Why do big thermoplastics (like Polytetrafluoroethylene) behave like thermosets?

Thermosets are liquids or soft solids which irreversibly 'set' or 'cure' when heated. This 'setting' produces covalently  bonded cross-linkages between adjacent molecular chains, resulting in in their mechanical rigidity.

Thermoplastics too are formed from linear polymer chains, but their molecules typically have very long chains with high molecular weight, with very few cross-linkages. The weaker van der Waals bonds which link adjacent molecules in thermoplastics are easily broken when heated, allowing the chains to 'slide' past each other, thereby giving them the property of 'softening' when heated.

  Why does a big molecule break down before the van der Waals forces
  break?

In the case of polytetrafluoroethylene (PTFE), we have a very long, linear polymer chain of carbons covalently bonded with fluorine. PTFE is quite unique because it doesn't have any hydrogen atoms. Instead, it has these large highly electronegative fluorine atoms.

The bonds within the PTFE molecule are very strong (C-C bonds are 607kJ/mole and C-F bonds 552kJ/mole, much stronger than the typical C-H bonds found in other plastics such as polyethylene).  In effect, the PTFE molecule is like a 'rod' of carbons bonded to a surrounding 'blanket' of fluorine atoms (arranged in a slight helical twist).

Although the van der Waals bonds which form between adjacent the adjacent linear 'rod' molecules of PTFE are much weaker than the C-C and C-F bonds within the molecules, these van der Waals forces are able to combine in an additive manner, so the long straight PTFE 'rods' are able to be 'packed' together in a hexagonal arrangement, a bit like pencils in a box. This results in PTFE having a relatively high density and melting point of $327^o C$.

But when PTFE melts, its extremely long molecular chain have a high probability of entanglement, resulting in a extremely high melt viscosity. Although it's melt viscosity varies with the shear stress applied and temperature, in the temperature range between $360^o C$ and $380^o C$ and shear stresses of $0.1MPa$, commercial samples of the granular polymer display viscosities of the order of $10^{10}Pa.s$. This is exceptionally high compared with most other engineering thermoplastic melts. In fact, at atmospheric pressure, it hardly seems to 'melt' in the conventional sense of 'becoming a flowing liquid', although it becomes slightly softer, it forms a near-clear gel with very high elasticity.

Even below it's melting point, PTFE will start to decompose at $260^o C$, releasing toxic compounds through sublimation. When heated above $360^o C$ thermal decomposition is more rapid and the PTFE degrades noticeably. As a consequence, PTFE can not be processed by many 'conventional' thermoplastic techniques, such as injection molding. Instead, it is processed in a finely powdered or granular form, using modified powder metalurgical techniques such as cold-molding, sintering, compression and extrusion, to fabricate rods, tapes, wire, tubing, sheeting and other profiles which are typically suitable for further machining.

Above $480^o C$ PTFE thermally decomposes in air to release a mixture of vapours and particulates, including carbonyl fluoride, hydrogen fluoride, tetrafluoroethylene and other fluorinated hydrocarbon products, including a small amount of some highly toxic species such as perfluoroisobutylene.

The precise nature of the reactions responsible for this decomposition is still subject to research, but it appears likely that it is initiated by random-chain incisions in the PTFE chain, followed by depolymerization.


The short answer is stated on the Wikipedia page for trigonal bipyramidal:

  … there is no geometrical arrangement with five terminal atoms in equivalent positions.

This website from the School of Mathematics and Statistics at UNSW explains why distributing points on a sphere is tough and that perfect distribution is possible only in certain cases $(n=1,2,4,6,8,12,\&\ 20)$. These cases have been known for a long time — they correspond to the platonic solids (plus a point and a line segment). Even trigonal planar does not represent an even distribution on the sphere.

Trigonal bipyamidal then just happens to be the minimum energy conformation. The repulsive interactions between the chlorine atoms cannot be optimized but they can be minimized.


Which is the currently accepted mechanism of a Wittig reaction?

tl;dr: I don't think there is any mechanism that is 100% correct, and, in cases like this especially, I think it would completely depend upon what set of carbonyl/ylid/base/solvent etc. was used. But, of course, we like being able to generalise, and to my knowledge theres a lot more evidence to support a concerted type mechanism.
General Background
The question is hopefully summarised in the scheme below (no geometry, that, in itself, is a pretty lengthy discussion). Reaction of a carbonyl (in this case a ketone) with a phosphorus ylid is able to give rise to two species, either an oxaphosphetane directly or a betaine (an internal salt for the benefit of Loong) which then goes on to form the oxaphosphetane which itself is able to eliminate to afford the olefin with generation of triphenylphosphine oxide as a thermodynamic driving force.
This answer will present the case for a cycloaddition mechanism, and evidence against the betaine pathway. Importantly, only Wittig reactions of unstabilised ylids, as with stabilised ylids the rules of the game change due to reversibility of intermediate formation (see a Horner–Wadsworth–Emmons).

Evidence against the initial formation of a betaine
Betaines have never been observed during the course of a 'normal' Wittig reaction, that is, spectroscopically, we are unable to see it (this may, of course, just be due to the fact that the formation of the oxaphosphetane is so rapid relative to the timescale of the methods used).
Many reports of betaines have been from accidental (or otherwise) opening of oxaphosphetane intermediates, and indeed this seems to have been what led Wittig down the betaine pathway in the initial reports (though in the first ever paper he suggested that oxaphosphetane was the sole intermediate, but subsequently couldn't find any evidence for it back in the days before NMR etc).
Wittig[2] and others[3] reported the isolation of crystalline betaine salts. Phosphonium bromides were treated with PhLi to give a ylid, to which a carbonyl was added; subsequent addition of hydrogen bromide afforded crystalline solids which were able to be elucidated, proving the formation of a betaine.
Whilst this was convincing at the time, we now know that the betaine isn't observed, but rather the oxaphosphetane is. Vedejs (whom we'll come back to), has gone on to show that the earlier findings could be explained by quenching of the oxaphosphetane rather than as a result of directly trapping the betaine[4], which is more consistent with the other data available to us.

Scheme 1: Generation of the crystalline betaine derivative from the oxaphosphetane.

There are more recent reports of 'stable' betaines for very specific substrates where subsequent collapse isn't possible, again generated from oxaphosphetane intermediates.
Stefan Berger's group at Leipzig reported a betaine stabilised by a bipyridyl group, allowing NMR data to be reported for the first time[1].

Scheme 2 Reversible formation of a stabilised betaine

In this case, the lithium holds a chelated structure together, preventing it from forming the alkene (and also to a certain extent preventing reformation of the oxaphosphetane on steric grounds). Interesting, with addition of 18-crown-6 to sequester the lithium, the reaction proceeded normally in the forward direction with alkene signals observed.
The fact Berger was able to do this, does of course not imply that they are real intermediates along the Wittig pathway, but it was a nice piece of mechanistic work that I thought deserved mention.
In addition to this mechanistic work, there are some empirical issues with invoking a betaine intermediate that cannot be easily explained. Namely, all Wittig reactions using non-stabilised ylids should be under kinetic control and irreversible (hence highly (Z)-selective due to the inability of the initial intermediates to reverse and hence equilibrate to the thermodynamic product), however this is frequently observed not to be the case, and in certain cases, high (E)-selectivity can be achieved.
Evidence for direct oxaphosphetane formation
Vedejs was the one of the first to propose direct (irreversible) cycloaddition of the ylid and carbonyl to give rise to the oxaphosphetane, quickly followed by cycloreversion to form the desired alkene and a phosphine oxide.

Scheme 3: cycloaddition/cycloreversion mechanism for Wittig

In his early reports, direct observation of the oxaphosphetane by P-NMR was conducted[5], the paper reasons that the high-field chemical shift observed for the phosphorus containing intermediate ruled out charged, 4-valent species such as betaines, but was instead more consistent with formation of the 5-valent neutral species such as an oxaphosphetane.

Scheme 4: Vedejs' observation of oxaphosphetane.

This NMR analysis isn't really too convincing by itself, however in the scheme above, compound 4 had previously been characterised by X-ray crystallography[6] and various NMR methods and as such gave some weight to the species observed by Vedejs being similar to the known compound. More recent work[7] has also observed directly the cis/trans oxaphosphetanes formed via competing cycloadditions.
Summary
Overall, direct observation of the oxaphosphetane but not the betaine does seem to suggest that mechanistically the cyclic intermediate is formed directly rather than via the betaine closing.
As stated previously, we can't 100% rule out the possibility that the betaine closure is just sufficiently rapid to appear invisible to our currently spectroscopic methods however every report of isolated/observed betaines that I've came across thus far has either intentionally been formed from an oxaphosphetane or can be explained by this without the authors recognising it.
One thing that is hugely missing from the story is computational work, whilst some studies have calculated the relative energies of the two intermediates, to my knowledge, the entire reaction coordinate hasn't been fully explored.
In conclusion, I think the cycloaddition mechanism is safest, and indeed it was the one I was taught as an undergraduate (and the one I know several other leading universities use), so if in doubt i'd go for that explanation, but on the firm understanding that more mechanistic work is needed.

References
 Generally: Modern Carbonyl Olefination, Carey Advanced Organic A, and Comprehensive Organic Synthesis I.
[1] Neumann, R. A.; Berger, S. Observation of a Betaine Lithium Salt Adduct During the Course of a Wittig Reaction. Eur. J. Org. Chem. 1998, 6, 1085. Subramanyam had reported similar chemistry earlier, but the NMR data was incomplete.
[2] Wittig, G.; Haag, A. Über Phosphin-alkylene als olefinbildende Reagenzien, VIII. Allenderivate aus Ketenen. Chem. Ber. 1963, 96 , 1535. DOI: 10.1002/cber.19630960609. A very old paper in German.
[3] Schlosser, M.; Christmann, K. F. Olefinierungen mit Phosphor-Yliden, I. Mechanismus und Stereochemie der Wittig-Reaktion. Liebigs Ann. Chem. 1967, 708, 1. DOI: 10.1002/jlac.19677080102.
[4] Vedejs, E.; Meier, G. P.; Snoble, K. A. J. Low-temperature characterization of the intermediates in the Wittig reaction. J. Am. Chem. Soc. 1981, 103, 2823. DOI: 10.1021/ja00400a055.
[5] Vedejs, E.; Snoble, K. A. J. Direct observation of oxaphosphetanes from typical Wittig reactions. J. Am. Chem. Soc. 1973, 95, 5778. DOI: 10.1021/ja00798a066.
[6] Mazhar-Ul-Haque; Caughlan, C. N.; Ramirez, F.; Pilot, J. F.; Smith, C. P. Crystal and molecular structure of a four-membered cyclic oxyphosphorane with pentavalent phosphorus, PO2(C6H5)2(CF3)4C3H2. J. Am. Chem. Soc. 1971, 93, 5229. DOI: 10.1021/ja00749a044.
[7] Maryanoff, B. E.; Reitz, A. B.; Mutter, M. S.; Whittle, R. R.; Olofson, R. A. Stereochemistry and mechanism of the Wittig reaction. Diasteromeric reaction intermediates and analysis of the reaction course. J. Am. Chem. Soc. 1986, 108, 7664. DOI: 10.1021/ja00284a034.


Why does having identical groups on either side of the pi bond lead to no geometrical isomerism?

Hmm, you seem to be having a problem visualizing the structures in the case of 2 identical groups on one carbon. But never fear, 3D molecules are here.
Check out chloroethene:

It has two identical groups (hydrogens) on the left side. If you tried to make a geometrical isomer for this, you'd get:

But hey! That's just the previous image flipped upside down! That's not an isomer, it's the same thing!
Indeed it's the same molecule. These molecules can't have a geometrical isomer because it possess an plane of symmetry:

So there you go. Go ahead and visit this link if you want an interactive 3D model that you can spin around.

Added based on comments from @Zhe, there's a plane of symmetry in following angle,


Does electron shielding increase or stay constant moving LEFT to RIGHT across a period?

To answer this question, it's important to define what you mean by shielding. Generally, shielding refers to a reduction in the effective nuclear charge experienced by an electron in a given orbital due to the other electrons on the same atom.

The quantitative degree of shielding for a given electron can be approximated by Slater's rules. According to those rules, electrons within the same group of orbitals (for example 4d) have a coefficient of 0.35 (except for 1s electrons, which have 0.30). So valence electrons do shield each other, just not as much as the lower level electrons shield the valence electrons.

For example, let's consider the elements with increasing numbers of 2p electrons (B, C, N, O, F, Ne). Going from left to right, each addition of a 2p electron reduces the effective nuclear charge experienced by another 2p electron by 0.35. So the amount of shielding is increasing as we move left to right.

The apparent contradiction with the ionization energy comes about because you have not considered the increase in the actual nuclear charge. Each time we add a 2p electron, we also add a proton to the nucleus. So even though 0.35 of that charge is shielded from a valence 2p electron, that electron still experiences an increase of 0.65 effective charge and is therefore bound more tightly than a 2p electron on the element to its left.

As a specific case, let's compare the shielding experienced by a 2p electron in carbon versus neon. The carbon electron configuration is $1s^22s^22p^2$. Using Slater's rules, we can calculate that the shielding coefficient for a 2p electron is 2(0.85) + 3(0.35) = 2.75. So the effective nuclear charge experienced by the 2p electron is 6-2.75=3.25.

For neon, the electron config is $1s^22s^22p^6$, so the shielding coefficient of a 2p electron is 2(0.85)+7(0.35)=4.15. So there's more shielding. But the effective charge is 10-4.15=5.85, so the effective charge is still greater and the energy of the coulombic interacation of the electron with the nuclear charge is greater in magnitude. Hence, the ionization energy is greater.

EDIT: Original version of this answer did not address the question about anion formation (presumably mostly interested in anion formation by addition to an already partially filled shell). When an anion is formed, the additional electron increases the shielding of the other valence electrons, but there is no change in the actual nuclear charge. Thus, each valence electron experiences a lower effective nuclear charge. Since the size of an orbital is a function of the energy of interaction with the nuclear charge, the atomic size increases.


How to balance this redox reaction taking place in a basic medium?

Writing an answer since I can't comment yet.

You are using the ion-electron method for basic medium, but are balancing using $\ce{H+}$ and later adding $\ce{OH-}$? Sounds suspicious to me, but the outcome must be the same.

Generally, I add $\ce{OH-}$ ions on the oxygen deficient side and then add water to compensate the hydrogen deficiency on the other side.

Note: your first half-reaction balancing the conversion of permanaganate to $\ce{MnO2}$ must be regarded incorrect, as $\ce{MnO2}$, being soluble in acidic conditions (presence of $\ce{H+}$, as the equation indicates!) would yield to further reduction, and end up as $\ce{Mn^2+}$.

Here's how I balanced it:

Oxidation half

$$
\begin{align}
\ce{Al &-> Al(OH)4-}\\
\ce{Al + 4OH- &-> Al(OH)4- + 3e-}
\end{align}
$$

Reduction half:

$$
\begin{align}
\ce{MnO4- &-> MnO2}\\
\ce{MnO4- &-> MnO2 + 2OH-}\\
\ce{MnO4- + 2 H2O &-> MnO2 + 4 OH- + 3 e-}
\end{align}
$$

Adding:

$$
\ce{Al + MnO4- + 2 H2O -> Al(OH)4- + MnO2}
$$

Your method is fine as well. The only mistake is the wrong $\ce{H+}$ coefficient in $\ce{Al}$ oxidation half-reaction, which will be 4 instead of 8.

P. S. (only concern): note the differences in half reactions in basic and acidic media. Conclusion: when products differ in acidic/basic conditions, it is best not to balance in the asker's method, as it cannot be considered entirely correct. Nothing more, nothing less.


What is the vapor pressure of a liquid if it is not in a sealed container(i.e., a river or lake)? Is it just the atmospheric pressure?

From what I understand, the boiling point of a liquid is when its surrounding pressure is equal to its vapor pressure.

This is backwards. The boiling point is when the vapor pressure of the liquid is equal to the atmospheric pressure. So given the average pressure of 1 atmosphere at sea level water would boil at $\pu{100 ^\circ C}$. However on some mountain top water might boil at $\pu{95 ^\circ C}$ since the air pressure is lower.

  The vapor pressure is the pressure of the gas form of the liquid in a container when the rate of condensation equals the rate of evaporation.

For a particular temperature, in a sealed container, when the rate of condensation equals the rate of evaporation then the system is at equilibrium. The partial pressure of the gaseous form of the liquid is then the equilibrium vapor pressure of the liquid.

Without being at  equilibrium the partial pressure could be anything.

  If this is the case, what is the vapor pressure of a liquid that is not placed in a container(i.e., a river or lake)? Is it the same as the pressure of the atmosphere? If so, then how does this affect its boiling point?

If I have an open beaker of benzene on an outside picnic table then the system isn't at equilibrium and benzene has no defined vapor pressure. There just isn't enough benzene to ever reach an equilibrium with the whole atmosphere of the earth.

This wouldn't effect the boiling point of the benzene. Assuming that the air pressure is 1 atmosphere, benzene would boil at the same temperature whether if it was being boiled  in the open air or refluxed in a lab.

This bring up another point. If we say that the vapor pressure of water is $\pu{0.0231 atm}$ at $\pu{20 ^\circ C}$, we are referring to "local conditions." In other words, that gravity can be assumed to be  a constant within the system. Obviously if I had a cylinder 200 miles high at $\pu{20 ^\circ C}$, the vapor pressure wouldn't be the same at the top and bottom of the cylinder due to the difference in gravity.


Why do we use coke instead of coal in order to reduce FeO as basic reaction in ironmaking process?

The single most important factor is strength ( mechanical compressive ); coal is heated to make coke, the resulting coke is stronger than the original coal. Also, coke helps to make the charge of iron oxides and limestone more porous to permit gas flow up and droplets of liquid iron and slag down. The coke oven heating drives off volatiles from the coal, which are a valuable source of chemicals, and contain much of the undesirable $\ce{S}$ and $\ce{P}$, which would otherwise go into the pig iron. Minerals like silica (ash) remain in the coke but they are easily collected by the slag. A bit off topic, but powdered coke and gasses are blown into the bottom of the blast furnace to provide more $\ce{C}$, which makes modern blast furnaces much more productive than 60+ years ago.


Typical reaction yield for combustion of hydrogen

The equilibrium constant for the reaction $\ce{2H2 + O2 <=> 2H2O}$  is colossal, equal to $\mathrm{2.4 \times 10^{47}}$ at $\mathrm{500\ K}$ (source). This would ensure virtually 100% yield for the combustion of a stoichiometric mixture of oxygen and hydrogen in equilibrium conditions. However, a rocket engine is a reaction vessel that is far from equilibrium, and therefore oxygen could remain unreacted. This is in fact a problem, as it could damage the engine at its operating temperature. This, among other reasons, is why oxygen/hydrogen mixtures for rocket engines are actually very hydrogen-rich, containing up to twice the stoichiometric amount of hydrogen required.

So in practice, taking everything into account, rocket engines running on oxyhydrogen fuel are designed to have a  practically 100% combustion yield based on the limiting reagent, oxygen, while possessing a substantial excess of hydrogen fuel. There is a strong engineering pressure to maximise energy output by ensuring the oxygen is completely consumed; unreacted oxygen is a waste of lift capacity, storage space, materials, cost, etc.


Although voltage can be calculated from the electromotive series, maximum current is harder to predict due to a number of factors, some of which you've mentioned.

Electrolyte conductivity, which varies with components, concentration, temperature, etc. As you state, weak electrolytes are less conductive. Resistance of a composite electrolyte would be the sum of its components.
Cell polarization, which is an effect at the electrode/electrolyte interfaces. The causes are buildup of gas bubbles, concentration gradients that develop in the electrolyte, etc.
Deterioration of the electrodes and/or electrolyte, particularly under high current drain.
Of course, as @MaxW states, a major factor is cell geometry: area of electrodes and separation due to insulators and electrolyte.

For these reasons, though the maximum current can be calculated from the resistances and EMF, in actual use, cells often fall far short of their theoretical potential (no pun intended).

The resistance of a battery is primarily the sum of that of the cells. Since cells are joined with metallic connections, the cell-to-cell resistance is usually negligible (though I've used poorly designed battery holders with high-resistance spring contacts that severely limited the current, and that damaged the plastic holder due to ohmic heating).


What kind of chemicals would exist at the heat death of the universe?

There is a lot of uncertainty regarding the far future of our Universe, but it seems that chemistry as we know it will be gone long before the end. Both free and bound protons (and neutrons) are expected predicted to decay through at least one of several mechanisms, with a half-life somewhere between the range of $10^{35}$-$10^{200}$ yr (far shorter than the timescale for nuclear reactions to convert all matter into iron, ~$10^{1500}$ yr). Therefore, at some point the statistically expected number of atoms inside our observable bubble could potentially dip below 1.

Should proton decay happen relatively quickly, perhaps the last elements in the Universe will be present near the end of the black hole era (~$10^{100}$ yr), when the last supermassive black holes decay in a flash of subatomic particles, possibly creating small amounts of new protons/neutrons. These will decay in a comparative blink of an eye at this timescale, however, and the Universe will soon become a barren place once more.

The last structures with any semblance to matter in the Universe will likely be positronium "atoms", where an electron and a positron (anti-electron) orbit each other in a pseudo-hydrogen atom. Though ground-state positronium decays in a matter of nanoseconds or less, it will be very long-lived in the future due to the extreme sparsity of subatomic particles in the hugely expanded observable bubble, with a mean distance between a bound positron and electron measured in trillions of trillions of trillions of trillions of... billions of lightyears or more. Even in these conditions, positronium too shall eventually decay, in a timescale of anywhere between ~$10^{140}$ yr and ~$10^{800}$ yr.

Before all of these nuclear processes take place, there may also be a significant amount of unusual chemical reactions happening solely due to tunnelling effects, even at arbitrarily low temperatures, since the kinetic barriers for chemical reactions are far smaller than that for nuclear reactions. For example, most molecules containing nitrogen atoms are thermodynamically unstable with respect to release of $\ce{N2}$, and so might spontaneously break up even at $\mathrm{0\ K}$ given enough time. I don't know what the timescale for these processes is, however.


Why is sodium potassium tartarate used in Fehling's solution?

Fehlings soluition is alkaline. Mixing cupric ions with an alkaline solution would result in formation of cupric hydroxide of very low water solubility. By adding tartrate ions, the cupric ions will form a complex with the cupric ions which will keep them dissolved in the alkaline solution. The type of cation of the tartaric acid is of no importance other than for solubility reasons.

The same principle is used in Benedicts´s solution, where tartaric acid is replaced by citric acid or a salt thereoff.


How to determine whether the enthalpy of solution is positive or negative by calorimetry?

The sample takes in the heat from the water, so it takes in energy to "form" the solution and the water thereby loses it, becoming a lower temperature. You need to realize the reference point is not the water, but rather the sample in this case. The heat transferred (lost) by the water is to the sample, so it is the negative of the heat lost by water, making it positive.


Picric acid (2,4,6-trinitrophenol) is an aromatic compound has been used as an etchant in the past. Picral is an ethanol solution of picric acid and could etch steel. Look up www.metallographic.com/etchants/etchants.htm


Why is there a z² label for d orbitals, but no x² and y² labels with corresponding shapes?

TL;DR The $\mathrm{d}_{z^2}$ orbital is a result of solving the Schrödinger equation for the hydrogen atom in the most mathematically convenient way.

To properly understand this, it is necessary to go back to the fundamentals. The complex d-orbitals are obtained by solution of the Schrödinger equation. In general, these d-orbitals are made up of a radial part $R(r)$ and an angular part $Y(\theta,\phi)$. The radial part is not relevant to the question, because its value does not depend on the direction in which you look, so it cannot explain why there is a $\mathrm{d}_{z^2}$, but not a $\mathrm{d}_{x^2}$. So, it is instructive to look at the angular component (the spherical harmonics), which are tabulated below. The normalisation factors are ignored here:

$$\begin{align}
\mathrm{d}_{+2} &\propto R(r) \cdot \sin^2\theta \cdot \mathrm{e}^{\mathrm 2i \phi} \\
\mathrm{d}_{+1} &\propto R(r) \cdot (-\sin\theta \cos\theta) \cdot \mathrm{e}^{\mathrm i \phi} \\
\mathrm{d}_0 &\propto R(r) \cdot (3\cos^2\theta - 1) \\
\mathrm{d}_{-1} &\propto R(r) \cdot \sin\theta \cos\theta \cdot \mathrm{e}^{-\mathrm i \phi} \\
\mathrm{d}_{-2} &\propto R(r) \cdot \sin^2\theta \cdot \mathrm{e}^{\mathrm -2i \phi}
\end{align}$$

These are combined in an appropriate manner to generate the real d-orbitals. It turns out that the $\mathrm{d}_{z^2}$ orbital is simply the same as the original $\mathrm{d}_0$ orbital, which is already real (the radial component $R(r)$ is real). Since $\cos\theta = z/r$ and $r^2 = x^2 + y^2 + z^2$ (spherical coordinates), the angular component can be rewritten as

$$\begin{align}
3\cos^2\theta - 1 &= 3\left(\frac{z}{r}\right)^2 - 1 \\
&= \frac{3z^2 - r^2}{r^2} \\
&= \frac{2z^2 - x^2 - y^2}{r^2}
\end{align}$$

and conventionally we denote this as $\mathrm{d}_{z^2}$. There is no analogous $\mathrm{d}_{x^2}$ orbital because the conventional way of deriving real d-orbitals does not yield any orbital with an angular form of $(2x^2 - y^2 - z^2)/r^2$.

But that still does not properly answer the question. To do so, we should look more closely at how the complex d-orbitals are obtained. The Schrödinger equation, $\hat{H}\psi = E\psi$, only prescribes that we need to find eigenstates of $\hat{H}$, but does not say how. It turns out that the easiest way to accomplish this is to look for solutions which are simultaneous eigenstates of $\hat{H}$ (energy), $\hat{L}^2$ (total orbital angular momentum), and $\hat{L}_z$ (projection of orbital angular momentum onto the z-axis). The eigenvalues of the solutions under these operators are directly tied to the quantum numbers $n$, $l$, and $m_l$, respectively.

Fundamentally, the $\mathrm{d}_{0}$ orbital (i.e. the $\mathrm{d}_{z^2}$ orbital) arises because we have chosen it to be an eigenstate of the operator $\hat{L}_z$. If we were to solve the Schrödinger equation in a different way, by searching for simultaneous eigenstates of $\hat{H}$, $\hat{L}^2$, and $\hat{L}_x$ (instead of $\hat{L}_z$), we would indeed get a $\mathrm{d}_{x^2}$ orbital. This is an equally valid solution of the Schrödinger equation (as it is an eigenstate of $\hat{H}$), but it is simply not the conventional solution.

So, why choose $\hat{L}_z$ and not $\hat{L}_x$ when solving the Schrödinger equation? The answer is simply mathematical convenience: the Schrödinger equation is most easily solved in spherical coordinates, and in spherical coordinates, $\hat{L}_z$ has a particularly nice form of being equal to $-\mathrm{i}\hbar(\partial/\partial \phi)$. The eigenstates of this are none other than $\mathrm{e}^{\mathrm{i}m\phi}$, which is exactly what we observe in the complex d-orbitals at the beginning of this post ($m$ is, of course, quantised, so there are only five valid solutions with $l = 2$, i.e. d-orbitals).

But why does $\hat{L}_z$ have a nice form and not $\hat{L}_x$? Ultimately, this mathematical convenience is a result of the definition of spherical coordinates. Because $\theta$ is chosen to be the angle made by the vector with the z-axis, the z-axis has a special status in spherical coordinates:

$$\begin{align}
z &= r\cos\theta \\
x &= r\sin\theta\cos\phi \\
y &= r\sin\theta\sin\phi
\end{align}$$

So, at the end of the day, it is a completely arbitrary convention. And it should be, too, because no axis should have an inherently special status in a spherically symmetric system like the hydrogen atom. As mentioned previously, you could obtain a $\mathrm{d}_{x^2}$ orbital by solving the Schrödinger equation in a different way. You could even choose to redefine the axes in such a way which gave you a $\mathrm{d}_{x^2}$ orbital (it's quite easy; simply rename $(x,y,z)$ to $(y,z,x)$).

But there is no way you can have both at the same time. Because $\hat{L}_x$ and $\hat{L}_z$ do not commute with one another, it is not possible to "solve" the Schrödinger equation in a way which gives you both $\mathrm{d}_{x^2}$ and $\mathrm{d}_{z^2}$ orbitals at the same time.


We can.

But I see few reasons why it is not used:

Iron is much cheaper than zinc.
There can be remaining residue of iron/zinc, coated by copper, or just being excessive. While copper can be melted away and iron stays, zinc would melt together with copper, causing unwanted impurity (unless wanted for making brass alloys)
If we remove copper for environmental concerns, then by using zinc, we would just replace one evil by another. While both metals are essential for life in small amounts, both are toxic in larger amounts. ( So does iron, but at much higher level)
Removing dissolved iron from waste is easier than removing zinc.


Enantioselectivity of Noyori Asymmetric Hydrogenation

I think this is easy to answer than you think. Let's look at the stereochemistry of compounds in the question, 1g and 1h (I also put the common product, 1a, for convenience):

The stereospecific reduction is on carbon-3 ($\ce{C}$3) in all cases. Accordingly (see diagram above), if you apply Cahn–Ingold–Prelog (CIP) sequence rules for each, you'd find (3R)- for 1a (methyl group is priority 3 based on $\ce{CHH}$ vs $\ce{HHH}$), (3S)- for 1g (isopropyl group is priority 2 based on $\ce{CHH}$ vs $\ce{CCH}$), and  (3S)- for 1h (phenyl group is priority 2 based on $\ce{CHH}$ vs $\ce{CCC}$). All these assignments are $\ce{C}$2 versus $\ce{C}$4 since priory 1 for all three groups on $\ce{C}$3 is $\ce{OH}$.

Thus, we can conclude that the change of stereochemistry is mainly due to the priority rules (not due to the change in stereochemistry on transition states or intermediates). This is also supported by the example given in the same reference (Ref.1) in the box in depicted diagram. Even when bulkiness at $\ce{C}$2 was increased by adding a methyl group, the reaction (with R-BINAP) has undergone same manner to give (R)-configuration to $\ce{C}$3 center.

References:

Ryoji Noyori, Takeshi Ohkuma, Masato Kitamura, Hidemasa Takaya, Noboru Sayo, Hidenori Kumobayashi, Susumu Akutagawa, "Asymmetric hydrogenation of $\beta$-keto carboxylic esters. A practical, purely chemical access to $\beta$-hydroxy esters in high enantiomeric purity," J. Am. Chem. Soc. 1987, 109(19), 5856-5858 (https://doi.org/10.1021/ja00253a051).


High voltage AC water electrolysis not working

From the comments, mcocdawc:

  Was it DC or AC?

OP:

  AC. I was trying to see if it was possible.

There's your problem. Alternating current spends half its time pushing electrons forward through the circuit, and half the time pushing them in reverse through the circuit.  Thus, on average the net current pushed through the cell is zero.

So, leaving aside any metal dissolution/deposition, you were forming tiny, tiny amounts of electrolysis gases at the surface of your electrodes while there was current moving through the cell in one direction, but immediately consuming them again when the current reversed. The total number of coulombs passed in each direction was presumably so small that no gas bubble nucleation occurred before the polarity reversed.

Regardless: STOP USING $\mathbf{500V}$ FOR ELECTROCHEMISTRY EXPERIMENTS. It's way more potential than you need, and terrifically unsafe. In most cases, anything above $40~\mathrm V$ is overkill, and for most experiments that $9~\mathrm V$ battery you mentioned will more than suffice.


Is sodium chloride really odourless? If yes, what do I smell?

Since odour is generally coming from vapour / gas form of a material, and NaCl has zero vapour pressure, it should be odourless.
However:

Moving around solid materials, dust, small particles can fly around that can be dissolved in the nose and interact with the odour sensors.
You almost always have something else together with salt. And if it is not pure, you cannot be sure what makes the odour, the salt or the impurity. Eg. sea salt often contains remains of algae.


Why is the electronegativity of indium greater than that of thallium?

Going down any group of the Periodic Table the general trend is that electronegativity decreases. The relative electronegativities of In and Tl are simply a reflection of this fact. That's the end of the story; you don't need to bring in anything more complicated.

You are correct that the 4f electrons are poorly shielding, and thus the 6p elements (Tl, Pb, Bi, Po, At, Rn) would likely be more electronegative than otherwise expected. But there's no reason why this should make them so electronegative so as to reverse the general trend. None of these elements are more electronegative than their 5p counterparts.


$P 4_{2} 22$ tetragonal group choice of origin for symmetry of special projections

As stated in the International Tables of Crystallography vol A, as a rule the origin is locatedat a centre of symmetry or at a point of high site symmetry. Anyway, the choice of such a coordinate system is not mandatory since in principle a crystal structure can be referred to any coordinate system. In particular, all centrosymmetric space groups are described with an inversion centre as origin, whereas for noncentrosymmetric space groups, the origin is at a point of highest site symmetry.
In the space group n°93 the origin is at the intersection of 3 different 2-fold rotation axes (site 2a: 0,0,0), not at 0,0,1/4. I think this you're confused on this point, since the origin is always at 0,0,0 by definition.
From the trasnformation that you have assumed (y,x,-z+1/2), I guess you are referring to the operation number (7), that is a 2-fold rotation along the crystallographic [x,x,1/4] axis (not a plane! rotations are along an axis). As stated above, you can choose another coordinate system with a new origin shifted in respect the standard one, but any case the new origin will have again 0,0,0 coordinates in the new coordinate system. If you shift the origin, the coordinates of all the symmetry operations will be obviously redefined in the new coordinate system, since they are not independently distributed in space.


If you accumulate 100 identical spheres how many will end up on the surface?

Experimental approach
Here are 100 marbles in a polyethylene bag:

If a marble touches the bag, it is counted as being on the surface. I did not count, and the bag was not perfectly round. A balloon might be better.
Theoretical approach
This works for large $N,$ and the result is as Ivan predicted (from scaling arguments in 2D and 3D, I would imagine). We neglect the packing defects in the bulk, and treat the surface as having no curvature and no defects. The maximal packing density of spheres is well-known by chemists,
$$P_\mathrm{3D} = \frac{π \cdot \sqrt{2}}{6} = 0.74046\ldots\tag{1}$$
The maximal packing density of circles is less well-known on this site, but you can look it up on Wikipedia:
$$P_\mathrm{2D} = \frac{π \cdot \sqrt{3}}{6} = 0.90687\ldots\tag{2}$$
If you have $N$ marbles with radius $r$, you can estimate the radius $R$ of the enclosing sphere, get its area ($4 \pi R^2$), and calculate how many circles (of area $\pi r^2$) fit in that area. The resulting expression scales with $N$ to the power of two-thirds, as Ivan said:
$$N_\mathrm{surface} = 4 \cdot P_{\mathrm{2D}} \cdot \left(\frac{N}{P_{\mathrm{3D}}}\right)^\frac{2}{3}\tag{3}$$
For example, for a million spheres, about four percent are on the surface. For a single sphere, all are on the surface. The expression for $N = 1$ yields $4.4$, pretty bad. For $N = 100,$ it yields $95.2$ That is a not a bad estimate; from the experimental approach, I see more than $30$ marbles on the surface, and the picture only shows one side.
Approximating a sphere with a cube
If you have 125 spheres that pack in a primitive cubic fashion, you would have a cube with 5 spheres in each direction. The top and bottom, right and left, front and back layer would be on the surface, leaving a 3 by 3 by 3 cube in the center. So 27 spheres would be on the inside and the remaining 98 would be on the surface. When we take those same 125 and pack them tighter and such that they fit in a minimal sphere, the surface area would get smaller, and the packing on the surface would get denser, leading to a similar fraction of spheres on the surface.


Adding non stoichiometric amount of NH3 to copper solution

Both your conclusions are absolutely correct.

I'll point out that you'll never get "all" the $\ce{Cu^{2+}}$ to the amine complex. You're trying to get greater than 99% of the $\ce{Cu^{2+}}$ to the amine complex. Because 99% and 99.9% and 99.99% would have virtually the same absorption reading.

To give even more detail, in mildly acid solution the copper will be in a octahedral complex $\ce{Cu(H2O)6^{2+}}$ which is a light blue. As you add ammonia at first $\ce{Cu(OH)2}$ forms, which is then dissolved as the various ammine complexes form.

$\ce{Cu(H2O)_6^2+ + NH3 <--> Cu(NH3)(H2O)5^2+}$
$\ce{Cu(NH3)(H2O)5^2+ + NH3 <--> Cu(NH3)2(H2O)4^2+}$
$\ce{Cu(NH3)2(H2O)4^2+ + NH3 <--> Cu(NH3)3(H2O)3^2+}$
$\ce{Cu(NH3)3(H2O)3^2+ + NH3 <--> Cu(NH3)4(H2O)2^2+}$

The tetraammine complex, $\ce{Cu(NH3)4(H2O)2^2+}$, is a much darker blue than the corresponding complex with water molecules alone. Thus colorimetry is much more sensitive measuring the tetraammine complex that the water complex.

The tetraammine complex has shorter Cu-N bonds than the C-O bonds. Thus the four ammine groups are one plane and the two water molecules at the apexes of the square bipyramidal complex.

Hopefully now the point about not converting all the copper to the tetraammine complex is clear. The various ammine complexes are in equilibrium with each other. Given that $\ce{[NH3] >> [Cu^{2+}]}$  then one can imagine the overall reaction as

$\ce{Cu(H2O)_6^2+ + 4NH3 <--> Cu(NH3)4(H2O)2^{2+}}$

In extremely concentrated solutions of ammonia the final two ammine groups will attach.

$\ce{Cu(NH3)4(H2O)2^{2+} + NH3 <--> Cu(NH3)5(H2O)^2+}$
$\ce{Cu(NH3)5(H2O)^{2+} + NH3 <--> Cu(NH3)6^2+}$


Why is the difference between the melting point and boiling point large for some compounds, and small for others?

well thinking that the same factors influence the melting and boiling point of materials is possibly wrong.
high melting point is due to strong packing(highly symmetric molecules) while  high boiling point is due to large surface area
so if a compound has good packing and less surface area will end up to have very less difference in melting and boiling point and sublime


Confusion between moles and molarity units

$\ce{M}$ is a shortcut for $\frac{\ce{mol}}{\ce{l}}$. $\ce{mol}\%$ on the other side is defined as $\frac{n}{n_{\ce{tot}}}$. In order to convert between the two you need to know the density $\rho$ and the molar mass $M$ (don't confuse this with the unit) of your solution/mixture. You can then do
$$
c_{\ce{mol}\%} = \frac{c_\ce{M} \cdot V_{\ce{mix}}}{V_{\ce{mix}} \cdot \rho \cdot M^{-1}} = \frac{c_\ce{M}}{\rho \cdot M^{-1}}
$$
$c_{\ce{mol}\%}$ is the concentration in $\ce{mol}\%$ and $c_M$ the one in $\ce{M}$.


Chocolate consists of tiny clumps of cocoa solids, sugar, and milk solids dispersed in cocoa butter.  Because the solids aren't soluble in the butter, they tend to clump up, which reduces the quality of the texture of the chocolate, it feels less "smooth."  When you add an emulsifier, the hydrophilic end tends to stick to the relatively more polar surface of the solids (the proteins in the cocoa solids, for example, or the carbohydrates) and then the hydrophobic end interacts favorably with the cocoa butter.  The coating of emulsifier tends to help the clumps stay small and well dispersed in the fat.  That makes it easier and faster to process the chocolate, and it also lets the chocolate last longer before the fat and chocolate solids start to phase separate on a macroscopic scale.


Why does CaCO3 react with HCl, but not with H2SO4?

Your marble chips react on the surface.

In the case of hydrochloric acid, the resulting salt, calcium chloride, is highly soluble in the acid, dissolves and provides further attack to the (new) surface.

With sulfuric acid, the highly insoluble calcium sulfate is formed on the surface of the marble chip. With other words:

Calcium sulfate acts like a protective layer.


Yes. A simple example is an ordinary balloon, which places the gas inside it under pressure (because the tension of the rubber surface adds to the air pressure from outside the balloon), yet the balloon still floats readily.

What you're probably talking about is a theoretical container that can hold highly-compressed hydrogen, like from a supply cylinder, yet has negligible weight. At what point, then, would the hydrogen weigh more than air and the container would settle to the ground?

To answer that, we'll look at the ideal gas law: $PV = nRT$, or equivalently, $P = \dfrac{nRT}{V}$. For our purposes, we'll ignore the temperature increase of the compressed gas by saying this container, while having negligible weight, also has an efficient cooling system so the temperature delta is negligible.

At STP (273*K, 100kPa), one liter of volume filled with hydrogen gas contains:

$$100 = \dfrac{n(8.3145)(273)}{1}$$
$$100 = 2269.8482n$$
$$n = .0441\text{ mol}$$

With hydrogen having a molar weight of 1.00794 g/mol (and $\ce{H2}$ gas having double that), the density in g/L of hydrogen gas at STP is .08890 g/L.

Borrowing from ashu's answer, since his math and mine are within 1% error, the density of air at STP is 1.225g/L (1 m3 = 1000 L). So, in order to have the same mass of gas in the same volume, and thus the same density, you would need 13.78 times the amount of gas than you'd have in the same volume at STP (pretty much 1.225 mol). Because, given a unit volume, double the gas equals double the pressure, the resulting pressure of a container of negligible mass holding enough compressed hydrogen to not float in air is 1377.953kPa.

In SAE units, that's actually not much in relative terms; only 199.855 psi (call it 200 so it would actually sink in air and not be neutrally buoyant). That's the water pressure at a depth of about 460 feet (140.72m), which human freedivers have matched using dive sleds.


What is the correct temperature probe for a wet etch solution?

Teflon is resistant against corrosive chemicals and stable up to temperatures of about 200 °C, so it is suitable for use in hot aqueous salt solutions. Using a coating or casing made of plastic which is unreactive and resistant against heat should be a good way to protect temperature probes.


Relation between the virial coefficients and van der Waals constants

You have to bring both equations into the same form and then compare the coefficients. You seem not to have used the tip with the Taylor expansion and I guess that's why you got an incorrect result.

As you wrote: The virial equation of state is

\begin{align}
p &=\frac{nRT}{V} \left(1+\frac{nB}{V}+\frac{n^2C}{V^2}+ \ldots \right) \\
  &= \frac{RT}{V_{\mathrm{m}}} \left(1+ B\frac{1}{V_{\mathrm{m}}}+ C\frac{1}{V_{\mathrm{m}}^2}+ \ldots \right)
\end{align}

and the van der Waals equation is

\begin{align}
 p &=\frac{RT(1-\frac{b}{V_{\mathrm m}})^{-1}}{V_{\mathrm m}}- \frac{a}{V_{\mathrm m}^2} \ .
\end{align}

Now, you have to bring the van der Waals equation into the same form as the virial equation by factoring out $\frac{RT}{V_{\mathrm{m}}}$

\begin{align}
 p &= \frac{RT}{V_{\mathrm{m}}}\left( \left(1 - \frac{b}{V_{\mathrm{m}}} \right)^{-1} - \frac{a}{R T V_{\mathrm{m}}} \right) \ ,
\end{align}

using the Taylor expansion $\left(1 - \frac{b}{V_{\mathrm{m}}} \right)^{-1} = 1 + \frac{b}{V_{\mathrm{m}}} + \frac{b^2}{V_{\mathrm{m}}^2} + \ldots$ and ordering the terms according to their power in $\frac{1}{V_{\mathrm{m}}}$

\begin{align}
 p &= \frac{RT}{V_{\mathrm{m}}}\left( 1 + \left(b - \frac{a}{R T} \right) \frac{1}{V_{\mathrm{m}}} + b^{2} \frac{1}{V_{\mathrm{m}}^{2}} + \ldots \right) \ ,
\end{align}

so that you can compare the coefficients in both equations to get the result

\begin{align}
 B &= b - \frac{a}{R T} \\
 C &= b^2 \ .
\end{align}


The form of the equation is correct, but one of the variables is wrong. It should be:

$$\mathrm{R} = N_{\mathrm A}k_{\mathrm B}$$

The Boltzmann constant ($k_{\mathrm B}$) is for one molecule (or atom); the universal gas constant ($R$) is for one mole of molecules (or atoms).

Therefore, in this relationship - and which is clear from dimensional analysis ($R$ is per mole) - $N_{\mathrm A}$ is the Avogadro constant, and not number of moles: in your argument, $n$ is a dimensionless integer, and thus the units do not work out. You need to use $N_{\mathrm A}$ instead of $n$. Note that the Avogadro constant can be denoted by either $L$ or $N_{\mathrm A}$.

The proper relationship is:

$$R = (\pu{6.022*10^{23} mol-1}) \cdot (\pu{1.380648*10^{-23} J*K-1}) = \pu{8.314 J*K-1*mol-1}$$

Note that I have included more significant digits for the Boltzmann constant here to get your value for $R$.


You have to consider the system. If energy is being added to the system the sign is positive, if energy leaves the system the sign is negative.


Quantitative expression for steric hindrance

There are two ways I know of to measure steric hindrance. Once such quantitative measurement are A values, which give the energetic preference of having the substituent on a substituted cyclohexane ring in the equatorial position versus the axial position. This preference is a result of 1,3-diaxial strain.

Additionally, the Taft equation is a linear free-energy relationship used to measure the steric effects of a substituent on the rate of reaction. The equation is given by:

$${\displaystyle \log \left({\frac {k_{s}}{k_{\ce {CH3}}}}\right)=\rho ^{*}\sigma ^{*}+\delta E_{s}}$$

where $\displaystyle \frac {k_{s}}{k_{\ce {CH3}}}$ is the ratio of the substituents rate of reaction relative to a methyl group, $\sigma^*$ is the polar substituent constant that describes the field and inductive effects of the substituent, $E_s$ is the steric substituent constant, $\rho^*$ is the sensitivity factor for the reaction to polar effects, and $\delta$ is the sensitivity factor for the reaction to steric effects.$^{[1]}$

Though A values are useful as a quick reference in deducing the major products of an unfamiliar reaction, the Taft equation would be used preferentially when doing an in depth mechanistic study of a reaction for publication.

$^{[1]}$ Wikipedia, Taft equation


How to calculate S² value of a broken-symmetry wave function?

The spin operators for the $p^\textrm{th}$ MO are defined as follows:
\begin{align}
S_{x,p} &= \frac{\hbar}{2} \sum_{\tau,\tau'} c_{p,\tau}^\dagger \hat{\sigma}^x_{\tau,\tau'} c_{p,\tau'}, \\
S_{y,p} &= \frac{\hbar}{2} \sum_{\tau,\tau'} c_{p,\tau}^\dagger \hat{\sigma}^y_{\tau,\tau'} c_{p,\tau'}, \\
S_{z,p} &= \frac{\hbar}{2} \sum_{\tau,\tau'} c_{p,\tau}^\dagger \hat{\sigma}^z_{\tau,\tau'} c_{p,\tau'}, \\
\end{align}
where $\sigma_{\tau,\tau'}^{x}$ denotes the matrix elements of the Pauli matrix $\sigma^x$ and so on. Any spin-observable can then be constructed from these basic blocks. For instance, $S^2$ operator takes the form
\begin{equation}
S^2 = \sum_{p \in MO} \frac{3}{4}\left( c_{p,\uparrow}^\dagger c_{p,\uparrow} +
c^\dagger_{p,\downarrow} c_{p,\downarrow} - 2c^\dagger_{p,\uparrow} c^\dagger_{p,
\downarrow} c_{p,\downarrow} c_{p,\uparrow} \right )
\end{equation}
All one need is to evaluate the expectation value of these operators over the wavefunction of interest.


Let me start by saying that aromatic compounds are not exceedingly stable, they are just much more stable than antiaromatic compounds. However, they can still be hydrogenated exothermically to the corresponding cycloalkanes.

$$\ce{C6H6 + 3 H2 -> C6H12}$$

But why are they more stable than antiaromatic ones? Let’s take a look at the orbital scheme for cyclopentadienyl $\ce{C5H5^{?}}$ (the scheme is the same for radical, anion or cation — at least theoretically). (Taken from this site).

The pictures of the molecular orbitals correspond to their energies. The e means that the two orbitals next to each other are degenerate, i.e. have the same energy. Let’s fill 6 $\unicode[Times]{x3c0}$-electrons ($\ce{C5H5-}$) into this scheme. We get three completely filled orbitals and two rather antibonding unoccupied orbitals. Looking good.

Now let’s extract a single electron to create the $\ce{C5H5^{.}}$ radical. The electron must come from one of the two HOMOs, so we end up with one degenerate orbital completely occupied, the other one only occupied by a single electron (usually drawn with ↑ or $\unicode[Times]{x3b1}$-spin). Now we’ll abstract another electron to arrive at the $4\unicode[Times]{x3c0}$ $\ce{C5H5+}$ cation. Two of the remaining four electrons can be filled into the lower-lying $\ce{a_1}$ orbital, the other two need to be in $\ce{e1}$ and since the latter is degenerate we arrive at a triplet state: Both electrons have $\unicode[Times]{x3b1}$-spin. This is rather unstable.

The triplet state and hence the instability can be avoided by distorting the molecule. If elements of symmetry ($\ce{C5H5^?}$: most importantly the plane of symmetry) are lost, the equivalence of the orbitals here labelled $\ce{e_1}$ is also lost and there is no reason not to pair the electrons into a single orbital. That’s why cyclobutadiene is not a square but a rectangle.

If I remember correctly, the results of simple approximations are actually mathematically equivalent to taking a circular $\unicode[Times]{x3c0}$-system, putting it on one corner and drawing an orbital energy on every carbon — the cyclic Hückel approximation. And since you will always get pairs of degenerate orbitals by this method using perfect shapes, that gives us the Hückel rule of $4n+2$ — two for the lowest orbital (the corner the circle is resting on) and $4n$ for every pair of orbitals that then have the same energy.


Mechanism for acylation of benzene with succinic anhydride

The mechanism shown in the question seems reasonable enough to me. I would note that ring opening before the attack on the $\pi$-system is likely, but not absolutely necessary. Loss of $\ce{AlCl3}$ may not occur until workup. Direct proton transfer seems unlikely, but is a inconsequential detail. Note however, that the electron pushing arrow is drawn opposite to what it needs to be for that transfer.


Why doesn't the expermintal molar solubility match the one calculated by the Ksp value?

The $\mathrm{p}K_\mathrm{sp}$ value of $15.69$ for $\ce{HgCl2}$ given by your reference must have an error. First of all, $\ce{HgCl2}$ is fairly soluble in water at $\pu{20 ^\circ C}$. Wikipedia lists it as $\pu{65.7 g L-1}$, yet lists solubility of $\ce{Hg2Cl2}$ as $\pu{3.25e-4 g L-1}$. Meantime, I found an University of Arkansas, Little Rock website, which has listed $K_\mathrm{sp}$ value of $\ce{Hg2Cl2}$ as $\pu{1.43e-18}$ ($\mathrm{p}K_\mathrm{sp} = 17.84$).

The molar mass of $\ce{Hg2Cl2}$ is $\pu{472.09 g mol-1}$. Thus, theoretically, the solubility ($s$) of $\ce{Hg2Cl2}$ can be calculated similar way you did:

$$s = \left({\frac{\pu{1.43e-18 mol3 L-3}}{4}}\right)^{\frac13}\times \pu{472.09 g mol-1} = \pu{3.35e-4 g L-1}$$

I think this is close enough. :-)


Its because the gold nanoparticles are the star of the show. The researchers were just looking at how surface modification using the silver would alter the morphologies of the gold nanoparticles. Probably the researchers wondered to themselves one day, "How could we possibly control the shape of these gold nanoparticles?" and decided to try with silver.
The bimetallic part would suggest that the researchers were aiming for gold-silver nanoparticles (agglomerated together, not one on the surface of another like in the research)


Number of total stereoisomers of a piperazine derivative

I believe this is a sophomore organic chemistry text book question, so I won't elaborate possible stereoisomers involving lone pairs on nitrogen atoms. Nonetheless, 3,6-dimethylpiperazine-2,5-dione has 2 chiral carbon centers so that it should theoretically have $2^2 = 4$ stereoisomers. If you look at the molecule, it has center of symmetry (i):

Therefore, two of the isomers are identical and we call it meso-isomer. Thus, total isomers of this molecule are three: ($3R,6R$)-, ($3S,6S$)-, and  ($3R,6S$)- (($3S,6R$)- is essentially ($3R,6S$)-).


Determining weight-percent methanol in water from specific gravity and temperature

You are looking for the excess molar volumes of water and methanol.  A quick Google search shows that these quantities have been determined and re-determined dozens of times.  The art is to find which paper has the best values.

The CRC Handbook of Thermophysical and Thermochemical Data and Landolt-Börnstein may have done the sanity check already.


How to calculate the charge on silicates?

In this answer, the basic concept of how to understand what figure they are and how much charge they have would be discussed.
As OP has well stated, the basic unit of any silicate is the $\ce{SiO4^4-}$ tetrahedron which has four oxygens that are connected to one silicon atom.  Each unshared oxygen provides -1 charge and each shared oxygen contributes 0.5 oxygens to one tetrahedral unit.
Now, depending on how these units interconnect and how oxygens are shared, we get multiple different structures. I shall discuss three of them here since that should be enough to get the idea across.
Type 1: Chain Silicates

In this type of silicates, as you can see above, the tetrahedron shares one oxygen on either side and so the basic unit here would be the $\ce{SiO3^2-}$ unit, since two oxygens are being shared.
Since two oxygens are shared per tetrahedral unit, the basic unit would have 3 (2 unshared + 0.5 * 2 shared) oxygens and 2- charge.
Type 2: Pyro silicates

Here, the one oxygen in the middle is shared by the two tetrahedral units. This means that there is a total of 6 free oxygen atoms that are only bonded to the silicon with a single bond. Therefore the structure arises as $\ce{Si2O7^6-}$.
Since one oxygen is shared per tetrahedral unit, the basic unit would have 3.5 (3 unshared + 0.5 * 1 shared) oxygens and 3- charge. However since the values need to be integers $\ce{SiO_{3.5}^3-}$ would be written as $\ce{Si2O7^6-}$
Type 3: 3D Silicates
In this case, all the oxygen atoms are shared leading to the formation of $\ce{(SiO2)_n}$ molecules.
Further Reading
https://www.adichemistry.com/inorganic/p-block/group-14/silicates/silicates-1.html


Find the Structural Formula of Hydrocarbon Compound X

Your reasoning is correct. Are you confused because acetone is not a hydrocarbon? That has to be a mistake in the wording of the question. Another common mistake is to forget that there are some reactions that aldehydes can do that ketones are inert to.

There are seven constitutional isomers for $\ce{C3H6O}$, not counting the enols, which are the minor tautomers of acetone and propanal:

Acetone - a ketone
Propanal - an aldehyde
Cyclopropanol - an alcohol
Allyl alcohol - an alkene and an alcohol
Methoxyethene - an alkene and an ether
Oxacyclobutane - a cyclic ether
Propylene oxide - an epoxide

Your reasoning is correct based on the chemical tests:

Sodium reacts with alcohols - a negative sodium test means no alcohol functional group is present, eliminating allyl alcohol (and the enols if you consider them).
Bromine reacts with alkenes and alkynes (and a small number of other functional groups, like phenols) - a negative bromine test means no alkene functional group is present, eliminating allyl alcohol, methoxyethene, and the enols.
Ammoniacal silver nitrate (Tollens' reagent) reacts with aldehydes - a negative Tollens test means no aldehyde function group is present, eliminating propanal.

The results of the qualitative test leave you with three possibilities, acetone, oxacyclobutane, and propylene oxide. With the evidence you have, you might not be able to distinguish between the three.

An iodoform test would confirm or eliminate acetone.

A test using silver nitrate and periodic acid would confirm or eliminate propylene oxide.


What actually happens in real life depends on a lot of things. Factors like the shape of the pot can make a big difference to how much of that 2500 W actually goes into vaporizing the water and how much of it is lost to the surroundings. If we assume that all 2500 W is going into the water, it makes things a lot simpler. If the water is already at the boiling point, then all we have to do is calculate the total amount of energy needed to vaporize all the water:

The enthalpy of vaporization of water is 40.65 $\mathrm{kJ/mol}$, which means 40.65 kJ of heat needs to be added to turn one mole of liquid water to steam. 50 L of water is 2775 moles:
$$\mathrm{\frac{(50\ L)\left(1000\frac{g}{L}\right)}{\left(18.02\frac{g}{mol}\right)}=2775\ mol}$$

So the total heat we need is:
$$\mathrm{(2775\ mol)\left(40.65\frac{kJ}{mol}\right)=112.8\ MJ}$$

1 watt is one joule per second, so for a 2500 W heat source:
$$\mathrm{\frac{112.8\ MJ}{2500\ J/s}=45\ ks}$$
Or, approximately 12.5 hours. In real life, it may be significantly longer for various reasons, but this is the fastest it could go.


How to quickly diffuse a dye through still water

If you want a homogeneous distribution of dye, you will have to stir the solution.
If you want to use electrophoresis, you need a charged dye.

Azo dyes used for hair coloring have colour and charge, for instance Basic Red 51 and Basic Yellow 87:

Of course, heat will increase the diffusion rate but will also disturb the water.


Site of first protonation of imidazole and aminoimidazole

For imidazole, the lone pair of pyrrole-type nitrogen (the one with hydrogen) is involved in aromaticity of the whole compound, as it contributes two π-electrons in addition to four of those from carbon-carbon or carbon-nitrogen bonds, which makes it 6, which is a classical sign of Huckel aromaticity. In fact, this NH-nitrogen tends more to act as an acidic center than as a basic one, as it loses its hydrogen when treated by strong bases, like $\ce{NaNH2}$.

The pyridine-type nitrogen has its lone pair out of ring conjugation, thus it can act as a acceptor of hydrogen ion, placing its two electrons into a vacant orbital of a $\ce{H+}$.

For reference, you can see any heterocyclic chemistry textbook. I prefer the one authored by Joule, Mills. Or this.

When it comes to 2-aminoimidazole, this publication  (which is unarguably very old) says all three nitrogens are involved in protonation, because the structure is a derivative of guanidine, a strong organic base. Which seems only legit to me if it said that the contribution of pyrrole-type nitrogen is negligible, as aromaticity is ruined in that case. I could also predict that exo-aminogroup gets protonated more likely, as its basicity is somewhat higher: its lone pair is on a sp3-orbital, rather than the lone pair of endo-nitrogen (the one without H), where it is on a sp2-orbital, thus, the contribution of s-orbital is higher and it is closer to the nucleus of nitrogen atom, decreasing its availability to act as a covalent bond acceptor.


Is change of pH of water with respect to temperature linear?

To summarize... in aqueous solutions:

Basic means $\ce{[OH−] > [H+]}$
Acidic means that $\ce{[H+] > [OH-]}$
Neutral means $\ce{[OH−] = [H+]}$

There is a for the Wikipedia article for the autodissociation of water. The  chemical equation for the autodissociation is:
$$\ce{H2O <=> H^+ + OH^-}$$
The mathematical relationship between $K_d^t$, $\ce{[H^+]}$, $\ce{[OH^-]}$ and $t$ is: $$K_d^t =\ce{[H+][OH-]}\quad\text{for }t > 32\text{ C}$$
Note that $K_d^t$ changes with temperature $t$, but for pure water at any temperature, $t$,  the water will be neutral, and $\ce{[OH−] = [H+]}$. The temperature dependency of $K_d^t$ also means that pH + pOH is not always 14.

The Wikipedia article shows that the relationship between $K_d^t$ and temperature is a curve. (The curve shown is a bit odd in that water boils at 100 C, so higher temperatures must be taken under pressure.)

Also regardless of temperature the $\text{pH}$ of an aqueous solution is defined as $\text{pH} = -\log{\ce{[H^+]}}$ so the pH does not depend on $\ce{[OH^-]}$.


If you are talking about the 1.13 and 1.14 equations, than whta he did was just a simple subistituction where $[A] = x,\ d[A] = dx$, if not, please identify the step.

And, yes, you are going to get diferent equations for $[A]_{(t)}$ depending on the order of reaction, $0^{th}$ order gives you a linear equation, $1^{st}$ order gives you an exponetial equation, and all the orders higher than 1 gives you a fraction diference like: $$\frac{1}{A_t^n}-\frac{1}{A_0^n} = n\ k\ t$$ where n is the order of reaction minus 2.

The step where they make $[A]^2$ is only because you have an elementary equation, it can not be assumed to be that if the equation was golbal or not elementary.


Why are rare earth metals and platinum group metals are often found clustered together in ores

The factors that generate mineral concentrations are complex and often only partly known
Introduction: geology is complicated
The one thing we can be very certain about is is that the distribution of minerals in the earth's crust has very little to do with the primordial origins of the component elements (that is where they came from in the early solar system and how they were originally generated). Most "heavy" elements are originally formed in the cores of supernovae and not in either the big bang or in normal stars.
The distribution of elements in the earth is mostly unrelated to the cosmic origins of elements because the earth's crust is not static but is frequently churned up by a variety of processes on a geological timescale. If we go back far enough in the history of the planet, everything was molten and this allowed some of the denser components to separate out before the surface cooled enough to be solid. The led to the core being mostly metallic (and consisting of mostly iron and nickel). Higher layers contain less dense minerals containing a lot of silicate minerals. At the top there is a thin layer, the crust, which is where we find useful minerals and it is even more concentrated in silicate minerals and even less dense.
But those early processes are mostly irrelevant to what we see in the crust. The crust is subject to a variety of processes that churn up the content, many of which concentrate specific components.
On a very large scale we have plate tectonics where large arts of the crust are both made and recycled on a geological timescale. To simplify greatly, new continental rock at emerges at one place (eg the mid atlantic ridge) and is consumed by subduction at other plate boundaries (eg the Andean belt in South America).
On a smaller scale (though often related to plate tectonic boundaries) volcanism takes molten rock from relatively deep in the crust and spews it out to the surface, bringing minerals of new compositions to the surface and altering the surrounding minerals by heat and pressure.
Also the surface is subject to weather which causes erosion (leading to both chemical and physical separation of the mineral content of rocks) and the uncovering of rock layers originally formed much deeper in the crust by the cooling of liquid mantle. Another product of erosion is sedimentary rocks where the things being eroded reform into new types of rock.
Plus life itself leads to the creation of some rocks. Some creatures collect carbonate minerals to make their protective bodies, for example, and if these concentrate when they dies, they may deposit layers which over time become new rock types (eg limestone).
To make things even more complicated these processes may interact. Volcanic heat or pressure caused by burial or major stresses may cause major changes to other types of rock, altering their mineral content in the process. Limestone may be recrystallised into marble; plant deposits may be transformed into coal or oil. And other separations are caused by related processes. Metamorphism is often associated with fluid flow which, depending on the composition of the fluid, may move specific minerals around, extracting specific components from some minerals and recrystallising them in voids left by the stresses associated with the metamorphic processes.
In short, things are pretty complicated in the crust and the dynamics will churn things up a lot over geological time. The one thing we can be sure of is that what we see now is not  primarily dominated by the primordial origin of the elements.
There are three major processes that concentrate specific ores
I'm going to greatly simplify some of the things that matter here but there are basically three important processes that concentrate things. Not that geologists can typically reach a consensus on what specifically happened with many deposits.
The three processes are:

separation due to erosion
separation due to differential crystallisation of liquid rocks
separation caused by metamorphic processes

Some major mineral deposits of economic importance (such as the Klondyke deposits in Canada) are caused by the first type of process. Some major platinum and related mineral deposits are from similar processes. What is basically happening is that the metals or minerals containing the metals are concentrated by flowing water because they are denser than the bulk of other minerals from the source rock. This is the same process that prospectors practice when panning for gold in rivers (fast flowing water tends to wash away the less dense clay minerals and leave the denser specs of gold). Over a geological time scale this process sometimes leads to very significant concentration of "heavy" minerals. The association depends on the density of the specific minerals and their specific presence in the rock being eroded. But, if the eroding rock contains many dense minerals, then the process can concentrate them all in the alluvial deposit.
But why would some rocks originally contain more than their fair share of specific minerals? One reason is the second geological process that leads to selective concentration of some minerals. This is that, as liquid rocks in the mantle cool, different minerals will crystallise from the mix at different times. a visible manifestation of this process can be seen in many of the polished granites used to decorate kitchen tops or the floors and walls of buildings. Granites consist of three key minerals: feldspars, quartz and mica each with very different mineral contents. The rock is usually formed deep in the crust when a large body of liquid rock cools. But the feldspars crystallise first, giving the large colourful crystals that make the polished surfaces so attractive. Sometimes the large crystals even show patterns of flow in the liquid source rock. The important general point is that the composition of the liquid changes as crystals form and this may concentrate some components. But this process of selective crystallisation is very general and explains why some particular concentrations exist.
The largest concentration of platinum group minerals is in a formation in southern africa called the Merensky "reef". The key concentrations of minerals appear to have been caused by separation as the rock crystallised (according to this book):

“Present opinion is that the Complex was intruded from a magma that was undergoing some differentiation, but was intruded in discontinuous phases. There is an overall trend in the mineralogy and chemistry of the basic rocks that is normal, and the layering of individual rock units is thought to be due to settling out of crystals according to density modified by connection currents flowing in the magma.”

The third process is also important for many valuable minerals. Metamorphic processes, for example, often involve heat and pressure but also hydrostatic processes. The pressure may crack existing rock leaving voids which are filled with high pressure liquid water containing various minerals that can selectively solubilise the contents of the rock and, later, deposit it in the voids. Once chemical property that can selectively separate gold and platinum group metals occurs when the water contains a lot of sulfur, selenium or tellurium and is a reducing environment (eg sufides rather than sulfates). Many precious metals will go into solution in such environments only to be deposited later from the solution as the state changes. Many gold deposits are like this as are some platinum group deposits.
Rare earth (lanthanide-containing) minerals are thought to be concentrated by the same sorts of processes but are, despite rare earths being pretty common in the crust (far more common that precious metals), much less common as economically viable deposits. This may be because the specific chemical process that concentrate them in the first place are less common. This is partially because they have large atomic radii and high charge and that means they don't fit well into the commonest silicate types of mineral. The key minerals they do occur in are carbonate and phosphate based. And these are often thought to occur near locations where continental crust is being consumed at a plate boundary where metamorphic processes may increase the presence of phosphate and carbonate in the resulting rock. See this presentation for some ideas. But they can also be concentrated by sedimentary processes where solutions eroded from source rocks are selectively absorbed into clay minerals elsewhere.
In general the reason why rare earths are found together is that they are chemically similar so there are few natural processes that will separate them (heck, the industrial processes are very hard and expensive).
Summary
There are many processes, both physical and chemical, that can cause concentrations of particular minerals in the earth's crust. Some of these depend on chemistry but others are simple physical processes. But geological chemistry is pretty messy and complicated and, even now, geologists can't always agree what particular process created a deposit. But there are broad types of process that probably contribute some of which depend on the similar chemistry of particular groups of elements. Rare earths are very chemically similar, for example, and precious metals have some chemical similarities (especially solubilisation in reducing environments with sulfur and other group 16 elements) that enable geological processes to concentrate them in some locations.
Some material here is from The Atlas of Economic Mineral Deposits which, though fairly old (1979) is worth a read to get a feel for how complex this subject is.


Electrolysis of aluminum in a solution of magnesium sulfate

Aluminum is not a good idea for carrying out electrolysis, because this metal is covered by a thin layer of aluminum oxide (or alumina) that is insulating. The current passing between these electrodes must be really weak, because the resistance of this layer is high. So that you need a high tension to break this layer and make an electrolysis. But, as soon as aluminium is in contact with water, a new aluminum layer is formed. The majority of the electric energy is wasted and lost for heating the solution. This is non-sense. The powder that you get is probably aluminum oxide coming from the broken alumina layer I told you about earlier. And magnesium is not involved in this reaction.


How should a triple bond be depicted in skeletal structures?

My guess, is that you forgot to take the linear disposition of the carbon-carbon triple bonds into consideration.

So the bond lines cutting through four carbon atoms will be linear, and not zig-zagged as in the rest of the molecule (the latter implies tetrahedral geometry accorded by the $sp^3$ carbon-carbon bonds)

The first diagram may feel more "graceful" (the term I used when I defended my incorrect depiction of the triple bond to my teacher when he asked me a similar question a while back), but it is incorrect.


What does this property of transition metals mean?

Take the electronic configuration of copper (atomic number $29$) $\ce{[Ar] 3d^10 4s^1}$. According to the first portion of the definition, copper doesn't have incompletely filled subshells. But, copper readily gives rise to cation $\ce{Cu^{2+}}$ that has the electronic configuration $\ce{[Ar] 3d^9 4s^0}$ with incompletely filled d subshells. So, copper is considered as a transition metal.

Now, you can guess why zinc (atomic number $30$) is not considered as a transition metal.


What reagents can be used to distinguish pentose and hexose sugars?

I'm not going to address all of the reagents, but I want to highlight a specific approach in this partial answer.

Occasionally, knowing some history of chemistry is quite useful when thinking about how to solve problems. In this specific case, I direct you to the work of Emil Fisher (chemistry Nobel 1902), who determined the structure of all hexose sugars. He did this by building up the sugars from smaller subunits while keeping track of symmetries.

This immediately suggests a solution where one performs reduction (or oxidation) of the aldehydes. In other words, make the both ends of each molecule look the same. This is a valid strategy because the underlying symmetries of glucose and xylose are different.

The penta-ol is meso and therefore achiral, while the hexa-ol is chiral and optically active since you started with one enantiomer.

Thus, any reagent that can produce the bis-acid or bis-aldehyde without epimerizing any stereocenters along the way or any reagent that can reduce the aldehyde to an alcohol will provide a suitable method of differentiation in this specific example.


Find the velocity constant K given the speed

You are really confusing yourself. The problem is very simple. First of all, remove the fact that the rate of reaction depends on all reactants from your mind. It is not the case. Sometimes, even though multiple reactants are involved, only some of them actually matter when we need to find the rate. Let me explain it without chemistry.

Consider a factory where a car is made. The first level makes the engine in 1 hour. The second level makes the tyres in 1 hour. The third level makes the rest of the parts in 1 hour. The fourth level puts them all together in 1 day. Now, even though all the 4 levels are involved in the process, it is the slowest level (level IV) that determines the rate of the process because the rest of the levels can keep sending the parts, but the finished car won't exit the factory unless level IV puts them all together. This is the case with chemical reactions too! The slowest step decides the order and rate of the reaction.

Now, back to the problem. Since they have given that the reaction follows first order kinetics in $[A]$, the rate law can be written as $R=k[A]$. After a little bit of integration and other stuff, we have the equation, $k=\frac{ln(2)}{t _{1/2}}$ , where $t _{1/2}$ is the half-life, i.e. time taken for concentration (actually, it's no. of molecules) of reactant to become half its original value. In this problem, $t _{1/2}$=10.0 seconds. Substitute and find the answer.


How or why can carboxylic acids be protonated?

A general acid-base reaction is one of the kind $(1)$.

$$\ce{\underset{\text{acid}}{HA} + \underset{\text{base}}{B} <=> \underset{\text{conj. acid}}{HB+} + \underset{\text{conj. base}}{A-}}\tag{1}$$

All that the acid needs to react as an acid is some hydrogen atom bonded to something, and all that the base needs to react as a base (although it is not shown here) is a lone pair which can accept the acid’s proton. Depending on the relative strength of the acid and the conjugate acid, this equilibrium will be shifted. If the acid is stronger, it will be shifted towards the product; if the conjugate acid is stronger, towards the reactant. If the acidity are within reasonable range of each other, you can always expect a catalytic amount of the stronger acid to be protonated.

The above was the general introduction; now how does this apply to your question?

The ‘acid’ you are talking about is a carboxylic acid. These have the functional group $\ce{-COOH}$ and each of the oxygens in there has two lone pairs. Thus technically, a carboxylic acid could be protonated up to four times.[1] We only need one further protonation, so we can define the carboxylic acid to be compound $\ce{B}$ above — the base in the acid-base reaction.

In fact, acid-catalysed esterification is typically catalysed by the addition of a small drop of sulfuric acid. An average carboxylic acid has a $\mathrm{p}K_\mathrm{a}$ value of $\approx 5$, sulfuric acid $-3$. The hydronium ion $\ce{H3O+}$, that forms if water takes the place of the base in $(1)$, has a $\mathrm{p}K_\mathrm{a}$ of $-1.7$ or $0$, depending on who you ask. Other alcoholic oxonium ions have similar $\mathrm{p}K_\mathrm{a}$ values. The lower the $\mathrm{p}K_\mathrm{a}$, the stronger the acid. Thus, we can assume that most carboxylic acid molecules remain as $\ce{-COOH}$, while the catalytic amounts of sulfuric acid (practically completely deprotonated once) protonated a few solvent molecules at equilibrium. The stronger acid — sulfuric acid — practically governs that none of the weaker acid — the carboxylic acid — can lose its proton.[2]

Even though it is equilibrium, a few of the acid molecules — the ‘next-weakest links in the chain’ — will also be protonated at a given point in time, even though we can expect the $\mathrm{p}K_\mathrm{a}$ value of a protonated carboxylic acid to be rather low. This small proportion is enough to get the reaction going.

Notes:

[1]: However, each successive protonation introduces a positive charge which always makes further protonation less favourable. A positively charged compound is, generally, also more inclined to transfer hydrogen to something else, i.e. is more acidic.

[2]: This can actually be extended. There are a number of superacids with even lower $\mathrm{p}K_\mathrm{a}$ values — in fact, those values are so low that they can no longer accurately be calculated and instead a different value, the Hammett acidity is used. Some of the strongest superacids are even able to protonate methane which is unsuspiscious of providing a lone pair an acidic proton could interact with. Hence Ivan’s comment of:

  Everything can be protonated, given enough effort.


Repeating your line of reasoning, if $C_{\mathrm m,p}<C_{\mathrm m,V}$ you would conclude based on the expression you provide ($C_{\mathrm m,p}-C_{\mathrm m,V} = T V_\mathrm m \alpha^2/\kappa$) that $\kappa<0$ which in turn would mean that

$$\left(\frac{\partial V_m}{\partial P}\right)_T > 0,\label{eqn:1}\tag{1}$$

which is not physically reasonable except presumably in some very unusual circumstances. Therefore, either the question statement contains a fallacy, or the above analysis contains a fallacy, i.e. $\eqref{eqn:1}$ is reasonable. My bet is on the first scenario, as you suggest.

What seems to happen is that near $\pu{4 °C}$

$$C_{\mathrm m,p} = C_{\mathrm m,V}\label{eqn:2}\tag{2}$$

because the change in the density or molar volume with $T$ (and therefore also the thermal expansion coefficient $\beta$ - usually designated with the symbol $\alpha$) becomes zero, that is there is a (well known) minimum in the dependence of the density or molar volume on $T$ at constant $P$, illustrated in the following figure (source: wikipedia)

In agreement with this, if you inspect available heat capacity data you will see that the isochoric and isobaric heat capacities converge as the temperature of liquid water nears $\pu{0 °C}$.

The International Association for the Properties of Water and Steam has released a document (IAPWS R7-97(2012)) containing expressions for the heat capacities of water parameterized based on experimental data (see Table 3).


Rate equation for an elementary reaction

The reaction order with respect to a reactant species cannot be determined from the chemical equation alone. You would have to be given more information. In real life, orders of chemical reactions are found experimentally.


When we add soap in low concentrations to water it reduces the surface tension. Why? What is the purpose of this?

The lowering of surface tension is a side effect, though it does relate to the reason why we use soaps when cleaning things.

Many of the things we want to clean are not naturally soluble in water (e.g. things like vegetable oil). The structure of soaps and detergents usually consist of molecules with two parts: one is water-loving and one is oil-loving. When soaps are added to mixtures of oily things and water, they align so the oil-loving parts of the molecules surround the oil and present only a water loving surface to the bulk water. Thus the oil is made much more soluble than it would be otherwise (so it is possible to remove it from your clothes, hands or dishes despite most of the liquid being water).

In reality the situation is quite complicated but the general principles of how soaps and detergents work is captured by this basic view.

The effect on surface tension is related to this. The soap molecules tend to cluster at the surface of the water (to get their oil-loving ends away from the water environment). This breaks up the surface structure of water molecules that normally create strong surface tension. But this effect is basically the same effect that enables soaps to solubilise normally insoluble things in bulk water.


If polar non-polar intermolecular forces are stronger, why do polar compounds elute first in GC?

I think you're incorrectly remembering something you've been told. People speak of the adage "like-dissolves-like" which really means that polar things interact best with polar things and nonpolar interact best with nonpolar. There are always exceptions and fringe cases, but that adage works pretty well here. Nonpolar molecules interact with other nonpolar molecules pretty much exclusively through dispersion forces. These are the so-called "instantaneous dipoles" you have probably heard of. Polar molecules usually have some way of interacting through dipole-dipole interactions, hydrogen bonding, or simple electrostatics.

Note that the mobile phase, i.e. the carrier gas, does not have a large effect on the retention time in GC, so it really is mostly due to the simplified explanation of intermolecular forces given above.

One might ask why the polar molecules don't also have large dispersion interactions with the stationary phase (they certainly have some), and this is due to two effects. First, polar molecules tend to interact fairly strongly with each other in the gas-phase, and so there is less of an energetic gain for interacting with the stationary phase. This is a problem because interacting with the stationary phase is entropically unfavorable. Second, many polar molecules are smaller than a lot of the nonpolar systems that are studied using GC, and dispersion is correlated with surface area, so the polar molecules tend to have weaker dispersive interactions even ignoring the first point.


Salts will lower the melting point of snow or ice.
$\ce{NaCl}$ is chosen because it is cheap.

Edited: Salt with more total ions is more effective. $\ce{CaCl_{2}}$ is the one of the most available salt with 3 total ions. $\ce{MgCl_{2}}$ is another option.
Some crude $\ce{NaCl}$ produced from sea water directly contain $\ce{Ca}$ and $\ce{Mg}$ salt and can be used to melt ice.


Effect of stereoisomerism and racemization on solubility

Indeed a racemic mixture can be considered - at least respect to collective properties - as a "third stereoisomer", so much that it deserves the special status of raceme.

Not only the 1:1 ratio of the two enantiomers results in optical inactivity, but its physicochemical properties differ from those of the two separated compounds.

As for the solid state, any single stereoisomer behave as usual. This means that the presence of a second compound lowers the melting point.

However, a stereoisomer can have a higher (lower) affinity and bound better to its specular companion as compared to itself. This could be due to a better filling of the space or the formation of H bonds at opposite sites of the molecules, for instance.

In this case, the addition of the L form to a pure D sample initially lowers the melting point. Nothing special here.

However, because of lattice symmetry, once the ratio becomes 1:1 a more stable structure is achieved and the sample has now a melting point higher than the two stereoisomers alone.

This is right the case for tartaric acid,

Now consider the Delta G for the solubilisation process. As correctly suggested in a comment, the equilibrium involves not only the solvated molecules but the solid lattice as well.

Standard thermodynamic consideration tell us that as the latter gets more stable, solubility at a given T gets lower

according to the experimental data that you have posted.

(Note that for both enantiomers and for all their mixtures the entropic terms are surely comparable and primarily dictated by the disruption of the solid lattice, especially so given the specular structure of the molecules).


Can ethanol be both a Lewis base and a Lewis acid?

Lewis acidity and basicity require some amount of context because they can't function in those roles without at least a putative base or acid, respectively. Certainly the oxygen here has a lone pair so it could be basic, and the $\sigma^{*}$ orbital of the O-H bond (on the hydrogen side) can accept an electron pair. What matters more is the context of what acids and bases are present to interact with these moieties. There is no reason why these pieces can't separately fulfill their roles independent of each other.


Why do the boiling and melting points decrease as you go down group 1 and vice versa for group 7?

The group 1 elements are the so-called alkali-metals. The bonding between the atoms is caused by the interaction of the nuclei with the delocalized electrons. With increasing number of electrons and protons, the atomic radii get bigger and hence this interaction becomes weaker as the average distance between nuclei and electrons increases as you go down in group 1.
The group 7 elements are the so-called halogens. They exist under normal circumstances in their molecular form ($\ce{F2}$, $\ce{Cl2}$ and so on). In contrast to the group 1 elements, the dominant intramolecular force here is London dispersion (or van der Waals forces if you prefer). This attraction is caused by the correlated motion of electrons. With increasing amount of electrons, there can be more correlated motion and hence there is a stronger interaction between the molecules and an increasing melting / boiling point when you go down in group 7.


calculate cell potential for the voltaic cell where its involving the two cobalt species with Copper

The equation noted corresponds well to that of the first half-cell: on the other hand, the potential indicated is not the good one but just that of the standard potential.
The Nernst relation is written $\ce{E=E^0 +0.06*log \frac{[Co^{3 +}]}{[Co^{2 +}]}}$: the potential therefore depends on the concentrations
You can therefore calculate the potential of this half-cell.
You do the same thing with the second to arrive at its potential: from the 2, you can calculate the potential of the cell


It has been claimed that skin has a pH of about 5.5. Of course, this must be the sweat coming out of your skin, since skin isn't a liquid. So many body washes add citric acid to their formulas to be milder to tender skin.

For washing after poison ivy or car grease, pH 11 is much more effective.


"Chemicals" released in headphone with burned cable

Burning or Incineration of Plastic cause carcinogenic chemical products if without suitable catalysts... To pyrolyze plastics such as polyethylene(PE) and polypropylene(PP) of olefin series, the catalysts with high cracking property are
needed.

Dioxin- carcinogenic and antibiotic effect to the environment.
Flyash- may produce respiratory illnesses.
Sulfur dioxide
Nitrogen oxide- green plastics.
Toluene- aromatic compound that is related in the burning of cigarettes.
Carbon monoxide- main product of anerobic/ lack of ventilation combustion.
carbonaceous char, oils and gases.
paraffins, olefins, naphthenes and aromatics

Plastic and Diesel
http://www.chemtec.org/proddetail.php?pr...
http://www.p-fuel.com/index_files/faq.ht...

Thermofuel
http://www.cynarplc.com/thermo-fuel.asp

Mael is right that a  headphone set is made of some plastics (which are mainly made of basic carbon polymers) and the metallic part for the electronic conduction.so even i agree it can not be dangerous for human health to a great extent.


Alkene Electrophilic Addition Reaction Stereoselectivity When Two New Asymmetric Centers Form

Stereoselective becomes a hand-wavy term as soon as you may end up with more than one type of stereoisomer. If you want to actually state what is going on, you would most likely use either of the terms enantioselective (selects for a specific enantiomer) or diastereoselective (selects for a specific diastereomer; this may or may not include both enantiomers of the diastereomer).

In the first case, there is only one selectivity that can be observed; you can only get diastereomers, no enantiomers. Therefore, there may be some reason in shortening the term to stereoselective. In the second case you should not resort to using stereoselective because it is not specific enough — and that is irrelevant of the question whether stereospecificity is observed or not.

The second reaction would likely indeed be diastereoselective to a certain extent while definitely not being enantioselective (because the product is symmetric). So calling it diastereoselective — and thus, by intrapolation, stereoselective — may be justified.

In practice, the selectivities of both reactions would be so low that no practicing organic chemist would call either ‘selective’.


I suppose the author was talking about the internal overlap which takes place while the formation of sp and sp3 hybrid orbitals.
What I mean is when one s and one p orbital overlap, the extent of overlapping is less than the overlap that happens when one s and three p orbitals overlap.
To understand it more easily, suppose a ball represent a s orbital while a pencil represent p orbital. Think in mind the extent overlap of three pencils and one ball will be more than the extent of overlap of one pencil and one ball.
But the situation changes, when you overlap two sp3 orbitals or two sp orbitals. sp having more s character is relatively stronger then sp3, thus c-c bond is stronger in ethyne than ethane.


Convergence of spectra calculated from molecular dynamics trajectories

There are multiple ways in which the VACF, along with other properties derived from molecular dynamics trajectories like the dipole ACF to give IR spectra, can be considered "converged".

The molecular dynamics trajectory itself must be converged by some time-dependent statistical metric, such as measuring fluctuations in the energy or temperature over a number of steps and ensuring the average is below some threshold. This will be related to the ensemble chosen, the thermostat and barostat and associated parameters (if relevant), the integrator used for time evolution, and the size of the simulation, among other things.
Molecular simulations are layers of approximations piled on top of each other. In the case of purely classical dynamics, this usually refers to the force field parameters for the solute and solvent. Force fields are designed for different systems and may or may not be transferable, but the goal is usually to accurately predict dynamics. In the case of the VACF, if the dynamics of a system are described correctly, then the resulting power spectrum should be as well. However, in the case of the dipole ACF, the force field parameters enter directly as the nuclear charges (rather than just indirectly through nuclear velocities), which places stricter requirements on the quality of the force field. One hopes that newer force fields are better, but improvement over time isn't necessarily systematic and there is no quality hierarchy.
As an extension of 2, if the simulation is quantum mechanical in any way (such as in QM/MM, Born-Oppenheimer MD, or Car–Parrinello MD), then there are approximations in how the electronic structure is treated. For spectroscopic properties, it is more clear how to systematically improve results with regard to electronic structure than force fields due to well-established quantum chemical method hierarchies. Benchmarks of this hierarchy have been performed on many properties, including geometric parameters, dipole moments, and vibrational frequencies. In theory, the error in both dynamics and spectra should decrease up going from Hartree-Fock to MP2 to CCSD to CCSD(T), or by using more advanced density functionals (say, BP86 to PBE0 to M06-2X to B2PLYP). In practice, this would require benchmarking to draw any conclusions.
As an extension of 3, there are approximations in QM/MM of how the classically-treated and QM-treated regions interact with each other. Rather than go into more detail, here is a review of different QM/MM methods covering different definitions of the boundary and how one region may interact with the other.

Regarding estimation of the statistical error in resulting spectra, I am only aware of direct comparison (absolute error) between experimental and computed spectra, because it is assumed that the MD trajectory is converged (point 1). Performing multiple independent MD simulations and calculating the VACF from each should give an error measure closer to true statistical error, though I am unsure of how this would be computed.

Here is good almost-review (with good examples) of using Born-Oppenheimer molecular dynamics to calculate power spectra, IR spectra, and Raman spectra. I am not aware of any comparisons between classical MD and BOMD for computing spectra.

On the surface, it may appear as though the introduction of quantum mechanics makes things more difficult; certainly things become more time-consuming, however it is much clearer how results will change by treating electronic structure explicitly and with less severe approximations, rather than with a force field.


What is the Gibbs free energy difference between conformational states

Your question implies that you define the Gibbs free energy as $G=-RT\ln(Z)$ where $Z$ is the partition function, in which case there will be a difference in $G$ only if there is a difference in the partition function for your two configurations, i.e differences in energy levels and/or in the population in these levels. That will depend entirely on your system (and also the temperature).