Is it correct to talk about an empty orbital?

"The properties of an orbital are those of an electron contained in it. It is normal practice, illogical though it may sound, to talk of 'vacant orbitals'. The properties of vacant orbitals are those calculated for electrons occupying them."

I consider this a bit of a truism, especially the last sentence. IUPAC defines as an orbital:

  Wavefunction depending explicitly on the spatial coordinates of only one electron.

which as noted in this answer falls short of considering spin. So let's use that definition for spatial orbitals. (For a spin-orbital, the analogous definition would be: depending explicitly on the spatial coordinates and spin coordinate ...) At this point, let us point out that a spin-orbital is a one-electron wavefunction. It is not an observable. One can observe the density or even the spin-density by means of X-ray diffraction, indirectly by NMR, ESR etc. (Another truism: orbitals of a single-electron systems, such as the hydrogen atom, are an important, but also somewhat trivial exception.) The density can be calculated from a many-electron wavefunction (WF) and a popular way of obtaining such a WF is combining several orbitals. This involves a lot more theory and mathematics than introductory chemistry courses can show.

So one can decide that if no electron is present, there is no orbital and stand on sound mathematical ground. Then again, I can just put an electron there by means of mathematics. Depending on the "computation-chemistry-method", for instance when employing a basis set (which is by far the most common approach), one can calculate properties even for empty orbitals such as the orbital energy (not an observable), which can be used to (approximately) describe electronic excitations (an observable). Computational chemists have resolved to call the empty orbitals "virtual" to bridge the gap between the two opposing views and use them as mathematical tools to describe excited states or to improve the quality of the description of the ground state.

  Since an orbital isn't defined until it is occupied by an electron, would it still be correct to say that an empty orbital (such as a LUMO) can interact with other filled orbitals?

Let's consider the formation of a bond between two hydrogen atoms in a special way: reversing a heterolytic splitting, which we will briefly compare with one reversing a homolytic splitting.
In order to decide this question, one needs to split the non-existent hair dividing the following two views: a) Since there are no empty orbitals, in the heterolytic case, the proton will distort the orbitals on the anion until the bond is formed. Does the orbital on the former proton now magically appear? b) There is an empty orbital on the proton. However, at large distance, its effect on the anion can be simulated by an electric field, which should not carry orbitals (one can assume that the metal plates creating the field are far away and crank up the charge). If a situation involving an empty orbital cannot be distinguished from one where there isn't one, is it really there?

Of course, the end result of either way of looking at things is the same hydrogen molecule that we also get from combining two hydrogen atoms the standard way. I thus suggest to abandon the view of orbitals carried around by atoms (except in the computational chemist way, as will be outlined below). Rather, I suggest to think of the effective potential felt by a newly added electron - where would it go? Regardless of how the nuclei got to where they are now, where do the electrons go? "Unoccupied orbital" is then a useful shorthand for the relevant regions.

(At this point, I consider the original question answered. I will elaborate a bit on how I got here.)
The last question of the previous paragraph is one way of looking at the algorithm of the Hartree-Fock procedure (the first step of wavefunction-based quantum chemistry, which is comparable to density functional theory, DFT, in this regard). We have some "basis set" (and we do not care what is looks like right now) containing candidates for orbitals, nuclei (i.e. charges and positions) and a number of electrons. The first step is to evaluate the potential/the forces acting on the candidate orbitals, which is nuclei only at this point$^1$. One then linearly combines the candidate orbitals (that's LCAO right there) to form the best choice one can make. Obeying the aufbau principle, one fills the candidate orbitals until all electrons have got an orbital. One then updates the potential (which now also considers the electrons), combines again, fills the new candidates, updates the potential again and so forth until the changes between iterations are small. The result is then evaluated in terms of energy, and possibly electron density or other observables. By doing it this way, one arrives at the familiar MO picture without assuming an electron distribution or specific orbitals filled in a certain way on certain atoms.

Let's get back to the basis set. In the previous paragraph, no assumption was made on its nature. The initial candidate orbitals could be any shape (as long as they are reasonable from a mathematical point of view, for instance, they must be twice differentiable, square-integrable etc.) and many different choices exist. Experience shows that using functions resembling the exact orbitals of hydrogen are very useful for molecular chemistry$^2$. This is because they combine results of decent accuracy with computational efficiency and chemist-friendly starting points of interpretation - such as HOMOs and LUMOs that can be related to individual atoms, flawed though as that line of reasoning may be (as I have hoped to show with this wall of text).

Footnotes:
$^1$ A bit of a white lie: this is only the special situation of the so-called core guess. But done here for illustration.

$^2$ Solid state chemistry is a different matter, where standing waves are the norm.


Substituents are listed alphabetically by the name of the substituent. The N in front of the substituent is a locant, and thus is not used when alphabetizing:

  3-chloro-N-ethyl-4-methyl-N-propylhexanamide is correct because the substituents are in correct alphabetic order:

chloro, ethyl, methyl, propyl


The procedure you are describing sounds like fractional freezing. Acetic acid is frequently purified by this method, hence 100% acetic acid is often called "glacial" acetic acid.

  As an example of how this works, let's consider a mixture of acetic acid and water.

Acetic acid has a freezing point of 16.6 oC. Water has a freezing point of 0  o. They form a eutectic at -26.7 oC (although I do not know the composition). The phase diagram probably looks something like (but is not) the following.

Phase Legend:

A = Liquid solution
B = Solid solution of acetic acid in water PLUS liquid solution
C = Solid solution of water in acetic acid PLUS liquid solution
D = Solid solution of acetic acid in water
E = Solid solution of water in acetic acid
F = Solid solution of acetic acid in water PLUS Solid solution of water in acetic acid

A mixture of acetic acid and water that is approximately 72% acetic acid is cooled until it begins to freeze (solid blue vertical line), which happens around 2 oC. The phase changes from A (liquid solution) to C (Solid solution of water in acetic acid PLUS liquid solution). The solid solution is enriched in acetic acid compared to the original liquid solution. The composition of the remaining supernatant liquid solution is enriched in water compared to the original liquid solution. The composition of the solid solution can be determined by drawing a horizontal (red) line to the phase boundary between C and E and reading down to 91% acetic acid. The composition of the remaining supernatant liquid solution can be determined by drawing a horizontal (green) line to the phase boundary between A and C and reading down to 63% acetic.

The supernatant phase is removed and discarded, and the solid phase is allowed to melt. The process is repeated. Each successive freezing yields a new solid solution that is enriched in acetic acid, until the purity is as high as you want it.

Note, that since water and acetic acid have a eutectic point, if we has started on the left side of the eutectic point, the phases would have been reverse. The solid would be enriched in water and the supernatant would have been enriched in acetic acid. The supernatant would be remove and kept, and the solid would be discarded. The supernatant would then be cooled and fractionally frozen again to further enrich the supernatant in acetic acid until the eutectic point is reached (at which point the composition of the supernatant is equivalent to the composition of the solid phase). If you want to isolate acetic acid from a solution that is mostly water, I would distil the solution until the composition is mostly acetic acid, and then fractionally freeze it.


What does "less hindered hydrogen" mean?

Consider these following two structures from the same compound:

The two Newman projections are along the horizontal $\ce{C-C}$ bond which is, of course, freely rotatable. The two structures represent two possible syn-elimination transition states. Note that in the left-hand structure there is a steric clash between the tert-butyl group and the methyl group. The right-hand structure has no such clash. It is therefore the preferred transition state for this reaction.

‘Less hindered hydrogen’ basically means that hydrogen that leads to the more stable transition state.


Number of gauche interactions in an alkane

When we sight down a carbon-carbon single bond and draw a Newman projection of a particular conformation, any two groups on the adjacent (vicinal) carbons that have a dihedral angle of 60° between them are said to have a gauche relationship.

In the following figure, the $\ce{X}$ and $\ce{Y}$ have a gauche relationship.

$\ce{X}$ and $\ce{Y}$ can be the same (they can both be methyl groups, other alkyl groups, other atoms such as halogen, etc.) or they can be different from each other. In most organic classes, the gauche interaction between methyl groups or alkyl groups is frequently discussed.

The answers you included (2,3,3) are correct.  Looking at the middle molecule (b) in your figure, there are 3 gauche interactions.  They are between

the front methyl and the back ethyl
the same front methyl and the back methyl
the other front methyl and the back methyl


Do you refer to a bonded atom as an "ion"?

Answer: Hydrogen atom.
Remember, the Hydrogen in this case is polar covalently bonded to the Oxygen. The Oxygen has not completely stripped the Hydrogen of it electron but it DOES have primary ownership.
Take a look at NaOH. In this case you would consider the Sodium portion of this to be an ion. Because the Oxygen has completely stripped the Na of its electron so the two are now held together by opposite charges. [Na+][OH-]
At least that's how i like to think about it.

All the best,
D


What would be the most suitable solvent for a single-solvent recrystallization?

The ideal properties for a recrystallization solvent are that the compound of interest is soluble at high temperature but insoluble at low temperature. You have addressed the first part, so you just have to assess the second part.

Given that acetanilide is a non-polar compound, I would expect that it would have the lowest solubility in the most polar solvent. For this reason, water is probably the best, then ethanol. I would expect acetanilide to be readily soluble in ethyl acetate at room temperature, making it unsuitable for this recrystallization.


Avogadro Molecule Editor: How to render image from command line?

There is no way to currently do this using Avogadro, although Python bindings have been "in the works" for a while. The problem is, with volunteer developers, there's not always enough free time to tackle all features.

You can create pseudo-3D images using Open Babel. I developed this for the Pitt Quantum Repository to give SVG and PNG images in the search interface. It's not perfect, but I like the effect.

obabel beta-carotene.xyz -O beta-carotene-2d.svg -xS -xd -xb none

The -xS indicates this 3D "ball and stick" style, -xd prevents the molecule name from being included, and -xb none gives  transparent background to the SVG (useful for presentations).


How to determine polarity of components in TLC

I am assuming this is TLC run on silica. Silica is a mildly acidic surface so basic compounds such as amines do not come off the baseline unless you use a very polar eluent with added base (e.g. ammonia/triethylamine). Starting material B is a basic amine. Starting material A is an epoxide and fairly non-polar (kinda similar to an ether) and will run high in most systems, certainly any system where B comes off the baseline. Your product C is more basic (tertiary v secondary), this is the strongest effect. The second non-polar aromatic group raises the likelihood that it will run above B, but this is a weak effect. The alcohol is a polar group and will also pull it down thus the sum of these effects is that it runs lower

This missing information is the eluent. From what what is given in the question it is clear that the eluent is something like DCM/MeOH/NH3 (that would be my first attempt to separate such basic components). As B is in excess what you are really looking for is loss of A. When you can no longer see any A by TLC, the reaction can be worked up.

The groups that make for low running are those that interact strongly with the silica surface: basic amines, acids, alcohols. This is a general rule of thumb, but in my experience trying to guess where a multi-functioned molecule is going to run is quite a lottery. I do not believe that it is a safe assumption that a product similar to C would always run below a SM similar to B.


How does water deionization process work?

Most cost/energy efficient method to make deionized water is reverse osmosis. The mechanism is to apply external pressure to water and force it to go through a layer or layers of semipermeable membrane.


Shouldn't it take more energy to break CO2 compared to CO?

The bonds are not equivalent, therefore they have different bond energy.

  Carbon monoxide consists of one carbon atom and one oxygen atom, connected by a triple bond that consists of two covalent bonds as well as one dative covalent bond.It is the simplest oxocarbon and is isoelectronic with other triply-bonded diatomic molecules having ten valence electrons, including the cyanide anion, the nitrosonium cation and molecular nitrogen. In coordination complexes the carbon monoxide ligand is called carbonyl.

Source: Wikipedia page on Carbon Monoxide


Predicting the spontaneity of non-isothermal reactions

For an adiabatic spontaneous process in a closed system (like the one in your example), according to the Clausius Inequality, the change in entropy is greater than zero.  If you have a closed system in which a chemical reaction can occur, if the system is adiabatic, the same criterion applies.

The criterion that, at constant temperature, if a reaction is spontaneous, $\Delta G^0$ in going from pure reactants to pure products at 1 bar will be negative is just a rule of thumb which, over the years, has confused students to no end.  If you start out with pure reactants and mix them, spontaneous reaction will always occur to some extent.  It's just that, if $\Delta G^0$ is positive, the reaction will tend to proceed  to a lesser extent when equilibrating, and if $\Delta G^0$ is negative, it will tend to proceed to a greater extent when equilibrating.  So, in this sense, every reaction is spontaneous.


What you did in your explanation is correct. First find the difference in the weight:
$$20.3102-20.2367=0.0735 \ \mathrm{g}$$

We know that from your density of water:
$\mathrm{\frac{1\ g}{1\ mL}\ \text{that }\ 0.0735 \ g = 0.0735\ mL}$

Then the difference between the unknown and the vessel is calculated as:
$$20.300-20.2367 = 0.0933\ \mathrm{g}$$

and then density is expressed as mass over volume, so:
$$\mathrm{\frac{0.0933\ g}{0.0735\ mL}=1.2693_8\ g\ mL^{-1}}$$


pH effect in the hydrophobic interactions between two polypeptide chains

Here is one possible mechanism:

In tightly folded, compact, "globular" protein structures, the most hydrophobic bits of the protein tend to be in the core of the structure, away from the surface.

If the pH changes, then by mechanisms you mention in your question, this structure can be disrupted.  If it is side chain interactions among surface-oriented, polar residues that "hold together" the globular structure, then a pH change can disrupt those interactions, allowing the previously isolated hydrophobic "core" to become  exposed.  This core can now bind more surfactant molecules, aggregate with itself, or do any number of other "hydrophobic" things.


Are there other Density Functional Theories other than Kohn Sham DFT?

KS-DFT is the basis for all orbital-based DFT Formulations. In it you have different specialities (constricted DFT, TD-DFT, rs-DFT to name a few).

KS-DFT is used for molecular problems, because of atom bonds etc.

TD-DFT is used for time-dependend problems like excitation energies, polarizabilities etc.

RS-DFT is using range-separated functionals (long-range, short-range) to handle effects better in the long-range domain.

Addition by @pentavalentcarbon:
Current cDFT implementations are based on KS-DFT and perform a real-space partitioning of the electron density.

Plane Wave DFT (no orbitals, electron density calculated with plane waves, was the first DFT) is used for periodic problems like metals etc.

DFTB (Density Functional Tight Binding) is good for Molecular Dynamics since its an approximate DFT and allows handling of larger systems and greater timescales.
Addition by @pentavalentcarbon:
A good analogy for DFTB would be that it is similar to semiempirical methods (PM7, other NDDO parametrizations) where certain integral matrix elements are replaced with precomputed, fitted parameters, but there is a minimal basis set present.

KS-DFT is dominant because as computational chemists we are mostly working with molecules. As that we need a simulation of atom bonds. There is a good Book about DFT which I can only recommend (A chemists guide to Density Functional Theory): http://www.goodreads.com/book/show/70706.A_Chemist_s_Guide_to_Density_Functional_Theory


The real definition of the $\text{pH}$ is not in terms of concentration but in terms of the activity of a proton,

\begin{equation}
  \text{pH} = - \log a_{\ce{H+}} \ ,
\end{equation}

and the activity is a dimensionless quantity. You can think of the activity as a generalization of the mole fraction that takes into account deviations from the ideal behaviour in real solutions.
By introducing the (dimensionless) activity coefficient $\gamma_{\ce{H+}}$, which represents the effect of the deviations from the ideal behaviour on the concentration, you can link the activity to the concentration via

\begin{equation}
  a_{\ce{H+}} = \frac{\gamma_{\ce{H+}} c_{\ce{H+}}}{c^0} \ ,
\end{equation}

where $c^0$ is the standard concentration of $1 \, \text{mol}/\text{L}$.
If you ignore the non-ideal contributions you can approximately express the $\text{pH}$ in terms of the normalized proton concentration

\begin{equation}
  \text{pH} \approx - \log \frac{c_{\ce{H+}}}{c^0} \ .
\end{equation}

In general, there can be no logarithm of a quantity bearing a unit. If however you encounter such a case it is usually due to sloppy notation: either the argument of the logarithm is implicitly understood to be normalized and thus becomes unitless or the units in the logarithm's argument originate from using the mathematical properties of logarithms to divide the logarithm of a product which is by itself unitless into a sum of logarithms: $\log(a \cdot b) = \log(a) + \log(b)$.


Biochemical conversion of ethanol to acetic acid

The principal conversion of ethanol to acetaldehyde takes place in the cytosol of liver cells, no ETC involved directly. This is mediated by the enzyme alcohol dehydrogenase (ADH). However, the NADH produced as a consequence would indeed deliver its electrons to the ETS, an electron transport chain.

The acetaldehyde is then metabolized to acetate in the mitochondria. Again, the involvement of the ETS is in accepting electrons from the NADH that is also produced in this step.

Here's the details: https://pubs.niaaa.nih.gov/publications/arh294/245-255.htm


Geminal coupling constant of formaldehyde protons in NMR

This is an extrapolation of an experimentally measured coupling constant. It is possible to determine the coupling constant, $J_{\ce{H-H}}$ based on the $J_{\ce{H-D}}$, where the following relationship is valid:

$$ \left |J_{\ce{H-H}} \right | = (\gamma_\ce{H}/\gamma_\ce{D}) \cdot \left |J_{\ce{H-D}}\right |$$

Shapiro, Kopchick and Ebersole reported the coupling constant for $J_{\ce{H-D}}$ in formaldehyde-d1 in 1963 (The Journal of Chemical Physics 39, 3154 (1963); doi: 10.1063/1.1734163), and calculated the $J_{\ce{H-H}}$ from this. Changes in coupling constants due to the influence of isotopically-modified electron density are apparently very modest for H/D, and so this relationship appears to hold well.

There is no experimental method to measure coupling between magnetically equivalent spins. In order to break the redundancy of the energy transitions, one must remove the equivalence.

You can actually check this isotope relationship very easily with common NMR solvents and the $\ce{C-H}$ and $\ce{C-D}$ coupling if you run carbon experiments ($\gamma_\ce{H}/\gamma_\ce{D}=6.51439$).

$$\begin{array}{ccc} \hline
\text{Solvent} & J_{\ce{C-D}} & J_{\ce{C-H}}\\[1.1em] \hline
\ce{CHCl3} & 32 & 208\\
\ce{DMSO} & 21 & 137 \\ \hline \end{array}$$


This Wikipedia article gives a good, if brief, discussion of a slow-release form of iodine called "povidone-iodine" in which iodine is the only active ingredient.

  Free iodine, slowly liberated from the povidone-iodine (PVP-I) complex
  in solution, kills eukaryotic or prokaryotic cells through iodination
  of lipids and oxidation of cytoplasmic and membrane compounds. This
  agent exhibits a broad range of microbicidal activity against
  bacteria, fungi, protozoa, and viruses. Slow release of iodine from
  the PVP-I complex in solution minimizes iodine toxicity towards
  mammalian cells.

  PVP-I can be loaded into hydrogels, which can be based on
  carboxymethyl cellulose (CMC), poly(vinyl alcohol) (PVA), and gelatin,
  or on crosslinked polyacrylamide. These hydrogels can be used for
  wound dressing. The rate of release of the iodine in the PVP-I is
  heavily dependent on the hydrogel composition: it increases with more
  CMC/PVA and decreases with more gelatin.


What is molecular system in general terms? What is a supramolecular system?

Molecular chemistry is based predominantly on interactions through covalent bonds. It looks at molecules as individual entities, and is limited in its scope of accomodating intermolecular interactions. Your solution of sugar in water could easily be studied as a molecular system, if all you are interested in is the sugar molecule itself. Subtle differences in molecular conformation of the sugar in different solvent systems, for instance, would simply be written off as 'solvent effects'.

Supramolecular chemistry is based on intermolecular interactions. Typically, these are based on weak interactions (non-covalent forces), and include:

ion-ion interactions
ion-dipole interactions
dipole-dipole interactions
hydrogen bonding
van der Waals forces
pi stacking
hydrophobic interactions
and probably some others

It is through these types of interactions of that we can explore phenomena such as host-guest association, self-assembly, metal-ligand chelation, micelles, and even the structure of DNA, as all of these systems rely on the weak intermolecular interactions outlined above. You could also study your aqueous sugar system as a supramolecular system, and here you could consider what predominant intermolecular interactions are at work, and how they affect the overall chemistry of the bulk system.

Water is supposed to exist as water clusters, weak oligomeric and polymeric forms of $\ce{(H2O)_{n}}$


Does vibrating an NMR sample produce the same benefits and artifacts as spinning an NMR sample?

When you spin a sample you make sure that orientation effects are averaged faster than the relaxation time. This holds true if you spin a sample at 8 s^-1 and H1 relaxes at 1 s^-1. Now, wobbling (aka shaking) will keep orientation of the molecules for the most part (unless it creates spinning motion in the solution). It will change several %, but the general direction will be the same. So, no, shaking will not do the molecule averaging like spinning does.

Btw, why do we spin samples. I mean, if the tube is half decent and you spend 5 minutes on shimming you get about as good of a result. All spinning does is hiding poor XY shims by averaging them.


Why do α-amino acids have a C-H bond at the α-carbon?

In the most general sense, the term α-amino acid does encompass doubly substituted structures like those you mention. After all, the name doesn't specify any particular substitution, or lack thereof, at the α-carbon. The name only really means that there is an amino group α to a carboxylic acid, i.e. one carbon away. Similar nomenclature is used to describe molecules where the amino group is 2, 3, ... carbons away from the carboxylic acid, also without any specification of substitution.

The disubstituted compounds you described, $\ce{H2N-C(R)(R')-COOH}$, can be synthesised in the laboratory. Pharmaceutical companies seem to have an interest in these compounds. Here is one example from a collaboration between Eli Lilly and IUPUI1 (there are many more - pharmaceutical companies like Pfizer2 even have some patents on these compounds):

Wikipedia actually has an example of a naturally occurring α-amino acid without a hydrogen at the α-carbon. Apparently, it is produced by some fungi:

However, the term α-amino acid is nearly always used to refer to the proteinogenic amino acids, which do happen to have a hydrogen at the α-carbon.
These proteinogenic amino acids were discovered many years ago. Generally, they are obtained by treatment of proteins with acid (which hydrolyses proteins into their constituent amino acids). Their chemical formulae were painstakingly identified by a series of decomposition reactions - back in the old days, there was no NMR. The history of the discovery of the proteinogenic amino acids is described in a review article3 from 1931(!) - which tells you how long ago these were discovered.
So, part of the reason why these aren't disubstituted is simply because we found them to not be so. However, I can understand if one complains of this conclusion being intellectually unsatisfying. So, here is some extra speculation.
One major step involved in the biosynthesis of amino acids is transamination. I don't have a list on me right now, but I think most amino acids go through such a step at one point in time.

The mechanism (described on Wikipedia, or in any biochemistry textbook) necessitates that one hydrogen is added to the α-carbon. [For the organic chemists out there, this is essentially an enzyme-mediated reductive amination.]
It turns out that, on top of this, it is also possible that such disubstituted amino acids are actually harmful. According to Weber and Miller, writing in the Journal of Molecular Evolution:4

Replacement of the α-hydrogen by a larger substituent, such as a methyl group, would also increase significantly steric hindrance around the amino and carboxyl groups. Steric difficulties have been encountered in the chemical synthesis of peptides with α-aminoisobutyric acid.
Also, the severely constrained stereochemical behavior of peptides of α-methyl-α-amino-n-butyric acid suggests that peptides composed of α-methyl substituted amino acids would not have the structural versatility of peptides derived from α-hydrogen substituted amino acids.
An additional factor involves the ribosomal peptidyl transferase, which would develop a specificity for either an α-hydrogen or an α-methyl of the amino acid. This would prevent the synthesis of proteins containing both types of amino acids.

References

Green, J. E.; Bender, D. M.; Jackson, S.; O'Donnell, M. J.; McCarthy, J. R. Mitsunobu Approach to the Synthesis of Optically Active α,α-Disubstituted Amino Acids. Org. Lett. 2009, 11 (4), 807–810. DOI: 10.1021/ol802325h.

Graham, S. R.; Mantell, S. J.; Rawson, D. J.; Schwarz, J. B. (Pfizer, Inc.) Amino acid derivatives. U.S. Patent 7612226 B2, Nov 3, 2009.

Vickery, H. B.; Schmidt, C. L. A. The History of the Discovery of the Amino Acids. Chem. Rev. 1931, 9 (2), 169–318. DOI: 10.1021/cr60033a001.

Weber, A. L.; Miller, S. L. Reasons for the occurrence of the twenty coded protein amino acids. J. Mol. Evol. 1981, 17 (5), 273–284. DOI: 10.1007/BF01795749.


Compensating for hardness in aqueous NaOH solution?

The short answer to your question is that the dGH of your water is negligible compared to the amount of lye you are using, so no adjustment needs to be made on the basis of hardness. 15 dGH corresponds to 150 mg/L or 0.015% as CaO, which is trivial compared to your 2.5% lye solution.

Other possible reasons for the slow cleaning process include the way in which the oil has polymerized (as a result of what has been cooked and under what conditions) and the quantity of polymerized oil present. Regardless, the means by which you can speed up the cleaning using your method are to increase temperature and increase lye concentration. If heating really is not an option for you, then you could try a 10% solution (as noted by Waylander in the comments). Disposal of this solution is not a problem, just dilute the solution first or during disposal by running water and pouring slowly. Many household cleaning products use lye so this really isn't an issue.


Well, according to this Wikipedia article, it might be an alloy of tantalum carbide and hafnium carbide, specifically $\ce{Ta4HfC5}$.  A melting point of 4215 °C (4488 K) is listed.  But who really knows the answer to your question with so many new, proprietary alloys with complex compositions being fabricated these days.


There is a trend that follows an arrow from the lower left of the periodic table to the upper right.  As the atomic number increases along a period, the radius for each element decreases.  Here is a useful image of this:

Notice that, with the exception of the noble gases, you can draw an arrow from the lower left to the upper right.  The justification for what you see here is that as the principal quantum number increases, the probability that the electrons are more distance from the center of the atom also increases.  The higher the principal quantum number, the higher the energy of the electron and therefore its probable distance from the center of the atom.  Therefore, as you move down a group, the radius increases.

However, another factor is that as the number of protons increase in an atom, so does the positive charge which happens to pull the electrons toward the center of the atom.  So, there is the impact of the protons causing a decrease in radius (probability of the electron being closer to the center) and the probability that the electrons are in an orbital that is farther from the center.

The transition metals, however, have d orbital electrons.  These electrons enter into a shell that is closer to the nucleus and the change is less noticeable.  That is part of the reason why they do not seem to change as much as the other atoms, according to my understanding.


cooling/refrigeration essentiality for carbonated softdrinks

It has a longer shelf life because the bacteria, (which is the reason it will go bad), cannot reproduce as fast in cold temperatures.

As far as taste goes with carbonated soft-drinks, I think it is mostly just about preference.


Mechanism of chlorine disproportionation

According to this paper,
the attacking species is $\ce{OH-}$ rather than $\ce{H2O}$ on $\ce{Cl2}$

The paper also says that:

  The proposed mechanism also provides a good explanation for the high initial rate of reaction. At the instant the strong chlorine solution is mixed with the distilled water the hydroxyl ion concentration is considerably higher than that calculated from $\ce{kw}$ and the hydrogen ion concentration, for it takes a finite time for the hydrogen and hydroxyl ions present to react to reduce the ion product to the equilibrium value. During this period chlorine molecules also react with hydroxyl ions at approximately the same rate as hydrogen ions are reacting. So for this initial period a higher rate of hydrolysis of chlorine is expected than that found after the hydroxyl ion has reached its equilibrium concentration.

It also says that:

  During the course of the reaction the equilibrium concentration of hydroxyl ion is only about 10^(-12) M. Since about 10^-(3) mole per liter of chlorine must react before final equilibrium is reached, new hydroxyl ions must be supplied by the dissociation of water molecules. This could easily complicate the reaction kinetics, if the dissociation rate were not rapid enough. However, calculations assuming that the dissociation of water is a normal bimolecular reaction with an activation energy equal to the endothermicity of the reaction show that the total hydroxyl ion required may be formed in 10^(-5) sec. This is so rapid that the steady state concentration of hydroxyl ion should not be measurably different from the equilibrium value.

What this stuff basically means is that using $\ce{OH-}$ as the nucleophile, as it can explain the initially high rate of reaction(at the back of our minds, we know that $\ce{OH-}$ is a better nucleophile than $\ce{H2O}$). When $\ce{Cl2}$ is added to the distilled water, we find that $\ce{[OH-]}$ is more than that what would be expected at equilibrium value, so $\ce{OH-}$ and $\ce{H+}$ keep combining to reduce the ion product, while at the same time, $\ce{OH-}$ also reacts with $\ce{Cl2}$. So at this time, rate of reaction will be quite high.
Also, the ionization of water to provide the $\ce{OH-}$ and $\ce{H+}$ will be keeping up with the consumption. Also, the reaction rate tells us that collisions in this reaction are very effective, and lead to reaction almost all the time

So, essentially, to explain the initial high reactivity, we take the reaction mechanism to be a coupled equilibria of $\ce{Cl2 + OH- <--> HOCl + Cl-}$ and $\ce{ H2O <--> H+ + OH-}$


Dissolving silver braze from tungsten carbide

Most common silver brazes are 50 to 90 % silver with copper or maybe zinc.  A complication may be the WC; it is sintered with 2 to 20 % cobalt ( usually).


This answer is a summary of the string of very good comments to the OP's original set of short questions.  This is both my take on the questions and an attempt to produce a more readable "net" answer than those contained within the set of individual comments.

Regardless of the intended context, "A is oxidized on B" means that A looses electrons.  It's seems this context likely implies that B is an electrode surface "on" which A is being oxidized.

Regardless of context, if A is oxidized by B, then B is reduced.

If B is an electrode surface, then the electrons from A needn't reduce B itself, but could flow through that part of the cell (the anode).

I hope this (correctly!) answers each of the OP's questions.


This is more a comment than an answer, but too long for the former....
There are two things for you (and your company) to consider:

The SDS of the chemicals being handled - that will give you guidance for how best to handle a chemical. Flavouring chemicals like benzaldehyde, cinnamaldehyde, limonene all have irritation warnings, and need to be handled in well ventilated areas; others like diacetyl and allyl hexanoate are straight up nasty. And I don't mean to be patronising with this list of flavouring chemicals 101. I'm sure your list of chemicals is extensive, but they will all have SDS that will dictate how you should be handling them. Use that to base your arguments for change, not the opinions of an online discussion group.
Relevant building codes for laboratories that will specify the required number of air turnovers per hour. If your workplace is old, there is a good chance that ventilation is not quite what it should be. This is not an easy (cheap) problem to solve, but if the company is aware of any potential problems it will be easier to improve workplace practices of working in fumehoods.


Resonance contributors and IR stretching frequency

Yes, it is true and applicable. In fact, the resonance structure you've drawn with the double bond between the carbonyl carbon and the amide nitrogen is so significant that one can actually observe restricted rotation about this bond on the nmr timescale.  That is to say, if a methyl group was attached to the amide nitrogen, you can see distinct resonances for the syn and anti isomers.  The rotational barrier about the C-N bond is around 20 kcal/mol, depending upon the substituents, etc.  In the case of the ester, for the reason you provided, the barrier to rotation about the carbonyl carbon - ester oxygen bond is much lower, typically in the range of 10-12 kcal/mol.

So in the case of the amide, the second resonance structure is more important than it is for the ester.  Consequently, the amide carbonyl bond is weaker than the carbonyl bond is for the ester.


Why is a silent electric discharge used for the preparation of ozone from oxygen?

According to this Wikipedia article:

  Dielectric-barrier discharge (DBD) is the electrical discharge between two electrodes separated by an insulating dielectric barrier. Originally called silent (inaudible) discharge and also known as ozone production discharge or partial discharge, it was first reported by Ernst Werner von Siemens in 1857. On right, the schematic diagram shows a typical construction of a DBD wherein one of the two electrodes is covered with a dielectric barrier material. The lines between the dielectric and the electrode are representative of the discharge filaments, which are normally visible to the naked eye.

  DBDs can be used to generate optical radiation by the relaxation of excited species in the plasma. The main application here is the generation of UV-radiation. Those excimer ultraviolet lamps can produce light with short wavelengths which can be used to produce ozone in industrial scales. Ozone is still used extensively in industrial air and water treatment. Early 19th-century attempts at commercial nitric acid and ammonia production used DBDs as several nitrogen-oxygen compounds are generated as discharge products.


Converting an Alcohol to Alkyl Chloride (Ibuprofen)

The presence of the aromatic ring next to the position where the reaction is happening it what would allow the reaction to take place under the mild conditions described.

The HCl first protonates the alcohol, causing it to leave (i.e. the first step of an SN1 reaction). This is made 'favourable' by the fact that a cationic charge can be stabilised at the benzylic position (you can push the arrows so that the benzenes pi-electrons move, delocalising the positive charge around the ring). The chlorine anion (produced when the alcohol is protonated) may then attack the cation at the benzylic position, giving the product.


What're compounds like $\ce{CoBr2,CoCl2,AgI,AgBr,AgCl,AgF,MnBr,Co(CN)2}$, supposed to be?

That depends.

In a lot of cases, especially if we are talking about large anions with low charge density, such as nitrate and perchlorate, and moderately small cations, like $\ce{Co(II)}$, obtaining an anhydrous salt is a long, tiring quest, if possible at all. For example, anhydrous metal nitrates for small cations in general are obtained in anhydrous solvents only.

In case we are talking about medium anion with moderate charge density, like $\ce{Cl-}$ and the cation binds with its neighborhood strongly, like $\ce{Cr^{3+}}$, a lot of structures with different hydration degree may be obtained, such as
$\ce{[Cr(H2O)3Cl3], [Cr(H2O)4Cl2]Cl, [Cr(H2O)5Cl]Cl2, [Cr(H2O)6]Cl3}$ in addition to anhydrous form $\ce{CrCl3}$ (and again, obtaining the anhydrous salt from water solutions is virtually impossible)

Sometimes a salt is unsolvable in water. This typically means that it is anhydrous, like silver halides and many sulfides, or at least low-hydrated, like $\ce{Cu(CH3COO)2\cdot H2O}$

In some cases a salt may have even more weird structure. For example, $\ce{MoCl2}$ actually contains $\ce{[Mo6Cl8]^{4+}}$ unit with $\ce{Mo}$ atoms forming an octanedron and $\ce{Cl}$ atoms occupying positions on top of each of the octahedron's faces. This unit is rather stable and the remaining chlorine my be exchanged with the unit intact.

Similarly, a series of oxoacetates is known, built around $\ce{M3O}$ unit. And so on....

Unfortunately, most transition metals are rather unique.  There is a strong vertical similarity between second and third rows of d-elements, but even it is not perfect.

  If they're not, how do they bond with halogens like Cl or ligands like OH/CN? Is it similar to ligand interaction and d-orbital splitting?

Typically yes.

  If they're neither complex ions nor have split d-orbitals, what determines their magnetic properties?

d-orbital splitting happens for any atom with said orbtials in neighborhood with symmetry below spherical, i.e. in all chemically viable cases.


What salts (or other substances) are available for increasing water's boiling point?

For relatively dilute solutions of non-volatile solutes, the following equation (which assumes ideal behavior) can be used to predict boiling point elevation with reasonable accuracy:

$$\Delta T_b = K_b \cdot b \cdot i$$

Here, $K_b$ is the solvent-specific ebullioscopic constant, $b$ is the molality of the solute, and $i$ is the so-called van 't Hoff factor, which accounts for the extent of solute dissociation. For example, if one were to dissolve $\ce{NaCl}$ in water and assume total dissociation, $i = 2$, since two ions are produced in solution. It follows that, ceteris paribus, a salt that dissociates to yield a greater number of ions will produce a greater boiling point elevation. In practice, the magnitude of boiling point elevation is capped by solubility and deviation from ideal behavior at concentration extremes.

Water also forms negative (i.e., higher-boiling) azeotropes with various substances. Addition of a miscible higher-boiling liquid compound can also effect elevation of the boiling point (assuming the mixture does not form a positive azeotrope), a colligative property similar to the above case for salts. Various polyols (e.g., glycerol, propylene glycol, etc.) are commonly used for such applications.


Variations in protonation in reaction mechanisms

English Wikipedia is more correct. Under acidic conditions, carbonyl and imine additions should happen via the oxocarbenium and the iminium ions, respectively.

Also, the in German version, the step between 5 and 6 is not a step; that's a resonance structure. There's also sloppiness in 3 and 10. Those attacks should come from the lone pair. Technically, those steps should've happened before the addition anyway, as I've already pointed out. Bottom line is that, again, the English version looks better.


Acid catalysed bond hydrolysis in 3-bromobut-1-ene

Without looking up data, I would firstly assume that the $\ce{-CHBrCH3}$ group acts like any other alkyl group, and that path I would be preferred (by Markovnikov's rule). Bromine is electron-withdrawing via the inductive effect, but only very weakly.

I did go and look up data for this particular substrate. The reference is J. Org. Chem. 1936, 1 (4), 393–404. On page 399 the authors describe the addition of $\ce{HBr}$ to the above substrate.

  In the absence of air and peroxides, however, 3-bromo-1-butene gives mostly (60 per cent.) 2,3-dibromobutane.

2,3-Dibromobutane is the product arising from path I, which confirms my assumption.


Why is it so hard to turn plastics into something useful or decompose it?

Most plastics are reusable . The problem is wastes are mixes of various different plastics and other materials . The cost to reliably separate them into specific materials is greater than the cost to make new plastic.


Why are radical intermediates are more stable on tertiary carbons?

Before thinking about why tertiary radicals are 'stable', you need to have some kind of understanding about the structure of radicals.

Carbon centered radicals adopt similar geometries to carbocations: three coordinate and planar. Whereas carbocations have an empty p-orbital, radicals have a singularly occupied p-orbital. Overall the geometry is usually described as sp2 if we're thinking about hybridisation. (*)

With the structure of the radical now defined, we can think about what gives the tertiary radical such stability, and to do that we need to discuss what hyperconjugation is.

A good place to start (other than a textbook) might be the IUPAC definition from the Gold book:

  ...The concept of hyperconjugation is [...] applied to carbenium ions and radicals, where the interaction is now between σ-bonds and an unfilled or partially filled π- or p-orbitals...

Essentially, in the sp2 geometry, adjacent C-H bonds are able to interact with the vacant p-orbital, donating a small amount of electron density in order to stabilise the electron deficient p-orbital.

Individual C-H bonds to not exert a particularly strong stabilising effect, however, in a t-butyl cation/radical, there are 9 adjacent C-H bonds, each donating a very small amount of electron density, the sum of which amounts to a net overall stabilisation relative to primary radicals.

Whilst this explains the stability of the tert-butyl radical, there are other mechanisms of stabilisation for radicals which carbocations are not able to participate in. Due to the single electron in the p-orbital, radicals can also be stabilised by resonance, in a similar (but not identical) way to lone pairs or carbanions. A classic example is a radical generated next to a C=O group or C=C group.

Both of these effects discussed (hyper conjugation and resonance) are thermodynamic effects. The stabilisation is lowering the energy, making the radical less reactive.

(*) In the case of radicals, this is a bit misleading, theres a lot of spectroscopic and computational evidence to suggest that radicals deviate from planarity towards a pyramidal structure (the idea of hybridisation and sp2 doesn't hold up well)- especially when electron withdrawing groups such as fluorine are bonded to the radical centre. From the point of view of undergraduate chemistry this is largely irrelevant, however, given that the planar model explains the features of radical reactions that we're usually interest in, as organic chemists rather than theoreticians (EPR studies are one place in which the planar model falls down, the measured values don't correlate well to a completely flat carbon centre).


Iron(III) nitrate and hydrogen peroxide reaction

Nitrates tend to lower the temperature on dissolving, having a positive enthalpy change of solution. For example, $\ce{NH4NO3}$ has an enthalpy of solution in water of ~26 kJ/mol at ~300 K, and it is used in self-cooling cold-packs for that reason.

Sorry, I could not locate the enthalpy change of solution for $\ce{Fe(NO3)3}$, so this is not a definitive answer. You could measure that value, though, with some  $\ce{Fe(NO3)3}$, some $\ce{H2O}$ and a calorimeter.


Why must there be degeneracy in the orbitals to have a Jahn-Teller effect?

In the abstract to the original Jahn-Teller paper it is stated:[1]

  We shall show that stability and degeneracy are not possible simultaneously unless the molecule is a linear one.

Thus the Jahn-Teller effect is specifically about the removal of a degeneracy, and that that will always lower the total energy of the system. Thus you can not have a Jahn-Teller effect without a degeneracy in the reference, fully symmetric state.

Reference:
[1]: H. A. Jahn, E. Teller, Proc. Royal Soc. A 1937, 161, 220–235. DOI: 10.1098/rspa.1937.0142.


How to name this quintuply-bonded chromium dimer?

It was a great effort. However, in the proposed name

  bis[(μ-2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl-1κC1,2κC7)chromium](5 Cr—Cr)

I can see several issues:

In you point 12: Forcing the metal-metal bond multiplicty in the name is interesting, however I wouldn't go against the rules. See the example with the multiple (triple) Cr-Cr bond, in the rules:

  IR-10.2.5.1 The eta (η) convention
  (…)
  Example 26.

  (μ-2η4-buta-1,3-diene-1,4-diyl-1κ2C1,C4)carbonyl-1κC-bis[(η5-cyclopentadienyl)chromium](Cr—Cr)

not "…(3 Cr—Cr)".

The purpose of the rule seems to be to just specify (only) the connectivity, i.e. the count of such and such metal-metal connections.

  IR-9.2.5.6 Trinuclear and larger structures
  (…)
   Example 6.

  octacarbonyl-1κ4C,24κC-bis(triphenylphosphane-3κP)-triangulo-diironplatinum(Fe—Fe)(2 Fe—Pt)

One could speculate that they, to denote the bond order as well, intend to use a multiplied Em dash character variants, which are non-existent presently.

Another motivation might have been that the bond nature / bond-order is often disputed in organometallics.
Your name can denote a different isomer (though maybe impossible), with the two ligands oriented the same way, not say antiparallel, or head-tail, but head-head. I.e. one of the ligand flipped, and connected to chromiums the other way.

From the rules:

  IR-10.2.5.1 The eta (η) convention
  (…)
  Example 24.

  (μ-1η5-cyclopenta-2,4-diene-1,1-diyl-2κC)(μ-2η5-cyclopenta-2,4-diene-1,1-diyl-1κC)bis[(η5-cyclopentadienyl)hydridotungsten]

not bis(μ-1η5-cyclopenta-2,4-diene-1,1-diyl-2κC)bis[(η5-cyclopentadienyl)hydridotungsten].

I.e. repeating the ligands name, with the 1/2 metal locants swapped. Which, however, is obviously inconvenient, maybe the rules will be improved in the future to somehow support relative locants, or so.
2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl part (you construction in points 3, 4) is weird.

The standalone ligand residue would be named according to the organic chemistry nomenclature:

2,6-bis[2,6-di(1-methylethyl)phenyl]phenyl, more correctly
2,6-bis[2,6-di(propan-2-yl)phenyl]phenyl

Inorganic nomenclature does not have so strict "preferred name" concept. Table IR-10.1 lists isopropyl as an acceptable alternative name for the propan-2-yl. So I would go with 2,6-diisopropylphenyl instead of 2,6-di(1-methylethyl)phenyl or , to not complicate the name further.
Problematic locant for the kappa notation in the ligand 2,6-{[2,6-di(1-methylethyl)]diphenyl}phenyl-1κC1,2κC7. Maybe a prime symbol / apostrophe could be used like in biphenyls, terphenyls..

I think it should be rewritten to more verbose, but unambiguous 2-(2,6-diisopropylphenyl-1κC1)-6-(2,6-diisopropylphenyl)phenyl-2κC1 (both locant superscripts after κC's are now 1, but its not a problem).

And one note:

Even though the rules state

  IR-9.2.4.3 Comparison of the eta and kappa conventions

  The eta convention (Section IR-10.2.5.1) is applied in cases where contiguous donor atoms
  within a given ligand are involved in bonding to a central atom. Thus, it is used only when
  there is more than one ligating atom, and the term η1 is not used.

in the literature(web) for the structure with a more complicated ligand

featuring the similar 2,6-di(iPr)Ph···Cr bond, the following shortcut is used

  Ar′Cr(μ2:η1,η3-N3(1-Ad))CrAr′

which contains the η1 (and, it interestingly relates to the NNN part of the ligand on the right, not to the phenyl). But I'm not gonna use it in the proposed name.

So my candidate name is
[μ-2-(2,6-diisopropylphenyl-1κC1)-6-(2,6-diisopropylphenyl)phenyl-2κC1][μ-2-(2,6-diisopropylphenyl-2κC1)-6-(2,6-diisopropylphenyl)phenyl-1κC1]dichromium(Cr—Cr)

Finally, lets mention the paper in which this interesting compound was described: Nguyen, T.; Sutton, A. D., Brynda, M.; Fettinger, J. C.; Long, G. J.; Power, P. P. Synthesis of a Stable Compound with Fivefold Bonding Between Two Chromium(I) Centers. Science 2005, 310 (5749), 844–847.


Why is strontium(II) ion bigger than krypton atom?

Sr2+ is exactly the same as Kr, in terms of electrons and orbitals.

I think the appropriate word for this relation is isoelectronic.
According to Ptable and LumenLearning, $\ce{Kr}$ has an atomic radius of 88 pm, while $\ce{Sr^2+}$ has an atomic radius of 132 pm. Although it is true that $\ce{Kr}$ and $\ce{Sr^2+}$ have the same electron orbital configuration and $\ce{Sr^2+}$ has two more protons than $\ce{Kr}$, $\ce{Sr^2+}$'s greater nuclear attraction is likely countered by the nature of metal ions being more willingly to lose electrons, which implies the electrons are more loosely held by the nucleus than the noble gas $\ce{Kr}$.
Another reason I conjecture is that $\ce{Sr^2+}$ has unfilled 5s orbitals to readily accept promoted electrons, while $\ce{Kr}$'s electrons have less tendency to fill the 5s orbitals because the property of being a noble gas.


Effective nuclear charge of oxygen atom (O) vs oxygen anion (O2-)

The size of an anion is greater compared to its parent atom because former's effective nuclear charge is lesser than that of latter.

This is not the best explanation. The size of the anion is larger because, as Poutnik wrote in the comment, the effective nuclear charge is almost the same (if you account for nuclear charge and inner electrons only), and there is more repulsion among the valence electrons.

The size of fluorine atoms is smaller than the size of sodium atoms because the effective nuclear charge is larger in fluorine  than in sodium.

This is the typical explanation for the biggest jump in atomic sizes. It makes sense that as you add electrons to a given nucleus, the particle gets larger. It makes sense that going down a group, atoms get larger. What needs an explanation is the dip in size going within one period.

So comparing F and Na (or any other group 17 element and the following group 1 element), we go from a high effective nuclear charge (nucleus and inner electrons combined have a +7 charge) to a low effective nuclear charge (nucleus and inner electrons combined have a +1 charge). On top of that, the outer electrons are also in the next higher shell in Na (3rd period) compared to F (2nd period).

Why are we switching explanations?

Depending on which two species you compare in size, some things change and some things stay the same. Comparing $\ce{O, O- and O^2-}$, the nucleus and inner electrons don't change, so you focus on the number of electrons in the outer shell. Comparing isoelectronic species such as $\ce{O^2-, F-, Ne, Na+, Mg^2+}$, the number of electrons stays the same, but the nuclear charge (and the effective charge) change, so you would focus on that. Comparing atoms in a group, e.g. $\ce{Li, Na, K, Rb}$, the effective nuclear charge is similar (corresponding to similar chemistry of the valence electrons) and you would focus on the valence electrons occupying higher and higher shells.

Comparing neutral atoms in order of atomic number, e.g. $\ce{O, F, Ne, Na, Mg}$, everything changes. Within a period the increasing effective nuclear charge dominates (making the atoms smaller and smaller). Jumping from one to the next period, the drastic decrease in effective nuclear is the most pronounced effect (in addition to using the next shell), resulting in a larger radius.


Can gaseous hydrogen and gaseous oxygen be compressed without reaction?

The study Explosion Characteristics of Hydrogen-Air and Hydrogen-Oxygen Mixtures at Elevated Pressures includes data for pressures up to 200 bar.

Data were collected for Hydrogen-Oxygen mixtures at both 20 and 80 degrees C at pressures ranging from 1 to 200 bar.  A high voltage spark was still needed to cause explosion.

Storing a compressed mixture of hydrogen and oxygen still seems extremely unsafe to me.  I definitely wouldn't do it.


Mixtures readily oxidzing organic matter

In biophysics, when you want to destroy anything organic on a platinum probe(1), you usually use sulfochromic acid (mixture of chromic acid and concentrated sulfuric acid). I do not know anything stronger.

(1) A platinum probe is used just because it can handle being subjected to sulfochromic acid. Most other metals would not.


How to make something solid that dissolves faster than bar soap in water

Potassium soap is soft.  Vary the sodium-potassium ratio to get what you want.  Leave some glycerin in.  Make the soap from polyunsaturated fats.


At equilibrium are the rates of the forward and backward reactions equal to zero?

In equilibrium, the forward and backward rates are equal to each other. The net is zero, but the individual rates are not zero.

Consider something as simple as water — we know that the water molecule can dissociate in to an $\ce{OH-}$ and $\ce{H+}$, and the pH is a direct measure of the amount of $\ce{H+}$. Once equilibrium is achieved, it isn't like $\ce{H2O}$ stops splitting, or $\ce{H+}$ and $\ce{OH-}$ get back together — it is just at a macro level nothing changes.


Boiling sodium hydroxide in stainless steel cup: Solution turning to a blue color

Do you say stainless steel? Stainless steel is an alloy of $\ce{Ni, Cr, Fe}$ with other trace elements, and owes its apparent resistance to corrosion to a protective, adherent, coating of mixed chrome, nickel, and iron oxides. A large amount is probably $\ce{Cr^3+}$, which is amphoteric and will dissolve in hydroxide solution. Once the protective coating is breached chromium will react in base similarly to aluminum. The potential to $\ce{[Cr(OH)4]^-}$ is about $\pu{+1.2 V}$. Stainless steel can be much more reactive than pure iron if the protective layer is continually disrupted.


How do thermodynamics and kinetics control this reaction?

Initially no B and C exist. At short times as $k_1 > k_2$ production of B is favoured over C. (The initial amounts of B and C are so small that back reactions can be ignored wrt. forwards ones.) At long times  equilibrium constants matter and as the equilibrium constant $K_2$ is greatest this is favoured to produce C over B.


Why polar aprotic solvents don't surround negatively charged nucleophiles?

Found a good explanation in Oxford University presentations.

It appears that partially positive side of aprotic solvents is hindered in the middle of the molecule. Only negative charge is on the surface. I've looked at DMSO, DMF, acetone, acetonitrile - all of them share this feature. So in case of a negatively charged ion - such solvents are not good at surrounding it.

It looks like elements that end up partially positive (except for $\ce{H}$) are those that can have 3+ bonds ($\ce{C}$, $\ce{S}$), so they are mostly surrounded by other atoms. It's only $\ce{H}$ who has 1 bond and therefore ends up at the surface of the molecule and can interact with negatively charged nucleophiles.


Do bond lengths vary across molecules of the same kind?

The atoms within a molecule are not stationary, but are constantly vibrating, twisting, etc. Thus, the value given for a bond length between two atoms must be an average. Specifically, the average distance from the nucleus of one atom to the nucleus of the other is what constitutes the definition of bond length.


When is it permissible to use the stoichiometry of a reaction to determine rate laws?

Your reasoning is correct. When the reaction you are investigating is an elementary reaction step, you can write down the rate law using the stoichiometry of the reaction. Mathematically, this is done as follows.

Consider a reaction as

$$
\sum_{i} \nu_{i} R_{i} \rightarrow \sum_{j} \nu_{j} P_{j}
$$

Where $R$ and $P$ are reactants and products respectively and $\nu_{i}$ and $\nu_{j}$ are the stoichiometric coefficients.

If this reaction is an elementary reaction step, than the rate law in the forward direction is given by:

$$
r = k \prod_{i} [R_{i}]^{\nu_{i}}
$$

where $[R_{i}]$ is the concentration of reactant $R_{i}$ and $k$ is the rate constant.

For example, if you consider the reaction

$$
A + 2B \rightarrow 3C,
$$
then the rate in the forward direction is

$$
r = k[A][B]^{2}.
$$


Hydroboration oxidation reaction on trans,trans,trans-1,5,9-cyclododecatriene with carbon monoxide

You guessed it right: a ketone will be formed.
This reaction involves migration of alkyl groups from boron to the carbon atom of CO. There will be three migrations after which you will get your alcohol:

Image taken from National Programme on Technology Enhanced Learning (NPTEL). Lecture 26 : Boron Containing Compounds. Image link: Scheme 9

See also Handouts for Organic Chemistry Lectures given at Imperial College London, Chemistry, by Henry Rzepa.


The Hoffmann product is formed as a major product generally in four cases. They are:

When the ${\gamma}$ carbon is 4°
When the leaving group is poor leaving group
When the leaving group is bulky leaving group
When the elimination is internal elimination (${E_i}$ mechanism)

Case 1:

It happens in the example given in the question the hindrance of base in abstraction of proton favors Hoffman product

Case 2:

It happens with poor leaving groups such as flourine (based on the carbanion character on the carbon from which proton is abstracted in the trasition state)

Case 3:

It happens  in  the Hoffmann's exhaustive methylation reaction where the bulky leaving  group  is a 3° amine group

Case 4:
It happens in the treatment of a 3° amine with Caro's acid followed by heating and in heat treatment of xanthate esters

Lastly I want to add that these are general results the yields may vary based on reactant.


What is the electron count in this nickel complex?

The bromide ions contribute an electron pair on each side of the square they form with the nickel centers (thus the bromine has a positive formal charge like a bridged bromonium ion), so in your fraction you need $4×2$ in the numerator where you have $2×2$.  Then it will come out to $16$.


Does symmetry lower the energy of molecules?

Symmetry is preferred if it lowers energy, but it does not always.

My high school chemistry teacher had a good rule of thumb. A chemical system will [almost] always tend towards the lowest energy configuration. (I insert "almost," because there are some cases where this isn't true - usually because of entropy.)

So yes, many things are symmetric and this usually lowers the energy.

You raise a good counter-example. Cyclobutadiene is not square. A square cyclobutadiene would be a di-radical, so it undergoes symmetry breaking (i.e., a Jahn-Teller distortion) and becomes a rectangle.

We see here that the rectangular distortion has lower total energy - one orbital goes down in energy and becomes doubly-occupied. The other goes up in energy and is now empty.

So what about benzene? Why is it symmetric, vs. a double-single-double.. bond alternation?

Well, in this case, the symmetric (aromatic) version is lower in energy.
(Figure, Chemwiki)


Why is decomposition of ammonia gas on quartz a 1st order reaction?

Firstly a correction and general clarification: your first general assertion is not true. The metal surface acts as a heterogenous catalysts (supported by paper linked later), and the kinetics can be sufficiently modeled by the Langmuir-Hinshelwood model for a monomolecular reaction in which the rate determining step is the surface reaction

$$ \ce{A_{ad} -> B_{ad}} $$

so

$$ r = \frac{\mathrm{d}[B]}{\mathrm{dt}} =k_r [A_{ad}] = k_r \theta_A $$

$\theta_A$ is the surface coverage of A. Applying the steady state approximation to $\theta_A$ over the adsoprtion, desorption and reaction steps and using the Langmuir adsorption model gives

$$ \theta_A = \frac{k_{ad}\:p_A}{k_{ad}\:p_A+k_{de}+k_{r}}$$

so the rate is given by

$$r = k_r \; \frac{\bar{k} \: p_A}{1+\bar{k} \: p_A}$$

where $\bar{k}$ is a complex rate constant. For low partial pressures of A, the reaction thus approximately follows first order kinetics and zeroth order for high $p_A$. Here is a nice paper verifying this for several metals.

Now lets get to the quartz. The authors of the paper also mention a possible bimolecular reaction mechanism, of which the key step is the formation of hydrazin from two adsorbed ammonia molecules. They disregard this mechanism due to experimental evidence, but lets apply some metal organic chemistry here.

$\ce{NH3}$ is a great $\sigma$ donor, so a $\sigma$ acidic metal center will happily bond. A simplistic view might be that the nonbonding lone pair is being donated to the metal - taking a look at a MO diagram of $\ce{NH3}$ you will however find that the HOMO (from which we donate) is slightly bonding.

Source

Therefore, this particular $\sigma$ donation will decrease N-H bond strength, which supports the monomolecular reaction mechanism for such a surface (this does put us somewhere in the grey zone between heterogenous and homogenous catalysis).

$\ce{SiO2}$ is not a substance that comes to mind when we are looking for a $\sigma$ acid. So maybe, under these conditions the bimolecular mechanism becomes the preferred pathway. Bimolecular reaction kinetics of this sort can be modeled by Langmuir-Hinshelwood (both reacting species adsorbed) or Eley Rideal (one adsorbed, one in gas phase). Langmuir-Hinshelwood would give us the boundary cases of second order rate at low $p_A$ and zeroth order at high $p_A$, and I could find no experimental results supporting that anywhere. An Eley-Rideal model would give us a rate law of the form

$$ r = k_r \; \frac{\bar{k} \: p_A^2}{(1+\bar{k} \: p_A)}$$

which would give us first order behaviour for high partial pressures of ammonia! Here is a paper finding experimental data supporting this kinetic model. Possibly the reaction product in the rate-determining step is not hydrazin, but an intermediate hydrazinium which then decomposes.

It thus seems plausible that ammmonia decomposition on a quartz surface follows a bimolecular Eley-Rideal mechanism, compared to the monomolecular mechanism on a metal surface - as a result of the different interactions between surface and adsorbent.


Chemical equation that can be balanced in infinite number of ways?

We first note that your reaction (4) can be broken down into two reactions by splitting up those chemical species that depend on $n$ and those that don't: $$\begin{align}\ce{CH4 + H2O &-> CO + 3H2}\\
\ce{CO2 + H2 &-> CO + H2O},
\end{align}$$
so your reactions (1) through (3) are just statements about the relative rates of these two reactions.

  Which one is true? If all of them are true, why my textbook mentions only one of them?

I presume none of them are really true, in the sense that the ratio of the reaction rates isn't something as nice as 1:1 or 1:2 or 1:3, etc., and we haven't accounted for side reactions either. But assuming a 1:1 ratio is certainly a useful simplification.

  What really happens in reality?

The ratio of the rates will depend on the temperature and pressure and composition of the feed. There will be side reactions that occur. You can perform experimental measurements of the product composition to determine what happens.

  Is equation 2 just a simplification of more complicated equations?

Yes.

  Does this have to do anything with the reaction mechanism?

No. The overall reaction does not, in general, tell you anything about the reaction mechanism.


Risks from overheated phenolphthalein in green/orange silica gel drying

I believe you are trying to reactivate moisture absorbing silica gel with a indicator dye. However, you are blaming the internet for the mishap you have faced. Yet, the website you have directed is ScinceDirect, which provides subscription-based access to a large database of scientific and medical research (Producer: Elsevier). Therefore, I find they give wrongful information on scientific research (considering almost all of these papers are peer-reviewed). I think you made a mistake where considering the indicator dye in green/orange silica gel desiccants is phenolphthalein. My conclusion is based on the inventors' remark in their patent application (Ref.1), which states that color change is from orange (when the gel is dry) to creamy white (when the gel is saturated with moisture). My best guess is it contains something different. Even with something else as an indicator dye, it should be able to stand at $\pu{150 ^\circ C}$ temperature in order to be able to reactivated. What that suggest is, your silica gel might be absorbed unknown chemical without your knowledge, which must be thermally unstable.
According to the patent, there are three kind of materials they are suggesting as indicator dyes: (1) phenolphthalein (color change Orange-red to creamy white); (2) thymolphthalein (color change dark-brown to creamy white); and (3) mixture of phenolphthalein/thymolphthalein (color change Orange-is not provided). All three types are able to get reactivated under $\pu{150-250 ^\circ C}$ temperature range (Ref.1).
References:

Waclaw Lanowy, Krystian Lukaszczyk, Przemyslaw Kasprzycki, “Silica gel with moisture indicator, a method of obtaining them, and use of phthalein dyes,” International Patent 2006, WO2006/130029 A2 (PDF).


Can grignard reagent contain secondary amino group?

I guess it wouldn't even exist. Grignard's reagent is basic enough to easily take off an imine proton, so as soon as you try to make such a thing, it will turn upon itself.

Probably they have a suitable protecting group for this, but here I'm stepping outside my expertise.


What is the benefit of high surface area electrodes?

A current is simply the rate of flow charge (and hence the rate of flow of electrons); the more electrons flowing, the higher the current.

Anions migrate across the potential gradient between electrodes, carrying the electrons to the anode and the anions are then oxidised.  The oxidised species (often uncharged) must then diffuse across a concentration gradient to move away from the electrode (via Fick's laws).  Until the oxidised species moves away from the electrode surface, they are blocking the surface for more anions to migrate in, and hence the current is eventually limited when the rate of migration of anions is equal to the rate of diffusion of the oxidised species.  This exchange happens at a fixed rate per unit area at the surface of the electrode.

When we use a larger surface area electrode, there is more area for this exchange to happen, therefore more species involved in the exchange, therefore more electrons can be deposited in a given timeframe, hence permitting a higher current to flow.


When discussing about same environment and different environment, not only the different position should be noted (axial/equatorial), but also the different electronic effects implied.

In the first case:

In short: the molecule has a plane of symmetry, and the substituents are "mirrored" through the plane.
"Long" explanation: This means that they "see" the same electronic environment: the
equatorial substituents have the same electronic effects on the two
axial substituents. This means that their resonance frequency, is the same.

In the second case:

Short answer: no symmetry exists between the substituents
"Long" answer: every group neighboring the substituents is different. One of the two substituents feels the effects of $\ce{CO}$, the other one feels the effect of -$\ce{O}$-. Except in some unlikely and unlucky cases, this means two different signals.

Note to the reader: shielding, resonance frequencies and NMR behavior arise from (way more) complex interactions: I did not mention, above, any possible effect of non-neighbouring substituents, for instance, neither I talked about the shielding or deshielding effects of near substituents.

Nonetheless, a perception of the "symmetry" of the involved groups might give you a "dirt cheap" and operational insight on what could be the outcome of an NMR experiment, in cases as simple as this.


Do all the angles change in cyclopropane and cyclobutane or it is only the C-C-C angles?

The other bond angles can't all be $109.5°$ if the angles in the ring are different from that.  Bond angles around an atom are not independent variables because the amount of space to be spanned is fixed.


Why is pressure negative in force and work

In this context we are talking about the work done by a gas.

Rather than representing it as $w = -PAD$, think of the formula as representing an area (that the gas is pressing against) moving a distance $D$. Since this a change in volume, we can set $w = -P\Delta V$

If you are putting work into the system (say pressing a piston into a cylinder containing a gas at a given pressure), the volume is becoming smaller, so $\Delta V$ is negative.  The negative of the pressure times a negative $\Delta V$ is a positive value, which matches with the sign convention for putting work in.  Vice versa, an expansion of a compressed volume at a given pressure will push the piston back out of the cylinder and allow the gas to do work, so $w$ is negative which follows the sign convention for work output.


Are coordination numbers of elements fixed?

No. Coordination numbers of no element are fixed. For example, if you add chloride ions to a pink solution of octahedral $\ce{[Co(H2O)6]^3+}$, a blue, tetrahedral complex is generated according to the following equation:

$$\ce{[Co(H2O)6]^3+ + 4 Cl- <=>> [CoCl4]- + 6 H2O}$$

In fact, if you study the mechanisms of different metal-catalysed organic reactions, you will notice that different coordination numbers of the central metal and specific switching between them is the key to understand these mechanisms.

Finally, for some complexes even establishing their coordination is hard. Zinc(II) forms aquacomplexes like any other cation. But while those complexes of transition metals with any number of d-orbitals unequal to zero or ten are often well-defined and established, that is not the case for zinc. Zinc(II) aquacomplexes rapidly equilibrate from tetracoordinated and pentacoordinated to hexacoordinated species.


How do I know which atoms react during hydrolysis?

I think you mean which atoms react with the water, as clearly the molecule that reacts is propanamide (it's the only non-water molecule in your question).

The product of the reaction is propanoic acid and ammonia.

$$\ce{CH3CH2CONH2 + H2O -> CH3CH2COOH + NH3}$$

The carbonyl carbon in amides (and esters and acyl halides) has a partial positive charge because of the electron withdrawing effect of the carbonyl oxygen.  (Oxygen is very electronegative.)  This makes the carbon atom is activated toward nucleophilic attack.

This Wikipedia page has a good explanation and different mechanisms for acid vs. base conditions.


Conservation equation in an enzyme-catalysed reaction

The reaction scheme is:

$$\ce{S + E ->[k_1] C1} \tag{1} $$ $$\ce{C1 ->[k_2] E + P} \tag{2}$$ $$\ce{S + C1 <=>[k_3][k_4]C2} \tag{3} $$

Let's define a bit of nomenclature.

S = compound S in chemical reaction equation
s = $[S]_x$, the concentration of S at t=x.
$S^*$ = ds/dt

The whole set of differential equations has 5 equations with 5 unknowns.

\begin{align} \\
S^* = \frac{\mathrm{d}s}{\mathrm{d}t} &= -k_1se-k_3sc_1+k_4c_2  \tag{4} \\ E^* = \frac{\mathrm{d}e}{\mathrm{d}t} &= -k_1se+k_2c_1\tag{5} \\
C^*_1 = \frac{\mathrm{d}c_1}{\mathrm{d}t} &= k_1se-k_2c_1-k_3sc_1+k_4c_2 \tag{6}
\\ C^*_2 = \frac{\mathrm{d}c_2}{\mathrm{d}t} &= k_3sc_1-k_4c_2 \tag{7}\\ P^* = \frac{\mathrm{d}p}{\mathrm{d}t} &= k_2c_1 \tag{8}\end{align}

What is a conservation equation

In mathematical terms a conservation equation is some linear combination of the time derivatives which add up to zero on the right hand side.

The gist is that the various conservation equations allow some simplification in the various interactions in the system.

Rather than just search blindly, there are a couple of relationships that can be exploited to find time derivatives which add up to zero on the right hand side. First is a sort of mass balance. The initial concentration of the reactants has to be spread as the various intermediate species and the final product. A second line of attack is to exploit and equilibrium conditions such as equation (3).

First conservation equation from "mass balance"

The first conservation equation is found noting that at t=0 then all is [S], a constant. Noting that it takes 2 $\ce{S}$ to make one $\ce{C_2}$, at some later time t=x, thus:

$$[\ce{S}]_{t=0} = [\ce{S}]_{t=x} +  [\ce{C_1}]_{t=x} + 2[\ce{C_2}]_{t=x} + [\ce{P}]_{t=x}\tag{9} $$.

Now for some intermediate time we can take derivative of the mass balance equation with respect to time. All the dt's get cumbersome so lets use just $S^*$ from now on to mean ds/dt. From the mass equation then since the initial concentration is a constant we get

$0 = S^* + C^*_1 + 2C^*_2 + P^*\tag{10} $

Most likely solving for either $S^*$, or $P^*$ would give an interesting result.

A second conservation equation from mass balance

The second conservation equation comes from noting the enzyme participation in the reaction.

$$[\ce{E}]_{t=0} = [\ce{E}]_{t=x} +  [\ce{C_1}]_{t=x} + [\ce{C_2}]_{t=x}\tag{11}$$.

Taking the derivative we get:

$ 0 = E^* + C^*_1 + C^*_2 \tag{12}$

which can be rearranged to

$E^* = - C^*_1 - C^*_2\tag{13}$

Substituting equation 6 for $C^*_1$ and equation 7 for $C^*_2$ this gives us equation 5 again which isn't helpful.

A third conservation equation from the equilibrium condition

From the third reaction we can derive the equation.

$ \frac{k_3}{k_4}= \frac{[S]_x[C_1]_x}{[C_4]_x} = \frac{sc_1}{c_4}\tag{14}$

Taking the derivative:
$ 0 = \frac{c_1}{c_4}S^* + \frac{s}{c_4}C^*_1 - \frac{sc_1}{c_4^2}C^*_4\tag{15}$

rearranging

$  S^* = -\frac{s}{c_1}C^*_1 + \frac{s}{c_4}C^*_4\tag{16}$

Infinite time as a boundary condition

Obviously at $t=\infty$, all the S has been converted to all P which is a constant. This factoid isn't very useful though in predicting how long that an "effective infinite time would be. Rather it is just a boundary condition.

$$[\ce{P}]_{t=\infty} = [\ce{S}]_{t=0}\tag{17}$$


When the balloon is fully inflated with helium its density (mass/volume) is less than that of the air around it, allowing it to float. Over time, some of the helium will "leak" out of the balloon. As the helium is released the mass of the balloon decreases slightly, but the volume decreases at a much faster rate, resulting in a balloon with a density greater than that of the air around it, which is why it falls to the ground.


Composition and pH of water produced by air conditioners

Pure condensed water is neutral (pH 7). However, water in contact with air is slightly acidic (approximately pH 5.6) due to dissolved carbon dioxide $\ce{(CO2)}$. Remarkably, the same applies to rainwater.

Furthermore, water condensed from air is not pure. The condensing water collects small amounts of various airborne substances (soluble gases and vapours, dust, and even radioactive particles from radon decay). (Note the large volume of air that is passed through the air conditioner while a certain amount of water is condensed.) Usually, these substances do not significantly affect the pH of the water. However, when the air contains large amounts of pollutants, such as sulfur dioxide or nitrogen oxides (e.g. produced in industrial processes), the condensed water can become more acidic (like acid rain, which can have pH values of about 4).

By way of comparison, typical values for black coffee are about pH 5.0–5.1, and for orange juice about pH 3.3–4.2.


Why the boiling point decreases with increase in branching in alkyl groups?

It's not just how much surface area, but, crucially, the surface area in contact with other molecules. Bear in mind that we are looking especially at van der Waals' forces here (or London dispersion forces, or instantaneous dipole forces, or whatever other name you have for them).

If you have a lot of long chain molecule together, almost all of their surface area is very near the next molecule. Examine these sausages. See how almost their entire length on the left and right side of each sausage is in contact with the next sausage. Going back to molecules, this represents a large surface over which the intermolecular forces can act.

On the other hand, branched molecules do not fit as tightly together. See this image (no pun intended)

Extra branches on the molecules mean that they do not stack as well, so for a given molecule, there is less surface area of that molecule in close contact with another molecule (therefore weaker van der Waals).


Preparation of oxalic acid from acetylene

Ozonolysis of acetylene won't produce oxalic acid.  The triple bond holding the two acetylenic carbons together will be attacked as shown below.  The reaction is complex and depending upon reaction conditions, different products can result (ref_1, ref_2).

Oxalic acid can be produced from acetylene by reacting acetylene with alkaline $\ce{KMnO4}$ (ref_3).

$\ce{H-C#C-H ->C[KMnO4/OH^-]\ HOOC-COOH}$

EDIT: Response to OPs comment
Here is a rough idea of what is going on mechanistically:

1) the permanganate forms a cyclic intermediate

2) that produces a diol

3) since a double bond is still present the reaction can be repeated to yield 2 gem-diols that can equilibrate to an alpha-diketone

4) now the standard alkaline permanganate oxidation of a carbonyl occurs

5) hydroxide can add to a carbonyl and

6) the manganese ion complexes with the carbonyl oxygen

7) the alpha-proton donates its electrons to reform the carbonyl which is now part of a carboxylic acid

8) repeat steps 5-7 on the other carbonyl and you have oxalic acid


To practically dilute this cream from $10\ \%$ to $5\ \%$, you would need to dilute it to a by a factor of $2$. You can see this by assuming an initial volume for your cream, in this case, $100\ \mathrm{ml}$,
and plugging into the $c_1V_1=c_2V_2$ equation. In this case that means
$$c_1:=10~\%~~~~~\\V_1:=100~\text{mL}\\c_2:=5~\%~~~~~~~$$
Solving for $V_2$ renders
$$~~~~~~~~\frac{c_1V_1}{c_2}=V_2$$
$$\frac{10\ \% \cdot 100~\text{mL}}{5\ \%}=V_2~~~~~~~~~~$$
$$\frac{10\ \% \cdot 100~\text{mL}}{5\ \%}=200~\text{mL}$$
Which means that the volume needed in the final solution needs to be $200~\text{mL}$. It is however, important to note that this does not mean that you need to add $200~\text{mL}$, but rather you need to add $100~\text{mL}$ of solvent to your $100~\text{mL}$ of cream for a final solution volume of $200~\text{mL}$.

Further you may calculate the dilution factor that this correlates to by using the equation

$$DF=\frac{V_\text{final}}{V_\text{initial}}$$

In this case $V_\text{final}=V_2$ and $V_\text{initial}=V_1$, so plugging into this equation renders

$$DF=\frac{200~\text{mL}}{100~\text{mL}}=2$$

You may apply these same equations to making a cream at $2.5~\%$ active ingrediant concentration and should find that if you start with the $100~\text{mL}$ of the $10~\%$ stock, you would need to dilute to a final volume of $400~\text{mL}$, which correlates to a dilution factor of $4$. If you start with $100~\text{mL}$ of the $5~\%$ solution you just made, you would need to dilute to a final volume of $200~\text{mL}$, which correlates to a dilution factor of $2$.

If you want to do this practically, then water may not be the best solvent – to continue to be able to use this as a cream, you may consider using a cream without any active ingrediant in it as your solvent – maybe something like a hand cream?


What is the reaction between Magnesium chloride and Sodium bicarbonate?

Having actually perform this reaction, I can say with certainty that when you mix aqueous sodium bicarbonate with aqueous magnesium chloride, there is no apparent reaction. This is consistent with an equilibrium reaction (actually, just ions):

$$\ce{2 NaHCO3 + MgCl2 <=> 2 NaCl + Mg(HCO3)2}$$

However, upon brief heating in a microwave (for example), a white precipitate of magnesium carbonate develops indicating the reaction has moved to the right.

$$\ce{2 NaHCO3 + MgCl2 ->[Heat] 2 NaCl + MgCO3(s) + CO2 (g) + H2O}$$


An F-center (or any color center) is a point defect in the material that generates localized electronic states. When these states are in the band gap, they will result in strong peaks in optical absorption. An F-center in an alkali-halide material is an electron bound to a negative ion vacancy.  A brief overview may be found in Ashcroft and Mermin's Solid State Physics, Chapter 30.


Why does HNO3 not give off H2 when reacting with Cu?

Whenever any metal is added to an acid, the metal is oxidised to one of its positive ions, at the cost of either the acid which is reduced or water which is reduced to hydrogen. The product formed by reduction is what is referred to as the reduction product.

The reduction potential of $\ce{NO3-}$ is $0.8$ which makes the reduction of nitrate instead of hydrogen ion $\ce{H+}$ ($E^\circ=0.0$) more favourable. Thus it is the nitrate in nitric acid which is reduced instead of $\ce{H+}$ or $\ce{H2O}$. But at very low concentration of nitrate ion and low temperature, as explained by Nernst equation, the reduction potential decreases significantly and combined with overpotential effects, it is $\ce{H2}$ which is formed and not the reduction product of nitrate which is $\ce{NO2}$.

In the case of halic acids, the halogens cannot be further reduced. In the case of the sulfate anion, although it has a reduction potential of $0.17$, owing to overpotential effects, sulfate is not reduced, instead, water is reduced yielding $\ce{H2}$. At high concentration, the reduction potential rises sufficiently to overcome the overpotential effects, allowing the reduction of sulfate to form $\ce{SO2}$ at higher concentrations of sulfuric acid. Hence under normal conditions, only nitric acid gives a product other than hydrogen gas.

Reduction potential data

$$\ce{SO4^{2-} + 4H+ + 2e^- -> SO2 + 2H2O}(E^\circ=0.17)$$
$$\ce{NO3- + 2H+ + e- -> NO2 + H2O} (E^\circ=0.8)$$
$$\text{Nernst equation } E=E^\circ- \frac{RT}{nF}\ln Q$$
$Q$= reaction quotient, $E$=redox potential, $E^\circ$=Standard redox potential

EDIT
The oxidation of metal at the cost of reduction of another substance occurs whenever it is thermodynamically favourable. Usually, (except silver, copper) metals have negative reduction potentials, which makes their oxidation at the cost of hydrogen thermodynamically favourable as the $\Delta G<0$. This leads to a spontaneous redox reaction. This is not true if there is a kinetic or physical barrier to reaction. For example, Iron is rendered passive by nitric acid due to the formation of an oxide layer at the surface which prevents any further contact and hence stops the reaction.


Determining stoichiometry of an unsaturated hydrocarbon via combustion

First step: balance your equation.

$\ce{(CH2)$_n$ + O2 -> CO2 + H2O}$ is not balanced, this is:

$$\ce{(CH2)_{\it n} + \frac{3 }{2}{\it n}~O2 -> {\it n}~ CO2 + {\it n}~ H2O}$$

Second step: perform mole balances on relevant species.

The balanced equation has $1 + \frac{3}{2}n$ moles of gas reacting, and $2 n$ moles of gas formed.  However, the question states that the products are cooled to room temperature, which would eliminate the water from the gas phase.  Thus the reaction were are really interested in is

$$\ce{(CH2)_{\it n}_{\rm(g)} + \frac{3 }{2}{\it n}~O2_{(g)} -> {\it n}~ CO2_{(g)} + {\it n}~ H2O}_{(l)}$$

This means that $1 + \frac{3}{2}n$ moles of gas react, and $n$ moles form for each mole of reaction $\xi$.  Thus, the reaction causes a loss in moles of gas of is $-(1 + \frac{1}{2}n)$ moles per mol $\xi$ of reaction.  If we assume the ideal gas law applies, then we could also say that the loss in volume from the reaction is $(1 + \frac{1}{2}n)~{\mathrm{{cm}^3}}$ per $\xi~{\mathrm{{cm}^3}}$ of reaction.

Thus a total gas balance is:

$$\rm GAS~IN - GAS~OUT = GAS~REACTED $$
$$(60~\mathrm{{cm}^3} + 10~\mathrm{{cm}^3}) - (45~\mathrm{{cm}^3})=(1 + \frac{1}{2}n)~\xi$$

A $\ce{CO2}$ balance is

$$\rm \ce{CO2}~IN - \ce{CO2}~OUT = \ce{CO2}~REACTED $$
$$ 0 - x = - n \xi $$

Third step: assess the resulting equations

These equations are not solvable for $x$ because they depend on both $\xi$ and $n$.  That is, there are two equations, and three unknowns.  Writing additional balances on oxygen or on fuel will not help because we will not know the gas remaining for either species, so each new equation would introduce new variables. Thus, we will need additional assumptions to solve the problem.

Fourth step: determine additional assumptions.

Probably we should assume that either (a) fuel was in excess or (b) oxygen was in excess, which is equivalent to assuming that (a) oxygen was completely consumed or (b) fuel was completely consumed.  At least one of these assumptions makes sense because combustion reactions are essentially irreversible.  In my answer I'll assume oxygen was in excess, meaning that the fuel was completely consumed.  You should make the alternate assumption and repeat the procedure.

Fifth step: apply additional assumptions.

Assuming fuel was completely consumed, a fuel balance is

$$\rm FUEL~IN - FUEL~OUT = FUEL~REACTED $$
$$\rm (10~\mathrm{{cm}^3}) - 0=\xi $$

This conveniently gives us a solution for $\xi$, the extent of reaction.  It is $10~\mathrm{{cm}^3}$.

Sixth step: use new information to solve existing equations

We can substitute that solution for $\xi$ into the total gas balance to get that $1 + \frac{1}{2}n = 3$, which implies that $n=4.$  That and our solution for $\xi$ allows us to solve the problem: if oxygen is in excess, then 40 $\mathrm{{cm}^3}$ of $\ce{CO2}$ were formed.


Reaction of vinylacetylene (but-1-en-3-yne) with HCl

Beyond a shadow of a doubt, the second explanation, involving an allene rearrangement is the correct one. As you said, alkenes are more reactive towards electrophilic addition and hence the addition of HCl happens first on the double bond to give 3-chloro-but-1-yne. What follows next is an intramolecular rearrangement, through a cyclic five-membered transition state, which is catalysed by CuCl with NH4Cl. It encompasses the abstraction of a hydrogen from the C4, the migration of the chloride from C3 to C2, following the formation of a double bond at C3 and C4. The product is referred to as chloroprene, which gives neoprene on polymerisation, a useful polymer. However the other explanation offered by your facilitator is used by many as a trick to remember the final product, (including, me!). Hope that helps!


Is there a certain amount of titrant that is ideal?

In a perfect world there would be no optimal range. We do however live in a world full of errors and practical considerations.

Why use over a minimum amount?

Let's assume for demonstration, that each drop has a volume of 0.05 mL. Of course this is not an accurate value as various factors would affect this.

The titration has reached its end point when one last drop changes the colour. If your titre value is very small e.g. 1 mL, then each drop is 5% of your titre.

If only part of the drop is needed to reach the end point then all the extra part of the drop is inaccuracy, which in this scenario could be up to 5%! If you have the misfortune of accidentally adding one drop too many you are introducing an even larger error. This will impact concentration calculations based on the experiment.

Why not just use 'a lot' then?

There are a couple of things here:

1) As noted by Mithoron, this would be a waste. The accuracy gained from requiring additional titrant will level off rather rapidly.

2) If you use more than one full burette's worth there is an error penalty.

3) Using extra titrant is a waste of your time. Consider a 50 mL burette, if it needs to be filled up after each titration, that would be inefficient. I'm assuming you're doing repeats so this has real merit.


Oxidation of water by permanganate: Does this reaction happen?

You have indeed gotten it terribly wrong. The key hint is that you have more than one element that would be reduced: hydrogen also goes from $\mathrm{+I}$ to $\pm0$ in your sample equation. Furthermore, in neutral or weakly basic solutions permanganate gives $\ce{MnO2}$ after oxidation, not manganese(III).

What you do want to do is to reduce permanganate to something (which I have already established would be manganese dioxide) while oxidising something in water to something. Looking at the oxidation states of hydrogen and oxygen in water, the only element that can be oxidised is oxygen, since hydrogen is already at its highest possible oxidation state $\mathrm{+I}$. Oxidising water’s oxygen leads to a few possibilities of which oxygen gas does indeed sound most likely. Therefore, we have the redox pairs:

$$\begin{align}\text{Red: }\ce{\overset{+VII}{Mn}O4-}/\ce{\overset{+IV}{Mn}O2}&&\text{Ox: }\ce{H2\overset{-II}{O}}/\ce{\overset{\pm0}{O2}}\end{align}$$

We can then determine the appropriate half-reactions by the standard procedure. Instead of walking you through it, I will present you with the final result for comparison.

$$\begin{align}\ce{MnO4- + 3e- + 4 H+ &-> MnO2 + 2 H2O}\tag{Red}\\
\ce{2H2O &-> O2 + 4e- + 4 H+}\tag{Ox}\\[1em]\hline
\ce{4 MnO4- + 4 H+ &-> 4 MnO2 + 3O2 + 2 H2O}\tag{Redox}\end{align}$$

It turns out that the final reaction equation formally doesn’t even need water. However, potassium permanganate is a shelf-stable solid under normal laboratory conditions (if light is excluded), so it is reasonable to assume that the reaction will only proceed in aqueous solution.

To determine whether the redox reaction is feasible in aqueous solution, it is easier to go with the two half-reactions we already have; thermodynamically, they must correspond to the same process with the same reaction enthalpy etc. The Nernst equation together with the inspection of the cell potential allows us to estimate whether a process will be spontaneous or not by mere inspection of the standard potentials.

$$\begin{align}E^0_\text{cell} &= E^0_\text{Red} - E^0_\text{Ox}\\
&= \pu{+1.70V} - (\pu{+1.23V})\\
&= \pu{+0.47V}\\
E^0_\text{cell} &> 0 \\&\Longrightarrow \text{spontaneous process}\end{align}$$

Thus, thermodynamics already predict the decomposition of permanganate in aqueous solution to be spontaneous under standard conditions. As mentioned, the Nernst equation will allow us to estimate whether it also is spontaneous given general laboratory conditions; I will assume $\mathrm{pH} = 7$ and $p(\ce{O2}) = \pu{0.2bar}$

$$\begin{align}E &= E^0 - \frac{0.059}{z} \lg \frac{Red}{Ox}\\[0.5em]
E &= \pu{+0.47V} - \frac{\pu{0.059V}}{12}\lg\frac{p(\ce{O2})^3}{[\ce{MnO4-}]^4[\ce{H+}]^4}\\[0.5em]
E &= \pu{+0.47V} - \frac{\pu{0.059V}\times28\times2.1\times(-4)}{12}\lg {[\ce{MnO4-}]}\\[0.3em]
E &= \pu{+0.47V} + \pu{1.16V}\times\lg {[\ce{MnO4-}]}\end{align}$$

We can calculate the minimum concentration from which the process will be spontaneous:

$$\begin{align}\pu{0V} &= \pu{+0.47V} + \pu{1.16V}\times\lg {[\ce{MnO4-}]}\\
\pu{-0.47V} &= \pu{1.16V}\times\lg [\ce{MnO4-}]\\
-0.405 &= \lg [\ce{MnO4-}]\\
10^{-0.405} &= [\ce{MnO4-}]\\
0.393 &= [\ce{MnO4-}]\end{align}$$

Thus, at any concentration higher than $\pu{0.393M}$ permanganate should start formally oxidising water or decomposing.

Whether or not this happens at exactly this concentration depends on kinetic factors, i.e. can oxygen actually be produced under these conditions, but thermodynamically it would be favourable.


Solubility of potassium nitrate and potassium chloride

Recently, I answered a similar type of question. The reasoning for that question is also applicable for this answer also.

Electrostatic force between potassium cation and chloride ion is way stronger than that with nitrate ion. So, more energy is needed to break the  bond between potassium chloride than potassium nitrate. Hence, lattice enthalpy is more in case of potassium chloride and thus more water is needed to dissociate the bonds. So, potassium chloride is less soluble in water.

You can compare the solubilities of potassium nitrate and potassium chloride by observing the graph taken from this site:

You can observe that solubility of both potassium chloride and potassium nitrate is same at approximately 23℃ but then as temperature is increased, solubility of potassium chloride increases steadily but solubility of potassium nitrate increases drastically.

So solubility order:

Potassium nitrate > potassium chloride


How to calculate the partial pressure of oxygen in this mixture of oxygen and argon?

How is the mixture composed? Find expressions for mass, volume and amount of substance.
\begin{align}
\tag1\label1
  V_\text{total} &= V_{\ce{O2}} + V_{\ce{Ar}}\\
\tag3\label2
  n_\text{total} &= n_{\ce{O2}} + n_{\ce{Ar}}\\
\tag3\label3
  m_\text{total} &= m_{\ce{O2}} + m_{\ce{Ar}}
\end{align}

Transform $V_\text{total}$ with the ideal gas and solve for $n_{\ce{Ar}}$
\begin{align}
  \tag4\label4
  V_\text{total} &= \frac{RT}{p_\text{total}}(n_{\ce{O2}} + n_{\ce{Ar}})\\
  \tag5\label5
  n_{\ce{Ar}}  &= \frac{p_\text{total}V_\text{total}}{RT} - n_{\ce{O2}}
\end{align}

Use total mass \eqref{3} and transform it with the with molar mass to amount of substance equation ($m = nM$), then plug in \eqref{5}:
\begin{align}
  \tag3
  m_\text{total} &= m_{\ce{O2}} + m_{\ce{Ar}}\\
  \tag6\label6
  m_\text{total} &= n_{\ce{O2}}M_{\ce{O2}} + n_{\ce{Ar}}M_{\ce{Ar}}\\
  \tag7\label7
  m_\text{total} &= n_{\ce{O2}}M_{\ce{O2}} + \left(\frac{p_\text{total}V_\text{total}}{RT} - n_{\ce{O2}}\right)M_{\ce{Ar}}
\end{align}

Now there is only one variable left, which you do not know:

\begin{align}
  \tag8\label8
  m_\text{total} &= n_{\ce{O2}}\left(M_{\ce{O2}} - M_{\ce{Ar}}\right) + \frac{p_\text{total}V_\text{total}}{RT}\cdot M_{\ce{Ar}}\\
  \tag9\label9
  n_{\ce{O2}} &= \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{RT} \cdot  M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1}
\end{align}

We need the following data (It would be by far better to only use SI [derived] units):

\begin{align}
M_\ce{Ar} &= \pu{39.9 g//mol}\\
M_\ce{O2} &= \pu{32.0 g//mol}\\
R &= \pu{8.314 L kPa //K mol}\\
\pu{1 mmHg} &= \pu{133.322 Pa}\\
p_\text{total} &\approx \pu{90 kPa}\\
T &= \pu{316.5 K}\\
m_\text{total} &= \pu{19.3 g}\\
V_\text{total} &= \pu{16.2 L}
\end{align}

Plugging this into \eqref{9}:
\begin{align}
  \tag9
  n_{\ce{O2}} &= \left(m_\text{total} - \frac{p_\text{total}V_\text{total}}{RT} \cdot  M_{\ce{Ar}}\right)\cdot \left(M_{\ce{O2}} - M_{\ce{Ar}}\right)^{-1}\\
  n_{\ce{O2}} &= \left(\pu{19.3 g} - \frac{\pu{90 kPa}\cdot\pu{16.2 L}}{\pu{8.314 L kPa //K mol}\cdot\pu{316.5 K}}\pu{39.9 g//mol}\right)\cdot \left(\pu{32.0 g//mol} - \pu{39.9 g//mol}\right)^{-1}\\
  n_{\ce{O2}} &= \pu{0.36 mol}
\end{align}

Using the ideal gas with the data provided yields:
\begin{align}
  n_\text{total} &=\frac{pV}{RT}\\
  n_\text{total} &=\frac{\pu{90 kPa}\times\pu{16.2 L}}{\pu{8.314 L kPa K-1 mol-1}\times\pu{316.5 K}}\\
  n_\text{total} &=\pu{0.55 mol}
\end{align}

Therefore the mole fraction is
$$x_\ce{O2} = \frac{n_\ce{O2}}{n_\text{total}} \approx 0.64.$$

And the partial pressure is therefore $\pu{57.7 kPa}$ (or $\pu{433 mmHg}$ which is close enough to the solution, given that I rounded the medieval witchcraft units).