According to Wikipedia, this type of isomerism is known as Isotopomerism

  Isotopomers or isotopic isomers are isomers with isotopic atoms, having the same number of each isotope of each element but differing in their positions. The result is that the molecules are either constitutional isomers or stereoisomers solely based on isotopic location. For example, $\ce{CH3CHDCH3}$ and $\ce{CH3CH2CH2D}$ are a pair of constitutional isotopomers of propane

For a different source, you can also check out the IUPAC Gold book's definition of an isotopomer.


Change in inductive effect of a susbstituent when -R group replaces -H atom

There are several examples where the negative inductive effect of a substituent gets increased when a hydrogen atom on that substituent is replaced by an alkyl group.
Some particular examples:
\begin{align}
\ce{-NR3^+} &> \ce{-NH3^+}& \ce{-OR} &> \ce{-OH}
\end{align}

About comparing $\ce{-NR3^+}$ and $\ce{-NH3^+}$, at first, the electron-donating inductive effect of the methyl groups (+I) makes us feel that the electron-withdrawing inductive effect trend should be the opposite. Neither, could I find any reliable sources with an explanation. So, I came up with a few of my own.
Part I: The Bent's Rule Interpretation (More Reliable)
To analyse the inductive effect, let's take the groups attached to a hydrocarbon chain and then see.

The $\ce{C-H}$ bond orbitals will have more s-character than in the $\ce{N-C}$ bond orbitals because there is more electron-density in the former due to the greater electronegativity of carbon over hydrogen and nitrogen over carbon, which requires the presence of more, lower-energy s-character in them to lower the system's overall energy. Thus, an implication of this would be the increase of p-character in the three $\ce{N-C_{Me}}$ bond-orbitals and the subsequent decrease of p-character in the $\ce{N-C_{chain}}$. This results in an increase of electronegativity of $\ce{N}$, thus the (-I) trend is:
$$\ce{-NR3^+} > \ce{NH3^+}$$
Part II: I don't know what to call this (Need help with this)
Let's consider the examples of phenol and anisole. It is considered that phenol donates more electron density to the benzene ring than anisole because it exists in equilibrium with the phenoxide ion, which is a far-more electron-donating group, despite anisole's $\ce{-OCH3}$ group being a better electron-donating group.
If we apply this to our case, we can assume that $\ce{-NH3^+}$ may be in equilibrium with $\ce{-NH2}$ which makes it a better (+I) group and poorer (-I) group, while $\ce{-NMe3^+}$ can't deprotonate itself and establish an equilibrium like that, making it a better (-I) group.
(somebody please help here)
Part III: Should we really consider $\ce{-CH3}$ as a (+I) group here? (Just thinking, very reliable*)
It is considered that the inductive effect of $\ce{-CH3}$ is mainly attributed to its hyperconjugative effect. However, hyperconjugation requires an empty p orbital carrying a positive charge into which it donates its $\sigma_{\ce{C-H}}$ bond-electron density. However, nitrogen's orbitals are sp³ hybridized here, with no empty orbital carrying the charge as such. So, I think (not really sure*), we should only consider carbon's greater electronegativity over hydrogen, thus making $\ce{-NMe3^+}$ more inductively electron-withdrawing than $\ce{-NH3^+}$. You can also apply this to $\ce{-OR}$ and $\ce{-OH}$.
* Just saw Jan's answer in the middle of my writing process. That makes this a valid point. Great answer, a must check-out.


Why is black-body radiation curve smooth without a sharp cutoff?

The key to understanding the curve is thinking about how energy is distributed among components of the system
The mathematics that produce the curve involves some not-very-simple statistical mechanics some of which Planck didn't understand when he first developed his theory.
But it isn't that hard to get the intuitive idea. Consider the case of a gas where molecules have some kinetic energy. The individual molecules don't all have the same energy: some are moving faster than others. The temperature of the gas is function of the average kinetic energy of the molecules. But individual molecules are exchanging energy very frequently as they bang into each other. But those random collisions exchange random amounts of the kinetic energy between the molecules. Sometimes one molecule will gain a great deal of energy so it ends up with far more than the average amount. But the probability of a series of collisions giving a single molecule a very, very large kinetic energy is low and the higher that energy, the lower the probability. There is no sharp cut-off just an exponentially decreasing probability of getting a higher and higher energy.
Take that picture and apply the appropriate statistical probability theory and you get the overall Planck distribution: a curve where the average kinetic energy is given by the temperature but where individual molecules have some probability of having much lower energy (with slowly decreasing probability) and others have some probability of having much higher energy (but with a sharply decreasing probability). There is no sharp cutoff at high energies just sharply decreasing probability of reaching those levels.
In reality the picture is far more complex as molecules have vibrational and electronic energy as well as kinetic energy, but these details don't matter much for the intuitive picture.


Making carbon monoxide gas from formic acid and sulfuric acid?

The formation of carbon monoxide already takes place when mixing formic acid and sulfuric acid at room temperature, but the reaction is relatively slow. Since the temperature of the mixture may significantly increase during the course of the reaction, the reaction rate may increase, which makes the reaction difficult to control.

For laboratory applications, concentrated sulfuric acid (or phosphoric acid) is usually heated to about 80 °C. Formic acid is added slowly, drop by drop. The formation of gas should start upon addition of the first drop, otherwise the temperature should be increased.

The generated gas contains various impurities, which can be removed using gas washing bottles. Usually, the gas is passed first through $\ce{KOH}$ solution; however, you should take care to prevent any backflow of $\ce{KOH}$ solution to the sulfuric acid.


What is the net ionic equation for rection of liquid bromide with sodium iodide crystals?

Technically you are correct. How can there be any ions when the reactants are not in aqueous medium ? But here it is a matter of convention and maybe an illustration of how to write the ionic equations from molecular ones. Likewise, if you have done balancing of redox reactions , you might notice half of the reactions given to balance are completely impossible according to the laws of chemical combination. You'll always be at a loss when you overthink in chemistry, even if your argument is right.  Taking it as just a convention would be a better idea .

Hope this helped !


How to calculate the potential via Hess's law? Is my textbook wrong about this?

You have to take into account, that the given reactions are only reductions, while you need to have an oxidation, too. So one of the equations has to be reversed.

\begin{aligned}
(A)E^\circ = -0.10~\mathrm{V}&&\ce{CO3^{2-} + 7H2O + 8e- &-> CH4 + 10 {}^{-}OH}  \\
(B)E^\circ = +0.55~\mathrm{V}&&\ce{O2 + 2H2O + 4e- &-> 4 {}^{-}OH}  \\\hline
\color{\red}{-1}\cdot(A) + 2\cdot(B)
   &&\ce{CH4 + 2O2 + 10{}^{-}OH + 4H2O&-> CO3^{2-} + 7H2O + 4 {}^{-}OH}\\
\implies
   &&\ce{CH4 + 2O2 + 2{}^{-}OH &-> CO3^{2-} + 3H2O}
\end{aligned}

Equation $(A)$ gives you the reduction of the carbonate ions, hence the oxidation of methane. Equation $(B)$ is the reduction of oxygen. You have a galvanic cell of the form $\ce{CH4|CO3^{2-}||O2|{}^{-}OH}$, hence the standard potential difference is
\begin{aligned}
\Delta E^\circ(\ce{CH4|CO3^{2-}||O2|{}^{-}OH})~
               &= E^\circ(\ce{O2|{}^{-}OH}) -E^\circ(\ce{CH4|CO3^{2-}})\\
               &= E^\circ(\ce{O2|{}^{-}OH})
                  -\left[\color{\red}{-1}\cdot E^\circ(\ce{CO3^{2-}|CH4})\right]\\
               &= +0.55~\mathrm{V}
                  -\left[\color{\red}{-1}\cdot (-0.10~\mathrm{V})\right]\\
\Delta E^\circ(\ce{CH4|CO3^{2-}||O2|{}^{-}OH})~
               &= +0.45~\mathrm{V}
\end{aligned}

Just to address the second edit of the OP and to clarify:
\begin{aligned}
 (1): &&\ce{3 Ag+(aq) + Fe (s)   &-> 3 Ag (s) + Fe^{3+} (aq)}\\
 (2): &&\ce{3 Ag+(aq) + 3 e-     &-> 3 Ag (s)} & E^\circ &= 0.80~\mathrm{V}\\
 (3): &&\ce{Fe^{3+} (aq) + 3 e- &-> 3 Fe (s)} & E^\circ &= -0.04~\mathrm{V}\\
\end{aligned}

The sign of the second equation has to be flipped and then subsequently subtracted from the first equation, hence the potential that should be found is $\Delta E^\circ = 0.74~\mathrm{V}$. Therefore, I believe your textbook is wrong.


Sulfuric Acid - Bond Lengths and Charge Separation

Good questions, let's step through them one at a time.

1) "So I know that the best Lewis Structure representation of sulfuric acid is one with only single bonds - i.e. sulfur isn't hypervalent. "

Look at the resonance structures below.  The one on the left with sulfur hypervalent is the only uncharged resonance structure.  I suspect that that counts for something.  Here is a link that says it is the most important of the structures (don't get confused when you view the link, they use equilibrium arrows (wrong) in their diagram instead of  resonance arrows (correct).

2) "why are the S-O and S-OH bonds not equivalent then? Why do their lengths differ?"

This is a very fundamental question.  My answer is, they are not equivalent because they are different.  Just as you wouldn't expect the same $\ce{C-O}$ bond length in $\ce{CH_3-O-CH_3}$ and $\ce{CH3-OH}$, you wouldn't expect the same bond lengths in an $\ce{S-O}$ bond and an $\ce{S-OH}$ bond.  In fact, since these bonds are not superimposable one on the other by a symmetry operation, we can say that they MUST be different (i.e. have different lengths).  As one of my favorite professors used to say, now if were to actually measure the $\ce{S-O}$ and $\ce{S-OH}$ bond lengths and our measurements gave rise to the same numeric bond length out to 10 decimal places, we could still step back and say that both we and God know that the lengths are not the same, we just need to measure out to more decimal places!  Let me say the converse, the only bonds that can be identical are those that can be interconverted one with another through a symmetry operation.  For example in chloroform, $\ce{CHCl3}$, there is a $\ce{C_3}$ axis containing the $\ce{C-H}$ bond.  Rotation about this axis interconverts the 3 $\ce{C-Cl}$ bonds, therefor they are equivalent bonds.  Their bond lengths MUST be the same.


What does A. R. in A. R. sodium nitrite mean?

I don't know if I have completely answered, but here's my go:
AR : (ANALYTICAL REAGENT) : Reagents essential for analytical purpose and research work having high purity. If reagent meets the requirements of the American Chemical Society Committee on Analytical Reagents, it will be denoted as an AR (ACS) reagent.
Also see:

https://www.researchlab.in/?product-details/grade-of-purity/
https://www.avantorsciences.com/pages/en/grade-definitions#:~:text=AR%E2%80%94The%20standard%20Macron%20Fine,an%20AR%20(ACS)%20reagent.

EDIT:And note that it isn't a Analytical ring


Equilibrium constant for a solid-solid equilibrium

How do I even write equilibrium constant for solid solid equilibrium?

First, let's write the expression for the reaction quotient:

$$Q = 1$$

Q is a constant, always one, as long as all reactants and products are present. If you run out of one, it is zero or infinity.

Now, there are two cases. If the reaction is at equilibrium, K is also 1. If not, you can't measure the equilibrium concentrations (the reaction will go to completion, that is not an equilibrium). If you want to obtain the equilibrium constant, you would have to determine the Gibbs energy at standard state, and calculate K from that.

This is certainly quite confusing when we contrast it with other reactions (like R1 and R2), which will reach equilibrium. It gets less confusing if we rewrite R3 as:

$$\ce{A(s) -> C(s) +D(s) +E(s)}$$

An example you might be familiar with is ice turning into water. The reaction quotient is always one. Depending on the temperature, the equilibrium constant is smaller than one (all water freezes), is larger than one (all ice melts) or is exactly one (water and ice coexist in an equilibrium). You might think it is unlikely that we adjust the temperature exactly right to hit the melting point, but in fact it self-adjusts because melting ice is endothermic.

  Since B doesn't appear in equilibrium constant, is it safe to say that the concentration of B doesn't affect the equilibrium concentrations of any other species?

It is best to un-ask the question, for two reasons.

There is no equilibrium
The concentration of the other species are constant, so they are not affected by the reaction at all (unless the species is used up)

For reactions where at least one species is a solute or a gas, i.e. a species whose concentration changes continuously and where there is some entropy of mixing, the Gibbs energy will have a minimum somewhere along the way between all reactants and all products. In this special case, everything is a bit different.


Why can air accommodate more water when the air has a higher temperature?

For a substance to be a gas, the kinetic energy of the molecules that make it up must be higher than the electrostatic potential energies that hold it together. When we talk about air "holding" water we are really talking about the amount of water that would be in the vapor phase relative to the amount that would be liquid at equilibrium.

Since temperature is a measure of the average kinetic energy of molecules, as the temperature increases, the fraction of water that can exist as vapor increases. Since air is already a gas under any conditions where water vapor can exist, it looks like the air "holds" more water as temperature increases.


Brown ring from the anion detection of NO3-

The brown ring formed in the confirmatory test for the nitrate ion is due to the complex called nitroso ferrous sulphate, or pentaaqua nitrosyl iron (II) sulfate with the formula $\ce{[Fe(H2O)5NO]SO4}$.
The concentrated sulfuric acid decomposes this complex and brown ring disappers in a few seconds when $\ce{H2SO4}$ moves to the upper layer. This is why slow addition of the acid (through the sides of the test tube) is advised.


Quaternary Ammonium salts and peroxy acid

I believe this is an example of a Cope elimination mechanism here

The EtI is making the tertiary amine. The persulfate is taking the tertiary amine to the N-oxide. Heating this produces ethene and hydroxylamine of the starting amine.

This is consistent with the Cope as "The sterically demanding amine oxide function reacts preferentially with the more easily accessible hydrogens"


Gaussian parameters for IP and EA calculation

If you can afford to do so, it is always a good idea to compare calculations between at least two methods. Good results with B3LYP/6-31G(d) in particular are based upon favorable error cancellation, and B3LYP was not even parameterized against any data in particular (B3PW91 was parameterized for thermochemistry and the PW91 correlation functional was replaced with LYP without changing any other parameters). So, double-checking is wise!

Modern perturbation theory-based approaches have become much more affordable due to the density fitting (DF) and resolution of the identity (RI) approximations (same thing, different name), combined with spin-component scaling which may or may not improve things. The scaling is now roughly $O(N^4)$ and a small prefactor with basis set size; compare this with B3LYP at $O(N^3)$, exact MP2 at $O(N^5)$, CCSD at $O(N^6)$, and CCSD(T) at $O(N^7)$.

Energies are one of the most obvious improvements of range-separated or long-range-corrected (LRC) density functionals. Here the authors find that LC-$\omega$PBEh deviates little from $G_{0}W_{0}$, which is a workhorse for IE and EA calculations. Perhaps the most common new functional that is seeing quite a bit of use in the computational organic chemistry literature is M06-2X, which is shown here to perform even better than most approximate coupled cluster methods for this set of molecules (which are all closed shell!):

With regards to the basis set size, there is only one correct direction to go and that is up. You may be able to afford CCSD(T)/6-31G(d) calculations, but the results will be meaningless due to a severely deficient basis set for anything but the simplest calculations. See here for a recent discussion about the use of basis sets in DFT versus wavefunction methods. Wavefunction methods have always required larger basis sets for quantitative results, especially when not performing a basis set extrapolation. This would mean something like Dunning's correlation-consistent basis sets, augmented with diffuse and polarization functions, such as aug-cc-pVDZ or aug-cc-pVTZ if one can afford it. An interesting and effective truncation of these basis sets (they become very large) is given by Truhlar's "calendar" basis sets.

Less well-known are the requirements for LRC-DFT, which due to modern parameterization schemes are larger basis sets and larger integration grids for the XC contribution, often larger than what is default. Even Pople's triple-zeta sets based on 6-311G will offer some improvement, but the Ahlrichs def2 and Jensen's pcseg-n are probably better.

Finally, a word about the cost of LRC-DFT. Although its formal scaling cost is identical to traditional global hybrids such as B3LYP, in practice range-separated functionals may be 2-3x as expensive due to how the different contributions are calculated and then combined, depending upon the package being used.


Isomerisation of alkanes under Lewis acidic conditions

The mechanism for butane is shown here, instead of a general n-alkane.

Initiation

As the comments have mentioned,  it probably begins by the protonation of butene impurities.

The Lewis acid, $\ce{AlCl3}$, stabilizes the chloride ion from $\ce{HCl}$ when the $\ce{H}$ is used to protonate the butene.

A secondary carbocation is formed, which undergoes methyl shift to form a primary carbocation. I have no idea why a primary carbocation can be formed.

Propagation

An alkane attacks the carbocation, and supplies a "hydride" to the carbocation to complete the carbocation.

Meanwhile, another carbocation is formed which also undergoes methyl shift to regenerate the carbocation at the beginning of this step.

Termination

I am going out of my way here to suggest a isobutyl chloride (1-chloro-2-methylpropane) termination product. I can't find other Lewis bases to complete the carbocation for the termination step.


Does Bent's Rule only apply to molecules where there is hybridisation?

Bent's rule is not as much a rule as it is an observation. It was first applied in the context of valence bond theory and therefore used for hybrid orbitals only. At the initial stages, VBT only used s, sp, sp2, sp3, and p type orbitals. Bent proposed to allow also anything in between aiming at more flexibility for the wave function and therefore better accuracy of the description. As such it is only a tool for an already approximate view on bonding.

Most importantly, and in any and all general cases, the hybridisation state of the orbitals is determined by the molecular structure. For any case in which hybrid orbitals are used, there is also a true solution which does not use them.

Though Bent's rule has some predictive character for anything that is close to hybrid orbitals, it is basically just a shorthand for a much deeper rooted phenomenon.

However, Bent's rule is not useful in any case where there is a low s-p mixing, i.e. anything below period 2 is a stretch.


Thermodynamics: Relation between different types of energy: My Analogy

This is not an attempt at a complete answer but I think its relevant to share this great explanation from "An Introduction to Thermal Physics" by Schroeder:


What does the Pi stand for in the reaction for biological nitrogen fixation?

It stands for inorganic phosphate (Pi). When ATP is broken down into ADP, energy is released along with a phosphate. You can visualise this if you look at the structure of ATP. ADP is then also able to be broken down into AMP and more Pi


The half-life of $\ce{^{89}Sr}$ is 50 days compared to a half-life of 29 years for $\ce{^{90}Sr}$. So a large enough dose of $\ce{^{90}Sr}$ to be useful would leave the patient radioactive for the rest of his/her life.

Edited for use of $\ce{^{89}Sr}$. In a comment Karl pointed out that the isotope is used to treat bone cancer.


Calculating ratio of bicarbonate ion to carbonic acid in blood

For these types of problems, the general approach would be to use the Henderson-Hasselbalch equation. For a general weak acid/conjugate base system, this has the form: $$\mathrm{pH}=\mathrm pK_\mathrm a+\lg\left(\frac{\ce{A-}}{\ce{HA}}\right)$$ Since you are given $\mathrm{pH}$ and have found $\mathrm pK_\mathrm a$, it won't require much more work to obtain the ratio.


What do the stereochemical labels D and − mean in tartaric acid?

As far as I remember, small d always means the same as (+), not small cap ᴅ. So, a dextrorotatory compound is prefixed with either (+) or d while a levorotatory compound is prefixed with either (−) or l.

Small caps ᴅ and ʟ prefixes have different meaning: if an enantiomer can in principle be synthesized from naturally occurring (+)-glyceraldehyde without affecting its configuration, it is considered the ᴅ-form. Otherwise, it is considered the the ʟ-form. And, say, the ᴅ-form of a molecule is not necessarily a dextrorotatory one.

So, the answers to your question are options 2 (the (−) sign indicates that it rotates plane-polarized light anticlockwise), and option 4 (the ᴅ label indicates that it is related to ᴅ-glyceraldehyde).


How do I measure out these salts using the proper method?

You are dealing with very low concentrations and for those you will need very accurate equipment and a very clean lab. I am not sure that reagent purity is enough. In some parts of the world ordinary tap water might already have higher concentrations.
Unfortunately you have not given enough context for what you are doing, so I will outline some procedures which should guarantee quite good accuracy.

First thing you will need is bidistilled water probably lots of it.

For very accurate (and reproducible) results I would recommend purchasing stock solutions and the doing serial dilutions.

If the above is no option you will have to do some more research by yourself, because it depends on the quality and variation of the chemicals you buy. (And preparing a list of all that for your list of chemicals is beyond the scope of this summary.)
For example calcium chloride comes on merck in at least four variations: $\ce{CaCl2}$, $\ce{CaCl2.xH2O}$, $\ce{CaCl2.2H2O}$, $\ce{CaCl2.6H2O}$, and solutions. Depending on your starting material you will need to look up the decomposition temperatures. For calcium chloride you may find them on Wikipedia.

You first need to dry your container with which you will measure the mass of your reagent. Make sure you use gloves as fats and skin can change the result at these low concentrations. Measure it multiple times until you have a constant value (within the limitations of your scale). You need to dry the container at the same temperature which is appropriate for drying your chemical.
Then you follow the same procedure with the chemical. Obviously you will not hit the concentration exactly, but in science that hardly matters, as long as you know the correct value. Once you started the drying process you should keep the container closed until you are ready to transfer it into the solution.
You shouldn't store your prepared stock solutions for a long time and at below room temperature in appropriate flasks.

The other way would be checking the concentration against a primary standard. For calcium chloride you can probably use silver nitrate, which you have to dry like explained above. (It doesn't have the nasty crystal water, which is a plus as $\pu{50 - 90 ^\circ C}$ should be sufficient.)

For setting up the solutions you will need a good assortment of volumetric flasks, pipettes, micro pipettes, thermometers, and probably a bit more I just can't think of right now.

You might want to check with this open access publisher what is common in the field and adjust appropriately. It very much depends on your goal; approximately can cover a pretty large range of desired accuracy.


Why does cyclobutadieneiron tricarbonyl behave aromatically?

In the paper "(Cyclobutadiene)iron Tricarbonyls - A Case of Theory before Experiment", Organometallics 2003, 22, 2-20 https://pubs.acs.org/doi/pdf/10.1021/om020946c, at the page 12 nice explanation of bonding in this compound is provided:

  The electronic structure of (cyclobutadiene)iron tricarbonyl has been the subject of many papers. He I and He II low-energy photoelectron spectra provided useful information.77,78 The eight observed bands in the lowenergy PE spectrum of  (cyclobutadiene)iron tricarbonyl in the range 7.65-20.31 eV all were assigned.78 The
  calculations (ab initio SCF MO) showed that there is a net negative charge on the cyclobutadiene ligand that results from π back-bonding from the iron atom into an
  antibonding MO of the ligand (δ bond). Better agreement between the experimental assignments and theoretical calculations was obtained by Chinn and Hall using generalized MO calculations with configuration interaction.79 A simple textbook approach to the bonding is shown in Figure 8.80 It is assumed that it is triplet state cyclobutadiene that is involved with unpaired electrons in the two degenerate ψ2 and ψ3 molecular
  orbitals. These interact with two singly occupied iron orbitals, generating two covalent bonds.


What is meant by electrons moving to lower potential energy?

In a chemical reaction, the electronic states in the reactants are different from the electronic states of the products because the atoms are arranged differently into molecules. Lower potential energy of the electrons is associated with stronger bonds. When the products have stronger bonds than the reactants, energy is released in the reaction (exothermic). Conversely, when the bonds of the reactants are stronger than the bonds of the products, energy is absorbed in the reaction (endothermic).

  do they move to an even lower ground state?

Yes. For example, the ground state in the hydrogen molecule $\ce{H2}$ is lower than in the hydrogen atom $\ce{H.}$ (because there is more room for the electrons as they are interacting with two nuclei instead of one). If you want to break the molecule into atoms, you have to provide energy to move the atoms apart (i.e. break the bond) because separating them will result in a higher ground state. For molecules reacting to other molecules, the energy balance depends on the relative strength of the bonds (and the number of bonds).

Follow-up questions

  [OP's comment]: Could you just elaborate on the part where you say "because there is more room for the electrons as they are interacting with two nuclei instead of one"? Why does the ground state get lower because of that?

This might be jumping ahead a bit too much, but electrons don't get closer and closer to the nuclei (even when that would increase interactions) because "they don't like to be pinned down" (Heisenberg's uncertainty principle). If an electron can keep the same average distance to the nuclei but occupy a large space, it will lower its energy. This is reflected in the color of molecules with conjugated double bonds (carotene in carrots, for example), and also in a model called particle in a box. One analogy is strings of different length in music, where longer strings produce a lower pitch (lower energy) and short strings produce a higher pitch (higher energy).

  [OP's comment]: I read this online. It talks about how the bonds formed in the products after exothermic reactions have lower potential energy. Is it the electrons which have lower potential energy or the new molecule? This answer states that it's the electrons which have lower potential energy, which one is more correct?

Typically, adjectives for bonds are strong or weak. If you are actually measuring an energy, it is the energy of the entire molecule. The reason energies are different is because the electrons have different energies (nothing happens in the nuclei). So saying electrons have lower potential energy makes the most sense to me. Distinct bonds don't really exist anyway, we just use them for making sense of the molecular properties.

  [OP's comment]: So are bond strength and energy of the entire molecule both dependent on the electron energy?

Yes, much of chemistry is about the electrons. Chemical bonds and intermolecular interactions both are a reflection of "what the electrons are doing", and chemical reactions go along with "electrons doing something new", so knowing the electronic states of the molecules carries enormous predictive power.


What does the slope of a Levich plot represent?

if you mean the following plot:

(Electrochemical Methods. fundamentals and applications, p. 341).

If a bulk concentration is the same as the concentration at the surface of the electrode (the mass-transfer is very high) than the current will be independent on $\sqrt\omega$ and will be constant at the value $i_K$.

The slope of the bold line represent deviation form Levich line and tend towards the limit $\lim_{\sqrt\omega \to \infty} i = i_K $. The higher the speed of rotation is, the less is the electrode reaction limited by mass-transfer effects.

The Levich line (dashed; $\frac{i}{\sqrt\omega}$ ) is constant only when $i_K$ is much larger.


The answer is that cookware transfers the energy from the flame into whatever is being cooked. For instance a pot of boiling water will stay unmelted until all the water has evaporated, after which the pot will start heating up towards the melting point of the metal it is made from.

This is called heat conductance and is used in both cookware and liquid bi-propellant rocket engines (which burn closer to 3000°C). In fact some rocket engines have been made from aluminium with a melting point as low as ~460°C depending on which alloy is used.


This is the general MO diagram you need to fill with the valence electrons of BN

Boron has 3 valence electrons, and nitrogen has 5 valence electrons, this makes 8 electrons.

You have to start filling the orbitals from those with lowest energy to those with higher energy.

So, 2 electrons on $\sigma_\mathrm{2s}$, two electrons on $\sigma_\mathrm{2s}^*$, two electrons on $\sigma_\mathrm{2p}$. You have now 2 electrons left, and two orbitals of the same energy.

In this case, you need to follow Hund's rule, which states that  if two or more orbitals of equal energy are available, electrons will occupy them singly before filling them in pairs. So you will have to put one electron on the $\pi_x$ orbital and another one on the $\pi_y$ orbital.
So you end up with 2 umpaired electrons, and paramagnetism of the molecule is explained.


Is there a material that reduces copper melting point and completely dissolves?

The usual metal 3D-printing processes usually involve metal powder and a plastic binder.  After printing, the green part is sintered.  See this.


The rate of reaction depends on how stable the transition state is. In both E1 and E2 good leaving group is present along with presence of a weak and strong base ,respectively. This leads to an alkene type of transition state in E2 mechanism and a cationic type transition state in E1 mechanism . Both these transition states gets stabilized by the electron donating ability of the attached alkyl groups (i.e By Hyperconjugation and Inductive Effect).This leads to the increase in rate of reaction and thus the secondary and tertiary alcohols are easier to dehydrate than the primary alcohols.
Hope this helps!!


Which elements have no liquid form at atmospheric pressure?

Have a look at a generic phase diagram, for a compound with one solid phase, one liquid phase and a gas phase:

For the solid to go directly into the vapor phase (sublimation) at ambient pressure upon heating, you thus need its triple point to have a higher pressure than atmospheric pressure, i.e. ~1 bar.

Looking at this partial table of element triple points, arsenic is another element that sublimes at ambient pressure. Iron would seem to qualify, but it has more phases that reported in the table, and the full phase diagram indicates that it does not sublime at ambient pressure.


What does "Anal. Calcd." stand for written out?

They are two distinct terms, that just happen to frequently be found next to each other. Their definitions can be found in the Journal of Organic Chemistry Guidelines for Authors (see page 29).

Anal. - abbreviation for combustion elemental analysis.

Calcd - abbreviation for calculated.

It is not actually a single phrase, as you've suggested here. Combustion elemental analysis is what people often mean when referring to 'getting an analytical done'. Writing Anal. signifies the type of experimental data that is about to be listed, just as you might specify 1H NMR, IR or MS. A common alternative to combustion elemental analysis is high resolution MS, which is often written in experimental sections as HRMS (ESI) m/z calcd

The format for presenting elemental analysis is journal specific, and therefore varies. The format you're asking about is consistent with the ACS publication recommendations. RSC guidelines accept the the data in a slightly different format.


Why are fluorides of transition metals unstable in low oxidation states?

You can take the example of $\ce{CuF}$ which is not known.

Reason: $\ce F$ is such an electronegative element that it will always oxidise $ \ce{Cu}$ to $\ce{Cu^2+}$ and not $\ce{Cu^1+}$. Hence, whenever $\ce{Cu}$ reacts with flourine, copper(II) flouride is formed. The following reaction illustrates this:

$$\ce{Cu + F2 -> CuF2}$$

$\ce{CuF2}$ loses fluorine at temperatures above $\pu{950 °C}$.  $\ce{CuF}$ will be formed initially and as your question says, it is highly unstable so it will be transformed to some other stuff and the reactions are:

$$\ce{2CuF2  -> 2CuF (unstable) + F2}$$

$$\ce{2CuF  -> CuF2 + Cu}$$

Similarly, you can illustrate this for $\ce{VF2}$.


Does adding solutes to water alter its triple point?

Triple points are typically defined as points on the phase diagram of a pure substance where three phases coexist. Their existence requires that the phase rule be observed, which is $$f=C+2-P$$ where $$\begin{align}f &: \text{number of degrees of freedom} \\ C &: \text{number of components} \\ P &: \text{number of phases}\end{align}$$ The number of degrees of freedom is the number of intensive variables that can be altered independently while remaining in a given phase.

In general, when three phases coexist, $$f=C-1$$ If the system consists of a pure substance, $f=0$, as expected (the coexistence region corresponds to a point). If you have a mixture of $C$ components (labeled $i$), you will observe a different triple point for each choice of $C-1$ independent mole fractions $\chi_i$. But for each combination of mole fractions, coexistence of three phases will still occur at regions corresponding to single points.


This law bears Avogadro’s name because he implicitly provided circumstantial evidence that it is true, by successfully applying it as explained below. But, strictly speaking, Avogadro’s law has not been discovered neither by experiments nor by reasoning since it had already been in the air if one had accepted the permanent chaotic movement of particles in gases as suggested by Daniel Bernoulli in 1728.

Also Dalton reflected about the equal volumes−equal numbers hypothesis before Avogadro, but discarded it as impracticable because he did not factor in the idea that gaseous elements might be polyatomic molecules; in consequence he thought that the equal volumes−equal numbers hypothesis were not useful to explain the known physical and chemical facts about gases. (According to Avogadro, Dalton had additionally some strange caloric objections against this view, see p. 59 of the link in the next but one paragraph.)

Hence, the idea that elements might be polyatomic molecules was not accepted at that time (1811). It was even declined thereafter for a long time (by Berzelius who dominated chemistry then).

Being a physicist and mathematician, Avogadro did not himself accomplish chemical reactions with gases, but he could built on existing gas density measurements and the following two results of Gay-Lussac (1808), which he discussed in his article in the Journale de physique, de chimie …, 1811 (all the following page numbers refer to this text) with the equal volumes−equal numbers hypothesis as starting point (!):

Gay-Lussac's combining law: “Not only do gases combine in very simple proportions” by volume, “but the apparent contraction of volume, which they experience on combination, has also a simple relation to the volume of the gases, or at least to that of one of them” (p. 58), and
his experimental combining volume ratios for different gas reaction, e.g. of the formation of water (1 + 1 = 2), ammonia (1 + 3 = 2) from the elements (p. 60).

When comparing all these data, Avogadro proceeded in two steps:

he discussed that, as a consequence of the equal volumes−equal numbers hypothesis, one could easily bring into line the densities of elementary gases with the relative molecular weights of the gases; for instance the density of oxygen being about 15.074 times larger than that of hydrogen (p. 59).
Then he had (p. 60) the idea that gaseous elements could consist of polyatomic molecules (“molecules composées”). Subsequently he discussed extensively the known combining volume experiments of gases in this light and showed that so all hitherto contradictions thereof could be solved.

In this way, Avogadro had simultaneously also proved by circumstantial evidence that the equal volumes−equal numbers hypothesis, from which he started his eventually successful reflections, must be true because he had demonstrated that it is one of the important factors for the causal explanation of all those gas data by particle theory.

Hence, the so called “discovery” of Avogadro's law consists merely in the success to bring all these gas data into accord, which vice versa substantiates that it is a law, whereas before this application, it was unclear how the equal volumes−equal numbers hypothesis could be generally made use of.


Specific heat capacity and the periodic table

Either you have a periodic table that features the specific heat capacity of elements in their standard state at STP (standard temperature and pressure), in which case you merely have to look it up, or you don't, and there's nothing you can do.

Specific heat capacity of materials, even elemental solids, are physical properties resulting from complex behavior of the systems, and cannot be determined by simple laws or equations. They have to be determined experimentally, or by first principles calculations.

PS: water, methanol, carbon tetrachloride and dichlorodifluoromethane have got nothing to do with it.


Why does water evaporate at room temperature?

First, I think I should make it clear that when water boils, the bonds in the water molecule linking the hydrogen and oxygen atom are not broken. During boiling, the intermolecular bonds in water are the ones that get broken, that is the bonds that link the water molecules together.
At room temperature, there is evaporation (I wouldn't call it excitation). This is because there are a few molecules of water which can manage to muster enough energy to escape from the large body of molecules and escape into air.
This can be explained through a graph depicting the distribution of speed among water molecules worked out by Maxwell and Boltzmann.

As you can probably see, there are a lot of water molecules with lower kinetic energy than with higher kinetic energy. Those that have the higher kinetic energy are the ones that are able to break through the water surface to become vapour.
Even at low temperatures, there are some water molecules are have enough energy to escape and that's why evaporation in water can occur at any temperature (yes, even if the water is in ice).
When the temperature increases, there are more molecules with higher kinetic energy and thus, more water can evaporate.


Here you can find a phase diagram for $\ce{Si/C}$ system. It does not have zones with homogeneous non-stohiometric solids. So, there is not thermodinamically stable Si/C isomorphic alloys. However, since the liquid likely to be homogeneous, it is likely for fast cooled liquid to form amorphous alloys. Indeed, google search provides plenty of links for amorphous $\ce{Si/C}$ alloys. However, it is also possible for cooling liquid to form polycrystalline alloy, say, with $\ce{SiC}$ and $\ce{C}$ grains, similar to pig iron.


Determining the temperature at which 90% of the reactants in a equilibrium have reacted

Assuming $\ce{FeO}$ and $\ce{Fe3O4}$ are solid throughout the range of temperatures we are interested in, they won't appear in the expression for the equilibrium constant $K$ (which in this case is equivalent to $K_p$). If $n$ is the total number of moles of $\ce{CO}$ that you start with,

$$K_p = K = \frac{p_{\ce{CO2}}}{p_{\ce{CO}}} = \frac{n_{\ce{CO2}}}{n_{\ce{CO}}} = \frac{0.9 n}{0.1 n} = 9$$

Plugging this into the formula previously derived, we find $T = 1388~\mathrm{K}$.


Hyperconjugation accounts for the somewhat greater negativity of the 4-nitro-group compared with the 2-nitro-group in 2:4-dinitrotoluene; in consequence, selective reduction of the former by alkaline sulphides produces mainly 2-nitro-4-aminotoluene, although some 4-nitro-2-aminotoluene is also formed. In acid medium the 2-nitro-group becomes relatively negative in comparison with the 4-nitro-group, owing to the inductive (+I) effect having the chief role, with resultant preferential salt formation to give the very positive $\ce{-NO2H}$ group at 2-position and preferential reduction of it by acid stannous chloride.

Quoted from Coloration Technology 62 (1946) 114

As for how ammonium sulphide reacts see this answer:  Selective nitro reduction of poly nitro compounds


Nitrogen was originally called 'azote' by Lavoisier. This name persists in many nitrogen containing species such as azide, hydrazine, diazonium etc.

The '-ide' ending is the standard ending for anions of a single element such as carbide, oxide, sulfide etc.


After bit research i got finally one link where they mentioned the colour. The other name of(Synonym) of Nickel thiocyanate is "Nickel Thiocyanate Greenpowder" which suggest it is Green in colour.

Reference:ChemicalBook


Finding equilibrium concentrations, trouble simplifying equation

Everything you do is correct and it is only the final algebraic transformation missing:
\begin{align}
0.068 &= \frac{1-x}{(1.25+x)(2.00+x)}\\
0.068 &= \frac{1-x}{x^2 + 3.25x + 2.5} &&|\cdot (x^2 + 3.25x + 2.5)\\
1-x   &= 0.068(x^2 + 3.25x + 2.5) &&|-1 +x\\
0.0   &= 0.068x^2 + 1.221x -0.83\\
\end{align}

Now you can solve for $x$ according to
\begin{align}
0&=ax^2+bx+c & x_{1,2}&=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\
\end{align}

And that gives me the values $x_1\approx0.656$ and $x_2\approx-18.6$. The latter one of course makes no sense.


Prove that idempotency is a necessary and sufficient condition for a densiy matrix to be N-representable

A density matrix $D$ is positive semidefinite, hermitian, and has trace one.
Because of hermiticity we may assume that it is diagonal.
Let's denote the eigenvalues with $\lambda_i$ .
Because it is positive semidefinite, we have $0 \leq \lambda_i \leq 1$.

The matrix $D^2$ has eigenvalues $\lambda_i^2$. Since it is idempotent ($D = D^2$) and has trace one, we may write:
$$\sum_i \lambda_i = \sum_i \lambda_i^2 = 1$$
The only possibility for this to be true is, if there is one and only one $j$
for which $\lambda_j = 1$.

So $D$ has one $1$ in the diagonal at the $j$-th column and all other values in the matrix are zero.
You can easily represent this matrix with:
$ e_j e_j^T$. ($e_j$ being the $j$-th unit vector.)

If you transform back to you original nondiagonal form, you obtain your $\Psi$.


What is the softening temperature of styrofoam?

There are a number of different temperatures one could consider the softening point of atactic (randomly distributed phenyl groups) polystyrene, used in making expanded polystyrene :

Softening point: ~90 °C
Heat Deflection Temperature: 95 °C
Typical Injection Molding Temperature: from 170 °C at back to 230 °C at nozzle


How do I determine the molecular vibrations of linear molecules?

As Tyberius noted, the projection formula does not work for infinite order groups (this is because the Hermitian form on characters is defined to be G-invariant by averaging over all elements in a group. That is, it works because it is possible to hit all the elements of the group in some order).

How can one work in the infinite dimensional groups then? One property that still holds for these particular infinite groups (though not in general 1) is Maschke's Theorem, that every representation is the direct sum of irreducible representations. So we can still determine what irreps make up a rep by adding them up.

First we need to find the representation we wish to reduce. I find that the easiest way to do this is to start with just the atoms (instead of the vectors) and then include the vector contribution afterwards. For the sake of a new example, I will do the linear $\ce{I4^{2-}}$ anion instead of $\ce{CO2}$.

$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma_\text{atoms} & 4 & 4 & \cdots & 4 & 0 & 0 & \cdots & 0 &\\
\end{array}$$

Now, we multiply by the vector's contributions to the character.
$$\begin{array}{c|c}E&3\\\hline C_2&-1\\\hline \sigma&1\\\hline i&-3\\\hline C_n&1+2\cos(\frac{2\pi}{n})=1+2\cos\theta\\\hline S_n&-1+2\cos(\frac{2\pi}{n})=-1+2\cos\theta\end{array}$$

These come from taking the trace of the matrix representations of each symmetry operation. For example, a rotation about the z-axis by an angle $\theta$ is given by:
$$\left[\begin{array}{ccc}\cos\theta&\sin\theta&0\\-\sin\theta&\cos\theta&0\\0&0&1\end{array}\right]$$

$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma_\text{xyz} & 12 & 4+8\cos\theta & \cdots & 4 & 0 & 0 & \cdots & 0 &\\
\end{array}$$

Now, we are only interested in the vibrational modes so we must subtract off the translational and rotational degrees of freedom. We can check the character table for these (they are the ones which correspond to $z$, $(x,y)$ and $(R_x, R_y)$, $R_z$ doesn't count because we define the molecular axis to be the $z$-axis.
$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma_\text{xyz} & 12 & 4+8\cos\theta & \cdots & 4 & 0 & 0 & \cdots & 0 &\\
-\Gamma_\text{rot} & 2 & 2\cos\theta&\cdots&0 & 2 & -2\cos\theta&\cdots&0 \\
-\Gamma_\text{trans} & 3 & 1+2\cos\theta&\cdots& 1 & -3 & -1+2\cos\theta&\cdots&-1 \\
\hline\Gamma_\text{vib} & 7 & 3+4\cos\theta&\cdots& 3 & 1 & 1&\cdots&1 \\
\end{array}$$

So now we have the actual representation of the vibrations. How do we go about decomposing it? We want to eventually subtract off all of the irreps until we have the zero vector so we consider what effect each irrep has and try to move our remaining rep closer to zero. The first thing to note is that only $\Pi_g$ and $\Pi_u$ irreps have $2\cos\theta$. $\Pi_u$ adds $2\cos\theta$ under both proper and improper rotations, while $\Pi_g$ adds to proper and subtracts from improper. We note that in our rep, we only have $\cos\theta$ terms under proper rotations and so we need to subtract off enough $\Pi_u$'s until there are the same number of $\cos\theta$ (with different signs) under proper and improper rotations, at which point we should subtract off $\Pi_g$s until there are no $\cos\theta$ terms.

$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma_\text{vib} & 7 & 3+4\cos\theta&\cdots& 3 & 1 & 1&\cdots&1 \\
-\Pi_u & 2 & 2\cos\theta&\cdots&0 & -2 & 2\cos\theta&\cdots&1 \\
-\Pi_g & 2 & 2\cos\theta&\cdots&0 & 2 & -2\cos\theta&\cdots&1 \\
\hline\Gamma & 3 & 3 &\cdots& 3 & 1 & 1&\cdots&1 \\
\end{array}$$

Similar to the $\Pi_g/\Pi_u$, the $\Sigma_g^+$ shift adds 1 to all positions, while $\Sigma_u^+$ adds 1 to all positions before $i$ and subtracts from all after. We subtract enough $\Sigma_g^+$ until all the positions before and after $i$ have equal magnitude.
$$\small \begin{array}{c|cccccccc} \hline
D_{\infty\mathrm{h}} & E & 2C_\infty^\phi & \cdots & \infty\sigma_\mathrm{v} & i & 2S_\infty^\phi & \cdots & \infty C_2 & \\ \hline
\Gamma & 3 & 3 &\cdots& 3 & 1 & 1&\cdots&1 \\
-2\Sigma_g^+ & 2 & 2 &\cdots&2 & 2 & 2&\cdots&2 \\
-\Sigma_u^+ & 1 & 1 &\cdots&1 & -1 & -1&\cdots&-1 \\
\hline \Gamma& 0 & 0 &\cdots&0 & 0 & 0&\cdots&0
\end{array}$$

Thus, we have found that the vibrational modes of $\ce{I_4^{2-}}$ to be $2\Sigma_g^+ +\Sigma_u^+ + \Pi_u + \Pi_g$.

We can check this with the formulas given in Group Theory And Chemistry by David M. Bishop as noted by Feodoran. One can also follow Tyberius's suggestion and work in $D_{2h}$ which will give the decomposition $2A_g+B_{1u}+B_{2u}+B_{3u}+B_{2g}+B_{3g}$.

It is important to note that correlation from the $D_{2h}$ point group to $D_{\infty h}$ works fine for vibrations because you cannot have terms higher than $\Pi$ and so there is no ambiguity. For things like electronic states, though, in which direct products must be decomposed, you will run into some problems. If the $D_{2h}$ result includes $A_g+B_{1g}$ does that correlate with a $\Sigma_g^+ +\Sigma_g^-$ or a $\Delta_g$ state in $D_{\infty h}$? Working in the parent infinite group may be more difficult without the reduction formula, but these problems are avoided.

The reason going to lower symmetry works is because all characters under the infinitely many proper and improper rotations are the same for each irrep (namely $2\cos n\theta$) and so we can represent the behavior of all of them by the behavior of one of them. Hence, we pick $\theta=\pi$ and ignore all the others and this produces the $D_{2h}$ point group (the two improper rotation axes are equivalent to $i$ and the two proper axes are equivalent to one $C_2$ about $z$). This is origin of the ambiguity. In $D_{\infty h}$, $\Sigma_g^+ + \Sigma_g^-$ are equal to $\Delta_g$ at $\theta=\pi$, but not in general.

Having now identified the origin of the ambiguity, one can tune how fast these ambiguities occur though. The problem occurs when the $\theta$ chosen to represent all $\theta$ has the property $1=\cos n\theta$ where $n$ represents the period of the ambiguity. So for $D_{2h}$, $\theta=\pi$ and $1=\cos2\theta$ and the period of the ambiguity is 2. That is, every other term will be ambiguously identical ($\Sigma, \Delta, \Gamma, \ldots$ are the same and $\Pi,\Phi,\ldots$ are the same). By dropping to $D_{4h}$ instead, $\theta=\pi/2$ and $1=\cos4\theta$. The ambiguity period is longer and we can distinguish $\Delta$ from $\Sigma$ by correlation from the $D_{4h}$ point group. We can thus select the point group to descend to based on how much resolving power we expect to need in the parent point group (though we should stick to even numbers to maintain inversion symmetry for $D_{\infty h}$). Of course we could also stay in the parent point group and never have a problem, but also have no reduction formula.


Kevlar is an A,B co-polymer where the monomers are terephthalic acid and 1,4-diaminobenzene. The amide linkage, together with the aromatic rings in the polymer, makes a very rigid polymeric structure that stacks the chains in an organized way, maximizing the effectiveness of hydrogen bonding between amide linkages, together with pi-stacking of the rings:

http://pubs.acs.org/cen/coverstory/8005/8005olympics.html

The chains form sheets that stack together along the axis of the fiber. In order for projectiles to break these fibers, these strong, inter-strand attractions must be overcome.

Standard Nylon is a similar polymer, where the aromatic moieties are replaced by aliphatic chains. These are much more flexible and give more degrees of freedom to the polymer chains. This results in a more random arrangement of chains that still hydrogen bond, but not in such an organized way, giving a weaker fiber.

https://www.researchgate.net/publication/305313239_Nylon_66_Nonwoven_Fabric_Separates_Oil_Contaminates_from_Oil-in-Water_Emulsions


Calculate the pH of acetic acid for a 0.02 M solution

This type of problem is best solved using the ICE (Initial Change Equilibrium) method.
First, write down the chemical equation and the law of mass action for this reaction.
$$\ce{CH3COOH <=> H+ + CH3COO-}$$
$$K_a = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}$$
Second, determine the initial concentration of each of the three species. Make a little table to keep everything straight:
\begin{array}{lc} \hline
&\ce{CH3COO-} &\ce{<=>}&\ce{H+} &+&\ce{CH3COO-} \\ \hline
\text{I}& \pu{0.02 M} &&\pu{0 M} &&\pu{0 M} \\ \hline
\end{array}
At equilibrium, some of the acetic acid will be ionized. How much? We don't know right now, so let's have that be a variable; $x$. At equilibrium, the concentration of acetic acid will have decreased by $x$. How much will the concentration of $\ce{H+}$ and acetate anion increase? Since the stoichiometric coefficient of all species is 1, then those concentrations will increase by $x$. Let's add that to our table. Let's also write an expression for the equilibrium concentrations in terms of the initial concentrations and $x$.
\begin{array}{lc} \hline
&\ce{CH3COO-} &\ce{<=>}&\ce{H+} &+&\ce{CH3COO-} \\ \hline
\text{I}& \pu{0.02 M} &&\pu{0 M} &&\pu{0 M} \\
\text{C}& -x &&+x  &&+x \\
\text{E}& 0.02-x  &&x &&x \\ \hline
\end{array}
Now, plug these expressions into the law of mass action:
$$K_a = \dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \dfrac{(x)(x)}{(0.02-x)}$$
As long as $x\ne 0.02$, which is very unlikely at equilibrium, we have a solvable quadratic equation:
$$K_a=1.8\times 10^{-5}= \dfrac{(x)(x)}{(0.02-x)}$$
$$3.6\times 10^{-7}-1.8\times 10^{-5}x-x^2=0$$
Some texts will encourage you to make the simplification $0.02-x \approx 0.02$ since $x<<0.02$, however, WolframAlpha can solve the original equation (and so can you).
The equation has two roots: $x=5.91\times 10^{-4}$ and $x=-6.09\times 10^{-4}$. The second one is nonsense (we cannot have negative concentrations).
If you make the assumption that $0.02−x \approx 0.02$, then the positive root is $x=6.0 \times 10^{−4}$, not a big difference given the significant figures. If you knew your initial concentration more precisely, this assumption would be invalid.
Now, we know what $x$ is. What do we still need to know to determine pH? How is $x$ related to pH?


What is the main contributor of the color of honey?

I found that the color of honey depends on the presence of plant
pigments including carotene, xanthophylls, anthocyanins, flavonoids,
polyphenols, as well as amino acids and mineral salts

Yes, you are right about that. The color differences is due to the difference in levels of minerals, anti-oxidants, and other trace elements that are present. In particular, levels of polyphenols, flavonoids, and carotenoid have been observed to have high correlation with color. Polyphenols are found in high concentration in darker honeys, whereas flavonoids are found in higher concentrations in lighter honeys.
References

Antioxidant and antimicrobial capacity of several monofloral Cuban honeys and their correlation with color, polyphenol content and other chemical compounds by Jose M.Alvarez-Suarez Sara Tulipani, Daimy Díaz, Yadiley Estevez, Stefania Romandini, Francesca Giampieri, Elisabetta Damiani, Paola Astolfi, Stefano Bompadre, Maurizio Battino, Volume 48, Issues 8–9, August–September 2010, Pages 2490-2499, DOI: https://doi.org/10.1016/j.fct.2010.06.021
Total Phenol, Antioxidant Activity, Flavonoids, Anthocyanins and Color of Honey as Affected by Floral Origin Found in the Arid and Semiarid Mediterranean Areas by
Taha M. Rababah  Mohamad Al‐Omoush  Susan Brewer  Mohammad Alhamad  Wade Yang  Mohammad Alrababah  Abd Al‐Majeed Al‐Ghzawi  Muhammad Al‐ u'datt  Khalil Ereifej  Fawzi Alsheyab  Ranya Esoh  Ali Almajwal, Volume38, Issue3, June 2014, Pages 1119-1128, DOI: 10.1111/jfpp.12071


Yes NaCl3 is possible.

Think of the more common iodine analog.

$\ce{KI + I2 -> KI3}$

For a comprehensive analysis see:

Unexpected stable stoichiometries of sodium chlorides

  At 25-48 GPa, NaCl3 is stable in the Pnma structure, which has 4 formula units in the
  unit cell. Unlike all the other new phases predicted here (which are metallic), this phase is
  a semiconductor. Its structure (Fig. 2(B)) contains almost linear asymmetric Cl3 groups.
  Bader analysis (19) shows that the middle atom in the Cl3-group is nearly neutral, with
  most negative charge on the side atoms (Table 1) and the total charge of this anion group
  being ~-0.8. This result is reminiscent of well-known triiodides and azides with I3
  - and
  N3
  - ions, and of recent calculations on hypothetical H3
  - ions (20), which were found to
  have charge configuration [H-0.81H+0.72H-0.81]-0.9. Pnma-NaCl3 still can be viewed as an
  octet compound, with covalent bonding within [Cl3]--groups and ionic bonding between
  Na+ and [Cl3]-. At 48 GPa, with the formation of a new phase of NaCl3, the octet rule
  breaks down.

This research group compressed chlorine gas between NaCl plates, while heating with lasers and observed formation of $\ce{NaCl3}$.


Platinum is chemically inert ("noble") meaning it resists forming oxides or other compounds. This stability is desirable when doing a flame test because it ensures that we only see the spectrum of what we are burning. (Think how much noise we would have in our spectrum if we used a Magnesium wire to hold our sample)
Of the chemically inert metals (e.g. Au, Ag, Pd, Pt), Platinum has the highest melting point (2041º C), so it's the most useful for holding in a flame.
Glass, on the other hand, melts at a temperatures ~500º C lower than Platinum.


Taken from the website of a chemical supplier, the following solubility data can be easily obtained. Historically benzene and chloroform are commonplace for rotation measurements, however it can be any solvent in which the sample is soluble (or mixtures of solvents). You might be best to find a literature value for cholesterol (scifinder/reaxys) and then measure in the same solvent so at least you have a reference point to compare to.

One gram dissolves in 2.8 ml of ether, in 4.5 ml of chloroform, and in 1.5 ml of pyridine. The product is soluble in acetone, dioxane, ethyl acetate, benzene, petroleum ether, oils, fats, and in aqueous solutions of bile salts. Solutions should be protected from light.
Source Sigma Aldrich

Generally speaking, optical rotation isn't a method commonly used to determine whether or not you have a given compound. Methods like NMR, IR, MS, or even just TLC comparison are far more suited to this.
Optical rotation is useful if you already know you have the right thing, and want to know what your optical purity is like. Without having already confirmed that the sample is 1. clean and 2. the right thing, you'll struggle to make sense of the data since the polarimeter will simply give you a number.
Depending on where you bought your cholesterol from, and various other things, this number may be close to the literature value (maybe some confidence in it being the right thing), however even then you cant be sure its cholesterol as many other things will have similar values.


Converting Lactic Acid to Potassium Lactate

OP wasn't clear about OP's procedure. Yet, OP is clear that OP needs a table to calculate amounts of lactic acid (MW: $M_\ce{LA} = \pu{90.08 g mol-1}$) and potassium hydroxide (MW: $M_\ce{NaOH} = \pu{50.11 g mol-1}$) need make certain amount of potassium lactate. The chemical equation regarding this calculation is as follows:

$$\ce{CH3CH(OH)COOH + KOH -> CH3CH(OH)COO^-K+ + H2O}$$

This means one mole of lactic acid reacts with one mole of potassium hydroxide to give one mole of potassium lactate.

Suppose OP needs $x$ moles of potassium lactate. Thus, OP needs $x \ \pu{mol}$ of $\ce{KOH}$. However, I believe OP has only $90\%\mathrm{(w/w)} \ \ce{KOH}$, meaning if you weigh $\pu{100 g}$, you get only $\pu{90 g}$ of $\ce{KOH}$.  Thus, mass ($m$) you need to measure to get $x \ \pu{mol}$ of $\ce{KOH}$:

$$m_{\ce{KOH}} = \frac{\pu{56.11 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{90}$$

Similarly, since OP has only $88\%\mathrm{(w/w)}$ lactic acid, meaning if you weigh $\pu{100 g}$, you get only $\pu{88 g}$ of lactic acid.  Thus, weight ($m$) of lactic acid OP needs to measure to get $x \ \pu{mol}$ of lactic acid:

$$m_{\mathrm{LA}} = \frac{\pu{90.08 g}}{\pu{1 mol}}\times x \ \pu{mol}\times \frac{100}{88}$$

Now, using these equations, OP can measure any amount of reagents in order to make desired amount of potassium lactate. For example if OP need $\pu{5 mol}$ of potassium lactate ($x=5$ in both equation), then OP can simply weigh following amounts of $\ce{KOH}$ and lactic acid respectively, and react them together:

\begin{align}
m_{\ce{KOH}}
  &= \frac{\pu{56.11 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{90}
 = \pu{311.7 g}\\
m_{\mathrm{LA}}
  &= \frac{\pu{90.08 g}}{\pu{1 mol}}\times \pu{5 mol}\times \frac{100}{88}
 = \pu{511.8 g}
\end{align}

Keep in mind that these calculations are valid only if purity of both reagents is given in $\%\mathrm{(w/w)}$. Otherwise, additional information such as density is needed.


Recall that $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$
Where $\ce{[OH-]}$ is the concentration of hydroxide ions.

Suppose a solution has $10^{-7}$ moles of $\ce{[OH-]}$ ions. Then this implies $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ and $\mathrm p\ce{OH} = -\log_{10}\space [10^{-7}]$ which is equal to $7$.

Suppose a solution has $10^{-4}$ moles of $\ce{[OH-]}$ ions. Then this implies $\mathrm p\ce{OH} = -\log_{10}\space [\ce{OH-}]$ and $\mathrm p\ce{OH} = -\log_{10}\space [10^{-4}]$ which is equal to $4$.

So as you can see, lower the $\mathrm p \ce{OH}$, the higher the concentration of $\ce{OH-}$ ions. Le Chatelier's principle is not violated.
(How I like to remember this is - the strongest acids have the lowest $\mathrm pK_\mathrm a$. So that extends to $\mathrm p\ce{OH}$ too. This is all due to the fact that $\mathrm p\ce{OH}$ is the negative logarithm of the $\ce{[OH-]}$ concentration in solution.)


How is Humic substance destroyed in nature?

Sunlight is aggressive, as is microbiology.  Ozone, hydroxyl radicals (including Fenton chemistries with peroxides), air pollution in general eventually degrade and destroy near anything organic.  Examine sediments downstream of rivers, including outflows, especially for heavily colored waters.  Sampling vs. linear distance and depth will give you histories of degradative chemistries and biologies.  Note that phenolics are overall good microbiocides re lignin.

Terra preta soils demonstrate the survivability of charcoal.  Humic substances are more functionalized and more reactive.  One can start at the other end, too - weathering of coal.  Coal contains dispersed iron pyrite and other sulfide minerals that speed degradation.  Powdered coal heaps can self-ignite from cumulative oxidation heat.  Carbon dioxide, water, and elemental nitrogen are deep thermodynamic holes.


Ionic Radius of Iodine in High Temperature Form

Considering by cubic lattice you mean a Simple Cubic structure, you could do the following:

Reference image1
Reference image2

(could not put it here because of copyright)

In a cubic lattice the atoms touch each other along the edges(see image1 and image2), hence the edge length/parameter will be equal to twice the radius (see image2).
$$\therefore\ \ce{Edge\ Length(a)=2\times radius(r)}$$
$$\therefore\ a \ = \ 2\times r$$

This method can be adopted for any lattice, you only need to find the axis along which the atoms touch each other(image1 bottom row).

If they exhibited an fcc close packing structure, you would notice that the atoms touch each other along a face-diagonal.

$$
\begin{align}
\therefore\ce{Length\ of\ face\ diagonal&\ =\ radius\ of\ top\ left\ atom +\\
                                        &\quad\ diameter\ of\ atom\ at\ face-center + \\
                                        &\quad\ radius\ of\ bottom\ left\ atom}\\
\therefore\ce{Length\ of\ face\ diagonal&\ =\ r\ + ( 2\times r ) + r}\\
\therefore\ce{\sqrt{2}\times a&\ =\ 4\times r}\\
\therefore\ce{a&\ = 2\sqrt{2}\times r}
\end{align}
$$

Similarly for bcc structure the atoms touch each other along the body-diagonal
$$\therefore\ce{\sqrt{3}\times a\ =\ 4\times r}$$


Use of repeated numerical prefixes for substituents on methane

According to Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013,

  P-14.3.4.2 The locant '1' is omitted in substituted mononuclear parent hydrides.

For the compound $\ce{CHCl3}$ (chloroform), the parent hydride is methane, which is mononuclear. Hence the locant '1' should be omitted in the preferred IUPAC name (PIN); it is simply trichloromethane.


You are wrong. Assuming that you are doing ppm in terms of mass: There are $0.0055$ grams of $\ce{Fe(NO3)2}$. Its molar mass is $179.8548$ grams/mol. Thus, you have $3.058022\!\times\! 10^{-5}$ moles. Multiplying this by the molar mass of iron gives you $0.00170775$ grams of $\ce{Fe}$. Thus, you have $0.0037922$ grams of $\ce{NO3}$. The concentration of iron is
$$\mathrm c = \ce{\frac{Fe}{Water+Fe(NO3)2}} = \frac{0.00170775}{0.0055+2000}= 8.538726518 \times10^{-7}$$
Thus, $\ce{Fe}$ has a $\pu{ppm}$ of $0.85$


Is the first ionization energy in oxygen slightly more than nitrogen?

Oxygen has a lower first ionization energy as the electron that is removed is coming from a paired orbital.

Electrons within the same orbital experience maximum repulsion as the distribution of their wavefunctions is the same, so the probability density distribution is the same and the electrons can be thought of as occupying the same space. This maximizes their repulsion and increases the potential energy of the electrons in that orbital, making the electrons easier to remove. This is despite the increased effective nuclear charge experienced by the electron in oxygen and the decreased radius of the orbital.

See: "Physical Chemistry", Atkins, P.W. Section 13.4, p.p.370 (4th edition) - sorry, I have an old one!


The notion of orthogonality in the context of the question referrers to the very well-known general concept of linear algebra, the branch of mathematics that studies vector spaces. Instead of going deep into the mathematics (that requires at least 50 textbook pages) let's just clear some OP's doubts.

First, a small (but important) correction: two wave functions $\psi_1$ and $\psi_2$ are called orthogonal to each other if
$$
  \int \overline{\psi}_1 \psi_2 \, \mathrm{d} \tau = 0 \, ,
$$
where the first function is complex-conjugated as indicated by a bar on top of it. Wave functions are complex-valued functions and complex-conjugation of the first argument is important.1

So, yes, orthogonality is a not a property of a single wave function. It either refers to a pair of them being orthogonal to each other as described above, or, in general, to a set of them, being all mutually orthogonal to each other, i.e. to a set $\{ \psi_i \}_{i=1}^{n}$ such that for any $i \neq j$
$$
  \int \overline{\psi}_i \psi_j \, \mathrm{d} \tau = 0 \, .
$$
In the last case it is said that the whole set $\{ \psi_i \}_{i=1}^{n}$ is orthogonal.

1) More precisely, one of the two functions has to be complex-conjugated in this expression, where which one is the matter of convention: in physical literature it is often the first function, while in mathematically oriented literature it is usually the second one.


The very term "covalent compounds" is confusing, so you may just as well stop using it. The elementary substances you specified are covalent crystals (AKA "covalent solids"). A single crystal of any of these is in fact one huge molecule linked by covalent bonds. To melt that crystal, you have to break the covalent bonds, and they are pretty strong.

On the other hand, we have $\ce{N2}$, $\ce{O2}$, and other covalent molecules. They may be quite strong (in that each molecule by itself is pretty hard to tear apart), but the molecules are linked together by the so-called van der Waals forces, and these are quite weak, so these compounds are gases, and it takes some effort to see them turning into a liquid or a solid.


About Fe(III) and Fe(II) coordination complexes

Octahedral complexes are somewhat of a ‘standard’ for transition-metal complexes. Not only can they happen if there are too few electrons to satisfy the 18-electron rule (e.g. $\ce{[Ti(H2O)6]^3+}$, a $\mathrm d^1$ complex) but also if there are too many electrons to satisfy the 18-electron rule (Jahn-Teller distorted $\ce{[Cu(H2O)6]^2+}$, a $\mathrm d^9$ complex).

So how are they still stable? Well, coordination compounds quickly exceed the simplistic bonding pictures that students are taught at the beginning of their chemistry classes. And the ligand field (or crystal field) model—while it does a good job of explaining some observations at a low level—also doesn’t exactly capture what we consider the current knowledge of coordination compounds. A much better description of how they are bonded lies within molecular orbital theory and scheme 1 below.

Scheme 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

On the left of the scheme, the metal’s 3d, 4s and 4p orbitals are placed, while the right-hand half contains the ligand orbitals; simplified as one σ-type orbital per ligand in an octahedral complex. After taking a look at the scheme and finding it perplexing, we can attempt to find the stabilising and destabilising factors in forming a complex.

The most obvious stabilising effect is that all ligand orbitals form bonding interactions that lower their energy. For this, they make use of the typically unoccupied metal’s 4s and 4p orbitals alongside the often occupied 3d orbitals. As we are stabilising occupied orbitals at the expense of unoccupied orbitals, this already results in a net stabilisation.

The second bit of stabilisation results from the interaction of the ligands with parts of the 3d orbitals: precisely the $\mathrm d_{x^2-y^2}$ and $\mathrm d_{z^2}$ orbitals, often collectively termed $\mathrm{e_g^*}$ orbitals according to their symmetry group. These may be occupied, but these also have better overlap with the ligand orbitals so the bond formed is stronger than in the case of 4s and 4p.

Finally, the electrons in the other three metal d orbitals ($\mathrm d_{xy}, \mathrm d_{xz}$ and $\mathrm d_{yz}$; collectively termed $\mathrm{t_{2g}}$ according to their symmetry group) are, in fact, non-bonding in this simplified picture as their symmetry does not match any ligand orbital. Thus, any electrons in here do not incur any stabilisation but also—arguably more importantly—are not destabilised.

The 18-electron rule derives from this simplified picture and assumes that all bonding and non-bonding orbitals are occupied (that means: all ligand orbitals and the $\mathrm{t_{2g}}$) while all antibonding orbitals (most notably: $\mathrm{e_g^*}$) remain unoccupied. If you count electrons and add up, this gives 18.

However, as I have established the $\mathrm{t_{2g}}$ electrons don’t actually destabilise the complex (under the simplified scheme). Thus, it is relatively irrelevant if there are 0, 2 or 4 of them; it doesn’t change the general scheme. Likewise, even though any $\mathrm{e_g^*}$ orbitals are antibonding, their overall effect is not extreme, so there can be electrons occupying them while still allowing for a stable complex.

The picture on the metal is further complexified (pun intended) by considering high spin versus low spin. In an uncomplexed metal ion, all five d orbitals are energetically equivalent, so they should be filled according to Hund’s rule. In a complex, you have two options:

Hund’s rule is observed across all five d orbitals regardless of the energy difference
the energy difference is large enough that spin pairing in the $\mathrm{t_{2g}}$ orbitals occurs before the $\mathrm{e_g^*}$ orbials are filled according to Hund’s rule.

The first case is termed high spin and should be treated as the standard case for 3d metals, while the second case is termed low spin and is the standard case in 4d and 5d metals. Note however, that clever choice of ligands may cause a 3d complex to adopt a low-spin configuration and some complexes are even capable of flipping between high and low spin.

While $\ce{Fe(H2O)6]^2+}$ may look like a prime example of the 18 electron rule, it is actually a high-spin complex so the outline I drew above is no longer valid. On the other hand, high-spin $\ce{[Fe(H2O)6]^3+}$ has 5 d-electrons which allows for every d orbital to be populated by exactly one electron—a ‘slightly more favourable’ state. Therefore, high-spin iron(II) complexes are easily oxidised in aquaeous solution but high-spin iron(III) complexes are not as easily reduced. (Obviously, the picture changes when you consider low-spin complexes such as $\ce{[Fe(CN)6]^3-}$, which is a good oxidising agent.)

tl;dr:

The 18-electron rule is more of a guideline than an actual rule; it states that more stable adducts can be formed, if, but that’s all;
All ligand–metal interactions can and should be considered covalent bonds;
Technically, the orbitals and the ligands do need to line up but because electrons are volatile we can assume they have lined up automatically;
Essentially, the crystal field model isn’t entirely wrong when it predicts the $\mathrm{e_g^*}$ orbitals to be raised to a higher energy level. It just provides a bad explanation. The orbitals end up being antibonding; that should equally explain.


Removing all of the debris from the stones without damaging them can be difficult, the following is what I've done in the past to clean geodes with debris and deposits.  Your mileage may vary, especially since we don't know exactly what is on your stones.

Soak the stones in soapy water for a day or so, then scrub them thoroughly with a stiff brush to clean dirt and most debris from them.  Be cautious if the stone has delicate structure that you're trying to preserve.  If some debris remains that appears to be washable, you can try scrubbing with a toothbrush and toothpaste to get rid of it.
If the stone is acid safe, soak it in vinegar for a couple of days to dissolve carbonate deposits.  You can check to see if it will react with vinegar by adding a few drops of vinegar to the stone's surface and waiting.  If it hasn't started bubbling within 10 minutes or so, it should be safe.
If the stone has rust deposits and is acid safe, you may be able to remove them with something like CLR Rust Stain remover.  If you need to go this route, skip the vinegar step, as this will also clean carbonate deposits.
Finally, it's possible that your crystal has something like granite attached.  If that is the case, nothing practical, short of cutting it off, is likely to clean the crystal.

Once you've got the crystal to the cleanliness you're happy with, you can polish it with powdered polish and a cloth, given time.

If instead of nice, angular crystals you're just trying to produce something like a smooth, rounded stone, after cleaning with water you can use a series of sandpapers (start at 60-80 grit and work up to ~400 grit) and polish to gradually smooth, then polish the stone by hand.  This will clear away any deposits, but it will also change the surface of the stone itself.  That might be acceptable to you if you've got something like a quartz river rock, and it will make a very nice end product.  Take care to keep the stone wet while working, you don't want to breathe in rock dust if you don't have to.


What happens when an explosion can't lead to volume increase of the thermodynamic system?

I'm not an expert in thermodynamics and gas physics so there might be some errors in this calculation but here goes:

According to several sources (wikipedia included) TNT decomposes according to this equation:
$$\ce{2C7H5N3O6 -> 3N2 + 5H2O + 7CO + 7C}$$

If 100kg of TNT was placed in an indestructible $1m^3$ container containing an inert gas (we will say helium) at standard temperature and pressure here is a rough estimate of what might happen if it was detonated.

Using a Hess's Law cycle we can calculate the enthalpy of decomposition of TNT from formation enthalpy data: $$\Delta_{decomp}H = 63.2*2 - 285.8*5 - 110.5*7 = -2076.1~ kJmol^{-1}$$

Then we can calculate the energy released by the decomposition: $$M_r~of~TNT = 227.13~gmol^{-1}$$ $$n = \frac{100}{0.22713} = 440.3~mol$$ $$Energy~released = 440.3 * 2076.1 *1000 = 914.1~MJ$$

I'm not sure how to go about specific heat capacities for gas mixtures so I won't put a figure on the temperature rise that this will cause but I suspect it will be in the thousands of degrees due to the amount of energy and so the resultant pressure rise will also be very significant, probably enough to liquefy the gases. Also as @Klaus Warzecha hinted at in his comment these kind of temperatures and pressures will probably cause some interesting chemical reactions to take place.

http://courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf


Is Sweetwater, Missouri, named due to water contamination with lead compounds?

According to The State Historical Society of Missouri:

  Sweetwater is the name of a little trading-point established on Sweetwater Branch just after the Civil War. It was named for the branch.

and

  Sweetwater Branch was named by the pioneers for the unusually good water in the stream.

The place (and branch) are both in Newton County, Missouri. The same reference linked above indicates that what was once called Sweetwater is now referred to as Boulder City (because of the boulders nearby).


Composition of Portland Cement before Hydration

According to Ullmann’s Encyclopedia of Industrial Chemistry (sixth edition), Portland cement consists mainly of tricalcium silicate, dicalcium silicate, tricalcium aluminate, and calcium aluminoferrite.

$$
\textbf{Phase composition of Portland cement clinker} \\
\begin{array}{llll}
\hline
\text{Constituent} & \text{Formula} & \text{Content in wt %} \\
& & \text{Range} & \text{Average}\\
\hline
\text{Tricalcium silicate} & \ce{3CaO.SiO2} & 46{-}79 & 61 \\
\text{Dicalcium silicate} & \ce{2CaO.SiO2} & 5{-}30 & 15 \\
\text{Calcium aluminoferrite} & \ce{2CaO.(Al2O3,Fe2O3)} & 4{-}16 & \phantom{0}8 \\
\text{Tricalcium aluminate} & \ce{3CaO.Al2O3} & 6{-}18 & 12 \\
\text{Free calcium oxide} & \ce{CaO} & 0.1{-}4 \\
\text{Free magnesium oxide} & \ce{MgO} & 0.7{-}1.5 & \phantom{0}1.5 \\
\hline
\end{array}$$

Büchel, Moretto, Woditsch Industrial Inorganic Chemistry (second edition) lists the same six constituents as important clinker phases. This reference also includes a diagram for the three component system $\ce{CaO}$–$\ce{SiO2}$–$\ce{Al2O3}$ with composition ranges for various cement types, which shows that calcium monosilicate (CS) is not necessarily within the typical range for Portland cement.

Furthermore, this reference includes a table with a typical chemical analysis of a Portland cement, which shows that the $\ce{CaO}$–$\ce{SiO2}$ ratio is too high for calcium monosilicate and also indicates that small amounts of phosphate could be present:

$$
\textbf{Typical chemical composition of a Portland cement clinker} \\
\begin{array}{ll}
\hline
\ce{CaO} & 58{-}67\ \% \\
\ce{SiO2} & 16{-}26\ \% \\
\ce{Al2O3} & \phantom{0}4{-}8\ \% \\
\ce{Fe2O3} & \phantom{0}2{-}5\ \% \\
\ce{MgO} & \phantom{0}1{-}5\ \% \\
\ce{SO3} & \phantom{0}0.1{-}2.5\ \% \\
\ce{Mn2O3} & \phantom{0}0{-}3\ \% \\
\ce{P2O5} & \phantom{0}0{-}1.5\ \% \\
\ce{TiO2} & \phantom{0}0{-}0.5\ \% \\
\hline
\end{array}$$


Is the voltmeter in a galvanic cell in parallel?

Short answer, to the question title:

Yes, always.

Long answer:

  Often, a galvanic cell is illustrated as having two electrodes, two electrolytes, a salt bridge, and wires bridging the electrodes to a voltmeter.

The piece missing from your list of components, in terms of a galvanic cell being put to practical use, is the load.  You're correct, if a load were attached to the cell, it would be wired in parallel with the voltmeter.

A quick search turns up the following schematic along the lines of your description, from the UC Davis ChemWiki:

From a brief scan, the construction of the half-cells and salt bridge look okay to me. I have not double-checked the half-cell potentials for accuracy.

If I read the schematic correctly, there is no load attached, and so the only thing completing the circuit with the galvanic cell is the voltmeter.  In this case, the appropriate cell voltage would be read on the voltmeter. However, the current through the system would be negligibly small, due to the extremely high resistance of the voltmeter. The $\leftarrow e^-$ notation here indicates only the direction that current would flow through a lower-resistive load that were connected (again, yes, in parallel) with the voltmeter.


Pay attention to the prefix "hydro-", it can definitely change the meaning of the name.

Quinone is the common name given to cyclohexa-2,5-diene-1,4-dione:

However, hydroquinone is not a ketone. Hydroquinone is basically hydrogen added to ketone. As a result, it's not a ketone, but rather it's a phenol.

I must agree, the naming does cause confusion, as this isn't the systematic nomenclature. The common nomenclature doesn't follow any strict rules, and are often named after the compound they were formed from (such as this case). Don't rely on such names too much.


Stereoselectivity of hydroboration of dihydrofuran

Generally, the regioselectivity (and stereoselectivity, depending on the situation) of hydroboration reactions is determined by steric factors. This is a prime example of when it isn't.

Regioselectivity

Borane, $\ce{BH3}$, is electrophilic on boron because of the vacant p orbital (low-lying LUMO). It will therefore preferentially add to the alkene carbon that is more nucleophilic. The enol ether you have drawn is clearly nucleophilic on the carbon β to oxygen:

and consequently it is this carbon that the boron atom becomes attached to:

Reading this:

  Also if the OH group had to be attached to one of the carbons wouldn't it be more likely to attach to the carbon next to the oxygen since it will be more electrophilic?

I suspect you need to check the mechanism of hydroboration-oxidation again. There is no attack of $\ce{OH-}$, or indeed any oxygen-based nucleophile, on an electrophilic carbon. The C-O bond is formed by migration of carbon from boron to oxygen. Which carbon is more electrophilic is inconsequential.

Stereoselectivity

It is very common practice in organic chemistry to only draw one enantiomer when a racemic mixture is implied. The purpose of this is to save time: if you were discussing the diastereoselectivity of a particular reaction, you don't want to have to keep on drawing two enantiomers at every single step of the reaction.

Since attack from the top and bottom faces is equally likely to happen, an equal mixture of enantiomers is formed. As described above, the single enantiomer drawn therefore refers to a racemic mixture:

There is no diastereomerism present in this compound, but I would guess that the other reason why stereochemistry is indicated is to alert the reader to the fact that the product is chiral. Sometimes, this can be considered to be misleading. The most unambiguous way to indicate "undefined stereochemistry" is therefore using the wavy line as I did above.


Why is Polonium the only metal to crystallize forming Simple Cubic Cell?

The inorganic crystal structure database (ICSD), by specifying that
      the structure contains only a single element, you can reduce the
      number of hits to 55. Structure examples are Cr, Po, Mn, Sb, Se, P,
      Ca, Li, As. Some are specific phases occurring at different
      temperature or pressures.

So Po is not the ONLY metal to have a simple cubic structure, but
the other cases require special conditions.

  The origin of the stabilized simple-cubic (SC) structure in Po is
  explored by using the first-principles band calculations. We have
  found that the prime origin is the inherent strong spin-orbit (SO)
  interaction in Po, which suppresses the Peierls-type structural
  instability, as usually occurs in p-bonded systems. Based on the
  systematic analysis of electronic structures, charge densities, Fermi
  surfaces, and susceptibilities of Se, Te, and Po, we have proven that
  the stable crystal structure in VIA elements is determined by the
  competition between the SO splitting and the crystal-field splitting
  induced by the low-symmetry structural transition. Our study suggests
  that the large SO interaction would suppress the Peierls instability
  which is generally expected to occur in one-dimensional conductors.

  In fact, Po becomes airborne with ease: if a sample is heated in air
  to 55 °C (131 °F), 50% of it is vaporized in 45 hours to form diatomic
  Po2 molecules. (The melting point of polonium is 254 °C and its
  boiling point is 962 °C.)

Putting it all together, Po is in the column with Se and Te, and just to the right of Bi, none of which are typical of "normal" metals. The formation of $\ce{Po2}$ molecules at near ambient temperatures suggests that the outer electrons of Po form bonds of a highly covalent nature.

But it is indeed unusual and fascinating, and if Po were not 5000 times more radioactive than radium, I might be tempted to do a little chemistry.

References

Quora:
"What are examples of simple cubic crystal structure?" Gervais Chapuis, Nov 3, 2016
B. I. Min, J. H. Shim, Min Sik Park, Kyoo Kim, S. K. Kwon, and S. J. Youn, Origin of the stabilized simple-cubic structure in polonium: Spin-orbit interaction versus Peierls instability, Phys. Rev. B,  73, 132102 – Published 7 April 2006
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.73.132102
https://en.wikipedia.org/wiki/Polonium


What is the physical explanation driving current flow in a concentration cell?

Start with a simple thought experiment: pour 100 mL of 1 M nickel (II) sulfate solution into a beaker and very carefully layer 100 mL of 0.01 M nickel (II) sulfate solution on top of the more concentrated layer. Then, even without convection or deliberate mixing, diffusion will, sooner or later, result in the solution having concentration of 0.55 M. In what follows, it is assumed that evaporation is negligible, even on long time scales.
Now consider Fig. 1 below:

In this concentration cell, it is assumed, for simplicity, that the solution volumes are equal in the two electrode solution reservoirs. The salt bridge is 0.55 M nickel (II) sulfate solution, so its concentration is exactly half-way between that of the anode solution and cathode solution.
With the concentrations shown in the figure, the cell potential is +0.02958 V, and the anode is the nickel electrode in the more dilute solution at the left. On the cathode side, the log term is zero, in the Nernst ewuation at upper right in Fig. 1, so the potential is -0.236 V. But on the anode side, the potential is lower by 29.58 mV, due to the log term. So the anode is more negative, the cathode is more positive and electron flow is always from more negative to more positive. Therefore, Ni is spontaneously oxidized at the anode, yielding the necessary electrons, and nickel ions are spontaneously reduced at the cathode.
As a direct consequence of having an external connection, i.e., load or voltmeter, between the electrodes, net oxidation will spontaneously take place at the anode, resulting in an increase in the nickel ion concentration in that solution reservoir. Likewise, nickel ions in cathode solution reservoir will be spontaneously reduced at the nickel cathode and the nickel ion concentration will decrease. The external connection affords the opportunity for something useful to happen, i.e., using the cell as a power source. Otherwise, only diffusion happens (see below).
The cell will be "dead", i.e., have no more free energy to tap, when all three solutions are 0.55 M. There will be no concentration gradient left.
Now consider Fig. 2:

This is the same as Fig. 1 except that the cell is open circuited, i.e., there is no way for electrons to transfer from one electrode to the other. So all that is going to happen is spontaneous diffusion through the salt bridge. Eventually, there will be just one final concentration at 0.55 M.
During this process, no net oxidation occurs at the left electrode and no net reduction occurs at the right electrode. If a voltmeter is subsequently attached, the reading will be less than +0.02958 V because the concentrations are no longer the original ones: both concentrations are closer, thanks to diffusion, to the 0.55 M center value.
The measured voltage, even using a high impedance voltmeter, is not actually the open circuit voltage. It may be very close, but there is a conceptual issue. In the open circuit situation, no electrons can flow between the electrodes and no net redox processes occur at the electrodes. So the open circuit potential is an ideal potential and the measurement goal is to estimate accurately the open circuit potential while allowing negligible current to flow.
With a high input impedance voltmeter attached to the electrodes, a very small current, i.e., nA or pA, can flow. This only slightly loads down ("perturbs") the cell potential, so it provides an accurate estimate of the true open circuit potential. And it means that oxidation occurs at the anode, to a very slight extent, and reduction at the cathode, likewise to a very slight extent. Solution concentrations are negligibly perturbed.
So there are two parallel processes taking place. First, diffusion spontaneously acts to equalize the solution concentrations. Second, the concentration cell provides a way of hurrying the equalization process along while extracting useful electrical energy. So the former wastes the available free energy while the latter extracts much of it.
Finally, consider Fig. 3 below:

Now the salt bridge is gone and nothing happens: there is no DC cell potential and the solution concentrations are constant. In figures 2 and 3, there is no net redox happening: for every Ni atom that might get oxidized, there is a nickel ion that gets reduced. Everything is balanced at both electrodes. (So if the electrodes start with highly polished "mirror" finishes, they may get surface roughened by this zero net redox process.)
But once a load or voltmeter is attached to the two electrodes in Fig. 2, this opportunity causes the equilibria to become unbalanced: the solution concentrations must change by increasing (in the anode reservoir) and decreasing (in the cathode reservoir), and that immediately implies that electrons flow from the anode (at the left), through the load or voltmeter, to the cathode (at the right).
Perhaps the following will also help clarify matters. Consider Fig. 4 below:

This shows two beakers, each made of solid nickel, and containing the indicated nickel (II) sulfate solutions. Obviously, this does nothing: it is just two fancy beakers of solutions. Next consider Fig. 5:

Now the two nickel beakers are touching, i.e., in electrical contact. Does this change anything? No. Nothing happens. Next, consider Fig. 6 below:

Now there is simply a wide glass beaker with a solid nickel partition that entirely prevents the two solutions from making contact. Does this change anything? No. Nothing happens. Finally, consider Fig. 7 below:

This differs from the previous figure in having a salt bridge. It is exactly equivalent to using a nickel wire to short the two nickel electrodes in Fig. 2. So the cell is shorted out, there is no cell potential, and the cell will try to supply its short circuit (i.e., maximum) current, limited by the kinetics at the electrodes. So the nickel ion concentration will increase in the anode reservoir and decrease in the cathode reservoir, as expected.


How do I find the thermal effect of this reaction?

Look at the difference between the enthalpies of formation. Not sure why you'd need oxygen, unless there's something I'm missing.


Is the Hollywood portrayal of the effects of ingesting HCN correct?

Is there anything realistic about that scene, or is it just another movie gaffe?

I have never heard about anything similar caused by $\ce{HCN}$ or $\ce{KCN}$.

However,

bleeding of the gums,
loss of teeth, and
osteonecrosis of the jaws (= the bone tissue literally rots away)

are known symptoms of certain intoxications:

Bisphosphonate-associated osteonecrosis of the jaw was observed in the treatment of patients with cancer
Radium jaw was observed in painters (Radium girls) that licked their paintbrushes while applying radioluminescent paint to dials
Phossy jaw was observed in (mostly female) workers in the matchstick industry of the late nineteenth century that worked with white phosphorus.

Back in the days, the lives of the patients often could only be saved by removing the jawbones surgically. (If you feel for it, look for some photographs online.) Since one of the tags is chemistry-in-fiction, you might be interested that this was partly covered in the episode Become Man of the BBC series Ripper Street.


In the Lewis structure for barium fluoride, why does barium lose its valence electrons?

First, a little bit of background. I assume you're already familiar with atomic orbitals — functions that describe the wave-like behavior of an electron in an atom. There's also molecular orbitals (MO), that describe its behavior in a molecule. You can ignore the odd-looking labels in images and such, since they're not entirely relevant to this question.
Atomic orbitals interact to form molecular orbitals (so called LCAO). The number of resulting molecular orbitals is the same as the number of the atomic orbitals that interact in the first place. To visualize this in simple interactions, you can use a molecular orbital diagram, in which orbitals of lower energy are drawn as lines placed lower than the others. There are three types of resulting MOs:

Bonding MO, which is usually much lower in energy than the starting atomic orbitals
Anti-bonding MO, which is usually much higher in energy
Nonbonding MO, which would have the same energy as the starting atomic orbitals

You can imagine that electrons are rearranged when a molecule is formed, and they fill the orbitals with the lowest energies first, so the order is bonding first and anti-bonding last. In a symmetric molecule, like the boilerplate example $\ce{H2}$, this is what the diagram looks like:

Image taken from this helpful page on Libretexts
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Back to your question, the energy difference between the atomic orbitals and the bonding orbital (presented as the space between the lower bonding MO and the higher atomic orbitals of each H atom) is the same, so they're said to 'contribute the same to the bonding MO'. Bonds between homonuclear diatomic molecules are therefore considered to be purely covalent (whoops, almost!). Covalent compounds are the ones you're taught to draw Lewis structures for, apply formal charges, etc., but it's only one extreme of the spectrum.
The next example would be the MO diagram of HCl.

Cl's 3p atomic orbitals are lower in energy (check the provided link for the reasoning); this time round, the atomic orbitals do not contribute equally to the molecular orbital, and Cl's orbital's contribution to the bonding orbital is greater. Electronegative atoms are generally said to have atomic orbitals of lower energy. This is probably more simply and familiarly stated as "the more electronegative atom pulls/attracts the bonding electrons". Note that, now, there's also an energy gap between the initial atomic orbitals. The $\ce{H-Cl}$ bond is, indeed, somewhere between covalent and the other extreme.
Now to the punchline, what happens when the gap is too large? This would be what an MO diagram of LiF would look like:

Thanks Geoff!
The gap is so large that, essentially, the energy difference between molecular orbitals filled with electrons, and fluorine's atomic orbitals would be very small compared to the energy difference between F's 2p orbitals and Li's 2s. Put in simpler terms, the very electronegative fluorine has such a stronger pull than Li that the electrons are almost fully 'localized' on the fluorine, and Li has approximately lost its electron. This is what we call an ionic bond, the other end of the spectrum.
Most bonds between very electronegative atoms and very electropositive ones are considered ionic, and there's the guideline that metal-nonmetal bonds are often ionic. Don't let it trip you up though, since it's a vague general rule, with plenty of exceptions. As you might have suspected, there's no truly, purely ionic bond either, and it's a spectrum.