also make your solution black by forming elemental mercury solution but that would be too toxic for your purpose.

Other ways include adding iodide to excess amount of bismuth(III) nitrate  in dilute nitric acid but be careful not to overdo as excess concentration of iodide will make solution orange-red(See here).

If you do not want any reaction to happen then just dump manganese(IV) dioxide in water. They are black and insoluble.


Little bit tricky question. As you say, the work is force acting along the path. But there is no force acting, as there is weightless piston with vacuum on the opposite side.

To visualize the process better, imagine an inflatable balloon in outer space, when you puncture it. All gas molecules continue to fly in the direction and speed they had in the moment of accident (therefore the temperature is the same) and they exert no work (no collision thereafter).


Averaging of dipolar coupling in solution due to tumbling

The $\sin(\theta)\,\mathrm{d}\theta$ comes from the surface element of a sphere of unit radius where $\theta$ is the polar angle measured from the north pole, as it were, to the equator. The surface element is $\mathrm{d}s = \sin(\theta)\,\mathrm{d}\theta\,\mathrm{d}\phi$ where $\phi$ is the azimuthal angle measured is if longitude, i.e around the equator. The integral is
$$\frac{1}{4\pi}\iint f(\theta)\sin(\theta)\,\mathrm{d}\theta\,\mathrm{d}\phi$$
where $f(\theta)$ is the function you want to average.

The sketch shows the idea


Lewis structures of molecules containing oxygen

If you consider the $\ce{ClO2-}$ as coming together from neutral atoms (plus the extra electron), all the bonding electrons "come from" the chlorine. This is commonly called a 'dative covalent bond' (or just 'dative bond' for short), and is sometimes drawn like this, instead of just a line:

As this image suggests, you can think of it as an atom with a lone pair bonding with an atom with no spare valence electron.


How to determine bond length and angle in a conjugated pi-system (polymethine)?

So what are the average bond lengths and bond angles in this molecule
  and how are they calculated?

1,3-butadiene should serve as a reasonable model for your work.

We can find the bond lengths in the literature.  The carbon-carbon double bond length is 1.338 Å, typical for a double bond (reference).  The carbon-carbon single bond length is 1.454 Å, shorter than expected due to resonance.  All of the carbons are $\ce{sp^2}$ hybridized, which means that all of the bond angles should be ~120°.

Using this information and vector addition you can determine the length of any polymethine.


Which one is the best notation for sodium acetate

There is nothing wrong with either formula. And you can use even more:

$\ce{NaC2H3O2}$
$\ce{C2H3NaO2}$

It really depends on which point you want to bring across.

$$\ce{NaCH3COO}$$

This formula, being analogous to formulae like $\ce{NaCl}$ stresses the inorganic salt view more. It shows that there is a cation ($\ce{Na+}$) and an anion ($\ce{CH3COO-}$) which form a salt crystal together much like the anion $\ce{Cl-}$ would do with the same cation. Inorganic nomenclature prefers cations to be written first.

$$\ce{NaC2H3O2}$$

Is basically the same except for saying ‘I don’t care how those seven atoms combine to form the anion, all I care for is what’s in there.’

$$\ce{CH3COONa}$$

This one stresses a more organic-chemical point of view where it’s relevant where the cation is actually bound to within the molecule. Oftentimes, organic chemists would even write structural formulae with a bond between oxygen and sodium as if it were hydrogen. The reasoning behind this is ‘I don’t care if the structure ends up being a salt, all I need to know is that the sodium somehow connects to the carboxyl group.’

Note that you can combine the first’s and the third’s idea to give $\ce{NaOOCCH3}$ which works, but is a lot less used than either of them.

$$\ce{C2H3NaO2}$$

This is the Beilstein-type lookup formula. You just know the elemental composition of your compound and now want to look it up — you can’t know whether it’s sodium acetate or sodium hydroxyethanalate (if that name is even correct). Or maybe you just don’t care. The reasoning behind this is $\ce{C}$ first, $\ce{H}$ second, the remaining elements following in alphabetic order.

Finally, there are also shortened formulae. The most common (the one I actually use most) would be $\ce{NaOAc}$. Here, $\ce{Ac}$ is an abbreviation for $\ce{H3C-C=O}$ where whatever follows or precedes is bound to the carbonyl carbon. Note that this formula is not considered standard, and would need to be included in a list of abbreviations or defined otherwise if you decide to use it.


How to find the band gap for a compound?

There are two ways in which the band gap of a material: using optical spectroscopy and using electrochemistry. The way you go about analyzing the data to determine the band gap is independent of material, but the process of collecting the data is very specific to your material, and you should consult papers where others have made similar measurements on similar systems.

Optical spectroscopy.

You need to collect an absorbance spectrum of your material. Given what I can gather from your fabrication description, you will probably need to collect said spectrum in the solid state. The exact means of doing that vary somewhat by spectrometer model. Find someone who knows your spectrometer very well and ask them how to do it (or read the manual).

The absorbance process measures the energy difference between the ground state and the excited state as electromagnetic radiation is absorbed. Since there are a number of vibrational states associated with both the ground state and the excited state, the absorbance spectrum appears as a blob instead of a nice line. The "true" band gap, or $E_{00}$, the energy difference between the lowest vibrational states of the ground and excited states, is difficult to measure. However, the most commonly accepted approximation is the onset of absorption from the low energy side. The onset value is the intersection of the extrapolations of the linear parts of the spectrum. See the image below.

Once you have the wavelength (x) at that point, you can convert it into energy using the Planck relation:
$$E(\text{J})=\dfrac{hc}{\lambda(\text{m})}$$
If you want your energy in electronvolts, you can convert directly:
$$E(\text{eV})=\dfrac{1239.84187 \text{ eV nm}}{\lambda(\text{nm})}$$

Electrochemistry, specifically voltammetry

Voltammetry gives you the ability to determine the peak oxidation potential, $V_{\text{ox}}$, and the peak reduction potential, $V_{\text{red}}$, of your material. Both cyclic voltammetry and differential pulse voltammetry can give you these data, and both should work on solids, though you may want to find someone who has experience in such things.

The assumption that we operate under when using electrochemistry to determine HOMO-LUMO or bad gap, is that when we oxidize a molecule or material, we take the first electron out of the HOMO (or valence band). When we reduce the molecule or material, we put an electron into the LUMO (or conductance band). Thus, $V_{\text{ox}}$ tells us something about the HOMO (valence band) and $V_{\text{red}}$ tells us something about the LUMO (conductance band). The energy gap (in electronvolts) is simply the difference between the two potentials (in volts) multiplied the charge on the electron $e=-1 \text{ eV/V}$.
$$E(\text{eV})=e(V_{\text{red}}-V_{\text{ox}})$$


Dissenter is correct about aromatic compounds and that the basic requirement for a compound to be detected by smell is for it to be volatile enough to reach the nose. This is why things like dry salt don't really smell like anything. Once they reach the nose, things become more complex.

Why many esters smell fruity while most thiols smell rather unpleasant is a rather interesting question. Just as most sugars taste sweet, many molecules that have structural similarities also smell similar. Humans have about 350 different olfactory receptors. These are G protein-coupled receptors that change conformation when they bind a compound and activate a signal pathway that is eventually sent to the brain for processing. However, since we only have 350 different receptors and there are far more than 350 odoriferous compounds, there is not a one-to-one mapping of receptors to compounds—there isn't an "acetone receptor". Instead, olfactory receptors have partial and overlapping selectivity. Each receptor type responds to more than one compound, thiols with a certain general shape, for instance, and different receptor types can respond to some of the same compounds. A given compound then has a more or less unique pattern of receptor activation. All this information gets sent to the brain and is processed into the perception of smell. Compounds that have similar structural properties activate many of the same receptors, so they smell similar.


Predict the Product(s) of a reaction of tertiary benzylic alkyl halide

$\ce{CN-}$ is a significantly stronger nucleophile than $\ce{EtOH}$ and will therefore primarily displace the chloride. The facts that substitution takes place at a tertiary benzylic carbon, and that a polar protic solvent (ethanol) is used indicate that the reaction follows the $\ce{S_{N}1}$ mechanism (sources: 1, 2, 3). Due to the achiral intermediate (carbocation), the nitrile will be obtained as a racemic mixture, i.e. 2 products.

Elimination is also a possible reaction pathway which needs to be considered, since cyanide is a good nucleophile but a weak base, which favors an E2 reaction unless an excess of $\ce{CN-}$ is present. In this case, the product with the more substituted double bond should be favored.


Balancing redox reaction of zinc and hot diluted nitric acid

The oxidation state of $\ce{Zn}$ as calculated by you is incorrect. The correct oxidation state of $\ce{Zn}$ is $+2$ as it is associated with two mono negatively charged $\ce{NO3-}$ ions.
Thus the unbalanced reaction with correct oxidation states would be:
$$\ce{\overset{0}{Zn} + \overset{+1}{H}\overset{+5}{N}\overset{-2}{O_3}\longrightarrow \overset{+2}{Zn}(\overset{+5}{N}\overset{-2}{O_3})2 +\overset{-3}{N}\overset{+1}{H_4}\overset{+5}{N}\overset{-2}{O_3} +\overset{+1}{H2}\overset{-2}{O}}
$$
Now the half-cell reactions are as follows:
$$
\ce{Zn -> Zn^{2+} + 2e-}\\
\ce{NO3- + 10H+ + 8e- -> NH4+ + 3H2O}\\
$$
Thus the balanced iconic reaction would be:
$$\ce{4Zn + NO3- + 10H+ -> 4Zn^{2+} + NH4+ + 3H2O}\\\\$$
Now adding nine $\ce{NO3-}$ spectator ions to the ionic reaction gives us the standard molecular reaction.
$$\ce{4Zn + 10HNO3 -> 4Zn(NO3)2 + NH4NO3 + 3H2O}\\\\$$
What did you learn from this?
Always try to write the half-cell reaction in ionic form and only involve the ions which undergo oxidation state change. Don't bother yourself with the spectator ions at first. After you balance the equation in an ionic  form adjust it to the molecular form by adding relevantly needed ions. Identify the oxidizing and the reducing species is the key.


Thermal decomposition of calcium propionate

There is a paper (1) where TG-GLC-MS with subsequent XRD analysis of solid products has been utilized for the analysis of evolved gases during the decomposition of calcium propionate at various heating rates. Detected products were $\ce{H2O}$, $\ce{CO2}$, $\ce{C2H5CHO}$, $\ce{C2H5COCH3}$, $\ce{C2H5COC2H5}$, $\ce{CaCO3}$.

The Discussion section of the paper suggests the following mechanism of formation:

  Following the free radical mechanism of Hites and Biemann (2), proposed for
  the decomposition of the calcium salts of monocarboxylic acids we suggest that the initial stage in the decomposition of calcium propanoate can be represented as:

  $$\ce{Ca(C2H5COO)2 -> CaCO3 + C2H^.5 + C2H5C^.=O,}$$

  Combination of the two free radicals results in the formation of the major product, 3-pentanone:

  $$\ce{C2H5C=O + C2H^.5 -> C2H5 - CO - C2H5}$$

  The other trace products observed, 2-butanone and propanal, can be accounted for by disproportionation reactions involving the ethyl radical:

  $$\ce{C2H^.5 + C2H^.5 -> C3H^.7 + CH^.3}$$

  and

  $$\ce{C2H^.5 -> C2H5 + H^.}$$

  The methyl and hydrogen radicals so produced may react with the large excess of the original propanoyl radicals to give 2-butanone and propanal respectively.

  $$\ce{C2H^.5=O + CH^.3 -> C2H5COCH3}$$

  and

  $$\ce{C2H5C=O + H -> C2H5CHO}$$

  The $\ce{C3H7}$ may combine with any of the other radicals or may disproportionate. In any event there was no evidence for compounds larger than 3-propanone. However on close examination of the GC curves there were indications of a very small peak before that due to $\ce{CO2}$ which could be attributed to low molecular weight alkanes and alkenes.

(1) Barnes, P. A.; Stephenson, G.; Warrington, S. B. Journal of Thermal Analysis 1982, 25 (2), 299–311. DOI: 10.1007/BF01912955

(2) Hites, R. A.; Biemann, K. J. Am. Chem. Soc. 1972, 94 (16), 5772–5777. DOI: 10.1021/ja00771a039


Do symmetric hydrogen bonds in neutral molecules exist?

It appears that perhaps water may have symmetric hydrogen bonds between oxygen atoms, but it would take at least 60 GPa of pressure to make water to bond like that.

Article named very descriptively "Compression of Ice to 210 gigapascals: Infrared Evidence for a Symmetric Hydrogen-Bonded Phase" proves it:

  Protonated and deuterated ices ($\ce{H2O}$ and $\ce{D2O}$) compressed to a maximum pressure of 210 gigapascals at 85 to 300 kelvin exhibit a phase transition at 60 gigapascals in $\ce{H2O}$ ice (70 gigapascals in $\ce{D2O}$ ice) on the basis of their infrared reflectance spectra determined with synchrotron radiation. The transition is characterized by soft-mode behavior of the $\nu_3$ $\ce{O-H}$ or $\ce{O-D}$ stretch below the transition, followed by a hardening (positive pressure shift) above it. This behavior is interpreted as the transformation of ice phase VII to a structure with symmetric hydrogen bonds.

On the other hand, article "The Hydrogen Bond in the Solid State" mentions:

  First example of an $\ce{O-H-N}$ hydrogen bond with a centered position of
  the proton: pentachlorophenol/4-methylpyridine at 100 K characterized
  by neutron diffraction studies (...)

Unfortunately:

  Even very small chemical changes in the system lead to loss of the
  symmetry (...)

so symmetric hydrogen bonds can probably exist also in very low temperatures.

References

Goncharov, A F; Struzhkin, V V; Somayazulu, M S; Hemley, R J; Mao, H K. Science 273.5272

Steiner, Thomas Angew. Chem. Int. Ed. 41 1 71


How to develop a retrosynthesis plan for 2-[(E)-2-(3-bromophenyl)ethenyl]-1,3-benzoxazole?

Extempore, this should be available by the reaction of 3-bromocinnamaldehyde with 2-aminophenol. Samarium triflate as a catalyst might be a good idea.

In any case, I'd start the retrosynthesis at the benzoxazole moiety.


test to differentiate between NaCl and zinc hydroxide

You can use the large difference in water solubility to determine whether the residue is sodium chloride or zinc hydroxide. Sodium chloride is very soluble in water and zinc hydroxide is insoluble in water. You can do the test as follows:

1) Brush or scrape a portion of the residue onto a piece of paper or card stock.
2) Transfer the residue onto a dark colored plate or bowl.
3) Add enough water to cover up the residue and swirl gently.

Any white solid that remains should be zinc hydroxide as the sodium chloride will dissolve away.


Defining an acid-base reaction from the perspective of the Arrhenius theory of acids and bases

I understand that ammonia counts as a base because it can react with water to give hydroxide ions, and HCl is an acid.

This is where you went wrong.
The Arrhenius theory is the oldest of the common acid-base theories. According to Arrhenius:

An acid is a compound that will dissociate in aquaeous solution forming $\ce{H+}$ ions.
A base is a compound that will dissociate in aquaeous solution forming $\ce{OH-}$ ions.

As per the Arrhenius theory, ammonia is not a base. Also, contrary to all later theories (Brønsted-Lowry, Lewis and Pearson), Arrhenius did not account for any acid-base reactions; instead, he intended to define only those substances whose solutions reacted acidicly or basicly. It is therefore meaningless to define an acid-base reaction according to the Arrhenius theory.
What Arrhenius did account for was neutralisation reactions. Indeed, if an Arrhenius acid and an Arrhenius base are mixed, the form water and a salt, which can be easily shown in the equation below:
$$\ce{(H+ + A^-) + (B+ + OH^-) -> HOH + A^+B^-}$$
But it was not until Brønsted’s and Lowry’s theory that it was realised how a reaction should always be acid-base and not one or the other.


Well, guess what, even I study the exact same textbook (class 11 NCERT, from India). I had asked my teacher the exact same doubt, and the answer was:

Due to its proximity with neighbouring substances, like Arsenic (which was already discovered by then - see here), Mendeleev was able to accurately calculate the average values of density and melting point of Germanium, or eka-Silicon.
The formula of its Oxide and Chloride was predicted with respect to the formula of the Oxides and Chlorides formed by other elements of the group IVA (as per old IUPAC numbering), such as $\ce{CO2}$, $\ce{SiO2}$, $\ce{CCl4}$ and $\ce{SiCl4}$. Their densities were calculated using the knowledge of Chlorine's and Oxygen's atomic weight, and the predicted density and/or atomic mass of the new element.

Although I don't know how the boiling point of a substance can be predicted, my best guess for it is that it was approximated using the boiling point of other similar compounds near it in the Periodic table. For more information on how Boiling point can be possibly approximated, here is a link.


What will be the name of CH2=CH-O-CO-CH3?

Most people would call it vinyl acetate. If you want the systematic name, then it is ethenyl ethanoate.

It is formally derived from "vinyl alcohol" and acetic acid, from which the name follows easily. See here (https://en.wikipedia.org/wiki/Ester#IUPAC_nomenclature), for example.

The compound is https://en.wikipedia.org/wiki/Vinyl_acetate.


How does the activity of Sn chemical potential vary as a function of composition at 900 C?

Just above the line, in the two-phase field where $\epsilon$ is in equilibrium with the liquid, the chemical potentials of both Cu and Sn are constant (very likely not equal to each other mind you).  This is a consequence of the tie-line on the Gibbs free energy diagram - check out a book on binary alloy thermodynamics such as Porter and Easterling, Phase Transformations in Metals and Alloys.

At the peritectic point itself, the tie line connects all three phases, $\epsilon$, $\eta$ and the liquid, so the chemical potentials for each element are equal. Dropping the temperature just a smidge leads to two separate two-phase regions, one of $\epsilon$ with $\eta$, and one with $\eta$ and liquid.  The chemical potential of Sn (or Cu) is different in each of those two-phase regions.

Again, Porter and Easterling would be a good reference.


Aqueous carbonate solutions contain four different solute species:

$\ce{CO2(aq)}$
$\ce{H2CO3}$
$\ce{HCO3-}$
$\ce{CO3^2-}$

The equilibrium for the hydration reaction

$$\ce{H2O + CO2(aq) <=> H2CO3}\tag1$$

lies to the left. The corresponding equilibrium constant $K$ at $25\ \mathrm{^\circ C}$ is about

$$K=\frac{\left[\ce{CO2(aq)}\right]}{\left[\ce{H2CO3}\right]}\approx650$$

(literature values of $K$ range from $350$ to $990$.)

The commonly used first acid dissociation constant of carbonic acid $\mathrm pK_{\mathrm a1}=6.35$ (at $25\ \mathrm{^\circ C}$) actually is a composite constant that includes both the hydration reaction $\text{(1)}$ and the protolysis of true $\ce{H2CO3}$ $\text{(2)}$.

$$\ce{H2CO3 <=> H+ + HCO3-}\tag2$$

The constants $K$, $K_{\mathrm a1}$, and the actual acid dissociation constant $K_{\ce{H2CO3}}$ of true $\ce{H2CO3}$ are interrelated according to

$$K_{\mathrm a1}=\frac{K_{\ce{H2CO3}}}{1+K}\approx\frac{K_{\ce{H2CO3}}}{K}$$

Calculated values for $\mathrm pK_{\ce{H2CO3}}$ range from $3.8$ to $3.4$. Thus, true $\ce{H2CO3}$ is stronger than acetic acid and formic acid.

The second acid dissociation constant of carbonic acid is $\mathrm pK_{\mathrm a2}=10.33$ (at $25\ \mathrm{^\circ C}$).

Therefore, the predominant species in aqueous carbonate solutions (at $25\ \mathrm{^\circ C}$) are

$\ce{CO3^2-}$ at $10.33\lt\mathrm{pH}$
$\ce{HCO3-}$ at $6.35\lt\mathrm{pH}\lt10.33$
$\ce{CO2(aq)}$ and $\ce{H2CO3}$ at $\mathrm{pH}\lt6.35$

Thus, true $\ce{H2CO3}$ primarily exists in acidic solutions. However, because of the hydration equilibrium $\text{(1)}$ and the corresponding equilibrium constant $K\approx650$, most of the true $\ce{H2CO3}$ is converted to $\ce{CO2(aq)}$.

(All values taken from: Stumm, W.; Morgan, J. J. Aquatic Chemistry, Third Edition; John Wiley & Sons: New York, NY, 1996; pp 150–152.)


Most favorable position of attack of an electrophile on 4-trifluoromethyl-1,2-diphenylbenzene

TL;DR - The ring with positions 1, 2, and 6 on it is more deactivated than the ring with positions 3, 4, and 5, so the correct answer cannot be position 2 (as marked) or positions 1 and 6 (as you determined). The correct answer must be on the other ring. Since the central ring imposes a -I effect, the correct answer is position 4.

There are two factors to consider in this question. The first is that the trifluoromethyl group ($\ce{CF3}$) is electron-withdrawing by induction. Fluorine is the most electronegative element, and the $\ce{C-F}$ bond is polarized toward $\ce{F}$. The $\ce{CF3}$ group has a -I effect on the ring, and that effect is strongest at ortho and para.
The second is that the three rings in this structure are not coplanar. The steric interactions between hydrogen atoms on the two rings with labeled positions in your image causes the rings to twist out of coplanarity. This feature of the structure means that the central ring cannot exert a resonance effect on either of the other rings. The central ring will is electron poor thanks to the $\ce{CF3}$ group, and so it will provide a -I effect to the other two rings. Since the -I effect on the central ring is stronger at the position para to the $\ce{CF3}$ group than at the position meta, the ring with positions 1, 2, and 6 will be more deactivated than the ring with positions 3, 4, and 5.

Since the lower ring is experiencing a -I effect, electrophilic attack at meta is most likely, so position 4 is the true correct answer.


Why does acetate aldol with Evans auxiliary give almost no diasteroselectivity?

1. The acetate aldol reaction
Before looking at selective variants, its worth pointing out what an acetate aldol is.
An acetate aldol (below), is the simplest kind of aldol in which the β-hydroxyketone product has no methyl group at the α-position to the carbonyl. This is in contrast to the (more usual?) propionate aldol reaction in which the α-position contains a methyl group (as is common in polyketides).

Source: Modern methods in stereoselective aldol reactions, Wiley.

Early studies into acetate aldols quickly established that it was far harder to set a single stereo centre than to introduce two stereocentres concomitantly. This was due to the cyclic transition states involved (Zimmerman-Traxler model) in which the methyl group bound to the enolate (in the propionate aldol) exhibits a level of control which isn't present in the acetate aldol.
2. The Evans' auxiliary and its applications in the aldol reaction
Whilst many aldol reactions exhibit diastereoselectivity, they show no inherent enantioselectivity. Several approaches were developed to fix this including chiral auxiliaries and chiral reagents. One of the most 'famous' of these was Dave Evans' oxazolidinone auxiliaries.
These are used to gain enantioselectivity, but can also be used to override substrate control to get the desired product.

Source: Dave Evans' CHEM206 Lecture notes, Harvard

With propionate aldols, this works excellently (see above), giving d.r.'s well in excess of 20:1 (about the maximum NMR can detect), with a 300:1 d.r. being quoted by Evans'. With acetate aldols however, the selectivity drops down to 1:1.
Let us consider the TS for the propionate aldol.

Source: Dave Evans' CHEM206 Lecture notes, Harvard

Evans' himself discusses the two diastereomeric TS structures. The favoured on top we now rationalise by opposing the dipoles between the auxiliary and enolate.
An alternate reason for why the disfavoured TS was disfavoured was provided by computational work, in which it appears that the methyl group in the propionate system clashes with the auxiliary.

Source: Evans' and Houk.

By disfavouring one TS (Steric clash) and favouring another (opposing dipoles), we gain good diastereoselectivity.
When the α-methyl group is taken away (as it is in the acetate aldol), we lose the unfavourable interaction (no more steric clash between the enolate and the auxiliary) leaving us only with the favourable dipole-dipole minimisation.
There is perhaps also another argument here which is that without the α-methyl group the enolate is considerably smaller, meaning that the TS doesn't even need to be quite so organised in order for the reaction to proceed.
3. Selective acetate aldols
As a closing point, switching from boron --> tin, and using a slightly modified auxiliary does allow for selectivity to be achieved.

Source: Unknown

This works because the sulfur and tin remain coordinated throughout the reaction rather than with the Evans' chemistry where the oxygen points into space to allow the dipoles to oppose. This results in a much more ordered TS.


Why is negative charge more stable on sulphur than on oxygen?

There is more to it than meets the electronegative eye.
Get into the habit of transferring attributes into energy.  In this case the relevant energies are the electron affinity of the base money ($\ce{RO}$ vs $\ce{RS}$) versus the bond energy of the moiety to hydrogen.  A strong anionic Bronsted-Lowry base will have a low value for the first versus a high value for the latter.
Surprise!  Sulfur has a higher electron affinity than oxygen!  Add the fact that the oxygen-hydrogen bond is stronger than the hydrogen-sulfur bond and by both measures, $\ce{RO^-}$ appears as the stronger base.
The higher electron affinity of sulfur and even selenium, compared with oxygen, has another curious effect.  We know about the aromaticity of certain ring ketones in which the electronegative oxygen atom imparts a positive charge to a three- or seven-membered carbon ring.  It turns out that despite what an elctronegativity table might suggest, the higher electron affinity of sulfur and selenium versus oxygen causes this characteristic to carry over to the corresponding $\ce{C=S}$ and $\ce{C=Se}$ compounds.


I am not sure if Silane ($\ce{SiH_4}$) can really be considered.

But there are silicides which silicon forms with strongly electropositive metals. In these compounds, silicon has a negative oxidation state.

For magnesium silicide - $\ce{Mg_2Si}$, the oxidation state of silicon would be -4.


The reaction you are proposing is very unlikely to happen under standard conditions. Taken average bond energies from UC Davis ChemWiki,[1] the approximated heat of reaction would be $\Delta_\mathrm{R}H = 307~\mathrm{kJ\,mol^{-1}}$.
$$\ce{HCl + HF <=> H2 + FCl}$$

The reaction mechanism you are proposing is also not quite feasible. If we would ignore the energetics of this reaction, and we speak hypothetical about this reaction, we would still have to consider an activation step. What we also need to consider is the high affinity of fluorine towards electrons and therefore have to exclude anything that would enhance the heterolytic bond cleavage of hydrogen fluoride. Hence this reaction can only take place in gas phase in an inert atmosphere.
A feasible starting point would be the thermolytical or photolytical cleavage of one of the bonds.
$$\begin{align}
\ce{HCl &->[\Delta T/ h\nu] H + Cl}\\
\ce{HF  &->[\Delta T/ h\nu] H + F} \\
\end{align}$$

Then we would need to consider propagation of the reaction.
$$\begin{align}
\ce{HCl + H/Cl/F &->{} [ HCl.(H/Cl/F)]^{\ddagger} &&
-> H(H/Cl/F)/Cl(H/Cl/F) + H/Cl}\\
\ce{HF + H/Cl/F &->{} [ HF.(H/Cl/F)]^{\ddagger} &&
-> H(H/Cl/F)/F(H/Cl/F) + H/F}\\
\end{align}$$
And then finally recombination of the remaining free radicals.

We are basically looking at the reverse mechanism of the reaction of hydrogen and chlorine.[2]

The practical approach is given in orthocresol's answer. Interhalgen compounds, i.e. halogen halides, are formed by reacting the pure halogen gasses. For the reaction of chlorine and fluorine we find the approximate enthalpy of reaction to be $\Delta_\mathrm{R}H = -113~\mathrm{kJ\,mol^{-1}}$.
$$\ce{F2 + Cl2 <=> 2FCl}$$
Side products also include $\ce{ClF3}$ and $\ce{ClF5}$. The reaction itself is exothermic and ignoring entropy should proceed spontaneously.
$$\begin{align}
\ce{Cl2 &->[h\nu] 2Cl}    &&\text{(start)}\\
\ce{Cl + F2 &-> ClF + F}  &&\text{(propagation)}\\
\ce{F + Cl2 &-> ClF + Cl}\\
\ce{F + Cl &-> ClF}       &&\text{(stop)}\\
\end{align}$$

All reactants are highly reactive and may decompose explosively when other reactants are involved. Even in diluted systems these compounds react with almot everything, see also Why does free chlorine in the stratosphere lose its ozone-depleting potential after about 100,000 reactions?

Conclusion

The formation of interhalogen compounds via their hydrogen halides is energetically not feasible. They can be synthesised from their pure compounds under moderate conditions.
The mechanisms for these reactions can only develop in a closed system as radical chains, which need an initiation step.

Appendix

Average bond energies (Bond dissociation energies; BDE) from UC Davis ChemWiki:
$$\displaystyle\begin{array}{lr}
&\text{BDE}/\mathrm{[kJ\,mol^{-1}]}\\\hline
\ce{HCl} & 427\\
\ce{HF}  & 565\\
\ce{H2}  & 432\\
\ce{FCl} & 253\\\hline
\ce{F2}  & 154\\
\ce{Cl2} & 239\\\hline
\end{array}$$
A. L. Marshall, J. Phys. Chem. 1925, 29 (7), 842–852.
Carl M. Furgason and John W. Moore, J. Chem. Educ. 1943, 20 (1), 41.


Is this Fischer Projection of (3R, 4S)-3-ethyl-3,4-dimethyl-1-hexene wrong?

Is there a mistake at the upper stereocenter?

Yes you are thinking correct, the R configuration will be the diastereomeric pair of this compund, the second chiral carbon is in correct configuration. Don't you want to ask why?


Is a Methoxy group meta directing in this problem?

Now that the general consensus of the Commenters is that meta cyclization is viable and that the answer is (d) and not (a), the OP and other readers should be concerned as to how an (E)-double bond can cyclize in a Friedel-Crafts (F-C) reaction. It doesn't! Prior isomerization to the (Z)-configuration is required.
(E)-Acyl chloride 1 must rotate about the red β-γ single bond and $\ce{AlCl3}$ must complex with the carbonyl oxygen forming species 2. [One may also invoke $\ce{AlCl3}$ complexation with the chlorine and rationalize the subsequent steps via an acylium ion and conjugated ketenes. The principle is the same.] Loss of a proton at the γ-position of conformation 2 provides the β,γ-(Z)-double bond of 3. Now deconjugated acyl chloride 4 is ready for F-C cyclization to intermediate 7 and onto 7-methoxynaphthalen-1-ol (9), aka (d). Alternatively, rotation about the red bond in 3 leads to conformation 5, which, upon γ-protonation leads to α,β-(Z)-conjugated acyl chloride 6. Subsequent F-C cyclization via intermediate 8 affords 9. Incidently, base is not required to produce 9 from 7 or 8. F-C conditions are sufficient.


In the usual substitutive nomenclature, simple halogen compounds are always expressed by prefixes (‘bromo’, ‘chloro’, etc.). Hence, the senior parent structure of such compounds is not determined by a principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name).

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the ring or the chain can be the senior parent structure; however, for the preferred IUPAC name, the ring is always selected as the senior parent structure.

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, the benzene ring is selected as the senior parent stucture for the preferred IUPAC name of the compound given in the question, which leads to ‘1-chloro-4-isopropylbenzene’. However, according to current recommendations, the preferred prefix for the isopropyl substituent is ‘propan-2-yl’; the prefix ‘isopropyl’ is retained for use in general nomenclature. Thus, the complete preferred IUPAC name is ‘1-chloro-4-(propan-2-yl)benzene’.


Singlet carbene is sp2 hybridised whereas triplet carbene is sp Hybridised. More s character, more nuclear force, more Electro negativity so more acidic hydrogen and hence less basicity.

So singlet carbene is more basic.


Calculating the pH of the endpoint in a titration of weak acid and strong base

Would it be correct to assume that the concentration of $\ce{CH3COO-}$ to be the concentration of ethanoic acid that was present at the start of the titration ($\mathrm{0.100~mol~dm^{-3}}$), given that at the equivalence point, all $\ce{CH3COOH}$ had reacted to form a solution containing only the conjugate base $\ce{CH3COO-}$?

You can easily rearrange the $\text{K}_\text{a}$ equation to solve for the information that you need to answer that question.

$$\text{K}_\text{a} =\frac{\ce{[H^+][A^-]}}{\ce{[HA]}} $$
or
$$\frac{\ce{[HA]}}{\ce{[A^-]}} = \frac{\ce{[H^+]}}{\text{K}_\text{a}} = 1.06\times10^{-4}$$

Since $1.06\times10^{-4}$ is insignificant compared to 0.100 you can assume that essentially all the HA is converted to $\ce{A^-}$.

NOTE: This does not mean that $\ce{[HA] = 0}$


Electrochemical characterization of benzenesulfonamide

I'd change the solvent to acetonitrile (don't forget the quarternary ammonium salt) and give a glassy carbon electrode a try.

It should be possible to observe the (irreversible) reduction of the sulfonamide.

And once you have optimized the conditions, please perform a final run with ferrocene as an internal reference.

Edit:
Let me explain the last suggestion. The literature is full of electrochemical data of organic compounds measured in non-aqueous solvents, mostly acetonitrile.

It was once common (bad) practise to report the potentials vs SCE (saturated calomel electrode), although the latter, an aqueous(!) reference electrode, was atually never used in the measurement. Often rather exotic conditions plus a shaky conversion of data turned good experimental work into a pile of data of limited value.

Using an internal standard is the obvious solution to obtain solvent-independent electrochemical data for publication.

Note that the use of ferrocene as such a standard isn't my idea! It was originally proposed by GRITZER and KUTA in Pure Appl. Chem. and is a IUPAC recommendation nowadays.


First, using the smell test for chemistry is dangerous - don't do it!

Second, for this an similar questions, consult a Safety Data Sheet (ex-MSDS) for exposure limits.

The copy that I looked at indicates:

TWA of 25ppm (from the ACGIH) or 50ppm (OSHA) -- these are based on a 8 hour/day, 40 hour/week exposure.

STEL of 35ppm (short term exposure limit). Exceeding the STEL may result in 'upper respiratory tract irritation, eye damage'.

Now, the yield from your lab bench chemistry experiment is unlikely to be very high, and as you may have noticed ammonia is very noticeable (although of the several SDSs that I consulted, none list an odor thershold, surprisingly).

But, the bottom line is, do not inhale your chemistry experiments.


What are the factors controlling the fate of post-adhesed volatile molecules of the olfactory epithelium?

Simply, the odorant particles/molecules never actually enter the cell through endocytosis. Instead, these olfactory receptors, belonging to the G-protein-coupled receptor family, are transmembrane receptors whose structure passes through the cellular wall (several times actually) to functional complexes on the other side, and they use simple and reversible ligand binding as their primary activation mechanism.

Odorant particles are dissolved in the mucous lining the nasal cavity, and in this form come into contact with the receptors in the cell membrane of the epithelial cells. These bind as ligand complexes to receptor sites, triggering a change in the receptor's structure which in turn causes the coupled G-protein to release its bound guanosine diphosphate, and instead bind guanosine triphosphate from the interior cell environment. When that happens, the G-protein is "activated"; it decouples from the receptor, splits in half into Ga and Gbg subunits, and those two halves bind to effector enzymes which trigger the cell signalling. As a result of the protein decoupling, the charge balance of the receptor changes, releasing the complexed ligand from its binding to the receptor, back into the extracellular environment.

To reverse all this, an enzyme called RGS (Regulator of G-protein Signalling) binds to the Ga subunit, triggering hydrogenation of the GTP molecule it still has bound into GDP, causing the protein's subunits to recombine and recouple to the receptor.

The released ligand, meanwhile, is washed away again by the mucous flow from the Bowman's glands. It may complex with other receptors along the way, but eventually it makes its way past the nasal membrane to the pharynx at the back of the throat, where the mucous is swallowed and the odorant broken down in the stomach. This is the final fate of virtually all odorant agonists.

A select few, however, bind more strongly, and are not released by the structure changes of the receptor. When this happens, the receptor is stuck in the "on" position. One of two things then happens; either the receptor continues to activate, causing neural impulses which your brain eventually ignores (odor fatigue), or mechanisms inside the cell notice the faulty receptor, bring it into the cell in a process similar to endocytosis, and attempt to break down its components and fix the damage to reuse it. More often than not, the bound chemical ends up poisoning the cell by binding to something more vital than an ordinary G-protein receptor, eventually triggering programmed cell death. That's fine; the basal cells behind the surface of the olfactory epithelium divide fast enough to replenish every cell in the nasal membrane about once every two days.

Sources: Wikipedia - Olfactory epithelium, Olfactory receptor and G protein-coupled receptor.


You can find plots and further explanations online (as I easily did). This is a homework-type question, yet I do feel that the information request for diatomic molecular spectroscopic constants warrants at least this partial answer. See this writeup for the "Vibronic Absorption Spectrum of Molecular Iodine" for more details and illustrations.

From NIST:

T$_e$ minimum electronic energy (cm$^{-1}$)

R$_e$ internuclear distance (Å)

ω$_e$ vibrational constant – first term (cm$^{-1}$)

ω$_e$x$_e$  vibrational constant – second term (cm$^{-1}$)

B$_e$ rotational constant in equilibrium position (cm$^{-1}$)

α$_e$ rotational constant – first term (cm$^{-1}$)

D$_e$ centrifugal distortion constant (cm$^{-1}$)

β$_e$ rotational constant – first term, centrifugal force (cm$^{-1}$)

From Wikipedia:

D$_0$ dissociation energy (cm$^{-1}$)


This leaves us with diatomic Selenium which I could not find anything on.

Selenium forms a dimer with bond energy of 79 kcal/mol, according to table 8.6 (page 451) of Comprehensive Handbook of Chemical Bond Energies, which provides 5 references.

Tellenium dimer is 61 kcal/mol

Polonium dimer 45 kcal/mol


Successive ionization energies (work functions?) for metals

In an infinite metal solid, there are an infinite number of electrons, so each one removed will be at the exactly identical work function with no change in energy.

Intuitively, this may seem a bit strange because .. well, it should be harder to remove an electron once positive charges start to accumulate.

If I have even a micron-sized piece of metal, that's ~$10^{13}$ atoms. So you really have to do a lot of ionization before you're even going to get 0.1% of the metal ions.


Reaction between chlorobenzene and sodium methoxide to produce anisole

Chlorobenzene is inert towards most forms of nucleophilic substitution. In particular:

A $\mathrm{S_N1}$ mechanism is not likely here because the carbon is $\mathrm{sp^2}$ hybridised, the $\ce{C-Cl}$ bond is strong, and the dissociation of $\ce{Cl-}$ is therefore not very likely.
$\mathrm{S_N2}$ substitution doesn't occur because 1) the $\ce{C-Cl}$ bond is strong 2) the $\sigma^*$ orbital is in the middle of the ring and is inaccessible to a nucleophile.
The $\mathrm{S_NAr}$ mechanism could be followed if there is an electron-withdrawing group ortho or para to the chlorine atom, but in chlorobenzene, there is no such EWG.

I would like to suggest the possibility of a benzyne mechanism, i.e. an elimination-addition mechanism. The elimination step follows an $\mathrm{E1_{cb}}$ mechanism:

The benzyne intermediate is very electrophilic and nucleophilic attack by methoxide ion proceeds rapidly to give anisole.

Although this reaction is well-documented with both $\ce{NaOH}$ and $\ce{NaNH2}$, I can't find any real-life examples with $\ce{NaOMe}$. Most examples of chlorobenzene → anisole reactions involve some kind of catalyst. However, if it were to happen, it would probably occur via this parthway.


Amount of substance of a molecule in a solute the same as amount of substance of constituent elements?

You are right thinking that the molar concentration of S will be the same as the molar concentration of SO4, unless there is no other molecule whith S in the sample.

If there is 1 mole of molecules of sulphate (SO4), there will be exactly 1 mole of atoms of sulfur (S), and 4 moles of atoms of oxygen (O).

On the other hand, though "theoretically" it is correct to say there will be 2 moles of molecules of O2, it is really not correct, because there is absolutely no molecule of O2 in your sample. The amount of oxygen atoms is the same in both cases, but how they are organized is not.

If you want to get the mass concentration of S, say in g/l, you would just multiply the concentration of SO4 in g/l by the mass ratium of sulfur within sulphate, (sulfur mass over sulphate mass, 32/96=0.33333). Since for every sulphate gram there is 1/3 sulfur mass and 2/3 oxygen mass, 1/3 of your sulphate mass is due to sulfur.
Another strategy would be to get the concentration of sulphate, which is exactly the same as that of sulfur (1 mol of atoms of S for every mol of molecules of SO4), and multiply that by the molar mas of the sulfur atom (32).

In brief, the molar ratio between S and SO4 is 1, and the mass ratio is 1/3 grams of sulfur for every gram of sulphate


Which functional groups are not reduced by DIBAL.H?

DIBAL-H does the following:

Acid chlorides to aldehydes (Fast)
$3^\circ$ amides to aldehydes
Nitriles to aldehydes
Ketones to $2^\circ$ alcohols
Aldehydes to $1^\circ$ alcohols. (Slower)

Also, it works best in these solvents: THF, toluene.
We need to avoid solvents like alcohols, amines, strong acids and halogenated compounds.


How does an electron identify whether to absorb a photon of a certain wavelength or not in absorption spectrum?

[OP..] How would an electron identify the specific photons of a certain wavelength which are to be absorbed in a Hydrogen atom while creating an absorption spectrum?

As discussed in the comments, this is an observed phenomenon. We can try to describe this observation in a way that we can make predictions, but that does not answer the questions "how?".

[OP..] Otherwise, it cannot find a shell and since electrons cannot exist in between shells, they simply re-emit that photon and we perceive the incident as if the photon weren't absorbed at all.

There are processes that do something similar as the one you describe, but they involve observable events. When an atom scatters light, it scatters light of the same wavelength as the incoming light, but in all directions. There are also processes where the light is absorbed and then emitted again after a while (sometimes at different wavelength). What you suggest is an explanation for observing no interaction between the electron and the photon, and explaining it with an interaction (absorption and instant re-emission at the same wavelength in the same direction) that we can't observe.
You could have explanation models of the same flavor for many things that don't happen. For example, on a day it does not rain, I could say drop form in the clouds but instantly are destroyed again, so we can't observe them ever.

Way-02: The explanation given above is false and electrons choose the photons via some other mechanism

As you can read in the comments, using the term "choose" for something electrons do is not so helpful. We know under what conditions the photon might get absorbed (not all of them are), but to go beyond that and explain the "how" or even "why" is difficult to impossible (and not taught in e.g. the 4-year college chemistry curriculum, as far as I know).


Molten or solid at the end?  It can be shaped in an induction field (liquid)  or dimagnetically levitated (liquid to solid).  Magnetoforming, hydroforming.  Spin molten into into an oblate ellipsoid (solid).

http://www.capturedlightning.com/frames/shrinkergallery.html
Z-pinch and such.

Munroe effect for jets and planes (angle charges), and self-forging munitions.


Hygroscopic Substances and Drying Agents

Read closer:

  Most of the hygroscopic substances are drying agents

not

  all drying agents are hygroscopic substances.

First  Terminology:
Hygroscopic: Absorbs water from air
Drying agent: Removes water from organic substance

Your book did not exclude other materials from being considered drying agents. It just said many of the substances that are hygroscopic are also useful as drying agents.

Take for example anhydrous calcium chloride calcium sulfate hemihydrate (also called plaster or DrieriteTM) and phosphorous pentoxide. All of these are hygroscopic (sequester water from air)

$$\begin{align}
\ce{CaCl2 (s) + H2O (fl) & -> CaCl2 (aq)}\\
\ce{CaSO4.\tfrac12 H2O (s) + H2O (fl)&  -> CaSO4.2H2O (s)}\\
\ce{P2O5 (s) + H2O (fl) &-> HPO3 (cd)}\
\end{align}$$

Calcium chloride is deliquescent (hygriscopic to the point that it forms a liquid) and calcium sulfate hemihydrate is hygroscopic (stays solid but still absorbs water) and both can be used as drying agents for most solvents. Phosphorous pentoxide on the other hand cannot be used as a drying agent for some materials such as alcohols and styrene.

$$\ce{P2O5 (s) + EtOH -> EtH2PO4 (cd) + HEt2PO4 (cd)}$$
$$\ce{PhCHCH2 ->[P2O5] [PhCHCH2]_n \tag{explosive}}$$


How were x-ray diffraction patterns deciphered before computers?

I agree with @andselisk that this question is quite broad. I will focus on two specific questions asked

  The only equation I know for x-ray diffraction is Bragg's Law but is this the only equation used to interpret the data? [...] How do you translate the spots on a detector to electron density plots using Braggs law?

Apart from Bragg's law (which tells you where the diffraction spots are for a known orientation of a crystal with know unit cell), it was also known that real space (electron density) and reciprocal space (diffraction pattern) are related by 3D Fourier transform. For structures containing one or two atoms, just knowing the unit cell parameters and the symmetry is enough to get the entire structure. For anything slightly more complicated, the Fourier transform of amplitudes or later, of intensities (Patterson methods, http://reference.iucr.org/dictionary/Patterson_methods), had to be used. The first crystal structure analyses were of crystals with centrosymmetry, where the phase problem is easier to solve.

  My question is what is the general process for doing this without computer software?

Diffraction data was measured on film, with gray-scales to assess intensity of signals. To calculate a Fourier transform, pre-computed tables were used, such as the Beevers-Lipson strips. As Andselisk commented, Fourier transform was used late in the 20s, and initially for problems that were one- or two-dimensional. There is a nice account by P.P. Ewald available here: https://www.iucr.org/publ/50yearsofxraydiffraction/full-text/structure-analysis
This was written at a time when the myoglobin structure had just been solved, after "50 years of X-ray diffraction".


How to conveniently separate Cd impurities from ZnO without contaminating it?

Since you actually considered the use of a calutron, I assume that you are interested in separation of relatively small amounts of pure ZnO - on the order of a few grams.

Cadmium and zinc are similar in many respects, which makes me think of rare earths. One of the separation techniques used there is ion exchange chromatography. Resins with special affinity for many ions (either to catch it or reject it) are available. Some contaminants are not easy to remove by conventional ion exchange resins. In many cases, very specific resins have been developed for these contaminants. Selective resins from Rhom and Haas, www.lenntech.com, are available today for the removal of:
• Boron
• Cadmium, mercury and other heavy metals
• Chromate
• Lead
• Nickel
• Nitrate
• Perchlorate
and some other contaminants.

On the more researchy side, the stability constants of the tetra ammonium complexes of zinc and cadmium differ by a large factor, unlike their complexes with EDTA and similar chelators. For Zn(NH3)4++, the dissociation constant is 9.8 x 10^-10; for Cd(NH3)4++, it is 2.5 x 10^-7. (Reference Book of Inorganic Chemistry, Latimer and Hildebrand, 1951, 3rd ed., pp 135-6). The impure ZnO could be dissolved in NH4Cl (which will also dissolve CdO) and the solution run thru a column. Here's the research: a column of what? Perhaps a column of the impure ZnO. Since the zinc complex is more stable (more soluble), it should concentrate in the first eluate, leaving CdO behind. Solid ZnO would rip NH3 off the cadmium complex and become soluble, so the ZnO will turn into CdO. I have no idea how useful this scheme could be, but the big difference in stability constants is an eye-catcher.


Why do spectra of elements never overlap?

My guess would be that every element is different. Going from the basic shell model of the atom to orbitals and transition probabilities there are huge lists for all different transitions. I think there is a green emission for boron where a $11f$-orbital or so is populated and relaxes to give off a green light. And this is so unlikely and weak that you don't even see this on a regular spectrum. So this may be one hint, there are so many possibilities. On the other side if we agreed on one electron, let's say the $1s$-electron, because this is present in all atoms, just exciting that would require different energies for all atoms because the number of protons increase and therefore the positive charge pulling it closer and strong increases as well. It may happen that one transition of element A and a completely different one  on element B have the exact same wavelenght, but the spectrum has so many possiblities and there are only very few elements in the PSE for comparison that I doubt that there will be two identical ones. And even if this happens you can strech that spectrum to the UV, X-ray region as well. What is more likely to happen is that the characteristic emission of an element, let's say the one you observe strongest may be similar, like for Lithium, Calcium and Strontium. I doubt that I'd be able to distinguish them by their red color, but then I'd just take a look at the rest of the spectrum.

But I once had a related case. I was preparing a Dysprosium compound and used a sodium based flux. After solving the structure we had the idea that traces of sodium might occupy long anionic-channel in the compound. So we analyzed the elements using energy-dispersive x-ray spectroscopy. And in the spectrum I got were the characteristic Dysprosium emission line and the one from Sodium pretty much on the same spot. So we had to do a wavelenght-dispersive test as well and then I was able to confirm the sodium. This was a case where the strong x-ray emission line of Sodium almost identical to the one of Dysprosium. But in regular spectra you have much more intense emissions so I guess the probability will be really small.


I have not heard of glucose being burned in chlorine, but I do know that glucose reacts with chlorine in aqueous-buffer medium. See reference 1.
The full paper is behind a paywall.

It says that oxidation of glucose by bromine water was studied previously and the reaction of glucose and aqueous chlorine also happens in a similar fashion although the reaction is slow and products yield is less. Nonetheless, major product of the reaction is D-gluconic acid in both acidic and alkaline medium but unbuffered chlorine may further react with D-gluconic acid to various products like 5-keto-D-gluconic acid and D-glucaric acid.

Reference

J. Am. Chem. Soc., 1955, 77 (7), 1875–1880. DOI: 10.1021/ja01612a050.


Most efficient level of theory for sulfur-containing proline analogues (one that will work)

You seem to be running Gaussian through another program or script. Consequently, I have no idea where the various arguments should go. Conventionally, you make a Gaussian input file with keywords, such as:

  %n B3LYP/6-311G** EmpiricalDispersion=GD3BJ Opt

  [... molecule goes here ..]

For dispersion-correction (i.e., GD3BJ) you would need an appropriately recent version of Gaussian and use the separate term as indicated. It is not part of the functional designation. (Although some functionals implicitly have dispersion correction, B3LYP does not.)
I don't know a priori that PBEPBE would give more accurate results than B3LYP or any other functional. Your best bet is to either consider results from papers in the field (e.g., molecular crystal prediction or runs with DMACRYS) or testing a few functionals and finding out for yourself. Consider reading this question and answers.
Intermolecular interactions are not likely to be handled well by B3LYP. This is the whole reason for range-separated functionals and dispersion corrections. As to a specific functional suggestion, I highly recommend trying a few and looking at how to choose a functional.


What does the reaction of chromium with ozone form?

Ozone is a strong oxidizing agent, even stronger than molecular oxygen. But it is still a donor of oxygen atoms. Thus, depending on conditions, you'll have different chromium oxides. The most stable and common chromium oxide is chromium (III) oxide $\ce{Cr2O3}$. It has numerous nicknames. Other oxides are much harder to get.

Substances named ozonides do exist and in two different flavours.

Inorganic ozonides are salts of big, inoxidizable cations, like $\ce{Cs}$ and have formulas like $\ce{KO3}$. I'm not sure which cations have ozonides that have been already produced, but to my knowledge the only reasonably easy-to-produce ozonides are $\ce{KO3}$, $\ce{RbO3}$ and $\ce{CsO3}$ by reaction of the corresponding hydroxides with ozone.

Another type of ozonides is organic ozonides that are quite weird heterocyclic compound, which are product of reaction between alkenes and ozone.

If you need a name based on word games, you may consider that chromium is a derivative of an ancient word for 'color' and ozone, as a strong oxidizer, has bleaching and antiseptic properties.