Does hyperconjugation and induction always have the same effect?

The inductive effect occurs through overlapping sigma bonds, whereas the hyperconjugative effect occurs through overlap of a sigma bond with a p orbital.

There are many different types of hyperconjugation and hyperconjugation has been used to explain many different chemical phenomenon.  A nice overview can be found here.

  Are there any cases in which the hyperconjugative effect and the
  inductive effect have opposite effects?

Since inductive effects are transmitted through the sigma system, they effect should be independent of rotation around the bond(s) involved.  On the other hand, the pi type of overlap used in hyperconjugation is very dependent upon rotation about the bond(s) involved.  In the diagram above, rotation about the $\ce{C-C}$ bond by 90° shouldn't affect any inductive effect of the methyl group upon the carbocation center, but such rotation would completely destroy the overlap required for hyperconjugation to occur.

As you might expect, hydrogen/deuterium isotope effects can be used to assess hyperconjugation $\ce{C-H}$

[Note: isotope effects are a subject unto themselves; a nice overview can be found here.  See p. 12 for their application to hyperconjugation.]

Here is a "thought experiment" where inductive effects and hyperconjugation should have different effects (I don't know if this or a similar experiment has been done).  Take the following two carbocations.

One would expect them to both have similar inductive effects upon the carbocation center.  The tris-ethyl carbocation, free to rotate about its various $\ce{C-C}$ bonds could also exist in conformations where the $\ce{C-H}$ bonds alpha to the carbocation center are correctly positioned for hyperconjugation to occur - it should show a secondary $\ce{K_{H}/K_{D} >1}$ since hydrogen is more electron donating than deuterium (see here for the reasoning behind this statement).  However, the $\ce{C-H}$ bonds alpha to the carbocation center in the rigid bicycloctyl carbocation cannot adopt a conformation that would allow effective hyperconjugative overlap of these $\ce{C-H}$ bonds with the cationic center - it should show a secondary $\ce{K_{H}/K_{D} =1}$.  Different secondary isotope effects would be expected in the solvolysis of these two systems.

Another case where inductive effects and hyperconjugation do not provide equivalent explanations relates to rotational barriers.  The origin of the rotational barrier in ethane has been a subject of investigation for a long time.  Inductive effects cannot be used to explain the barrier because, as discussed above, inductive effects are rotationally invariant.  An argument using hyperconjugation to explain the rotational barrier has been advanced and is the subject of much discussion (see the "Rotational barrier of ethane" section).  Whether the hyperconjugation argument is correct remains to be proven, but it is another case where hyperconjugation can be used to explain observations that the inductive effect cannot.


How does the addition of NaCl to the cathode side effect the potential of a voltaic cell: Zn(s) | Zn2+(aq) 0.1M || Cu2+(aq) 0.1M | Cu(s)

@JR19. If you modify the anode compartment, and if you then observe that this change produces a decrease in the cell voltage, it means that [$\ce{Zn^{2+}}$] has increased. It is probably an effect of "activity". At high concentration, concentration must be replaced by "activity". The voltage delivered by the electrodes is not really proportional to the log of the concentration, but it is proportional to the log of the so called "activity". I explain.
Let's consider the case of $\ce{HCl}$, which is best known. If you measure the pH of $0.01$ M HCl, you find pH = $2.0$. If now you add and dissolve some $\ce{NaCl}$ to this solution, the pH goes down to $1.5$ and even to $1.2$, as if the solution gets more and more concentrated. It is hard to believe, but easy to check. Do it in the lab, if you can. Here is an explanation taken from an old issue of J. Chem. Educ. (I don't remember which one). The measured "concentration" given by an electrode is the number of moles divided NOT by the volume of the solution, but divided by the volume of the "free water", namely the volume of water not engaged in adsorption around the ions. This sort of "new concentration" is called activity. By adding more $\ce{NaCl}$, the volume of free water decreases, so that the activity of $\ce{HCl}$ increases. In $1$ liter concentrated solution of $\ce{NaCl}$, there is only about $0.2$ liter of free water. The activity is much higher than the concentration. It must be the same phenomena in your cell.


I assume you are talking about a 2 phase system of vapor in equilibrium with liquid.  Strictly speaking, the term "vapor pressure" only applies to single-component systems.  If there is a second component present, like air and water in a closed sealed container, then the equilibrium of each species is described by the partial molar free energy (or fugacity) of the species in the vapor as being equal to the partial molar free energy (or fugacity) of the same species in the liquid.  Even in the case of water and air in a closed container at 1 bar (where air is regarded as insoluble), the partial molar free energy of the liquid water will be slightly different from its value in the single component species at the same temperature (Poynting correction).  Even if the gas phase were ideal, such that the partial molar free energy of the water in the gas phase were the same as its value for pure water vapor at the same partial pressure, the effect of the higher pressure on the liquid water would slightly disturb the equilibrium and the partial pressure of the water vapor would differ from its value for a single component system.

If total pressure in the two-component system were high enough (i.e., higher air pressure), the partial molar free energy of the water in the gas phase would also be affected (say by the non-ideal interaction between the water vapor and the air), but not in the same way as the partial molar free energy of the liquid water changes.  So here, even more strongly, the total pressure times the mole fraction of water vapor in the gas (which would typically be defined as the partial pressure of the water vapor for an ideal gas mixture) will not be equal to the equilibrium vapor pressure of water at the temperature of the system.  With all this said, in practice, particularly for air and water, the effect can be expected to be pretty small.


Is dihydrogen the only example of overlap between two s orbitals?

You don't need particularly exotic things to get a counterexample. Metal–metal bonding in mercury(I) compounds is quite prevalent, where the electronic configuration is $\ce{Hg+}$: $\ce{[Xe](4f^14)(5d^10)(6s^1)}$, and the bonding involves 6s–6s overlap. The prototypical example is the $\ce{Hg2^2+}$ cation, which features in mercury(I) chloride $\ce{(Hg2^2+)(Cl^-)2}$; this cation is valence isoelectronic to $\ce{H2}$.

Fundamentally, it is a relativistic contraction of the 6s orbital which makes such bonding common in Hg(I): see also Why does mercury form polycations?. The bonding in alkali metal dimers is qualitatively similar, but significantly weaker.

If we play the "every bond has orbital overlap" card (i.e. not just covalent bonds, which I think is perfectly fair), then the delocalised bonding in Group 1 and 2 metals arises from overlap of ns orbitals on each atom to form bands, as has already been alluded to in the comments. Even simpler would be lithium hydride, which likely has some degree of 1s–2s orbital overlap, even though the bonding may be primarily ionic.


LCs materials are dielectric. But they are not perfect insulators and have some non-zero conductivity, allowing a leakage current to flow, slowly discharging the capacitor.
This is caused by mobile ions in the liquid crystal and inside the polyimide layer [2].
Infact a liquid crystal cell can be modeled electrically as this circuit:

               Rl
         _____WWW_____
         |           |
         |     Cl    |
     O--------||---------O

References:
http://www.lintech.org/comp-per/03lcd.pdf p.10-11
http://www.emdmillipore.com/showBrochure?id=200908.037 p.9


How it is possible to make Aqua Regia in the lab?

Typically, 1 part of concentrated nitric acid is added to 3 parts of concentrated hydrochloric acid.

Personal protection (goggles, gloves, lab coat) is vital.
Prepare and use it under a well-vented fumehood only - hands inside the fumehood, head outside!

UPDATE

The comment of Crystal Lettuce is very important and points to a crucial aspect:

Aqua regia isn't a mixture to keep bottled on the shelf: it has to be prepared and used immediately!


What happens chemically to 5'-radiolabeled aqueous thyroxine (thyroid prohormone) when its I-131 undergoes beta+gamma decay?

When the beta decay occurs it is likely that the C-I (or now C-Xe) bond will break. . The electron emitted in the beta decay will fly away from the molecule, the positively charged xenon atom will now pick up an electron from the aqeuous solution.

The Szilárd-Chalmers effect needs to be considered, it is typcial that when a radioactive decay occurs that the bonds to the atom at which they decay is will be broken. The recoil energy from the beta emission should be able to break the C-I bond.

I suspect that the bond will break to form a carbon centred aryl radical. This is a very high energy organic radical which will be likely to react with dissolved oxygen to form a Ar-O-O. radical. This in turn will abstract a hydrogen from another organic molecule to form Ar-O-O-H.

When this decomposes the O-O bond will break thus forming ArO. and HO. The ArO. will abstract a hydrogen from another organic solute to form a phenol. I know that 8-hydroxy-20-deoxyguanosine (8-OHdG) in urine is associated with exposure to radiation in humans. See S. Kato et. al. JOURNAL OF PEDIATRICS, 2015, volume 167, pages 1369 onwards. Also see the work of Martin Tondel et. al., ARCHIVES OF ENVIRONMENTAL CONTAMINATION AND TOXICOLOGY, 2005, volume 48, pages 515-519.

This work on this DNA marker for radiation damage does to my mind support the hypothesis that the autoradiolysis of the thyroxine will create hydroxy compounds. When S. Cefova et. al. irradated emulsions of nitrobenzene and water they did observe the formation of nitrophenols which lends further weight to the idea that the product will be a phenolic compound.


Hydroxyl group to free amine in polyethylene glycol using CDI?

Conversion of Hydroxyl group to free amine in polyethylene glycol: This is a case of converting a primary alcohol to a primary amine

$$\ce{R-CH2OH -> -> -> R-CH2NH2}$$

To my understanding, converting hydroxyl group to imidazole carbamate as an active intermediate is not the way to do this. First thing come to my mind was a named reaction called the Mitsunobu reaction. The Mitsunobu Reaction allows the conversion of primary and secondary alcohols to esters, phenyl ethers, thioethers and various other compounds including amines. The reaction uses triphenylphosphine ($\ce{Ph3P}$) and diethylazodicarboxylate ($\ce{DEAD}$) to activate hydroxyl group. Then, activated alcohol can undergo substitution reaction with a suitable nitrogen nucleophile (either phthalimide or hydrogen azide can be used) followed by reduction (in the case of azide formation) or acid or base hydrolysis (if phthalimide were used as nucleophile) to give primary amine groups.


Why do solutions of electron in a box (and in a ring) predict coefficients for LCAO (linear combination of atomic orbitals) in 1D systems?

A short answer is that the functions that satisfy the Schrodinger equation for a system are largely determined by the potential energy part of the function, since that's the part of the Hamiltonian that varies with different scenarios. The particle-in-a-well (= 1D box) is defined by having infinite potential energy outside the walls and a constant finite energy within. That turns out to be a very good approximation of the potential energy function of a pi system or a line of identical metal atoms.

At the end of the line of atoms, an electron faces a steep potential energy curve. Not quite the vertical wall of the 1D box, but pretty close to it. The same is true on the sides, so the system can be treated as 1D. Within the system, the electron moves essentially freely (or exists in a broadly delocalized way) with very little change in potential across the system.

Since the potential functions are nearly identical, the solutions to the Schrodinger equation take the same mathematical form.


The formation of ice crystals on/in frozen food that has been defrosted and then frozen again

Industrial ice cream makers make it frozen much faster than household freezers. They freeze faster and ice cream is disturbed to take out heat and to keep the emulsion well dispersed. As result, frozen water based emulsion droplets are very small.
When melted and standing, emulsion slowly separates either and oily phases.
She put in household freezer, it is getting frozen very slowly. That helps further emulsion separation and ice have enough time to form big crystals.
Note that it happens for frozen vegetables as well, if melted and refrozen again. Without any added water.


How is bismuth(III) able to form 3 bonds with hydroxide molecules in bismuth(III) hydroxide?

There isn't a unique sturcture for bismuth hydroxide, but ions such as

$$\ce{[Bi6O4(OH)4]^{6+}}$$

are involved, where the 6 Bi atoms are at the corners of an octahedron, none of the Bi atoms are directly bonded to each other, and O atoms bridge pairs of Bi atoms.

See Solvation of the Bismuth(III) Ion by Water... Inorg. Chem. 2000, 39, 4012-4021 as a starting point.

But supposing the structure was $\ce{Bi(OH)3}$, Bi3+ has two electrons, each OH- has 8 electrons, so Bi could share 2 electrons from each OH- to form an octet.


Why do disproportionation reactions take place?

"Stability" is not really a driving force of a reaction. That would be the change in Gibbs Free Energy or ∆G. That takes into account both entropy and enthalpy, as well as concentrations of reactants and products, and temperature.

If you haven't yet discussed these in class, you should be able to read about these concepts by looking up "Gibbs Free Energy" in your textbook's index.


Is iron the most stable element in the periodic table?

Yes, $^{56}\ce{Fe}$ has the most stable nucleus, and $\ce{He}$ is the most chemically inert element. These are different and unrelated qualities, pretty much like physical fitness and intelligence in a man. As for structural stability, there is no such thing in chemistry (there is one in architecture and another in mathematics, but those are out of scope of this question).


Whats the symmetric, anti-symmetric and non-symmetric part of the direct product of irreducible representations?

The symmetric and antisymmetric part of the direct product here refers to the fact that certain irreducible representations in the direct product are symmetric while others are antisymmetric (w.r.t some operations like $\sigma_v$, $C'_2$). So, for instance, when we write:

$$\Pi \times \Pi = \Sigma^+ + [\Sigma^-] + \Delta$$

the $[ \ ]$ simply means that the irreducible representation $\Sigma^-$ is the antisymmetric part of the direct product. It doesn't mean that you are somehow decomposing $\Sigma^-$ into a symmetric and antisymmetric part, and then selecting the antisymmetric one. This won't be possible since $\Sigma^-$ is a one-dimensional representation.


Is one syn aldol product preferentially formed over the other?

Ignoring the difference between the SPh and Et groups, the two products shown are enantiomeric, not diastereomeric. Both of these reactions are not enantioselective in any manner, so both enantiomers will be formed in equal amounts, via both transition states. Presumably, in the book, only one enantiomer is shown to save space and to avoid redundancy.

The only thing that needs to be illustrated here is that the phenyl group is equatorial in the lower-energy transition state. That leads to the diastereoselective formation of the syn aldol product (but no enantioselectivity between the two syn products).

So, there is no difference at all between the two models you have used. Both are the same thing and are equally applicable. If you had a chiral auxiliary on boron or something of that sort, the story might be different. But that isn't the case here.


Answers to numbered questions:

1.  Yes (with the errata applied)
3.  Yes, however before the lowest locant set rule (P-45.2.2), maximum number of substituents as prefixes rule (P-45.2.1) is to be used. (See e.g. Alkane nomenclature - choosing the parent chain - maximize or minimize substituents?)
4–5. No, “prop-2-yl” is an incorrect spelling of propan-2-yl (P-29.3.2.2), which is preferred to both isopropyl and 1-methylethyl (P-57.1.1.1).


Deriving the partition function for a harmonic oscillator

I'm confused why you're interpreting the partition function as a count of states. It can't be a count; it's continuous.

The zero point energy doesn't actually matter because you can just shift the energy scale so that it starts at zero. The main point of zero point energy is that the ground state of the harmonic oscillator is such that there is energy, and the system is not stationary. I'm going to use it below anyway because you are.

You can get the answer you want, but you'll want to look at the probability $P_i$ of being in state $i$, where the partition function is used to normalize:

$$P_{i} = \frac{\epsilon_{i}}{q} = \frac{\mathrm{e}^{-\beta\hbar\omega (i + 1)/2}}{\sum_{i}\mathrm{e}^{-\beta\hbar\omega (i+1)/2}}$$

Substitution with your convergent sum:

$$P_{i} = \mathrm{e}^{-\beta\hbar\omega (i+1)/2} \frac{1 - \mathrm{e}^{-\beta\hbar\omega}}{\mathrm{e}^{-\beta\hbar\omega/2}} = \mathrm{e}^{-\beta\hbar\omega i}(1- e^{-\beta\hbar\omega})$$

For $T\rightarrow 0$, $P_{i} = \delta_{i0}$, which is exactly what you are looking for.


Which hydrogen of 3-methylcyclohex-2-en-1-one is the least acidic?

You are correct. Position 2 is the least acidic.

Position 3 is not the least acidic hydrogen. It might be the most acidic hydrogen. Position 2 is the least acidic hydrogen of the four marked.

What does position 2 have going for it? Why might it not have been marked as the least acidic? The lone pair that would be left in the conjugate base is sitting in a sp² orbital.

All else equal, sp² hybrid orbital containing $\ce{C-H}$ bonds are more acidic than sp³ hybrid orbital containing $\ce{C-H}$ bonds. The $\mathrm{p}K_\mathrm{a}$ of ethane $\ce{CH3CH3}$ is reported to be $\sim 50$, and the $\mathrm{p}K_\mathrm{a}$ of ethene $\ce{CH2=CH2}$ is reported to be $\sim 44$.

However, all else is not equal. Resonance, especially with a carbonyl group, stabilizes conjugate bases more. Acetone $\ce{CH2COCH3}$ has a $\mathrm{p}K_\mathrm{a}$ of $\sim 19$. Position 2 is not stabilized by resonance. The conjugate bases from positions 1, 3, and 4 are all resonance stabilized through the carbonyl group.

Position 1 has two resonance structures.

Position 3 has three resonance structures.

Position 4 has three resonance structures.

It is between positions 3 and 4 for the most acidic. I chose position three because alkyl groups (the rest of the ring) are electron donating by induction, but alkyl group also stabilize pi bonds, so position 4 is a good candidate for most acidic as well.


I was able to reach the manufacturer and ask her directly. USPXX is the "edition" of the US Pharmacopeia's "exams" that a product must pass in order to be considered pharmaceutical grade. As time goes on, these exams may get updated and improved. So a higher number USPXX means that a product passed a more recent (and likely, more demanding) USP set of exams.
Therefore, USP 26 is a cheaper PVP, that may not be able to pass the USP 42 standards. Or at least, wasn't tested on them.
On the other hand, USP 42 is a more expensive PVP,  that can and did pass the standard.
As for the actual differences between both standards, I don't know them. So I can't say if the difference between then it's significant or not. It seems USP guidelines are not available freely.
Edit: Todd Minehardt's comment had the right answer it seems.


What are the differences between the Hilbert space and real space representations in chemistry?

A Hilbert space is a kind of linear vector space.

In chemistry which encounter it when quantum mechanics, when we can represent wavefunctions by their contributions from different orthonormal single particle states.  It is these single particle states which we build up out our atomic orbitals (AOs).

Example:

Consider a two level systems - minimal basis $\ce{H2}$:

Each atom has a $1s$ AO on its hydrogen: $\phi_1(\textbf{r})$, $\phi_2(\textbf{r})$.  These AOs aren't orthogonal though:
$$
\iiint d^3\textbf{r}\ \phi_1^*(\textbf{r}) \phi_2(\textbf{r}) \ne 0
$$but we can construct orthonormal states:
$$
\psi_\pm(\textbf{r}) = \frac{1}{\sqrt{2}}\left(\phi_1(\textbf{r})\pm\phi_2(\textbf{r})\right)
$$
such that like functions integrate to $1$ and functions of unlike functions vanish:
$$
\iiint d^3\textbf{r}\ \psi_\pm^*(\textbf{r}) \psi_\mp(\textbf{r}) = 0 \\
\iiint d^3\textbf{r}\ \psi_\pm^*(\textbf{r}) \psi_\pm(\textbf{r}) = 1
$$.
These two orthogonal function can be represented as state vectors:
$$
\psi_+(\textbf{r}) = \pmatrix{1\\0},\; \psi_-(\textbf{r}) = \pmatrix{0\\1}
$$
This defines a Hilbert or vector space, where our single particle states $\psi_\pm$ are our vectors and the inner product of two vectors, $\psi_i$ and $\psi_j$, is defined by the integral over all space:
$$
\langle\psi_i\vert\psi_j\rangle = \iiint d^3\textbf{r}\ \psi_i^*(\textbf{r}) \psi_j(\textbf{r})
$$
and is equivalent to the dot product of two state vectors.

In a electronic structure (specifically DFT) code, a "real space code" means using atom orbitals as basis functions rather than plane waves.

The Hilbert space described above is complete - it describes all possible wavefunctions which can be made from the AOs included.  If you included the infinite number of hydrogenic wavefunctions on each atom you would be able to have an exact description of all possible wavefunctions for $\ce{H_2}$ - as the hydrogenic wavefunctions form a complete set, instead you include as many as practical.

An alternative complete set of functions is the infinite set of all plane waves of all frequencies.  In cases where you are performing calculations on a periodic system, such as a solid, plane waves can be a more natural choice as they themselves are also periodic.  In a box of width $L$, the set of waves with wavelengths:
$$\lambda = \frac{2L}{n}$$
for all integral n, form a complete set.  They are more commonly described by their wavevectors, $\textbf{k}$, a vector in what is called reciprocal space, which describes the momentum of the wave.

In DFT, when calculating integrals over density, in a plane wave code you manipulate the density in terms of wavevectors, $\textbf{k}$.  In a real space code, you compose your density from Kohn-Sham orbitals in real space and calculate integrals of the density over quadrature grids in real (Euclidean) space.

Hence "real space" is used as a descriptor, for DFT codes that work with AOs and basis functions defined on atoms in Euclidean space, in contrast to "plane wave" codes which used basis functions defined in position space.


Colour of metal complexes with the same type and number of ligands

When electrons come back down they emit the same wavelength of light and that's the colour we see.

I think this is a major misconception held by students. The color of these transition metal complexes is due to the absorption of light in the visible region. Very crudely, imagine you have six colors in the rainbow, if you remove a certain color portion from that rainbow, the transmitted light is no longer white to our eyes. For example, copper (II) absorbs red, that is why it appears blue to us. Yes, the electron has to relax back to its ground state, it does so non-radiatively.

The spacing (∆o) in the energy level determines the wavelength which will be absorbed.


Is molecular weight of proteins additive?

The molecular mass in Dalton, amu or whichever unit you prefer is equivalent to the molar mass $M$ in $\mathrm{g/mol}$ that chemists prefer to use for this discussion. Both are strictly additive. The molar mass of a larger complex is the sum of the molar masses of all its components.

This is because of how the unit is defined. Focussing on molecular mass, we could see ot as mass of a molecule. The mass of a molecule is put together by adding the masses of the atoms that this molecule is made up of. If that molecule now forms larger aggregates, the mass of the resulting thing (still a single ‘thing’ albeit larger) is the sum of the masses of its constituents. And so on.

You just need to beware in case you have aggregates that are not strictly $\ce{AB}$. Consider two proteins that form an $\ce{AB2}$ complex: In this case $M(\ce{AB2}) = M(\ce{A}) + 2 \cdot M(\ce{B})$, because we have two $\ce{B}$ in the final complex.


The Wikipedia article you quoted clearly says:

  The low toxicity is relevant to the widespread use of tin in dinnerware and canned food.

Moreover, tin cookware, which used to be very common in earlier centuries, is still sold for outdoor cooking purposes. It is thus without a doubt deemed safe to use under current health regulations.

As a side note, the pH of coffee is around 5, which is only weakly acidic (same as unpolluted rain, and less acidic than tomato juice or beer):


How did J. J. Thomson prove that the cathode rays were made of negative particles and not negative rays?

Thompson was able to measure the ratio of their mass to their speed

Cathode "rays" had been known for some time before Thomson. They were first observed as experiments in gas discharge tubes started to exploit better and better vacuums (the early experiments observed the varying forms of discharge in low pressure gases; cathode rays only become significant when there is very little gas left in the tubes).

At very high vacuums, people observed that the glass at the end of the tubes would glow even though the contents of the tubes did not (as there were too few gas particles left to cause a discharge). So the idea that some sort of ray was emitted by the cathode in the tube was developed. Initially the rays were seen to travel in straight lines.

Later, scientists such as Crookes observed that the rays could be deflected by magnetic fields (though he thought they were just negatively charged gas particles). Hertz, though, claimed electric fields would not deflect them (he was wrong).

The critical experiments done by Thompson relied on having an excellent glassblower who was able to construct an apparatus where electrodes inside the discharge tube which could be charged with an electric field (while retaining a very high vacuum). This enabled Thompson to refute the claim by Hertz that electric fields didn't cause deflection (maybe Hertz's glassblower wasn't as good). Once he had this apparatus further experiments with electric and magnetic fields could reveal a great deal more about the particles.

Thompson describes the critical experiments like this:

  As the cathode rays carry a charge of negative electricity, they are deflected by an electrostatic force as if they were negatively electrified, and are acted on by a magnetic force in just the way in which this force would act on a negatively electrified body moving along the path of these rays, I can see no escape from the conclusion that they are charges of negative electricity carried by particles of matter. The question next arises, What are these particles? Are they atoms, or molecules, or matter in a still finer state of subdivision? To throw some light on this point, I have made a series of measurements of the ratio of the mass of these particles to the charge carried by it (“Cathode rays,” Philosophical magazine, 44 (1897), 293-316, pdf of original available here)

[A good review and description of modern recreations of this experiment is given here]

The point of the experiments is that the deflection of the particle in a uniform electric field depends on its mass, velocity and charge. And its speed can be measured by carefully balancing the electric and magnetic fields to give zero net deflection (why this is so involves equations, but this is Chemistry.SE not Math.SE and you can read them in the paper).

So Thompson observed that cathode rays were best explained by a charged particle (a classical wave would not have been deflected) with an observable charge to mass ratio (which implied either a very large charge or a very small mass). And he observed that the particles were always the same regardless of the cathode material or the original gas in the tube.

Later experiments clarified the mass and charge. Which correspond to what we now call electrons.

Note that this happened well before any of the ideas of quantum mechanics and wave/particle duality appeared. Thompson was concluding, in contradiction to some theorists at the time, that the properties of cathode rays were better explained by particles obeying the known laws for electricity and magnetism than by a wave. And he was right. His discovery was the first time a key component of the atom had been identified.


What is the correct Lewis structure for calcium phosphate?

Calcium phosphate crystallises in a number of different modifications of which β-$\ce{Ca3(PO4)2}$ is the stable room temperature and ambient pressure modification. It has a rather complex structure not easily described, consisting of chains and rings with phosphate and calcium ions in specific patterns. It is orthorhombic, but $c \gg b \approx a$.

Any attempt to simplify its structure by drawing it in pseudocovalent depictions like your lower scheme is doomed to fail. Better just use the upper scheme; unless you are expected to know a certain salt’s crystal structure that one is safe.


Reference states for molecular orbital energies in DFT calculations

In molecular electronic structure, electronic/orbital energies are measured with respect to an energy zero of an isolated electron in free space.

Hence a lower LUMO energy does correspond to an orbital that is more apt to accept an electron/a more negative electron affinity compared to another molecule.

Correspondingly a higher HOMO energy corresponds to an orbital that is more liable to give up an electron/a less positive ionisation energy compared to another molecule.

This is the basis of Koopmans' theorem in Hartree–Fock theory and it's equivalent in DFT.


How is the Activation Energy of the Arrhenius equation related to the exponents in rate laws?

Rate Law

For the reaction:

$\ce{aA + bB -> cP}$

The rate law is as follows:

$R = k[A]^x[B]^y$

where:

  $a$,$b$,and $c$ are stoichiometric constants which have integer values
   $[A]$ is the concentration of species A
   $x$ is the order with respect to species A
   $[B]$ is the concentration of species B
   $y$ is the order with respect to species B
   $k$ is the rate constant.

(1) Neither $x$ nor $y$ depend on $a$, $b$, or $c$.

(2) Neither $x$ nor $y$ is required to be an integer value.

(3) k is measured for a specific temperature. The Arrhenius Equation would predict how k would change with temperature. So k does not depend on any of $x$, $y$, $a$, $b$, or $c$.

Say the reaction was for some sort of gas reaction. I now add a catalyst. I get a new rate:

$R_2 = k_2[A]^{x_2}[B]^{y_2}$

There isn't any way to use the first rate equation to predict what the second rate equation will be. (You could of course try to analyze the function of the catalyst to make predictions on what the best catalyst would be...)


The environments are the same only to within the limits of the spectrometer's resolution and other experimental details; remember that every measurement comes with uncertainties, some of which are baked into the very fabric of the universe.

If you had an "infinitely precise" NMR spectrometer, all the peaks would be distinct (except when symmetry demands they be the same). In fact, every single possible compound would be uniquely identifiable by its atoms' specific shifts, shared with no others. In principle, you'd be able to assign a carbon peak at 123.6207426[...]964109454164[...] ppm to riboflavin, whereas a peak at 123.6207426[...]964109454165[...] ppm would indicate monobrominated hexabenzocoronene intercalated into a strand of DNA, and so on. Of course this is just a thought experiment and would never be possible in practice, even though it would make characterisation of complex molecules and mixtures a dream...


Why there is hydrogen bonding involved in salt bridge formation between glutamic acid and lysine?

An electrostatic interaction is one between a centre of positive charge and a centre of negative charge. In your example, instead of a primary ammonium cation ($\ce{H3\overset{+}{N}R}$) you could also consider the fully methylated quarternary ammonium cation $\ce{Me3\overset{+}{N}R}$ which would still be able to build up an electrostatic interaction with the acid anion. The centre of positive charge is somewhere between the three ammonium hydrogens and the centre of negative charge somewhere between the two carboxylic oxygens.

A hydrogen bond, on the other hand, is a directional interaction of the type $\ce{X-H\bond{....}Y}$, where $\ce{X}$ and $\ce{Y}$ are electronegative elements and $\ce{Y}$ must feature a lone pair. Thus, hydrogen bonds always contain three atoms and only those three. One one of the three ammonium hydrogens can form a hydrogen bond with only one of the two oxygen atoms.


Why isn't ammonium chromate called diammonium chromate?

You don't need the prefix because you know (or rather are assumed to know) the charges on the constituent ions of an ionic compound, which this is. The ammonium ($\ce{NH4^+}$) cation has a charge of +1 and the chromate ($\ce{CrO4^{2-}}$) anion has a charge of -2. Since the formula unit has to have a net charge of zero, it follows that "ammonium chromate" must have the empirical formula of ($\ce{{(NH_4^+)}_2CrO4^{2-}}$). Similarly "ammonium nitrate" must have an empirical formula of $\ce{NH4NO3}$ because the nitrate anion has a charge of -1.  In both cases, you can deduce the formula from the name and your pre-existing knowledge of the properties of the ions.

The prefixes are used much more often in simple molecular compounds because the elements in periods 2 and 3 can often be in multiple oxidation states, and you need the prefixes to figure out the actual molecular formula, the names of the elements are not enough. The most famous example is probably nitrogen and oxygen, which form at least half a dozen neutral compounds. You need prefixes so you can distinguish between $\ce{NO2}$ (nitrogen dioxide) and $\ce{N2O}$ (dinitrogen monoxide).  "Nitrogen oxide" wouldn't cut it, because you wouldn't know the oxidation state of $\ce{N}$ and $\ce{O}$ and could not deduce the formula from the name.

Even when the constituents of ionic compounds can form more than one ion, e.g. for transition metals, we usually choose to write the oxidation state of the varying ion instead of using prefixes, e.g. $\ce{FeO}$ is "iron(II) oxide" instead of "iron monoxide" and $\ce{Fe2O3}$ is "iron(III) oxide" instead of "diiron trioxide."

You may observe that this is all shot through with logical inconsistency and redundancy. That is a reflection of the chaotic history of chemical nomenclature and the need to balance systematic naming conventions to make it easier for the newcomer with the ability to use the vast existing knowledge base that uses older and more ad-hoc or individually varying schemes.


This highly depends on what you are electrolyzing. When using solutions that are not very acidic, oxide-coated metals sometimes work, like lead-oxide electrodes, that can be harvested from old lead accumulators (attention, lead is toxic, wear lab coat and gloves when working with them. Lead accumulators contain ~20% sulfuric acid, so wear gloves as well and unassemble them in an acid-proof working space)

If you search the web for recipes of perchlorate preparation by electrolysis at home, you will find extensive discussions on the topic of anode materials and making oxidation-resistant electrodes.

Platinum is really hard to corrode, but even it corrodes in some cases.


How many basis functions used in STO-3G and 6-31+G** for the water molecule?

Notation

** is just a synonym for (d,p), so that 6-31+G** basis is just a different name of 6-31+G(d,p) one.
The meaning of (d,p) is that one set of d-type polarisation functions is added for heavy atoms and one set of p-type polarisation functions is added for hydrogens.
6-31G(3pd) is essentially 6-31G(3p1d), i.e. it is 6-31G plus 3 sets of p-type polarisation functions and 1 set of f-type polarisation  functions on heavy atoms.

How many functions are there in a set?

There are always 3 functions of p-type, but the situation with d- and f-type functions is a bit more involved.
There are either 5 or 6 functions of d-type and either 7 or 10 functions of f-type depending on is a calculation restricted to use only spherical harmonics or not.

Cartesian vs. spherical harmonics

Basis functions can be written in two forms: as spherical harmonics
$$
  N(n, \alpha) Y_m^l r^n e^{-\alpha r^2}
$$
or Cartesian ones
$$
  N(l_x, l_y, l_z, \alpha) x^{l_x} y^{l_y} z^{l_z} e^{-\alpha r^2} \, .
$$
And for $l \geq 2$ the number of Cartesian harmonics is greater than spherical harmonics. For instance, a set of d-type function can have five spherical harmonics ($d_{3z^2-r^2}$, $d_{xz}$, $d_{yz}$, $d_{xy}$, $d_{x^2-y^2}$) or six Cartesian harmonics ($d_{zz}$, $d_{xx}$, $d_{yy}$, $d_{xz}$, $d_{yz}$, $d_{xy}$). Note that these six Cartesian harmonics can be transformed into five above mentioned spherical harmonics of d-type plus one function of s-type that can be neglected. Similarly, there are 10 Cartesian f-type functions and 7 spherical ones, and the later  could be constructed from the former with 3 p-type functions as a by-product.

The choice which harmonics to use can be controlled through the corresponding option in an input file and default choice varies from program to program. If you're specifically interested in Gaussian, then, if I remember correctly, by default Cartesian d-type functions and spherical f-type functions are used for the Pople 6-31G basis set. Thus, there will be 6 functions in a set of d-type functions and 7 functions in a set of f-type functions. This might be the case that Pople basis sets were even originally defined to use Cartesian d-type functions and spherical f-type ones, but I'm not sure.


Ability to produce hydrogen gas on reaction with Na/K

First of all, this is not an acid-base reaction. Alkali metals are not bases, rather strong reducing agents. For example, in this site, you may find a excellent explanation. Also remember, metallic $\ce{Li}$, $\ce{Na}$, or $\ce{K}$   does not abstract acidic hydrogen from organic molecules, not necessarily. For example,  metallic $\ce{Na}$ reacts with acetylene ($\mathrm {p}K_a$ of terminal alkyne-$\ce{H}$ is $25$) to give sodium acetylide, but it does not undergo aldol condensation with acetone with more acidic $\ce{H}$, $\mathrm {p}K_a$ of which is $19.3$ (https://owl.oit.umass.edu/departments/OrganicChemistry/appendix/pKaTable.html). The reaction of acetone and metallic $\ce{Na}$ rather undergoes dimerization (e.g., see here). Another example is the reaction of triphenylmethane (the $\mathrm {p}K_a$ of $\ce{H}$ on the central carbon is $33$) and metallic $\ce{K}$ in 1,2-dimethoxyethane, which has given solutions of triphenylmethyl potassium containing various cleavage and reduction products derived from triphenylmethane (Ref.1), but authors stated that no hydrogen was evolved in the course of the preparation.

However, the given compound has potential to undergo redox reaction with alkaline metal and release $\ce{H2}$ gas as a byproduct because the resultant dianion stabilized by resonance and show hint of aromaticity (Ref.2). Tentative mechanism is illustrated in following scheme:

References:

R. O. House, V. Kramar, “The Chemistry of Carbanions. I. The Reaction of Triphenylmethane with Potassium$^1$,” J. Org. Chem. 1962, 27(12), 4146-4149 (DOI: 10.1021/jo01059a007).
G. Boche, H. Etzrodt, M. Marsch, W. Thiel, “The Dianion 1,2,3,4‐Tetraphenylcyclobutadienediide,” Angew. Chem. Int. Ed. Engl. 1982, 21(2), 133-133 (https://doi.org/10.1002/anie.198201331).


Best reagent to convert 4-chloro-4H-pyran into an aromatic compound

In order to choose the best reagent to convert the compound to an aromatic one, let's ask the question of what is preventing this compound from being aromatic? At carbon-4, the pi-system is disrupted by an sp3-hybridized atom. If carbon-4 could be converted to sp2 by some reaction, then the compound could become aromatic. At C-4, we see a chloride. We know that chlorides are good leaving groups, so let's look at the product of that step.

After chloride leaves, there's a cyclic, fully conjugated pi-system containing six electrons (two for each double bond, two from the oxygen, and zero from the carbocation). Drawing a resonance structure shows that this is an oxygen analog of benzene (known as pyrylium ion).

What kind of reagent would promote this process, i.e., a halide leaving? A Lewis acid, such as $\ce{AlCl3}$ could promote this in two ways. First, it could interact with the chloride in the starting material, building up positive charge on the chloride, making it a better leaving group. Secondly, the Lewis acid can coordinate the chloride on the product side, preventing it from attacking the pyrylium ion.


How does the anti-bonding eg orbital (eg*) affect the bonding of ligands?

As you would expect from increasing occupancy of antibonding orbitals, the result is weaker metal-ligand bonding. This is one reason why the octahedral geometry is not as strongly favored for metals with greater numbers of d electrons. For a d8 metal such as the Ni(II) in the example, square planar and octahedral complexes are both common, and planar structures tend to be favored if the ligand is strong enough to result in a low-spin planar complex (as is often the case with Ni(II)). Only a weak field ligand like H2O will tend to form octahedral complexes with Ni(II).
A semi-quantitative approach to this issue is found in "structure energy" approximations and determination of "structure preference energy". Although that approach is not rigorously quantitative, it will show you trends.
It is important to note, though, that even with full occupancy of the eg antibonding orbitals (for example in a d10 metal like Zn(II)), octahedral complexes are possible. One can think of the sigma bonding in that case as being essentially entirely via the s and p metal orbitals, similar to a p-block central atom such as S in SF6.


How do atomic orbitals interfere both constructively AND destructively to give two molecular orbitals?

Part of the issue is that you're treating orbitals as objects. They aren't objects, they are quantum mechanical wavefunctions. The electrons which occupy the orbitals are objects, but the orbitals themselves are decidedly not objects.
When we talk about quantum mechanical states, the overall phase of an individual state has no meaning. It is a fundamental postulate of quantum mechanics that the wavefunction $\psi(x)$ and the wavefunction $k\cdot\psi(x)$ represent exactly the same state, so the phase $k$ is completely meaningless. 1s orbitals are states as well, so it is meaningless to talk about the phase of a 1s orbital.
However, if you have a state that is made up of two different states (like a molecular orbital), then the relative phases of the two constituent orbitals matter. That's why the bonding and antibonding MOs are different: it's because the two constituent parts don't have the same relative phase in the two MOs (in the bonding it's the same, in the antibonding they're opposite).

There are very few, and possibly no, quick and dirty ways to explain MO theory, in my opinion. However, if you have studied linear algebra before,* then the transition from AOs and MOs can be viewed as a change of basis. A basis refers to a minimal set of vectors, from which any arbitrary vector may be constructed through linear combination. If you have a vector $(a, b)$, you can express this as: $$\begin{pmatrix}a\\b\end{pmatrix} = a\begin{pmatrix}1\\0\end{pmatrix} + b\begin{pmatrix}0\\1\end{pmatrix}.$$
It doesn't matter what $a$ and $b$ are, you can always express $(a, b)$ as a linear combination of the two basis vectors. You can think of the basis vectors $(1, 0)$ and $(0,1)$ as being the 1s orbitals on both individual atoms.
The MOs correspond to using the basis $(0.5, 0.5)$ and $(0.5, -0.5)$.† Notice that this is still a valid basis, because we can still express any arbitrary vector $(a, b)$ as a linear combination of these:
$$\begin{pmatrix}a\\b\end{pmatrix} = (a+b)\begin{pmatrix}0.5\\0.5\end{pmatrix} + (a-b)\begin{pmatrix}0.5\\-0.5\end{pmatrix}.$$
If we had rejected the "antibonding" combination $(0.5, -0.5)$, and only taken the "bonding" combination $(0.5, 0.5)$, then we would not be able to form a valid basis. For example, we could not express the vector $(1, 2)$ as just a linear combination of $(0.5, 0.5)$.
From this point of view, AOs and MOs are simply different bases.‡ The existence of both the bonding and antibonding MO is not only logical, but also mandatory in order to form a complete basis.
The familiar AOs that you have studied (1s, 2s, 2p, ...) are simply a series of basis states for the hydrogen atom. In the $\ce{H2}$ molecule, there are two sets of basis states, one for each atom. The transition from "2 sets of AOs" to "1 set of MOs" is just a change of basis: we go from two sets of atomic bases to one single molecular basis.

The proper answer is, of course, to pick up a QM textbook and study it. By QM I do indeed mean quantum mechanics, not just MO theory. It is a long journey, and you may not necessarily be fully prepared for it right now, but it will reward you with a better understanding of how this works. MO theory fundamentally relies on quantum mechanics, and to learn MO theory without understanding quantum mechanics is what leads to confusion and questions like yours.

* The analogy I use above is slightly simplified, and might look slightly contrived, but the link between QM and linear algebra is actually very deep. QM is somewhat analogous to linear algebra but on an infinite-dimensional vector space (technically, a Hilbert space).
† Technically this should be $1/\sqrt{2}$, not $0.5$. In QM we are nearly always concerned with orthonormal bases, i.e. ones where each basis vector has a length of $1$ and forms a scalar product of $0$ with every other basis vector. (Mathematically, we have a basis $\{|i\rangle\}$ for which $\langle i | j \rangle = \delta_{ij}$.)
‡ "Bases" being the plural of "basis".


Why is the cyclic voltammetry of my ferrocene compound not showing a standard shape?

Without a proper reference electrode, knowing what potential your ferrocene/ferrocenium couple will show up is quite impossible. You should sweep a larger potential window, until the solvent breakdown potential on either end.

As Michael Dryden suggested, you should sweep slower as well. That will help with the signal getting resolved easier.


What kind of bonds (σ or π) are formed by px, py and pz orbitals?

It's a question of

(1) labeling conventions (what you call x, y and z in the coordinate system defining the position of two bonded atoms)

and

(2) the orientation of the two overlapping orbitals from the bonding atoms

If an overlap maximum occurs on the internuclear axis between the atoms, you call the resulting bonding orbital a "sigma" orbital. If maximum overlap occurs on an axis parallel to the internuclear axis, you call this type of bonding "pi". Either s or p orbitals can form sigma bonding orbitals, but s orbitals generally form sigma orbitals, whereas of the 3 conventional p orbitals, only one has lobes oriented along the internuclear axis (this orbital is usually called $p_z$ because its principal symmetry axis coincides with the internuclear axis, which is also labeled the z axis) and forms sigma bonding orbitals. The remaining two (x and y) form pi orbitals. This is illustrated schematically in the following figure:

(note the AOs don't overlap significantly, focus is on showing their relative orientation and labels)


Why is fluorine the best oxidising agent, but also the most basic?

Fluorine is the most electronegative element.
But due to its high effective nuclear charge it holds up its electron density very tightly, moreover its a non-metal so preferably accepts electron to form stable (F–) ion which has configuration of Ne.
F2 ------> F- + e- is very favorable therefore it gets reduced very easily (reduced form F- is very stable than F2) and so it oxidises other elements and self gets reduced.


Titrating and the addition of acidic distilled water

Unfortunately I could not find any sources that specifically address your question so I will be answering using logic. If you notice any fallacies or points you disagree with, please point them out.

$\ce{CO2}$ dissolves in water at such low concentrations normally, the pH would not be greatly affected. Unless you are working in extremely small concentrations that require high accuracy, it would not be a very large problem. If you were, you could try to minimize the time the distilled water is opened, and perhaps work in a vacuum really quickly or find some way to remove that factor.
If we were adding the distilled water to another solution, it is possible that the other solution was already in equilibrium with the partial pressure of $\ce{CO2}$ in the atmosphere and thus the concentration would remain constant ( a very small concentration ).

I do agree that protons may be added, but they would be so few, especially if you are reacting the reaction with acid or some strong base that it would be insignificant. Often times, conditions are not perfect, but good enough to be estimated and be close enough. However, measures can be taken to prevent your worries if the experiment demands it.

Any $\ce{CO2}$ probably already dissociated and is in equilibrium, or comes to it very closely. Opening up a coke and watching it enter equilibrium may serve as a nice visual. Don't worry about it in a titration unless you are trying to do something really really unusual. The $\ce{CO2}$ is insignificant in most titrations.

Note, the solubility is 1.45 g/L at 25 °C, 100 kPa max, so even a saturated solution would be not greatly affected. Furthermore, your solution is probably already in equilibrium so you have already accounted for the carbon dioxide. Think about how when you do a titration, it doesn't matter if you begin at 0 or whatever. As long as you account for it.


How can I prevent the effervescence of a solvent during evaporation?

Break a ceramic utensil (mug, plate, any porous item will do) and add a few chips in the solution. This should lead to a smoother boil rather than the liquid jumping around.

This is a cheap alternative of anti bumping granules and works in a similar way by allowing the gas to accumulate on the rough surface and escape the solvent in a controlled fashion.


Approximated formula of acid dissociation constant

The acid dissociates in equal amounts of $[\ce{H^+}]$ and $[\ce{A^-}]$ hence we can say they are equal quantities.

We can approximate it like this because we assume the $[\ce{H^+}]$ from the dissociation of water to be negligible compared with the $[\ce{H^+}]$ from the dissociation of the acid.

We also assume the $[\ce{HA}]$ at the start is equal to $[\ce{HA}]$ at equilibrium because the dissociation of weak acids is small.

Here is some mathematics that you requested:

  $K_{\mathrm{a}}$ can be calculated using equilibrium concentrations:
  $$K_{\mathrm{a}} = \frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{equim}}=\frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{start}-[\ce{H+(aq)}]_\text{equim}}$$

Taken from my OCR A Level Chemistry Textbook (The specification under which this year's exams will be.)

Here is an extension to fully show how the two equate. Taken from the same textbook:

  $$K_{\mathrm{a}} =\frac{[\ce{H^{+}(aq)}]_\text{equim}[\ce{A^-(aq)}]_\text{equim}} {[\ce{HA(aq)}]_\text{start}-[\ce{H+(aq)}]_\text{equim}}\rightarrow\frac{[\ce{H^{+}(aq)}]_\text{equim}^2} {[\ce{HA(aq)}]_\text{start}}$$
  Therefore, $$K_{\mathrm{a}} =\frac{[\ce{H^{+}(aq)}]^2} {[\ce{HA(aq)}]}$$


How to temporarily hide hydrogen atoms in Molden?

As long as you are not using an ancient version of Molden, this should be no problem. (The oldest one I found in my software archive was Moleden5.1 from January 2014 and it already contains this feature.)

In the Molden control window there is a button with an italic H. This will open a menu with "Excl. All Hydrogens", "Excl. H except H-bond", "(Re)calculate H-bond", "H-bond parameters", and "Opt. H positions". I'll leave it to you to play around with it, below you find a small animation of the first two modes for the base pair GC.

And because it is a little bit too tiny in the gif, here is a crop of the Molden control window.


Atomic orbitals: how is bond strength related to stability, energy, bond length?

There are several factors that affect bond disassociation energies (bond enthalpies) but one quick correlation you can make is bond distance vs bond strength. The interatomic distance of the H2 bond is 74 pm vs 199 pm in Cl2. The H2 bond consists of 1S electrons that are very close to the nucleus, giving a short bond. The valence electrons in Cl2 are 3p (relatively far away). This leads to other properties such as polarizability.

In short your first assumption is incorrect, the trend you are seeing has nothing to do with s or p bonds, but on the bond lengths. Therefore comparing s-s bonds and p-p bonds is meaningless unless you take into account the atomic orbital number and therefore an estimate of bond length.

For your final questions, the stronger the bond, the more stable, the higher energy, and the shorter the bond.

For more info this UF link has good information:http://www.science.uwaterloo.ca/~cchieh/cact/c120/bondel.html


Counting valency of atoms, in a molecule with python

I was looking into this, and the results I was getting seemed very anomalous.  Consider this iteration from a single carbon atom at -4 through +4, this is the output I receive (python code at the bottom of the post):

[C+4]
C valence= 4 radicalE= 0 charge= 4 implicitH= 0
[C+3] |^1:0|
C valence= 4 radicalE= 1 charge= 3 implicitH= 0
[C+2] |^3:0|
C valence= 4 radicalE= 2 charge= 2 implicitH= 0
[C+]
C valence= 0 radicalE= 0 charge= 1 implicitH= 0
[C]
C valence= 0 radicalE= 0 charge= 0 implicitH= 0
[C-]
C valence= 0 radicalE= 0 charge= -1 implicitH= 0
[C-2] |^3:0|
C valence= 4 radicalE= 2 charge= -2 implicitH= 0
[C-3] |^1:0|
C valence= 4 radicalE= 1 charge= -3 implicitH= 0
[C-4]
C valence= 4 radicalE= 0 charge= -4 implicitH= 0

I printed them using the canonicalSmiles format.  After I realized something strange was going on (neutral, and +/- 1 charges were 0, but everything else was fine), I dug into SMILES documentation to find out that currently it appears only up to divalent radicals are supported.  Something like a lone carbon atom has technically a tetra-valent radical, so it defaults to nothing, because it cannot be interpreted by SMILES.

I pulled this from ChemAxon which explains what the notation after the atom means. I'm going to assume that |^1:0| is the notation for a mono-valent system.  The 0 indicates the index of the atom, which is always zero for this example.

  Radical numbers
  Atom indexes with:
  - divalent radical center are written after "^2:"
  - divalent singlet radical center are written after "^3:"
  - divalent triplet radical center are written after "^4:"
  - trivalent radical center are written after "^5:"

And from the Indigo website itself it states under the supported extensions

  Radical numbers: monovalent, divalent singlet, and divalent triplet

So, I don't think you've done anything wrong, you've just simply hit a limitation of the SMILES system and the interpreter of the indigo software.  Here is the code I used below:

from indigo import *
indigo = Indigo()

from indigo_renderer import *

renderer = IndigoRenderer(indigo)
mols = []
mols.append(indigo.loadMolecule("[C+4]"))
mols.append(indigo.loadMolecule("[C+3]"))
mols.append(indigo.loadMolecule("[C+2]"))
mols.append(indigo.loadMolecule("[C+1]"))
mols.append(indigo.loadMolecule("[C]"))
mols.append(indigo.loadMolecule("[C-]"))
mols.append(indigo.loadMolecule("[C-2]"))
mols.append(indigo.loadMolecule("[C-3]"))
mols.append(indigo.loadMolecule("[C-4]"))

i = 0
for mol in mols:
    print mol.canonicalSmiles()
    for atom in mol.iterateAtoms():
      print atom.symbol(),"valence=",atom.valence(),"radicalE=",atom.radicalElectrons(),"charge=",atom.charge(),"implicitH=",atom.countImplicitHydrogens()
    filename = 'mol'+str(i)+'.png'
    renderer.renderToFile(mol,filename)
    i += 1


A few questions about the conditions of the Diels Alder Reactions

Yes, conjugated dienes can react with other conjugated dienes in Diels-Alder reactions. A well known example of this is the dimerization of cyclopentadiene. All the standard caveats, restrictions, and other considerations still apply, namely:

The diene must not be rotationally locked in such a way that it cannot adopt an s-cis/synperiplanar conformation. As a logical corollary, cyclic dienes that are already locked into the correct s-cis conformation often have lower activation barriers.
The kinetics of the reaction are accelerated by electron-withdrawing substituents on the dienophile and electron-donating substituents on the diene.
The stereochemistry of the major product is highly substrate-dependent. A preference for the endo isomer exists in cases where the dienophile has unsaturated substituents. In those cases, the transition state is stabilized by an overlap of the $\pi$-electron system between said unsaturated substituents and the $\pi$ orbitals of the newly forming double-bond at the back of the diene. (Just as a side note, Carey & Sundberg refer to this as the "Alder rule" in Advanced Organic Chemistry, but I've never seen that term used anywhere else.)

In principle, there's no reason why two conjugated $\pi$-bonds of a polyene cannot react similarly, provided they satisfy the orbital symmetry and conformational requirements and the reaction kinetics are favorable. Even aromatic molecules can undergo Diels-Alder reactions under the right set of conditions (see, for example, the Diels-Alder reaction between benzene and furan via a benzyne intermediate). There are also numerous instances of well-documented Diels-Alder reactions involving heteroatomic dienophiles, alkynes as dienophiles, polymerization reactions, etc.

If you can get your hands on the aforementioned Carey & Sundberg book, they discuss the Diels-Alder reaction at length (including relatively rigorous explanations of the stereospecifity and reaction kinetics via molecular orbital theory). If you're looking for a comprehensive reference, I doubt there's anything superior to the venerable March's Advanced Organic Chemistry. There's a chapter on Diels-Alder reactions (along with virtually every other organic chemistry reaction known to science), as well as a massive list of accumulated primary literature references if further detail is required.


How to calculate the final temperature of a gas when it undergoes adiabatic expansion?

Rather than answer the question numerically I have outlined the four different cases, reversible / irreversible and isothermal / adiabatic.

In adiabatic changes no energy is transferred to the system, that is the heat absorbed or released to the surroundings is zero. A vacuum (Dewar) flask realises a good approximation to an adiabatic container. Any work done must therefore be at the expense of the internal energy. If the ‘system’ is a gas then  its temperature will not remain constant during any expansion or compression.

In expansion the work done is $dw=-pdV$ and the change in internal energy $ dU=C_vdT$.

The heat change is zero then $dq=0$ which means from the First Law
$$dU = dw$$
and so
$$ C_vdT = -pdV$$

Dividing both sides by T and for one mole of an perfect gas $p=RT/V$ thus
$$C_v\frac{dT}{T}=-R\frac{dV}{V}$$
If the gas starts at $T_1,V_1$ and ends up at $T_2,V_2$ the last equation can be integrated (and rearranged) to give
$$\ln\left (\frac{T_2}{T_1}\right)=-\ln\left(\frac{V_2}{V_1}\right)^{R/C_v}$$
or
$$\frac{T_1}{T_2}=+\left(\frac{V_2}{V_1}\right)^{R/C_v}$$

using the relationship $C_p = C_v+R$
$$\frac{T_1}{T_2}=+\left(\frac{V_2}{V_1}\right)^{(C_p-C_v)/C_v}$$
Using the gas law this van be rewritten in a more useful form as
$$ p_1V_1^{\gamma } = p_2V_2^{\gamma }$$
where $\gamma = C_p/C_v$. This is also written as $pV^\gamma = const$.

The change in internal energy in an adiabatic process is
$$\Delta U=C_v(T_2-T_1)$$

Adiabatic reversible

In a reversible adiabatic change we use the formulas above to work out what happens. If n moles of a gas fill a container of volume $V_1$ at $p_1$ atm. and is expanded reversibly and adiabatically until it is in equilibrium at a final pressure $p_2$ we can calculate the final volume and temperature. The values of $C_p$ and $C_v$ are assumed to be known and are constant with temperature.

The equation to use is
$$ p_1V_1^{\gamma} = p_2V_2^{\gamma}$$
and as only $V_2$ is unknown this can be calculated.  The work done is
$$w=nC_v(T_2-T_1)$$

Adiabatic irreversible

In an irreversible adiabatic change if n moles of an perfect gas expands irreversibly from a pressure of $p_1$ against a constant external pressure $p_2$ the temperature drops from $T_1$ to $T_2$. We can calculate how much work is done and the final volume.
The internal energy change is $\Delta U= nC_v(T_2-T_1)$
and the work done is $w=p(V_2-V_1)$ and as $\Delta U = w$ (as the system is adiabatic) the final volume can be obtained.  Similarly if the volumes are known the final temperature can be obtained.

Isothermal irreversible

In an isothermal irreversible change the work done on suddenly allowing a perfect gas to expanding from $V_1$ to $V_2$ is determined by the final external pressure $p_2$ and is
$$ q=-w =\int_{V_1}^{V_2} pdV = p_2(V_2-V_1)$$
thus expansion into a vacuum does no work.

Isothermal reversible

A reversible isothermal change performs the maximum possible amount of work, and assuming a perfect gas,
$$ q=-w_{rev} =\int_{V_1}^{V_2} pdV=nRT \int_{V_1}^{V_2} \frac{1}{V}dV$$

$$w_{rev} =- nRT\ln\left(\frac{V_2}{V_1}\right)= - nRT\ln\left(\frac{p_1}{p_2}\right)$$


What is the general phenomenon taking place here?

At low pressures the activity of a gas can usually be related in a straightforward way to its partial pressure as
$$a=p/p^{\circ}$$
Since the activity of a substance present in multiple phases is the same in all of the phases, for a volatile compound it suffices to determine the partial pressure in the gas phase to know its activity in condensed phases. This is how the activity of water in different materials including foodstuffs is usually determined.
For the case of a non-volatile solute in a condensed phase things are not so simple. For electrolytes, theories such as Debye-Hückel do a reasonable job of predicting activities at low concentrations, and other theories do better at higher ionic strengths. Electrochemical measurements can provide experimental values of activities. Because ions usually exhibit strong deviations from ideality this field has been studied extensively and even introductory physical chemistry or analytical chemistry textbooks will address it at some length.
One path to determine the activity of a non-volatile non-electrolyte - for which a vapor-pressure measurement is impractical - is to determine the activity of the solvent and to invoke the Gibbs-Duhem equation relating the activities of the solute and solvent. This is what is meant in the Libretext:

The activity coefficient for a nonvolatile, neutral solute is often estimated by non-linear curve fitting, taking into account the molality of the solute and the activity of the solvent (usually its vapor pressure).


What does it mean to say "the 222 space group"?

This is a crystallographer’s shorthand for a certain type of space group.

You may have heard of the 14 Bravais lattices, e.g. cubic, monoclinic, orthorhombic. These lattices all assume perfect sphere’s at the corners of the elementary cell and potentially also in certain central positions (if the lattice is not a primitive one). In actual X-ray structure determination, these perfect Bravais lattice spheres are typically not spheres. Instead, we should consider each sphere a ‘hook’ and a molecule hung to that hook. Since not all molecules are symmetric with respect to all symmetry operations a certain Bravais lattice has, the symmetry of the crystal is markedly reduced. For example, an enantiopure chiral compound cannot form crystals that contain planes of symmetry, so even if that compound were to crystallise in the orthorhombic lattice, there would be no planes in spite of them being required in simple orthorhombic crystals.

This reduction in symmetry leads to a large number of lower-symmetry sub-divisions of the Bravais lattices called space groups. Strictly speaking, the Bravais lattices are merely highly symmetrical space groups. There is a total of 219 or 230 — depending on whether you count enantiomeric space groups as two or one.

The shorthand to unambiguously label these space groups basically describes which symmetry operations are present in a certain direction. Orthorhombic is the easiest to understand: the primitive Bravais lattice has:

in $x$ direction:

a two-fold axis
a plane of symmetry perpendicular to that axis.

in $y$ direction:

a two-fold axis
a plane of symmetry perpendicular to that axis.

in $z$ direction:

a two-fold axis
a plane of symmetry perpendicular to that axis.

You could label this is $\mathrm{P\ \frac2m \frac2m \frac2m}$, but it is tiresome to always include the ‘2 over m’ part. Hence, it is shortened to $\mathrm{P\ mmm}$, the shorthand being ‘m replaces $\frac2m$’. If there is no plane of symmetry, then only the two-fold axis remains and that dimension is given a 2 label instead. Thus, $\mathrm{P\ 222}$ is the same lattice but there are no planes of symmetry in the actual crystal structure. The crystal could also include a screw axis, in which case the symbol $2_1$ would be used, or a glide plane in which a different letter, depending on the orientation of the glide plane, is used.


Does potassium aluminum sulfate have any effect on sexual performance/sperm counts or anything relating to the functionality of the sexual organs?

Oral ingestion of simple aluminum compounds is not absorbed.  Topical genital application is definitely not recommended.  There is no obvious connection between aluminum and tumescence, one way or another

Soluble aluminum is toxic, likely cumulatively neurotoxic.  It is not absorbed from the GI tract unless complexed.  Aluminum hydroxide antacids as foaming tablets( citric acid plus sodium bicarbonate) gave aluminum citrate that is bioavailable.  Bad idea.  Pickles are made with alum to keep them ever so crunchy. The soluble aluminum is claimed not to be absorbed. Antiperspirant contains aluminum chlorhydroxide or glycinate to stopper sweat glands.  Topical application is not absorbed.  Exposing mucus membranes to soluble aluminum is definitely not recommended.


Could Na+ ions in water used with a humidifier cause dust everywhere?

In the end, the solution was that instead of using an "ultrasound" based humidifier, we bought "evaporative humidifier".

delete this line if it cannot be said, but we bought a "trotec aw 10 s"


I've never tried to grow big crystals or anything, but we commonly use 30 mM silver nitrate in 1 M $\ce{NH4OH}$ for electrodepositing silver onto metal (usually gold) electrodes.
If you don't care what kind of surface finish you end up with, it will work, but it makes a fairly rough film that can be somewhat easily removed with abrasion.
The tricky part is trying to get very smooth and adherent films.
It seems that there are still not any good replacements for electroplating from cyanide-containing solutions, for decorative or engineering purposes where the deposit will be subject to wear or has to look nice.


One of the usual setup for the coordination system would be that $x$ axis coincident with a two-fold axis. If we use this setup, we can easily write out the matrix for rotation around this 2-fold axis:
$$\begin{bmatrix}
1&0&0\\
0&-1&0\\
0&0&-1
\end{bmatrix}.$$

Then you can combine this with the transform matrix of the 3-fold axis to get the transform matrices of the other two 2-fold axes. That's just matrix multiplication.

In this setup, $yz$ is one of the reflection plane and the corresponding matrix is
$$\begin{bmatrix}
-1&0&0\\
0&1&0\\
0&0&1
\end{bmatrix}.$$

Note that this is just the combination of the previous matrix with inversion. You can apply the 3-fold axis rotation to it (once again, matrix multiplication) to obtain the matrix with regard to other two reflection planes.


Finding change in activation energy from change in rate constant in catalysis

It appears that you are correct in your thought process.
Determining by how much the activation energy of the reaction is reduced requires you to basically find

$$Δ(\%) = \frac{E_\mathrm{a1} - E_\mathrm{a2}}{E_\mathrm{a1}} × 100\% = \left(1 - \frac{E_\mathrm{a2}}{E_\mathrm{a1}}\right) × 100\%$$

There is a system of three independent equations

$$
\begin{cases}
k_1 = A\mathrm e^{E_\mathrm{a1}/(RT)}\\
k_2 = A\mathrm e^{E_\mathrm{a2}/(RT)}\\
\frac{k_2}{k_1} = x
\end{cases}
$$

with five unknowns $k_1$, $k_2$, $E_\mathrm{a1}$, $E_\mathrm{a2}$, $A$, which requires either one more equation, or to know one more parameter of the system in order to find $E_\mathrm{a2}/E_\mathrm{a1}.$
The textbook provides you with activation energy in the previous illustration right next to your problem [1, p. 841], so I assume you are supposed to use $E_\mathrm{a1} = \pu{76 kJ mol−1}:$

At this point the problem can indeed be solved to a certain numerical value:

$$E_\mathrm{a1} - E_\mathrm{a2} = RT\ln x = \pu{8.3145 J mol-1 K-1} × \pu{298 K} × \ln{2000} \approx \pu{18.8 kJ mol-1}$$

$$Δ(\%) = \frac{\pu{76 kJ mol-1} - \pu{18.8 kJ mol-1}}{\pu{76 kJ mol-1}} × 100\% \approx 25\%$$

The question whether it's a textbook flaw, or authors assumed the reader would read the material more carefully still remains open.

References

Atkins, P. W.; De Paula, J. Physical Chemistry: Thermodynamics, Structure, and Change, 10th ed.; W.H. Freeman: New York, 2014. ISBN 978-1-4641-2452-5.


Evaporation of concentrated formic acid and semi-concentrated sulfuric acid at room temperature

I might sound like a kill joy but I strongly suggest that you do not attempt to make carbon monoxide. The reaction you describe is a not one for the faint hearted. I would like to point out that aerosols of sulfuric acid are very harmful to the lungs, eyes and the rest of your body.

I think that any work done in the lab with concentrated sulfuric acid should be done in a fume hood to protect you against aerosols of the acid. The fumes from the formic acid are going to be corrosive.

In my youth we used to work on the open bench with glacial acetic acid, this is not a good thing to do as the whole lab soon becomes like a viniger factory. I fear that with formic acid being handled outside the hood the lab will soon become rather unpleasnent.


Why are double bonds at more substituted carbon more stable?

Let's start by providing some useful background information that will help answer your question.

1,3-Butadiene is more stable than 2 separate ethylene molecules due to the extended conjugation possible in the diene.  We can illustrate this by drawing simple resonance structures for the diene that can't be drawn for ethylene.

Huckel calculations show the same thing, 1,3-butadiene is more stable than 2 individual ethylenes.  Here is the Huckel MO diagram for ethylene.  The two electrons in ethylene have a net stabilization of 2$\beta$ compared to an electron in an isolated p-orbital.  4 electrons in 2 ethylenes would have an overall stabilization energy of 4$\beta$

Here is the Huckel diagram for 1,3-butadiene.  The 4 electrons have a net stabilization of [(2 x 0.62) + (2 x 1.62)] = 4.48$\beta$.

Even though the HOMO in 1,3-butadiene is higher in energy then the HOMO for ethylene, overall 1,3-butadiene is more stable than 2 isolated ethylenes by 0.48$\beta$.  This is because the MO below the HOMO is stabilized much more (0.62 $\beta$ compared to ethylene).

Now let's return to your question.  Hyperconjugation is a reasonable way to explain why more highly substituted double bonds are more stable. Just like in 1,3-butadiene, when we delocalize electrons over a larger number of atoms, which is what hyperconjugation does, an overall stabilization of the molecule should result.  And, as the hyperconjugated resonance structure shows, electron density will shift from the alkyl group to the double bond.

Even though the HOMO may be raised in energy, (at least some of) the other occupied MO's below the HOMO will be lowered in energy (just like in 1,3-butadiene) and overall a net stabilization of the molecule will result.


Is there such a thing as a "minimal soap" molecule?

It boils down to the definition of soap. Wikipedia defines a soap as the salt of a fatty acid.
IUPAC claims the smallest fatty acid can be considered to have 4 carbons.
Therefore the simplest soap molecule would be a (generally sodium) salt of butyric (butanoic) acid, i.e. sodium butyrate.

Now apart from the chemical definition, a soap must adhere to its function in order to be defined as such. Therefore, before calling something a soap, we would have to know if the substance does in fact act as a surfactant in a oil-water surface (reducing the interfacial tension) - but that will depend on the nature of the oil in question and the purity of the water. Only for a well defined system we can then conclude that such molecular salt is acting as a soap in it.


When a molecule loses hydrogen atoms, does it become oxidised?

This phrasing is only true for organic molecules. If for example, sodium hydride loses hydrogen, the sodium ion will get reduced. But since you seem to come from a biochemical background, this simplification is okay since you will be dealing with organic molecules primarily.

The idea behind that statement is that hydrogen atoms in biomolecules are typically only bound to carbon, nitrogen, oxygen and sulfur. Hydrogen is less electronegative than all these elements and therefore any $\ce{X-H}$ bond will be polarised towards $\ce{X}$; the non-hydrogen atom.

When determining oxidation states, bonds are formally cleaved heterolytically in such a way that the electronegative partner gets both electrons. Then, the electrons on the formal atomic ions created this way are counted and subtracted from the number the compound should have. Bonds between the same element are cleaved homolytically and the same procedure applied. Thus, if we take ethene ($\ce{C2H4}$, structure see below) the $\ce{C-H}$ bonding electrons are formally attributed to carbon entirely while the $\ce{C=C}$ double bond is split equally. Doing this, each carbon atom ends up with six electrons, formally, and therefore a formal charge of $2-$ or an oxidation state of $\mathrm{-II}$. If we did the same for ethyne ($\ce{C2H2}$) — a compound with two hydrogen atoms removed — we arrive at five formal electrons and thus an oxidation state of $\mathrm{-I}$. Therefore, going from ethene to ethyne is an oxidation.

$$\begin{align}\ce{H2C=&CH2} & \ce{H-C&#C-H}\\
\text{ethe}&\text{ne} &\text{eth}&\text{yne}\end{align}$$

Having gotten the theory out of the way, what does this mean for the simplified rule? Well, as I mentioned any $\ce{X-H}$ bond is polarised away from the hydrogen and thus effectively adds an additional electron to $\ce{X}$. Carbon, which is the most common bonding partner of hydrogen, is less electronegative than all the other typical elements of biomolecules, and thus will formally lose electrons to them. When removing hydrogen from a compound chemically, we will always remove at least one $\ce{C-H}$ bond in favour of a $\ce{C-C, C-O, C-N}$ or $\ce{C-S}$ bond. Thus, the carbon atom will formally lose electrons and we will have an oxidation.

This can also be rationalised in another way to determine oxidation states: the additive method. Starting from a neutral compound, assign each hydrogen atom a $\mathrm{+I}$, each oxygen $\mathrm{-II}$. Add that up and invert the sign; this is the oxidation state of all the carbons (assuming a molecule of the formula $\ce{C_mH_nO_x}$). Divide by the number of carbons to get an average oxidation state of carbon.

$$\chi(\ce{C}) = -\frac{n - 2x}{m} = \frac{2x - n}{m}$$

As you see, each hydrogen atom in a compound enters into the formula with a negative sign. Thus, removing a hydrogen must increase the average oxidation state of the carbons. Unfortunately, there is no trivial way to extend this formula to compounds containing oxygen and nitrogen or oxygen and sulfur.


Is there a way to generate SMILES from ChemDraw files?

Of course openbabel can do this (at least with release 3.1.0 from here).  To rewrite a file here and then, the GUI on top of the program on top may suffice for you.  However, you would skip a lot of the program's functionality if not trying to interact from the command line: for one, the conversion basically follows the pattern of
obabel -icdx input.cdx -osmi -O output.smi

which actually may be faster than clicking in an interface. For the other, that openbabel offers multiple interfaces to automate file transformations programmatically, i.e. to interact with programming languages like Python and C.  See the current  documentation rather than the a bit outdated on the project's senior page.
Note 1: ChemDraw offers a structure to SMILES string conversion, too.  Be aware that a .cdx file may contain much more information, than a SMILES string may retain.  Seeking for a text-only representation of the molecular structure, axial chirality (P/M) is an example where a SMILES string is less suitable than, for example, an InChI string.
Note 2: Openbabel aims to interconvert chemical formats; on occasion with a visual output like .svg, or .png.  RDKit has a stronger emphasis on computing with the molecules (a.k.a. cheminformatics) as required while interacting e.g., with large sets of data and databases.  There is some overlap of the two programs, and «the trick» is rather to recognize where the two are complementary to each other.


Why are the plastic balls used to reduce evaporation in this reservoir black instead of white?

This piece in Bloomberg News is about Sydney Chase, the founder of XavierC, the company which makes the "shade balls" in question (there are two other suppliers, Orange Products, Inc. and Artisan Screen Printing).

From the section of the XavierC site on how they work:

  Two main causes of water evaporation in reservoirs are temperature and surface area. Our hollow plastic conservation balls address these two issues effectively and inexpensively.
  The black conservation balls provide a floating cover on the reservoir taking up approximately 90% of the surface area. Simultaneously the balls prevent UV rays reaching the water and therefore reducing algae.

From the above-mentioned Bloomberg piece:

  The Los Angeles Department of Water and Power has now dumped 96 million balls into local reservoirs to reduce evaporation and block sunlight from encouraging algae growth and toxic chemical reactions. The balls are coated with a chemical that blocks ultraviolet light and helps the spheres last as long as 25 years. Las Virgenes, north of L.A., now uses shade balls, too.

From the Los Angeles Times story on August 12, 2015:

What are they made of?

  The 4-inch-diameter balls are made from high-density polyethylene, which is the same material you would find in a one-gallon milk jug. This plastic is approved to come into contact with drinking water.

Why are they black?

  Carbon black was added to the plastic to stabilize the balls in UV light from the sun.

and, from another LA Times piece about the "shade balls" written in 2008:

  The balls, which cost 40 cents each, are made of polyethylene. The coating contains carbon. Black is the only color strong enough to deflect ultraviolet rays, said Paul Sachdev, president of Orange Products.


I think you are confusing unit of measurement with fundamental unit.  A unit of measurement is a definite magnitude of a physical quantity.  Consider a very common grouping unit, the dozen which is conceptually the same as a mole and is a little easier for us to grasp since it is a commonly used small number.  As far as I know, the only requirements for a unit of measurement to be valid is general consensus.  If you and I agree that we will measure distance in units of 'Chemistry Textbook Thickness', then it is a valid unit of measurement.

A fundamendal unit is a part of a particular set of units upon which all other units can be derived.  In this case, the mole is related to the fundamental unit of mass and is defined as an amount of items equal to the amount of carbon atoms in 12 grams of $^{12}C$.


Question: Can $\ce{SiO2}$ melt at $\pu{20 ^\circ C}$?

According to experimental and calculated data values, my answer is no. See the phase diagram of pure silica based on the experimental and calculated data given in Ref.1:

Reference 1 states that:

An internally consistent data set on the thermodynamic properties of the silica polymorphs stable up to $\pu{15 GPa}$ (α‐quartz, β‐quartz, tridymite, cristobalite, coesite, and stishovite) and the liquid phase is presented. The data set was produced through a computer‐based assessment of the properties in which the available thermochemical (calorimetric), physical (bulk modulus and thermal expansion), and solid‐state and melting transition data (including some newly determined data on the high‐pressure polymorphs cosite and stishovite) were considered. The data set can be used to calculate phase relations at pressures of $\pu{0.1 MPa}$ to $\pu{15 GPa}$ and temperatures of $300$ to $\pu{3200 K}$. The calculated phase diagram using these data agrees quite well with the phase equilibrium determinations except for the high‐temperature part of the coesitestishovite boundary. The properties of the liquid phase obtained are also in good agreement with the available data.

Keep in mind that $\pu{0.1 MPa} \approx \pu{1 atm}$. According to the phase diagram, cristobalite form is in liquid form at temperature grater than $\pu{1750 K}$ in pressure range of $\pu{0.1-0.7 MPa}$. The extrapolation of that boundary would show it would never cross $\pu{20 ^\circ C}$ point on $x$-axis without going to negative pressure. Therefore, it is impossible to have any form of $\ce{SiO2}$ in reduced or high pressusituations at room temperature or below.
References:

V. Swamy Surendra, K. Saxena, Bo Sundman, J. Zhang, “A thermodynamic assessment of silica phase diagram,” Journal of Geophysical Research 1994, 99(B6), 11787-11794 (https://doi.org/10.1029/93JB02968).


Can the oxidation of hydrogen peroxide be reversed?

Yes, if you use some quantum chemistry tool like density functional theory to find the enthalpy of the compounds at many different conditions then you can determine where hydrogen peroxide is the ground state.

You would probably find this at low temperatures and high pressures since it is denser than the other two phases and oxygen is condensed as a liquid or solid so the entropic contribution is smaller.


What temperature is needed for carbon monoxide to reduce iron oxide?

Based on an experiment, carbon monoxide dose reduce iron oxide at 1200°C (2192°F).

But probably, this is not the minimal necessary temperature.
A better answer is still welcome.


How a plane of symmetry makes compounds lose their optical activity

That's really simple. Imagine two different vessels which are precise mirror images of each other; so are all the molecules inside. Imagine that you shine polarized light on them. Being the mirror images of each other, they have to rotate light in opposite directions. But as the compound itself is symmetric, they are in fact identical, so they have to rotate light in the same direction. How can these conditions be reconciled? Just like that: don't rotate light at all.

So it goes.


Table salt is obtained from various sources from mines and seas. The world's second largest salt mine (Khewara salt mine) is in Pakistan and this is where this so-called Himalayan pink salt originates. All of it is mostly sodium chloride however it appears pinkish because of iron impurities. If you dissolve it in water, a very small amount of brown precipitate can be seen which is mostly iron oxides.

There is no dedicated study which shows which salt is better for health or not. Himalayan pink salt has recently become a craze in gourmet (cooking) world and among the pseudo-medical sciences which claim that this pink salt contains 99 minerals and it can cure everything.

  wonder do different types of salts have a different level in salinity when dissolved in water.

You cannot quantify your personal taste for salinity. As chemists, we only care how much sodium chloride is there by mass percentage. Whatever table salt ends up in grocery stores is highly refined and purified. Grocery store salts contain small amounts of additives such as potassium iodide, potassium iodate, silicates, prussiates, thiosulfate, and so on. So the taste may be slightly different from brand to brand.


I feel you are confused on atomic orbital and molecular orbital.

In the excited state of a single sulfur atom, electron do fill 4s first. You are right here. However, it is a different story in molecules.

In sulfur trioxide the bonding electrons are filling the molecular orbital, not the atomic orbital such as 4s or 3d. It is forbidden that both 3s and 4s participate any hybridization or formation of molecular orbital. So no matter how low is the energy of 4s, because 3s is already participating, 4s can not be part of the bonding.


How hazardous are peroxide-able ethers if stored properly?

Peroxides are no fun.  There have been explosions when ether solutions were concentrated to dryness.  What you really want to do is store your ether over potassium hydroxide - hydroperoxides are slightly acidic, and the base forms hydroperoxide salts and pulls them out of solution.

Also, if you keep solvents for more than a year you clearly aren't turning them over fast enough.  Keeping flammable material that you don't need just increases the danger in case of fire.


Why is the equilibrium constant increasing in this exothermic reaction?

An exothermic reaction has a reduced equilibrium constant at higher $T$ because while the contribution of the change in the entropy of the system is a fixed quantity (for a small $T$ change), the effect of transferring heat to the surroundings is reduced at higher $T$ (because it causes a smaller change in the entropy of the surroundings). Mathematically this is described by the van't Hoff equation:
$$\left(\frac{\partial \ln K_\textrm{eq}}{\partial T}\right)_p = \frac{\Delta H^\circ}{RT^2}$$
On the other hand the change in $\Delta G$ with temperature is given by $\Delta S$:
$$\left(\frac{\partial\Delta G}{\partial T}\right)_p = -\Delta S$$
Since $\Delta G =\Delta G^\circ +RT\ln Q = -RT\ln (K/Q)$ one can write
$$\left(\frac{\partial\Delta G}{\partial T}\right)_p = \left(\frac{\partial\Delta G^\circ}{\partial T}\right)_p +R\ln Q\\ = -\Delta S^\circ+R\ln Q$$
Therefore if you want to see how $ K_\textrm{eq}$ depends on $T$ you should focus on $\Delta H^\circ$. If you want to see how $\Delta G$ depends on $T$ you should inspect $-\Delta S^\circ+R\ln Q$.
It is somehow unfortunate that the temperature dependence of $\Delta G$ and the equilibrium constant are different, because it leads to endless mix-ups, but it reminds you that they describe different things.


Most commonly they work by pulling air through a UV absorption cell and simply measuring the absorption at 254 nm.  Often there is a pre-concentration step involved.  This usually involves a gold-plated region prior to the UV cell.  At room temperature mercury adsorbs to the gold, then after some period of time the gold-plated region is flash heated to release all of the mercury at once.


What is the definition of organic compounds?

So, what should be the redefined definition for organic compounds?

We should not define organic compounds. As F'x pointed out in the comments, the distinction between organic and inorganic is largely arbitrary and mostly based on a distortion of historical precedent.

Rather, we should classify compounds by the type(s) of bonding interaction present in the substance. By this criterion, we have several classes of compounds:

Molecular (or covalent) compounds - the particles that make up these compounds are molecules held together by covalent bonds. This class of compound includes traditionally inorganic $\ce{CO2}$ and organic $\ce{CH4}$ carbon compounds. Covalent compounds also the majority of organometallic compounds, so we do not need a separate category for these. Other traditionally "inorganic" compounds, like glass, also belong in this category.
Ionic compounds - the particles that make up these compounds are largely ions held together by Coulombic forces. For the sake of simplicity most people would include in this class of compounds those ionic compounds where one or both ions is a small molecular ion held together by covalent bonds $($e.g. the $\ce{NH4^+}$ ion in $\ce{NH4Cl}$ or the  $\ce{CO3^{2-}}$ ion in $\ce{Na2CO3})$.
Mixed covalent/ionic compounds - those in which there are some ionic functional groups, but mostly covalent bonds, like methyl orange. Many folks might list these substances in Category 1. This category would include most coordination compounds including such things as metal-organic frameworks and probably many composite and/or ceramic materials.
Metallic compounds - those substances with extremely delocalized bonding, typical of metals. Alloys, including steel, belong in this category.

Under this system, what we traditionally study as organic chemistry becomes redefined as the study of the structure, properties, and reactions of covalent compounds (Classes 1 and 3), especially the nature of the covalent bond and how it can be formed and broken selectively. What you might learn about the structure, properties, and reactions of  organic compounds can easily be extended to the structure, properties, and reactions of molecular inorganic compounds. For example, we do not need two different branches of chemistry to study benzene and borazine.


What is the origin of the transferred oxygendianion in redox reactions?

I think you are misunderstanding the meaning of superoxide and the rules.

Superoxide is $\ce{O_2^-}$

Superoxide is NOT $\ce{O^{2-}}$ as you have written.

The rules are referring to hydroxide which is $\ce{OH^-}$, not superoxide.

The reason hydroxide is used to balance equations in basic aqueous solution is:

$\ce{H_2O \leftrightharpoons H^+ + OH^-}$

There is hydroxide present in aqueous solution, but not superoxide, peroxide, hydroxyl radial, etc.


Superscript open circle notation in formula

According to The Manual of Scientific Style: A Guide for Authors, Editors, and Researchers 9.2.5.12

  Indicate a free radical by placing either a centered dot or a superscript dot, signifying the unshared electron, next to the chemical symbol or compound formula

This convention is also followed by the ACS.

So it would be better if the article used the solid symbol throughout, but the solid and open symbols mean the same thing in the article.


Why can hydroxide not be used to synthesise ethanol from ethyl amine?

Hydroxide may be a somewhat good nucleophile but the amide anion is one of the most lousy leaving groups you can imagine. If you want to remove an amino group in an $\mathrm{S_N2}$ reaction, it can only be eliminated as an ammonium cation — but protonating it means we can no longer use hydroxide as a base. Under basic conditions an amino group will never be a leaving group.

Instead, you must turn the amino group into something that is actually a good leaving group. If you include nitrogen and let your thoughts wander, you may arrive at the diazonium group $\ce{-N2+}$. Upon nucleophilic attack, this will be released as nitrogen gas which is great both thermodynamically and kinetically. This leaving group is so good that even relatively bad nucleophiles such as water will react with it.

However, you need to produce it from the amine. This is where nitrous acid comes in. Under acidic conditions, $\ce{HONO}$ will react with primary amines to give diazonium salts according to the following general equation:

$$\ce{R-NH2 + HONO + H+ -> R-N2+ + 2 H2O}\tag{1}$$

Since nitrous acid is produced by acidifying aquaeous solutions of nitrite salts, you have all three components in hand if you ‘just’ add $\ce{HONO}$ or $\ce{HNO2}$.


Resonance is a conceptual tool to help make sense of the real observed structures of molecules
Resonance is not a real physical phenomenon. That is the first important thing to realise. It is merely a tool to bridge the simple view where all bonds consist of integral numbers of electrons and the real structures of molecules.
The simple view of bonds often fails to represent the "real" observed structure of molecules. The archetypal example is benzene. A simple view of the bonding of benzene has the molecule consisting of alternating single and double bonds (which might be called cyclohexatriene). This theoretical molecule has the bonds neatly grouped into simple bonds that use up all the available electrons into either single bonds (one pairs of electrons) and double bonds (two pairs of electrons). But this would imply that the bond lengths alternate. But we know the structure of benzene and the bond lengths are all the same. We can explain this with molecular orbital theory when we recognise that some of the bonds are delocalised, but that gets fairly sophisticated fairly quickly and chemists prefer easier ways to keep track of the electron counts when writing structures on paper and explaining reaction mechanisms.
One halfway house that saves the ability to write structures where simple electron counts are possible is the idea of "resonance". Instead of needing to invoke molecular orbital theory all the time we can write the structure of benzene as a combination of two structures where the double and single bonds alternate. We pretend that the real structure is a "resonance hybrid" of the two hypothetical structures. This is like the mathematical average of the two competing structures.
More complex molecules can also be represented as combinations of (more complex) resonance structures. We often say that the structure consists of resonance hybrids. But resonance is not a real physical phenomenon. The different structures are not real independent structures that rapidly interchange: they are merely a convenient model so chemists can talk about electron counts and draw structures without having to invoke much more complex bonding theories.
So there is no such thing as "resonance" as a physical phenomenon. What drives the structures of molecules is the way orbitals in atoms combine together in molecules and that is more complicated than simple views of bonding can represent. The underlying driving force is quantum mechanics and energy minimisation of the possible ways atomic orbitals interact in molecules. And that is sometimes far more complicated than simple views of bonding can easily represent.


In infrared spectroscopy, why is there no C=C reading for E alkenes?

The $\ce{C=C}$ stretch is responsible for this ir peak. For an ir absorbtion to occur, the absorption must result in a change in dipole moment.  If we examine the $\ce{C=C}$ stretch in cis- and trans-2-butene we find that

image source

cis-2-butene has a dipole moment (0.33 D) and upon stretching the double bond the dipole moment will change; therefore we would expect to see an ir peak for the $\ce{C=C}$ stretch in this molecule
trans-2-butene does not have a dipole moment  (0 D) and upon stretching the double bond the dipole moment will remain zero, there is no change; therefore we would not expect to see an ir peak for the $\ce{C=C}$ stretch in this molecule

A general rule based on these dipole moment arguments is that, for symmetrically di-substituted olefins $\ce{C=C}$ stretches in trans olefins are non-existent or very weak, the absorption is much stronger in the cis isomers.

We can extend this reasoning to molecules with dipole moments, if the dipole moment is very small.  Trans-2-pentene does have a dipole moment but it is very small and it is less than the dipole moment in the cis-isomer and the dipole moment in the trans-isomer (being smaller to begin with)  will change less upon stretching the double bond than it will for the cis isomer.  Therefore we would expect a very weak absorption in trans-2-pentene, while the cis isomer should show a stronger absorption.


Why does one bicyclic diastereomer react faster than the other one?

Yes, the fast reaction is a syn-periplanar E2 reaction. If we are ranking the rates of E2 based on the orientation of the C-H and C-LG bonds, anti-periplanar would be the fastest, syn-periplanar would be slower, and perpendicular would be the slowest. The anti-periplanar arrangement is the fastest because it provides the best overlap of the (filled) bonding C-H orbital with the (empty) anti-bonding C-LG orbital. The syn-periplanar arrangement is slower because there is much less productive overlap between these orbitals. However, the small lobe of the bonding C-H molecular orbital does provide some overlap with the anti-bonding C-LG orbital. A perpendicular arrangement results in no productive overlap.

The slow reaction must occur by some other mechanism, such as an E1 or E2-conjugate base mechanism.


I'm pretty sure it just means the charge of the ligand is -1.

https://en.wiktionary.org/wiki/unipositive

Cyanide qualifies, ETDA does not.

Anyway, the word is on its way out. "Monoanionic" seems to be more popular.