Importance of higher order exchange terms?

Exchange is important and is present in any system. To be more specific, in MP2 you already have exact exchange for the ground state from the zeroth-term(HF) while in the second-term (MP2) you have a correction to correlation plus some more exchange from the excited state.

MP2 allows the electrons to be partly in the unoccupied MO however, it is limited to double excitations. It turns out that you end up having a small correction to exchange and a big correction to correlation which is completely missing in HF. So the exchange correction will be small in every system due to the type of excitation treated in MP2.


Part 1: The elements on the periodic table are arranged according to their ground-state electron configurations. He 2s is an excited state of He 1s. There are even He 3s and 4s depending on whether one gives enough energy for the atom to excite to such levels. In general, higher principal quantum number (n) indicates a higher energy state (less stable)

Part 2: Your reasoning is correct.


It is not that we need the identity operator. It is just that things (character tables, irreducible representations, etc.) work the way they do. As to why they do so, I refer you to any decent textbook on group theory.

I guess one could rebuild the entire theory from scratch (that is, reformulate all definitions, starting from the very definition of a group, reprove all theorems, etc.) without explicitly using $\rm E$. That would be a titanic amount of work, though, and quite pointless at that. Also, many things like "order of a group is divisible by..." would start to sound less natural.

Come to think of it, why do we need the number $0$ in math? It just sits there and does nothing, after all. Well, doing nothing may be quite important: it makes a lot of other things simpler.


Will an alloy leaching in an acid act as both anode and cathode?

You can setup a situation where one metal is acting as a cathode and the other metal acting as an anode. There will be a small potential difference due to their differences in Fermi Levels much like a thermocouple but the alloy will be at an electrochemical potential that is constant.

This is not that different than a single metal(like a piece of iron) in solution. In the corrosion reaction, the counter reaction may happen somewhere else on the iron. So the stick of iron may be the anode and the cathode at the same time.


I am posting a solution image regarding same question. 2nd hydrolysis of X2- is neglected as you did. important thing is degree of hydrolysis is not small and should not neglected.


Why does water favour nucleophilic substitution over elimination?

The nature of the substrate (in particular the degree of substitution of the α-carbon) is of principal importance in determining which mechanism predominates. Generally, $\mathrm{S_N1}$ and $\mathrm{E1}$ are both less common than $\mathrm{S_N2}$ and $\mathrm{E2}$, partially as a consequence of their first-order kinetics and the stability constraints imposed by the charged intermediate species the mechanisms generate. $\mathrm{S_N1}$ and $\mathrm{E1}$ both require tertiary, allylic, or benzylic α-carbons in order to be viable (although secondary carbons can occasionally react by these pathways as well, but slowly, and only if conditions disfavor competing reactions), mainly because the carbocation intermediate that forms during the rate-determining step can be stabilized by the neighboring electron-donating alkyl groups via hyperconjugation (or via resonance in the case of carbons in allylic or benzylic positions). Tertiary carbons also disfavor $\mathrm{S_N2}$ because the approach of the nucleophile is sterically hindered by the β-alkyl groups. Under those conditions, $\mathrm{E1}$ and $\mathrm{S_N1}$ reactions would be in competition. $\mathrm{E1}$ would be favored if there were no strong nucleophiles present, as well as by the application of heat. Obviously, $\mathrm{S_N2}$ reactions are favored by the presence of strong nucleophiles (ideally, ones that don't also act as strong bases) and a primary α-carbon substrate. $\mathrm{E2}$ reactions are favored over $\mathrm{S_N2}$ if strong bases, especially non-nucleophilic, are present. $\mathrm{E2}$ also prevails when either the substrate and/or the base is sterically hindered, since abstraction of a proton is vastly less impacted by steric factors than is the approach of a nucleophile attacking a carbon with simultaneous loss of the leaving group. Finally, $\mathrm{E2}$ reactions impose an additional restriction in that the β-hydrogen and the leaving group must be anti-periplanar to one another, in order for simultaneous loss of the leaving group and formation of the π-bond to occur in a coordinated fashion. If the molecule can't adopt such a conformation (for example, it may be rotationally locked due to the presence of π-bonds, or lack the kinetic energy necessary for a ring flip, etc.), then $\mathrm{E2}$ is categorically ruled out.

(Parenthetically, I should add that elimination reactions, in general, require higher temperatures than substitution reactions. The naive explanation for this is based on the equation for the free energy change of a reaction: $\Delta G = \Delta H\ -\ T\Delta S$. Elimination reactions, at the most simplified level, involve two reactant molecules [namely, a base and the substrate], and yield three products [the protonated base, the substrate molecule that has undergone elimination, and the substrate's leaving group]. Consequently, elimination reactions are entropically favorable [i.e., the $T\Delta S$ term is positive]. However, elimination reactions are also endothermic [i.e., $\Delta H$ is positive], because heat energy is required to break the two σ-bonds of the abstracted proton and the leaving group, while the resulting products form only one σ- and one π-bond, the latter of which is less stable. The implication of this is that, in order for the overall change in free energy, $\Delta G$, to be negative [a requisite condition for spontaneity of the reaction], $T\Delta S$ must be greater than $\Delta H$, so the temperature needs to be sufficiently high if the reaction is to take place.)

The solvent is of only slightly less importance. Polar protic solvents (water, alcohols, weak carboxylic acids, etc.) have partial-positive hydrogens, which serve to stabilize negatively charged species. Consequently, $\mathrm{E2}$ and $\mathrm{S_N2}$ mechanisms, which are aided by having negatively charged bases or nucleophiles in high concentration, are somewhat dampened by protic solvents, allowing otherwise disfavored $\mathrm{E1}$ and $\mathrm{S_N1}$ pathways to compete. Polar aprotic solvents (THF, DMF, DMSO, chloroform, ethers, ketones, etc.) cannot stabilize negative charges as effectively, making them the preferred choice when $\mathrm{E2}$ or $\mathrm{S_N2}$ mechanisms are desired.

Finally, to your specific question about haloalkanes. I've explained above why various combinations of conditions might favor one class of reaction over another. In water, specifically, a few distinct factors, beyond the general principles outlined above, might be relevant:

Due to the leveling effect, no base stronger than hydroxide can exist in water at any appreciable concentration (that is, any base much stronger than hydroxide will simply deprotonate the water molecules, forming hydroxide and the base's conjugate acid). However, strong, non-basic nucleophiles are not subject to this effect, and though they may be solvated and weakened by hydrogen bonding, this effect would be less drastic than the protonation that strong bases would undergo.
Hydroxide is actually a much stronger base in many other solvents than it is in aqueous solution, which is chiefly the result of it being very thoroughly solvated and stabilized by hydrogen bonding. Hence, competing nucleophiles become stronger by comparison, while stronger bases simply end up being consumed, producing more hydroxide.
Because elimination reactions require heat, substitution reactions will be heavily favored and immediately take place unless the reaction vessel is heated prior to mixing of the reagents. The problem is compounded by the fact that, with hydroxide as the nucleophile, an alcohol would form, which would be stabilized by hydrogen bonding with water molecules. Further, because haloalkanes are poorly soluble in water, and immiscible liquids in the same container will boil at lower temperatures than they would unmixed, it could become very difficult in practice to achieve the temperatures necessary to favor elimination reactions while also maintaining the agitation necessary to keep the haloalkane and the aqueous phase from completely separating.


How to figure out the most stable conformation of the diastereomer?

To answer your professor's questions, there is no need to determine the R/S configuration of the two carbon atoms. We have to check the clockwise/anticlockwise arrangement of the substituents around the carbon atom, but checking specifically whether it is R or S is unnecessary.
Before moving further, let's have a look at the following image so that we are clear on how to visualise Newman Projections:

Since in the options provided in the question the carbon atom having bromine is at the back, we'll have to look at the molecule in the following manner:

Hence the Newman Projection obtained would be this:

Now, in question 1, option A can be rejected as it's an entirely different molecule. Option 4 has the atoms in the same order as above, so it is the exactly same molecule. In option 2 and 3, the arrangement of atoms is same for the back carbon but different for the front carbon. Therefore, without finding R or S configuration, we can say that options 2 and 3 are the diastereomers!
Out of these two structures, option 3 is more stable as the bulky ethyl groups are at the farthest position from each other.
Moving on to question 2:
You are right in flipping the arrangement about both the carbon atoms and you should get the answer that way. Maybe you are doing some mistake in getting the right Newman projection. Look again at the Newman projection I've provided above and then compare the options with it to see which option has different arrangement of substituents on both the carbons.


Viewing a protein or bio structure in Chembio3D in its natural conformation

The site you are looking for is the Protein Data Bank.

You can get to it with one of these links.
www.wwpdb.org
www.pdbe.org
www.rcsb.org/pdb
www.pdbj.org


Reaction of glucose with 2,4-dinitrophenylhydrazine (DNPH)

The test is based on imine formation. The formation of imine from carbonyl compound occurs in equilibrium ergo the principle of Le Chatelier is applicable to the test. According to this principle the increase of the concentration of at least one of the products results in the decrease of degree of conversion of the reactant.
I guess that you are trying to perform the test in aqueous solution, because glucose has poor solubility in ethanol. The second product of the conversion of carbonyl to imine is water. So, performing the reaction in the tremendous excess of water, you are in fact pushing the degree of conversion toward zero in accordance with the principle of Le Chatelier.
That's why the expected precipitate is not formed.


How do you track Iodine on a black surface?

I would spray some lead nitrate solution to the surface if I didn't care about it at all. It would form nice yellow crystals of lead iodide that could be seen on a black surface. Never tried it though.


What aspects of a molecule's chemical makeup determine which portions of it will react?

Let me answer with a question (or two):

How do you identify in an electric circuit which parts are just the wiring and which are functional parts?

How's that for a mechanical engine? Can you tell cogwheels and cranks from "just the basic structure"?

It's all about pattern recognition in the first to identify the functional groups.

Next step is to know how they react under certain conditions. As a rule of thumb, a group with a high electron density will probably react with something more positively charged (think plus and minus attract each other).

But it takes quite a while and learning to predict exactly what will go on exactly in the reaction. But that not different from any other scientific/technical discipline.

I tried to keep this as general and simple as possible ;)

Edit
Let me work you comment in:

  Generally molecules don't contain ions though

Says who? :D Molecules may contain ionic units, think in zwitterions of amino acids, think in all the mesoionic sydnones, münchnones and whatnot. Well, no. Forget about the special cases ;) But if it has more than one atom and is electrically neutral in total it is a molecule. At least in the less rigid sense of the definition.

But yes, the electronegativity is a good point.

Forgive me that I'm not going into detail why the example with fluorine and sodium is a not the best one, but in principle, you're right: Such a setup would allow to form a new $\ce{C-C}$ bond between a carbon that is negatively polarized (the one bearing the sodium) and another one that is positively polarized (the one bearing the halogene).


How exactly does electrolysis occur if the molar conductivities of cation and anion are different?

You are completely right that ions migrate at different rates, and the relative rates can be quantified by a property of the electrolyte called the transport number, which is the fraction of the current that can be accounted by one of the ions:
$$t_\circ^+ = \frac{\Lambda_\mathrm{m}^\circ(\ce{A+})}{\Lambda_\mathrm{m}^\circ(\ce{A+})+\Lambda_\mathrm{m}^\circ(\ce{B-})}$$
$$t_\circ^- = \frac{\Lambda_\mathrm{m}^\circ(\ce{B-})}{\Lambda_\mathrm{m}^\circ(\ce{A+})+\Lambda_\mathrm{m}^\circ(\ce{B-})}$$
(here $t_\circ^i$ designates a limiting transport number ie one in the limit of dilute electrolyte solution).
In the early days of the study of electrolytes, Hittorf developed a way of determining the ratio of transport numbers by measuring the concentrations of electrolytes in different parts of an electrolytic cell. Say the cell contains a cathode, an anode, and a bridging chamber. Cations are attracted toward the cathode chamber and depleted at the cathode, and anions are similarly depleted at the anode but also migrate into the anode chamber. The change in the concentration of cations in the cathode chamber is therefore proportional to the current at the electrode minus the fraction of current carried by cations replenishing the cation chamber:
$$\begin{align} \frac{dc_+}{dt} &= \frac{I_+}{z_+F} - \frac{I}{z_+F} \\  &= \frac{(t_+-1)I}{z_+F} \\  &= \frac{-t_-I}{z_+F} \end{align}$$
A similar equation can be written for the anions. The resulting concentrations in the chambers can be related to the migration rates $v_i$ of the ions , which can in turn be related to the transport numbers as
$$\frac{t^+}{t^-} = \frac{|Z^+v^+|}{|Z^+v^-|}$$
Therefore from measurement of electrolyte concentrations in different parts of a cell it is possible to deduce the relative molar conductivities of the components of an electrolyte.
Because the counterions have different mobilities a concentration gradient will develop and this leads to a liquid junction potential. This results in a voltage drop within the cell that counters the EMF, such that the measured cell potential is not the ideal EMF. In Hittorf's experiment this is not a problem. Where it is a problem it might be avoided by using a salt bridge.
For a textbook that provides a reasonably clear introduction to electrochemistry see Atkins Physical Chemistry, from which I derived some of this answer.
Transport numbers are important in practical applications. See for instance Ref. 1 for an interesting thesis that focuses on the determination of transport numbers in cements.
References

J.A. Muka (1992). Transport Number Determination by Analysis of Transport of Aqueous Electrolytes Through Cementitious Materials Using Emf Method. University of Nairobi.


How do you calculate the correct charge on F after F2 dissociates?

If you have two fragments that are very far apart, certain quantum chemistry methods can cause spurious effects from including both fragments in the same calculation.  A well known example of this is dissociation curves for $\ce{H2}$ with Restricted Hartree Fock.

By using a density based method, we are less likely to localise the electrons on any one atoms in integer quantities.  A wavefunction method such as Hartree Fock, which places integer electrons in well defined molecular orbitals is more likely to create this localisation.  If you need higher accuracy, MP2 or coupled cluster may be of more use.

In fact I have achieved Mulliken charges of +2 and +3 on the two different atoms at 10 angstrom separation using Hartree Fock, with a cc-PVTZ basis set in Gaussian 09.

If you insist on using DFT you should use some chemical/physical intuition to partition this system.  From experiment, i.e. time of flight spectrometry, we know that the atoms have integer charge when they strike a detector, and dispersion interactions over that distance are going to very small compared to the coulomb attract of the other atom as a point charge.  So instead you can calculate the energies of the energies of the individual $\ce{F^n+}$ for $n=0,1,2,3,4,5$.  The combination of atoms with total charge $+5$ and the lowest energy, considering the classical coulomb attraction between them, is probably the product of the dissociation.

Gaussian input used:

%chk=FF_HF.chk
%nproc=1
%mem=4GB
# UHF/CC-PVTZ NoSymm

F2 5 plus Ion

5 2
 F      0.00000000    0.00000000   -5.00000000
 F      0.00000000    0.00000000    5.00000000


Facing a problem in enthalpy calculation

$\ce{P4}$ forms a molecular solid, whereas diamond is a covalent-network solid (see eg definitions here).

In a molecular solid, molecules containing a discrete number of covalently bonded atoms are arranged in a lattice stabilized by weak interactions such as van der Waals forces or hydrogen bonds between the molecules. In a covalent-network solid, the "molecules" are extended with no finite stoichiometry definining a molecular size.

When estimating the enthalpy change, you consider only covalent bonds, not weak intermolecular forces. Therefore, for $\ce{P4}$ you consider only the 6 P-P bonds, whereas in diamond you consider covalent bonds to nearest neighbours, and correct for the fact that when you rupture a bond to a neighbour, that neighbour also loses a bond, therefore the division by 2.


What will be the product ratio if I do chlorination of p-toluic acid?

I couldn't find any literature evidence for the nuclear chlorination reaction of 4-methylbenzoic acid (p-toluic acid). Instead, I found a patent in Japanese for the nuclear chlorination of 4-methylbenzoic acid chloride, of which, I believe close enough to your question (See below the translation given in Google).

According to the patent, when mixture of 99.7% pure 4-methylbenzoic acid chloride ($\pu{927.6g},\; \pu{6.0mol}$) and ferric(III) chloride ($\pu{2.9g}, 0.3 \:wt\%$) was heated at $\pu{50\!-\! 55 ^\circ C}$ for $\pu{22 h}$ in the presence of $\ce{Cl2}$ gas. After completion of the reaction, 84.9% of 3-chloro-4-methyl benzoic acid chloride was isolated. In addition, 0.1% of 2-chloro-4-methyl benzoic acid chloride and 2.2% of dichloro-4-methyl benzoic acid chloride were also isolated (dichloro-4-methyl benzoic acid chloride was not identified, but it should be 3,5-dichloro-4-methyl benzoic acid chloride, judging from the ratio of 3-chloro-4-methyl benzoic acid chloride to 2-chloro-4-methyl benzoic acid chloride).

Reference:

Inventor: 智規 加藤亮太 甲斐幸基 小林, "Isopropyl 3-chloro-4-methylbenzoate and method for producing same." Patent application # (2013): WO2013015203A1 (https://patents.google.com/patent/WO2013015203A1/en).


How do I calculate the molarity of the Na2CO3 solution?

The first part makes more sense writing moles explicitly. The first step is determining how many moles of sodium carbonate are in the original $1.00\ \mathrm  g$ sample:

$$\frac{1.00\ \mathrm{g}}{286\ \mathrm{g/mol}} = 0.003447\ \mathrm{mol}$$

The second step is determining the concentration of the first $20.0\ \mathrm{mL}$ solution, i.e. what concentration results from dissolving the $0.00345\ \mathrm{mol}$ of sodium carbonate calculated above.
$$\frac{0.003447\ \mathrm{mol}}{0.0200\ \mathrm{L}} = 0.174825\ \mathrm{M}$$

The third part is simply a dilution – what concentration results when we dilute the first $20.0\ \mathrm{mL}$ solution to $250\ \mathrm{mL}$:
$$c_2 = \frac{c_1V_1}{V_2}$$
$$c_2 = \frac{(0.174825\ \mathrm{M})(0.020\ \mathrm{L})}{(0.250\ \mathrm{L})} = 0.0140\ \mathrm{M}$$


The fact that the methoxycyclohexane carbocation is resonance stabilized greatly increases its stability over the the trifluoromethylcyclohexane carbocation.

The same power of oxygen’s ability to donate electrons can be seen in its effects on benzene rings. Both $\ce{-OH}$ and $\ce{-OR}$ groups are strongly activating groups, while $\ce{-CF3}$ is a ring deactivating group, meaning it actually pulls electrons from the benzene ring. This further shows that the methoxycyclohexane carbocation is the most stable of the 4.


Will supercritical nitrogen stay supercritical if we raise temperature and pressure high above critical values?

If we maintain nitrogen at 220 bar and 400 °C, what will be the state of $\ce{N2}$? Will it be still supercritical?

Yes, it will still be supercritical.  The only reasonable definition of "supercritical" is something along the lines of "a fluid phase without surface tension wherein the fluid pressure and temperature exceed the critical point".

Thus, compressed air is a supercritical fluid.  The same is true of compressed nitrogen.

You may be wondering how "supercritical" air is different from regular, non-supercritical, gaseous air.  The answer is, it isn't meaningfully different.

If you research "supercritical fluids", it will be very hard to discover this fact.  That is because the "exotic" properties of "supercritical fluids", such as having densities "in between that of a gas and a liquid", are really only properties of supercritical fluids that are near (i.e. slightly above) the critical point.

It isn't really much of an exaggeration to say that supercritical fluids are really gases.  They have no surface tension.  They expand to fill their container.  Just like gases.  Near the critical point, supercritical gases can have densities that are very unusual for a gas. At these high densities, the gas is very non-ideal, and intermolecular interactions in the gas lead to all sorts of interesting behavior.  Those high-density intermolecular interactions give rise to the interesting properties that people refer to when they say "supercritical fluid".  The strength and nature of those interesting intermolecular interactions depends strongly on density, which is why people refer to "supercritical fluids" having "tunable" properties.  Small changes in density (by changing temp. or pressure) can give rise to large changes in those interactions.

At conditions way past the critical point, those intermolecular interactions don't matter very much any more, and the "supercritical fluid" again starts to behave more and more like an ideal gas.

  I know that phase diagram says after critical temperature and pressure we have a supercritical phase, but with enough p and T, more exotic states of matter occur.

Yes, other answers and comments have done a good job of discussing this point.  If you increase the pressure enough, eventually you'll have a black hole!  Of course, long before that happens, you can get other less exotic phases of matter such as various solid phases.  In general, predicting the high-pressure phase behavior of even small molecules like nitrogen is very difficult, and as such, experimental data is required.  Unfortunately, measuring the high-pressure phase behavior of materials is also very difficult, at least past a few hundred gigapascals.  So I'm not sure where the high-pressure "exotic" phases of e.g. solid nitrogen start to form.  But it's not at 220 bar and 400 ° C, under those conditions nitrogen will just be a gas, albeit a "supercritical" gas.


Do polarizability and electron delocalization have analogous effects on leaving groups?

TL;DR: Nucleophilicity is primarily a kinetic phenomena.  Leaving group ability is a thermodynamic phenomena.

The stabilization of an anion by delocalization of charge means that the resulting products will have lower energies than alternate situations. In accordance with the Hammond postulate (a few logical steps skipped), transition state barriers with low energy products will be smaller than those with higher energy products.

However, for a bimolecular reaction to occur, the molecule must have 'proper collision speed and orientation'.  Diffuse charges (over multiple atoms) generally make this more difficult.  This extra facet of the of the geometry of the transition state is why the equivalent stability argument does not apply to both leaving group ability and nucleophilicity.


How much yeast is necessary to produce the same amount of carbon dioxide that I exhale?

I think this is a chemistry question, so I will answer it here.

First, I would question whether the amount of $\ce{CO2}$ is relevant or the concentration (or if more convenient, partial pressure). There is no way for the bed bugs to integrate over the entire trap vs. the bed. If you have a concentration gradient from bed to trap, it should work.

Second, yeast is a catalyst. The amount of $\ce{CO2}$ made depends on the amount of available food (the yeast will multiply to keep up). In the simplest terms, if you want the yeast to produce as much $\ce{CO2}$ as you do, you have to feed it the same amount of food that turns into $\ce{CO2}$.

Because yeast and humans have a slightly different diet, you have to interpret "same amount of food" a bit. In human metabolism, all carbon from foods with calories (i.e. not fiber) is used in biosynthesis or ends up as exhaled $\ce{CO2}$. In an adult who doesn't gain or lose a lot of weight, you can estimate the amount of $\ce{CO2}$ produced from the amount of food taken in. In yeast metabolism, depending on the conditions, either most carbon turns into $\ce{CO2}$ (when there is plenty of oxygen) or into ethanol (when there is a lack of oxygen). For bed bug traps or baking, you want the former, while for brewing, you would want the latter.

For practical purposes, though, I think the total amount of $\ce{CO2}$ is moot (see my first point). As long as you have a source of $\ce{CO2}$ in the trap and remove the source of $\ce{CO2}$ in the bed (i.e. sleep elsewhere) and keep the room warm, there should be a nice $\ce{CO2}$ concentration gradient for the bugs to follow.


Is food that burnt in an 18/10 stainless steel pot still edible?

Your stainless steel pot is thoroughly thermally stable well above the temperature of your stove, and there are no volatile compounds present that could migrate out of the pot and into your food. It is (was) perfectly safe to consume the tamales cooked in this pot.


Different solutes in the same solution indeed can affect each others' solubility. One example where this effect can be used to advantage is the "salting out" of a protein from water solution. Adding a salt (ammonium sulfate is especially good for this purpose) reduces the solubility of the protein in water, causing it to precipitate or crystallize out. Salting out can also be used to more fully separate an organic solvent from water. I imagine that in your example with the sugar alcohols, adding one component to water will reduce the solubility of the second component by a similar degree. But there may be cases where one solute increases the solubility of another.


No, that's not how it all works on more than one level.
First, the amount of $\ce{NH4+}$ out there is small. And by "small", I mean really, really small - about as small as you can imagine, only smaller. Many times smaller, I must reiterate. If some substance was to react with it (which is certainly not outside the realm of possible) and form a product in equally small amount, you won't be able to see it.
I intentionally avoid any numeric estimates here. You must have seen them already. They are too far outside the everyday experience to make an impression. Well, you may look up the auto-ionization constant of ammonia and see how it relates to that of water (which itself is not big). Guess what? It is many orders of magnitude smaller.
Now comes the second point, which kind of negates the first one. If something reacts with that $\ce{NH4+}$ (which, I repeat, is totally possible), then the auto-ionization reaction will produce more $\ce{NH4+}$ to replace the loss, and then more and more, probably until no more $\ce{NH3}$ is left. How so? Well, imagine a crowded theater emptying through one door. Obviously all those people can't fit in the door at once. Not even half of them can. But they don't have to. They just walk through it one by one, and before long, the theater is empty. Much like that, all $\ce{NH3}$ molecules will go through the form of $\ce{NH4+}$ (or $\ce{NH2-}$) to whatever products are there.
In a way, this will prove that the auto-ionization was there in the first place, but tell us nothing about its extent. To measure that, we need other methods; conductometry comes to mind.
As for pH indicators for water, they are definitely no good for liquid ammonia. They all will be far off-scale, much like your cooking thermometer in the middle of Antarctica. To work in liquid ammonia, one needs to develop a whole new set of indicators. Some seems to be known already, but I have no firsthand experience with them, and Google is the same over there as it is over here.
So it goes.


Just as hyperconjugation stabilizes carbocations and radicals by sharing the electron density of adjacent $\ce{C-H}$ bonds with the unoccupied/partially occupied orbital, hyperconjugation destabilizes carbanions. So more substitution means less acidic and more ionizable. I had a hard time finding a single resource listing bond lengths, but they seem to correlate with bond-dissociation energy. The wikipedia page for methane states that its $\ce{C-H}$ bond length is 108.7 pm and an answer by Jan says ethane’s is 109.4 pm.

$$ \small
\begin{array}{lcc}
\hline
\text{Hydrocarbon} & \mathrm{p}K_\mathrm{a} ^{[1]} & \text{BDE (kcal)}^{[2]}\\
\hline
\ce{CH3-H} & 48 & 104\\
\ce{CH3CH2-H} & - & 98\\
\ce{(CH3)2CH-H} & 51 &  95\\
\ce{(CH3)3C-H} & 53 & 93\\
\hline
\end{array}
$$

 $[1]$  Evan’s pKa Table
$[2]$ Wikipedia: Carbon-Hydrogen Bond


Nomenclature of common naming of alkynes vs alkenes; both -ene?

The official IUPAC name for ethyl with a double bond is ethene, but this is a relatively recent change.
Before the official name (as an exception) was ethylene. I’m not sure about propene/propylene but am fairly confident the official IUPAC name for propyl double bond is propene and not propylene. However, these compounds are often referred to with the unofficial name as per tradition.

For alkynes, the IUPAC name is, for ethyl, the name is ethyne.

Don’t remember to specify the location of the double/triple bond (ex: 2 - ethene)

For alkenes, the formula is $\ce{C_nH2_{n+2}}$ and for alkynes is $\ce{C_nH_{2n}}$.


How is volatility useful in the production of acids?

Yes, this is indeed the case. The reasoning behind it is using chemical equilibria to their fullest.

If you have a Brønsted acid and a Brønsted base in the same vessel, you will always have an equilibrium of the following kind:

$$\ce{HA + B- <=> A- + HB}$$

It depends on the nature of the acid and the base — i.e. their $\mathrm{p}K_\mathrm{a}/\mathrm{p}K_\mathrm{b}$ values — which side of the equation is favoured in a closed system (i.e. solution). For example, dissolving $\ce{NaHSO4}$ and $\ce{NaSH}$ in the same sample of water will result in the following set of equations:

$$\ce{H2SO4 + Na2S <=>> NaHSO4 + NaSH <=>> Na2SO4 + H2S}$$

Because the $\mathrm{p}K_\mathrm{a}$ values of both compare as follows:

$$\begin{array}{ccc}\hline
\text{acid} & \mathrm{p}K_\mathrm{a1} & \mathrm{p}K_\mathrm{a2}\\
\hline
\ce{H2SO4} & \approx -3 & 1.9\\
\ce{H2S} & 7.0 & 12.9\\
\hline\end{array}$$

We already have a way of producing $\ce{H2S}$ by means of protonating sulfide ions in solution here. However, $\mathrm{p}K_\mathrm{a1}\left (\ce{HCl}\right ) \approx -6$, which is more acidic than sulphuric acid, so adding $\ce{NaCl}$ to $\ce{H2SO4}$ should result in a net reaction of nearly nothing.

This is where vapour pressure comes in as a second factor. $\ce{HCl}$ is a gas under standard temperature and pressure while $\ce{H2SO4}$ is a liquid. ($\ce{H2S}$ is also a gas. If we added liquid sulphuric acid to sodium sulfide, $\ce{H2S}$ gas would evolve, but that’s what the $\mathrm{p}K_\mathrm{a}$ value predicts anyway.) If we have an open vessel — i.e. gases can diffuse away — then any $\ce{HCl}$ produced in equilibrium will dissociate away in the long run. This is due to a second equilibrium that we could write as follows:

$$\ce{HCl (g) <=> HCl (aq)}$$

The ‘equilibrium constant’ of this physical process is more or less what we call vapour pressure, and for hydrogen chloride both sides are significantly more balanced than for sulphuric acid. Therefore, we should describe the entire system in the following way:

$$\ce{NaCl + H2SO4 <=> NaHSO4 + HCl (diss) <=>> NaHSO4 + HCl (g) ^}$$

Where $\ce{(diss)}$ is a shorthand for dissolved in sulphuric acid, our reaction medium.

Drawing away the hydrogen chloride gas in the rightmost equation, for example by reducing pressure or inducing an air flow, will shift all equilibria to the right since that product is constantly being removed. And thus we can liberate $\ce{HCl}$ gas with sulphuric acid, even though the $\mathrm{p}K_\mathrm{a}$ of hydrogen chloride is lower than that of sulphuric acid.


Vegetable oil is extracted from plant. According to Wikipedia, the term "vegetable oil" can be narrowly defined as referring only to plant oils that are liquid at room temperature,

So, answer is no.

If it occurs naturally than we will not call it vegetable oil.


Under the right conditions, crystallization can be a very effective means of obtaining a pure solid form of the solute. Unfortunately, in your case one of those conditions is violated; you have a second solute with similar solubility characteristics.  You didn't quantify the solubilities of the solutes, but the phrase "very close solubility in a certain solvent" tells me that under the conditions of your system, crystallization will not be an effective means of obtaining a pure solid form of the solute.

One option, as mentioned in a comment to your question, is to choose a different solvent. Selecting the right solvent / solvent mix can be a bit of an art and will also require some research and experimentation on your part.  Because you didn't discuss the specific solvents or solutes used in your system, it is not really possible to give more specific advice on solvent selection.

Another option is to choose a different means of solute separation, like some type of chromatography. Again, not knowing anything about your compounds makes it difficult/impossible to give further advice regarding this technique.


In $\ce{[CrF6]^{3-}}$ the central atom is $\ce{Cr(III)}$ or $\ce{Cr^{3+}}$. Chromium is usually $d^6$, so now (take away 3 electrons) $d^3$.

When we start filling in the electron pairs from the ligands, we get the following:

The electrons from the metal are marked in red.

The electrons from the ligands (blue) are filled into two $d$, one $s$ and three $p$ orbitals: $d^2sp^3$ hybridization as per solution!

Now try to do the same with $\ce{[Co(NH3)6]^{3+}}$!


Finding the equivalent point of a weak polyprotic acid when reacted with a strong base

A simple way to determine whether an equivalence point in a titration is the last one is to figure out the volume of titrant added for the first titration point, then realize all subsequent equivalence points will be at exact integer multiples of this volume, under the assumption that the analyte contains only one acid (a mixture of acids won't behave the same). When an integer multiple of the first equivalence point volume fails to be an equivalence point itself, then there will be no further equivalence points.
For example, let us imagine a generic polyprotic acid with formula $\ce{H_{k}A}$, being titrated with a strong monoprotic base such as $\ce{NaOH}$. Starting from the fully protonated acid, let us neutralize each ionizable hydrogen in steps. Let the acid  $\ce{H_{k}A}$ being titrated have an initial concentration $c_{acid}$ in a solution of volume $V_{acid}$, and let the titrant have a concentration $c_{base}$ and a total volume added $V_{base}$.
By definition, to reach the first equivalence point ($V_{base}=V_{eq\ 1}$), we consider the reaction which represents the neutralization of only the first proton:
$$\ce{H_{k}A + OH- -> H_{k-1}A- + H2O}$$
Each one mole of acid  $\ce{H_{k}A}$ must surrender exactly one mole of hydrogen ions $\ce{H+}$, each of which will react with exactly one mole of hydroxide ions $\ce{OH-}$. Thus:
$$n_{\mathrm{released}\ \ce{H+}}=n_\mathrm {acid}=c_\mathrm {acid} \times V_\mathrm {acid}$$
$$n_{\mathrm{added}\ \ce{OH-}}=n_\mathrm {base}=c_\mathrm {base} \times V_\mathrm{base}=c_\mathrm {base} \times V_\mathrm {eq\ 1} $$
$$n_{\mathrm{released}\ \ce{H^+}}= n_{\mathrm{added}\ \ce{OH-}} \rightarrow\  c_\mathrm {acid} \times V_\mathrm {acid} = c_\mathrm {base} \times V_\mathrm {eq\ 1}$$
$$V_\mathrm {eq\ 1} = \frac{c_\mathrm {acid} \times V_\mathrm{acid}}{c_\mathrm {base}}$$
Now, to reach the second equivalence point ($V_{base}=V_{eq\ 2}$), we consider the reaction which represents the neutralization of the first and second protons:
$$\ce{H_{k}A + 2 OH- -> H_{k-2}A^{2-} + 2 H2O}$$
Each one mole of $\ce{H_{k}A}$ must surrender exactly two moles of $\ce{H+}$, each of which will react with exactly one mole of $\ce{OH-}$. The relevant equations are now:
$$n_{\mathrm{released}\ \ce{H+}}=2\times n_\mathrm {acid}=2\times c_\mathrm {acid} \times V_\mathrm {acid}$$
$$n_{\mathrm{added}\ \ce{OH-}}=n_\mathrm {base}=c_\mathrm {base} \times V_\mathrm {base}=c_\mathrm {base} \times V_\mathrm{eq\ 2} $$
$$n_{\mathrm{released}\ \ce{H+}}= n_{\mathrm{added}\ \ce{OH-}} \rightarrow\  2\times c_\mathrm {acid} \times V_\mathrm {acid} = c_\mathrm {base} \times V_\mathrm {eq\ 2}$$
$$V_\mathrm {eq\ 2} = 2\times \frac{c_\mathrm {acid} \times V_\mathrm {acid}}{c_\mathrm{base}} = 2\times V_\mathrm {eq\ 1} $$
Notice how the added volume to reach the second equivalence point is exactly twice that added to reach the first equivalence point. This pattern repeats, i.e. the third equivalence point will happen at $V_\mathrm {base}=3\times V_\mathrm {eq\ 1}$, and the $k$-th equivalence point will happen at $V_\mathrm {base}=k\times V_\mathrm {eq\ 1}$. The equivalence points are equally spaced.
In short, if the titration of a pure polyprotic acid shows the first sudden pH swing after adding $\pu{15 mL}$ of the base titrant, then $V_\mathrm {eq\ 1}$ equals 15 mL, and the only possible next equivalence points are at $\pu{30 mL}$, $\pu{45 mL}$, $\pu{60 mL}$, etc. If during the titration another sudden pH swing is observed at $\pu{30 mL}$, but not at $\pu{45 mL}$, then you can deduce that there is no third equivalence point, and hence the acid is diprotic.


Saturated vs unsaturated fats - Structure in relation to room temperature state?

This has to do with the chemical structure of the fatty acids. Single bonds create a straight molecule, which can be easily packed together to form a solid, while the double bond introduces a kink in the structure.

  Fats, which are mostly from animal sources, have all single bonds between the carbons in their fatty acid tails, thus all the carbons are also bonded to the maximum number of hydrogens possible. Since the fatty acids in these triglycerides contain the maximum possible amount of hydrogens, these would be called saturated fats. The hydrocarbon chains in these fatty acids are, thus, fairly straight and can pack closely together, making these fats solid at room temperature. Oils, mostly from plant sources, have some double bonds between some of the carbons in the hydrocarbon tail, causing bends or “kinks” in the shape of the molecules. Because some of the carbons share double bonds, they’re not bonded to as many hydrogens as they could if they weren’t double bonded to each other. Therefore these oils are called unsaturated fats. Because of the kinks in the hydrocarbon tails, unsaturated fats can’t pack as closely together, making them liquid at room temperature.

The source website is no longer accessible, but biochemistry textbooks will state the same thing.


What is the term for the amount of volume change at a phase boundary?

Phase transition can be distinguished by analysing their chemical potential, such classification has been introduced by Ehrenfest. For a given transition, lets call it $\phi$ transition:

$$ \ce{X_{\alpha} \rightleftharpoons X_{\beta}} $$

You can assess transition molar specific volume in the following way:

$$ \Delta_{\phi}V = v_{\beta} - v_{\alpha} = \left(\frac{\partial \mu_{\beta}}{\partial p}\right)_{T} - \left(\frac{\partial \mu_{\alpha}}{\partial p}\right)_{T}$$

Thus, by modelling your potential in term of pressure and temperature, you can assess molar volume change of a phase transition. This is the key to solve your question. In real life application, we use diagrams and tables to assess it. But building a model is always valuable for comprehension.

In a similar way, molar entropy is given by:

$$ \Delta_{\phi}S = s_{\beta} -s_{\alpha} = \left(\frac{\partial \mu_{\beta}}{\partial T}\right)_{p} - \left(\frac{\partial \mu_{\alpha}}{\partial T}\right)_{p} = \frac{\Delta_{\phi}H}{T_\phi} $$

Because $\Delta_{\phi}V$ and $\Delta_{\phi}H$ are not null for first order (in term of first derivative) transition, slope difference is not null and therefore potential slopes are different both side of phase transition when Gibbs parameters vary ($p$ and $T$). Thus there are discontinuities in specific volume, enthalpy and entropy when a first order transition occurs.

Ehrenfest did a classification of phase transition based on this criterion which led to: first order, second order (smooth, no slope discontinuity) and $\lambda$ (sharper) transitions regarding how such chemical potential discontinuity happens.

This discontinuity also impact heat capacity $c_p$. In case of water, whose phase transitions can be assumed as a first order, when a phase transition occurs, discontinuity leads to a virtually infinite heat capacity, explaining why fusion or vaporisation happens at constant temperature (when pressure is held constant). This is why we need an extra term called latent heat (temperature independent) which contrast with sensible heat (temperature dependent).

You will find detailed explanation (models, diagram and tables) of those phenomenons in Atkins, Physical Chemistry, Chapters 2-7.


Why do amino acids hydrogen bond rather than ionic bond when they are forming secondary structures in proteins?

(this should be a comment)
I think you are not quite understanding the concept of partial charged or charged. The N atom in the polypeptide chain (so the N in the peptide bond) is partially charged! to be in an ionic state it should have 4 bonds instead of 3.
 The peptide bond which is formed(above) just contains a N atom which is bounded to three other atoms. The oxygen atom is also partially charged but not charged! read this about H-bonding and partial charge


A low quantum yield would indicate that there are efficient intra-molecular non-radiative decay pathways, likely internal conversion or spin-orbit coupling. These decay pathways may be faster that FRET, rendering FRET essentially a non-competetive process.

A small extinction coefficient would indicate a small transition dipole moment. If the transition dipole moment is small it is difficult for the FRET partner to transfer energy since it is a weak transition resulting in a small coupling  between the two chromophores.


N-alkylation of amides with alkyl halides?

Sure, there are a number of ways to do this with copper, iridium, or palladium catalysts.  This link should hopefully be helpful.

http://www.organic-chemistry.org/synthesis/C1N/amides2.shtm


Deodorizing using baking powder instead of baking soda

Baking soda is an effective deodorant because of its basicity. Many foul-smelling compounds are acidic. Neutralizing the compound gives it an ionic character which reduces the vapor pressure (meaning it is less available for the nose to smell it) and makes it much more water soluble (it can be washed away). Odorous compounds like fatty acids (butyric acid is BO) and thiols  have much higher $pK_a$s (~4.8 and 7-10, respectively) than tartaric acid ($pK_a$ ~3) means that they would remain un-neutralized with baking powder. It would not be an effective  deodorizer.


Understanding the derivation of Gibbs Free Energy

The surroundings are treated as a constant volume reservoir, so the first law of thermodynamics,
$$
\mathrm{d}U_\mathrm{surroundings}
  = \mathrm{d}q_\mathrm{surroundings}
  + \mathrm{d}w_\mathrm{surroundings},
$$
yields
$$
\mathrm{d}U_\mathrm{surroundings}
  = \mathrm{d}q_\mathrm{surroundings}
  \text{, or }
\Delta U_\mathrm{surroundings}
  = q_\mathrm{surroundings}.
$$
Notice that $q_\mathrm{surroundings}$ is equal to a state function, so it behaves as one, that is, it is independent whether the process is reversible or not. If the process is reversible
$$
q_\mathrm{rev,surroundings}
  = \Delta H_\mathrm{surroundings},
  \tag1\label{eq:rev}$$
if it is not reversible
$$
q_\mathrm{surroundings}
  = \Delta H_\mathrm{surroundings},
  \tag2\label{eq:irrev}$$
from \eqref{eq:rev} and \eqref{eq:irrev} follows
$$q_\mathrm{surroundings} = q_\mathrm{rev,surroundings}.$$
Therefore
$$
\Delta S_\mathrm{surroundings}
  = \frac{q_\mathrm{rev,surroundings}}{T},
$$
in particular
$$q_\mathrm{rev,surroundings} = -q_\mathrm{sys},$$
then
$$\Delta S_\mathrm{surroundings} = \frac{-q_\mathrm{sys}}{T}.$$
If the process occurs at constant pressure
$$q_\mathrm{sys} = \Delta H_\mathrm{sys},$$
therefore
$$\Delta S_\mathrm{surroundings}  = \frac{-\Delta H_\mathrm{sys}}{T}.$$


Can nitrous dioxide develop when copper and dilute nitric acid react?

There are two ways to interpret an observation nitrogen dioxide.

Reactions between a metal and nitric acid rarely proceed by a single path to generate a specific nitric acid reduction product; in all likelihood there will be a combination of multiple reactions and thus a mixture of products.  With copper you could get both $\ce{NO}$ and $\ce{NO2}$.  And with a more powerful reducing metal like magnesium even more products could get into the mix, including ammonium nitrate.

You may have conducted the reaction in the presence of air.  $\ce{NO}$ is readily oxidized by the air to make $\ce{NO2}$ "after the fact".  This is in fact part of the natural production of nitric acid in thunderstorms.  Further complicating this is the fact that nitrogen dioxide in turn reacts with water, forming nitric oxide and regenerating some of the nitric acid (this is also indicated in the above link).  To get the true product mix out of the nitric acid you needed to use an environment free of air and unintended water.


Formation of thermodynamic product from 1,4-addition

Both of your products have a tetrasubstituted double bond.  This makes me think that they have similar stabilities, therefore there is no thermodynamically preferred product.

In both the top and bottom lines of your drawing you show the formation of allyl carbocations.  You use an arrow like this "$\ce{->}$" to connect the second structure from the left to the third structure from the left. The allyl carbocation is a resonance stabilized carbocation.  The second and third structures are really resonance structures and should be connected with a resonance arrow "$\ce{<->}$".  Both resonance structures taken together describe the allyl carbocation.

If you compare the allyl carbocation on the top line to the one on the bottom line, you'll notice that the one on the bottom has the positive charge delocalized onto 2 tertiary carbons while the allyl carbocation on the top line has the positive charge delocalized onto 1 tertiary and 1 secondary carbon. Therefore, the bottom allyl carbocation is more stable and the product derived from it will be preferred.


Height in HPLC simply refers to the height of the peak (= maximum value of the peak) above the baseline. The point by z1273 is right in the sense that this voltage to height conversion factor is instrument specific but this height cannot be in milliabsorbance units- it is too large for absorbance. This height is apparently "intensity". Anyway, you do not need to worry about the "true height" for peak detection. It will be fine to use the height of 20 V for the first peak at 7.867 min. In older instruments, the detector output was displayed as Volts or mV and the manufacturer provided a conversion factor from volts to light absorbance or sometimes as intensity in the manual. See this discussion here (https://www.chromforum.org/viewtopic.php?t=6442). You should really look at modern HPLC data. The one you are showing here is from an ancient instrument.
I am assuming you are using derivatives to detect peaks? For any chromatography program, I would suggest that you do a simulation first, because all variables will be in your hand. Make a chromatogram as sum of large and small exponentially modified Gaussians or simple Gaussians of various heights and areas and add some noise to test your program.


What is the oxidation mechanism of gunpowder?

$\mathrm{KNO_3(s)}$ can oxidize charcoal without sulfur. It needs a "hot" primer through to get the reaction started.  Sulfur's role is to lessen the ignition temperature.

See section in Wikipedia article.

Since the reaction isn't complete and has a number of possible products bthere isn't an exact chemical formula for black gunpowder burning. The Wikipedia article lists several approximations to black gunpowder burning. What is supposedly the best follows.

    Although charcoal's chemical formula varies, it can be best summed up by its empirical formula: $\ce{C7H4O}$. Therefore, an even more accurate equation of the decomposition of regular black powder with the use of sulfur can be described as:
    $\ce{KNO3 + C7H4O + 2S → K2CO3 + K2SO4 + K2S + 4CO2 + 2CO + 2H2O + 3N2}$


Number of stable molecules of artificial molecule XY4

There are 5 ways to choose the central atom.

Choosing the possible combinations of the other 4 items from 6 isotopes is complicated as this article explains. But it provides a formula for combinations with repetition that gives the answer of 126.

This implies that there are 630 distinct isotope combinations in the molecule (assuming that we can distinguish combinations of isotopes with the same atomic number which is possible with high resolution mass spectroscopy). If only integer atomic numbers count then the distinguisahble combinations will be lower. But the easiest way to work that out is to do a Monte-Carlo simulation not to look up probability formulae.


What would a (gas) chromatogram look like if two compounds have the same retention times?

] Complete overlap (= identical retention) is a regular feature of experimental chromatography with very different compounds and complex mixtures. It just happens when the selectivity happens to be equal to 1.00. You can have one, two, or three different compounds hidden under a single peak and this usually happens in early eluting peaks in all modes of chromatography of a complex mixture.
Peak overlap is undesirable, but it happens all the time. All you need to do is to change the method.
People have spent a portion of their life in something called as peak deconvolution to find out what is hidden in a single peak. Note how the area adds up.


How to detect if a beverage has Sucralose or aspartame?

Here's the structure of sucralose

and here is the structure of aspartame

They are very different one from the other and should be easily separable.  You can do a "poor man's" separation using paper chromatography.  Aldrich sells both of these materials, but I'm sure you could find them much cheaper at a local Health or Health Food store.  Get the samples and some filter paper.  You'll need some solvents, a tank (preferably glass) with a lid, some filter paper coated with a fluorescent material (as an easier, but more expensive alternative, you could always buy some TLC plates from a chemical supply house) and a black light.  Put one of the solvents in the tank so that it is 1/8 - 1/4" deep.  Put the top on the tank and let it equilibrate. Make dilute solutions of your known samples using the same solvent that's in the tank.  Using the tip of a needle or a toothpick place a small spot of each of the dissolved materials on the filter paper, put the spots about 1" apart and high enough above the bottom edge of the filter paper so that they will be above the solvent level when you place the filter paper in the tank.  Give the spots a moment to dry and then put the spotted paper in the tank.  Watch as the solvent level climbs up the filter paper (or tlc plate) and remove the filter paper when the solvent is about 3/4 of the way up the paper or plate.  Let the paper or plate dry and then examine it with the black light.  The paper should fluoresce under black light except where the two spots have migrated up to.  The spots cover the fluorescent material and will appear as dark spots under the black light.  If you got a good separation between the two spots stop here and repeat the experiment putting a spot of your diluted drink on a new piece of filter paper along with spots of the two known materials.  Rerun it in the tank and see if there are any spots in your drink that match either of the spots from the two known compounds.  If there is no match, then sucralose or aspartame are not in your drink.  If there is a match, then whichever spots matched might be in your drink - more advance experiments will be needed to further analyze the situation.  If you didn't get a good separation between the spots for the two known compounds to begin with, start over and repeat with different solvents until you do get a separation.


In practice distillation is not 100% efficient: You never get a perfect separation of a mixture.  But fractional distillation is more efficient, so you would expect the product to be of higher purity than that produced by simple distillation.

E.g., if you have a 50/50 mixture by mass of two liquids, you would expect a higher concentration of the higher-boiling-point liquid at the end of a fractional distillation apparatus, which means a smaller fraction of the original mass.


HOMO Odd, LUMO Even: Why does the affected orbital change depending on the carbon? How are carbons numbered?

To start with, I'll address the frontier orbitals. Lets look at what the HOMO and LUMO look like.

Notice the LUMO has no density on carbons $1$ or $3$ and (while it is a little difficult to see in this image) little density on carbons $5$ and $7$. Similarly the HOMO has little to no density on any of the even carbons. So, if we want to affect the energy of these orbitals, we have to add a substituent to a carbon that has electron density from that orbital. The odd-trend is likely not a general rule, but if you know the HOMO and LUMO reside mainly on different carbons, you can assume that selective functionalization of those carbons will affect the orbital of interest. I say "assume" because you should always keep in mind that the orbital picture is an approximation and that once you make a substitution you have to check to make sure you haven't perturbed the system enough where your new compound doesn't have frontier orbitals that look like this anymore.

In regards to the numbering, this is really just a matter of convenience for a fused ring system like this. We don't number these bridging carbons because they can't be functionalized; that is, we can't add a group to these carbons without turning the molecule into something other than azulene. So the numbering starts from next to these bridging carbons and skips over them as you move around the rings.


Why can't diamagnetic, and paramagnetic magnetize?

The concept of Ferromagnetism, Paramagnetism and Diamagnetism is completely different, though they all show magnetic properties.

FERROMAGNETISM: This is constituted of domain and each domain contains a large number of molecules (the number varies from material to material). These domains tends to align themshelves in a certain fashion in presence of magnetic field in order to achieve lowest energy and hence ferromagnetic materials show attraction to presence of magnets. Now these domains are not loose, but the arrangement can be made permanent in presence of some prolonged magnetic field. The amount of magnetism something can hold can be explained by hysteresis loop (NCERT Class 12 Physics Text Book). The arrangement of domains are shown below:
PARAMAGNETISM and DIAMAGNETISM depends on the number of unpaired electrons present in a system. The presence of unpaired electron to the system gives it a magnetic moment (given  by $\mu=\sqrt{n(n+1)}$, in Bohr Magenton , where n is the number of unpaired electrons). Hence the number of unpaired electrons, i.e. the excess number of electrons in the same spin, determines the magnetic moment. A material is called diamagnetic if the value of $\mu = 0$.

Thus paramagnetic materials are permanent magnets by their intrinsic property, but the magnetic moment is too weak to detect it physically. The increase or decrease in magnetic moment, or, the number of unpaired electrons in paramagnetic materials, is impossible by physical methods like keeping them in touch with magnet, or rubbing a magnet, or keeping it in strong magnetic field as done with ferromagnetic materials.

However, this is possible through chemical methods. For example, $\ce{[Fe(CO)6]^{3+}}$ is diamagnetic while $\ce{[Fe(H2O)6]^{3+}}$ is paramagnetic. However, this is not magnetizing, it is just changing what you are working with.


What is the entropy of mixing of two ideal gases starting out with different pressures?

A strategy that often works is to look at the initial and the final states and then compare them. A good place to start is usually the Gibbs free energy of the system (with the intention to use $\Delta_{\mathrm{mix}} S = - \frac{\partial \Delta_{\mathrm{mix}}G}{\partial T} \big|_{p, \{ n_{i} \}}$ later): for isothermal and isobaric conditions it is

\begin{equation}
  G = \sum_{i} n_{i} \mu_{i} + \mathrm{constant} = n_{\mathrm{tot}} \sum_{i} \underbrace{x_{i}}_{= \, \frac{n_{i}}{n_{\mathrm{tot}}}} \mu_{i} +  \mathrm{constant} \hspace{4em} (1)
\end{equation}

with $n_{i}$, $\mu_{i}$, and $x_{i} = \frac{n_{i}}{n_{\mathrm{tot}}}$ being the amount, the chemical potential, and the mole fraction of the $i$th component in the system, respectively, and $n_{\mathrm{tot}} = \sum_{i} n_{i}$ being the total amount of gas in the system.
So, now we need an expression for the chemical potential of an ideal gas.
It can be shown that this is given by

\begin{equation}
  \mu_{i} = \mu_{i}^{0} + R T \ln\biggl(\frac{p_{i}}{P^{0}}\biggr) \hspace{18em} (2)
\end{equation}

where $\mu_{i}^{0}$ is the chemical potential of the $i$th component in a given standard state, $R$ is the universal gas constant, $T$ is the temperatur, $p_{i}$ is the partial pressure of the $i$th component, and $P^{0}$ is the standard pressure (i.e. the pressure connected to the standard state).
For ideal gases we can apply Raoult's law stating that

\begin{equation}
  p_{i} = x_{i} P_{\mathrm{tot}}
\end{equation}

where $P_{\mathrm{tot}}$ is the total pressure of the system.
Using this with equation (2) we get

\begin{equation}
  \mu_{i} = \mu_{i}^{0} + R T \ln\biggl(\frac{x_{i} P_{\mathrm{tot}}}{P^{0}}\biggr) = \mu_{i}^{0} + R T \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{tot}}}{P^{0}}\biggr) \hspace{6em} (3)
\end{equation}

Furthermore, it is known that the partial pressures in a system must add up to the total pressure, i.e.

\begin{equation}
  P_{\mathrm{tot}} = \sum_{i} p_{i}
\end{equation}

Ok, now that the stage is set we simply have to apply equations (1) and (3) to the initial and the final state of the mixing process.
At the initial state the chambers are seperated (whereby chamber $i$ contains only one sort of gas labeled by $i$) and so these subsystems are completely independent from each other.
Thus, the Gibbs free energy of the total system of chambers is simply the sum of the Gibbs free energies of each chamber, whereby the gas in each of the chambers feels the pressure $P_{\mathrm{tot}} = P_{i}$ in the system (whereby $P_{i}$ is the ambient pressure in chamber $i$), since $x_{i} = 1$ (each chamber/independent subsystem holds just one kind of gas).
According to equation (1) the Gibbs free energy of the $i$th chamber is then given by

\begin{align}
  G_{i} &= n_{i} \mu_{i} = n_{i} \biggl( \mu_{i}^{0} + R T \underbrace{\ln 1}_{= \, 0} + R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) \\
        &= n_{i} \biggl( \mu_{i}^{0} + R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr)
\end{align}

and the Gibbs free energy of the total system at the initial state is

\begin{align}
  G_{\mathrm{ini}} &= \sum_{i} G_{i} \\
                 &= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln\biggl(\frac{P_{\mathrm{tot}}}{P^{0}}\biggr) \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + \sum_{i} x_{i} R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) \\
                 &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + \sum_{i} x_{i} R T \ln\biggl(\frac{P_{i}}{P^{0}}\biggr) \biggr) \ .  \hspace{7em} (4)
\end{align}

For a multicomponent system with more than two components I would simply use this equation. But for a system with exactly two components we can introduce some substitutions that will simplify the final equations a little. For that purpose we want to establish a relationship between the initial pressures $P_{1}$ and $P_{2}$ and the final pressure $P_{\mathrm{fin}}$. In order to get there we use the ideal gas law and the additivity relations for the volumes and amounts: From

\begin{equation}
  n_{\mathrm{tot}} = n_{1} + n_{2} \ , \qquad V_{\mathrm{tot}} = V_{1} + V_{2} \\
  P_{1} V_{1} = n_{1} R T \ , \qquad P_{2} V_{2} = n_{2} R T \ , \qquad P_{\mathrm{fin}} V_{\mathrm{tot}} = n_{\mathrm{tot}} R T
\end{equation}

it follows that

\begin{align}
  P_{1} V_{1} + P_{2} V_{2} = P_{\mathrm{fin}} V_{\mathrm{tot}} \ .
\end{align}

Now, we can express $P_{2}$ in terms of $P_{\mathrm{fin}}$ by introducing the deviation factor $\alpha$ which is defined by $P_{2} = (1 - \alpha) P_{\mathrm{fin}}$ and use this in the previous equation

\begin{align}
  P_{1} V_{1} + P_{2} V_{2} &= P_{\mathrm{fin}} V_{\mathrm{tot}} \\
  P_{1} V_{1} + (1 - \alpha) P_{\mathrm{fin}} V_{2} &= P_{\mathrm{fin}} V_{\mathrm{tot}} \\
  P_{1} V_{1} &= P_{\mathrm{fin}} \bigl( \underbrace{V_{\mathrm{tot}} - V_{2}}_{= \, V_{1}} + \alpha V_{2} \bigr) \\
  P_{1} &= \biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) P_{\mathrm{fin}}
\end{align}

which gives us the relation between $P_{1}$ and $P_{\mathrm{fin}}$. Introducing these equations into equation (4) yields

\begin{align}
  G_{\mathrm{ini}} &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl(\frac{P_{1}}{P^{0}}\biggr) + x_{2} R T \ln\biggl(\frac{P_{2}}{P^{0}}\biggr) \biggr) \\
  &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\frac{\bigl( 1 + \alpha \frac{V_{2}}{V_{1}} \bigr) P_{\mathrm{fin}}}{P^{0}} + x_{2} R T \ln\frac{(1 - \alpha) P_{\mathrm{fin}}}{P^{0}} \biggr) \\
  &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} R T \ln(1 - \alpha) + \underbrace{(x_{1} + x_{2})}_{= \, 1} R T \ln\biggl( \frac{P_{\mathrm{fin}}}{P^{0}} \biggr) \biggr) \\
 &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} R T \ln(1 - \alpha) + R T \ln\biggl( \frac{P_{\mathrm{fin}}}{P^{0}} \biggr) \biggr) \ .
\end{align}

At the final state the chambers are connected and the gases mix, so the chambers are not independent subsystems anymore and the Gibbs free energy cannot be broken down into independent parts like before.
Also, the ambient pressure in the system changes to $P_{\mathrm{tot}} = P_{\mathrm{fin}}$.
Now, the complete system of connected chambers must be treated via equation (1)

\begin{align}
  G_{\mathrm{final}} &= \sum_{i} \underbrace{n_{i}}_{= \, x_{i} n_{\mathrm{tot}}} \biggl( \mu_{i}^{0} + R T \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( \sum_{i} x_{i} \mu_{i}^{0} + \sum_{i} R T x_{i} \ln x_{i} + \sum_{i} x_{i} \underbrace{R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr)}_{= \, \mathrm{constant}} \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( \underbrace{\sum_{i} x_{i} \mu_{i}^{0}}_{= \, G_{\mathrm{ini}}^{0}} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \underbrace{\sum_{i} x_{i}}_{= \, 1} \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \sum_{i} x_{i} \ln x_{i} + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \\
                   &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \ .
\end{align}

Now, that we have the Gibbs free energies of the initial and final state we simply need to subtract the former from the latter one in order to obtain the change in Gibbs free energy during the mixing process:

\begin{align}
  \Delta_{\mathrm{mix}}G &= G_{\mathrm{final}} - G_{\mathrm{ini}} \\
                       &= n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + R T \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) + R T \ln\biggl(\frac{P_{\mathrm{fin}}}{P^{0}}\biggr) \biggr) \\
                       &\quad - n_{\mathrm{tot}} \biggl( G_{\mathrm{ini}}^{0} + x_{1} R T \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} R T \ln(1 - \alpha) + R T \ln\biggl( \frac{P_{\mathrm{fin}}}{P^{0}} \biggr) \biggr) \\
                       &= n_{\mathrm{tot}} R T \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) - n_{\mathrm{tot}}  R T \biggl( x_{1} \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} \ln(1 - \alpha) \biggr) \qquad (5)
\end{align}

Now you see why we went through those substitutions for $P_{1}$ and $P_{2}$. This way the $n_{\mathrm{tot}} R T \ln\Bigl(\frac{P_{\mathrm{fin}}}{P^{0}}\Bigr)$ terms cancelled in equation (5).

Differentiating equation (5) with respect to the temperature we get the entropy of mixing since $\frac{\partial G}{\partial T} \big|_{p, \{ n_{i} \}} = - S$

\begin{align}
  \Delta_{\mathrm{mix}} S &= - \frac{\partial \Delta_{\mathrm{mix}}G}{\partial T} \bigg|_{p, \{ n_{i} \}} \\
   &= - n_{\mathrm{tot}} R \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right) + n_{\mathrm{tot}}  R \biggl( x_{1} \ln\biggl( 1 + \alpha \frac{V_{2}}{V_{1}} \biggr) + x_{2} \ln(1 - \alpha) \biggr)
\end{align}

The first term $- n_{\mathrm{tot}} R \left( x_{1} \ln x_{1} + x_{2} \ln x_{2} \right)$ is just the usual entropy of mixing. The second term $n_{\mathrm{tot}}  R \Bigl( x_{1} \ln\Bigl( 1 + \alpha \frac{V_{2}}{V_{1}} \Bigr) + x_{2} \ln(1 - \alpha) \Bigr)$ describes the contribution of the pressure equilibration to the entropy.


Does reaction of alkali and alkaine earth metals' reaction with water an acid-base reaction?

Na and K react with the O-H bond of the water molecule. This removes the H atom that is reduced to an H atom and then to a H2 molecule. It is not an acid/base reaction. It is a redox reaction. Na or K atom is oxidized at the end. No metal oxides are produced as intermediating state.

Alkaline and alkaline earth metals act in the same way.


In an emission spectrum, the limit of convergence at higher frequency corresponds to the first ionization energy

This is an excellent question! Just to elaborate upon what @porphyrn and @M. Farooq have said, imagine that the universe is entirely empty except for one sodium atom and one photon (particle of light) that has a wavelength of $\pu{242 nm}$. If the photon bumps into the sodium atom, it might just bounce off: this is called scattering. But another big possibility is that the photon will be absorbed (i.e., ‘taken in’, as it were) by sodium’s 3s valence electron. The energy of a $\pu{242 nm}$ photon is $E_{\pu{242 nm}} = h\nu = \frac{hc}\lambda$, where $h = \pu{6.626 x 10^{-34} Js}$ and $c = \pu{2.9979E8 m/s} = \pu{2.9979E17 nm/s}$. This energy is $\pu{8.21E-19 J}$.

Then the sodium’s 3s electron would have just enough energy to escape, i.e., ‘break free’, from the sodium atom. The result would be a sodium ion, $\ce{Na+}$, and an electron, $\ce {e-}$, that were free from one another: the sodium atom has been ionized, i.e., has become an ion, so the energy of the $\pu{242 nm}$ photon is the ionization energy of the 3s electron in sodium. Using the $h$ and $c$ values, the required ionization energy, for one sodium atom, is $\pu{8.21 x 10^{-19} J}$. For a mole of sodium atoms, multiply by Avogadro’s number, resulting in $\pu{494 kJ/mol}$.

So where does the infinity enter the picture? Well, a positively charge sodium ion and the electron that escaped from it attract: opposite charges attract. The Coulombic force of attraction is an inverse square law, similar to gravity, so total absolute freedom would mean that the electron was infinitely far away from the sodium ion and had zero kinetic energy, i.e., zero energy of motion. But, as a practical matter, the separation can be relatively small, not infinity! As a practical matter, it is known experimentally that an isolated sodium atom, doing its thing in a vacuum, can be ionized by shining $\pu{242 nm}$ light on it.

In your example, suppose a sodium ion, in vacuum, happened to recombine with a free electron. Assume the sodium ion and electron are moving very slowly relatively, so there is negligible excess kinetic energy: we don't want the electron to 'harpoon' the poor sodium ion! The electron could drop down into any empty orbital of the sodium ion. Potentially, that is an infinite number of possibilities, hence M. Farooq's answer (and sweet Grotian diagram). The result would be release of a photon with wavelength longer (numerically 'larger') than $\pu{242 nm}$. This excited (it has more energy than a 'ground state') sodium atom must eventually drop to the 'ground state' (the only state where it is indefinitely stable) by emitting one or more additional photons or perhaps getting 'mugged' of its excess energy, via collision with something else. This is collisional deactivation.

But what if the electron dropped all the way down, in one fell swoop, to the 3s orbital? Then the resulting sodium atom would be in its 'ground state', i.e., lowest energy (and only indefinitely stable) state, and the emitted photon would be at $\pu{242 nm}$.


Are there any reducing agents for reduction of nitrobenzene to aniline other than Sn/HCl?

Hydrogenation of aromatic nitro groups (over, for example, $\ce{Pd/C}$) usually results in reduction to the corresponding aniline and is probably easier to carry out and work up as compared to $\ce{Sn/HCl}$.

Wikipedia has a list of several conditions for reduction of $\ce{ArNO2}$ to $\ce{ArNH2}$.  It also states that $\ce{LiAlH4}$ reduces it to the azo compound $\ce{ArN=NAr}$, which is corroborated by March's Advanced Organic Chemistry (7th ed.).

In any case, from a practical perspective, you would probably want to avoid using $\ce{LiAlH4}$ if there's a simpler alternative. It's dangerous and the workup is not fun.


Freezing point depression and organic molecules

To get a freezing point depression of $\pu{−1.86 °C},$ the solution should be made by adding $\pu{1 mol}$ organic substance to $\pu{1 L}$ water. And $\pu{1 mol}$ per liter is much. For glucose, it is $\pu{180 g}$ per liter. It is nearly a syrup. I doubt your lake has such a high concentration whatever the nature of the solute.
For salts, it is less, because usual salts are made of two ions. For example, $\pu{29 g}$ $\ce{NaCl}$ in $\pu{1 L}$ water is enough to get a freezing point depression of $\pu{−1.86 °C}.$ Anyway, $\pu{29 g}$ salt added to one liter is not negligible. Are you sure of your measurements?


in hybridization, can bond angles be different

You have to keep in mind that hybridisation doesn't really exist, its just a convenient way of explaining how x orbitals can form y bonds with a certain geometry even though the constituent atomic orbitals point in wildly incorrect geometries.

Take sp3 hybrid orbitals as an example, we combine s, px, py and pz to get 4 sp3 hybrids which we consider to be in a tetrahedral arrangement (i.e. bond angles of around 109.5).

Most molecules deviate (in some cases significantly) from these values, and we can explain that through a variety of steric and electronic effects. The carbon centre may still be considered sp3 due to the orbitals, even if the geometry doesn't exactly match what you'd expect.

Within various hybridisation states, theres also a situation where we would say that a centre has 'more s-character' or 'more p-character', for instance if you consider the bond angles of NH3 vs PH3 and keep on going down, the bond angles change progesively despite them all having the same hybridisation state.


Polar rearrangements: What is this reaction called?

It is a Wagner–Meerwein rearrangement. Your particular example has the following mechanism:

Edit: Regarding your comment, the best way to see what is happening is to draw the product so that it resembles the reactant (like in the image below) and then in another step draw it in a more reasonable conformation:


How to assign priority groups of tartaric acid?

First you have to rank the molecules around the two chiral centres of tartaric acid according to Cahn Ingold Prelog principle. (I have added the principle in below after this 3 points)
After ranking the molecules according to Cahn Ingold Prelog principle, then orient the molecule such that the lowest priority group is directed away from you. The three remaining groups then project toward you.

If you intend to find the configuration of the molecule (i.e. R or S), then let me add this point:

If the three groups projecting towards you are ordered from highest priority to lowest priority (i.e from rank 1 to rank 3) in clockwise manner, then the configuration is "R". If the three groups projecting towards you are ordered from highest priority to lowest priority (i.e from rank 1 to rank 3) in counterclockwise manner, then the configuration is "S". I mentioned "from rank 1 to rank 3" - because the 4th one is oriented away from us and is not considered while checking the order.

Cahn Ingold Prelog principle:

Rule to prioritize the groups attached to a chiral carbon:

Prioritize the four atoms, or groups of atoms, attached to the chiral center based on the atomic number of the atom that is bonded directly to the chiral center. The higher the atomic number, the higher the priority.

Like say you have carbon and lets the 4 atoms attached to it are: $\ce{H},\ce{OH},\ce{Cl},\ce{Br}$.
So here the carbon is chiral, as all the 4 atoms attached to it are different and now for ranking, we rank it on the basis of atomic number. Here as Br has highest atomic number, It would be ranked 1, then Cl would be ranked 2, then OH will be ranked 3 (oxygen will be considered as it is directly attached to the chiral carbon) and then hydrogen will be ranked last.

Atoms participating in double/triple bonds are considered to be bonded to an equivalent number of similar "phantom" atoms by single bonds. This is done just for the purpose of the rule. See this pic for better understanding:
If two or more of the atoms that are bonded directly to the chiral center are the same, then prioritize these groups based on the next set of atoms (i.e. atoms adjacent to the directly bonded atoms)

Lets take an example here. Say to the carbon I have $\ce{CH3},\ce{COOH},\ce{Cl},\ce{H}$ attached.
Here first we will compare the atomic number of the atoms which are directly attached to the chiral carbon i.e. C, Cl and H. From these as Cl has highest atomic number, hence it is ranked 1. Then, for the carbon, as for both the groups i.e $\ce{CH3}$'s and $\ce{COOH}$'s, it is the carbon atom which is directly connected to the chiral carbon. Hence, we need to look for the next set of atoms. $\ce{CH3}$ next set of atoms are H, H & H; while $\ce{COOH}$'s next set of atoms are O, H and O-C (see pic provided in rule 2). So from here easily we can say $\ce{COOH}$ is ranked 2, then $\ce{CH3}$ (i.e. rank 3) and then H.


Inter- means in between molecules (like the interstate highway - in between states)and intra- means within one molecule ( like an intrastate highway- within one state). So a covalent bond of CO2 OR an electrostatic ionic bond would be an intramolecular bond. The type of bond that keeps water molecules close to each other ( hydrogen bonding) is an example of an intermolecular attraction.


Why won't increasing the temperature increase the solubility of a gas in a liquid solvent?

A gas has bigger entropy when it's free, than in a solution. By increasing the temperature the molecules have bigger energy and they tend to be in a state with more freedom. Therefore, the solubility would decrease.


If heavy metals bind to sulfur why isn't sulfur used as a safer way to chelate them?

Sulfur-containing molecules are indeed used in chelation therapy: prime examples are dimercaprol and penicillamine, both of which contain thiol functional groups.


Is there any evidence, *any evidence at all*, that nascent hydrogen actually exists?

You should read the Solution to nascent hydrogen challenge by Juris Meija & Alessandro D’Ulivo Link. The authors vehemently deny the existence of any nascent hydrogen.

You will find that the concept of nascent hydrogen and nascent oxygen is still taught in some Indian chemistry textbooks. This term has literally vanished from US & European chemistry books. I just thought to search an elite journal JACS  and this term is present in 2011 in the abstract of

Highly Selective Ammonia Synthesis from Nitrate with Photocatalytically Generated Hydrogen on CuPd/TiO2, J. Am. Chem. Soc. 2011, 133, 5, 1150-1152

  "The continuous generation of nascent hydrogen atoms on the surface of
  the CuPd nanoalloy, where Cu and Pd are homogeneously mixed, led to
  the high selectivity for ammonia."

Well, this is exactly a classical example of the usage of nascent hydrogen when Zn could reduce nitrate ions in an alkaline medium. The texts of 1940s describe those experiments very well. See Mellor's Treatise of Inorganic Chemistry.

Another 2012 RSC journal Chemical reduction of an aqueous suspension of graphene oxide by nascent hydrogen


Casein is a mixture of phosphoproteins found in milk that accounts for approximately 80% of the total protein content of cow milk (cow milk contains 3.3% total protein). Hence, it is easy to prepare casein as a pure protein from cow milk.

Casein exists in milk as the water soluble calcium salt of a phosphoprotein (UC, Davis). Thus, it is usually prepared by acid precipitation. Acid treatment removes the calcium cation, leaving a water insoluble phosphoprotein. Note that casein is amphoteric, i.e., it forms salts with both acids and alkalines.

Generally, acetic acid is used in the casein precipitation, because its mildness and less harsh property compared to hydrochloric acid. Although, most people extract casein precipitation in the cold acid solutions, such practice would lead to the formation of protein with poor physical properties. Softness and rather incomplete separation of the casein are problems encountered.

If the precipitation is performed in warm solutions, the casein separates as a single large colloid, with complete separation, leaving a yellow solution of whey behind (which primarily consists of milk-sugar, lactose).

Keep in mind that usual casein precipitation is comparatively easy but delicate. If too much acid is added or if the acid is added too quickly, or if the acid is too strong, part of casein mass will redissolve in whey. Also, casein may be extracted from whole milk. Disadvantages are that whole milk contains a relatively large concentration of fat. This fat will be physically included in the casein curd as it separates. Thus, I'd recommend to use skim milk (if possible), which would be one solution for your co-precipitation of fat problem.

A procedure for casein precipitation in UC, Davis is as follows:

First, prepare a solution of $\pu{7.0 mL}$ glacial acetic acid in $\pu{50.0 mL}$ deionized water. This is your acid solution to be used for the precipitation.

Place $\pu{600.0 mL}$ of skim milk in a $1$-$\mathrm{L}$ beaker. Warm the milk to $\pu{110 ^{\circ}F}$ (approximately $\pu{43 ^{\circ}C}$; Note that temperature should not be $\gt \pu{55 ^{\circ}C}$). At this temperature, add the acetic acid solution (prepared earlier) drop-wise and with stirring. At a certain point, casein will separate as an amorphous mass leaving clear supernatant (light yellowish). Gather the solid from the beaker with a glass stirring rod or separate it using suction filtration.

Caution: Do not add any more acid than necessary to obtain a good curd. A few minutes must be given between acid additions to allow for stabilization.

Press the removed mass of casein between paper tissues to remove as much water as possible. Place the mass of casein in a food blender with $\pu{500 mL}$ of water. Suspend the casein in the water by blending as much as possible.

Separate the casein from the water with a filter and wash the cake with 2-propanol. If 2-propanol is not available use ethanol or some low molecular weight alcohol.

Let the casein powder air dry. You may wash the dried cake with two portion of ether if necessary. You may suspend the dry powder in ether and let it soak well before filtering.

Yield of casein in $\pu{600.0 mL}$ of skim milk is reported as $\pu{13.8 g}$.


Naming a molecule with both alcohol and aldehyde functional groups

You have correctly identified the principal characteristic group as well as the principal chain. Thus, the parent structure of the compound is pent-2-enal. You also correctly included the ‘3-methyl’ substituent.

The missing $\ce{-OH}$ group cannot be expressed as suffix ‘ol’ since it is neither the principal characteristic group nor situated on the principal chain. Therefore, it has to be indicated by the prefix ‘hydroxy’. The corresponding rule in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

  P-63.1.4 Substitutive nomenclature, prefix mode

  Hydroxy groups are indicated by the prefix ‘hydroxy’ when:

  (1) a group having priority for citation as the principal characteristic group is present; or

  (2) a hydroxy group cannot be denoted by a suffix.

In this case, the hydroxy group is placed on a methyl side chain, which results in the compound substituent group ‘hydroxymethyl’.

  P-29.4 COMPOUND SUBSTITUENT GROUPS

  P-29.4.1 Compound substituted substituent groups

  A compound substituted substituent group is formed by substituting one or more simple substituents into another simple substituent that is considered as the principal chain. The choice of the principal chain is fully discussed in Section P-44.3. The first criterion to be applied is that the principal chain is the longest chain; it will be applied in the following examples of acyclic compound substituent groups.

  (…)

Parentheses are used around such compound prefixes in order to avoid ambiguity.

  P-16.5.1 Parentheses (also called curves or round brackets)

  P-16.5.1.1 Parentheses are used around compound (see P-29.1.2) and complex (see P-29.1.3) prefixes; after the multiplicative prefixes ‘bis’, ‘tris’, etc.; to enclose a multiplied parent compound consisting of a parent hydride and a substituent suffix or a multiplied substituent prefix even though preceded by a basic numerical prefix, such as ‘di’, ‘tri’, etc.

Therefore, the complete systematic name for the compound that is given in the question is (2Z)-2-(hydroxymethyl)-3-methylpent-2-enal. (Also note the stereodiscriptor for the double bond.)


You probably want to drain its charge and put it into recycling.  It's probably a fire hazard at this point, and the damage that a fire from it could cause will cost more than a new one.


Does 'Stoichiometry' refer to the mole ratio of atoms in compounds and the mole ratio of compounds involved in chemical reactions?

That's right -- stoichiometry is a field which uses these numbers and the relationships between reactants and products, but not the numbers itself. The mole ratio is just the mole ratio, whereas stoichiometry is the entire discipline of using mole ratios to calculate relative quantities of reactants and products in reactions.


Which compound results in the following products when mixed with AgNO3 and Chlorinated water

The black precipitate can either be

metallic silver, formed by reduction of the $\ce{Ag+}$ ions (most likely)
an insoluble silver salt

Possible candidates for the sodium salt:

╔══════════════════╦═══╦═══╗
║      SALT        ║ 1 ║ 2 ║
╠══════════════════╬═══╬═══╣
║Sodium bisulfate  ║   ║   ║
║Sodium borate     ║   ║ ? ║
║Sodium citrate    ║   ║   ║
║Sodium chloride   ║   ║   ║
║Sodium fluoride   ║   ║   ║
║Sodium nitrate    ║   ║   ║
║Sodium nitrite    ║   ║   ║
║Sodium oxalate    ║   ║   ║
║Sodium sulfate    ║   ║   ║
║Sodium sulfite    ║ ✔ ║   ║
║Sodium thiosulfate║ ✔ ║   ║
╚══════════════════╩═══╩═══╝

The reaction with chlorinated water tends to show it is the first case (maybe with sodium thiosulfate).

Regarding the pungent odor, $\ce{H2S}$ has a strong rotten-egg odor.

A specific test for sodium thiosulfate is adding an acid (such as $\ce{HCl}$ to the unknown salt solution). A yellowish-white sulfur colloid should slowly appear.

A specific test for sodium sulfite is the decoloration of fuschine (should be available in a microbiology lab): add a buffer at pH = 7 to 1 mL of your unknown salt solution, then 1 mL of fuschine. After a few seconds, it should turn transparent.

Can you conduct those test to get more insight on this? If the first one is positive, no need to do the second one.


Relation between equilibrium constant and entropy change

Like the Van't Hoff equation, which relates change in enthalpy to equilibrium constant, is there a similar equation for the relation between change in entropy and equilibrium constant?

A relation between $K$ and $\Delta S^\circ$ can be obtained as follows:

$$\begin{align}T\log K &= -\frac{\Delta G^\circ}{R} \\
\left(\frac{\partial(T\log K)}{\partial T} \right)_p &= -\frac{1}{R}\left(\frac{\partial \Delta G^\circ}{\partial T}\right)_p= \frac{\Delta S^\circ}{R}\end{align}$$

As regards the multiple choice question, the best way to answer it is to understand that $\Delta G$ represents a sum of the change in the entropy of the system and of the surroundings, scaled by $T$:

$$\begin{align}\Delta G &= \Delta H - T\Delta S \\&= -T\Delta S_{surr} - T\Delta S_{sys} \\&= -T(\Delta S_{surr} + \Delta S_{sys}) \end{align}$$

Exchange of heat at constant pressure (equal to the enthalpy) alters the entropy of the surroundings, so when discussion turns to the heat exchanged at constant pressure, you don't need to consider the entropy of the system, only that of the surroundings (at least for this type of problem - this alone eliminates a and c as possible choices). Next you need to remember that endothermic means the surroundings lose heat to the system, which lowers the entropy of the surroundings. The opposite holds for an exothermic process (surroundings gain entropy).

To summarize: an endothermic reaction lowers the entropy of the surroundings. The decrease in the entropy of the surroundings decreases if you increase T. The result is therefore a net increase in the total entropy. Finally, since $$\Delta S ^\circ_{\text{total}} = -\frac{\Delta G^\circ}{T}=R \log K_{\text{eq}}$$
increasing the total entropy results in a higher equilibrium constant.


Which compounds react positively in the aniline dye test?

First, the aniline is converted to a diazonium halide salt in the presence of acid and sodium nitrite, per this reference.

Then, the diazonium salt reacts with the phenol to give an azo compound, per this reference (see the section entitled Coupling reactions of diazonium ions

Note that the diazonium ion is shown reacting with the para position of the phenol. That is because the -OH group in phenol is an ortho,para-director (reference), meaning that it enhances the nucleophilicity of the ortho and para positions on the phenol ring. However, the para position is significantly more nucleophilic because of the relative lack of steric hindrance in comparison to the ortho positions, which are hindered by the proximity of the -OH group.

So, having a free para position on the phenol means that the phenol is especially reactive towards electrophiles, including the diazonium salt in this example, meaning that it will react faster and in higher yields - increasing the sensitivity of the aniline dye test.


Can aged acetylcholinesterase be reactivated?

Until the late 2010s, it was considered impossible to reactivate acetylcholinesterase that has undergone aging. However, in recent years, new advancements have been made that show promise in this area.
A group of researchers investigated several sulfonium-containing alkylating agents that were supposed to alkylate the phosphonate moiety bound to the serine of aged AChE, in hopes that this would allow the enzyme to be reactivated by traditional methods.

Computational simulations suggested that all three of these compounds would be able to alkylate the aged enzyme.
Another group of researchers studied two aminoalcohols that would act by transferring a proton to the aged adduct. One of the compounds showed promise in reactivating aged AChE, as it was able to do so according to computer simulations. This is the molecule:

What is notable about this aminoalcohol is that it is predicted to be capable of reactivating the enzyme without alkylating the adduct.
Yet another series of compounds was shown to be successful in reversing the aging of acetylcholinesterase. These compounds reverse the aging reaction by alkylating the phosphonate/phosphate-AChE adduct, and also exhibit the potential to reactivate the enzyme entirely.

Unfortunately, they were quite slow to reverse the aging of acetylcholinesterase, with the most active compound, C8, resurrecting 12% of methylphosphonate-aged AChE in one day at a concentration of 4 mM.
In conclusion, while several compounds have shown promise in reactivating acetylcholinesterase that has undergone aging by traditional nerve agents or pesticides, therapeutic methods of administering those compounds have yet to be developed. Furthermore, these compounds have not yet been tested in vivo, under physiological conditions, or against other anticholinesterase agents such as Tabun or the Novichoks.
Sources:
Franjesevic, A. J., Sillart, S. B., Beck, J. M., Vyas, S., Callam, C. S., & Hadad, C. M. (2018). Resurrection and Reactivation of Acetylcholinesterase and Butyrylcholinesterase. Chemistry - A European Journal. doi:10.1002/chem.201805075
Khavrutskii, I. V., & Wallqvist, A. (2017). β-Aminoalcohols as Potential Reactivators of Aged Sarin-/Soman-Inhibited Acetylcholinesterase. ChemistrySelect, 2(5), 1885–1890. doi:10.1002/slct.201601828
Zhuang, Q., Franjesevic, A. J., Corrigan, T. S., Coldren, W. H., Dicken, R., Sillart, S., … Hadad, C. M. (2018). Demonstration of In Vitro Resurrection of Aged Acetylcholinesterase after Exposure to Organophosphorus Chemical Nerve Agents. Journal of Medicinal Chemistry, 61(16), 7034–7042. doi:10.1021/acs.jmedchem.7b01620


Calculating dipole moments from atomic partial charges

From a theoretical standpoint, that approach is correct:

$$
\mu_i = \sum_{a}^{N_{\text{atoms}}} \sum_{i\in\{x,y,z\}} r_{ia}q_{a},
$$

where the set of atomic partial charges $\{q\}$ could come from partitioning the density in AO space (Mulliken, Lowdin), partitioning the density in real space (Bader, Voronoi, Hirshfeld), partitioning the electrostatic potential (ESP; Merz-Singh-Kollmann or ChElPG), or other methods I'm certainly forgetting. Another interesting method is based on atomic polar tensors [1], called GAPT or IR charges, which are a natural byproduct of harmonic vibrational frequency calculations.

This doesn't mean the results will be of good quality. Those based on partitioning the AO density seem to be the worst overall. Anecdotally, those based on ESP calculations are better, however if your system has a lot of "buried" atoms that aren't solvent surface-accessible, results may be questionable. Real-space partitioning is an attractive method since it is independent of direct basis set effects and surface accessibility. GAPT results also seem to be good, but they require a harmonic frequency calculation, which can be expensive.

Milani, A.; Castiglioni, C. Atomic charges from atomic polar tensors: A comparison of methods. Journal of Molecular Structure: THEOCHEM 2010, 995 (1-3), 158-164. DOI: 10.1016/j.theochem.2010.06.011