Which d orbitals are involved in "d3s hybridisation"?

First off, I'll reiterate what's mentioned in the comments - that hybrid orbitals are not a particularly good approach for transition metal complexes. But with that said, if you are going to apply this approach to a tetrahedral transition metal complex, here's how it works. For simplicity, I'm going to assume that only sigma bonding is relevent. You can do the same analysis including pi bonding, but it's more complicated.
Short answer: only the off-axis d orbitals have the right symmetry to match the ligands in a tetrahedral point group.
Long answer: The mathematical basis for constructing both delocalized and localized hybrid bond orbitals comes from group theory. If you haven't gotten to group theory in your studies, this explanation might not make a lot of sense, but at least you'll have to starting point to look deeper.
A tetrahedral molecule with four identical ligands is said to be in point group Td, which is just an abbreviation of "tetrahedral". Every point group has symmetries associated with it based on what operations can be performed that return the original structure. In point group Td, these are named A1, A2, E, T1 and T2.
When we build molecular orbitals, we combine the orbitals on the four ligands and look for possible symmetry groups. Then we look at the symmetry of the orbitals on the metal center and match the metal orbitals and "ligand group orbitals" based on symmetry. (An alternative approach is to look at the symmetry transformation of the bond vectors, which is described here: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/01%3A_Chapters/1.09%3A_Td_Molecular_Orbitals. The result is the same.)
Making a long story short, the tetrahedral ligand group orbitals for a sigma-bonded complex fall into two symmetry groups. One orbital, in which all four ligand orbitals have the same phase, is A1. A second group of three orbitals in which the ligand orbitals are pairwise in opposite phases matches the T2 symmetry of the Td group.
On the metal center, our valence atomic orbitals are 4s, 3d and 4p. The 4s orbital matches the A1 symmetry, while the 4p and 3d (xy, yz and xz) match the T2 symmetry. The 3d(x2-y2) and 3d(z2) orbitals, on the contrary, match the E symmetry from group Td. Since the E symmetry doesn't match either of our ligand group orbitals, we can't use these two orbitals for bonding.
Since the 4s orbital is the only match for the A1 symmetry, we must involve that orbital in bonding. For the T2 symmetry, however, we have two choices. We can either use the three 4p orbitals and ultimately make four sp3 hybrids, or we can use the three T2 symmetry 3d orbitals and end up with d3s hybrids. The sp3 hybridization is much more familiar to most people because they have learned it for organic compounds such as methane, but that's not the right choice here. If we consider the relative energy levels of the orbitals on a transition metal, we see that 3d is lower than 4p and so is closer to the ligand orbitals (which are lower in energy than both). Furthermore, the off-axis d orbitals are pointed more directly at the ligands in a tetrahedral complex (which are also off-axis), so this intuitively makes more sense as well.
Finally, to your question about $\ce{CrCl2O2}$, that molecule belongs to point group C2v, so you can apply the analysis above to that point group to get the allowable hybridizations in the same manner as for the Td case.


Dissociation Energy for CO2 per molecule

$$\frac{44.01 \ \mathrm{grams}}{6.022 \times 10^{23}~\mathrm{molecules}} = 7.308 \times 10^{-23}\mathrm{\frac{grams}{molecule}}$$

$$\mathrm{\frac{joules}{molecule}} = \mathrm{\frac{joules}{gram} \cdot  \frac{grams}{molecule}} = (373.6 \times 10^{3}) \cdot (7.308 \times 10^{-23}) = 2.7303 \times 10^{-17} \mathrm{\frac{joules}{molecule}}$$

or,

$170.43\,\, \mathrm{eV/molecule}$


Is it possible to achieve a level of "truly zero" concentration?

It depends how you dilute it. If you take an aqueous solution of A and just add pure water (absolutely 100% water), the concentration of A will never quite be null. In this case however, you will reach a point where the concentration of A is so small that it can be considered null for your applications.
If, however, you dilute the solution, take a sample, then dilute that sample (and so on), you could reach a concentration of exactly 0M. Imagine you have diluted the solution enough so that it contains exactly 1 molecule of A. When you take your sample for the next dilution, if this molecule isn't in the sample, the concentration will be exactly null. If it does happen to be in the sample, it could be left behind when you draw the next sample, or the next, and so on.
In practice though, the water you use for the dilution will likely contain impurities. You will maybe not achieve exactly 0M, but the concentration could be so small that it is undetectable and have no measurable consequence.


Why does oxidation of the electrode take place in a solution?

Sorry to be so catty but the overall line of questioning indicates a profound lack of understanding. I can't explain all about of redox reactions in a few paragraphs to let me try to answer some of the questions in a general way and hope that those answers will provide some focus.

  What makes the electrode atoms start to ionize? Perhaps because of the solutes in the solvent?

Every redox reaction has both a reduction reaction and an oxidation reaction.

I'm going to assume two electrodes in the solution. Thus in a redox reaction one electrode will have an oxidation reaction and the other will have a reduction reaction. The two electrodes will be connected via a wire through which electrons flow. Without the wire no reaction will occur.

  So probably if you place it in distilled water it will not ionize, right?

There are two aspects to the answer to this question.

First distilled water doesn't conduct well. So even with our two electrodes in the distilled water we wouldn't see much current flow through the wire connecting the two electrodes. In this case it isn't the wire limiting current flow but the counter flow of ions through the distilled water.

Second chemists chose "electrodes" that will be inert in the solution being used. For example if a sodium metal "electrode" was dipped into water then a violent reaction would occur spontaneously on the contact surface between the electrode and the water. The products would be hydrogen gas and sodium hydroxide. So a sodium metal electrode in an aqueous solution just won't work.

  Follow up question: If the solutes in the water are the reason of ionizing the electrode, Do they rip off the electrons from the electrodes atoms? (Most of websites illustrate that both electrons (on the electrode) and ions (in solution) contribute to the voltage of the electrode).

An overall redox reaction can be written as two half-cell reactions. The gist is that we can use electrons to balance the formula. This makes it easy to balance an overall redox reaction since we just have to balance the electrons.

So for an oxidation half-cell for a zinc electrode we have:
$\ce{Zn -> Zn^{2+} + 2e^-}$

For a reduction half-cell for a copper electrode we have
$\ce{Cu^{2+} + 2e^- -> Cu}$

Since we have two electrons in both half cells we can just add the equations as is to get:

$\ce{Cu^{2+} + Zn -> Cu + Zn^{2+}}$

Now there is a quirk in this particular reaction. If the copper ions in solution, which are to be reduced, come in contact with the Zn electrode then the copper ions would plate on the Zn electrode. So a membrane of some sort must be used as a barrier which keeps the copper ions away from the Zn electrode but still allows anions to flow through the barrier. Of course if we used ordinary glass as the barrier then neither anions nor cations could flow through the barrier and the battery formed by the two half-cells wouldn't work.

  So why do I use electrode potentials if I have lets say a bowl made purely of Zn atoms and it contains lets say Copper ions. Even though I only have positive ions of copper. Does the copper ions generate the same voltage by their own without electrons?

Well a bowl can't be pure Zn and have Cu too.

So let me go to a hypothetical situation. I haven't checked Cu/Zn alloys (which would be possible but complicates the general idea here). So let's just assume a Cu/Zn alloy. When metals or alloys become solid they tend to form grains which are interlocked together by their 3D surfaces rather than one homogeneous material. The grains will vary in chemical composition. So the different grains have different electromotive potentials because of their compositional differences.  Thus corrosion will occur. Some grains will dissolve and other grains will plate the dissolved ions back to metallic atoms.

Now let me clarify that corrosion is a very complicated subject for which I have made a gross generality. The actual corrosion reactions will depend on the metals in contact, and the exact composition of the solution with which they are in contact.


Good sources to understand Stat Mech with concepts

I highly recommend Molecular Driving Forces by Dill & Bromberg.

https://books.google.com/books/about/Molecular_Driving_Forces.html?id=hdeODhjp1bUC

Contains many examples throughout to help with conceptual and mathematical understanding.


what wavelength should I use to measure the transmittance of iodine colouring?

You know that in first approximation the absorption recorded is a product of the optical path length $d$, the extinction coefficient $\varepsilon$, and the analyte concentration $c$ as $$\mathrm{abs} = c \cdot \varepsilon \cdot d$$

Now, depending on your specific task, consider the three variables to your advantage.  Because for the wavelength eventually elected to record your data, you want to stay in a linear relationship between analyte concentration and detected absorption as the read-out.

The pathlength $d$ is something typically fixed to $1\,\pu{cm}$, but be aware that there are optical cells offering a shorter path length (e.g., $5\,\pu{mm}$, $1\,\pu{mm}$ as here or even $0.1\,\pu{mm}$) or longer paths (e.g., $50\,\pu{mm}$ (example), or $100\,\pu{mm}$.  It depends on the design of the spectrometer if mounts are interchangeable, or not, to benefit from an intended adjustment to your needs here.
What is the range of concentrations $c$ to record?  As the extinction coefficient depends on the wavelength of recording, it may be beneficial to record at a wavelength maximizing $\varepsilon$ of your amylose iodine complex, provided you want to detect low concentrations of this specis.  (Of couse it shouldn't be in the range other molecules of your solutions absorb intensely.)

But: if you want to determine solutions with a high concentration of your analyte, than it makes sense to record the absorption at a wavelength where $\varepsilon$ of your analyte just is high (not maximal), while «all the rest» of the sample solution does not contribute significantly to the absorption.  This approach will guard you to encounter absorptions too high to be sensible for detectors in the spectrometer.  Look up the manual of the spectrometer or ask colleagues already using the instrument; for some, an absorption greater than $2.0$ equates to «skip it, there is no light on the detector at all».

The iodine-amylose complex is known to absorb intensely.  This equally may justify to record $\mathrm{abs}$ in function of analyte concentration $c$ a little «off» from $\varepsilon_\mathrm{max}$.  By recording the spectra of your blank, and of a typical solution with your complex to quantify, you may justify how many nanometers above or below $\lambda$ yielding $\varepsilon_\mathrm{max}$ of your analyte is fine.  If time and resouces permit, you may record at multiple wavelengths, too; eventually, the readings should lead to the same concentrations per sample.


Is there any difference between balancing redox reactions in acidic and basic medium?

The five points to pay attention are correct; and I would like to add a sixth one: depending on the reaction condition, namely if the reaction occurs in acidic, neutral, or basic solution, the outcome and the reaction product(s) of the half reaction may differ.

Take potassium permanganate, for example.  Under aqueous acidic conditions, this is a very powerful oxidiser, and formally $\ce{Mn^{7+}}$ is reduced all the way down to $\ce{Mn^{2+}}$.  This half reaction, starting from purple coloured solution of $\ce{KMnO4}$ to practical colourless $\ce{MnSO4}$ (assuming $\ce{H2SO4}$ as the acid present) may be written as

$$ \ce{MnO4^-  + 5e^- + 8H^+ ->  Mn^{2+} + 4H2O} $$

Under neutral conditions, the product however is manganese dioxide (formally $\ce{Mn^{4+}}$), which is poorly soluble in water and precipitates

$$ \ce{MnO4^-  + 3e^-  + 2H2O  ->  MnO2 + 4OH^-} $$

Eventually under (very) basic conditions, the reaction will stop already at a green colour in form of oxidation state $\ce{Mn^{6+}}$:

$$ \ce{MnO4^-  + e^- ->   MnO4^{2-}} $$

Paying attention to this has practical, preparative relevance, too.  It offers one way to attune the oxidation power of potassium permanganate.  A moderated reactivity often correlates with selectivity of the reactions.

This pH-dependence of oxidation-reduction reactions correlates with electrode potentials which often are already tabulated, like here or in the CRC Handbook of Chemical and Physics (keyword Nernst equation).

As a side note: These equations sometimes do not tell the complete story.  $\ce{MnO2}$, for example, may act / may be used itself as oxidiser. The simple sum formula however hides well that the oxidising power of this material depends how this material was prepared, too.


Term used to describe scalar coupling over a single bond?

I've never heard a special term but you can describe it as $^1 J$ or $^1 J_{CF}$ coupling using the systematic nomenclature.


What is the correct expression of pressure-volume work?

At the inside piston face, the pressure exerted by the gas on the piston is equal to the pressure exerted by the piston on the gas. This just follows from Newton's 3rd law of motion.  But, in an irreversible process, the pressure exerted by the gas on the piston face cannot be obtained from the ideal gas equation (or other real-gas equation of state).  This is because the equation of state applies only at thermodynamic equilibrium.  Otherwise, the gas pressure on the piston involves viscous stresses which depend not only on the volume but also on the rate of change of gas volume.  For a reversible process, the gas pressure throughout the gas, including at the inside piston face can be determined using the equation of state, since a reversible process consists of a continuous sequence of thermodynamic equilibrium states.
In using these work relationships, it is important to specify what you are calling the system. If the system includes only the gas, and the piston is massless and frictionless, the pressure on the outside piston face is equal to the pressure on the inside piston face (and the gas pressure).  If the piston is included as part of the system, and the piston is not massless or frictionless, then the external pressure is the pressure on the outside piston face, but this is not equal to the gas pressure on the inside piston face (or, equivalently, the piston pressure on the gas).


How can iodine oxidise hydrogen peroxide?

Harsh Jain, It may be a correct reaction, but it is a very complicated one.

Let us see what happens when iodine is added to a large volume of an aqueous base in the absence of $\ce{H2O2}.$ It forms iodide and iodate ion. If iodine is in excess, triiodide ion is also formed.

The complication arises from adding hydrogen peroxide because it can act as an oxidizing agent as well as a reducing agent. Similarly, iodate ion and iodide ion can react with each other as well as $\ce{H2O2}.$ For example, iodine can react with $\ce{H2O2}$ and form iodide ions.
On the other hand, iodide ion can be oxidized by $\ce{H2O2}$ to free iodine.

I would suggest that you read a paper by Bray and Liebhafsky [1].
Here is the summary by the authors in their table:

References

Bray, W. C.; Liebhafsky, H. A. Reactions Involving Hydrogen Peroxide, Iodine and Iodate Ion. I. Introduction. J. Am. Chem. Soc. 1931, 53 (1), 38–44. DOI: 10.1021/ja01352a006.


without using electronegativity, which one of these molecules is polar?

When we say something is polar, it has a very specific meaning. It means that there is a spatial separation of charges, or a dipole moment. In the simplest case, this is given by:

$$
\mu = Qr
$$

Where $\mu$ is the dipole moment, $Q$ is the charge on each pole (equal and opposite) and $r$ is the distance between the charges.

Anything with charges can have a dipole moment, and therefore be polar. In chemistry, we typically worry about whether or not two things are polar:

Bonds
Molecules

Bond Polarity

Chemists are concerned with the polarity of bonds because polarity affects the character of the bond - the more polar the bond is, the more it behaves like an ionic bond. The less polar, the more it behaves like a covalent bond. This has implications for everything from naming rules to reactivity, but all you need to know for this problem is:

  All bonds between two atoms are polar, unless the atoms are identical.

(This is including whatever else the atoms may be bonded to). The only question after that is "how polar is the bond," and that is where tables of electronegativity come in. Using an arbitrarily defined scale, we can measure the relative polarity of bonds by comparing the difference in electronegativity between the two elements.

Molecule Polarity

For molecules, polarity isn't quite so simple. We know that all bonds between dissimilar atoms are polar, but in a molecule, sometimes the dipole moments add up to form a net dipole moment of zero. This is a little bit difficult to explain if you don't have a mathematics and physics background that includes vectors and summations of forces/moments.

If you do, then when I say something like:

  The net dipole moment for a molecule is equal to the sum of dipole moments over each bond.

It will make sense to you. If you don't, then you have to get a little more creative. I tell my students to imagine that there are ropes connecting each outer molecule to the central molecule, along each bond. Then imagine pulling each rope towards the side that is more electronegative (has a higher electron density). The more polar the bond, the harder you pull on the rope. Then imagine whether the molecule moves. If the forces balance, it stays put, so the net dipole moment is zero and it is not polar. If it does move, there was a net dipole moment, so it is polar.

This works pretty well - as long as you can visualize the molecular geometry. That's the hard part. To know how the bonds are oriented in space, you have to have a strong grasp of Lewis structures and VSEPR theory. Assuming you do, you can look at the structure of each one and decide if it is polar or not - whether or not you know the individual atom electronegativity. This is because you know that all bonds between dissimilar elements are polar, and in these particular examples, it doesn't matter which direction the dipole moment vectors are pointing (out or in).

Let's look at each one, using wikipedia's geometric pictures.

$\ce{BrF5}$

As you can see, there is a lone pair at the bottom (this is a square pyramidal geometry). The net dipole moment will be pointing "up", which makes this a polar molecule.

$\ce{PBr5}$

This one has a trigonal bipyramidal geometry, with each bond symmetrically opposed to each of the others. If you add up the vectors, they result in a net dipole moment of zero. Therefore, it is non-polar.

EDIT: As ron points out in the comments below, this is not really a binary molecular compound, and in reality forms an ionic crystal structure with $\ce{PBr4+}$ and $\ce{Br-}$. I don't think the author of the question intended you to worry about that, but if they did, the answer is still "non polar," although we wouldn't really call it a "molecule" any more.

$\ce{CCl4}$

Carbon tetrachloride has a tetrahedral geometry, and all the dipole moment vectors cancel. Therefore, it is non-polar.

$\ce{XeF2}$

This one is linear - no net dipole moment, so it is non-polar.

$\ce{XeF4}$

I could only find the ball-and-stick model for this one. It is square planar - there is a lone pair of electrons (not shown) on each "face" of the square on Xe. There is no net dipole moment, so it is also non-polar.

Out of all of these, the only polar molecule is $\ce{BrF5}$ (the first one).

To review the steps:

Draw the Lewis structure
Figure out the geometry (using VSEPR theory)
Visualize or draw the geometry
Find the net dipole moment (you don't have to actually do calculations if you can visualize it)
If the net dipole moment is zero, it is non-polar. Otherwise, it is polar.


Why is pyridine so much more acidic than other similar imines?

When we discuss the acidity of a compound, we are really discussing where the equilibrium between the acid and its conjugate base lies.  Therefore, when we compare acidities we are really comparing equilibria, the relative stabilities of the reactant (acid) and product (conjugate base).

The following diagram illustrates the equilibrium between the iminium ion and the corresponding imine, it's conjugate base.

We can't draw any good resonance structures for the imine, but we can draw 2 resonance structures to describe the iminium ion, A and B in the figure. In the pyridinium ion - pyridine case we can draw 2 resonance structures for pyridine (just like in benzene), but we can't draw any resonance structures for the pyridinium ion that delocalize the positive charge without destroying the aromaticity of the ring.

So, in the iminium ion - imine case

we can stabilize the iminium ion through resonance - this stabilizes the iminium ion and shifts the equilibrium to the left making the iminium ion less acidic.

Whereas in the pyridinium ion - pyridine case

we can't draw good resonance structures that stabilize the positive charge in the pyridinium ion - this makes pyridinium ion more acidic.

Both factors operate in the same direction and tend to make the pyridinium ion ($\mathrm{p}K_\mathrm{a}=5.5$) more acidic than the iminium ion ($\mathrm{p}K_\mathrm{a}$ $\sim$ 9).


How can I measure carcinogenic risks of an M-ENDO-Agar LES substrate after heat destruction?

Most assays of carcinogenicity are designed to measure DNA damage. The 2 most common of those assays are the Ames test and the Comet Assay.

The Ames test uses a special set of strains of bacteria that have known phenotypes. You expose the bacteria to the suspected carcinogen and then look for changes in phenotype. Different strains will be susceptible to different types of mutations, deletions, insertions, frameshifts, etc.

The comet assay uses mammalian cell lines. You have to suspend the cells in an agar and break them up. This will create little pockets of cell debris in the agar. You fluorescently stain the DNA and run an electric current through it. If the DNA is damaged and broken up, it will migrate through the gel, creating tails, hence the "comet". If the DNA is intact, it will not migrate.

I've never done either assay, just heard about them in class. Wikipedia has more detail than I do.

http://en.wikipedia.org/wiki/Ames_test
http://en.wikipedia.org/wiki/Comet_assay


The rule of swapping the valencies of the atoms to get their ratios only works for a very specific set of compounds and is more of a convenient trick than a rule based on any sort of hard theory.

In particular, it works for binary 'ionic' compounds such as $\ce{NaCl}$, $\ce{CuCl2}$, $\ce{Al2O3}$ etc. because the valencies of the atoms correspond to the charges on the ions and the net charge has to be zero.

However, it obviously doesn't work for a huge range of other compounds where there is more complicated bonding. For example there are a whole range of covalent nitrogen oxides of varying degrees of stability which do not fit the rule ($\ce{N2O}$, $\ce{NO}$, $\ce{NO2}$, $\ce{N2O3}$, $\ce{N2O4}$ etc).

Ultimately the only way to determine the formula of any arbitrary substance is through experiment, typically with a mass spectrometer.

If you are interested in finding out more about the different types of substances that are possible I suggest that you start reading about different types of bonding, in particular ionic versus covalent, to begin to understand the different types of compounds that exist.


Electron Pair Geometry in Ibuprofen (Electron Pair Geometry in Organic Compounds)

You are right! There is no molecular geometry in any chart that has two lone pairs and one bond, because the only possible geometry with one neighboring atom is linear.
The central atom in a chemical moiety must have at least two neighboring atoms in order to predict the geometry around this central atom (according to VSEPR model).
Let's take the case of oxygen atom of the hydroxyl group (in Ibuprofen). The oxygen atom here, as a central atom,  has the steric number: 4 ( It has two lone pairs and two simple bonds with carbon and hydrogen atoms). So, The molecular geometry around oxygen atom is bent.


Crown ethers are typically not recycled because they remain in the organic phase where normally also all the byproducts are (so the purification is very complicated) and they are still complexing the alkali metal after the reaction.
You want to recycle them to again complex some alkali metal, but then you have first to break down the previous complex, which is not straightforward.

To recycle crown ethers you must use a different approach.
The trick is to immobilize them on some solid support.
Typically, you have some crown ether functionalized monomer and polymerize it, or you have some reactive polymer and introduce the crown ether at the end.
A review can be found here.

These solid phase crown ethers are used in the same way as the better known ion exchange resins.
The problem is that you will have to synthesize them, because they are not readily available for purchase.


If germanium is supposed to be bigger than aluminum, in fact, its empirical values correspond to $\pu{125pm}$ for each one respectively, but germanium being lower in the periodic table is bigger at the same time which is why we say that $\ce{Al < Ga}$. Now yes, the atomic radius of silicon is $\pu{118pm}$ while as we said before, the atomic radius of germanium is $\pu{125pm}$, in the same way the size is deduced based on its position in the periodic table, so $\ce{Si < Al < Ga}$.
This occurs because there is a relationship between the effective nucleus charge and the electronic levels of the atom, since the positive nucleus attracts the electrons, but in turn, the electrons repel each other (except those of the first level), it is by This means that the forces and counter forces end up having an effect on the valence electrons, and these determine the atomic radius of the atom. This difference of attraction by the nucleus and repulsion by the electrons increases as there are more electronic levels, that is, if I have a compound with Z = 11 and e = 11 that we will call "A", it will be greater than a compound "B" whose Z = 10 e = 10, leaving B <A since although both have the same number of electrons as protons, compound A has an electronic configuration $\ce{1s^2 2s^2 2p^6 3s^1}$ with 11 protons while compound B has an electronic configuration of $\ce{1s^2 2s^2 2p^6}$ with 10 protons, that is, compound A has a higher electronic level than compound B, which gives a larger atomic radius.
Reference
1.https://es.wikipedia.org/wiki/Anexo:Radio_at%C3%B3mico_de_los_elementos_qu%C3%ADmicos
Ah, I just realized that it was Galio and not Germanio, sorry.


Does the Fermi level correspond to the highest occupied energy level?

Be aware of Fermi level versus Fermi energy, as it seems to me the book speaks about the latter.
The quotes below are from both Wikipedia links.

In band structure theory, used in solid state physics to analyze the energy levels in a solid, the Fermi level can be considered to be a hypothetical energy level of an electron, such that at thermodynamic equilibrium this energy level would have a 50% probability of being occupied at any given time.

The Fermi energy is a concept in quantum mechanics usually referring to the energy difference between the highest and lowest occupied single-particle states in a quantum system of non-interacting fermions at absolute zero temperature. In a Fermi gas, the lowest occupied state is taken to have zero kinetic energy, whereas in a metal, the lowest occupied state is typically taken to mean the bottom of the conduction band.

The Fermi energy is only defined at absolute zero, while the Fermi level is defined for any temperature.

The Fermi energy is an energy difference (usually corresponding to a kinetic energy), whereas the Fermi level is a total energy level including kinetic energy and potential energy.

The Fermi energy can only be defined for non-interacting fermions (where the potential energy or band edge is a static, well defined quantity), whereas the Fermi level remains well defined even in complex interacting systems, at thermodynamic equilibrium.

Since the Fermi level in a metal at absolute zero is the energy of the highest occupied single particle state, then the Fermi energy in a metal is the energy difference between the Fermi level and lowest unoccupied single-particle state, at zero-temperature.

A collection of non interacting fermions, often called the Fermi gas is a simplified model useful in the solid state physics. There is supposed the considered fermions do not interact, so the energies of these fermion quantum states do not depend on the distribution of  these fermion quantum states.
A typical case would be the classical hydrogen atom model, applied on multielectron atoms. As electrons in the same atom interact quite a lot, it is not a good model.
For solid states, electrons and electron vacancies are often taken as  quasi fermions following Fermi-Dirac statistics for the Fermi gas,
as they are separated enough to be considered as non interacting.


Why is there only a slight change in pH when soda loses its fizz?

In addition to fizz (carbonic acid), Coca-Cola contains phosphoric acid (for an additional tangy taste). According to this leaflet put out by Coca-Cola, a 250 mL can of Coca-Cola contains 43 mg of phosphorus. With some quick calculation, you can find that this translates to $\ce{5.5538 \times 10^{-3} M}$ $\ce{H3PO4}$, which would have a pH of approximately 2.44. Pretty acidic! The other sodas are sure to contain other acids as well.


Do electrons have some volume, area or shape?

Quoting from the Nobel lecture of Hans G. Dehmelt (1989):

  With the rise of Dirac’s
  theory of the electron in the late twenties their size shrunk to mathematically
  zero. Everybody “knew” then that electron and proton were indivisible
  Dirac point particles with radius R = 0 and gyromagnetic ratio g = 2.00. The
  first hint of cuttability or at least compositeness of the proton came from
  Stern’s 1933 measurement of proton magnetism in a Stern-Gerlach molecular
  beam apparatus. However this was not realized at the time. He found for
  its normalized dimensionless gyromagnetic ratio not g = 2

...

  Similar later work at still higher energies found 3 quarks
  inside the “indivisible” proton. Today everybody “knows” the electron is an
  indivisible atomon, a Dirac point particle with radius R = 0 and g = 2.00....
  But is it? Like the proton, it could be a composite object. History may well
  repeat itself.

So there is no evidence that the electron is anything but a point particle, yet a non-zero size has not been ruled out either.

New Measurement of the Electron Magnetic Moment Using a One-Electron Quantum Cyclotron Phys. Rev. Lett. 97, 030801 (2006) (more-official link) shows that any radius of the electron is less than $10^{-18}$ meters.

See also the related webpage Limit on Electron Substructure


How does NMR spectrometer determine area under each peak?

The modern basic FT 1H NMR experiment involves irradiating and simultaneously exciting a wide bandwidth of proton frequencies. That is to say, all spin populations of the proton nuclei have their spin state reversed with a carefully calibrated Rf pulse that delivers the correct amount of power for the correct duration. As each spin relaxes back to equilibrium, it emits a photon of Rf energy. So, yes your assumption is pretty well correct. This energy packet has a specific frequency (a product of the chemical shift), and a specific intensity - it is a quantum packet of energy. If two nuclei of the same frequency each emit a photon of energy, the intensity of that frequency detected will be twice as large a single nucleus emitting a photon of energy.

Across the entire frequency range, the data collected in the free induction decay is the sum product of all frequencies,  and their intensities. The Fourier transform function then converts this time domain data into the preferred frequency domain spectrum.


Effective nuclear charge on an inner electron

Firstly, effective nuclear charge is just a model for predicting trends in atomic properties. The actual solution of the quantum mechanical equations does use the full nuclear charge and explicit calculation of the all electron-electron repulsions (and electron-nuclear attraction).

The point of $Z_\text{eff}$ is to approximate the value of the nuclear charge felt by electrons at a given distance. The answer to your question then is Gauss's Law,  which states that, to determine the field at a given point, you consider all and only the charge inside a certain surface (on which your point sits). Moreover, it states that the field there is equivalent to if all of the charge inside that surface was a point charge at the center.

So if we apply Gauss's Law to our atom, we add up all the charge inside a sphere centered on the nucleus ($Z-\text{effective number of core electrons}$), and regard the system as an effective nucleus with that charge, no core electrons and the electron of interest being the "first" electron and so we ignore the outer electrons.

Here are some physics discussions for why, intuitively, Gauss's Law allows you to ignore outside electrons from the physics stackexchange.


Hartree-Fock vs. DFT vs. Hohenberg-Kohn vs. Kohn-Sham

The Kohn-Sham approach is one on top of the original DFT idea proposed by Hohenberg-Kohn.
They propose – to find the energy functional and its equations, which, as you said, are unknown – to start with what is known: the functional if there is no interactions. So that is what they do: approximate the kinetic energy and Slater determinant from the known, non-interacting system and add all the terms that are additionally known (i.e. Coulomb, kinetic energies and the electron/nucleus potential) until you are left with the Kohn-Sham equation, which has "the smallest unknown", the exchange & correlation potential.
I recommend to read more in a book or on Wikipedia in case you want a more detailed picture on any/all of the methods.


Determine standard enthalpy of formation of salt given other reactions

The standard enthalpy of formation $\Delta H_\mathrm f$ refers to the formation of 1 mole of product, with the educts and product being in their respective standard states. The fourth equation is therefore correct, as it gives 1 mole $\ce{KNO3}$ as the product. Equation 3 is equation 4 multiplied by 2, which would give $2\Delta H_\mathrm f$. The first two equations can be ruled out because they do not contain the educts nitrogen and/or oxygen in their standard states, which are $\ce{N2}$ and $\ce{O2}$, respectively.


Change in internal energy for the heating and vaporisation of water

The easiest way to do this problem is to first determine the  change in enthalpy for this constant pressure process.  Do you think you can do that?  Then determine the change in PV between the initial and final states.


Coordination dilemma, change of energy related to ligand distance

The baricentre should probably be spelt barycentre. It is something analogous to a centre of mass. Simple models such as some flavours of the crystal field theory will assume a destabilisation only of the metal orbitals. However, for an energetic destabilisation to occur, something else must be stabilised — this is the conservation of energy. Thus, the logic says that the raise in energy of the $\mathrm{e_g}$ orbitals must be offset by a lowering of the $\mathrm{t_{2g}}$ orbitals.

When constructing a proper MO scheme of an octahedral coordination complex, this argument is revealed as bogus. As can be seen in this answer, the $\mathrm{e_g}$ orbitals are actually $\mathrm{e_g^*}$ orbitals, i.e. antibonding with respect to the $\ce{M\bond{<-}L}$ bonds which is the reason why they are raised. Disregarding π interactions, the $\mathrm{t_{2g}}$ orbitals are nonbonding and thus their energy is not affected.


Reducing pressure in a binary liquid solution

Consider a system with a variable volume (e.g., a syringe, but on a larger scale). As the volume increases, liquid becomes vapor without changing the amount of material present.

In a system where there is only one component, changing the volume of the container will change the fraction of material in the vapor vs the amount of material in the liquid. But as you concluded, the vapor pressure will remain constant (assume constant temperature, since it was not mentioned), because there is an equilibrium between liquid and vapor for the one material in each phase.

However, when there are two components of different volatility, the vapor composition varies, depending on how much vapor there is, compared to the fraction left as liquid. So the equilibrium is no longer between a vapor and a liquid of the same composition.

It might be easier to visualize the change going on in the liquid/vapor if you change the dimension of the vertical axis to read 1/Volume. Then the vapor over the solution is (intuitively) richer in A, always (again, intuitively), until the liquid is all gone, when the vapor attains the original composition. Then, BTW, you realize that the pressure was changing while you changed the volume, and pressure might be a more relevant parameter.


Finding (∂U/∂V )T = n^2a/V^2 for a van der Waals gas

The more general equation you are looking for is called the thermodynamic equation of state. You can start from the differential form of U according to the 1st + 2nd laws of thermodynamics for pV work only:

$$dU = TdS-pdV$$

Taking the derivative wrt V at constant T

$$\left(\frac{\partial U}{\partial V}\right)_T= T \left(\frac{\partial S}{\partial V}\right)_T-p$$

At this point you want to use the Maxwell relation

$$\left(\frac{\partial p}{\partial T}\right)_V= \left(\frac{\partial S}{\partial V}\right)_T$$

(which follows from the Helmholtz energy being a state function) so that

$$\left(\frac{\partial U}{\partial V}\right)_T= T \left(\frac{\partial p}{\partial T}\right)_V-p \tag{tes}$$

This last expression is the thermodynamic equation of state.

The final step is to evaluate $\left(\frac{\partial p}{\partial T}\right)_V$ for the van der Waals equation of state.


What is the amount of nitrate ions in 20 g of Fe(NO3)3?

The molar mass of iron (III) nitrate is $m(\ce{Fe(NO3)3}\approx 242~\mathrm{g\, mol^{-1}}$. Pay close attention to the unit.

You correctly calculated the amount of substance of iron (III) nitrate to be $n(\ce{Fe(NO3)3}= 0.0826~\mathrm{mol}$

Now you should ask yourself the question: How many nitrate ions are in one formula unit of iron (III) nitrate?

   There are three $\ce{NO3^-}$ per every $\ce{Fe(NO3)3}$.

What does that mean for the number of moles of nitrate ions?

   It means that $n(\ce{Fe(NO3)3} = \frac13\cdot n(\ce{NO3^-})$, so you have to multiply the number of moles by three. $n(\ce{NO3^-}) = 0.248~\mathrm{mol}$

Now you know the number of moles of nitrate ions and you simply have to multiply with Avogadro's constant.

   $N(\ce{NO3^-}) = n(\ce{NO3^-}) \cdot L = 1.49\cdot10^{23}$


Is there a possible effect of mechanical vibration on the stability of organic molecules?

What do you mean by "low frequency vibrations"?  The term "extemely low frequency" is used to describe earthquake type frequencies (3-30 Hz, give or take), whereas "low frequency" is used to describe frequencies in the 30kHz - 300 kHz range.  If you mean "low frequency" as defined in the previous sentence, then yes, that frequency range of vibration is known to have chemical effects.  The process involves the formation and collapse of bubbles (cavitation) in the solution.  High temperatures and pressures can be generated during the bubble formation and collapse process.  The typical lab apparatus can homogenize samples, produce catalytic effects by cleaning metal surfaces or increasing phase transfer, disrupt cells, and break some (weak?) chemical bonds.  This reference reports,"Ligand-metal bond cleavage in transition metal complexes to give coordinatively unsaturated species or modified complexes as well as complete strip off of ligands to produce amorphous metals disrupture of the solvent structure altering the solvation of reactants Sonolysis of molecules (homolytic fragmentation to radicals, rupture of polymers, generation of excited states, cell disrupture)."


First off this is SMARTS not SMILES.

Think of SMARTS as like a regular expression language for molecules (e.g., SMILES).

So let's break down some of those characters:

~ is "any bond"
! is "not"
$ indicates a recursive SMARTS expression.
@ indicates "any ring bond"

The $(*#*) means "two atoms with a triple bond between them". The !D1 means and not an atom with one explicit bond.

Taken as a whole, [*]~[!$(*#*)&!D1] means any atom with a bond to an atom that doesn't have just one explicit bond or a triple bond.

Basically, the pattern is trying to find a single bond in the middle with some rotatable side-groups:
[*]~[...]-&[..]~[*]

There's a very nice tool called SMARTS Viewer that can help to debug and construct SMARTS.

Seems like there are also now more polished commercial versions called SMARTStools through BioSolveIT including an editor.


Balancing chemical reactions, issues of Gauss's elimination method

Your "should-be" solution,

$$\ce{3H2O2 + 2NO3- + 2H+ → 3O2 + 2NO + 4H2O}$$

is a linear combination of

$$\ce{H2O2 + 2NO3- + 2H+ → 2O2 + 2NO + 2H2O}$$

and

$$\ce{2H2O2  → O2 +  2H2O}$$

Any other linear combination is valid as well.


Does water really 'go bad' after a couple of days?

First of all, it depends on how the tap water was treated before it was piped to your house. In most cases, the water was chlorinated to remove microorganisms. By the time the water arrives at your house, there is very little (if any) chlorine left in the water. When you fill you container, there is likely to be some microorganisms present (either in the container or in the water).
In a nutrient rich environment, you can see colonies within 3 days. For tap water, it will probably take 2 to 3 weeks. But that doesn't mean that the small amount of growth doesn't produce bad tasting compounds (acetic acid, urea, etc.).

BTW Nicolau Saker Neto, cold water dissolves more gas than hot water. Watch when you heat water on your stove. Before it boils, you will see gas bubbles that form on the bottom and go to the surface (dissolved gases) and bubbles that disappear while rising to the surface (water vapor).


What can this chemical be? (Qualitative Analysis)

What about a possible $\ce{F-}$ anion? While the nature of the cation might be guessed based on the flame test, since the $\ce{AgNO3}$ test gave no precipitate, it might have produced $\ce{AgF}$ (soluble in water).
This said, by looking at the densities of $\ce{M-F}$ systems with a cation with orange flame and density around $\pu{2 g ml-1}$, I'd say it's probably $\ce{NaF}$.
You might test for fluoride with a selective electrode, if you have one.
Otherwise, here are listed some other analytical methods for the identification of fluoride anions:PubChem: Sodium Fluoride-Analytical Methods


Ring currents of anti-aromatic and aromatic systems

Yes it has to do with the number of electrons. Consider a simple model of the electron on a ring, then the electron has to follow a circular path and its wavelength has to 'fit' exactly  round the ring. Thus the wavelength has to satisfy the condition $L=q\lambda$ where $q$ is a quantum number $q=0, \pm 1, \pm 2 ...$ characterising each eigenstate. The energy in this simple model comes to $E=(hq)^2/(2mL^2)$ ($h$ is the planck constant, $m$ the electron mass). Now in this model filling up the energy levels according to the aufbau principle, (Pauli principle & Hund's rules) leads to 2 electrons in $q=0$ (which is singly degenerate) and 2 in $q=+1$ , and $q=-1$ etc which are all doubly degenerate in energy.

Thus $4n+2 ~~ \pi$ electrons fill up all levels with two electrons in each level. The $4n~~ \pi$ contains only a single electron each. This is because it takes less energy to add one electron to each of the $q$ levels rather than pairing them up. The sketch shows this.

The unpaired electrons are shown in the $4n$ column,  however, the levels are also shown as being split in energy. This is due to the Jahn-Teller effect. This states that a slight interaction between degenerate levels can cause the degeneracy to be split and so lower the energy. The interaction could be due to slightly different bond lengths and the energy levels are split by only a small amount relative to the gap to other levels.

Although not shown in the diagram, (which is thus slightly inaccurate) both spins will occupy the same lower level so in this sense, [4n]annulenes do not really exhibit molecular paramagnetism. The electrons interact with the magnetic field $B_0$ and this leads to a mixing of levels and this causes a paramagnetic contribution to the shielding constant. This is larger than the diamagnetic contribution for  lower orbitals. Thus in [4n]annulenes we speak illustratively of a 'paramagnetic ring current' and molecules showing this effect are called paratropic and those with no ring current atropic.


Can there be active THC be trapped in the evaporator?

When burning a blunt, joint, doobie or whatever, THC evaporates (bp of tetrahydrocannabinoles is around 155 °C). Depending on

the length of the "object" , related to the temperature gradient between the front and the mouth piece, and
the type of the "filling" and its adsorptive properties

a part of the vaporized material might condense and/or get absorbed in a colder region of the filling. Similar effects, such as the enrichment of nicotine in the stump of cigarettes have examined.

It is however unlikely that any of the transported and reabsorbed components can be inhaled from a cold "reactor". The relevant components are no longer in the vapour phase and the vapour pressure of the tetrahydrocannabinoles will be too small.


How to measure the water content of opaque liquids?

I would add Karl Fischer titration.

Of course for your milk example, you would have to dilute your sample quite a lot in a dry (or known-dryness) solvent before titration. Also, the strong redox solution might interact in complicated ways with other materials in your sample (proteins, etc?).


Figuring out the theoretical mass of reactants

Try to answer to these questions, don't cheat, check the answer passing over the yellow box.
I think this software is confusing you always start from how much reactants you have and then calculate how much products you will have, suppose you have $\pu{100g}$ of oxygen and $\pu{44g}$ of propane.
How many mole of propane and oxygen ($\ce{O2}$) do you have?

 Oxygen moles$=\frac{\pu{100g}}{\pu{32g/mol}}\sim \pu{3mol}$,  Propane moles$=\frac{\pu{44g}}{\pu{44.096g/mol}}\sim \pu{1 mol}$

If this is your balanced equation:
$$\ce{C3H8 + 5O2 -> 3CO2 + 4H2O}$$
How many oxygen molecules do you need to react with one molecule of propane?

 You need 5 oxygen molecules for every propane molecule. So you need 5 moles of oxygen for every mole of propane.

Which is the ratio?

 $\frac{1}{5}$ if you have $x$ moles of oxygen you need $\frac{x}{5}$ moles of propane. e.g. if you have 5 moles of oxygen you need? 1 mole of propane.

But you have only 3 moles of oxygen to get the products, oxygen is a limiting reagent! So how many moles of propane will react:

 $$\pu{\frac{3}{5}mol} =\pu{0.6 mol} $$

So how many moles of the products will you have?

 $3\times0.6$ moles of $\ce{CO2}$ and $4\times0.6$ moles of water

And so how many grams?

 for $\ce{CO2}$, $\pu{1.8mol}\times \pu{44g/mol}$ for water, $\pu{2.4mol}\times \pu{18g/mol}$


First, a great source of information on this kind of problem is the website of Hans Reich at UW Madison, see e.g. here. Couplings can occur with protons at ring positions ortho, meta and para, with typical values of $\pu{7-9}$, $\pu{2-3}$ and $\pu{<1 Hz}$, respectively.

This  is (roughly) the information I extracted from your spectrum:

$\pu{  9.15 ppm, \space d,  \space 1.5 Hz}$
$\pu{  8.83 ppm,\space dd, \space 5 Hz, 1.5 Hz}$
$\pu{  8.3 ppm, \space dt (apparent), \space 2 Hz (t), 8 Hz }$
$\pu{  7.6 ppm, \space m (ddd),   \space 5 Hz, 8 Hz, < 1 Hz}$

You should report the field at which the spectrum was acquired since otherwise the magnitude of the J couplings is ambiguous. Assuming that the largest J is $\pu{8 Hz}$ leads to the guess that it is a 400 MHz ($\ce{^1H}$) magnet. You should also report the solvent.

Second, your spectrum is very nice, showing great resolution. Compare for instance to this spectrum (also at 400 MHz but presumably much higher concentration -  $\pu{100 mM}$). Either the concentration of the solvent and temperature may be reasons you obtain higher resolution, resolve very small couplings. The BMRB spectrum was acquired in D2O, pH 7.4, 298 K. Your spectrum was acquired perhaps in $\ce{H2O}$, and perhaps at low pH, resulting in protonation of the pyridine N, but probably not.

Based on the structure you expect 2 to have the weakest couplings to other H. It displays a small coupling, not easily quantifiable because you clipped the spectrum in the figure. 5 Is expected to be a doublet of doublets, with both couplings similar and large. It also displays a smaller third coupling. Therefore lets start by tentatively assigning 2 to the resonance at 9.15 ppm, and 5 to 7.6 ppm. The resonance at 8.3 ppm is an apparent doublet of triplets, the triplet splitting being small, 2 Hz, the larger 8 Hz matching up with that from 5, so this H is vicinal to 5. One of the smaller couplings is presumably to 2, the other to 6, explaining the apparent triplet multiplicity. The last resonance, at 8.8 ppm, has a coupling matching up to 5, and another smaller one to 4, so we assign this to 6. So, to summarize, here are the assignments:

$\pu{  9.15 ppm, \space d,  \space 1.5 Hz, \space }$  2
$\pu{  8.83 ppm,\space dd, \space 5 Hz, 1.5 Hz, \space }$ 6
$\pu{  8.3 ppm, \space dt (apparent), \space 2 Hz (t), 8 Hz , \space }$ 4
$\pu{  7.6 ppm, \space m (ddd),   \space 5 Hz, 8 Hz, < 1 Hz, \space}$ 5

This matches your assignments and those in the BMRB. So this analysis explains the small coupling for 2. The simplest explanation: that's a meta coupling to 4. It does not explain the small coupling for 5, though. That might be a trans coupling to 2.

The magnetization transfer pathways between the assigned H can be verified by inspecting the $\ce{^1H-^1H}$ TOCSY available at the BMRB. In particular, there is no pathway between signals at 9.1 and 8.8 ppm.


Is it possible to cut yourself on pyrite?

The Wikipedia page for Mohs scale shows that pyrite has a hardness of 6-7 (according to the page on pyrite, it's 6-6.5) while plate glass has a hardness of 5.5 and a pocketknife has 5.1. I've cut myself on both knives and glass before, so I'd wager that pyrite should be able to slice.

The catch is that I've never cut myself on a cube of glass or a cube of steel before--a 90 degree edge just doesn't generate enough pressure for that. So either your pyrite must have very nice serrations on the edge or it must have an edge sharper than 90 degrees.

Is that possible? I don't know. I suppose at this point you could invoke the leeway granted to you by fiction.


What does solubility limit mean in practical terms?

Q1) If 20 micrograms / ml is the upper solubility limit and you added less than 20 micrograms the solution could be less concentrated but a concentration higher than 20 micrograms / ml H2O can't be achieved.

Q2) No more than 20 micrograms of glycerol monolaurate will dissolve in 1 ml of water so if more than 20 micrograms was added the remaining undissolved powder would stay separated in some way (eg. precipitate) which could be filtered (as @MaxW has stated), or a colloidal suspension could also form. If more than 20 micrograms / ml were to be used the supernatant (GML/H2O layer) could be estimated to be 20 micrograms/ml) after filtration.


Why does hydrogen bonding in salicylic acid make it more acidic?

If the COO(-) group on salicylic acid's conjugate base didn't hydrogen bond, then wouldn't the COO(-) therefore be more inclined to pick up a proton and return the entire molecule to salicylic acid? In other words H-bonding is a stabilizing interaction. You can view H-bonding as diluting the negative charge density in the COO(-) group. This in turn makes proton abstraction less of a favorable move.

Structure of salicyclic acid's conjugate base:


The electrolyte in that clip is a solution of $\ce{NaCl}$ and acetic acid (vinegar). The zinc anode is oxidized, resulting in $\ce{Zn^2+}$ ions going into the electrolyte. The electrons went through the red LED, lighting it, and then on to the copper cathode. From there, they reduced some $\ce{H+}$ to hydrogen gas. So two $\ce{H+}$ ions were reduced to one $\ce{H2}$ molecule (which left: it's a gas) and the $\ce{Zn^2+}$ ion simply replaced the two $\ce{H+}$ ions.

Acetic acid is a weak electrolyte, so a solution of it (vinegar) is not a great conductor of electricity. The $\ce{NaCl}$ just makes the electrolyte a better electrical conductor: movable ions are necessary for a steady flow of electrical current in this battery. Without the $\ce{NaCl},$ most of the generated voltage would have been wasted on the electrolyte's resistance.

No sodium is produced (by reducing sodium ions) because the $\ce{H+}$ ions get the electrons. And no chlorine gas is produced (by oxidizing chloride ions) because the $\ce{Zn}$ anode is more readily oxidized.


Ea is the term for activation energy, what is the mean of a in this term?

The subscript $\mathrm a$ in $E_\mathrm a$ may refer to Arrhenius.
The IUPAC gold book states that:

  activation energy (Arrhenius activation energy)
  An empirical parameter characterizing the exponential temperature dependence of the rate coefficient, $k$, $E_\mathrm a = RT^2 \frac{\mathrm d(\ln k)}{\mathrm dt}$,
  where $R$ is the gas constant and $T$ the thermodynamic temperature.


What are the common products of the reaction: 2-bromobutane and NaOH

A chemical reaction is the result of competition; it is a race that is won by the fastest runner. A collection of molecules tend to do, by and large, what is easiest for them.

An alkyl halide with β-hydrogen atoms when reacted with a base or a nucleophile has two competing routes: substitution ($\mathrm{S_N1}$ and $\mathrm{S_N2}$) and elimination.

Which route will be taken up depends upon the nature of alkyl halide, strength and size of base/nucleophile and reaction conditions.

Thus, a bulkier nucleophile will prefer to act as a base and abstracts a proton rather than approach a tetravalent carbon atom (steric reasons) and vice versa.
Similarly, a primary alkyl halide will prefer a $\mathrm{S_N2}$ reaction, a secondary halide – $\mathrm{S_N2}$ or elimination depending upon the strength of base/nucleophile and a tertiary halide – $\mathrm{S_N1}$ or elimination depending upon the stability of carbocation or the more substituted alkene.

So, I think preferably in aqueous NaOH or KOH there should be formation of alcohol but that too by $\mathrm{S_N1}$ if there's a tertiary; and, if primary halide, then it would follow $\mathrm{S_N2}$. It's bit tricky for secondary; again it depends on the reagents and solvents used.


Chemically removing rust without leaving any unwanted residues

The common DIY rust remover is cola of some sort -- most colas contain a significant quantity of phosphoric acid, which reacts with the iron oxide to form iron (III) phosphate which you can then scrub off. Any remaining iron phosphate should come off with time and/or more water, and is harmless in domestic quantities.

(Wikipedia:Phosphoric acid - Rust Removal)


Calcium acetate, calcium hydroxide, calcium phosphate, cerium(III) selenate, cerium(III) sulfate, and lithium sulfate are more soluble in cold water than hot water.

Also, calcium selenate seems to have it's highest solubility at around 10 °C; and sodium sulfate has it's highest solublity at around 32 °C. These two chemicals have phase changes.

Gases (like oxygen) are also generally more soluble in colder water. The primary explanation is that colder temperatures and higher pressure results in gas becoming a liquid.


Why does being "bonded" decrease atomic size?

The first thing you should realize is that there are various definitions of the size of the atom. One distinguishes for example the covalent radius, the Van der Waals radius and the ionic radius. The covalent radius is based on the binding of atoms into molecules and on the resulting bond length. The van der Waals radius is half the minimum distance between two atoms that not form a covalent bond and the ionic radius is derived from crystal structures of ions in salts. The covalent radius cannot be determined for noble gasses as they do not bind under normal conditions. The Van der Waals radius however can be determined and since the noble gasses are not very polarizable, the Van der Waals radius does increase with respect to the preceding halogen. You'll find more in the wikipedia article as mentioned by ron.


Can ketones next to a nitrogen hetero atom tautomerise?

If I understand your question correctly, I think you're asking if an amide can enolize like a ketone.  The answer to that question is yes.

The enol content of most organic compounds is very low and special methods must be used to measure the enol content.  Perhaps the question becomes, "is there more or less enol content in an amide than in an ester, for example"?

We can draw resonance structures for an amide (or ester) showing that the nitrogen stabilizes the amide.  But above we also drew resonance structures showing how the nitrogen can also stabilize the enol.  We can draw similar structures for both an ester and its enol as well.  Although in the ester case we wind up putting a positive charge on oxygen in some of the resonance structures, so they might not count for as much as they do in the amide case - but both the starting ester and its enol would both be affected similarly.  So as a first approximation I might guess that the enol content in an ester and an amide would be similar.

Indeed the enol content in malonamide is similar to the enol content in diethyl malonate (see page 34 here)


It is likely to be due to the different amounts of partial positive charge which exist on the carbonyl carbon.

We can consider for example the rate of reaction of a given amine (such as aniline) with the acid chloride, acid anhydride and an ester such as the methyl ester. The partial positive charge argument would explain this reactivity series.


What are the different types of van der Waals forces?

The definition of van der Waals forces according to IUPAC. Compendium of Chemical Terminology, 2nd ed. (the "Gold Book"). Compiled by A. D. McNaught and A. Wilkinson. Blackwell Scientific Publications, Oxford (1997). XML on-line corrected version: http://goldbook.iupac.org (2006-) created by M. Nic, J. Jirat, B. Kosata; updates compiled by A. Jenkins. ISBN 0-9678550-9-8., is as follows:

  The attractive or repulsive forces between molecular entities (or between groups within the same molecular entity) other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules. The term includes: dipole–dipole, dipole-induced dipole and London (instantaneous induced dipole-induced dipole) forces. The term is sometimes used loosely for the totality of nonspecific attractive or repulsive intermolecular forces.

The phrase "other than those due to bond formation or to the electrostatic interaction of ions or of ionic groups with one another or with neutral molecules" would seem to imply that according to the IUPAC Gold Book, if a force is not an ionic or covalent chemical bond or electrostatic interaction, then it is a van der Waals force.


What dangers/risks are there in the production of carbon subnitride?

There is no way, dicyanoacetylene can be produced at room temperature. According to the paper "Syntheses of Dicyanoacetylene", decomposing 4,5-dicy ano-1,3-dithiol-2-one at 600-800°C, a 59-76% yield of dicyanoacetylene is obtained along with side-products like carbon disulfide, carbonyl sulfide and sulfur. Another scheme is pyrolyzing solid dicyanodiazomethane at 220°C, which polymerizes. The polymerised product is decomposed at 700°C to get desired product. Further, a note on the storage of dicyanoacetylene is mentioned:

Dicyanoacetylene is a hazardous material. Because of its toxicity and
high vapor pressure at room temperature, it should be handled only in
a hood. It is potentially explosive, both in the pure state and in
concentrated solutions, but its thermal stability in dilute solutions
in inert solvents is greater than previously re-ported. Pure
dicyanoacetylene turns dark slowly at room temperature, but it can be
kept almost indefinitely at Dry Ice temperature.

Dicyanoacetylene can only be kept at very low temperatures, in dry ice or liquid nitrogen, like a dewar, away from light and oxygen, as it may polymerize. Do not store it for long. Diluted solutions of dicyanoacetylene are much more stable, and can be safely kept at standard conditions.
Reference

Syntheses of dicyanoacetylene, Engelbert Ciganek and Carl G. Krespan, The Journal of Organic Chemistry 1968 33 (2), 541-544, DOI: 10.1021/jo01266a014


What could be causing noisy pH measurements?

Several factors may influence your sensor readings:

Electromagnetic noise. However, since EMF should be directly proportional to pH, it seems to be not your case. If it were the main reason, the same amount of noise would show up in both the nutrient and the storage solutions.
Temperature. I've included this one because it is unclear from your post whether you are measuring the actual solution temperature when you calculate pH from EMF, or just assume that the temperature is some constant value. The Nernst equation may be used to estimate the amount of "noise" (i.e. the difference in EMF) one may expect from the temperature fluctuations: $\Delta E=2.303\frac{R\Delta T}{F}pH$. That is, if we can neglect the temperature dependence of pH itself, i.e. when $\Delta T$ is relatively small (when $\Delta T$ is large we can no longer neglect the temperature dependences of equilibrium constants). If, for some reason, the daily temperature change in the nutrient bath is different than that in the storage solution, you may expect to get slightly different readings.
Math. We should not forget that the value measured (pH) is logarithmic, and that it is the concentration (well, activity) of protons that makes the difference. For example, to change pH from 6.45 to 6.05 (±0.2), you need to add just around $5.4\cdot 10^{-7}$ mol/l of protons, but to change it from 4.26 to 4.24 (±0.01) - $2.6\cdot 10^{-6}$, or almost 5 times as much. In other words, the same actual $[\textrm{H}^{+}]$ fluctuations cause different pH changes around different pH values. In turn, these $[\textrm{H}^{+}]$ fluctuations may be due to slight inhomogeneities of the nutrient solution, differences in dissolved carbon dioxide concentration, etc. The more neutral your solution is, the more sensitive, with respect to $\Delta [\textrm{H}^{+}]$, your pH electrode becomes. In this regard, you may want to prepare a few solutions with different pH and try to investigate the noise vs. pH dependence. Or do something even more simple - as a reference system, instead of the storage solution (which, in addition, probably possesses some buffer characteristics), try to use something more neutral, like water, and see if the noise is as large as in the nutrient solution.

Since my analytical chemistry is a bit rusty, no other potential cause comes to mind.


There are many ways to affect the transformation, as you have realised, and many of which Loong has detailed below in his post.

This answer gives one method which I would be confident of working practically, in the lab, using common reagents. A full search of Reaxys of SciFinder would reveal conditions/methods more suitable for a specific aldehyde.

A Pinnick oxidation will convert an aldehyde into its corresponding carboxylic acid, which may then be treated with a reagent such as oxalyl chloride to give the desired acyl chloride.


This is similar to pinacol-pinacolone rearrangement. Only difference is in pinacol-pinacolone rearrangement, carbocation forms by dehydration while here, it forms through diazonium reaction. Following is the tentative mechanism of the rearrangement:

Since reaction progress through carbocation intermediate, the final product is a racemic mixture. $\ce{CH3-C+-CH3}$ (in pinacol) or $\ce{CH3-C+-H}$ (in given compound) can rotate about middle horizontal bond before alkyl or aryl group migration.


Are the 2s and 2p subshells degenerate in the He+ ion?

Yes, because this system is identical to the original if we simply rescale the charge (or, if you'd like, modify the units of our new system). On a more fundamental level, degeneracies are caused by symmetries of the system, and these are affected not by the magnitudes of various constants or parameters, but rather by the form and structure of the relevant interactions.

The degeneracy of the subshells in the hydrogen atom corresponds to the SO(4) symmetry of the hydrogen atom, which leads to the conservation of the quantum-mechanical form of the Laplace-Runge-Lenz vector. This symmetry depends on the inverse-square law nature of the Coulomb force and appears also in celestial mechanics, where the corresponding inverse-square law force is now gravity.

Changing the charge of your hydrogenic atom will not break this symmetry. On the other hand, fine structure corrections to the hydrogen atom will: if you start looking at relativistic and spin-orbit corrections, you change the form of the Hamiltonian, break the symmetry, and hence resolve the degeneracy.


What is the black slag formed when sterling silver is melted with borax?

"Indestructible" by what means exactly? How did you try to "destroy" it? Provided that sterling alloy is an alloy of silver and copper, the first guess is that your slag is primarily a copper (II) oxide. Depending on the actual composition of the alloy, it may contain other 3d-metal oxides. The formation of borates (not of silver, but of less noble metals) cannot be excluded, for borax is frequently used as a constituent of the high-temperature fluxes for spectroscopic and calorimetric applications. To what extent the borates are formed depends on the temperature and duration of your melting experiment.

You should be able to dissolve your slug in a boiling mineral acid of your choice. Personally, I'd try the $\ce{HCl}$ solution, or $\ce{HNO3}$ if you do not care that silver dissolves as well.

The following may be irrelevant to the question asked, but, based on my experience and some textbooks, people first try to purify silver and then melt the pure metal, not the other way around. The methods of purification are aplenty, but most of them are based on dissolving the silver-containing alloy in $\ce{HNO3}$. Then $\ce{AgNO3}$ can be transformed into $\ce{AgCl}$, which, in turn, can be reduced by metallic zinc in acidic environment or formaldehyde - in basic.


Kinetics of air evolving from water on heating

Is there a simple relationship describing the kinetics of air evolving from water on heating?

There is no simple relationship. The problem is that you can't heat the water absolutely uniformly. Thus all the convection currents in the water would create a mathematically chaotic relationship. There is also the problem of the water being supersaturated - especially if you heat from the bottom.

  Is it fast or slow compared to a reasonable heating rate?

This would seem to be sort of a judgement call, but I'd say that it would be faster to dissolve a gas in a liquid (say via bubbling) than heating to expel it. The bubbles mechanically stir the solution rather than just equilibration by diffusion from the surface. Also think of liquid nitrogen. It boils so fast when spilled on the floor that the drops skitter around. The escaping gas insulates the liquid drop limiting the heat transfer to the liquid. So there is a limit as to how fast you can heat a solution.

  Does it depend on just how far the system is from equilibrium

Yes. The further from equilibrium, the less likely a solution could be super-saturated.

  Conditions like the amount of air-water surface boundary

The time to reach equilibrium surely depends on the surface area to volume. Think of a cylinder of water. A cylinder which has water one inch deep would certainly come to equilibrium faster than cylinder of the same diameter which has water a mile deep.


Even if it is possible to calculate periodic structures in Gaussian, I would not suggest it.
Better option is to use codes devoted to calculation of solids, among others the ABINIT you mentioned, or e.g. Quantum Espresso.

Gaussian does not use plane wave basis set, will be slower and in general less powerful than codes devoted to solids.


$$
\begin{align}
\ce{HNO2 + H2O &<=> NO2- + H3O+} &\quad K &= K_\mathrm{a} \tag{1}\\
\ce{NO2- + H2O &<=> HNO2 + OH-} &\quad K &= K_\mathrm{b} \tag{2}\\
\ce{H2O + H2O &<=> H3O+ + OH-} &\quad K &= K_\mathrm{w} \tag{3}
\end{align}
$$

Above are the reactions associated with equilibrium constants commonly called $K_\mathrm{a}$, $K_\mathrm{b}$, and $K_\mathrm{w}$. If you reverse the first reaction, you get:

$$\ce{NO2- + H3O+ <=> HNO2 + H2O} \qquad K = 1/K_\mathrm{a}\tag{1a}$$
(This looks a bit different from your equation because it shows the hydronium ion instead of the hydrogen ion, but it refers to the same reaction.)

As you might know, the three equilibrium constants are related:

$$K_\mathrm{a} \times K_\mathrm{b} = K_\mathrm{w}$$

This is because when adding up the first and the second reaction, you get the third.

  When to use $1/K_\mathrm{a}$ vs $K_\mathrm{b}$

Use the equilibrium constant that matches the reaction you are working on.


a) yes, those are equivalent structures of 2,2-dimethylpropane

b) yes, those are equivalent structures of 3,3-diethylpentane


How do I calculate the normality of NaBrO3?

To me your answer is correct. It seems that the book result have a '2' dividing the final result. Could you post all the text of the book exercise? Two possibilities, the result from the book is wrong (could be) or there is a mistake in one of the values (75 ml instead 150, 0.375 N instead 0.75 N...).


Why is the zero of standard enthalpy of formation a convention?

It starts with this:  The standard enthalpy of formation of any substance, $\Delta H^\circ_\mathrm{f}$, is (by convention) defined as the enthalpy change for the reaction at 1 bar and a specified temperature (usually $\pu{298.15 K}$), in which the product is 1 mole of that substance, and the reactants are its component elements in their respective standard states.
For instance, $\Delta H^\circ_\mathrm{f}$ for $\ce{H2O_{(l)}}$ is equal to $\Delta H^\circ$ for the following reaction at standard state:
$$\ce{H2_{(g)} + 1/2O2_{(g)}->H2O_{(l)}},$$
since $\ce{H2_{(g)}}$ and $\ce{O2_{(g)}}$ are the respective standard states for hydrogen and oxygen.
Once you've accepted this definition as the convention for determining $\Delta H^\circ_\mathrm{f}$ for any substance, it directly follows that the value of $\Delta H^\circ_\mathrm{f}$ for any element in its standard state must be zero.  For instance,  $\Delta H^\circ_\mathrm{f}$ for $\ce{H2_{(g)}}$ is equal to $\Delta H^\circ$ for the following reaction:
$$\ce{H2_{(g)} -> H2_{(g)}},$$
which is necessarily zero.
To give an analogy: Suppose you define the "altitude of formation", $\Delta z_f$, of any location on earth as the change in altitude necessary to reach that location from sea level.
Hence $\Delta z_f$ for the summit of Mt. Everest is $\Delta z$ for the altitutude change associated with:
$$\ce{sea level -> summit of Everest},$$
which is 29,029 feet.
It necessarily follows, from this convention, that $\Delta z_f$ for any location at sea level is zero, since that would be equal to $\Delta z$ for:
$$\ce{sea level -> sea level}$$.


Electronic configuration of chromium(II)

Chromium’s standard electron configuration as given by the aufbau principle is actually more of an exception than you may think initially. See for example this question to catch a small glimpse of the truth behind the simplifications.

The aufbau principle tells you to fill in the order $\ce{1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, }\dots$ That order, however, is only valid for neutral atoms (and not always as the linked question implies). As soon as you add charge to an atom, the relative orbital energies shift. The $\ce{3d}$ orbitals, belonging to the M shell, are stabilised more than the $\ce{4s}$ orbital which belongs to the N shell. This is a general rule: charges typically stabilise orbitals belonging to lower shells more than those in higher shells.

Thus, while in neutral chromium the $\ce{4s}$ and the $\ce{3d}$ shell are very close in energy, this balance completely tips towards $\ce{3d}$ (which then has a lower energy) upon oxidation. Therefore, one should always assume an empty $\ce{4s}$ orbital when discussing the electronic configuration of charged d-block elements.


Converting cyclic compounds to their corresponding aliphatic compounds

The reaction you want to know about is "Cleavage reaction" in particular oxidative cleavage.

In this dihydroxylation, if the solution is warm or acidic or too concentrated, oxidative cleavage of the glycol may occur.

Then they can be reduced to aldehydes by reducing agents.Reduction of carboxylic acids to aldehydes is difficult because aldehydes are more reactive than carboxylic acids toward most reducing agents. Almost any reagent that reduces acids to aldehydes also reduces aldehydes to primary alcohols.

Lithium tri-tert-butoxyaluminum hydride, LiAlH(O-t-Bu)3 is a weaker reducing agent than lithium aluminum hydride. It reduces acid chlorides to aldehydes because acid chlorides are strongly activated toward nucleophilic addition of a hydride ion.

Under these conditions, the aldehyde reduces more slowly and can be isolated. Therefore, reduction of an acid to an aldehyde is a two-step process: Convert the acid to the acid chloride, then reduce using lithium tri-tert-butoxyaluminum hydride.

And now we can reduce it by the Wolff Kishner reduction.

Ethylene glycol, diethylene glycol, or another high-boiling solvent is used to facilitate the high temperature (140–200 °C) needed in the second step. I am assuming you can convert the cyclohexane to cyclohexene by any radical halogenation and then dehydrohalogenation.


How hot will a carbon electrode get in an electrolysis setup?

Supposing there is a powerful DC source, then quite hot, unless small enough electrode current density is managed.

6 V is quite high voltage for electrolysis. But it is possible the source will not be able to provide the current for the voltage 6 V, unless the electrode area is small.

Consider also the graphite electrodes may get deteriorated by the released gas, releasing small carbon particles to the solution (remembering my own first electrolysis experiments with carbon electrodes from old Leclanché batteries, when I was 12).


How is knowing about the symmetry of a crystal useful

Properties that are directly dependant on crystal symmetry are

birefringence
piezoelectricity
(obviously) x-ray and electron diffraction pattern
pyroelectricity
nonlinear optical properties
...


What is the difference between "spontaneous" and "energetically favourable"?

Here is a very simple example of an exothermic, nonspontaneous process at room temperature:

$$\ce{H2O_{(l)} -> H2O_{(s)}}$$

You need to extract heat from water to freeze it, but it is obvious that at $25\ ^{\circ}\mathrm{C}$, this does not happen spontaneously. Only if you decrease the temperature to decrease the contribution of the negative $\Delta S$ term will this reaction occur spontaneously.


Why do weak acids like ethanol not act as a strong base in water?

There seems to be a misunderstanding: The conjugate of a strong base like EtONa can't be a weak acid, because weak acids, like acetic acid for example, are still quite acidic.

Strong acid (HCl) - very weak conjugate base (chloride anion)
strong base (EtONa) - very weak conj. acid (Ethanol)

A strong acid or base reacts strongly (with water), the resulting conjugate is rather inert, with the equillibrium fully on the side of the latter.

weak acid (acetic acid) - weak conj. base (acetate anion)
weak base (ammonia) - weak conj. base (ammonium)

A weak acid or base just reacts partly towards its conjugate, so does the pure conjugate: They are both rather weak. The equillibrium in an equimolar mixture is rather stable: A buffer solution.

The ideal weak base or acid has a $pKa$ of 7 (in water), that is $1/2$ of the -log ion product of the given solvent. If it's even weaker, the conjugate becomes stronger again.


Ethanol discolored yellow when stored with a rubber stopper. Forming fine white precipitate when added to water

It’s not easy to know unless you decide to throw the full set of analytical chemistry at your yellowish ethanol. But we do have a few clues.

It is most likely that something from the rubber stopper got dissolved in the ethanol. This solution turned out to be yellowish so we can assume the compound itself to be yellow to brown-ish. Dissolution either came by the ethanol soaking the stopper while the flask was lying down or by evaporation/condensation.

We also cannot determine from the information you have given what the precipitate is that forms when you add water. But my first guess would be ions from the water being less soluble in an ethanol-water mixture and precipitating. If that is the case, it would be made up mainly of sodium chloride. It could also be that some component of the ethanol is insoluble in water and precipitates out. Considering your alcohol is denatured, it might also be the additive that makes it undrinkable. However, once again there are too many variables and you need the full set of analytical chemistry thrown at the precipitate to find out what it is.

Most certainly whatever is happening is not a good indicator of anything.


The first thing that came to mind is oxygen itself, as the term "oxide" suggests one other element in its chemical formula. Merriam-Webster suggests a more strict version: "a binary compound of oxygen with a more electropositive element or group". So technically pure oxygen is not an oxide.

Putting oxygen aside, helium $\ce{He}$, neon $\ce{Ne}$ and argon $\ce{Ar}$ are probably the only three conventional elements that come to mind are not yet proven to form oxides. I find it rather tricky to back up this statement properly as the lack of knowledge about an object doesn't prove its non-existence; ideally one might want to do a stability simulation at extreme pressures and temperatures to prove the existence of $\ce{(He|Ne|Ar)O_x}$ or lack thereof. There are algorithms like USPEX that are capable of that – by the way, with the help of USPEX a family of krypton oxides has been recently predicted.


The walls of a plastic vial with Mercury inside are darkening. What is causing this?

Mercury doesn't react with PET under normal conditions, but mercury vapor will diffuse into plastics.  The diffusion will change the optical density of the plastic in the area where diffusion occurred.

Here's a link to a nice study where the mercury was doped with a radioactive isotope of mercury.  This doped mercury was placed in a glass chamber which also contained plastic discs held at a known distance above the mercury.  The plastic discs were removed after various times and then microtomed.  The various layers of the microtomed plastic disc were measured for radioactivity.  Analysis of the resulting data allowed the investigators to measure the rate constant for diffusion of mercury into the plastic.

One final point, plastics are sometimes contaminated with small amounts of various impurities.  Sometimes these impurities were intentionally introduced, such as radical chain terminating materials (sulfur is an example) in order to produce polymers with a certain molecular weight distribution.  So while the plastic itself might not react with mercury, an impurity might.  The above link also provides analysis of plastic contaminated with sulfur and finds a much higher mercury diffusion constant in such materials.


Handling and storing sodium hydroxide, hydrochloric acid, hydrogen peroxide and isopropyl alcohol

You probably don't need a mask - vapours from hydrochloric acid can be a little noxious, but if you're diluting it down and using it in a reasonably ventilated area, you shouldn't get much of that. Isopropyl alcohol vapour is toxic like most alcohols, but I don't think an ordinary mask will prevent that, and you'll be better off just making sure you have enough ventilation.

Standard white cotton labcoats are inexpensive and can be quickly removed if you spill a nontrivial quantity of something on them.

Hydrogen peroxide will probably come in an opaque container, but if not, it does break down in sunlight. Everything else is pretty stable to light, although you want to avoid storing the alcohol somewhere that's going to get too warm, because it evaporates readily and can pressurise the container more than is a good idea.

All these materials are pretty safe to store indoors in unspecialised storage, although for commercial purposes or quantities you may need to take regulatory precautions that will obviously vary by locale.

Edited to add: The copper chloride that comes out should be regenerated as many times as possible, but when you eventually end up with too much of it, you'll want to already have thought about what you're going to do with it - it's pretty toxic and you can't (or, well, shouldn't) just wash it down the sink.


How is octane rating defined for negative values and those over 100?

I recommend the @Stanley Yu own answer, as I admit I was not particular familiar with the shape of the ON number curve.

It is done by an extrapolation of the critical compression ratios of mixtures 0-100% of isooctane and 100-0% of n-heptane for ON values <0 and >100.

If, just illustratively, 100% isooctane had the critical compression ratio 1:20, 90% isooctane 1:19 ,... 90% n-heptane 1:11, 100% n-heptane 1:10, then the fuel with the critical compression ratio 1:21 would have the octane number 110.

The octane number/rating differs for different procedures. See Research octane number RON, Motor octane number MON,  Anti-knock index AKI=(RON + MON )/2 and Observed Road octane number RdON.

As seen in the OP own answer, the extrapolation of the octane number (ON) scale  above 100 is done by linear  extrapolation by the empirical aviation formula for performance number $$PN=100 + 3 * ( ON - 100 )
where on the graph is coordinate ON substituted by tetraethyllead addition.


Yes, salinity affects the pH of seawater.

See RECOMMENDATONS FOR THE
DETERMINATION OF pH IN SEA WATER
AND ESTUARINE WATERS Pure & Appl. Chem., Vol. 60, No. 6, pp. 865-870, 1988.

An approximation of pH as a function of temperature and salinity is given:

pH = (2559.7 + 4.5 S)/T - 0.5523 - 0.01391 S

Where "T" is temperature in kelvins and "S" is salinity in units of parts per thousand.

As seen above, in the temperature range of Earth's oceans, increasing salinity increases pH.

See also pH of seawater Marine Chemistry Volume 126, September 2011, pages 89-96.

Fig. 2 plots pH versus salinity at 25 degrees C and 0.000334 atmospheres of CO2 and shows pH increasing significantly with salinity.


How does pH depend on temperature, when taking into account volume expansion?

Your qualitative understanding is correct.  As you increase the temperature, you have two effects:

The dissociation reaction, which is endothermic, shifts to the right.  This effect increases the concentration of hydrogen ions.
The concentration of hydrogen ions is expressed as molarity, which is number per unit volume.  As you increase $T$, water expands, resulting (everything else being equal) in all concentraitons decreasing.

And you're asking which one wins.

The answer is the first, overwhelmingly.  For instance, as you go from $\pu{25 ^\circ C}$ to $\pu{50 ^\circ C}$, $K_\mathrm w$ increases by five times.  By contrast, the increase in volume over this temperature range is only $1\%$.


Through bond interaction in biradical dioxo compound and its correlation diagram

I would like to help here, but I am not sure where to begin. First of all, I want to point out that the reaction you have shown is not an electrocyclic reaction, because it does not involve the central saturated carbon atom. It is rather a simple homolysis reaction. The terms "conrotatory" and "disrotatory" are misapplied, since the two oxygen atoms do not have functional groups on them, and the two modes of ring opening are indistinguishable.

Dioxirane does have an electrocyclic ring opening reaction available to it - namely cleavage of a carbon-oxygen bond to form "acetone-O-oxide". Perhaps you should consider that as an alternative pathway. Does that help?


Finding [A] when looking for the velocity constant

For a first-order reaction, the half-life is constant and is given by the equation

$$t_{1/2} = \frac{\ln 2}{k}$$

In $10~\mathrm{s}$, two half-lives have passed (the concentration of $\ce{A}$ drops to a quarter), so $t_{1/2} = 5~\mathrm{s}$. Accordingly,

$$\begin{align}
k &= \frac{\ln 2}{5~\mathrm{s}} \\
&= 0.1386~\mathrm{s^{-1}}
\end{align}$$


Would a loaded cartridge ignite/explode when hit from the side or front by a bullet?

Maybe

Unfortunately, the correct answer is “it depends.” Small arms cartridges struck by small arms projectiles generally do not ignite – but they do sometimes.

The issue is quite important in the armed forces. If you come under enemy fire, the last thing you want is for a near miss to cause your own ammunition to explode. There has been formal research on the subject for decades. However, it can be quite difficult to prevent.

There are formal test protocols to assess the safety of new ammunition types, and they can involve munitions struck by a range of different “stimuli” from various distances, points of impact and angles. Depending on the exact set-up, the response can vary from 100% explosion to barely scratched.


Why rinse flask with distilled water when the flask will be used for storing aliquots of HCl for titration?

The dilution with excess water won't affect the total amount of $\ce{HCl}$ in the solution, which is what you will be titrating. You still know how much $\ce{HCl}$ you have from your quantitative addition of the $\pu{1 mL}$ aliquiot. If you rinsed with $\ce{HCl}$, you would have no idea how much $\ce{HCl}$ you really had and therefore you couldn't calculate the concentration of your $\ce{NaOH}$.


You overlooked the strain from the sp2 carbon in a small ring. The ring strain you have listed is based on sp3 carbon. sp2 supposed to have a planar triangular geometry (120°, compared to 109° for sp3) and will be more strained in a 3 or 4 membered ring. 6 membered ring will be best followed by 5 membered ring.

Also, you have discussed a lot about the stability of carboncations. Acid catalyzed elimination/rearrangement reaction is a thermodynamic condition. The stability of product is more important than the stability of intermediate.


Can you meassure a negative pH value with a pH meter?

You are most likely getting an inaccurate value. As anticipated in Poutnik's comment above is difficult to reliably measure a $\mathrm{pH}$ of very acidic solution but the $\mathrm{pH}$ scale is indeed open, and negative $\mathrm{pH}$ are a real thing.
It is easy and more meaningful to state a very high concentration or activity rather than reporting inaccurate $\mathrm{pH}$. This, together with typical figures found in textbooks, has spread the misconception that $\mathrm{pH}$ is bound in the range 0 - 14, but it is not.
A simple example of a solution with negative $\mathrm{pH}$ is in every lab, namely conc. $\ce{HCl}$, 37% by mass, $\mathrm{pH} = -1.1$.
Edit: For sake of clarity, as for its definition, pH can exceed 14, too. As an example, a saturated NaOH aqueous solution has pH = 15.


Is it possible to form bubbles of ionized hydrogen and ionized oxygen under the right pressure/temperature conditions during electrolysis of water?

The electrolysis will not form any ionized oxygen or ionized hydrogen, but instead proceeds through radical species. Some experiments have attempted to show the presence of $H^{\bullet}$ and $OH^{\bullet}$ radicals at the anode / cathode respectively (see P.H. Kasai, D. McLeod J. Am. Chem. Soc (1978)). Provided that the surface concentration is sufficient, these radical species will very quickly recombine to lead to $H_{2}$ and $O_{2}$. The surface of the electrode is very likely to influence the process, as the radicals species are likely to interact with it and not be released in high concentration in the solution.

Other factors are also influential, and the radical species might also react with other constituents of the solution (see for instance T. Mokudai,  K. Nakamura,  T. Kanno,  Y. Niwano, PLOS One (2012) on the formation of hydroxyl radical vs formation of hypochlorous acid through electrolysis).

Heating the solution is likely to increase radical concentration, but the temperature should be extremely high to prevent radical recombination within the gas. As for pressure, my guess would be that a reduced pressure is likely to improve radical isolation by displacement of the solvation equilibrium.


Strong acids or bases will hydrolyze polycarbonate resin--specifically poly(bisphenol A carbonate)--at the ester linkages, first giving shorter polycarbonate chains, and eventually down to bisphenol A and carbonate ion or carbon dioxide.


Combining acid dissociation constants to determine pH of diprotic acid

This should be broken into:
\begin{align}
\ce{H2A + H2O &-> HA- + H3O+}\\
\ce{HA- + H2O &-> A^{2-} + H3O+}
\end{align}

With equilibrium equations:
\begin{align}
K_\mathrm{a1}&=\frac{[\ce{HA-}]\cdot [\ce{H3O+}]}{[\ce{H2A}]}\tag1\label{Ka1}\\
K_\mathrm{a2}&=\frac{[\ce{A^{2-}}]\cdot [\ce{H3O+}]}{[\ce{HA-}]}\tag2\label{Ka2}\\
\end{align}

Let us solve for a general case. From logic and general knowledge we know the water dissociation constant, that the masses must balance, and that the charges must balance. These are shown here:
\begin{align}
K_\mathrm{w}&=[\ce{H3O+}]\cdot[\ce{OH-}]\tag3\label{Kw}\\
I&=[\ce{H2A}]+[\ce{HA-}]+[\ce{A^{2-}}]\tag4\label{Mass Balance}\\
[\ce{H3O+}]&=[\ce{OH-}]+[\ce{HA-}]+2[\ce{A^{2-}}]\\
&=\frac{K_\mathrm{w}}{[\ce{H3O+}]}+[\ce{HA-}]+2[\ce{A^{2-}}]\tag5\label{Charge Balance}\\
\end{align}

Using equations \eqref{Ka1} and \eqref{Ka2}, we can solve for $[\ce{HA-}]$ and $[\ce{A^{2-}}]$:
\begin{align}
K_\mathrm{a1} &= \frac{[\ce{HA-}]\cdot [\ce{H3O+}]}{[\ce{H2A}]}
&\Rightarrow &&
[\ce{HA-}] &= \frac{K_\mathrm{a1} \cdot [\ce{H2A}]}{[\ce{H3O+}]}\tag6\label{HA-}\\
K_\mathrm{a1} \cdot K_\mathrm{a2} &= \frac{[\ce{A^{2-}}]\cdot [\ce{H3O+}]^2}{[\ce{H2A}]}
&\Rightarrow &&
[\ce{A^{2-}}] &= \frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot [\ce{H2A}]}{[\ce{H3O+}]^2}\tag7\label{A2-}\\
\end{align}

Substituting into \eqref{Mass Balance}:
\begin{align}
I &= [\ce{H2A}]+\frac{K_\mathrm{a1} \cdot [\ce{H2A}]}{[\ce{H3O+}]}
     +\frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot [\ce{H2A}]}{[\ce{H3O+}]^2}\\
  &= [\ce{H2A}] \cdot \frac{[\ce{H3O+}]^2 + [\ce{H3O+}]
\cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}{[\ce{H3O+}]^2}\tag9\label{Sub Mass Balance}
\end{align}

Solving \eqref{Sub Mass Balance} for $[\ce{H2A}]$:

$$[\ce{H2A}]=\frac{I \cdot [\ce{H3O+}]^2}{[\ce{H3O+}]^2 + [\ce{H3O+}]
\cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}} \tag{10}\label{H2A}$$

Substituting \eqref{H2A} into \eqref{HA-} and \eqref{A2-}:
\begin{align}
[\ce{HA-}] &= \frac{K_\mathrm{a1} \cdot I \cdot [\ce{H3O+}]}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}
\tag{11}\label{Sub HA-}\\
[\ce{A^{2-}}] &= \frac{K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot I}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\tag{12}\label{Sub A2-}
\end{align}

Substituting equations \eqref{Sub HA-} and \eqref{Sub A2-} into equation \eqref{Charge Balance}:
\begin{align}
&&
[\ce{H3O+}] &= \frac{K_\mathrm{w}}{[\ce{H3O+}]}+\frac{K_\mathrm{a1} \cdot I \cdot [\ce{H3O+}]+2K_\mathrm{a1} \cdot K_\mathrm{a2} \cdot I}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\\
\Rightarrow &&
[\ce{H3O+}]^2 &= K_\mathrm{w}+\frac{(K_\mathrm{a1} \cdot I) \cdot ([\ce{H3O+}]^2+2K_\mathrm{a2}\cdot[\ce{H3O+}])}{[\ce{H3O+}]^2 + [\ce{H3O+}] \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}\tag{13}\label{Final Eq}
\end{align}

The above equation exactly calculates the $[\ce{H3O+}]$ for a solution of any diprotic acid. The same method used to solve for this equation can be applied to polyprotic acids as well. To make this into a more manageable equation, however, you may choose to assume $K_\mathrm{a1} \cdot K_\mathrm{a2}=0$ (but only in the case of weak acids).

As for the case you proposed, using $K_\mathrm{w}=1 \times 10^{-14}$, $I=1\ \mathrm{M}$, $K_\mathrm{a1}=5.9 \times 10^{−2}$ and $K_\mathrm{a2}=6.4 \times 10^{−5}$ in the equation $y=K_\mathrm{w}+\frac{(K_\mathrm{a1} \cdot I) \cdot (x^2+2K_\mathrm{a2})}{x^2 + x \cdot K_\mathrm{a1} + K_\mathrm{a1} \cdot K_\mathrm{a2}}-x^2$, and using a graphic calculator to find where the equation is equal to 0, we find $[\ce{H3O+}]=0.215508\ \mathrm{M}$ and $\mathrm{pH}=0.666537$.


Can absolute enthalpy values be used to determine reaction enthalpy?

Assuming that 3 are true, the only self-consistent set are 1,2 and 3.

Importantly, statements 3 and 4 contradict one another, so one of them must be the incorrect statement.

Consider a reaction with a change in enthalpy ($\Delta H$) of $\pu{-500 kJ/mol}$, and statement 3 is true:

\begin{align}
  \Delta H &= H_\mathrm{f} - H_\mathrm{i}\\
  \implies H_\mathrm{i} &= H_\mathrm{f} - \Delta H\\
  H_\mathrm{i} &= H_\mathrm{f} + \pu{500 kJ/mol}\\
  \implies H_\mathrm{i} &> H_\mathrm{f}
\end{align}
We've shown that statement 1 is true also.

As wolphram says in the comments, you can't measure an absolute value of enthalpy, but you can measure it relative to an chosen zero enthalpy.  If you chose zero enthalpy such that $H_\mathrm{i} = \pu{1600 kJ/mol}$, then by the first equation above we can say that:
\begin{align}
  H_\mathrm{f} &= \Delta H + H_\mathrm{i}\\
  \implies H_\mathrm{f} &= \pu{-500 kJ/mol} + \pu{1600 kJ/mol}\\
\implies H_\mathrm{f} &= \pu{1100 kJ/mol}
\end{align}
We've shown that statement 2 is true as well.

Statement 4 contradicts statement 3 (as well as the other two), so statement 4 must be incorrect.


You are indeed correct that the elemental molecule $\ce{H2}$ is the most abundant molecular species in the universe by far.[1]. In order to answer your question

Is water the most common multielemental compound?

I only focus on neutral species, of which around $50\%$ are in molecular, i.e. multiatomic form.[2]
In this category, second only in total abundance to molecular hydrogen $\ce{H2}$ is carbon monoxide $\ce{CO}$.
This has become evident during the extensive work that has been conducted in the last few decades as large-scale $\ce{CO}$ surveys of the Milky Way galaxy and other galaxies have been performed, further corroborating $\ce{CO}$ as the second most abundant neutral species and technically[3] as most abundant interstellar compound.
To conclude, it might be interesting to realize how substantial the difference in mass is still between the two most abundant molecular species in interstellar space:[2]

Estimates of the galactic mass of $\ce{H2}$ and $\ce{CO}$ can now be given with reasonable accuracy $m(\ce{H2}) = 2 \cdot 10^9 \; M_\odot$, and $m(\ce{CO}) = 2 \cdot 10^6 \; M_\odot$.

So in order to estimate $m(\ce{H2})$ which constitutes the bulk of the gas mass, carbon monoxide $\ce{CO}$ is typically used as tracer or proxy for $\ce{H2}$, as further reading I suggest a very informative post on the physics.SE: Why is $\ce{CO}$ a good tracer for $\ce{H2}$? How are those molecules correlated?
The mass unit $M_\odot$ is commonly used in astrophysics and represents the mass of the Sun, i.e. $M_\odot \cong \pu{1.98E30 kg}$.
The water molecule $\ce{H2O}$ is probably next, i.e. in third place, of the most abundant molecular (neutral) species in the universe which can be directly underpinned by the fact that it made up of the first and third most abundant elements in the universe, hydrogen $\ce{H}$ and oxygen $\ce{O}$*[4]. But I am not aware of any reputable source that explicitly states or confirms this assumption.
Footnotes & References

vide, e.g. Why is $\ce{CO}$ a good tracer for $\ce{H2}$? How are those molecules correlated? or Wikipedia: List of interstellar and circumstellar molecules respectively

Winnewisser, G. The chemistry of interstellar molecules. Topics in Current Chemistry,Cosmo- and Geochemistry 1981, 39–71. DOI: 10.1007/3-540-10920-X_14.

vide ut supra, comment by Karsten Theis:

$\ce{H2}$ is not a compound, it is an element. It is a molecule rather than an atom, though.

van Dishoeck, E. F.; Herbst, E.; Neufeld, D. A. Interstellar Water Chemistry: From Laboratory to Observations. Chem. Rev. 2013, 113 (12), 9043–9085. DOI: 10.1021/cr4003177.


$$\ce{C(graphite) + O2(g) -> CO2(g)}$$

$$\Delta H^{\varnothing} = -393.5\ \mathrm{kJ}\,\mathrm{mol}^{-1}$$

That is, the reaction is exothermic.

Since the reactants are elements in standard state, the heat of formation for carbon dioxide is the same as the enthalpy change of the above reaction.