Why does ice melt slower at higher altitudes?

A simplified attempt.

For a simplified answer that attempts at a solution using Le Chatelier's principle, you'll need a constraint: Ice and water are at an equilibrium at $\pu{0^\circ C}$.

  $$\ce{Ice <=> Water}$$

Simple enough, now pull up the values of densities of ice and water, $\rho_{ice}= 0.9167~\pu{g/cm3}, \rho_{water}= 0.9998 ~\pu{g/cm3}$. Mass being the same, you can conclude that volume of ice is greater than that of water, so in accordance with Le Chatelier's principle, when you move at higher altitudes, the atmospheric pressure falls, thus the equilibrium is favored towards a side where the species occupy a higher volume, i.e. towards the formation of ice. This works antagonistically for $\ce{Ice->Water}$.

Alternatively, have you heard of phase diagrams? This might help you to understand things a bit more clearly after you go through its definition from other sources: https://chemistry.stackexchange.com/a/72732/43942

The phase diagrams for water are different though.

https://d2gne97vdumgn3.cloudfront.net/api/file/m1GFJVbQSBmwkDXRvglV

This, too indicates that for water, at lower pressures, the formation of solid (i.e. ice) is more favored.

A realistic insight.

To be more realistic, as MaxW pointed in the comments, the pressure difference is rather small. The summit pressure at Mt. Everest is $\pu{33.7 kPa}$, while at the sea level it is $101.3$, which gives us a difference of $\pu{67.6 kPa}$ which converts to a mere $\pu{0.676}$ Bar.

Although this is quite a significant pressure (see Jon's comment) for our perception, it doesn't affect freezing that much. For a comparison, this answer https://chemistry.stackexchange.com/a/8883/43942 states that the pressure required to solidify $\ce{CO2}$ to make it into dry-ice at room temperature, you'll need a pressure of about $\pu{56 bar}$.

Freezing at higher latitudes, requires us to also ponder on the ambient temperature. You probably wouldn't want to talk about latitudes as high as the Everest, because then the temperatures will be extremely low (averaging about $\pu{-19^\circ C}$ in summers!) and you'll have to apply an external pressure to adjust the melting point. Latitudes where the ambient temperature is (say) $\pu{3^\circ C}$ have a less temperature gradient when compared to a place (probably at the sea level) with an ambient temperature of (say) $\pu{25^\circ C}$. A higher magnitude of a gradient means a high rate of heat transfer, which hints at a slower melting of ice at higher latitudes.

Also, airhuff suggested that thermal conductivity of thin air at high latitudes may have a role to play in this. I searched this up, you might want to look at this paper: http://nist.gov/data/PDFfiles/jpcrd269.pdf


What is the electronic configuration of Cr (+1) ion

I assume we are talking about the free ions in the gas phase.

I checked my notes and it seems that the electronic configuration of $\ce{Cr+}$ should be $[\ce{Ar}]\mathrm{3d^5}$, the reference being a rather old book: Phillips & Williams' Inorganic Chemistry (1966). I tried to look in slightly newer textbooks, but I couldn't find any information on the monovalent ions.

I would not claim to be 100% certain on this, but it can certainly be rationalised by the increase in $Z_\mathrm{eff}$ in the $\ce{Cr+}$ ion as compared to neutral $\ce{Cr}$ (one less electron means less shielding and greater effective nuclear charge). That would serve to stabilise the 3d orbital more than it stabilises the 4s orbital, and therefore favour the $\mathrm{3d^5}$ configuration over the $\mathrm{3d^44s^1}$ configuration.

In general, the electronic configurations of transition metals and their ions are a balance of delicate factors. It's not easy to rationalise the electronic configurations fully and there are several factors that will affect the configuration adopted, e.g. 1) nuclear charge and relative energy of 3d and 4s orbitals, 2) relative e-e repulsions in 3d and 4s orbitals, 3) exchange energy.

Some interesting discussion on the topic can be found here:

J. Chem. Educ., 1978, 55(1), 2
J. Chem. Educ., 1989, 66(6), 481
J. Chem. Educ., 1994, 71(6), 469


Building a Voltaic Cell Without Metal Ions

If you just put a copper strip and a zinc strip into some NaCl solution and try to write the Nernst equation for the whole system, you will notice that the electrochemical potential will be undefined. Since the electrochemical potentials involve the cation concentrations which are all zero by preparation.

What this means is that there is infinite driving force for some metal to get oxidized into the solution until there is some equilibrium. Therefore, as one of the comments to your question indicate, you will be dissolving some of the metals into the solution by the very act of putting them into such an arrangement.


Lifetime measurement: double peak in instrument response function

I am suspecting some sort of reflection causing this. Does anyone get what I mean here?

The time between the two peaks looks like 0.4 nanoseconds.

This time (0.4 ns) should be compared to the dimensions of the vessel (chamber, cuvette, I don't know the right word) and the speed of light in the medium being observed (water, air, diamond).

So for example in vacuum or gas the light would travel 0.4ns x (0.3m/ns) = 0.12m.

In water where the index of refraction is 1.33 more like 0.09m.


Why does water dissociate to H3O+ and OH- rather than H+ and OH-?

You have to analyze the reaction mechanism behind the self-ionization of water to understand better what happen. In fact is not strictly self but a matter of "couple-ionization": you know the water atoms have different electronegativity, the more electronegative oxygen attracts the less electronegative hydrogen of another water molecule forming an hydrogen bonding.

From there the electron from Hydrogen forms a lone pair with the oxygen of the same molecule forming $\ce{OH^-}$. Hydron ($\ce{H^+}$) is in fact a proton, a very reactive specie so is not so simple found it uncoupled and moreover the hydrogen is suppose to be bonded with an hydrogen bonding with the other water molecule so is more accurate use hydronium ($\ce{H_3O^+}$) instead of hydron...

However you don't have to worry about it for your pH calculation:

$\mathrm{pH=}-\log_{10}[\ce {H^{+}}]= -\log_{10}[\ce {H_3O^{+}}]$

Because, in fact, you have the same ratio between basic species ($\ce{OH^-}$) and acid species( $\ce{H_3O^+}$ or $\ce{H^+}$ as you prefer) in both cases!


Can a single monomer be a macromolecule? If so, when?

I can't think of an example where a biological monomer would be a macromolecule.

Definitions of macromolecule vary, usually by molecular weight or number of monomers (repeat units).

Personally, I'd go with ~1000 Dalton for a minimum, but the original definition of 1000 atoms is a good start too.

In any case, no biological monomer, including glucose will function the same as a macromolecule.

Consider starch - a macromolecule of sugars. It doesn't dissolve as quickly as simple sugar and has different physical properties.

Update

To clarify my comments.. Macromolecules or polymers are made up of monomers the way words are made up of letters. So no, a glucose molecule isn't really the same as a macromolecule, just like "R" is not a word.

Yes, macromolecules can be used to make larger assemblies like microtubules, filaments, etc., much the same way that words can form sentences and paragraphs.

In the polymer literature there's even the concept of a "macromonomer" referring to a monomer that is already large in size.

In my opinion though, the basic constituents (monomers) are still amino acids, nucleic acids, sugars, etc. These are not macromolecules.


What makes carbon special and versatile?

There are two key factors that account for the ubiquity of carbon compounds.

Bond strengths: Look at the following table of bond strengths and notice how the strength of both carbon-carbon single and double bonds is much greater than the bond strengths found in other molecules.

\begin{array}\hline
Typical~Single ~Bond~ Dissociation ~Energies~ (kcal/mole)\\ \hline
\end{array}
\begin{array}{|c|c|c|}\hline
C-C & 83    \\ \hline
N-N & 38    \\ \hline
O-O & 35    \\ \hline
Si-Si & 52    \\ \hline
P-P & 50   \\ \hline
S-S & 54    \\ \hline
\end{array}

\begin{array}\hline
Typical~Double ~Bond~ Dissociation ~Energies~ (kcal/mole)\\ \hline
\end{array}
\begin{array}{|c|c|c|}\hline
C=C & 146    \\ \hline
N=N & 109    \\ \hline
O=O & 119    \\ \hline
P=P & 84   \\ \hline
\end{array}
link to reference

These numbers tell us that carbon-carbon bonds are significantly more stable than other bonds and that the driving force for their formation will be more exothermic.  Therefore, large collections of carbon atoms can be bonded together to produce very stable systems.

Hybridization: Saying that carbon is tetravalent doesn't fully convey the versatility carbon can display in forming bonds. Hybridization involves a re-mixing of s- and p-orbitals in response to bond formation and this leads to the formation of more stable bonds. Carbon is capable of extensive variation in the hybridization of the bonds it forms, much more so than most other atoms.  This wide range in hybridization that is possible for carbon allows it to form much more stable bonds in a wide variety of different structural situations.
\begin{array}\hline
Typical~Carbon-Carbon ~Bond~ Strengths~ (kcal/mole)\\ \hline
\end{array}
\begin{array}{|c|c|c|}\hline
alkane & sp^3 & 84    \\ \hline
alkene & sp^2 & 150    \\ \hline
alkyne & sp & 200    \\ \hline
\end{array}


Is rotation of a mirror image allowed in order to determine chirality? Doesn't rotation alter the mirror image of the molecule?

When checking out chirality or achirality you are always allowed and supposed to rotate the molecule freely in all dimensions. Essentially, what you are trying to prove, is that the mirror image of a molecule can be achieved by rotation alone.

The thought behind this is simple. Chirality is macroscopically proven by optical activity. If you reduce the experiment to its bare basics, you have a polarised light beam that passes through a solution and hits a single molecule to be deflected either to the right (dextrorotatory) or to the left (levorotatory). A second molecule it hits which has the same chirality will move it in the same direction by the same amount. If, however, it hits a molecule which is exactly the mirror image of the first, the effect will be cancelled and the plane of polarisation is moved back into the vertical. In a solution containing only one enantiomer of a chiral substance, the beam will never hit a mirror image and thus always be deflected in a certain direction.

If we continue this thought experiment, we realise that molecular motion in solution is random. Thus, a molecule can — if we give it conciousness — move around freely and let it try to assume exactly the mirror image of the previous one. Therefore, we are allowed and encouraged to rotate (but not to use another mirror image or inversion!) our molecule until it ‘fits’.

By the way, most if not all achiral compounds can also be rotated before you apply the plane of symmetry. Oftentimes if not always, you can rotate them in a way that you see the plane of symmetry present in the molecule itself. If you managed to do that, you won, because just use a plane parallel to that one for mirror imaging and the image will automatically be identical to its template without further rotation.

The plane of symmetry in isopropanol cuts through the $\ce{CHOH}$ bit of isopropanol, dissecting each of those four atoms into two half-spheres. One methyl group will be transformed exactly onto the other by this plane.


In the old fashioned process, we used buttermilk sugar and a bit of vanilla. We put them into a churn (which is just a small bucket with a mixing blade in it so you could hand mix the contents) and then added ice around the churn (outside the churn) to cool it down. And then you mixed and mixed and mixed and mixed. But there was a problem. As the ice melted, the water it created was at $\pu{32°F}$ and even though the ice was colder (a commercial freezer is often about $\pu{15°F}$ (but a home freezer is warmer)) the water kept the churn at around (usually above) $\pu{32°F}$ ($\pu{0°C}$). And ice cream (being an "impure" water mixture) is a liquid, not solid, at $\pu{32°F}$. So, this wouldn't work. Enter salt. If you added salt to the ice, then the ice would melt well below $\pu{32°F}$ and be able to cool the churn to the freezing point of the ice cream mixture (I think its about $\pu{25 °F}$, but my memory shouldn't be trusted). So, the salt and ice form a salt water mixture which can be well below $\pu{32°F}$, and so can cool to below $\pu{32°F}$. If you were to add salt to the recipe, then you'd have to cool it even MORE to get it to freeze (although a little salt makes it taste better, imho). So, salt is added to the cooling mixture and (supposedly) not to the recipe's ingredients. I guess I should point out that you need to consider walking over hot rocks vs walking in say an inch of water at the same temperature. Which would burn you faster? There's no question that your feet would have more contact with the water than the solid rocks, so your feet would be more damaged by the water (water at $180$ or $\pu{200 °F}$ WILL burn you). My point is you actually WANT the ice to melt into a salt solution, because that will better cool the churn - as long as there's enough salt so that the liquid is cold enough.


Why do alkenes with more surface contact have greater London forces?

You are correct that London forces are dependent on the size of a molecule, but think about why this is. One reason is that a larger molecule tends to have a greater surface area than a smaller molecule, and thus there is more surface area available to interact via London (and other) forces.

Also as stated in this Wikipedia article:

  London forces become stronger as the atom in question becomes larger, and to a smaller degree for large molecules. This is due to the increased polarizability of molecules with larger, more dispersed electron clouds.


Boiling and melting point of trans- and cis-but-2-ene

Yes . You are right that structural symmetry comes into play .

Boiling point depends upon intermolecular interactions which over here is more in cis due to its net dipole moment .  The dipole moment enables electronic interactions which hold molecules together .  This shows some general factors of boiling point . Also the below link to the google book covers the structural aspect of a molecule towards b.p on pg47.

For melting points , let me quote Joseph Hornback from Chapter 2.6 Organic Chemistry pg46 (google books) :

  The shape of a compound is very important and has a dramatic effect on its melting point . More symmetrical molecules pack into the crystal lattice better , allowing closer approach and larger attractive forces , resulting in higher melting points .

Melting points of alkenes depends on the packaging of the molecules. Alkenes have similar melting points to that of alkanes, however, in cis isomers molecules are package in a U-bending shape, therefore, will display a lower melting points to that of the trans isomers.(source)

Trans 2-Butene has more symmetry than its cis isomer which results in better packing and hence higher melting point .

For further reading I suggest you visit 'Molecular Shape' over here.


Chemical potential of mixtures of ideal gases and the Gibbs-Duhem equation

The chemical potential of a the two species in your ideal gas mixture are $$\mu_A=\mu^0_A(T)+RT\ln\left(P\frac{n_A}{(n_A+n_B)}\right)$$
$$\mu_B=\mu^0_B(T)+RT\ln\left(P\frac{n_B}{(n_A+n_B)}\right)$$
and the total free energy of the mixture is:
$$G=n_A\mu_A+n_B\mu_B$$So, the chemical potential of A depends on the number of moles of B and the chemical potential of B depends on the number of moles of A.  Do these equations satisfy the Gibbs-Duhem equation?  Try them out and see.


Helium doesn't form any compounds, being a noble gas. That is, you can't extract it from any reaction, because you don't have anything that it reacted with already. Also, it would've probably taken much more energy to form that helium compound than it would've to actually extracted it from the atmosphere.

Now, you can produce helium, technically. That's what an alpha particle is: A loose helium nucleus bouncing about. But mucking around with radioactive things is dangerous and can be quite expensive.

As a note, you could probably extract it yourself if you have a natural gas well that you can access directly and a way to separate it out with fractional distillation. Almost all the gasses will condense out before helium does, which means you'll end up with very pure helium left behind, with a minor impurity of neon. This process is dangerous, however, as you'll both be working with very cold temperatures and have liquid methane running amok.

You could probably get a tank from it from a chemistry supply store somewhere. (Party supply stores add oxygen to prevent people from suffocating off of their tanks. The impurity is not insignificant.)

TL;DR: No, unless you're willing to irradiate or freeze yourself.


Cahn–Ingold–Prelog priority of CN vs CH2NH2

Each multiple bond is considered to be equivalent to an individual bond. So $\ce{-CN}$ has the same priority as (N,N,N), making it higher than (N,H,H). Read more about it here.


No doubt it is aromatic.

The $4n+2$ rule is intended for a single cycle of π-electrons, which may possibly include multiple rings but can't have any interior atoms or pendant π- bonds (which would not fit in a cycle). Here, as Kartik indicates in a comment, there are evidently two such cycles and you have to consider them separately. One cycle has six π-electrons and the other has ten. You have a molecule (or more accurately, a molecular ion) with two aromatic cycles.


After rediscovering my own question, I'm posting it as an answer for anyone else to reference or to correct.

Electron affinity (EA) refers to "the energy released when an atom in the gas phase accepts an electron." (Burdge, Atoms First, p.118)

Lower EA means that the process of accepting an electron is not energetically favorable. In other words, lower EA = less likely to accept an electron.

Groups 4 and 8, in this case, already have filled subshells. Adding an electron would place it in a new subshell that is higher in energy. The EA would be lower for these groups because of the "reluctance" of the atom to accept an additional electron.

Groups 3 and 7 have the anomalously high EAs, which might have been where the confusion was.


Why are halogens ortho para directing even though deactivating?

Let's take as an example a typical halogen, chlorine. Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through the inductive effect, chlorine destabilises the intermediate carbocation formed during the electrophilic substitution

Image courtesy: NCERT Chemistry Class 11th

Through resonance, a halogen tends to stabilise the carbocation, and the effect is more pronounced at ortho- and para- positions. However, the inductive effect is stronger than resonance here, and causes net electron withdrawal and thus net deactivation. The resonance effect tends to oppose the inductive effect for the attack at ortho- and para- positions and hence makes the deactivation less for ortho- and para- attack.

Reactivity is thus controlled by the stronger inductive effect and orientation is controlled by resonance effect.


Measuring pH using pH paper makes solution change color

What about the volume of the analyzed mixture? I doubt that anyone can guess what the substance is, but my guess about the color is that the indicators on the pH paper might have been partially dissolved in the mixture. And, since the color of litmus paper at a pH = 8 is blue, it could be a possible explanation for the color, especially if you are dealing with a small volume of liquid.


Why does Brady's reagent include sulfuric acid and methanol?

2,4-DNPH is usually dissolved in an alcoholic solvent, together with sulfuric acid, when used as a reagent to detect aldehydes and ketones (Brady's reagent). The sulfuric acid is a source of $\ce{H+}$, which catalyses the reaction between the hydrazine group and the carbonyl compound, as depicted in the mechanism below:

The final hydrazone product is the precipitate which signals a positive test.

The solvent used is typically methanol or ethanol. You are correct that these could theoretically react with sulfuric acid: dimethyl ether is the most likely product. However, the concentration of sulfuric acid is typically not sufficiently high, and these side reactions do not take place.


Carcinogenicity of polynuclear hydrocarbons and benzene rings

1.Why are they carcinogenic:

From my point of view as a medicinal chemist, polycyclic aromatic hydrocarbons compounds tend to be metabolized by specific cytochrome P450 (CYP 450) isofroms to form highly electrophilic species, which are highly reactive and can react with the DNA of the cell and damage it. Specifically, they can react with the nucleophilic nitrogens in the DNA bases such as adenine.
To see more detailed mechanism of the formation of the electrophilic species from the compounds, check this article: https://socratic.org/questions/how-do-polycyclic-aromatic-hydrocarbons-cause-cancer

2.Is there a reason for a given organic compound to be carcinogenic?

Mechanism for carcinogenicity can vary, in general, any compound damaging or altering the DNA can be carcinogenic. For example, highly reactive electrophiles (such as formaldehyde).


How to identify the asymmetric carbon in [4-(1-bromoethylidene)cyclohexyl]acetaldehyde?

There is probably little reason in discussing the hydrogen and the $\ce{CH2CHO}$ group which are definitely different residues. The question remains whether the sides of the ring are identical or not.

If you go along the top side of the ring, you will encounter:

$\ce{CH2}$
$\ce{CH2}$
$\ce{CR=CMeBr}$

Looking into it a little deeper, we see that we are coming from the (Z) side of the double bond.

Going along the bottom side you will encounter:

$\ce{CH2}$
$\ce{CH2}$
$\ce{CR=CBrMe}$

Here, however, we are arriving from the (E) side of the double bond.

Since (E) is not equal to (Z) the two sides of the ring are not equal, therefore the carbon has four different residues, therefore it is chiral.


Do ligands in ionisation isomerism interchange?

Actually, in the ionisation isomerism the ligands do not interchange. They get displaced. Thus one ion becomes counter ion for the other. When one has look at the molecular formula it seems the ligands in the coordination sphere are interchanged with the ligands outside the sphere. So, this does not cross out Werner’s theory as there is displacement of ion and not just interchange.


Why is a Grignard reagent not protonated by chloramine?

The $\ce{N-H}$ protons are relatively weakly acidic and this makes deprotonation kinetically less favored if an alternative nucleophilic attack mechanism is available.  With $\ce{NH3}$ no such alternative is there, due to lack of a decent leaving group, so deprotonation ultimately prevails; but $\ce{NH2Cl}$ allows displacement of a chloride ion and thus enables nucleophilic attack.


It's all about the Pauli exclusion principle. Assume the available MOs in each of two approaching molecules are fixed, and that each molecule is in a closed shell configuration (all electrons are paired). The electrons in closed shell systems cannot accept additional electrons that share principal, azimuthal and magnetic quantum numbers, since all the spin QMs are taken up. In fact, other than filling higher energy (LUMO) orbitals there is no way to accept additional electrons while retaining the current structure. The solution is to deform the structure of the molecule so as to generate new orbitals that allow formation of novel MOs. The deformation is associated with a potential energy barrier, since it is assumed that the colliding molecules are in low energy electron configurations. Therefore the instantaneous response to the collision is repulsive.


What's the most efficient way to find crystal structures of drugs?

I think my first question is why you need the crystal structure?

For organic molecules, the definitive resource is the proprietary (and fairly expensive) Cambridge Structure Database.

The Crystallographic Open Database which you linked is fairly new and expanding, but does not have the many, many molecular crystals in the CSD. For example, American Chemical Society journals required submission to the CSD for publication.

My suggestion would be to start with the NIH Chemical Resolver or PubChem. The latter has a 3D computed structure of (R)-fluoxetine, which I'm sure is pretty decent. Jmol should be able to load these directly, or you could save SDF files and open them in Jmol.

I'll point out that Avogadro can also do these tasks, but since I'm the creator, I'm biased.


The EPR experiment measures the interaction of unpaired spin density ($\rho(\alpha)$) at the nuclei.  This is quantified by an equation called the Fermi contact interaction.

If you have some prior knowledge of "where to put the dot" in your structure (highly localized), then generally you can assume that the EPR pattern is split most by that nucleus.  Splitting patterns are determined by standard rules in accord with the Zeeman effect.

The nucleus must have a non-zero nuclear spin to be observable. Certain nuclei such as hydrogen-1 and many transition metals have spin. It is very common that isotopic impurities (e.g. carbon-13) appear though the most prevalent isotope is EPR-silent.

(There are other complications, such as nuclear quadrupole moment and broadening due to exchange of chemical sites.)

An additional issue is that the spin density is generally spread over the molecule, and interacts with multiple nuclei.  It turns out to be easier to measure the spectrum, and then rationalize it given the particular structure and nuclear composition.

Accurate prediction of EPR spectra is a rather difficult problem, as quantum chemical calculations generally don't emphasize basis function placement at the nucleus, as they tend to focus on the "valence" and not the cusp-like nature of the radial distribution.  (Here, Slater-type orbitals would be better in principle than Gaussian-type, but STOs of sufficient number and of high-enough quality of Hamiltonian are virtually impossible).  Finally, solvent and powder effects can render the gas-phase ab initio calculations virtually irrelevant.

EPR simulation/interpretation of fine structure is much more difficult than NMR spectra, because in NMR spectra, one has the chemical shift to conveniently move splitting patterns out of the way.  In EPR, the splittings are often "piled" on top of each other, and make it necessary to use computer simulation to obtain precise fits.

In your specific example, I would identify the nuclei with spin (probably Cu, certainly phosphorus-31, and then hydrogen-1).  Then, identify where you think the unpaired density will be, and then take the nuclear spin + 1 and that will be the "guesstimated" isotropic spectrum.  The pattern will follow some variant Pascal's triangle if I=1/2, otherwise, you have to figure it out by writing out the number of possible configurations.

Edit: I have not even started on the characteristics of triplet spectra, as here the electrons interact.  One thing you can hope for is that all of your electrons are paired (singlet), and so it's EPR silent.


Do there exist aromatic compounds similar to azulene?

There is this compound which is I believe the closest to anything which you've asked for.
Except the fused part this satisfies all your conditions.
In a polar solvent the double bond breaks to form two highly stable aromatic rings(tropylium and dicyclopentene) which act as two separate poles. It also lowers the rotation barrier of this compound. Hence it even becomes soluble. Along with azulene this is as close as an organic compound gets to being ionic.


Trouble understanding interplanar spacing using miller indices

The lattice planes are a property of the lattice based on dividing up the unit cell. The cell can be divided up indefinitely to give all the Miller indices for example $(\bar{10}\;20\;0);(10\;23\;20)$ etc. Atoms do not need to be on a lattice plane to add to the diffracted signal. The structural information is in the intensity of the diffracted spots not their position. (2 2 4) is parallel to (1 1 2) etc. as shown in the figure. In the general case it is usual to use vectors to find the perpendicular distance using formulas for finding a point at a given distance from a plane.
(Edit. I added (2 2 4) to indicate that it is a different plane but parallel to (1 1 2), in crystallography it is conventional to make all parallel planes have the same Miller indices, (h k l) (as they all scatter in the same way) by removing any common factor. )


Reaction an ROH-resonance-stabilized Group with Hydroxide/Acid-Base Reactions

The conjugated base of the depicted alcohol is not resonance-stabilized. The negative charge cannot be delocalized into the aromatic ring because the alcohol group is not a phenolic OH group, whose oxygen lone pairs are conjugated to the aromatic $\pi$ electrons, like shown below. Instead, it is attached to an $sp^3$-hybridized benzylic carbon which is not part of the conjugated system.

You might have mistaken the depicted compound with 2-naphtol, whose $\ce{pK_{a}}$ (9.51) is much lower than that of aliphatic alcohols and which can therefore be deprotonated by $\ce{NaOH}$. Aliphatic alcohols typically have $\ce{pK_{a}}$ values of about 16-18 (source), so there will be no deprotonation by sodium hydroxide.


Chitin and protein are completely unrelated.
The only common thing is that they are polymers.
Chitin is a polymer of amino sugars while protein is a polymer of amino acids.
Both monomers are very different and are not converted one to the other.

In fact, chitin, like cellulose, is not fragmented by animals, so it is not absorbed.
Moreover, chitin is a homopolymer, so even if it where absorbed and transformed to amino acids, it would yield only one (or maybe a few ones) but all the others would have to be obtained through a different source.

So chitin is not a source of protein or amino acids.


What is the colour of chromium(III) chloride solution?

The color of the solution may vary as long it is not in equilibrium.

According Bjerrum and Bjerrum1 a $\ce{1\mathrm{M}}$ chromium (III) chloride solution in equilibrium at 25° C contains about 13% monochloro and 85% hexaqua ions.

The absorption spectrum published shows minima at $\ce{480 \mathrm{nm}}$ (blue) for the hexaqua and $\ce{510 \mathrm{nm}}$ (green) for the monochloro complex.

When we integrate over the spectrum we get the following colors.

However, it should be noted that the colors vary with spectral optical depth (concentration, thickness of the medium).

1 Bjerrum, M.J. and Bjerrum, J., 1990. Estimation of Small Stability Constants in Aqueous Solutions. The Chromium(III) Chloride System. Acta Chem. Scand. 44: 358-363.


Determining Rate Law and Reaction Mechanism from Method of Initial Rates

Empirical rate law is:

$$\mathrm{Rate} = \frac{[\ce{A}][\ce{B}]}{[\ce{B}] + [\ce{C}]}$$

In the first block of data you can notice that $\ce{C}$ and $\ce{D}$ are not necessarily required for reaction to proceed because otherwise the rate would be zero. You can also deduce that their concentration must appear as a sum in the rate law for the same reason. The sum will appear in denominator because $\ce{C}$ slows down the overall reaction.

Next step is to guess elementary reactions. From the law's denominator sum you can conclude that intermediate will competitively react with $\ce{B}$ and $\ce{C}$.

$\ce{A ->[k1] I}$ (Probably some kind of excitation)

$\ce{I + C ->[k2] D}$

$\ce{I + B ->[k3]Product}$

Then you derive the overall rate via steady state approximation (assumption).

$$\mathrm{Rate} = \frac{k_1k_3[\ce{A}][\ce{B}]}{k_3[\ce{B}] + k_2[\ce{C}]}$$

and ${k_1}={k_2}={k_3}$


Why does a chemical have more than one oxidation number?

Because a lot of what you have been taught in school is a lie.

In this particular case, you've probably been taught about the octet rule, a rule which makes it seem as if any given element should have exactly two oxidation states; 'n' and '8-n', where n is a simple function of it's location in the periodic table.

The truth, as always, is a little more complicated:

Compared to the (somewhat) well-behaved non-metals in the upper right corner of the periodic table, most metals have a lot for electrons. At 26, iron is a lightweight, but already there you have 4 (common!) oxidation states, the two most common ones being iron(II) and iron(III).

One reason for this is that while iron atoms have a "the lowest energy state", it's difficult to get at without first coming across several "a (locally) lowest energy state".

Its four oxidation states represent local minima, and moving between them requires an activation energy, making each of the states relatively stable, but possible to dislodge, i.e. metastable.


Why is benzyl alcohol a slightly stronger acid than cyclohexanol?

(This CW answer is adapted from an earlier comment by ron.)

In both cases, the carbon directly attached to oxygen is $\mathrm{sp^3}$ hybridised. However, the next carbon in the two compounds have different hybridisations. In cyclohexanol, it is also $\mathrm{sp^3}$ hybridised; but in benzyl alcohol, it is $\mathrm{sp^2}$ hybridised.

In general, $\mathrm{sp^2}$ hybridised carbons are more electronegative than $\mathrm{sp^3}$ hybridized carbons. This is because the electron density in an $\mathrm{sp^2}$ orbital has more s-character (50%) compared to one in an $\mathrm{sp^3}$ orbital (33%), and is consequently more strongly attracted to the carbon nucleus. Therefore, the carbon attached to oxygen in benzyl alcohol will be more electron deficient (more positive) and will be better able to stabilise the negative charge on oxygen in the alkoxide conjugate base, via the inductive effect.


Based on the IUPAC Goldbook definition, the only two requirements for a pericyclic reaction are:

Concerted mechanism
Cyclic transition state

Hydroboration reactions (and similar reactions) are pericylic reactions by these criteria, as are all cheletropic reactions.

The language about fully conjugated transition states can be used to explain some pericyclic reactions, but not all, and thus IUPAC uses the phrasing may be viewed as.... Hydride shifts, (and all sigmatropic rearrangements ene reactions) involve non-$\pi$ orbitals (even so, the transition states do tend to follow the Hückel rule).


How are the Van der Waals constants a and b related to each other?

Yes, they are certainly correlated as shown in the following plot based on data from the Wikipedia Van der Waals constants data page

The correlation should not be entirely surprising: the covolume b is a measure of the size of the gas molecules, whereas a is a measure of the strength of intermolecular interactions which is expected to depend through polarizability and dispersion interactions on the size of the molecules (molecular volume).

For a given gas parameters a and b tend to increase together. The shown correlation is empirical but since both a and b depend on molecular size and apparently in a very similar functional way, the result is a linear correlation.

Following the hint from Karl, here is a version of the plot with log(a) along the abscissa, and with superimposed labels.


Nomenclature in case when same fucnctional group is present more than once

The IUPAC rules do not seem to have an explicit rule for general cases when more than two cyano groups are attached to the (linear) alkane chain, like it does for aldehydes (see Is propane-1,2,3-tricarbaldehyde the correct IUPAC name of CHO-CH2-CH(CHO)-CH2-CHO?), but do have a related one:

  P-66.5.1.1.2 If an unbranched alkane is linked to more than two terminal cyano groups, all cyano groups are named from the parent hydride by the substitutive suffix ‘carbonitrile’, preceded by appropriate multiplying prefix and locants.
  Example:

  $\ce{\overset{4}{C}H3-[CH2]2-\overset{1}{C}(CN)3}$
  butane-1,1,1-tricarbonitrile (PIN)

I think that similar usage in general is implicitly allowed in the rule

  P-66.5.1.1.4 When a group is present that has priority for citation as the principal characteristic group or When all $\ce{-CN}$ groups cannot be expressed as the principal characteristic group, the $\ce{-CN}$ group is designated by the preferred prefix ‘cyano’.

  (…)

(Three examples follow, two of them have another higher-priority group $\ce{-COOH}$; the other one has three cyano groups, not on linear chain but one of them is attached via a chain branch, $\ce{NC[CH2]2CH(CH2CN)[CH2]2CN}$, 4-cyanomethyl(heptanedinitrile).)

I our case, we can express all the groups as principal characteristic group, so the correct name should be propane-1,2,3-tricarbonitrile.


Is there a simple way to predict the molecular pi orbitals in conjugated pi systems?

This should work for any even number of conjugated carbons (and with slight modification, for cations/radicals/anions of uneven numbers of carbons). You can follow a simple procedure to generate all the $\pi$ orbitals.

Generate the lowest orbital with no nodes
Generate next orbital by creating a node through the middle bond
For all subsequent orbitals, generate by taking the previous orbital, placing nodes to the left and right of the original node, and then removing the original node.
End when all bonds are separated by nodes

We can write these as binary strings where a 1 is one orientation and 0 is the other. So for example, you could write the lowest orbital of N=10 chain as 0000000000, the next as 0000011111, the next as 0000110000, and so on.

The Huckel method mentioned in the comments is a more formal way of obtaining the orbitals by solving for the eigenvalues and eigenvectors of an adjacency matrix. The adjacency matrix is formed by numbering the carbon chain and forming an NxN matrix with the rows and columns numbered like the carbons. In each row, put a 1 if the carbon of the row and column are adjacent and put an x if they are the same. A simple online calculator with a few examples and instructions for this can found here (this only gives the eigenvalues, you would probably have to download or code your own version to give the eigenvectors). You only really need these eigenvectors to get an equation for the orbitals, but for your case you can do that just as well with the simple procedure by simply writing the MOs as linear combinations of the AOs.


Can a Boc protecting group be removed with trifluoroacetic acid?

If you do a literature search (SciFinder/Reaxys), it seems that HCl in polar aprotic solvents (dioxane, THF, ether) are the 'standard' conditions for this cleavage.

You mention that you require the TFA salt, not the HCl salt, but even if you use HCl to carry out the reaction, its fairly trivial to get rid of the HCl and then convert the free amine into whatever salt you desire.

(Though as vapid said, your reaction conditions are probably quite important, and you might find if you just push the TFA reaction a little harder you can force the Boc off).


Is hydrogen (on its own) considered an acid?

Well, you can never have H+ by itself. In aqueous media it's always associated with the lone pairs of water molecules and exists as H3O+. But yes, theoretically speaking H+ is a Brønsted–Lowry acid as the definition of such is to donate a proton and it donates itself to water in the case. It's also a Lewis acid as by donating itself it accepts electrons in its 1s emptry orbital.

But be aware that in water, you never have H+, it's always H3O+ which is the strongest acid that can exist in water solutions, all acids are limited by the strength of H3O+ when dissolved in water. Therefore, a species is an acid in water if it generates H3O+. In other solvents, for example liquid NH3, the corresponding species is NH4+ and an acid in ammonia is defined as species that generates NH4+. This whole paragraph relates to the solvent definition of acids and bases.

I hope all that answers your question, please comment if you want anything clarified.


What is the difference between configuration and conformation in stereochemistry?

I recommend you this webpage: http://worldofbiochemistry.blogspot.com/2012/09/conformation-vs-configuration.html.
Configuration and conformation are two concepts are often used interchangeably, but represent very different things. The conformation regards the relative spatial orientation of a portion of a molecule relative to another. Thus, it is an aspect that is not directly related to the covalent bonds that are established within the molecule, but with their possible rotation.

When we talk about rotation around covalent bonds, we are only referring to the single bonds, as they are the only ones that can suffer rotation. Basically this concept is easily understood if we think that the bounds work as an axis...

It should be noted that when we speak of different conformations, it does not necessarily involve all the covalent bonds of a molecule, it can account only for one or few of them.
 Taken together, it is possible to convert one conformation to another without cleaving or forming chemical bonds, simply by rotating some simple covalent bonds. Thermal energy at room temperature is sufficient to rotate some simple covalent bonds.

The configuration is a concept that is related to the order by which different substituents linked to the same central atom establish covalent bonds. That means, in this case it is clearly an aspect that is a direct consequence of the covalent backbone of molecules.

To change the configuration, you must always cleave and form new covalent bonds...This costs a lot of energy!!

In conclusion, the concept conformation encompasses portions of a molecule which are not directly linked to the same atom and do not involve the covalent backbone of the molecules, while the configuration comprehends parts of the molecule which are bound to the same atom, which means that there is a direct involvement of the covalent bounds of the molecule.


Actually, most of linear imines and exocyclic imines display $E/Z$-isomerism (commonly known as $cis/trans$-isomerism) with a lone pair on the nitrogen, which is a part of the double bond. The isomerism follows various mechanisms. For instances, $cis/trans$-isomerism of azobenzene a noted elsewhere is operated by UV/vis-irradiation is a good example. The mechanism of this transformation has been depicted as follows (Ref.1):

Similar to cis/trans-isomerism in alkenes, extensive studies has been done on azo-compounds and imines. For example, acid-catalyzed $E/Z$-isomerization of imines has been reported frequently (Ref.2,3). The abstract of Ref.2 states that:

The kinetics and mechanism of acid-catalyzed Z/E isomerization of O-methylbenzohydroximoyl chloride (1Za and 1Ea), methyl O-methylbenzohydroximate (1Zb and 1Eb), ethyl O-methylbenzohydroximate (1Zc and 1Ec and five para and meta substituted derivatives), O-methylcinnamohydroximoyl chloride (2Za and 2Ea), and methyl O-methylcinnamohydroximate (2Zb and 2Zb) have been investigated. The kinetics of Z/E isomerization of these imines have been studied in glacial acetic acid (1Ea and 1Zc) and in dioxane solutions containing $\ce{HCl}$, trifluoromethanesulfonic acid, or tetrafluoroboric acid (1Ea, 1Zb, 2Ea, and 2Zb). The isomerization takes place by either (a) rotation about the carbon−nitrogen double bond of the protonated imine (iminium ion rotation) or (b) nucleophilic attack on the protonated imine to form a tetrahedral intermediate that undergoes stereomutation and loss of the nucleophile (nucleophilic catalysis). The hydroximoyl chlorides 1Ea and 2Ea only isomerize by the nucleophilic catalysis mechanism. The hydroximate 1Zb appears to be capable of isomerizing by either mechanism. The hydroximate 2Zb may be isomerizing only by iminium ion rotation. Theoretical calculations support the notion that increased conjugation in the protonated imine increases the rate of iminium ion rotation.

Photochromic properties of a series of novel pyrrolidene imines bearing functionalized aryl or naphthyl moieties was studied by UV spectroscopy (Ref.4):

UV irradiation of these heterocyclic Schiff bases at room temperature promotes the trans–cis photoisomerization of the $\ce{C=N}$ double bond with formation of a variable amount of the cis-isomer that, in the absence of light, returns thermally to the original form in few seconds. The thermal cis–trans re-isomerization of these molecular switches is much more slower than for the common benzylidene aniline-type imines, allowing the observation of the photochromic phenomena at room temperature.

Another interesting studies have been done with indigo derivatives (Ref.5) noting that:

Substituted indigo derivatives undergo photoisomerization of the central double bond if both nitrogen atoms are functionalized. Indigo itself however does not photoisomerize because of a competing and highly efficient excited‐state proton transfer...

References:

H. M. Dhammika Bandara, Shawn C. Burdette, "Photoisomerization in different classes of azobenzene," Chem. Soc. Rev. 2012, 41(5), 1809-1825 (https://doi.org/10.1039/C1CS15179G).
James E. Johnson, Nora M. Morales, Andrea M. Gorczyca, Debra D. Dolliver, Michael A. McAllister, "Mechanisms of Acid-Catalyzed Z/E Isomerization of Imines," J. Org. Chem. 2001, 66(24), 7979–7985 (https://doi.org/10.1021/jo010067k).
James E. Johnson, Nancy M. Silk, Elizabeth Ann Nalley, Mohammed Arfan, "Mechanism of an acid-catalyzed geometric isomerization about a carbon-nitrogen double bond," J. Org. Chem. 1981, 46(3), 546–552 (https://doi.org/10.1021/jo00316a013).
Paulo J. Coelho, M. Cidália R. Castro, M. Manuela M. Raposo, "Reversible trans–cis photoisomerization of new pyrrolidene heterocyclic imines," Journal of Photochemistry and Photobiology A: Chemistry 2013, 259, 59-65 (https://doi.org/10.1016/j.jphotochem.2013.03.004).
Ludwig A. Huber, Peter Mayer, Henry Dube, "Photoisomerization of Mono‐Arylated Indigo and Water‐Induced Acceleration of Thermal cis‐to‐trans Isomerization," ChemPhotoChem 2018, 2(6), 458-464 (https://doi.org/10.1002/cptc.201700228).


Is it possible to create a salt with multiple anions with only one cation(e.g. CaBrCl)?

A cursory search showed some interesting cases.

The oldest reported examples seemingly are mercury halides $\ce{HgXY}$.

$\ce{HgClI}$ was prepared by iodination of solid $\ce{HgCl2}$
$\ce{HgBrCl}$ was obtained from the chlorination of an ethanolic suspension of $\ce{HgBr2}$
$\ce{HgBrI}$ was prepared by mixing ethanolic solutions of $\ce{HgBr2}$ and $\ce{HgI2}$

The crystal structure of $\ce{Cs6Cl3[Fe(CN)6].H2O}$ has been reported.

Passing $\ce{H2S}$ through solutions of oxidosulfido molybdates and tungstates let to various mixed salts with halide, hydrogen sulfide and thiosulfate as additional anions beside the trisulfidometalates $\ce{[MoOS3]^{2−}}$ and $\ce{[WOS3]^{2−}}$.

But the concept isn't limited to inorganic chemistry.

The crystal structures of mixed salts of amino acids with different anions were reported.

From the field of organic conductor research, salts of BEDT-TTF (bis(ethylenedithio)tetrathiafulvalene) with mixed anions of $\ce{[CuBr2]-}$ and $\ce{[CuCl2]-}$ are known.


UV-Vis Spectra: Multiple peaks for one component

Different functional groups may show different absorptions and even a single functional group may exhibit two signals in your UV-VIS spectrum, e.g. corresponding to $\pi,\pi^*$ or $n,\pi^*$ transitions.

While HPLC grade methanol is typically much cheaper than acetonitrile, it might not be the best choice here. You might want to have a look at the cutoff wavelengths of different solvents.

I'd definitely give measurements at $\lambda = $ 280 nm a try.


Many volatile liquids are not combustible

Dichloromethane (DCM) is a widely used solvent by chemists. It boils at around 40°C (the same as diethyl ether) but is not remotely combustible or flammable. Ether is both very volatile and very flammable, so much so that most labs would prefer not to have it used anywhere where flames or sparks could be present. HFC-134a is a refrigerant gas at room temperature and used to be preferred to alternatives like propane (also a gas) because it didn't risk creating an explosive atmosphere when it leaked.

Combustibility has almost noting to do with volatility (other than in the trivial sense that things that are hard to vaporise are hard to set alight even if they are inherently flammable–like bitumen). Combustibility is determined by the chemical properties of a substance, usually whether the reaction of the substance with oxygen releases energy easily. Ether is both volatile and very combustible as that reaction releases a lot of energy and is easy to start. DCM can be burned but releases far less energy and the reaction is very hard to start or maintain as the chlorine in the molecule inhibits the combustion reactions.

Other compounds, like water, are certainly volatile enough to combust (how would we get rain if water were not volatile?) but there are no possible reactions that release any energy from reacting water with oxygen.

So two factors matter in combustibility: is there an energy releasing reaction from burning the compound?; is there a reaction mechanism that makes it easy to cause that reaction to happen? Neither are true for water; many other compounds either don't release enough energy from reacting with oxygen or have chemical components that inhibit the reaction.

Volatility has little or nothing to do with either.


Calculate the number of atoms of oxygen present in 1.3 mol of sulfuric acid

The key about the unit mole is that it does not reference which type of entity is specified. Using the mole only makes sense if you define what you are using the mole for. This could be:

protons
atoms
ions
molecules
formula units of an ionic or metallic solid
aggregates
apples
stars
…

The mole is basically just a large number that you can use instead of e.g. the number 1,000,000 or the number 0.5.

So in your question, $\pu{1.3 mol}$ of $\ce{H2SO4}$ means that you have $1.3 \times 6.022 \times 10^{23}$ molecules of $\ce{H2SO4}$, each of which contains one sulphur atom, two hydrogen atoms and four oxygen atoms. Thus, in one mole of $\ce{H2SO4}$ you can find:

$\pu{2mol}$ of hydrogen atoms
$\pu{1mol}$ of sulphur atoms
$\pu{50 mol}$ of protons
$\pu{50mol}$ of electrons
and $\pu{4mol}$ of oxygen atoms.

The correct solution is thus:

$$n(\ce{O}) = 4\times n(\ce{H2SO4}) = 4 \times 1.3 \times 6.022\times 10^{23} = 3.13 \times 10^{24}$$


Dyes and pigments work by absorbing certain wavelengths of light and reflecting or transmitting the rest. When a dye molecule absorbs a photon, an electron is excited to a higher energy state. Most of the time (neglecting fluorescence), the molecule de-excites by giving off heat and returns to the ground state intact. However, because the excited state is a high energy state, it has the potential to undergo a chemical reaction, breaking a covalent bond or otherwise irreversibly reacting with another molecule. This changes the electronic structure of the molecule which changes its absorption properties: e.g. many dyes that absorb visible light have large systems of conjugated double bonds and if these are broken, the absorbance can shift to much shorter wavelengths. How likely this kind of destructive chemistry is depends on the nature of the dye. Organic dyes tend to be more susceptible to photobleaching than things like quantum dots and inorganic pigments.

Touching on Nicolau's comment, red paint can often degrade faster because it absorbs higher energy (shorter wavelength) light. The more energy that is available, the more likely it is that reactions will occur. This is more pronounced as we enter UV wavelengths, which is why UV-B can cause thymine dimers in DNA, which can lead to skin cancer. For outdoor signs and such, UV-blocking coatings are often applied to extend the life of the dyes.

It's also not necessary for the light to interact directly with the dye molecules to cause bleaching: depending on what other molecules are present with the dye (solvents, etc), these other molecules may produce reactive species that then react with the dye.


Balance Repeatability Sample or Population for SD

Standard Deviation Formulas in Maths is fun is a nice article to review. It explains why we have use two different formulas to find standard deviation for the population and sample. However, I'd like to point out that your explanation of calculation is a bit off. The correct formula to calculate standard deviation is:

$$ \bbox[yellow,5px,border:2px solid red]
{
\sigma = \sqrt{\left(\frac{1}{N} \sum_{i=1}^N (x_i-\mu)^2\right)}\qquad (1)}$$

where $\sigma$ is the standard deviation of the population and $\mu$ is the Mean (the simple average) of the population of numbers:

$$\mu = \frac{1}{N} \sum_{i=1}^N x_i$$

According to the Math is Fun website:

  To calculate the standard deviation of those numbers:

  Work out the Mean (the simple average of the numbers)
  Then for each number: subtract the Mean and square the result
  Then work out the mean of those squared differences.
  Take the square root of that.

The correct formula to calculate standard deviation of a sample is:

$$ \bbox[white,5px,border:2px solid blue]
{
s = \sqrt{\left(\frac{1}{n-1} \sum_{i=1}^n (x_i-\chi)^2\right)}\qquad (2)}$$

where $s$ is the standard deviation of the sample (say you choose $n$ numbers out of total of $N$ numbers population) and $\chi$ is the mean (the simple average) of the chosen numbers, hence: $$\chi = \frac{1}{n} \sum_{i=1}^n x_i$$.

Note: In our old days when I was taking analytical chemistry, we were instructed to use equation (1), if the $N \gt 10$, but to use equation (2) whenever we have to deal with data where $N \lt 10$.


How to tell/control whether sodium ethoxide will react by a substitution or elimination pathway?

The above reactions belong to Williamson Synthesis of Ethers, probably the best of the alternatives for prepartion of ethers!

The Williamson Ether synthesis reactions follow SN2 mechanism. Since the SN2 mechanism proceeds through a single step where the nucleophile performs a “backside attack” on the alkyl halide, the only thing stopping this, is steric hindrance.

Methyl and primary alkyl halides are excellent substrates for this synthesis, as they provide no/very less hindrance for the approaching nucleophile.

Since alkoxides are strong bases, competition with elimination (E2) becomes a concern once the alkyl halide becomes more sterically hindered. This is exactly why, the first reaction that you’ve mentioned, proceeds by SN2 but in the second both elimination(major) and substitution (minor) takes place.

Also note that, when tertiary alkyl halides are used, there is no hope for substitution because of the bulky groups present around carbon. Hence, only elimination takes place in this case.

So the reactions are,

With primary alkyl halide,
$$ \ce{CH3CH2CH2CH2Br + C2H5ONa -> CH3CH2CH2CH2-O-CH2CH3}$$
With secondary alkyl halide,
$$\ce{CH3CH2CHBrCH3 + C2H5ONa -> CH3CH2CH=CH2(major)}$$

Moving onto conditions to favour a particular type of reaction:

One way to attempt to get the SN2 to be favoured over elimination is to use a polar aprotic solvent (such as acetonitrile or DMSO) that will increase the nucleophilicity of the alkoxide.
Increasing the temperature favours elimination over substitution.

Source for the above conditions: chem.ucalgary

  I believe sodium ethoxide being bulky should always do elimination

In most cases, rather than the incoming nucleophile, the hindrance in the the alkyl halide plays a more important role.

The rule #3 mentioned in the link you stated clearly says that the elimination takes place for a tertiary carbon in presence of a strong base like $\ce{CH3CH2-O-Na}$.


What nuclei need to be NMR active in an HMBC experiment?

It is incorrect to say that oxygen block the HMBC coupling because it be magnetically inactive.

The theoretical explanation is that the coupling of spins $^nJ_{\ce{XY}}$ is mediated by the electrons i.e. transferred completely independently of the nuclei on the way. And I can provide you with multiple examples that show how coupling crosses over magnetically inactive nuclei:

I myself in my bachelor’s thesis have distinguished the two different ortho-nitrobenzyl ethers of 1-(4-bromophenyl)-1,2-ethandiol by HMBC coupling. It was always possible to locate a coupling from one side’s hydrogen to the other side’s carbon in HMBC in spite of the magnetically inactive oxygen in-between.
During my Ph.D. thesis, I performed a number of $\ce{^29Si}$-HMBC experiments to determine which hydroxy groups were protected by the silyl group used. Again, this transferred across a magnetically inactive oxygen.
Remember that $\ce{^12C}$ itself is magnetically inactive. In your standard $^3J_{\ce{CH}}$ 3-bond-HMBC-experiment, the coupling and thus the cross-peak is transferred along $\ce{H-C-C-^13C}$. I didn’t spell it out explicitly but considering each carbon has only a $1~\%$ chance of being $\ce{^13C}$, it’s much more likely for the chain to be $\ce{H-^12C-^12C-^13C}$, i.e. involving only two magnetically active nuclei.
Finally, your standard hydrogen spectrum exhibits a lot of $^3J_{\ce{HH}}$ coupling. Again, this crosses over two carbon nuclei; the probability of both carbons being $\ce{^13C}$ is $0.01~\%$. If only magnetically active nuclei were able to transfer coupling, then the uncoupled hydrogen signals should amount to $99.99~\%$ of a signal with the coupling signal having only $0.01~\%$ of the intensity. Clearly not what is being observed. In fact, $98~\%$ of the coupled hydrogen signal is due to molecules without magnetically active carbons between the vicinal hydrogens.

The final node has to be NMR-active, otherwise you wouldn’t be able to observe magnetisation.


The short answer is that the equations you describe as graphs are the solutions of the differential equations that define the rate laws in question.  If you don't understand what a differential equation is, any sensible answer is going to be problematic.

Assuming you do know what a differential equation is, we proceed!

For a first order rate law, the differential equation is something like:

$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt}=k[\ce{A}]$$

Rearrange the concentration and time to get:

$$-\frac{\mathrm d[\ce{A}]}{[\ce{A}]}=k\,\mathrm dt$$

You now integrate both sides, with $[\ce{A}]$ as a function of $t$. This is a well-known integral, which is:

$$\ln [\ce{A}] = \ln [\ce{A}]_0 - kt$$

where $[\ce{A}]_0$ is an initial value for $[\ce{A}]$ at the start of the reaction.

Similarly, the differential equation for the second order rate law is:

$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{A}]^2$$

Rearrange the concentration and time to get:

$$-\frac{\mathrm d[\ce{A}]}{[\ce{A}]^2} = k\,\mathrm dt$$

And again integrate to get:

$$\frac{1}{[\ce{A}]}=kt+\frac{1}{[\ce{A}]_0}$$

where $[\ce{A}]_0$ is an initial value for $[\ce{A}]$ at the start of the reaction.

This all comes from the following Wikipedia page, which has a lot more details: Rate equation


Why can't a balance that can weigh to the nearest milligram be able to weigh 23.8 milligrams of something?

Couldn't agree less, but check my math.

First $\ce{C18H38}$ has a molecular weight of $254.5~\mathrm{g/mol}$.

Second $\mathrm{5.64\times 10^{18} ÷ 6.02214 \times 10^{23} = 9.3654 \times 10^{-6}}$ and so is $2.4\ \mathrm{mg}$.

A balance with an ability to weigh to the nearest mg would, in a perfect world give you a mass of 2 or possibly 3 mg. This wouldn't be very useful, but it wouldn't give you zero information, either. So, since I don't understand the context in which your book asked the question, I'd have to say it doesn't seem to be completely correct if it states "it cannot" weigh that amount. It certainly cannot weigh that amount with adequate precision for most purposes I can think of. But if, say, you were trying to determine whether there was enough of a sample to use in a mass spec., or to calculate the yield of a reactor-on-a-chip, then it would suffice.


How to run electrochemical tests on Microorganism (bacteria in a medium)?

From what I understand, your main issue is connecting a solution in an electrical circuit without changing the composition of the solution due to electrochemical redox reactions.
So, there are two problems here.  Firstly, passing direct current
(DC) changes the composition of the solution. Secondly, a solution cannot
be connected to the bridge like a metallic wire or any other solid conductor.
The first difficulty is resolved by using an alternating current (AC) source
of power. The second problem is solved by using a specially designed
vessel called conductivity cell.


Why do we have the concentrations powered to the coefficients in the equilibrium expression?

This is due to the way the equilibrium constant is defined.

A chemical equilibrium is a stable state at some point in the reaction (characterized by the extent of reaction $\xi$). Because it is a stable state the free enthalpy $G$ has to have a local minimum at exactly that place.
Therefore we can say
$$
\left( \frac{\partial G}{\partial \xi} \right)_{p,T} = 0
$$
On the other hand the following equation applies:
$$
\Delta_rG_m = \left( \frac{\partial G}{\partial \xi} \right)_{p,T} = \sum_i \mu_i \cdot \nu_i
$$
With the free, molar reaction enthalpy $\Delta_rG_m$, the chemical potential of the component $i$ $\mu_i$ and it's reaction-coefficient $\nu_i$ (To understand this consider this general reaction equation: $\ce{\nu_1A + \nu_2 B <=> \nu_3 C}$ - note that the coefficients for the educts are negative by definition).

Therefore we can say that the following equation applies for the equilibrium state:
$$
\Delta_rG_m = 0
$$

we can rewrite above formula with the aid of the activity-dependency (basically a concentration-dependency) of the chemical potential ($\mu_i = \mu_i^* - RT\cdot\ln(a_i)$):
$$
\begin{split}
\Delta_rG_m &= \sum_i \mu_i \cdot \nu_i\\
&= \sum_i \nu_i\mu_i^* - RT \cdot \sum_i \nu_i\ln(a_i)
\end{split}
$$
Because $a \cdot \ln(b) = \ln(b^a)$ this is the same as
$$
\Delta_rG_m = \sum_i \nu_i\mu_i^* - RT \cdot \sum_i \ln(a_i^{\nu_i})
$$

Now we define the standard, free, reaction enthalpy $\Delta_rG_m^* = \sum_i \nu_i\mu_i^*$ and because we found out (in the beginning) that for the equilibrium state $\Delta_rG_m$ has to be zero we can write:
$$
\Delta_rG_m^* = -RT \cdot \sum_i \ln(a_i^{\nu_i})
$$
Because $ \ln(a) + \ln(b) = \ln(a\cdot b)$ we can rewrite this as
$$
\Delta_rG_m^* = -RT \cdot \ln\left(\prod_ia_i^{\nu_i}\right)
$$

The last step is now to define the equilibrium constant $K$ as
$$
K = \prod_ia_i^{\nu_i}
$$

And that is how the coefficients in the recation equation end up in the exponent when calculating the equilibrium constant.


Connection between Gibbs Energy of Formation and Gibbs Energy of Mixing

Thanks for the links in the comments above. The second one J Chem Thermo, 37, 97 (2005) seemed to be clearer. Eqn (12) reads
$$\Delta_M G=\Delta_f G^\circ /4
$$
but is preceded by the sentence

  According to the definition of the molar free energy of mixing,
  $\Delta_M G$ for Ni$_3$Al would refer to the formula
  Ni$_{3/4}$Al$_{1/4}$, and thus the equation ... was obtained.

In the first link J Electrochem Soc, 128, 1181 (1981) it is harder to spot your equation. However, I think I see it, in comparing eqns [7] and [10] which relate $\Delta G_m$ and $\Delta G_f^\circ$ to a Coulombic titration expression involving an integral over the compositional variable $y$. This is for the alloy Li$_y$Sn, and the expressions look identical except for a factor $1/(1+y)$. So I guess that's it. However there's no helpful explanation.

Bearing in mind the sentence I quoted from the (2005) paper, I suspect that this factor is simply a question of how these quantities are defined, in relation to the formula of the alloy under consideration (i.e. per atom or per formula unit), and not intimately connected with the expression for the free energy of mixing in terms of mole fractions and activities (except in that it is written per atom, or more accurately, per mole of atoms).

[Edit following OP comment]

I think that this is less complicated than it may have looked. $\Delta G$ is just the difference in Gibbs energy between two well defined states. I see, for instance, on p43 of Statistical Thermodynamics of Alloys by NA Gokcen (1986)
the definition

  The molar Gibbs energy of mixing refers to the process prescribed by

  $x_1$ (pure component 1) $+$ $x_2$ (pure component 2) $\rightarrow$ $1$
  gram atom of alloy

where, of course, "gram atom" is old terminology; but the meaning is clear. If the original metals are in their standard states, this is essentially the same as the definition of the standard Gibbs free energy of formation of the alloy, except that the latter is usually expressed in terms of one mole of the compound. I think this difference is what is intended by the numerical factor, irrespective of any particular ways of writing the Gibbs energy of mixing in terms of other variables (for example, one can sometimes use regular solution theory to estimate $\Delta S$ and $\Delta H$ of mixing, but I don't think that's relevant here).


Explanation of the orbital energies resulting from Jahn-Teller distortion of octahedral complex using MO theory

I disagree with your statement that

"The dxz and dyz don't seem to have the correct symmetry to sigma bond with any p orbitals of the ligands. They are nonbonding in a perfect octahedron"

Consider the case of a halide ligand like Cl-, which has filled p orbitals perpendicular to the axis of the M-Cl sigma bond. If these are aligned with the coordinate axes, they can each interact with proper symmetry with one of the off-axis t2g orbitals. In the axial position, these orbitals match up in symmetry with dxz and dyz.
In the perfect octahedron, all three t2g orbitals are affected equally (each interacts with the four ligands in its plane), so whether or not the ligand has p orbitals (and therefore pi bonding) does not qualtitatively change the splitting diagram. It just changes the magnitude of the splitting.
When the octahedron is distorted, however, the interactions with each of the t2g are no longer the same in magnitude. Just as with the sigma bond, as the axial ligand moves away from the metal center, the pi bonding interaction also weakens, so the d orbitals involved in pi bonding (which as you noted are anti-bonding orbitals) lower in energy.
The reason that the dxy appears to be raised in energy is likely just an artifact of centering the orbitals on the average energy in both cases. Alternatively, the diagram could be based on the assumption that as the axial ligands move away from the metal, the equatorial ligands move closer, increasing the strength of the sigma and pi bonding of those ligands. In the pi case, this would raise the energy of all three t2g orbitals because all of them can engage in pi bonding with equatorial ligands, but the dxy would go up the most since it can pi bond with all four equatorial ligands, whereas the other two can only pi bond with two of the four equatorial ligands.
UPDATE: Here's an image from some old lecture notes showing the difference in convention of alignment between the "MO" view and "d-orbital view" in which the "d-orbital" diagram recenters the energies around the average, even though only the dxz and dyz MO (or more properly the antibonding orbitals primarily composed of them) energies actually changes:


The most common industrial method is one of various types of liquid chromatographies, which is usually how glucose-fructose solutions are enriched in fructose.
I did see a couple other possible methods though:

This one from 1965—before good LC methods existed—relies on the difference in solubility in partially water-miscible solvents like n-butanol to perform a liquid-liquid extraction.
Unfortunately, the pdf seems to be incomplete, but basically the water-solvent-sugar system separates into two phases with each sugar partitioning more into a different phase.
They calculated that it would take around 40 cycles to get 90% fructose from this method, but it doesn't seem to be very efficient due to slow diffusion of the sugars (and since LC is the dominant method today, it doesn't sound too promising on any reasonable scale).
In the introduction, they also note that some of the other attempts of the era involved chemical means—specifically precipitation of the fructose as calcium fructosate, getting the fructose back out by reaction with carbon dioxide, oxalic acid, or sulfuric acid in what is known as Dubrunfaut's method.
Again, this did not result in an industrially useful process.

This one involves careful fractional crystallization of glucose-fructose solutions, as fructose is far more soluble in water than glucose.
The solution is cooled from room temperature to -20 °C fairly slowly (0.5 K/min) while being agitated and in the presence of small glucose seed crystals.
Some of the glucose crystallizes out of the solution, which is separated by centrifugation, and the process is repeated going to -35 °C.
The crystals they get are 89% glucose and 11% fructose, but a big red flag is they don't actually measure the concentrations in the liquid left behind (separating fructose is far more important industrially because pure glucose is easy to get by hydrolysis of starch).
They say that "a concentrated solution of fructose containing a small amount of glucose was obtained", but don't actually prove it, and apparently it only works properly if you can maintain a total sugar concentration near 50%, so I'm pretty sceptical of the method's utility.
You could probably do something similar at home if all you're interested in is getting some somewhat purified glucose.

Unfortunately, if there were a simple and effective way to do this, industrial production probably wouldn't bother with LC.
If you just want to play around with separations, these might be some things to try, but if you just want glucose and/or fructose, I have seen both at bulk food stores. (glucose is often labelled as dextrose and fructose is sold as a low glycemic index sweetener)


Methane seems to dissolve more readily in cold water. Why?

The enthalpy of solution going from the ideal gas state to the solution at infinite dilution is exothermic.  However, more importantly, the Gibbs energy of solution is positive meaning that dissolution of methane into water is not a favored process.  Most of this comes from the relatively large negative entropy change associated with the dissolution at room temperature.  This positive Gibbs energy is manifested in the very small solubilities observed and gives rise to the whole concept of something being "hydrophobic."  The fact that any methane dissolves in water is due to equilibrium effects based on a finite positive Gibbs energy.  Since methane is dissolved to a greater extent at lower temperatures must mean that the Gibbs energy is less positive at the lower temperatures.


There are several complexes where methanol acts as ligand. Some examples are:

methanol coordinated vanadium(V) chalcogenido complexes: $\ce{V(Chal)Cl2(OMe)(HOMe)2}$ where $\ce{Chal = O, S, Se}$
bis(μ-methoxo)dichromium(III) complexes, $\ce{[(L^{Se})2Cr2(μ-OCH3)2(CH3OH)2]}$ and $\ce{[(L^{Se})2Cr2(μ-OCH3)2(CH3OH)(CH3O)]}$, where $\ce{L^{Se}}$ represents the dianion of 2,2‘-selenobis(4,6-di-tert-butylphenol)
di-µ-chlorido-bis[dichloridobis(methanol-κO)iridium(III)] dihydrate  $\ce{[Ir2Cl6(CH3OH)4]2H2O}$
$\ce{M(CH3OH)3SO4}$ where $\ce{M = Ni, Co, Zn}$

Reference

Google: Methanol ligand complex. First 4 search results.


What happens if water is treated to AC power supply of 12 kV and 14 kHz?

Using up to 30 kilovolts is not uncommon with aqueous systems in a technique called capillary electrophoresis. Similarly, applying kilohertz current is also common in a technique called conductometry to measure the resistance of aqueous solutions.
In general, when you pass high frequency current, electrolysis does not occur.

You have to check the dielectric breakdown voltage of water, perhaps with high voltage and high frequency current, severe Joule heating would be observed, if there is no dielectric breakdown (which means internal sparking).


Is it possible to calculate the 'overall' redox potential of a system?

No, it is impossible. You can have a situation where a cellular compartment contains a high ratio of $\ce{NAD+:NADH}$ (i.e. favoring oxidation) and a low ratio of $\ce{NADP+:NADPH}$ (i.e favoring reduction). This is advantageous for the cell because it can run oxidation of some metabolites in parallel with reduction of other metabolites, just by employing enzymes with distinct co-factor requirements.

This is different from pH. A solution is either acidic or basic, not both. The difference arises because acid/base reactions are usually fast, and water acts as acid or base, "communicating" acid/base status across the solution.


Generally speaking, percentage aqueous solutions like that are either weight for weight or weight for volume. Either way, in this example, you are comparing how many osmoles there for the same weight of urea ($CO(NH_2)_2$) and glucose ($C_6H_{12}O_6$).

Mole fraction as expressed in a percentage does exist though I have not seen that notation much myself and is usually designated n/n%

From the wording your question I'm guessing that it's a physiology or biochemistry question, in which case they're almost certainly referring to the 5% aqueous solution of glucose as the common IV hydration fluid 5% Dextrose Solution, which is w/w.

Otherwise, it's hard to guess as it's ambiguous!


Why do some molecules show non-symmetric absorption and emission bands?

Mirror image spectra are only observed in solution and then only if the ground and excited state potential energies have almost exactly the same shape. Thus in rigid molecules such as anthracene a good mirror image is observed but not in more flexible molecules such as stilbene ($\phi-C=C-\phi$). The spectra of each vibrational level is wide in solution due to solvent interactions altering energy levels v. slightly, this does not happen in the collisionless gas phase where lines are sharp. In solution the width of lines from different vibrations can overlap making a broad largely structureless spectrum.

There are different effects, a stokes shift due to solvent causing lowering of excited state energy shortly after absorption, a displacement in position of minimum energy, due to different bond lengths in each state and anharmonicity in one potential vs the other. Even slightly different Franck Condon factors between states can mean that a phosphorescence spectrum is similar, but different to its fluorescence spectrum, see figure below, when correction is made for detector sensitivity vs wavelength.

The figure below (not to scale) shows how ideal mirror image spectra are formed. Notice that displacement of potential energy is needed and that vibrational frequency in ground and excited state can be slightly different.


Is there a heat capacity of an object when it is melting?

Yes there is a heat capacity even at the melting point.  The reason the temperature doesn't change is that you need to add the enthalpy of melting to convert solid to liquid in equilibrium.  However, under the right conditions you can supercool the liquid under the melting point, and the temperature will behave as you would expect, needing to pull a certain amount of energy out to supercool by a degree.  But this merely delays having to deal with the enthalpy of melting until solid uncleared - it is a kinetics issue, not a thermodynamics issue.  So, it is clear that the heat capacity of a phase is continuous through the phase transition (and beyond).


I can't choose the set of 3 chemicals with each of which Si will react (out of 5 possible sets)

$\ce{Si}$ does in fact react with $\ce{HF}$ and $\ce{NaOH}$. The reactions would be:

$$\ce{Si + 4HF -> SiF4 + 2H2}$$

$$\ce{Si + 4NaOH -> Na4SiO4 + 2H2}$$

$$\ce{Si + 2MgO -> SiO2 + 2Mg}$$

You can conclude these, since $\ce{SiO2}$ is quite stable, and also reacts with $\ce{HF}$ and $\ce{NaOH}$ (You may have heard that $\ce{HF}$ dissolves glass), so definitely $\ce{Si}$ will react way better than $\ce{SiO2}$.


How to evolve hydrogen from aluminum when the household drain cleaner is safe to use on aluminum?

First, any drain cleaner that is at all effective is probably not safe for a child to use, even under adult supervision.
Second, you can easily produce enough hydrogen to inflate a plastic bag, perhaps one to four liters, using electrolysis. Unlike an elastic balloon, the bag will not resist inflation appreciably until full. Try carbon electrodes (e.g., from zinc-carbon cells) with $\ce{MgSO4}$ (Epsom salt) in water as the electrolyte.
You can also capture oxygen generated and test it with a glowing wooden splint.
It is an exercise for the experimenter to calculate how many ampere-hours it would take to inflate the specific bag.
BTW, you might want to perform this outdoors, in case a spark sets off the hydrogen!


Neighbouring group participation with and without retention of configuration in nucleophilic substitution of 2,6-dichloro-9-thiabicyclo[3.3.1]nonane

The only difference between substrates A and C is disulfide vs sulfone nature at 9-position. Actually, that is good enough to display this major stereochemical outcome. As explained in Ref.1, the disulfide nature at 9-position of A has an advantage of having two lone pair on sulfur, one of which is proximate enough to attack backwardly on $\ce{C-Cl}$ bond ($\mathrm{S_N2}$ in nature) to give an episulfide ring (see following diagram and the extended comment of ron). The reaction center must have an inversion of configuration since the mechanism is $\mathrm{S_N2}$.

This episulfide intermediate undergoes ring opening backward attack by the intended nucleophile (e.g.,$\ce{N3-}$) to give the final product under 2nd $\mathrm{S_N2}$ mechanism. This stepwise replacement of chlorine by a nucleophile, the first substitution being considerably faster than the second, was first observed by E. J. Corey and E. Block (Ref.2). The nucleophile of their reaction was methoxy group of methanol (solvolysis). However, although they are the first group to suggest the 1,2-epi-sulfonium ion intermediates in these reactions, it was suggested for the initial dichloro compound A formation, not for the solvolysis by methanol. These types of neighboring group participation (NGP; or anchimeric assistance) have been observed previously for bridging carbon, nitrogen, and boron, all for the bicyclooctane series. In mechanistic point of view, this double inversion process preserves stereochemistry in substitution reactions on 2,6-Dihalosubstituted-9-thiabicyclo[3.3.1]nonanes (Ref.3).

Finally, C cannot follow the fast double inversion process due to the lack of a lone pair on $\ce{S}$ bridge, yet undergoes usual $\mathrm{S_N2}$ transformation (slow because the leaving group is $\ce{Cl}$) under the given conditions.

References:

D. D. Díaz, A. Converso, K. B. Sharpless, M. G. Finn, “2,6-Dichloro-9-thiabicyclo[3.3.1]nonane: Multigram Display of Azide and Cyanide Components on a Versatile Scaffold,” Molecules 2006, 11(4), 212–218 (doi: 10.3390/11040212).
E. J. Corey, E. Block, “New Synthetic Approaches to Symmetrical Sulfur-Bridged Carbocycles,” J. Org. Chem. 1966, 31(6), 1663–1668 (DOI: 10.1021/jo01344a001).
A. Converso, K. Burow, A. Marzinzik, K. B. Sharpless, M. G. Finn, “2,6-Dichloro-9-thiabicyclo[3.3.1]nonane: A Privileged, Bivalent Scaffold for the Display of Nucleophilic Components,” J. Org. Chem. 2001, 66(12), 4386–4392 (DOI: 10.1021/jo015632y).


Completely symmetrical diatomic molecules like $\ce{O2}$ and $\ce{N2}$ are not IR-active because they have a zero dipole moment, even when oscillating within their one vibrational mode. Hence, there is no change in dipole moment possible, so these molecules will not be detectable by IR. The other molecules you mention, however, are IR-active, and I was able to find spectra of all three in the NIST database.

There are other spectroscopic methods that can detect IR-inactive molecules, however. Raman spectroscopy may be applicable, and I'm dimly aware of sophisticated methods based on absorptions in electronically excited states, but I don't know nearly enough about the subject to offer any concrete information on those techniques.


LeChatelier's Principle - Phosphofructokinase in Glycolysis

PFK plays a central role in the control of glycolysis because it
  catalyses one of the pathway’s rate-determining reactions. The control
  of Phosphofructokinase (PFK) is exquisitely complex.

To analyse this complicated enzymatic reaction pathway, we have to revisit important aspects of metabolic control mechanism:

Thus the flux (rate of flow) of intermediates through a metabolic pathway is constant; that is, the rates of synthesis and breakdown of each pathway intermediate maintain it at a constant concentration.
Regulation of the steady state (homeostasis) must be maintained in the face of changes in flux through the pathway in response to changes in demand. In an open system, such as metabolism, the steady state is the state of maximum thermodynamic efficiency.

This explanation provides a somewhat agreement of Le Chatlier’s principle. (i.e. where a dynamic equilibrium is affected by changing conditions, the position of equilibrium moves to counteract the change)

Consider this simple hypothesis:

  J = vf – vr

  Since a metabolic pathway is a series of enzyme-catalysed reactions,
  it is easiest to describe the flux of metabolites through the pathway
  by considering its reaction steps individually. The flux of
  metabolites, J, through each reaction step is the rate of the forward
  reaction, vf, less that of the reverse reaction, vr:

  At equilibrium, by definition, there is no flux (J=0), although vf and
  vr may be quite large. At the other extreme, in reactions that are far
  from equilibrium, vf>> vr, so that the flux is essentially equal to
  the rate of the forward reaction, J~vf.

The flux throughout a steady-state pathway is constant and is set (generated) by the pathway’s rate determining step (or steps).

Consequently, control of flux through a metabolic pathway requires:

that the flux through this flux-generating step vary in response to the organism’s metabolic requirements and
that this change in flux be communicated throughout the pathway to maintain a steady state

Phosphofructokinase is a non-equilibrium enzyme of glycolysis

One important thing to take note of is that the enzyme phosphofructokinase, a highly controlled enzyme functioning far from equilibrium (but ironically itself operates via an equilibrium of two conformational states), evidently is the major target for regulating glycolysis.

PFK is a tetrameric enzyme with two conformational states, R and T, that are in equilibrium. The substrate site binds ATP equally well in either conformation, but the inhibitor site binds ATP almost exclusively in the T state. The other substrate of PFK, F6P, preferentially binds to the R state. Consequently, at high concentrations, ATP acts as a heterotrophic allosteric inhibitor of PFK by binding to the T state, thereby shifting the T <=> R equilibrium in favour of the T state and thus decreasing PFK’s affinity for F6P.

Increase in concentration of PFK does not increase the rate of glycolysis.

In this scenario I believe the Le Chatlier’s principle applies little here. The reason is because the control is very complicated Although PFK has a major role in regulating the flux through glycolysis, it is controlled, in vivo, by factors outside the pathway.

An increase in the in vivo concentration of PFK will therefore not increase the flux through the pathway because these controlling factors adjust the catalytic activity of PFK only to meet the needs of the cell.

Let’s consider another illustration showing enzyme activity versus substrate concentration:

The inhibition of PFK by ATP is relieved by AMP. This results from AMP’s preferential binding to the R state of PFK. If a PFK solution containing 1 mMATP and 0.5 mM F6P is brought to 0.1 mM in AMP, the activity of PFK rises from 10 to 50% of its maximal activity, a 5-fold increase. when [F6P] =0.5 mM (the dashed line), the enzyme is nearly maximally active, but in the presence of 1 mM ATP, the activity drops to 15% of its original level (a nearly 7-fold decrease).

However, measurements of [ATP] in vivo at various levels of metabolic activity indicate that [ATP] varies <10% between rest and vigorous exertion. Yet there is no known allosteric mechanism that can account for a 100-fold change in flux of a nonequilibrium reaction with only 10% change in effector concentration.

Thus, some other mechanism, or mechanisms, must be responsible for controlling glycolytic flux. Let’s discuss the action of adenylate cyclase; adenylate cyclase catalyses the following reaction:

2ADP <=> ATP + AMP (K=0.44)

as a result of the adenylate kinase reaction, a 10% decrease in [ATP] will cause over a 4-fold increase in [AMP]. Consequently, a metabolic signal consisting of a decrease in [ATP] too small to relieve PFK inhibition is amplified significantly by the adenylate kinase reaction, which increases [AMP] by an amount sufficient to produce a much larger increase in PFK activity.

In summary the control of the PFK (non-equilibrium enzyme) has limited Le Chatlier’s principle application as other crucial regulation mechanism exist through the reaction pathway. Even if you were to consider temperature changes, these control enzymes operate far from equilibrium (these have very negative delta G°) and the actual free energy changes in vivo differ from experimental ones making it difficult to simulate actual effects in equilibrium states. It is more likely based on supply-demand basis. One evidence of this is seen through the variation of metabolic flux  throughout glycolysis (may vary by 100-fold or more), depending on the metabolic demand for ATP.

However, for pathways which operate close to equilibrium it may be applied, for example, the delta G° for aldolase is endergonic, whereas under physiological conditions in heart muscle it is close to zero, indicating that the in vivo activity of aldolase is sufficient to equilibrate its substrates and products. Their forward and reverse rates are much faster than the actual flux through the pathway (although their forward rates must be at least slightly greater than their reverse rates). Consequently, these near-equilibrium reactions are very sensitive to changes in the concentration of pathway intermediates.

References

Biochemistry Voet and Voet


rate determination by 1H NMR spectroscopy

As mentioned in the comments, J. Chem. Ed is a good resource for NMR kinetics experiments. I have had success with the experiment published by J. Orvis in J. Chem. Ed 80(7), 2003, 803 link.  This publication describes the synthesis of trans-dichlorotetraamminecobalt(III) and its subsequent reaction with water to produce cis-aquachlorotetramminecobalt(III).  There are several advantages to this experiment as a model study:

The synthesis is straightforward and does not require materials unlikely to be already in the laboratory, although it is critical that the directions are followed precisely.
The aquation reaction occurs within about one hour - slow enough to easily monitor via NMR but not so slow as to take up precious instrument time
The reaction rate is temperature sensitive, which makes it a good model for extracting thermodynamic properties from the reaction

My undergraduate students are able to complete the synthesis about 50% of the time, and the yields using the published procedure are about 400 mg, so one successful synthesis is suitable for a number of NMR experiments.


In Finkelstein reaction, how does precipitate formation favor forward reaction?

$\ce{NaCl} $ and $\ce{NaBr }$formed as products are polar compounds which are vey less soluble in non- polar solvents like dry acetone, but $\ce{NaI } $due to high polarizability of $\ce {I^-} $ is soluble in acetone. So the products precipitate.  If a reaction is in equilibrium in constant temperature  and it is distorted somehow, it goes in the direction to nullify the effect of distortion.  This is the Le-Chattelier's principle.  If some of the products are removed to keep the $\ce{K_c}$ constant , some more reactants will react and convert into products. Thus constant removal of products by precipitation helps to dissociate reactants into products and thus favour the forward reaction.


Molality of solution to be used when finding water separated from solution as ice

The question: $\pu{1kg}$ of a solution of cane sugar is cooled and maintained at $\pu{-4 ^\circ C}$. How much ice will separate out if molality of solution is $0.75$?
Given: $K_\mathrm{f} \text{ (Water)}= \pu{1.86 K kg/mol}$

The solution is given assuming you understand the concept. It seems you did not completely understand the colligative properties of solutions. Thus I'm giving some clues you how to approach this problem.
The textbook equation you are going to use is:
$$\Delta T_\mathrm{f}= K_\mathrm{f}\times i \times \text{molality of the solution} = K_\mathrm{f}\times i \times \frac {\text{amount of sucrose in }\pu{mol}}{\text{weight of solvent in }\pu{kg}} \tag 1$$
where $i = 1$ for sucrose (sucrose does not ionize in water).
The solution is kept at $\pu{-4 ^\circ C}$. The freezing point of pure water is $\pu{0 ^\circ C}$ ($\therefore \ \Delta T = 0 - (-4)=4$). First what you should do is find the freezing point of given solution using equation $(1)$. If $\Delta T_\mathrm{f} \gt 4$, the game is over. That means, this solution does not freeze at $\pu{-4 ^\circ C}$. If $\Delta T_\mathrm{f} \lt 4$, the game is on. That means, some water will freeze out until the final molality reached the molality of solution, which freeze at $\pu{-4 ^\circ C}$.
For given solution (no units was used):
$$\Delta T_\mathrm{f}= K_\mathrm{f}\times i \times m = 1.86 \times 1 \times 0.75 = \pu{1.395} $$
Thus, this solution freeze at $\pu{-1.395 ^\circ C}$.
Now, you have to find out at what molality of a solution would be just about to freeze at $\pu{-4 ^\circ C}$. That means, $\Delta T_\mathrm{f}= 4$ (still, $i = 1$):
$$\Delta T_\mathrm{f}= K_\mathrm{f}\times i \times m \ \Rightarrow \ m= \frac{\Delta T_\mathrm{f}}{K_\mathrm{f}} = \frac{4}{1.86} = 2.15$$
This means, water in original mixture freeze out until the molarity of solution becomes $\pu{2.15 mol/kg}$.
Put it this way:

The original solution has $\pu{0.75 mol}$ of sucrose in $\pu{1 kg}$ of water. The amount of sucrose would not freeze.
Thus, how much water would remain with $\pu{0.75 mol}$ of sucrose to keep molality at 2.15?

I'm sure it is given in your solution manual. Regardless, I'll show you the calculations. Suppose the weight of water is $W \ \pu{kg}$ (ignore the units in equation}:
$$ \frac{0.75}{W} = 2.15 \ \Rightarrow \ W = \frac{0.75}{2.15} = 0.349$$
Thus, the weight of water in final solution after freezing at $\pu{-4 ^\circ C}$ is $\pu{0.349 kg}$ (the amount of sucrose is still $\pu{0.75 mol}$).
However, this calculation is in significant deviation from actual amount since the given amount of original solution is $\pu{1.0 kg}$. That means it does not contain $\pu{0.75 mol}$ of sucrose. Now, molecular weight of sucrose is $\pu{342.3 g/mol}$. If amount of sucrose in $\pu{1.0 kg}$ of original solution is $x$, then:
$$\frac{\frac{x}{342.3}}{(1000-x)\times 10^{-3}} =0.75  \ \Rightarrow \ 1.257x = 0.75 \times 342.3   \ \Rightarrow \ x = \frac{256.725}{1.257} = \pu{204.2 g}$$
Therefore, the amount of water in that solution is $\pu{1000-204.2) g} = \pu{795.8 g}$ or $\pu{0.796 kg}$. Thus, amount of sucrose in the of original solution is: $\frac{\pu{204.2 g}}{\pu{342.3 g/mol}} = \pu{0.597 mol}$.
Using these values, we can recalculate the water amount in final solution. Suppose the weight of water in final solution is $W_f \ \pu{kg}$:
$$ \frac{0.597}{W_f} = 2.15 \ \Rightarrow \ W_f = \frac{0.597}{2.15} = 0.278$$
Thus, the weight of water in final solution after freezing at $\pu{-4 ^\circ C}$ is $\pu{0.278 kg}$ (the amount of sucrose is $\pu{0.597 mol}$ making it $\pu{2.15 molal}$).
Thus, the amount of water freeze out from the original $\pu{1.0 kg}$ of solution is:
$$\pu{(0.796-0.278) kg} =\pu{0.518 kg}$$


All mentioned terms -- classical Coulomb, exchange and correlation energy -- have the very same physical origin: the electron-electron Coulomb repulsion.

The split into these contributions is only an artifact arising from Hartree-Fock theory.
Here the electronic wave function is approximated by a Slater determinant (the anti-symmetrized product of orbitals). This results in a mean-field description of the electron-electron interaction, and the error with respect to the exact solution is the correlation energy.

Plugging the Slater determinant into the Schrödinger equation eventually leads to an expression for the Hartree-Fock energy:

\begin{equation}
  E_\mathrm{HF} = \sum_i^{N_\mathrm{elec}} \langle i|\hat h|i\rangle + \sum_{i>j}^{N_\mathrm{elec}} [ii|jj] - [ij|ji]
\end{equation}

The three terms in that equation use a short hand notation. Explicitly written those are the one-electron integrals
\begin{equation}
  \langle i|\hat h|i\rangle = \int\chi_i^*(\boldsymbol x_1)\hat h\chi_i(\boldsymbol x_1) \mathrm{d}\boldsymbol x_1
\end{equation}
which cover the kinetic energy and electron-nuclear Coulomb interaction.

The other two terms originate from the electron-electron interaction.
The first one is called Coulomb integral, and gives the contribution of the classical Coulomb energy
\begin{equation}
    [ii|jj] = \int\chi_i^*(\boldsymbol x_1)\chi_i(\boldsymbol x_1) \frac{1}{r_\mathrm{12}}\chi_j^*(\boldsymbol x_2)\chi_j(\boldsymbol x_2) \mathrm{d}\boldsymbol{x_1}\mathrm{d}\boldsymbol{x_2}
\end{equation}
The only reason for its name is, because with $\chi_i^*(\boldsymbol x_1)\chi_i(\boldsymbol x_1) = \rho_i(\boldsymbol x_1)$ and $\chi_j^*(\boldsymbol x_2)\chi_j(\boldsymbol x_2) = \rho_j(\boldsymbol x_2)$ one obtains the formula for the coulomb interaction of two classical charge distributions $\rho_i(\boldsymbol x_1)$ and $\rho_j(\boldsymbol x_2)$. However, we are doing quantum mechanics here, so do not over interpret that analogy.

The last term is the exchange integrals
\begin{equation}
    [ij|ji] = \int\chi_i^*(\boldsymbol x_1)\chi_j(\boldsymbol x_1) \frac{1}{r_\mathrm{12}}\chi_j^*(\boldsymbol x_2)\chi_i(\boldsymbol x_2) \mathrm{d}\boldsymbol{x_1}\mathrm{d}\boldsymbol{x_2}
\end{equation}
and simply named like that because it is essentially the Coulomb term but two indices have been exchanged.

In DFT these three terms show up, because it borrows some concepts from HF and but tries to add the missing electron correlation.

Interpretation of Electron Correlation

I always struggle with the "correlated motion" thing to explain what electron correlation is. I don't think it helps anyone to get an intuitive understand what this means. I do like however the following pictures from the book by Helgaker, Jorgensen and Olsen.

They consider the electron density in $\ce{H2}$ in HF (mean-field) approximation in the top plots and the exact solution below.
To the left we have the 1-electron density which is obtained by integration (=averaging) of the 2-electron density the in middle and right plots.
The positions of the two Hydrogen atoms are obvious by the peaks in the 1-electron density.

The 2-electron density of the HF solution has 4 peaks:
The 2 at ($r_1\approx 1$, $r_2\approx -1$) and ($r_1\approx 1$, $r_2\approx -1$) are the probability of finding both electron at different atoms. The other 2 at ($r_1\approx 1$, $r_2\approx 1$) and ($r_1\approx -1$, $r_2\approx -1$) are the probability of finding both electrons at the same atom. The latter 2 peaks are missing in the exact solution, because the electrons try to avoid such situations.
HF however is unable to consider this correlation. Note that this effect is not visible in the 1-electron densities. HF only works based the 1-electron wave functions (orbitals) and constructs the 2-electron wave function by simple multiplication the 1-electron wave functions.


Why can a wave-function be written as a product of 1-electron wave-functions?

To start with your first sentence: "Why can a wavefunction be written as a product of 1-electron wave-functions"? Well, you can write whatever you want to approximate a wavefunction; physics police is not going to come and take you away. But the two important questions are: How useful will the results from your approach be, and how hard is it to work with mathematically? These criteria obviously reduce your number of options quite a bit.

A wavefunction that is constructed as a product of one-electron orbitals definitely has that second point going for it. As you have already experienced yourself, it is straightforward to write down and calculate with. Arguably the most important implementation of this approach is Hartree-Fock theory where the electrons never see each other as distinct point charges in space, only as fuzzy charge distributions that each one has to wade through individually. Hartree-Fock gives you fairly straightforward equations to work out the optimal way to build the independent-particle wavefunction; and while it cannot accurately capture the energetics of interactions among the electrons "out of the box", it's a clean and simple starting point to describe the quantum mechanics of your molecule. And it's often not all too bad, either - for typical well-behaved cases, Hartree-Fock and its one-electron orbitals can already land you within a few percent of the full energy value. The remainder is still way too large for it to be practically useful, but you can reuse this initial wavefunction in more refined algorithms to reduce the error further.