In the last step of Kjeldahl Method, is the reaction of ammonium to NaOH not taken into account?

The Kjeldahl method has two versions, using $\ce{HCl}$ as trapping solution or a weak acid like boric acid. Depending which one is used, the nitrogen content is either determined by a back titration with $\ce{NaOH}$ (your case) or a direct titration with $\ce{HCl}$.

The goal of the titration is to figure out how much ammonia (gas) has been trapped in the acid. The equation for the trapping step is:

$$\ce{NH3(g) + HCl(aq) -> NH4+(aq) + Cl-(aq)}$$

So the trapping solution will get a bit less acidic because some of the $\ce{HCl}$ is neutralized by the weak base ammonia. To quantify the ammonia, you now titrate the remaining $\ce{HCl}$ with the strong base $\ce{NaOH}$:

$$\ce{HCl(aq) + NaOH(aq) -> H2O(l) + Na(aq) + Cl-(aq)}$$

The solution is titrated to about $\mathrm{pH} = 5$ as endpoint (at the equivalence point, the strong acids and bases are neutralize, and $\ce{NH4Cl}$ solution remains). At $\mathrm{pH} = 5$, very little ammonium is deprotonated to form ammonia ($\mathrm{p}K_\mathrm{a}$ is about 9.3). In other words, ammonium/ammonia does not act as a buffer, so it does not affect the titration. You might even call it a spectator ion at this $\mathrm{pH}$.

After reaching the endpoint, the amount of $\ce{HCl}$ originally present in the trapping solution is equal to the sum of the amounts of $\ce{NaOH}$ and ammonia added. The overall reaction is (non-standard notation):

$$\ce{HCl(aq) + x NH3(g) + (1-x) NaOH(aq) ->}$$
$$\ce{Cl-(aq) + x NH4+(aq) + (1-x) H2O(l) + (1-x) Na(aq)}$$

From that, you can calculate the amount of nitrogen in the sample.

  However, other than $\ce{HCl}$ there also exist an acid in the solution that can be titrated with $\ce{NaOH}$ which is the protonated form of $\ce{NH3}$.

As explained above, if the $\mathrm{pH}$ does not approach the $\mathrm{p}K_\mathrm{a}$ of ammonium, ammonium will not react with $\ce{NaOH}$ to an extent that matters. If you choose to end the titration when the solution reaches a $\mathrm{pH}$ of 11 (to give an example), however, the ammonium will be deprotonated to ammonia, invalidating the result. Worse, it might come out of solution, so it becomes a problem that can't be fixed.

Sources:

http://www.brooklyn.cuny.edu/bc/ahp/SDKC/Chem/SD_KjeldahlMethod.html
J Martín, Fernández Sarria L and Agustín G. Asuero (2017) DOI: 10.5772/intechopen.68826

Wikipedia on the other hand I think got it wrong (https://en.wikipedia.org/wiki/Kjeldahl_method, retrieved before I edited it):

  Ammonium ion concentration in the standard acid solution, and thus the amount of nitrogen in the sample, is measured via titration. If boric acid (or some other weak acid) was used, direct acid-base titration is done with a strong acid of known concentration. $\ce{HCl}$ or $\ce{H2SO4}$ can be used. Indirect back titration is used instead if strong acids were used to make the standard acid solution: strong base of known concentration (like $\ce{NaOH}$) is first added in excess, and then the excess is titrated with a strong acid of known concentration. Titration of ammonia absorbed in boric acid solution thus has an advantage that only one standard solution is needed. One of the suitable indicators for these titration reactions is Tashiro's indicator.

They say to add an excess of $\ce{NaOH}$ and then titrate with $\ce{HCl}$. This would drive ammonia out of solution. Also, this is not a back titration but a double-back titration (which is not a thing, I think).


Curly arrow mechanism for reaction between sodium borohydride and water?

Their mechanisms are easily searchable online.
This is from 2009 from art-xy.com:

Notice that the reaction would be different in the presence of metal catalysts:
$$\ce{NaBH4 + 2H2O ->[metal catalyst] NaBO2 + 4H2}$$
This is from 2007 from chembook.co.uk:

Which reaction is more violent is already explained by Jan:

In one case [$\ce{NaH}$], you have free hydrides, which go ‘Oh my god, I see a proton, I’m going to jump at it!’ and in the other case [$\ce{NaBH4}$] you have hydrides thinking the same thing, but being pulled back by a boron saying ‘No, stay here, I need your electrons! I’m poor!’

Therefore the former is more violent.


Why can hydrogen ions be added to balance a reaction?

You should really only add $\ce{H+}$ to balance if a reaction is explicitly said to take place in acidic solution. For example, in your link it says.

  The reaction is done with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulfuric acid.

So the sulfuric acid allows you to add these $\ce{H+}$. Alternatively, if it is in basic solution, $\ce{OH-}$ atoms can be added instead.


No, there is no such thing as chemical reaction are not that easily predictable. You can get result right in some cases, but it is impossible to count all factors.


Specific conductance of 0.1 M acetic acid

With the data provided with your problem I suspect you are only supposed to carefully use provided formula with the corresponding units of measurements. You need to convert the concentration from $[\pu{mol * L^{-1}}]$ to $[\pu{mol * cm^{-3}}]$ considering $\pu{1 L = 10^{3} cm3}$:

$$\kappa=\Lambda_m \times c = \pu{4.6 S * cm2 * mol^{-1}} \times \pu{0.1 * 10^{-3} mol * cm^{-3}} = \pu{4.6e-4 S*cm^{-1}}$$


Are all (pure) alkanes, alkenes and alkynes - tasteless, odourless and colourless?

Why would a mixture of two odourless alkanes have an odour?

As already pointed out by Greg E and Uncle Al, numerous hydrocarbons without any (heteroatom-based) functional groups do show characteristic smells.

Naphthalene is always remembered for the typical smell of grandma's mothballs.

More pleasant smells are found in the group of monoterpenes, i.e. in those molecules with the sum formula $\ce{C10H16}$ that are derived from the combination of two isoprene units.

These volatile compounds were/are used as fragrances for perfumes. Typical examples are myrcene (1), $\alpha$-terpinene (2), limonene (3), $\alpha$-pinene (4), $\beta$-pinene (5), $\alpha$-thujene (6), sabinene (7), and camphene (8).

As far as the colour of hydrocarbons is concerned:

A sufficiently conjugated system is far from being colourless, take lycopene (9), the deep red coloured dye in tomatoes as an example.


How does the bite angle of a ligand affect the efficiency of its catalyst?

The bite angle of a bidentate ligand often has a pronounced effect on reactivity, but it can be difficult to predict a priori whether a given reaction will be enhanced by increasing or decreasing bite angle - you can see this if you ever read a paper where new ligands / catalysts have been developed, which are often accompanied with large tables of screening in order to find an optimised set of reactions.

In general however, bite angle does correlate with the rates of oxidative addition and reductive elimination.

Reductive elimination

The rate of reductive elimination is often correlated to the bite angle of a bidentate ligand. Upon reductive elimination, the two coordinate Pd0 complex would preferentially adopt a linear geometry however with the bidentate ligand coordinating (and the chelate effect preventing it from 'falling off') this is not possible.

Larger bite angles therefore favour reductive elimination as the L-Pd-L (P-Pd-P) bond angle is closer to the desired 180o. The classic example of this was an experiment carried out by Brown in which the rate at which PhMe was eliminated from a variety of complexes was measured.

Pd(dppf)PhMe with a bite angle of 99.1o underwent rapid reductive elimination, even at low temperatures, while Pd(dppe)PhMe with a bite angle of 90.6o was stable at low temperatures, with elimination needing elevated temperatures.

Oxidative addition

The effect of bite angle on oxidative addition is less straightforward, as it depends largely on the mechanism by which the oxidative addition takes place. In general however, complexes with larger bite angles undergo faster rates of oxidation despite the fact that a smaller bite angle increases the rate of the C-X bond cleavage.


Still struggling with Huckel rule and aromaticity

Advice
Take a closer look at the lone pairs and double bonds that participate only and only in the delocalization of electrons.
1-methoxy-1,3,5,7-octatetraene
As you can see, 1-methoxy-1,3,5,7-octatetraene has four double bonds in an octagonal cycle. The lone pairs of the adjacent oxygen atoms, even though they directly influence the delocalization by means of the resonance effect, aren't actually part of the cycle and cannot be included amongst the pi electrons of the double bonds. Therefore, your molecule is antiaromatic. Its structure is not planar, in fact it looks something like this (the picture below includes only the cyclic backbone):

For aromatic molecules, planarity is absolutely necessary. But Huckel's rule which I'm sure you're familiar with, provides the reasoning behind the planarity of the molecule.
The imidazole
As for the imidazole, your assumption is absolutely correct. Here we get to know a scenario similar to pyridine:

The lone pair does not participate in the aromatic cycle. This can be easily checked by taking a look at the hybridization of the nitrogen atom. If the lone pair were to be part of the aromatic system, it should have been perpendicular to the other orbitals of the cycle. However, this is not the case. The 2p orbitals of the nitrogen are actually involved in the aromatic cycle, while the aforementioned paired electrons get to be embedded in an sp2 orbital, placed perpendicular to the plane of the cycle. This actually has significant consequences on the basicity of pyridine.
On the other hand, in order not to interrupt the aromatic cycle, the lone pairs of the other nitrogen atos must come into play when counting pi electrons. Thus, the molecule is aromatic.


Why is the force constant for double and triple bonds approximately double and triple of that of a single bonds?

Force constant is not the same as bond strength. Remember that an approximate measure of bond energy from the bond as a simple harmonic oscillator would be $$E=\frac{1}{2}kx^{2}$$ where $x$ is the bond length. In this case, you would also need to account for differing lengths of single, double, and triple bonds to recover the bond energy trend.

Note that this is not how a quantum mechanical harmonic oscillator is normally described. There, you would have quantized energy levels of $(n+\frac{1}{2})\hbar\omega$, where $\omega = \sqrt{\frac{k}{m}}$. But since the higher order wave functions "concentrate" at the classical turning points, you might still be able to use the classical harmonic oscillator as a model for what is going on, with strong caveats.

For reference, consult: https://en.wikipedia.org/wiki/Quantum_harmonic_oscillator


Thermodynamic vs kinetic reaction control with radical substitution

Thermodynamic control typically implies reversibility of all steps. Say you had three reactants $\ce{A, B}$ and $\ce{C}$ which could react either to product $\ce{A-B-C}$ or to product $\ce{A-C-B}$ (equation $(1)$). Then thermodynamic control would assume that the ratio of products depends only on the difference in Gibbs free energy between them, signified by the equilibrium constant (equation $(2)$).

$$\ce{A-C-B <=>[$\Delta G_1$] A + B + C <=>[$\Delta G_2$] A-B-C}\tag{1}$$

$$\frac{\ce{[A-C-B]}}{\ce{[A-B-C]}} = K = - \frac{\Delta\left(\Delta G\right )}{RT}\tag{2}$$

On the other hand, kinetic control implies that the reactions are basically irreversible and that it depends on the transitions states and their energies which product will be formed preferentially. This is often expressed as a difference in rate constants $k$ as can be seen in equation $(4)$ for the reaction $(3)$.

$$\ce{A-C-B <-[$k_1$] A + B + C ->[$k_2$] A-B-C}\tag{3}$$

$$\frac{\ce{[A-C-B]}}{\ce{[A-B-C]}} = \frac{k_1}{k_2}\tag{4}$$

One (simple) method of achieving thermodynamic control is to heat the reaction to oblivion, until one can assume that all transition states are sufficiently populated and the energy differences of products can take effect. Kinetic control, on the other hand, is often achieved by cooling down and higher dilution, since both allow the difference of transition states to take greater effect.

To apply this to the radical halogenation of hydrocarbons, Hammond’s postulate must be invoked, which basically states that the lower-energy intermediate (or product) will be reached by a lower-energy transition state. Since radicals are electron-deficient species, a tertiary radical is more stable than a secondary one, which in turn is more stable than a primary one. Thus, kinetic control implies that the lower-lying transition state determines the dominating product which would be the one leading to a tertiary radical.

A secondary factor under kinetic control is the number of hydrogens that can react. Of course, given a molecule such as isobutane, there are much more primary hydrogens ($9$) than there are tertiary ones ($1$). Yet, the reaction rate of the tertiary hydrogen is still much faster. Kinetic control still favours the tertiary halide.

Under thermodynamic control, the heats of formation of the different products must be examied. But conincidentally, these lead to the same conclusion. Since the $\ce{C-Br}$ bond is polarised towards bromine, the molecule is more stable if the ipso-carbon is tertiary. However, since we are no longer comparing rate constants but Gibbs free energy values, the ratio will be a different one.

Thus, both control types would favour the same product — a tertiary halide — albeit to different extents. This is the standard case! Only rarely, such as in certain Diels-Alder reactions does the main product differ when changing from thermodynamic to kinetic control.


Why do we use external pressure to find force exerted by gas on a piston in a closed vessel during its expansion?

For a frictionless, massless piston, the force per unit area that the gas exerts on the inner face of the piston is always equal to the force per unit area applied on the outer face of the piston (since the piston is massless, ma = 0).  However, for an irreversible process, the force per unit area exerted by the gas at the interface is composed both of the "thermodynamic pressure"  that one could determine from the equation of state for the gas (e.g., the ideal gas law) applied locally at the interface,  plus "viscous stresses" in the gas that are determined by the rate at which the gas is deforming.  Furthermore, if the expansion or compression is rapid enough, the conditions within the cylinder are not even uniform spatially.  So figuring out the force per unit area that the gas exerts at the interface for irreversible expansion is problematic.  However, fortunately, if the force per unit area on the outside is specified by manually imposing a force per unit area (e.g., if the atmosphere is imposing the pressure outside), the work done by the gas on the surroundings can still be determined.  This is because the force per unit area exerted by the gas at the interface will adjust to always match the force per unit are on the outside (again, since the piston is massless and ma = 0).

In the case of a reversible process, the situation is much simpler, since here, viscous stresses are negligible, and the force that the gas exerts per unit area on the inner face of the piston can be determined by applying the ideal gas law (or real gas equation of state) to the spatially homogeneous gas in the cylinder.  In this case, however, the force imposed on the outside must be controlled very carefully, so that it differs only slightly from the average inside gas pressure during the entire expansion or compression (and the gas pressure within the cylinder remains essentially uniform spatially).


Reaction of dihalides with excess of NaNH2

this once was my doubt...terminal alkynes are preferred because the acidic hydrogen at the terminal carbon is acidic(sp hybrid of Carbon) and which when taken up by the base gives a species with a negative charge on sp hybridized carbon which is indeed stable...so this provides the incentive for forming a terminal alkyne as the final product.

Trusted source: Jerry March Organic Chemistry 6th edition.

  When the base is NaNH2 1-alkynes predominate (where possible), because this base is strong enough to form the salt of the alkyne, shifting any equilibrium between 1- and 2-alkynes. When the base is $\ce{OH-}$ or $\ce{OR-}$, the equilibrium tends to be shifted to the internal alkyne, which is thermodynamically more stable.


Could we possibly discover new extraterrestrial elements and minerals?

One thing which can be said with absolute certainty is that no species  in the Universe (or even all their combined efforts) will even come close to exhausting the daunting enormity of chemical space. When you drag combinatorics into a problem, you can easily stumble on massive numbers, and this is the case here; it has been roughly calculated that there are on the order of $10^{60}$ "chemically reasonable" distinct molecules below a molar mass of $\mathrm{500\ g\ mol^{-1}}$! This means that almost all chemical substances which can exist, never will exist, be it from natural or artificial production.

No matter how far chemistry develops, we will always be interested in investigating the chemical composition of new samples, and will always find many new compounds doing so. Most of the time we'll find the same of what we do on Earth; there's no escaping the fact that $\ce{N_2}$ and $\ce{Al2O3}$ are very stable substances, for example, and can exist in a wide range of conditions. But there will always be something new for us waiting out there. Organic substances are tremendously varied thanks to the concatenability of carbon atoms. One might think inorganic substances are far less variable, but just the class of silicate minerals is enormous, and many polymorphs of the same substances can exist.

Having said that, at least one area of chemistry seems far closer to being exhausted; the elements themselves. There is no possibility of us having "missed" any element among the first 112, and we're still filling in the gaps at higher atomic numbers, but each heavier element discovered involves ever more complicated procedures and lower yields. Much of the experimentally determined properties of elements above $\mathrm{Z=104}$ come from the analysis of a tiny amount of atoms, as little as tens of them. Each new element is significantly more difficult to synthesize, and is less stable, which is hampering progress. We don't yet know very well how to get past $\mathrm{Z=122}$ or so even in principle, and are unsure how accessible the ultraheavy elements will prove to be. There is still great academic interest in several aspects of the elements and their isotopes, but current research deals with the production of very shortlived species, which have limited applicability.

In all, we will always find something new to explore, but whether we will always find something strange depends on what you mean by the word. There are still many, many substances to find which will display impressive biochemical properties, for example, and many lethal or crippling diseases today might still become curable in the future. But there is zero chance of finding a substance which makes magic real, which unlocks faster-than-light travel, which has endochronic properties or which can be used to make overunity engines and perpetual motion devices. Ultimately, Chemistry is a form of applied quantum electrodynamics, a subsection of Physics, and quantum electrodynamics is a very developed theory with incredible precision. Thus, there is little room for surprises, and they likely will not come from a chemical system unless it is very specially tailored, rather than randomly found in nature.


What would be the reaction equation between ammonium nitrate and a powdered metal?

I'm betting they mean that you can inadvertently make an explosive (technically a blasting agent) or highly flammable mixture by mixing powdered metal and ammonium nitrate.  I used to be a hard rock miner many centuries ago and was well read on explosives.  Aluminum powder and ammonium nitrate is a classic solid rocket fuel and industrial blasting agent/military explosive, for instance ammonal in bombs.  Ammonium nitrate/aluminum powder mixes (magnesium works, too) have better explosive properties than the usual mining ANFO (ammonium nitrate/fuel oil) mixtures, but the latter are preferred usually since ANFO is less expensive.

While blasting agents technically need a booster explosive to detonate, they will burn ferociously.  I'm guessing when a writer of a SDS sheet talks about powdered metal and AN, they are thinking about the potential of creating an extremely flammable mixture.

I looked at some MSDS for ammonium nitrate, and I see powdered metal listed along with reducing agents as incompatible mixtures.  Could you post a link to the MSDS you are reading?  I'm interpreting 'incompatible' as referring to explosive potential and not to a slow spontaneous reaction.

EDIT:  Well.  The usual metal powders that react with AN used in mining and explosives are aluminum and magnesium, so that is all I considered.  However, a bit of research shows that powdered zinc does react with ammonium nitrate, as in this video (note that it didn't actually ignite until set afire, but it definitely reacted).


What is the methodology to discover thousands of compounds from a complex mixture?

Most compounds can only be analytically quantified individually, so the first step in analyzing a complex mixture is typically to separated the different compounds from each other. For something like coffee, this would typically be done via different types of chromatography (i.e. gas (GC), liquid (HPLC), ion (IC)).

As the separated compounds exit the chromatography column, they can then be analyzed. This is usually done in real time using a number of techniques; the most relevant in this case is mass spectrometry (MS) following GC or HPLC.

In short, the primary techniques used to analyzed the compounds found in an organic sample like coffee would be GC-MS and HPLC-MS.


Dimers and polymers of electron deficient compounds

First of all, you have it all backwards.

You sound like the $\ce{AlCl3}$ says, "Oh, I am in solid state now! Let me turn to crystal (polymer) structure! [a while later] Oh, I am in vapour state now! Let me turn to dimer structure!"

Where in reality it says, "I don't have enough kinetic energy to break the bonds faster than they form, so I can only be in the crystal structure, in which I cannot move around, so I am in solid state now. [a while later] I have enough kinetic energy now! I can now break bonds faster than they form, but still not enough to break the bonds between the dimer, so let me exist as dimer. Also, I can move around now, since the intermolecular forces are not that strong. I am now a gas."

Almost everything polymerizes when the kinetic energy is low, forming a solid. The difference is only whether the bond responsible for the polymerization is van der Waals' forces or covalent bond or ionic bond.

For $\ce{BeH2}$, the bond is dative covalent as that makes the most sense.

For $\ce{AlCl3}$, the bond is ionic as that makes the most sense.

When the energy is high enough, $\ce{BeH2}$ decomposes, so all discussions henceforth are meaningless.

For $\ce{AlCl3}$, dimer has less energy than monomer, so dimer is formed. Remember that things like to be in states of low energy. Dimer enables it to move around, hence it is a gas.

When the energy is even higher, even the dimer bonds break and you have $\ce{AlCl3}$ monomers.


Is there any theory more advanced than MOT?

MO theory is more of a descriptive guide than a rigorous mathematical construct. The main idea of molecular orbital theory is that the electrons in atoms can be described by linear combinations of atomic orbitals (LCAOs), which themselves are the exact solutions to the Schrodinger equation for single electron atomic systems.
This idea can be applied in a more mathematically quantitative way by use of the Hartree-Fock (HF) method. HF is essentially MO theory put into practice. HF takes as input the geometry of a system (i.e. the spatial coordinates of all atoms in your molecule) and a set of "basis functions" which are usually sets of 3D gaussian functions centered on the atomic nuclei of the system. These basis functions are your "atomic orbitals" in MO theory. The HF method takes linear combinations of the atomic orbitals to generate a set of molecular orbitals which are usually more delocalized. These are the molecular orbitals of MO theory, and we can calculate them using the HF method running on a computer.
There are several approximations involved in HF theory, which the descriptive MO theory inherits.

The Schrodinger equation (SE) itself is an approximation. The SE is not exact as it fails to account for so-called relativistic effects. These are effects originating from Einstein's theories of special and general relativity. Some examples of relativistic effects are spin-orbit coupling and relativistic contraction of orbitals, especially in heavy atoms. The straightforward HF method doesn't take relativistic effects into account. Starting instead from the Klien-Gordon or Dirac equations will recover (some of) the missing relativistic effects.

The HF method assumes a complete separation between the movement of electrons and nuclei, known as the Born-Oppemheimer approximation. Basically, electrons are assumed to equilibrate instantaneously to the positions of atomic nuclei. This usually isn't a problem because electrons are much lighter than atomic nuclei so they do move much faster. Not a bad approximation in most cases.

Electron correlation. Even if we ignore the previous two approximations, the HF answer (and therefore the MO theory prescription) misses out on a very chemically important effect known as electron correlation. Electron correlation arises because a single set of molecular orbitals combined into a Slater determinant (in order to satisfy the Pauli exclusion principle) is not enough to fully describe the true exact solution to the Schrodinger equation for a molecular system. One must take linear combinations of many Slater determinants to improve on the HF answer. Electron correlation is responsible for effects such as the Van der Waals dispersion force. Another way of thinking about Electron correlation is that it is the energy that arises due to the dynamic Coulomb repulsion between moving electrons (although this picture is incomplete).

In summary, an exact molecular theory would need to account for Einstein's theory of relativity, combine the movement of atomic nuclei with electrons, and account for the dynamic correlations between different electrons in a system, in addition to being a proper quantum mechanical theory.
There are some quantum chemistry theories out there that incorporate relativistic effects into them. For example, the Douglas-Kroll-Hess approach is similar to HF theory, but begins from the Dirac equation and so incorporates relativistic effects explicitly. I believe there are also approaches that don't use the Born-Oppemheimer approximation, though I'm unfamiliar with them.
There is a massive amount of research and development into the so-called post Hartree-Fock methods. All of them attempt to account for electron correlation in various different ways. The main issue with the post HF methods is that the results (energies and wavefunctions) are often incredibly computationally demanding to actually compute, except on very small molecules. The most exact of the post-HF methods is the Full Configuration Interaction (FCI). With a sufficiently large input basis set, FCI produces an exact answer to the Schrodinger equation under the Born Oppenheimer approximation. Other post-HF methods include: Møller-Plesset perturbation theory, Coupled Cluster theory, and Multi-Configurational Self Consistent Field theory. All of these post HF methods are more accurate than the MO theory picture, but they are also way more difficult to visualize and understand for new students.


What is the nature of interstitial hydrides?

Nonstoichiometric compounds are simply compounds that cannot be represented with integer number of atoms. What the book probably is trying to say is that the hydrogen atoms do not simply fill the empty space between the atoms in the crystal. In other words, the hydrides will cause a change in the structure of the lattice. In some compounds, the atoms that are being inserted in the free space in the crystal will not cause any noticeable change in structure. The crystal may become slightly distorted. But it will retain its original shape before insertion of the new atoms. What the book is trying to say is that this not true for hydrides. Scientists used to think that this was the case, but it is not true.


How to find and use the Clausius-Clapeyron equation

See Integration by substitution, see also this question and another question at math.SE. This is the rule we have to apply here:

$$\int_{x_0}^{x_1} f(u(x))u'(x)\,\mathrm dx = \int_{u(x_0)}^{u(x_1)} f(u)\,\mathrm du $$

So we go from this equation:
$$\frac1p \frac{\mathrm dp}{\mathrm dT}=\frac{\mathrm d\ln p}{\mathrm dT}=\frac{\Delta H}{RT^2},$$
and integrate over temperature:
$$\int_{T_1}^{T_2}\frac1p \frac{\mathrm dp}{\mathrm dT}\,\mathrm dT = \int_{T_1}^{T_2} \frac{\Delta H}{RT^2}\,\mathrm dT,$$
Now we use the first equation, with $f=\frac{1}{p}$, $p=u$, and $x=T$:
$$\int_{T_1}^{T_2}\frac1p \frac{\mathrm dp}{\mathrm dT}\,\mathrm dT = \int_{p_1(T_1)}^{p_2(T_2)}\frac1p\,\mathrm dp=\int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\,\mathrm dT$$

Bonus question: Its just the application of the chain rule, namely
$$\frac {\mathrm d}{\mathrm dx}z = \frac {\mathrm dz}{\mathrm dy} \frac {\mathrm dy}{\mathrm dx} $$
which, if applied leads to
$$\frac{\mathrm d}{\mathrm dT}\ln p = \frac{\mathrm d\ln(p)}{\mathrm dp} \frac{\mathrm dp}{\mathrm dT} = \frac{1}{p}\frac{\mathrm dp}{\mathrm dT} .$$


Rust removal ended with black oxide on the metal?

In the absence of oxygen iron reacts with water forming $\ce{Fe(OH)_2}$, which can react with rust forming $\ce{Fe_3O_4}$ via the Schikorr reaction.  See https://en.m.wikipedia.org/wiki/Schikorr_reaction.

You should use a stronger acid, which will dissolve the rust and bring the iron ions into solution.  You will, of course, lose the original metal that rusted instead of recovering it, but you can't make such a recovery by "home remedies" anyway.  With such an aggressive approach rethink soaking overnight.


Determining Kohn-Sham and Hartree Fock virtual orbitals: The underlying field

That is true: in the Kohn–Sham model electrons both in occupied and in virtual orbitals are "moving" in the field $n-1$ electrons, while in the Hartee-Fock model electrons in occupied orbitals are "moving" in the field $n-1$ electrons, but electrons in virtual orbitals are "moving" in the field all $n$ electrons. And this is simply "by construction": you just have to look at the very definitions of these models. So, in short:

In the Kohn–Sham model, it is the exchange-correlation hole that by definition "excludes" a single electron from equations describing each and every orbital.

If you look at the Kohn–Sham equations
$$
\left(-{\frac  {\hbar ^{2}}{2m}}\nabla ^{2}+v_{{{\rm {eff}}}}({\mathbf  r})\right)\phi _{{i}}({\mathbf  r})=\varepsilon _{{i}}\phi _{{i}}({\mathbf  r}) \, ,
$$
where
$$
v_{\rm {eff}}(\mathbf {r} )=v_{\rm {ext}}(\mathbf {r} )+e^{2}\int {\rho (\mathbf {r} ') \over |\mathbf {r} -\mathbf {r} '|}d\mathbf {r} '+v_{{{\rm {xc}}}} \, ,
$$
it is last term of the Kohn–Sham effective potential, the exchange-correlation potential,
$$
v_{{{\rm {xc}}}}({\mathbf  r})\equiv {\delta E_{{{\rm {xc}}}}[\rho ] \over \delta \rho ({\mathbf  r})} \, ,
$$
that causes the above mentioned feature of Kohn–Sham orbitals.

For the case of the Hartree-Fock equations
$$
  \hat{F} \phi_i({\mathbf r}) = \epsilon_i \phi_i({\mathbf  r}) \, ,
$$
the summation in the Fock operator by the very definition of the model runs over all occupied orbitals only,
$$
  \hat{F}
  =
  \hat{H}^{\text{core}}
  +
  \sum_{j=1}^{n}[\hat{J}_{j} - \hat{K}_{j}] \, .
$$
For occupied orbitals, in the Hartree–Fock equation defining some particular (occupied) $i$-th orbital the Coulomb and exchange contribution from this orbital itself perfectly cancel each other, i.e. $\hat{J}_{j} \phi_i = \hat{K}_{j} \phi_i$ when $j=i$. Thus, in the Hartree-Fock model an electron from any occupied orbital is also moving in the field $n-1$ electrons. And this is of course physically more than reasonable: at the end of the day electron should not interact with itself.

But for virtual orbitals the situation in the Hartree-Fock model is different from that in the Kohn-Sham one: for any virtual orbital $\phi_k$ the sum in the Hartree-Fock equation defining it still runs over occupied orbitals only, but none of the terms $\hat{J}_{i} \phi_k$ and $\hat{K}_{i} \phi_k$ this time cancel each other since $k$ is simply grater than $n$.


Those rippling transparent patterns are caused by the variations in refractive index, which in turn are caused by the variations in concentration of the solute. As soon as the dissolution and mixing is over, and the solution is uniform again, the patterns are gone. I don't know the English word for it, or whether there is one; "Schlieren" seems to be much more general.

It's a physical phenomenon indeed, but chemists might be more used to seeing it.

You may observe a remotely similar pattern around a heated object in the air (see "heat haze"). In that case, it is caused by the variations of temperature and not concentration.


What is the E/Z configuration of cyclic compounds?

What is the meaning of cis and trans?
cis means on the same side (or face), trans means on the opposite side (or face).
Can I use cis and trans for cylic compounds?
Yes, you can describe the relative orientation of substituents on a ring using cis and trans, such as in the case of 1,3-dibromcyclohexane.

Can I use cis and trans for alkenes?
Yes, in priciple. But you should rather use E and Z. Note that these descriptors are taken from German.

E means entgegen and refers to opposite directions (or sites), while Z means zusammen, which is together, that is the same direction (or site).
What about the compounds in my question?
E/Z or cis/trans are not applicable here.
In the first case, the $\ce{Br}$ atom is in the ring plane, the methyl group can be above or below it. You could describe the absolute stereochemistry at the centre bearing the $\ce{CH3}$ substituent using R or S.
In the second case, there's no alkene at all but a disubstituted benzene. The relative orientation of the substituents on these systems can be described by assigning the proper numbers or by using the ortho- meta- or para-prefix. For this compounds, 2-isopropylnitrobenzene or o-isopropylnitrobenzene would be ok.


Why isn't CH3OH basic if it contains OH?

Unfortunately, chemistry is not as simple as going

  Oh, I spotted $\ce{OH}$, it must be a basic compound!

I would argue that the vast majority of all compounds one could write with $\ce{OH}$ inside are either neutral or acidic and only a minority is basic.

Take, for example, sulphuric acid. As Ivan correctly mentioned in a comment, rather than writing the traditional $\ce{H2SO4}$, you could also write it as $\ce{O2S(OH)2}$ which lays more emphasis on the molecular structure (to hydroxy groups and two oxy groups all bound to sulphur). We all know that $\ce{H2SO4}$ is acidic, not basic, even though it has $\ce{OH}$ groups.

Or take alcohols such as methanol ($\ce{H3C-OH}$) or ethanol ($\ce{H3C-CH2-OH}$). These also feature $\ce{OH}$ groups, they are even commonly written that way, but are about as acidic or basic as water ($\ce{HOH}$) is.

So the question may be why did you get taught that $\ce{OH}$ is basic? Well strictly speaking, it is not the $\ce{OH}$ hydroxy group that is basic but the $\ce{OH-}$ hydroxide ion. Only those compounds that dissociate into some cation and a hydroxide ion can be considered basic. For that to happen, ‘some cation’ typically needs to be metallic, at least at your level. So recognise a metal bonded to $\ce{OH-}$ and you should most likely be right calling the compound basic.

Some final notes:

You should always check the directly corresponding conjugate acid. So when analysing $\ce{K2SO4, Na2CO3}$ and $\ce{(NH4)2CrO4}$, trace them back to $\ce{HSO4- , HCO3-}$ and $\ce{HCrO4-}$, respectively. That still leaves us with two rather strong acids but the third ($\ce{HCO3-}$) is even weaker.
The conjugate acid to methanol $\ce{H3C-OH}$ would be the methyloxonium ion $\ce{H3C-OH2+}$, which is a strong acid. This goes for pretty much all organic hydroxy groups.
The $\mathrm{p}K_\mathrm{a}$ of water — hydroxide’s conjugate acid — is $15.7$, so yes, that is a very weak acid.


What does it mean for a reaction to favor the reactants/products?

You may not have had kinetics (reaction rates) yet, but I find it really useful in understanding equilibria.

Kinetics of a reaction

Molecules in solution (gas or liquid) move randomly. Because they need to bump into each other to react, the probability of reaction increases as concentrations of reactants increase. The rate of the reaction is then proportional to the concentrations of reactants:

$$\ce{A + B \rightarrow C}$$
$$\mathrm{rate} = k_\mathrm{forward}\ce{[A][B]}.$$

Likewise, the reverse reaction has a rate dependent on concentration of product:

$$\mathrm{rate} = k_\mathrm{reverse}\ce{[C]}.$$

From Kinetics to Equilibrium

At equilibrium, the forward rate is equal to the reverse rate, and concentrations of both reactants and products remain unchanged. $$k_\mathrm{forward}\ce{[A][B]}=k_\mathrm{reverse}\ce{[C]}$$ We can rearrange to get: $$\frac{k_\mathrm{forward}} {k_\mathrm{reverse}} = \ce{\frac{[C]}{[A][B]}}.$$ We can substitute the right side of the equation to be the equilibrium constant, and we get $$\frac{k_\mathrm{forward}} {k_\mathrm{reverse}}=K_c$$ Thus, when we say that $K_c>1$ we are also saying that $k_\mathrm{forward} > k_\mathrm{reverse}$ and inversely if $K_c<1$ then $k_\mathrm{reverse} > k_\mathrm{forward}.$ To favor either the reactants or the products in equilibrium is to say the formation of either the reactants or products is favored, as indicated by the rate constants.


I think it would go through a manganese-oxalate complex (oxalate is a good ligand for many metals), and that would fragment via a reaction such as:

The Mn(V) species would likely disproportionate to manganese dioxide and permanganate.


Why is activation energy drawn in a potential energy diagram in reactions?

The activation energy is not the average kinetic energy that the colliding particles need in order to successfully react. Instead, it is the energy barrier between reactants and products. Often, the energy required to cross the barrier comes from the kinetic energy of collisions.

To describe the rate of a reaction there are two models, the Arrhenius model and the more sophisticate Eyring model:

The Arrhenius model uses the activation energy, defined as the energy required to reach the transition state; the "energy" here is of various nature, not only kinetic an includes the chemical potential.

The Eyring model uses the Gibbs energy of activation for the transition state that includes an enthalpy and an entropy term ($ΔG^‡= ΔH^‡ – T ΔS^‡$). In comparison, for a unimolecular, one-step reaction, the activation energy can be approximately expressed as $E_a = ΔH^‡ + RT$.

There are different y-axis in these diagrams as explained in a related answer:
Y-axis of the reaction co-ordinate graph. Depending on the model and reaction you want to study you can use the Gibbs energy (Eyring model), energy in general (Arrhenius model) or enthalpy in case you want to show the heat absorbed or released in a reaction at constant pressure.

Why does your graph show potential energy?

In simple terms, potential energy is stored energy while kinetic energy is the energy of motion. Potential energy is converted into kinetic energy, therefore you can account for the latter including it in the overall potential energy.


Why does a smaller amount of propanol burn slower than a much larger amount?

For "burning" you need oxygen because actually when a compound is burning it's what is called an oxidation following this reaction

$$\ce{C3H7OH + 5O2 -> 3CO2 + 4H2O}$$

with propanol called the combustible and oxygen is called the comburant. Both are required as well as a power source to start the combustion process.

So in a plate you have the air surrounding the compound then you have practically an endless amount of oxygen available.

So the reaction will stop when all the Propanol will be consumed.

Hence, in a jug, this time the limiting reactive is the oxygen because you have less oxygen supply. Then the combustion will stop when all the oxygen will be consumed in the jug.

So, to sum up, in the jug the oxygen will be consumed faster than the propanol in the plate, that's why the combustion last longer in the plate even if you have more combustible.


Can aluminum and hydrochloric acid be used to generate hydrogen gas?

Yes. See this video for a demonstration.

The reaction is:
$$\ce{ 2Al (s) + 6 HCl (aq) -> 2AlCl3 (s) + 3H2(g)}$$

You can capture the hydrogen and proceed with your experiments.


IUPAC rules on omission of locants in carboxylic acid

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name is 2-methylpropanoic acid. The trivial name ‘isobutyric acid’ is no longer recommended (see Subsection P-65.1.1.2.4).

Concerning your objection

  You can't attach a methyl in carbon 3 (because it will become butanoic acid)

IUPAC nomenclature doesn’t work like that. Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information or hidden logic involving other structures.

The given parent structure is propanoic acid. It includes two kinds of substitutable hydrogen. Therefore, the locant cannot be omitted in substituted propanoic acid, e.g. 2-chloropropanoic acid or 3-chloropropanoic acid.

This principle applies to any substituent. The special situation that replacing the chloro substituent in 3-chloropropanoic acid with a methyl substituent would actually lead to the name butanoic acid instead of 3-methylpropanoic acid does not change the preferred name for 2-methylpropanoic acid.

Nevertheless, there are a few exceptional cases where locants are omitted, but only when there is no ambiguity. For preferred IUPAC names, the nomenclature rules explicitly stipulate when locants are omitted. Most of these rules apply to the omission of the locant ‘1’. One case is the example given in the question:

  (It's a bit like ethan-1-ol and ethanol.)

Homogeneous chains consisting of only two identical atoms such as ethane only have one kind of substitutable hydrogen. Therefore, the current IUPAC recommendations include a rule (in Subsection P-14.3.4.2) that the locant ‘1’ is omitted in monosubstituted ethane; i.e. ethanol (not ethan-1-ol) is the preferred IUPAC name. However, if any locants are essential for defining the structure, then all locants must be cited for the structure, e.g. 2-chloroethan-1-ol (not 2-chloroethanol; see Subsection P-14.3.3).

Coming back to carboxylic acids, acetic acid also has only one kind of substitutable hydrogen, i.e. all substitutable hydrogen atoms have the same locant. Therefore, explicitly according to Subsection P-14.3.4.6, this locant is omitted, e.g. chloroacetic acid (not 2-chloroacetic acid).


$\ce{C^{4+}}$ ions:

Single ions are really only observed in the gas phase.  There is absolutely nothing that prevents a C4+ ion from being generated in the gas phase, given sufficient energy.  Each successive electron removal requires additional energy though, so by the time you get to 4 electrons removed, you're looking at quite a lot of energy required.  Note that this does not mean that $\ce{C^{4+}}$ ions are seen in compounds, just that they can be made in the gas phase.  Any cation you can imagine can be produced in gas phase.

The difficulty arises when we want to do something with that carbon cation.  If I want to make a compound with it, it can bond with other substances and make new bonds.  Once that is done we can evaluate those compounds to tell whether the bond is truly ionic or not.  In condensed phases or in solution, we look at how the parts of a compound interact with the particles surrounding them to decide whether to treat them as ions or not.

The best chance at making a C4+ ion would be to bond it to fluorine, which is an extremely electron-hungry element.  When I actually try this, I find that the C-F bond is very polar, but the $\ce{CF_4}$ molecule that forms behaves as a covalent compound normally would.  For example, it does not dissociate into ions when dissolved or melted and it doesn't form an ionic crystal in its solid form.  As much as fluorine wants electrons, it doesn't want them enough to completely steal 4 of them from carbon, because the energy required to remove each electron becomes greater than the energy required to remove the one before it.  It never gets to the point of $\ce{C^{4+}}$ and $\ce{F^{-}}$ ions being bonded.

Lead can form a +4 ion, $\ce{Pb^{4+}}$, in compounds such as lead fluoride.  The reason that this can happen but carbon cations cannot is that the outermost electrons in a lead atom are much farther from the nucleus than those of a carbon atom and are easier to remove as a result.

$\ce{C^{4-}}$ ions:

In gas form this does not happen.  Carbon's first electron affinity is positive, so we would expect to see it form $\ce{C-}$ ions in the gas phase if given free electrons to attract.  Adding a second electron to this substance would require it to attract an electron against its already negative overall charge, and that is not energetically favorable.

In order to make the $\ce{C^{4-}}$ion in a compound, a species would have to donate 4 electrons to carbon.  The attraction of carbon's nucleus to other atoms' electrons is just too low for that to happen.  The best candidates for a cation in an ionic compound of this type would be either lithium or cesium, however, when we try to actually make this compound we get another class of ions, the carbides, which feature $\ce{C2^{2-}}$ ions.  In short, it doesn't happen.

Phillip's comment regarding the carbides is a good one.  There are a few metal carbides that feature carbon atoms bonding to a metal in the ratios we would expect if it were purely ionic, like $\ce{Mg2C}$.  In these compounds the carbon does have a formal charge of -4, but the properties of the substance are such that it behaves like a covalent network compound rather than an ionic compound.

What is the limit of this process?  Anions with -3 charge, like nitride or phosphide, do exist, though they require a very weakly electronegative species, like an alkali metal or alkaline earth metal, to form an ionic compound.

Carbon atoms can carry a charge as part of an ion, but only when they are part of a larger species (again, like $\ce{C2^{2-}}$) and they do not carry 4 additional electrons per atom.  Check out cyanide as another good example of this type of polyatomic ion.


Newer ionic radii data than Shannon's 1976 compilation?

tldr; it's very difficult to determine consistent additive ionic radii
For various reasons, I've gone back to look at covalent radii recently (e.g., for estimating bond lengths).
In that case, there seem to be fairly useful recent work, e.g. by Pekka Pyykkö:
"Additive Covalent Radii for Single-, Double-, and Triple-Bonded Molecules and Tetrahedrally Bonded Crystals: A Summary"
J. Phys. Chem. A 2015, 119, 11, 2326–2337
For covalent radii, it's clear that one can fit experimental and theoretical interatomic distance r(A-B) to a sum of two covalent radii with systematic accuracy. Notably, Pyykkö defines not only single-bond but double-bond and triple-bond radii for almost all elements.
(N.B.: I'm not sure it makes sense to consider a triple bond to $\ce{Br}$ or $\ce{I}$ but if you want them, it's in the article.)
There are also recent systematic van der Waals radii, e.g., Cramer and Truhlar:
"Consistent van der Waals Radii for the Whole Main Group"
J. Phys. Chem. A 2009, 113, 19, 5806–5812
So what about ionic radii?
There's a really nice article by Liu, Schwartz, and Li, cited by Pyykkö about the problems with ionic radii.
"On Two Different Objectives of the Concepts of Ionic Radii"
Chem (Eur. J.) 2013, 19, 44, 14758-14767
The Liu, Schwartz, and Li point out the competing problems with ionic radii:

Since “bonded atoms” are evidently neither rigid nor spherical, quantum
computational chemists apply “valence multiple zeta and
polarized” basis sets in their molecular calculations. Conversely, X-ray ED researchers use the procrystal model of independent rigid spherical atoms for starting and screening purposes in their software for the processing and analysis of the observed diffraction data.

In other words, we like to consider atoms as having consistent, rigid spherical shapes. For ionic compounds, that's hard to determine, because of charge transfer. By definition, $\ce{A+B-}$ indicates cation $\ce{A+}$ has given electron density to $\ce{B-}$ .. but how much? Is there any covalent character and shared electron density? How do you divide?
Pyykkö also mentions the problem in his covalent radii work. His atomic radii work:

… provided that the A–B bond is not too ionic, or the coordination numbers of the two atoms too different from the original input data, used in the fit.

In short, covalent radii are easy to define because most of the electron density is shared and covalent bonds are thus fairly predictable. In contrast, almost no compound is purely ionic so it's hard to perform the division for ionic species.
Here are two examples from Liu, Schwartz, and Li:

Consider interatomic distances from alkali halide crystals. (The portion is performed by Bader's AIM analysis.)

Note that there's no "vertical" or "horizontal" line for well-defined anion or cation radii. Maybe you can come up with a correction based on electronegativity difference .. but the cation radii particularly depend on the anion.

The same problem shows up for monomeric alkali halide molecules in the gas phase.

It makes some sense. It's difficult to rigorously know how much charge transfers from anion to cation (vs. shared), which is why calculation methods define multiple algorithms to determine it. (Atomic partial charges are something of an eternal debate.)
I wouldn't be surprised to find someone creating a correction using electronegativity differences or covalent character. And it's possible someone will sweep this away with a compelling article - that's the point of science. But it won't be easy, because we expect ionic radii to serve a lot of different purposes.


Can an atom have more than 8 valence electrons? If not, why is 8 the limit?

2017-10-27 Update

[NOTE: My earlier notation-focused answer, unchanged, is below this update.]

Yes. While having an octet of valence electrons creates an exceptionally deep energy minimum for most atoms, it is only a minimum, not a fundamental requirement. If there are sufficiently strong compensating energy factors, even atoms that strongly prefer octets can form stable compounds with more (or less) than the 8 valence shell electrons.

However, the same bonding mechanisms that enable formation of greater-than-8 valence shells also enable alternative structural interpretions of such shells, depending mostly on whether such bonds are interpreted as ionic or covalent. Manishearth's excellent answer explores this issue in much greater detail than I do here.

Sulfur hexafluoride, $\ce{SF6}$, provides a delightful example of this ambiguity. As I described diagrammatically in my original answer, the central sulfur atom in $\ce{SF6}$ can be interpreted as either:

(a) A sulfur atom in which all 6 of its valence electrons have been fully ionized away by six fluorine atoms, or

(b) A sulfur atom with a stable, highly symmetric 12-electron valence shell that is both created and stabilized by six octahedrally located fluorine atoms, each of which covalently shares an electron pair with the central sulfur atom.

While both of these interpretations are plausible from a purely structural perspective, the ionization interpretation has serious problems.

The first and greatest problem is that fully ionizing all 6 of sulfur's valence electrons would require energy levels that are unrealistic ("astronomical” might be a more apt word).

A second issue is that the stability and clean octahedral symmetry of $\ce{SF6}$ strongly suggests that the 12 electrons around the sulfur atom have reached a stable, well-defined energy minimum that is different from its usual octet structure.

Both points imply that the simpler and more energetically accurate interpretation of the sulfur valence shell in $\ce{SF6}$ is that it has 12 electrons in a stable, non-octet configuration.

Notice also that for sulfur this 12- electron stable energy minimum is unrelated to the larger numbers of valence-related electrons seen in transition element shells, since sulfur simply does not have enough electrons to access those more complex orbitals. The 12 electron valence shell of $\ce{SF6}$ is instead a true bending of the rules for an atom that in nearly all other circumstances prefers to have an octet of valence electrons.

That is why my overall answer to this question is simply "yes".

Question: Why are octets special?

The flip side of whether stable non-octet valence shells exist is this: Why do octet shells provide an energy minimum that is so deep and universal that the entire periodic table is structured into rows that end (except for helium) with noble gases with octet valence shells?

In a nutshell, the reason is that for any energy level  above the special case of the $n=1$ shell (helium), the "closed shell" orbital set $\{s, p_x, p_y, p_z\}$ is the only combination of orbitals whose angular momenta are (a) all mutually orthogonal, and (b) cover all such orthogonal possibilities for three-dimensional space.

It is this unique orthogonal partitioning of angular momentum options in 3D space that makes the $\{s, p_x, p_y, p_z\}$ orbital octet both especially deep and relevant even in the highest energy shells. We see the physical evidence of this in the striking stability of the noble gases.

The reason orthogonality of angular momentum states is so important at atomic scales is the Pauli exclusion principle, which requires that every electron have its own unique state. Having orthogonal angular momentum states provides a particularly clean and easy way to provide strong state separation between electron orbitals, and thus to avoid the larger energy penalties imposed by Pauli exclusion.

Pauli exclusion conversely makes incompletely orthogonal sets of orbitals substantially less attractive energetically. Because they force more orbitals to share the same spherical space as the fully orthogonal $p_x$, $p_y$, and $p_d$ orbitals of the octet, the $d$, $f$, and higher orbitals are increasingly less orthogonal, and thus subject to increasing Pauli exclusion energy penalties.

A final note

I may later add another addendum to explain angular momentum orthogonality in terms of classical, satellite-type circular orbits. If I do, I'll also add a bit of explanation as to why the $p$ orbitals have such bizarrely different dumbell shapes.

(A hint: If you have ever watched people create a two loops in a single skip rope, the equations behind such double loops have unexpected similarities to the equations behind $p$ orbitals.)

Original 2014-ish Answer (Unchanged)

This answer is intended to supplement Manishearth's earlier answer, rather than compete with it. My objective is to show how octet rules can be helpful even for molecules that contain more than the usual complement of eight electrons in their valence shell.

I call it donation notation, and it dates back to my high school days when none of the chemistry of the texts in my small-town library bothered to explain how those oxygen bonds worked in anions such as carbonate, chlorate, sulfate, nitrate, and phosphate.

The idea behind this notation is simple. You begin with the electron dot notation, then add arrows that show whether and how other atoms are "borrowing" each electron. A dot with an arrow means that that the electron "belongs" mainly to the atom at the base of the arrow, but is being used by another atom to help complete that atom's octet. A simple arrow without any dot indicates that the electron has effectively left the original atom. In that case, the electron is no longer attached to the arrow at all but is instead shown as an increase in the number of valence electrons in the atoms at the end of the arrow.

Here are examples using table salt (ionic) and oxygen (covalent):

Notice that the ionic bond of $\ce{NaCl}$ shows up simply as an arrow, indication that it has "donated" its outermost electron and fallen back to its inner octet of electrons to satisfy its own completion priorities. (Such inner octets are never shown.)

Covalent bonds happen when each atom contributes one electron to a bond. Donation notation shows both electrons, so doubly bonded oxygen winds up with four arrows between the atoms.

Donation notation is not really needed for simple covalent bonds, however. It's intended more for showing how bonding works in anions. Two closely related examples are calcium sulfate ($\ce{CaSO4}$, better known as gypsum) and calcium sulfite ($\ce{CaSO3}$, a common food preservative):

In these examples the calcium donates via a mostly ionic bond, so its contribution becomes a pair of arrows that donate two electrons to the core of the anion, completing the octet of the sulfur atom. The oxygen atoms then attach to the sulfur and "borrow" entire electrons pairs, without really contributing anything in return. This borrowing model is a major factor in why there can be more than one anion for elements such as sulfur (sulfates and sulfites) and nitrogen (nitrates and nitrites). Since the oxygen atoms are not needed for the central atom to establish a full octet, it is possible for some of the pairs in the central octet to remain unattached. This results in less oxidized anions such as sulfites and nitrites.

Finally, a more ambiguous example is sulfur hexafluoride:

The figure shows two options. Should $\ce{SF6}$ be modeled as if the sulfur is a metal that gives up all of its electrons to the hyper-aggressive fluorine atoms (option a), or as a case where the octet rule gives way to a weaker but still workable 12-electron rule (option b)? There is some controversy even today about how such cases should be handled. The donation notation shows how an octet perspective can still be applied to such cases, though it is never a good idea to rely on first-order approximation models for such extreme cases.

2014-04-04 Update

Finally, if you are tired of dots and arrows and yearn for something closer to standard valence bond notation, these two equivalences come in handy:

The upper straight-line equivalence is trivial since the resulting line is identical in appearance and meaning to the standard covalent bond of organic chemistry.

The second u-bond notation is the novel one. I invented it out of frustration in high school back in the 1970s (yes I'm that old), but never did anything with it at the time.

The main advantage of u-bond notation is that it lets you prototype and assess non-standard bonding relationships while using only standard atomic valences. As with the straight-line covalent bond, the line that forms the u-bond represents a single pair of electrons. However, in a u-bond, it is the atom at the bottom of the U that donates both electrons in the pair. That atom gets nothing out of the deal, so none of its bonding needs are changed or satisfied. This lack of bond completion is represented by the absence of any line ends on that side of the u-bond.

The beggar atom at the top of the U gets to use both of the electrons for free, which in turn means that two of its valence-bond needs are met. Notationally, this is reflected by the fact that both of the line ends of the U are next to that atom.

Taken as a whole, the atom at the bottom of a u-bond is saying "I don't like it, but if you are that desperate for a pair of electrons, and if you promise to stay very close by, I'll let you latch onto a pair of electrons from my already-completed octet."

Carbon monoxide with its baffling "why does carbon suddenly have a valence of two?" structure nicely demonstrates how u-bonds interpret such compounds in terms of more traditional bonding numbers:

Notice that two of carbon's four bonds are resolved by standard covalent bonds with oxygen, while the remaining two carbon bonds are resolved by the formation of a u-bond that lets the beggar carbon "share" one of the electron pairs from oxygen's already-full octet. Carbon ends up with four line ends, representing its four bonds, and oxygen ends up with two. Both atoms thus have their standard bonding numbers satisfied.

Another more subtle insight from this figure is that since a u-bond represents a single pair of electrons, the combination of one u-bond and two traditional covalent bonds between the carbon and oxygen atoms involves a total of six electrons, and so should have similarities to the six-electron triple bond between two nitrogen atoms. This small prediction turns out to be correct: nitrogen and carbon monoxide molecules are in fact electron configuration homologues, one of the consequences of which is that they have nearly identical physical chemistry properties.

Below are a few more examples of how u-bond notation can make anions, noble gas compounds, and odd organic compounds seem a bit less mysterious:


Additives to allow wax to melt in the microwave?

Microwave heating works best on polar molecules, such as water and alcohols, as it then causes the molecules to rotate or "jiggle" violently. Waxes are nonpolar, so heat poorly, but if your application can use fatty acids in the mixture, such as palmitic or stearic acid, you might try mixing some into the microcrystalline wax. The dielectric constant is not significant in your application.

However, the longer the fatty acid molecule, the less effective it is in absorbing microwave radiation and converting it to heat, and on the other hand, the shorter molecules have strong odors and do not mix well with wax... do not use butyric acid for that reason. The suggestion of using an absorbent container, by @Li Zhi, seems to be more effective.


Is it safe to use lab-grade calcium nitrate as a fertilizer?

Usually you can buy chemicals in different purity grades.
If you buy something that has the grade purum >95% will be your desired ingredient. However there is not usually any information about what the other <5% will consist of.

Personally I would never ingest (or use in food production) anything that is not food grade. Food grade does not necessarily mean very pure, it just means that there is nothing in the other x% that is potentially harmful. Look out for an FCC number, that is the Food Chemical Codex, when you order it and you should be fine.


How do I determine crystal field splitting (for example, for [Ni(CO)4])?

Firstly, what geometry does the complex have?

Once you know the geometry, find the spin-state of the $d$-electrons, as well as the number of $d$-electrons applicable.

Find the crystal field diagram relevant to your geometry. Fill in the $d$-electrons according to the Aufbau principle and to Pauli's exclusion principle.

For your example, tetracarbonylnickel has a tetrahedral geometry and is a $d$10 compound. Carbon monoxide is a strong field ligand causing a high spin complex (note that since $d$ = 10, all of the $t_{2}$ and $e$ orbitals are full and hence spin-state is irrelevant).

As per the Aufbau principle, you start building up from the bottom, placing two electrons in the bottom $t_2$ orbitals and three in the top three $e$ orbitals, giving:

$e$: ↑ ↑ ↑

$t_2$: ↑ ↑

The remaining five electrons go in each half-filled orbital to give 2 filled $t_2$ orbitals and 3 filled $e$ orbitals:

$e$: ↑↓ ↑↓ ↑↓

$t_2$: ↑↓ ↑↓

To find the CSFE, the $t_2$ orbitals each stabilise by $\frac{3}{5}$ and the $e$ destabilize by $\frac{2}{5}$. Since $2 * \frac{3}{5} - 3 * \frac{2}{5}$ = 0, the CFSE = 0.

For octahedral complexes, the $e_g$ and $t_{2g}$ orbital populations can be used analogously to find the stabilisation energy.


How to determine age of stones (and other inorganic matter)?

The question being asked, I believe, is the age of the inscription or statue rather than that of the material from which it is made.

The answer, though, is that it is quite difficult to do on the basis of a single stone object. If it is found in situ, then the age of surrounding items may be a guide.

There are some clues to the age of stone objects, though. For example, if a stone object (e.g. the walls of Siloam's Tunnel) has soot on its surface, such as from torches, then radiocarbon dating shows a minimum age (because the soot could have been deposited some time after the tunnel or object was made).

Weathering of a surface can provide information, providing the object has been exposed to known atmospheric conditions; but a buried statue would erode differently from one on the surface.

For stones, or better, for ceramics, that have been heated, thermoluminescence provides an accurate way to measure age (or at least time since last being heated).

One indication for the age of the Egyptian pyramids is their alignment to a calculated geologic reference that changed with time due to the precession of the Earth.


Why would the specific heat capacity of water be used as the specific heat capacity of zinc sulfate?

If the substance is dissolved in water, it's very typical (and more correct) to use the heat capacity of water in lieu of the pure salt, acid, base, etc.

The reason is simply that, when in solution with water, there is much more water than the solute. It's an approximation, but a decent one. This is common practice in freshman-chemistry lab courses which I've taught.

It's difficult to find heat capacities for pure salts such as zinc sulfate because the means by which we measure it is typically through a reaction, and typically a salt is dissolved in solvent (e.g. water) in order to perform a reaction. That is to say, reactions of solids are just less common than liquids, gases, and solutions.

That said, the NIST database reports a $c_{p}$ for zinc sulfate of 99.01 Jmol$^{-1}$K$^{-1}$ at 298 K.

Looking at the original paper, I found this plot of $c_p$ as a function of temperature. At room temperature, you can see it is roughly 100 Jmol$^{-1}$K$^{-1}$.

Stuff like this can be easily obtained from citrination.com search engine.


As already said, ion pair is a pair of two ions like NaCl. When the pair separates completely into its constituent ions in the solvent (like Na+ & Cl- in water), it is called dissociation.

After dissociation, if the solvent associates itself and surrounds the ions like this-

(Sodium ion surrounded by water molecules), it is called Solvation.
Solvation stabilises the compound and makes it either more reactive or less reactive.


Buffers: what does the strong base react with?

The reaction NH$_4^+$ $\Leftrightarrow$ NH$_3$ + H$^+$ is in a dynamic equilibrium that means the reaction will happen in both directions all the time and the acid dissociation constant will tell you how much on average you have all components in solution so by that you can also calculate the pH value. So of course since you have NH$_4^+$ in solution as well as H+ both can in principle react with the OH$^-$ in providing the proton to form water. The question is, which of the two reactions happens more. This will be determined by the two equilibrium constants

$K_{1} = \frac{[NH3][H2O]}{[NH4+][OH-]}$
$K_2   = \frac{[NH3][H2O]}{[NH3][H+]}$

You see the only difference are the concentrations of the educts, so how large are these? Suppose you prepare a 0.05M NH$_4$Cl solution. After a short time there will be thermodynamic equilibrium and by using the $pK_\mathrm{a}=9.25$ value for NH$_4$Cl you can calculate the respective concentrations of NH$_4^+$ and NH$_3$ in equilibrium:

$K_a=\frac{[H+][NH_3]}{[NH_4^+]}$

By using $[H^+]=[NH_3]$ we get

$10^{-9.25}=\frac{[x][x]}{0.05-[x]}$
gives $[x]=5 \cdot 10^{-6} M$, so actually not much has dissociated. So you can be nearly completely sure that the reaction takes place with the NH$_4$ rather than protons or hydronium ions. And that is what a buffer is for! The pH should not change even when you add $OH^-$ until you reach the stability window of the buffer, that is when the NH$_4^+$ concentration gets too low compared to the $H^+$ concentration. Free protons will also never exist in solution, they can nearly always be associated to some water molecule.


What is the theoretical quantum ground state energy of helium hydride positive ion?

You could get a quick estimate of the ground state energy with paper and pencil by using Heitler-London method, since all you need to do is change the charge on the second nucleus.

I would assume the vibrational energy absorption energy quoted referred to the He-H bond vibration.


What are the magnetic quantum numbers for the three real p orbitals?

The answer is... it is not so simple. Some quantum mechanics follow, but the TL;DR version is that while $m_l=0$ corresponds to $p_z$, the orbitals for $m_l=+1$ and $m_l=-1$ lie in the $xy$-plane, but not on the axes. The reason for this outcome is that the wavefunctions are usually formulated in spherical coordinates to make the maths easier, but graphs in the Cartesian coordinates make more intuitive sense for humans. The $p_x$ and $p_y$ orbitals are constructed via a linear combination approach from radial and angular wavefunctions and converted into $xyz$. Thus, it is not possible to directly correlate the values of $m_l=\pm1$ with specific orbitals. The notion that we can do so is sometimes presented in introductory courses to make a complex mathematical model just a little bit simpler and more intuitive.

From Physical Chemistry by Atkins and DePaula, the three wavefunctions for $n=2$ and $l=1$ are as follows.

$$\begin{align}
&\Psi_{2,1,0}&&=r\cos{\theta}f(r)\\
&\Psi_{2,1,+1}&&=-\dfrac{r}{\sqrt{2}}\sin{\theta}\mathrm{e}^{\mathrm{i}\phi}f(r)\\
&\Psi_{2,1,-1}&&=\dfrac{r}{\sqrt{2}}\sin{\theta}\mathrm{e}^{-\mathrm{i}\phi}f(r)\\
&\end{align}$$

The notation is $\Psi_{n,l,m_l}$, $r$ is the radius, $\theta$ is the angle with respect to the $z$-axis and $\phi$ is the angle with respect to the $xz$-plane.
$$f(r)=\sqrt{\dfrac{Z^5}{32\pi a_0^5}}\mathrm{e}^{-Zr/2a_0}$$

in which $Z$ is the atomic number (or probably better nuclear charge) and $a_0$ is the Bohr radius.

In switching from spherical to Cartesian coordinates, we make the substitution $z=r\cos{\theta}$, so:
$$\Psi_{2,1,0}=zf(r)$$

This is $\Psi_{2p_z}$ since the value of $\Psi$ is dependent on $z$: when $z=0;\ \Psi=0$, which is expected since $z=0$ describes the $xy$-plane.

The other two wavefunctions are unhelpfully degenerate in the $xy$-plane. An equivalent statement is that these two orbitals do not lie on the $x$- and $y$-axes, but rather bisect them. Thus it is typical to take linear combinations of them to make the equation look prettier. Linear combinations are allowed by the maths of quantum mechanics. If any set of wavefunctions is a solution to the Schrödinger equation, then any set of linear combinations of these wavefunctions must also be a solution. We can do this because orbitals and the wavefunctions that describe them are not real physical objects. They constitute a mathematical model.

In the equations below, we're going to make use of some trigonometry, notably Euler's formula:

$$\mathrm{e}^{\mathrm{i}\phi}=\cos{\phi}+\mathrm{i}\sin{\phi}$$
$$\sin{\phi} = \frac{\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi}}{2\mathrm{i}}$$
$$\cos{\phi} = \frac{\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi}}{2}$$

We're also going to use $x=\sin{\theta}\cos{\phi}$ and $y=\sin{\theta}\sin{\phi}$.

$$\begin{align}
\Psi_{2p_x}=\frac{1}{\sqrt{2}}\left(\Psi_{2,1,+1}-\Psi_{2,1,-1}\right)=\frac{1}{2}\left(\mathrm{e}^{\mathrm{i}\phi}+\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)=r\sin{\theta}\cos{\phi}f(r)=xf(r) \\
\Psi_{2p_y}=\frac{\mathrm{i}}{\sqrt{2}}\left(\Psi_{2,1,+1}+\Psi_{2,1,-1}\right)=\frac{1}{2\mathrm{i}}\left(\mathrm{e}^{\mathrm{i}\phi}-\mathrm{e}^{-\mathrm{i}\phi} \right)r\sin{\theta}f(r)=r\sin{\theta}\sin{\phi}f(r)=yf(r)\\
\end{align}$$

  So, while $m_l=0$ corresponds to $\Psi_{p_z}$, $m_l=+1$ and $m_l=-1$ cannot be directly assigned to $\Psi_{p_x}$ and $\Psi_{p_y}$. Rather $m_l=\pm1$ corresponds to $\{\Psi_{p_x},\Psi_{p_y} \}$. Put another way, I suppose we could say that $m_l=+1$ might correspond to $\Psi_{p_{x+y}}$ and $m_l=-1$ might correspond to $\Psi_{p_{x-y}}$.


Yes it works, according to result they have with organometallic chemistry. You may not know $e_g$ and $t_{2g}$ orbitals. Have a look at this diagram of orbitals :

Its diagram for an octaedric complex, where M is the metal. Positions of $e_g$ and $t_{2g}$ depends of what L is.


How can inconsistencies in these sodium chloride solutions be explained?

example, 4 g/L $\ce{NaCl}$ is coming out as 0.68 M/L...hopefully I'm missing something very basic here

Yes, you are off by a factor of 10.

4 g/L $\ce{NaCl}$ is 0.068M (0.068 moles per liter of solution).

As far as the journal article, I would used the 4 grams per liter value (and other grams per liter values) and convert to moles per liter properly and not rely on the 0.08 (small capital "M" l-1) value.

If the value is critical to your research, email the corresponding author of the paper and and ask about it in a polite way.


Since the formation of diamond from Graphite is endothermic and the entropy decreases for this process, how is it possible to make diamond?

Diamond is significantly denser than graphite.

atomic densities:

$1.14  \times 10^{23}$ cm$^{-3}$ for graphite
$1.77  \times 10^{23}$ cm$^{-3}$ for diamond

This suggests that higher pressure would favor the formation of diamond.  Indeed, as the following phase diagram shows, diamond is the most stable allotrope of solid carbon at high pressures.  Temperature has little effect, as long as the pressure is high, diamond is favored over graphite at all temperatures up to the point of liquefication.


Superscript and subscript together after the same atom

From IUPAC Green Book [1, p. 51]:

In writing the formula for a complex ion, spacing for charge number can be added (staggered arrangement), as well as parentheses: $\ce{SO4^2-}$, $\ce{(SO4)^2-}$. The staggered arrangement is now recommended.

Also, as you are apparently a $\mathrm{\LaTeX}$-user, the same convention is used by default in chemformula package. From the manual:

If a compound does not start with a sub- or superscript and there is both a sub- and a superscript, the superscript is shifted additionally by a length determined from the option charge-hshift = {❬dim❭}, also see page 13f.
The second point follows IUPAC’s recommendations.

References:

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.


Help with missing output from cubegen utility after density calculation in Gaussian16

You aren't getting the density data in your cube file because you haven't specified the kind keyword in your cubegen command. Gaussian allows a variety of different types of cube files to be generated, including electronic densities/spin-densities/gradients, MOs, and electrostatic potential, along with magnetic current and shielding densities. Without specifying the type of cube file you want, the program doesn't know what to generate.

The reason it can't determine what you want by parsing the chk file is that the file in general will contain information that could be used for multiple types of cubes. The program could hypothetically use the first valid piece of information or some specific type of cube data by default, but this would often produce an undesired output. It could also make all valid cubes files, but these take up a lot of memory so this would be wasteful. They presumably decided the default case should be to do nothing if a type of cube file isn't specified so as not to produce unnecessary or incorrect output.


Donation by inductive effect and withdrawal by conjugation

I guess a $\ce{-BH2}$ group attached to a benzene ring would have this effect. It has an empty $\ce{p}$ orbital that would withdraw electrons via conjugation, but $\ce{B}$ is less electronegative than $\ce{C}$ and so it would have a positive inductive effect.


Looking for a chemical to calibrate a temperature sensor

The standard point recommended in that range by the CIPM's ITS-90 calibration standard is the melting point of gallium at 29.7646 °C. Gallium is not known to be particularly toxic, but it can embrittle metals so is ideally used/stored in glass or plastic containers. It's used as a standard because it is relatively easy to make very pure and is stable in storage. Also, by picking an ITS-90 standard, you can ensure your thermometer is calibrated to the same standard as many other high precision thermometers. N.B. ITS-90 specifies specifically the melting point of gallium, not the freezing point, unlike some of the other calibration standards.


Boltzmann factor vs. graph of Maxwell–Boltzmann distribution

Rationale for posting an "answer" to my own question

[Comment from OP] I’m looking for an explanation and a graph that is better than what is currently in the first year college textbooks .

The comments to the question contain some excellent insight and references but there is no comprehensive answer to far, so I am collecting what was posted here. Maybe the depth of the comments and the lack of a highly rated answer reflects that it is impossible (at the first year undergraduate level at least) to make a meaningful quantitative connection between Maxwell-Boltzmann distribution of kinetic energy in a ideal gas on the one hand and the Arrhenius equation on the other hand.
Textbook treatment of the question
In first-year college textbooks, the Arrhenius equation is often rationalized for gas phase reactions via the collision theory. The quotes here are from OpenStax Chemistry as hosted by Libretext.org, but other textbooks have similar material:

Both postulates of the collision theory of reaction rates are accommodated in the Arrhenius equation. The frequency factor A is related to the rate at which collisions having the correct orientation occur. The exponential term, $e^{−E_a/RT}$, is related to the fraction of collisions providing adequate energy to overcome the activation barrier of the reaction.

This is great because it references collisions providing sufficient energy.
Then they go on to explain:

The Arrhenius equation (Equation 12.5.6) describes quantitatively much of what we have already discussed about reaction rates. For two reactions at the same temperature, the reaction with the higher activation energy has the lower rate constant and the slower rate. The larger value of $E_a$ results in a smaller value for $e^{−E_a/RT}$, reflecting the smaller fraction of molecules with enough energy to react. Alternatively, the reaction with the smaller $E_a$ has a larger fraction of molecules with enough energy to react. This will be reflected as a larger value of $e^{−E_a/RT}$, a larger rate constant, and a faster rate for the reaction.

This is pretty much correct if you don't over-interpret "fraction of molecules with enough energy to react". What is relevant is the energy made available by collisions, not the energy of the molecules, but if the molecules have higher kinetic energy, the collisions will be of higher energy. (There might be some energy associated with rotation and vibration, but since collision theory starts with a mono-atomic ideal gas and we are shooting for the simplest explanation, those will be ignored).
Then they show a figure (and this figure is fairly representative of the corresponding figures in other textbooks and those found online from various sources):

This figure suggests that the fraction of molecules with kinetic energy higher than the activation energy will react. It also suggests that if you integrate over the shaded area, you get the Arrhenius equation.
If you take a close look at the curve, you will see that the slope at the origin is too small. It looks like the x-axis is the velocity rather than the energy, and the graph is accordingly distorted.
As an aside, there are two aspects of this figure that are not representative of what first-year students will experience and observe in the lab.

The activation energy shown is on the order of a couple RT. This would be a reaction too fast to measure in an undergraduate lab, and it gives the impression that maybe 1 out of 10 collisions leads to a chemical reaction.
The two temperatures differ by almost a factor two. If $T_1$ is room temperature, the second experiment would have to be done at 586 K, i.e. in an oven where most solvents would evaporate.

Relevant distribution for collision energy
The comments were very insightful. Ivan Neretin addressed the relevant degrees of freedom:

When the particles are about to react, in fact you are only interested in the distribution of one component of the velocity vector (say, vx). This distribution is a simple exponential decay.

I made a diagram to illustrate what happens in an inelastic collision:

So you can't simply take the kinetic energy of one molecule (or add both kinetic energies together) but instead have to consider one degree of freedom each. The maximum energy provided by a collision is not the sum of the kinetic energies of the particle because the center of mass of the two particles combined continues on its path, so part of the kinetic energy is not available.
The wikipedia article on the Maxwell–Boltzmann distribution states:

Since the energy is proportional to the sum of the squares of the three normally distributed momentum components, this distribution is a gamma distribution; in particular, it is a chi-squared distribution with three degrees of freedom.
By the equipartition theorem, this energy is evenly distributed among all three degrees of freedom, so that the energy per degree of freedom is distributed as a chi-squared distribution with one degree of freedom.

So maybe the distribution of energies in a collision is a problem with two degrees of freedom ($v_x$ of either particle). The chi-square distribution with two degrees of freedom happens to be a pure exponential lacking a T-dependent pre-exponential factor, matching the Arrhenius equation.
Orthocresol gave a reference to the rigorous derivation of the Arrhenius equation from the Maxwell-Boltzmann distribution in one of his comments:

The TL;DR is that porphyrin's first comment is pretty much correct, the area under the graph does generate an exp(−E/kT) form, although the maths is slightly more nuanced than simply doing the integral from EA to +∞.

In this answer, orthocresol sketches the derivation, with two references that show the details.
Relevance to teaching at the first-year college level
Most reactions students will see and do will be in solution, where things are different than in the gas phase, but the Arrhenius equation still is a good description of rate vs. temperature. On the other hand, as porphyrin state in one of his comments, things are usually more complicated than a derivation of rates vs. temperature via the collision model suggests:

The Arrhenius eqn is only a v crude model, sure molecules have to collide in the correct way, even if this is inline the wrong part may still collide depending on the molecule's size. Also vibrational energy can aide or impede collisions even if every thing else is ok. Have a look at RRKM, Tolman's stat mech model or better still Kramer's model of reactions.

So maybe it is best to treat the Arrhenius equation as a definition of the activation energy in an empirical sense.

A graph to illustrate collision energy distribution at multiple temperatures might be helpful.

Following the logic of deriving the Arrhenius equation from the collision model, this would simply be a graph of $e^{−E_a/RT}$ at different temperatures. For values of temperature relevant to first year experiments (eg. room temperature vs water bath), the two traces would be very close to each other, and almost zero at energies equal to a typical activation energy obtained in a first year lab (e.g. 80 kJ/mol for iodination of acetone). To see anything, you have to add an inset zooming in near the activation energy of interest.

This graph is very different from the textbook graph. It shows the distribution of collision energies rather than the kinetic energy of individual particles, and the two temperatures chosen are explicitly given and more relevant to an undergraduate lab experience. It shows that the energy distribution is not so different between room temperature and body temperature (the average energy of a molecule is proportional to T, after all, and this is just a 6% difference in temperature). However, at typical activation energies (insets at 40 kJ/mol and 80 kJ/mol), this smaller difference results in a large difference in rate constants. It also shows that of the collisions of sufficient energy, most have an energy quite close to the activation energy (fast drop-off of probability with increasing energy).


Why do unboiled eggs last longer than boiled eggs?

An eggshell is a complex structure.  From Wikipedia:

Boiling the egg removes the waxy cuticle from the outside of the egg, dissolves a small but not insignificant amount of calcium carbonate from the shell, damages the protein matrix that holds calcium carbonate crystals in place in the shell, and can disrupt or destroy the two shell membranes.  All of these contribute to allowing bacteria a means to enter the egg and spoil it.

Note: Though calcium carbonate is actually less soluble in hot water vs cold, the rate of dissolution is substantially greater.  Unless the egg is boiled in a saturated solution of calcium carbonate, it will lose material to the solution when boiled, and any pores in the shell become enlarged as a result.


Why does NO have a residence time of 4 days?

It could be possible that you're thinking of the overall lifetime of $\ce{NO_x}$ ($\ce{NO2 + NO}$), which indeed does have a longer atmospheric lifetime than $\ce{NO}$ individually and which even, at around 5 km altitude, has a lifetime of ~4 days. The two interconvert during the day due to photo-chemistry, and the ratio of the two is dependent on the photolysis rate of $\ce{NO2}$ ($j_{\ce{NO2}}$) and on the rate coefficient of the reaction between $\ce{NO}$ and $\ce{O3}$.

As $j_{\ce{NO2}}$ is constant with altitude, the ratio between $\ce{NO}$ and $\ce{NO2}$ is dependent upon the reaction rate coefficient between $\ce{NO}$ and $\ce{O3}$. This, in turn, is governed by the temperature, and under lower temperatures, the reaction rate is slower. Combined with a decrease in the number concentration of ozone (due to the decrease in the number concentration of air with increasing altitude, if we assume a constant mixing ratio of ozone throughout the troposphere, of course), the ratio favours $\ce{NO}$ over $\ce{NO2}$.

At higher altitudes, then, you do see a situation where there is more $\ce{NO}$ than $\ce{NO2}$, and where the overall lifetime of $\ce{NO_x}$ can be higher than the 1 to 2 days found at the surface (reaching, sometimes, almost 12 days in the upper troposphere).


From the abstract of an article titled "The crystal structure of galactaric acid (mucic acid) at −147°: An unusually dense, hydrogen-bonded structure"

[...] D$_m$ = [1.790] $\pu{g cm−3}$ [...] The crystal structure has a system of strong, intermolecular hydrogen-bonds, which accounts for the high crystal density and low solubility in water.

The crystal structure was solved at very low temperature, but the density D$_m$ is quoted for room temperature. For comparison, the density of sucrose or cellulose is around 1.5 $\pu{g cm−3}$.

For glucaric acid, the volume of the asymmetric unit is 421.9 Å$^3$ / 2, while for mucic acid it is 194.0 Å$^3$. As the molar mass of the two stereoisomers is the same, there is a difference in density of about 9%.

A lower density implies more attractive interactions or less repulsive interactions in the crystal state. For a complete analysis, you would also have to compare how well these compounds are solvated in aqueous solution. Typically, it is impossible to predict solubility by just looking at structures because both solvation and crystal structure are hard to predict.


Electrical conductivity of organic solvents

Ethanol and methanol dissociate into ions, just as water does.  For water:

$$\ce{2 H2O <=> H3O+ + OH-}$$

For an alcohol $\ce{ROH}$:

$$\ce{2 ROH <=> ROH2+ + ^-OR}$$

For water the self-ionization constant is $K_\mathrm{w} \approx 10^{-14}$.  For ethanol or methanol, I was unable to find data on the self-ionization constant, which is essentially the same as the $\mathrm{p}K_\mathrm{a}$ of the alcohol in a solution of itself.  The $\mathrm{p}K_\mathrm{a}$ of ethanol in water is slightly higher than the $\mathrm{p}K_\mathrm{a}$ of water in water.  Also, the lower dielectric constant in alcohols will further disfavor self-ionization relative to water.  So for methanol and ethanol the self-dissociation into ions will be weaker than in water.

These ions will still be present, though, and will conduct electricity even in pure ethanol and methanol solutions, but the conductivities will be very low.  Very pure water has a resistivity of $\pu{18.2 MΩ cm}$.  If they self-ionize to a lesser extent than water does, the resistivities of methanol and ethanol will be even higher than water's.


Chemical Reaction Alternative to O2 Respiration in H2/N2 Atmosphere

The obvious approach to is to just do the direct analog of what Earthling metabolism does: exactly reverse the photosynthetic reaction (on net, anyway; the intermediate processes may be rather different, as they are on Earth).

In this case, that means reacting hydrogen with carbohydrates to get back methane and water. Based on this paper on photosynthesis in hydrogen-dominated atmospheres, building carbohydrates from methane and water requires considerably less energy than building them from carbon dioxide and water, so we'll equally get less energy out by decomposing them again--I'd estimate somewhere around a quarter to a third as much power density as Earthling animals relying on aerobic respiration, if we assume organisms in this environment will be building biomolecules optimized for this environment, rather than just copying ours. Given how much time I spend every day not eating, and not hyperventilating, I'm pretty sure that would still be enough to power complex animals; they just might need to eat more / breathe more / be more lazy than equivalent Earth animals. Even if you only get one fifth the power density (based on the Bains paper's estimate that hydrogenic photosynthesis can proceed with 20% of the energy required by oxygenic photosynthesis), that's still 7-8 alien-equivalent-of-ATPs per glucose, as opposed to the 2 ATPs per glucose provided by anaerobic respiration (cf. 38 ATPs-per-glucose for aerobic respiration).

As Andrew noted, however, land animals will have an extra energy supply available in the form of reducing N2 into ammonia--i.e., they can supplement their energy supply simply by breathing, and only really need to eat for the materials. Incidentally, autotrophs could also make use of excess ammonia production to power chemosynthesis in the absence of light.

Despite the relatively low abundance of ammonia in the atmosphere, however, it has much higher solubility in water than hydrogen does, and thus may be more easily available to aquatic organisms than free hydrogen is. That gives you a slightly modified "respiration" reaction that takes 3 formaldehyde units and 4 ammonia molecules to produce methane, water, and free nitrogen. Decomposing ammonia into hydrogen and nitrogen is endothermic, however, so this would be less energetic than hydrogen respiration. Not sure how well that would be balanced by the reduced energy needs for pumping water over the gills compared to processing enough water to get the equivalent quantity of free hydrogen. This is a Good Thing for the long-term stability of the atmosphere, as otherwise you'd expect life to just keep producing ammonia until either nitrogen or hydrogen is exhausted.

Now, that is not the most energetic set of reactions imaginable. Obviously, if they had access to oxygen, that would be way better. But, it's the only option that is consistently renewable. Other potential oxidizers, which are more oxidized than the plant and animal biomass that serves as food in this scenario, won't be produced by plant life (by definition--anything that is produced by plantlife is biomass available as food, and thus not more oxidized than itself!) and thus can only be regenerated in small quantities by geochemical processes, and will have very short environmental half-lives. E.g., if creature could eat lithium metal, they could react that with nitrogen, ammonia, water, or complex organic biomolecules and get energy out of any of those reactions--but you just aren't going to find large quantities of free lithium to power your biosphere!

For some back-of-the-envelope estimated specifics, let's initially suppose that plants are hydrogenically producing glucose and animals are hydrogenating glucose. That leads to the following metabolic reaction:

${C_6H_{12}O_6}^{174 kJ} + 12{H_2}^{0 kJ} \rightarrow 6{CH_4}^{-74.85 kJ} + 6H_2O^{-285.83 kJ}$

Which provides approximately 2338 kJ/mol of glucose, based on the heats of formation for each reaction component as listed in the NIST Chemistry WebBook; the net energetics for aerobic respiration are found just by replacing methane with ${CO_2}^{-393.51kJ}$ in that equation (diatomic oxygen, like diatomic hydrogen, conventionally has a heat of formation of 0, so that doesn't make any difference to the calculation), which gives us a theoretical output of 4250kJ/mol. (Looking up the actual empirical energy yield of aerobic respiration of glucose, we get 2880 kJ/mol--so, assume that all other energy calculations here are only good to within about a power of two!) So, if hydrogen-breathing animals can just manage to evolve a slightly more efficient energy system than we have (which seems plausible, especially if they are living at lower temperatures which permit better thermodynamic efficiency), they may even get up to a full half the power density of oxygen-breathing Earthlings. Given the seriously speculative and highly imprecise nature of these calculations, though, I'm more comfortable sticking with my earlier 1/3 estimate as a plausible ceiling.

Plants in a hydrogen-dominated atmosphere aren't necessarily going to be producing as much glucose as our plants do in our highly oxidized environment, so let's look at some other options for biomolecules that might be available for animals to eat. Suppose, for example, that plants produce acetylene as an energy storage molecule; hydrogenating that provides

$C_2H_2^{226.73 kJ} + 3H_2^{0 kJ} \rightarrow 2CH_4^{-74.85 kJ}$

376 kJ/mol of acetylene -- or about 14.5 kJ per gram. About half as much per gram as glucose.

And if we look at the above-mentioned ammonosynthesis reaction, using hydrogen and nitrogen directly from the atmosphere, we get

$3H_2 + N_2 \rightarrow 2{NH_3}^{-45.9 kJ}$

about 91 kJ/mol of nitrogen, or 5.3 kJ/gram of inhaled hydrogen/nitrogen mix. That's not a ton compared to glucose (or acetylene), but there's just so much more of it available, for cheap! After all, it takes quite a lot energy just to consume and digest food--and practically none to just breathe it in.

Also consider that, largely because of the above reaction, nitrogen is far more easily bioavailable in this world than on Earth, so maybe they use a lot more cyanide groups or even azides. Hydrogenating hydrogen cyanide gets you

${HCN}^{135.14 kJ} + 3{H_2}^{0 kJ} \rightarrow {CH_4}^{-74.85 kJ} + {NH_3}^{-45.9 kJ}$

255.75 kJ/mol, or about 9.5 kJ / gram.

Simply decomposing hydrogen azide, even before you consider the energy gains from hydrogenation, gets you

$2{HN_3}^{264 kJ} \rightarrow {H_2}^{0 kJ} + 3{N_2}^{0 kJ}$

about 6 kJ/gram.

Those last few reactions (somewhat surprisingly, given the notorious explosivity of azides) would tend to drive the average energy density of food down a smidge from the optimistic third-to-half estimates, so once again about a quarter of the total energy density of aerobically-metabolized Earthling food seems reasonable.


Are there atoms with different numbers of protons, but the same number of electrons, with different orbital occupations?

Yes, the number of protons in the nucleus does affect how the electrons are filled into orbitals. There are a number of real-world examples that usually stem from transition metals since they do not only have a rather rich oxidation state chemistry but also two energy levels (3d and 4s) which are rather close by each other.

To give you a concrete example, take $\ce{Fe^2+}$ and $\ce{Cr^{\pm 0}}$. They both have the same number of electrons. However, chromium’s ground state electronic configuration is given as $\mathrm{3d^5 4s^1}$, while iron(II) typically has $\mathrm{3d^6 4s^0}$.

The general explanation is that the higher nuclear charge in the iron nucleus attracts the electrons better than the lower nuclear charge in chromium. This difference causes the orbitals to contract. The d-orbitals which are closer to the nucleus, are contracted more and thus experience more stabilisation than the s-orbitals of the next shell.


Are polyhydroquinones considered to be polyphenols?

The WBBSH definition states that polyphenols are defined as:

generally moderately water-soluble compounds
with molecular weight of 500–4000 Da
with >12 phenolic hydroxyl groups
with 5–7 aromatic rings per 1000 Da

This is the most widely accepted definition. However, due to the increase in research, there is a proposed Quideau definition stating

“The term "polyphenol" should be used to define compounds exclusively derived from the shikimate/phenylpropanoid and/or the polyketide pathway, featuring more than one phenolic unit and deprived of nitrogen-based functions.”

However, this is not an official definition (yet)