Why does the mixture's temperature change while phase transition?

The concentration of water decreases as the mixture cools, so one is left with fairly pure ice, freezing at ~273 K, and pure(r) ethanol, freezing ~159 K (well, the eutectic mixture, about 93% ethanol). Since the ratio of the components are changing as it freezes, there is no distinct freezing point (unless you start with the eutectic mixture). See the phase diagram for ethanol-water mixture. Instead, the slush mix has a range of solidus and liquidus temperature. BTW, this is a method to make apple jack or brandy by freezing out water from cider or wine.

The same is noticed with solder, where it crystallizes gradually and dully, except for the sudden freezing of a eutectic mix (~63% Sn, ~37% Pb), making a shiny joint.


Is Sodium Hydroxide a Bronsted-Lowry Base?

All Arrhenius acids and bases are also Bronsted-Lowry bases.
According to this website,

  The Bronsted-Lowry theory doesn't go against the Arrhenius theory in any way - it just adds to it.

Remember, when we write NaOH (aq), what we really mean is Na+ (aq) and OH- (aq). Sodium hydroxide is a Bronsted-Lowry base because the hydroxide ions that are a part of sodium hydroxide accept protons.


Exact meaning of constant volume heat capacity

(a) is correct. (b) is wrong. $U$ depends on $V$ only in the combination $v=V/N$. $v$ is the specific volume, an intensive quantity that does not depend on the size of the system.

We can derive this from first principles. Take $U = U(T,V,N)$. Because $U$ is extensive, $$U(T,\lambda V,\lambda N) = \lambda U(T,V,N) \quad \Longleftrightarrow \quad U(T,V,N) = \lambda^{-1}U(T,\lambda V,\lambda N),$$ by taking $\lambda = N^{-1}$, we can write this expression as $$U(T,V,N) = NU(T,V/N,1) =: Nu(T,v),$$ which produces the desired functional relationship.


What does it even mean to calculate or measure solubility of something down the sea?

Why, there is a precise and well-defined meaning, and it is hinged on equilibrium between the gas and the solution, which supposedly should be established at the said conditions. Our imaginary gas bubble, like everything else, must withstand the immense pressure of all those layers of water up above.

So no, the partial pressure is not just the atmospheric one.

Deep abyss of the seas is undersaturated in all gases, including oxygen, and that by a wide margin.


pH of a solution of acetic acid and ammonium acetate

You have some mistaken assumptions.

$\ce{NH4OH}$ and $\ce{CH3COONH4}$ are not molecules that exist.

The assuption "all $\ce{CH3COOH}$ is consumed" is incorrect.

Why do you say more $\ce{NH4OH}$ will be formed? How is this possibe? What can it be formed from?

Review that Henderson–Hasselbalch equation


How do we know that hydrogen gas consists of two hydrogen atoms?

Assuming ideal gas behavior we can determine the molar mass $M$ of the hydrogen gas from its density $\rho$ (which can be measured) by the following formula
$$
    M = {{RT\rho}\over{p}}\ \, .
$$
At STP $\rho \approx 0.09 \, \mathrm{g} \, \mathrm{L}^{−1}$ so the following back-of-the-envelope calculation
$$
    M = {{8.31 \, \mathrm{L} \, \mathrm{kPa} \, \mathrm{K}^{−1} \, \mathrm{mol}^{−1} \cdot 273.15 \, \mathrm{K} \cdot 0.09 \, \mathrm{g} \, \mathrm{L}^{-1}}\over{101.325 \, \mathrm{kPa}}}\ = 2.02 \, \mathrm{g} \, \mathrm{mol}^{−1} \, .
$$

shows that $M$ is indeed close to 2 g/mol which means that hydrogen gas consist of $\ce{H2}$ molecules rather than $\ce{H}$ atoms


Is it possible for already charged metal ions to form a metallic bond?

Your reasoning is correct. Such things do exist, and they came to be known as clusters. The link above does not quite do them justice; in fact, they are vastly more numerous and diverse.
Come to think of it, the clusters go beyond the scope of your question, in that many of them are not charged. To pick a few examples, see the osmium carbonyl (which, arguably, is too small to consider the bond really metallic, but its bigger brothers are known), or gold clusters which range all the way up to the immense five-layer icosahedron $\ce{Au561}$ with some ligands on top.
To pick something charged and at the same time big enough to be called metallic, have a look at this: $\ce{Ta6Cl18^4-}$.

[source]

Upd. I thought I'd better stress this again. Yes, metal ions that still have some valence electrons can use those to form bonds, given the right circumstances. Whether those bonds qualify as metallic is a different question altogether, and depends on the number of atoms (or ions) in the cluster. Two is clearly too few. You can't possibly have a metallic bond between two atoms. A bond with electrons delocalized between two atoms is called covalent. That's the case of $\ce{Hg2^2+}$. Then you have clusters with 3 atoms, then 4, and so on, all the way up to 100 and beyond.
So it goes.


How is it possible to dissolve zinc oxide safely for skin use, preferably without acid?

As I previously stated in a comment, I think that the zinc oxide is not actually dissolved. That would just leave ions which you would hardly consider zinc oxide any more. $$\ce{ZnO + 2HCl <=> Zn^2+ (aq) + 2Cl- (aq) + H2O}$$

Instead I think that there are nanoparticles in use, that appear to be dissolved. In deodorant or other skin products it is likely to have quite a low concentration.

An admittedly short search turned up a patent by Marc Paye, in which is stated:[1]

  [...]
  (f) 0.01% to 1%, more preferably 0.03% to 0.5% of Nanoparticulates of zinc oxide which functions as an anti-irritant system for anionic surfactants,
  [...]

Or from another patent of the same company:[2]

  [...]
  (c) 0.5-10% (particularly 0.5-7.0, and more particularly 0.5-5.0%) of a small particle size zinc oxide, particularly a micronized zinc oxide or a nanoparticle size zinc oxide having a particle size in the range of 20 nanometers-200 microns;
  [...]

I also found, but have not completely read an article that confirmed my theory.[3]

  As an ingredient in dry deodorants to reduce wetness under the arm, ZnO can be used between 0.05 and 10% by weight with average particle size in the range of 0.02–200 microns. ZnO may be used to provide a pH range desirable for deodorants designed for use on sensitive skins [174].

At least in the last article you should also find some information about how to synthesise these particles. Spoiler: It's nothing you would try at home.

Paye, M.; Zinc oxide containing surfactant solution. US6774096 B1, 2004.
Hall-Puzio, P.A.; Gale, A.E.V.; Brahms, J.C.F.; Deodorant with small particle zinc oxide. US 6358499 B2, 2002.
Moezzi, A.; McDonagh, A. M.; Cortie, M.B.; Chem. Eng. J., 2012, 185–186, 1-22..


Cleaning up dispersed potassium hydroxide (KOH) spill?

You just encountered a situation where you did something without sufficient protective gear; in this example, if you had known the battery contains corrosive material, you would have used protective gloves (for your hands) and a pair of safety glasses (for your eyes).

Your estimate that the Leclanché and alkaline battery contains lye is correct. The sensation of the skin turning slippery is expectable, since $\ce{KOH}$ started to degrade the protective coating of your skin (fat) into soap.  Rinsing promptly with plenty of water was a good action.

The lye equally attacked your cloths (one reason chemists wear lab coats), and as your skin, it is beneficial to rinse them right after the skin with plenty of water, too; and then to wash in a washing machine.  On contrast, bringing them as they are into the washing machine only will spread unnecessarily the contamination to other clothes.  Expect that the areas touched by the lye eventually will become brittle, especially if i) they "dried" for some time prior to rising or washing and ii) upon action of heat (e.g. ironing).

Regarding your study, I recommend to go the extra mile, and and clean all the possible surfaces; albeit hosing sounds a bit overkill (esp. inside an appartment).  You want to eliminate the pollution quickly, as it may attack wood and wood stain; and imagine you accidentally glide a book over such a forgotten droplet, it will be attacked just as your skin was attacked, too.  (Even more, as in contrast to human skin, the dead book does not have the ability to re-cover itself by the thin film of protective glycerides.)  In addition, droplets of lye do not dry away like water, $\ce{KOH}$ is hygroscopic and will constantly remain somewhat wet (and corrosive).


K is properly defined as:

$$K = \dfrac{(a_C)^{c}(a_D)^{d}}{(a_A)^{a}(a_B)^{b}}$$

Activities involve relation of concentrations or parcial presures so K is dimensionless.

Let's find activities.

Ideal gases

$$a_{gas} = \dfrac{P}{P^o} $$  $P^º$ is 1 atm.

Solute in very dilute solutions

$$a_{solute} = \dfrac{C}{C^º}$$ $C^o$ is 1 M

Activities for solvents and pure solids are considered unity. $\ce{H2O}$ and $\ce{CaCO3}$ have unitary activity.

Answer

$$K = \dfrac{\ce{(a_{CaCO{_3}})(a_{H{^+}})^{2}}}{{\ce{(a_{H{_2}O})(a_{CO{_2}{_g}})(a_{Ca{^{2+}}{_{aq}}})}}}$$

$a_{CaCO{_3}}, a_{H{_2}O}$ are unity. It rests to use ideal gas equation and get $C_{CO2}$. $a_{Ca{^{2+}}{_{aq}}}$ and $a_{H{^+}}$ are usual concentrations.

EDIT

The way $K_p$ is converted to $K_c$ is:

$$K_p = K_c(RT)^{\Delta{n}_{gas}}$$

R: gas constant (0.082 L atm mol$^{-1}$ K$^{-1}$). T: temperature in Kelvin (K) at which the equilibrium reaction takes place.

To find $\Delta n_{ gas}$ , simply subtract the sum of the coefficients of the reactants from the sum of the coefficients of the products.
You can deduce that using $P=\frac{nRT}{V}$, $\frac{n}{V}$ is concentration.


Cahn–Ingold–Prelog Rules, cyclic structures, and breaking ties

You don't need a digraph for this one. Your structure is A. Remove the double bond and add duplicate carbon atoms to each of the double bond carbons (B). These duplicate carbon atoms each bear three phantom atoms of atomic number zero. Clearly at C5 hydrogen has the lowest priority and methyl is the next lowest. The methylene groups at C4 and C6 are a tie. [There is no need to consider C2 since it comes into play for both routes. It is a wash.] The red carbon of the ethyl group bears C,H,H while the red duplicate carbon at C1 bears 0,0,0. Place your right hand on C5 and point your thumb in the direction of the hydrogen. Now your remaining fingers will point as follows: C4>C6>CH3>H.  C5 is of the R-configuration. See, Twisting stereoisomers with rings to determine R/S.  .


What are the GDB-13 criteria for "synthetically accessible organic molecules"?

What you have provided is a definition of all the terms involved. This is fine, but there remains the (rather interesting) question of exactly how this is defined for a molecule in silico, especially because your question asks specifically about the GDB-13 database. Of course, one could evaluate "synthetic accessibility" by trying to make it in the lab: if you succeed, then it is synthetically available, and if not, then it isn't. Likewise, one could measure "stability" by (potentially) making it and studying its thermodynamic properties.

However, for a database of 1 billion compounds, this is obviously unfeasible. So the designers of the database have to resort to using some heuristics, or rules, to determine whether a given molecular structure is "synthetically accessible" and "stable". The rules are outlined in the original paper in which the GDB-13 database was published,[1] as well as the first GDB paper from slightly earlier.[2] Both are very interesting papers and well worth a read if you are interested in cheminformatics in general.

Because all organic molecules have a carbon chain or ring as their backbone, the first step is to construct all possible frameworks of carbons and hydrogens using graph theory (up to a maximum number of carbons, of course). Here there is not yet any consideration as to what other atoms may possibly exist in a typical organic molecule. However, there are some computer-generated frameworks which are almost certainly impossible to make because of strain or other considerations (such as Bredt's rule). On top of that, it appears that they have intentionally removed three- and four-membered rings, which are stable but would have dominated the database, thus making it highly unrepresentative:[2]

  The vast majority of these graphs (99.8%) contained three‐ and four‐membered rings and was excluded to avoid generating a database consisting almost exclusively of such small rings.

The next step is to introduce heteroatoms, which can be done fairly straightforwardly because in organic chemistry atoms have very specific bonding patterns. Thus, for example, because carbon forms four bonds and nitrogen three, you could replace a $\ce{CH2}$ group in a molecule with $\ce{NH}$. This is not trivial to do in the lab, but is very easy for a computer:[2]

  [...] all possible atom‐type combinations by introducing carbon, nitrogen, oxygen, and fluorine (as a model halogen) at each node

although for GDB-13, they have ignored fluorines:[1]

  We also eliminated fluorine because it was rarely found and never considered in our group for synthesis in virtual-screening guided drug discovery applications of GDB-11.

and also added chlorines (replacing $\ce{OH}$ groups in molecules which have them) and sulfurs (replacing $\ce{O}$ atoms).

The problem with doing that exhaustively is that too many heteroatoms tends to make a molecule very unstable. For example, directly joining heteroatoms via single bonds is (generally) a good recipe for making explosive compounds. So, essentially all such molecules were removed. The authors found that most of this work could be very quickly automated by using an even simpler heuristic: simply removing any compound with a high heteroatom:carbon ratio.[1]

  Because most of the rejected molecules contained multiple heteroatoms, we reasoned that it might be possible to accelerate the database computation using a very fast “element-ratio” filter. Analysis of databases of known compounds suggested cutoff values of (N + O)/C < 1.0, N/C < 0.571, and O/C < 0.666

and specifically disallowed cases, which have to be filtered out in another step, included:[1]

  The following functional groups are discarded as too unstable to be considered: hemiacetals, hemi-aminals, aminals, acyclic imines, non-aromatic enols, orthoesters and analogs, carbamic acids, non-aromatic enamines (except acylated enamines and vinylogous enamines), beta-keto-carboxylic acids and beta imino-carboxylic acids, and all compounds containing both a primary or secondary amines and an aldehyde or ketone.

This is also helpful for preparing a database of "druglike" molecules because molecules with too many heteroatoms are very polar, which makes it almost impossible to diffuse across cell membranes (see e.g. https://en.wikipedia.org/wiki/Lipinski%27s_rule_of_five).

What you get at the end is probably not likely to be 100% "synthetically accessible". However, the filters applied mean that if you were to pick a random molecule from the database, there is an exceptionally good chance that you could make it in the lab if you so desired. Quoting from the authors one final time:[2]

  The database construction strategy chosen also ensures that the majority of GDB, although presently unknown, should be synthetically accessible.

References

Blum, L. C.; Reymond, J. 970 Million Druglike Small Molecules for Virtual Screening in the Chemical Universe Database GDB-13. J. Am. Chem. Soc. 2009, 131 (25), 8732–8733. DOI: 10.1021/ja902302h.
Fink, T.; Bruggesser, H.; Reymond, J. Virtual Exploration of the Small-Molecule Chemical Universe below 160 Daltons. Angew. Chem. Int. Ed. 2005, 44 (10), 1504–1508. DOI: 10.1002/anie.200462457.


How to change the chirality of amino acids?

Temperature is likely not going to work. To racemise something that has a chiral center, you need to break a bond: if you want to supply that much energy by heating, you are going to fry your molecule.

The easiest things to do are usually extraction of an acidic proton with a strong base, formation of a double bond involving the formerly chiral carbon, and then you go back to the initial form, but obtaining a racemate because the attack on the two sides of the double bond is equally likely.


Particle in a box with different barriers between two molecules

for the infinite barrier I'd put an electron on just one side of the box with length L/2, meaning that electron is trapped on one side/face of the ammonia.

For the no barrier part treat the electron as a box with length L meaning it can easily go from one side to the other without barrier (for example maybe the ammonia is at the planar transition state).

For the finite barrier, I'd show tunneling from one side to the second, however, normally for PIB the tunneling region has finite length.

Assuming you have a tetrahedral geometry, according to the Born-Oppenheimer separation of electron and nuclear motion the finite barrier would extend from L/2 to L (or equivalently from 0 to L/2). Therefore the tunneling would be from the middle to the edge.

Image source


Basic behavior of ammonia in aqueous medium

All what Karl and Poutnik are saying is right. But the real question is : Why ? Why is $\ce{NH3}$ more reluctant to loose one proton than $\ce{H2O}$, at the atomic level ? Why is $\ce{NH2^-}$ such a strong base ?
Let's compare this behaviour in the series $\ce{N-O-F}$. Starting with $\ce{F}$, we can observe that $\ce{F}$ holds one $\ce{H}$, only one $\ce{H}$, and it does not hesitate to give its unique $\ce{H}$ atom (as a proton) to a great variety of bases. The next in line, the oxygen atom $\ce{O}$ can hold two $\ce{H}$ atoms. This is one more than what does $\ce{F}$. Well, $\ce{O}$ in $\ce{H2O}$ is also more reluctant to loose one $\ce{H}$, and accepts to give it (as a proton) only to strong bases. The third in line is the nitrogen atom $\ce{N}$ which can hold three $\ce{H}$ atoms. It's still more than what does Oxygen. But $\ce{N}$ in $\ce{NH3}$ is still more reluctant to loose one of its three $\ce{H}$ atoms as a proton $\ce{H+}$. It even prefers stealing one $\ce{H+}$ ion from the neighborhood rather than giving it. Apparently the more $\ce{H}$ atoms the central atom has, the more it wants to increase this number. It looks like the human society : the more money you've got, the more you want to increase your fortune.
Edit :
Of course it would be interesting to see if this tendency $\ce{F-O-N}$ can be extrapolated to the carbon atom. If my reasoning is correct, the Carbon atom in $\ce{CH4}$ should be extremely reluctant to loose one $\ce{H+}$ ion for producing $\ce{CH3^-}$. Indeed I am not sure that the ion $\ce{CH3^-}$ does exist, and has a measured $p\mathrm{K_a}$ value. It should be extremely high. On the other hand, $\ce{CH4}$ should be eager to get one more Hydrogen. However the electronic situation is not similar to $\ce{F, O}$ and $\ce{N}$. In $\ce{HF, H2O, NH3, F^-, OH^-, NH2^-, H3O+, NH4^+}$, the central atom has always the electronic state of Neon. It is known that $\ce{CH4}$ does produce $\ce{CH5^+}$ ion in high pressure mass spectrometry. But the electronic structure of the carbon atom in $\ce{CH5^+}$ is unknown, to my knowledge. So I am not sure whether the tendency, valid for $\ce{F-O-N}$, can be extrapolated to the Carbon atom.


How many atoms of Lithium are there in 14.3 grams of Lithium Fluoride?

Convert 14.3 grams of Lithium Fluoride to moles which is the same for lithium and lithium fluoride since there is one atom of lithium per molecule of lithium fluoride (carry an extra significant digit to round later to 3 significant figures).

  $\mathrm{moles(Li) = \dfrac{g(LiF)}{mw(LiF)} = \dfrac{14.3}{25.939} = 0.5513\text{ }moles(Li)}$

Convert moles to atoms by multiplying with Avogadro's number, and round to 3 significant figures.

  $\ce{atoms(Li) = moles(Li)\text{ }(N_{a})} \mathrm{ = 0.5513(6.022E23) =  3.32E23}$


What chemical properties make LSD so psychoactive?

It's the indole portion of the molecule. the lowest two rings...if you remove those, the compound has no such properties. Indoles, the side chains of tryptophan , an amino acid, are suitable for binding to receptors that affect emotion, perception, cognition, sleep/wake cycles etc. You can see how the indole looks similar to both DMT (dimethyltryptamine--another hallucinogen) and the benzene ring bears similarity to psilocin, the major constituent of psilocybin (the active principle in magic mushrooms). Tryptamines are common hallucinogens.


How do I decide whether ring will expand in this reaction or not?

Explanation

In case 1. You can see it is a 2° carbocation (or secondary carbonation) with angle strain of ~25° on cyclopropane

In case 2. You can see it is 3° carbocation (or tertiary carbocation)
with angle strain of ~10° on cyclopropane

It is obvious that case 2 will be major intermediate yielding $\ce{1-chlorobicyclic [3,2,0] heptane}$ as main product that is C.

Edit1: After reading your question more thoroughly I found that I missed one thing about 'exceptional' stability of cyclopropyl methyl carbocation.

The stability that cyclopropyl provides to carbocation is due to interaction of p-orbitals which is not possible in your case because p-orbitals of carbon having '+' charge becomes out of plane with respect to p-orbitals of cyclopropyl.

Edit2: See image below for reference

Then in case of cyclopropyl methyl carbocation the p-orbitals of both carbon containing '+' charge and the carbon on ring are parallel

In your case the p-orbitals of carbon containing '+' charge is perpendicular to plane of cyclopentyl ring but the p-orbitals of carbon on cyclopropyl ring is parallel to plane cyclopentyl ring, so stability is not provided by cyclopropyl ring.

To sum up you can say that the cyclopentyl ring does not let cyclopropyl ring to provide stability to carbocation which makes ring expansion a preferred option.

Edit3 (Bonus): I asked to my professor about this, he told me that there are two conformational isomers of cyclopropyl methyl carbocation. One is bisected conformation and other is perpendicular conformation. The bisected one allows the carbocation to be exceptionally stable but the perpendicular one isn't that much stable (like in your case) due to less interaction of p-orbital and the partial double bond character cannot be achieved.


How does molecular orbital theory explain CH4?

Methane doesn't illustrate the difference very well, but the difference is nicely illustrated with benzene. Valence bond theory would have the electrons in the "double bonds" isolated between atoms as so:

But the electrons in the "double bonds" are not isolated in three double bonds but distributed equally over all 6 carbon atoms. Valence bond theory tries to wiggle out of this dilemma by posing the Kekule resonance structure:

which still belies the situation. The double bonds don't "resonate." The extra 6 electrons are distributed in one aromatic molecular bond (with three degenerate orbitals) evenly over the 6 carbon atoms.


Filling a buret tip before an experiment

You fill it as high as you fill it and you can't put anything in without getting the tip filled up too,

No I don't think it will work like that. Usually you have the stock or valve closed at the bottom. Consequently, the tip will be empty until you bleed the buret. Further, you will often bleed (allow the titrant to flow) a buret while thumping the tip to remove any air bubbles that may form directly under the stop-cock when your titrant flows initially through the tip. Once you have titrant in the tip and the tip is devoid of air bubbles you will then close the valve and fill the buret to the desired level.

At this point you can better determine just how much you have used.

The measure marks on the buret itself take into account what is in the tip below the stop cock. But if you do not bleed the tip initially, filling it with your titrant then your first addition will be a bit off because the tip does not fill entirely with liquid before you allow some to drip through.

Here is a good write-up on titrations with burets.

  A bubble in the nozzle of a buret will produce an inaccurate volume reading if the bubble escapes during a titration. Bubbles may be large and visible as shown above left or so small as not to be seen, above center. During a titration such small bubbles begin to move in the direction of the nozzle but may remain in place even though there is a moderate flow of titrant (above right). Even when the buret valve is wide open some bubbles remain in place until you take your eyes off them. Then they sneak through the nozzle and ruin your titration. Also if you let a an air bubble stay because you think it will remain and then later it leave to the tip you will have to start over because you have no idea what


Distance between successive tetrahedral voids in FCC

As shown in figure, if we divide a FCC unit cell into 8 small cubes, then each small cube has 1 Tetrahedral void at its own body centre.
Thus, there are total 8 Tetrahedral voids in one unit cell.
It can also be seen from the figure that the nearest distance between two Tetrahedral voids is a/2.


Why do ionic compounds dissociate whereas coordinate complexes won't?

First of all, who says coordination complexes don't dissociate? The interesting chemistry happens right there, and without dissociation no catalytic coordination compounds would be possible!

Now, the way I see it, when you move your compound, be it ionic, covalent or coordination, from air to a condensed phase, it will automatically interact with said phase. I think that you assume this phase is water, but it might as well be dichloromethane or ammonia or diethyl ether or anything else that is liquid (basically).

What makes a bond happen (or not) is the overall energy of the system. See, in chemistry (as in physics) the energy minimum is the goal of the system, since it is the most stable. Using this line of reasoning we can derive that bonds are formed because the overall energy (of the bonding electrons) is lower than if they were not bonded. Note that it doesn't matter where these electrons come from (they might be shared by both bonding partners or made available by only one of them). A good and easy way to visualize this is if you scratch the surface of Molecular Orbital Theory.

So why do ionic compounds dissolve? Mostly due to the solvent-compound interactions. See more here.

(I also think that your approach using Coulomb theory doesn't work that well, since the change of the dielectric constant around the compounds has nothing to do with what is between the bonding partners, thus affecting attractive forces between the two partners.)


$1000 \, \text{g}$ of water occupy a volume of (approximately) $1 \, \text{L}$ because the density of water is (approximately) $1000 \, \text{g/L}$.
If you had $1000 \, \text{g}$ of mercury, whose density is $13534 \, \text{g/L}$, it would occupy a volume of $0.0739 \, \text{L}$.

Usually you can only dissolve so much of a solute into a solvent until a certain threshold is reached.
When this threshold is reached you have a saturated solution.
After that point, it is thermodynamically more favourable (under the given conditions, e.g. temperature, pressure, etc.) for any extra solute not to get dissolved in the solvent but clump together and precipitate.
But if you use some special techniques you can prepare a supersaturated solution that contains more solute than it should, thermodynamically speaking.
A supersaturated solution is a meta-stable state, meaning that it is only a local energy minimum for the system, not the global minimum.
But it would like to get to the global minimum if it gets the chance.
The only thing preventing that, is a small energy barrier that has to be surmounted.
So, if you disrupt the supersaturated solution in any way, like shaking it or adding a crystal seed, the solution can surmount the energy barrier and attain its global minimum state, i.e. the saturated state.


Liquids and solids form because intermolecular forces hold together different molecules/atoms/ions/biscuits in a strong enough fashion to prevent them from drifting away from each other. In a very basic sense you could say that gases are non-ordered, liquids have near-range ordering but not far-range, and solids are well ordered into crystal lattices.

Noble gases have very little intermolecular forces acting between them since they are monoatomic and unpolarised. Thus, only London dispersion forces act and these directly depend on the number of electrons in a compound. If you compare each noble gas with the element to its left in the periodic table (i.e. to the neighbouring halogen; hydrogen for helium) and take $n$ to be the number of electrons the noble gas has, then the next-door neighbour will always form a diatomic molecule with $2(n-1)$ electrons. In every case except for hydrogen, this is substantially more than in the noble gas, thus there is a substantial increase in melting/boiling point when going from the noble gas to the corresponding halogen.

Note that radon’s boiling point is already at $-61~\mathrm{^\circ C}$, higher than that of the highly polar molecule $\ce{HCl}$ ($- 85~\mathrm{^\circ C}$). And note that ununoctium has been predicted to have a boiling point of $(+80 \pm 30)~\mathrm{^\circ C}$ and would thus better be termed a noble liquid. Unfortunately, ununoctium has a half-life of about $0.89~\mathrm{ms}$ so actually observing the noble liquid could prove tricky.

(All values taken from the respective gases’ Wikipedia articles.)


Is there a chemical reaction that speeds up when the temprature gets lower?

That's a good question.

A reaction slowing down on a rise in temperature seems counter intuitive but it's not so always.

The Arrhenius equation works for only a few limited elementary reactions and the transition state theory is a much better explanation of rate laws and their variation with temperature.

The dependence on temperature of the reaction rate cannot always be analysed purely using the Arrhenius equation. It can be better understood by knowing the mechanism of the reaction.

For instance
$$\ce{2NO + O2-> 2NO2}$$

Has its first step as
$$\ce{2NO <=> N2O2}$$

And the second step as:
$$\ce{N2O2 + O2->2NO2}$$

The first step is exothermic in nature.
And the equilibrium shifts to the left on increasing temperature based on the Van't Hoff equation of gaseous equilibrium.

So the effective rate constant of the overall reaction decreases on increasing the temperature.

Thus, we see a reaction whose rate slows down with the rise in temperature.

There are in fact many such reactions.

To give another example, I could take an enzymatic reaction in a living cell and show that the rate of reaction falls down drastically with a rise in temperature. Ever wondered why your body rises its temperature during a fever? The aim there is to slow down the reactions in the cells of the invading germ and hence kill it.

From these examples, one could conclude that the famous Arrhenius equation works in only a few limited(and ideal) cases.


There are several properties you can infer from the compound's name.

The compound is ionic, which the structure cation [...]-ate informs us. In this case, we're dealing with the potassium salt of peroxymonosulfuric acid. This should therefore dissolve pretty well in water.
The compound will be an oxidizer, in the conditions present in your bathtub. peroxy means the compound contains an $\ce{O-O}$ bond, which breaks up homolytically (one electron per oxygen) to give radicals (quite reactive species, electronically unhappy).

As per both hints, the compound is expected to be pretty effective at "eating" through organic deposits, by oxidizing them, rendering some soluble. It should also decrease pH, but not so drastically as it is already the conjugate base of sulfuric acid (which can let go of two protons).

About the bubbling: it most probably is $\ce{CO2}$ coming from the oxidation of the organic matter.


Why is the polarity of the B-N bonds in borazine the reason for its reduced delocalisation of π electrons?

Let us understand and try to answer your question with a few simple points.
First it is important to understand why exactly is borazine aromatic. Boron has an empty p-orbital and Nitrogen has a lone pair of electrons.
According to requirements for pi-backbonding, there needs to be a sigma bond between both the atoms and requirement of one empty orbital for allowing the electrons of the side atom to delocalize into it needs to be satisfied. Along with this, the lone pair which needs to be delocalized should be in kept in a pure p-orbital. Boron and Nitrogen both facilitate these requirements and thus the lone pair delocalization takes place.
Thus in a way, we could say that the aromaticity in the borazine moeity is due the pi-backdonation of unpaired electrons.

Understanding that the main reason is sideways pi-backdonation for the cyclic delocalization of electrons, one also needs to keep in mind that Boron and Nitrogen have a significant difference in their electronegativities. The polarity occurs due to the pi-backdonation but keeping in the mind the electronegativity of nitrogen is higher than boron, the delocalisation isn't as pronounced as in the case of benzene where the bonds are non polar. Thus leading to comparatively lesser delocalisation in broazine than in benzene. You could similarly compare h-BN and grpahite; in the former case the delocalisation is less pronounced due to difference in electronegativity than in the later case where a very well pronounced delocalisation is observed.


What does it mean by 'photons of higher intensity'?

The assertion is incorrect. Intensity is related to the number of photons emitted and has nothing to do with Bohr orbit transitions.

The reason is correct. Rather than higher intensity, photons of higher energy are emitted when an electron falls from the 2nd to the 1st Bohr orbit as compared to when it falls from the 4th to the 2nd orbit.  This is because the energy emitted or absorbed in a Bohr transition is given by the following equation where Z is the atomic number of the atom, R is a constant and n represents the quantum number of the final and initial Bohr orbits.

$\mathrm{\Delta E=-Z^2R \left ( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right )}$

since

$\mathrm{\frac{1}{1^2}-\frac{1}{2^2}=3/4}$

is greater than

$\mathrm{\frac{1}{2^2}-\frac{1}{4^2}=3/16}$

more energy is emitted in the first case than the second.


Why is the halogenation of an alcohol a multistep process?

If the chlorine left straight away I.e after step 1 in the mechanism you linked to then you would have three charges on the structure - charges on two oxygens and a charge on sulfur. Doesn't seem too stable to me.

Instead what happens after step 1 is a proton transfer. Proton transfers are generally rapid. This proton transfer is also thermodynamically favored as it relieves the oxygen of its positive formal charge.


Do I really need 4 different water treatment systems with this water analysis?

You shouldn't need 4. The only dangerous thing is the barium, and permeakra is right that you should keep an eye on the equipment so that it doesn't silently fail on you. You should check perhaps once a month. It's only ~3.5 times the EPA limit, so a failure won't be immediately be a hazard (unless the sequestered barium is released all at once, but that seems unlikely).

The other 3 problems and the hardness are, at those levels in the water, are simply nuisances. So they will discolor your water, produce a weak sulfurous odor, and eventually clog the pipes, but they will not harm anyone. If you fixed the iron and sulfur, that would probably improve the color, so only 3 systems are needed. And you don't have to worry about the system failing silently since you will see it.

Check these statements before making a decision, this is too important to trust only one source!


How to choose the correct volumes when preparing a dilution series?

To your questions:

  I do not understand where she got $50~\mu\mathrm L$ of dye and $50~\mu\mathrm L$ of $\ce{H2O}$ if it's $1\!:\!2$.

It's always problematic expressing dilutions as ratios, because it's ambiguous as to what the numbers are referring to.  Here, the "$1$" is referring to the volume of the solution to be diluted, and the "$2$" is referring to the final volume after dilution.  Thus, the $50~\mu\mathrm L$ ("$1\times$") is being diluted to $100~\mu\mathrm L$ ("$2\times$"), and in order to do that you need to add $100~\mu\mathrm L-50~\mu\mathrm L=50~\mu\mathrm L$ of the RO $\ce{H2O}$.  This could just as validly be expressed as a $1\!:\!1$ dilution; thus the ambiguity.

  I also don't understand where she got the $30~\mu\mathrm L$ from in the second step.

Per the same $\mathit{Initial~Volume : Final~Volume}$ ratio convention, a $1\!:\!100$ dilution to a final volume of $3~\mathrm{mL}=3000~\mu\mathrm L$ will require starting with $3000\div 100 = 30~\mu\mathrm L$ of the intermediate solution.

More generally:

I prefer to express dilutions using an "n-fold" notation, e.g.:

Dilute $50~\mu\mathrm L$ of dye solution $2$-fold with RO $\ce{H2O}$
Dilute $30~\mu\mathrm L$ of the solution from step ($1$) $100$-fold with RO $\ce{H2O}$

Though there is still room for ambiguity in this phrasing, to me diluting $n$-fold indicates much more clearly that the final volume $V_f$ should be $n\times$ the initial volume $V_i$, and thus that the volume of diluent needed is:

$$
V_d = V_f-V_i = nV_i-V_i = V_i\left(n-1\right)
$$


Can an excessive reactant amount result in a longer reaction time?

Let me first say that I don't know this specific reaction, so I am not sure whether my answer holds for it, but let me answer the general question: can a high concentration slow down the reaction rate?

The answer is, yes it can. The mechanism by which this happens is called inhibition and it is often (although not necessarily) connected to reactions on surfaces. I will explain inhibition in terms of a surface reaction here, but keep in mind that it can also happen in solution due to 'crowding', in which there is so much of 1 reactant that it basically 'cannot find' the other reactant.

For surface reactions the story is as follows. You will get a competition between adsorption on the surface and reaction on the surface.

Qualitatively, if you have a large amount of one of the two reactants than many of the surface sites will have this reactant adsorbed to it meaning that some of the reactant on the surface has neighbors that it cannot react with (because it is the same molecule), this can slow down the reaction rate.

Quantitatively, the reaction order will be concentration dependent. A famous example of this is Langmuir-Hinshelwood kinetics
$$r =k C_s^2 \frac{k_1 k_2 C_A C_B}{(1+k_1 C_A+k_2 C_B)^2} $$
where $C_s$ is the concentration of available adsorption sites and $C_A$ and $C_B$ are the concentrations of your reactants.

There are many possible ways in which this reaction rate equation can behave, depending on the values of the rate constants $k$, $k_1$ and $k_2$, but let me single out two cases that will explain a shift from increasing to decreasing reaction rate with increasing concentration:

a very high concentration of $C_A$ will result in $k_1C_A>>>1, k_2C_B$. This allows us to neglect the $1$ and $k_2C_B$ terms in the denominator, turning the rate equation into this:
$$r =k C_s^2 \frac{k_2 C_B}{k_1 C_A} $$
which indeed decreases with increasing concentration of $C_A$.
a very low concentration of $C_A$ will result in $k_1C_A<<<1, k_2C_B$. Then the rate equation becomes:
$$r =k C_s^2 \frac{k_1 k_2 C_A C_B}{(1+k_2 C_B)^2} $$
which has an increasing rate with increasing concentration of $C_A$

So what you see here is that a change of $C_A$ from low to high concentration can change the reaction order in $C_A$ and therefore indeed result in an initial increase of the reaction rate with increasing $C_A$, but later a decrease if you increase it even further. When this crossover occurs depends on the values of the rate constants ($k$, $k_1$ and $k_2$) and the concentration of the other reactant.


When drawing structures and taking into consideration formal charge, negative charges are assigned to the most electronegative atoms. Since the electronegativity of chlorine is less than the electronegativity of oxygen, it follows that the model with 2 double bounds is less favorable.


How to find number of moles of HNO3 behaving like acid in a redox reaction?

It is always easy to balance redox equations by considering two half reactions. The equation to be balanced is:
$$\ce{Sn + HNO3 -> Sn(NO3)2 + NH4NO3 + H2O}$$
Thus, the two relevant half reactions to this redox equation are:
$$\ce{Sn (s) -> Sn^2+ (aq)} \\
\ce{NO3- (aq) -> NH4+ (aq)}$$
First balance elements other than $\ce{O}$ and $\ce{H}$. In this case they are $\ce{Sn}$ in first equation and $\ce{N}$ in second equation. Then, balance $\ce{O}$ and $\ce{H}$ in both equations with water and $\ce{H+}$, respectively. This is because the reaction is in aqueous acidic medium:
$$\ce{ Sn (s)  -> Sn^2+ (aq)}\\
\ce{NO3- (aq) + 10 H+ (aq) -> NH4+ (aq) + 3 H2O (l)}$$
Finally, balance the positive charges by electrons so that each equation has no net charges:
$$\ce{ Sn (s)  -> Sn^2+ (aq) + 2e-} \tag1$$
$$\ce{NO3- (aq) + 10 H+ (aq) + 8 e- -> NH4+ (aq) + 3 H2O (l)} \tag2$$
These are your mass and charge balanced oxidation ($(1)$) and reduction ($(2)$) half-reactions. Now, you can add these two equations in order to cancel electrons. To do so add $4 \times (1) + (2)$:
$$\ce{ 4 Sn (s) + NO3- (aq) + 10 H+ (aq) -> 4 Sn^2+ (aq) + NH4+ (aq) + 3 H2O (l)} \tag3$$
This is your balanced ionic equation. As evidence, you need nine $\ce{NO3-}$ ions to balance as counter ions in right hand side, so you need to add $\ce{9NO3-}$ to both side to keep the mass and charge balance.:
$$\ce{ 4 Sn + NO3- + 10 H+ + 9 NO3- -> 4 Sn^2+ + NH4+ + 9 NO3- + 3 H2O} \tag4$$
When simplify, the equation $(4)$ would look like:
$$\ce{ 4 Sn +10 H+ + 10 NO3- -> 4 Sn(NO3)2 + NH4NO3 + 3 H2O} \tag5$$
Since $\ce{ +10 H+ + 10 NO3- = 10 HNO3}$, your complete balance equation is:
$$\ce{ 4 Sn +10 HNO3 -> 4 Sn(NO3)2 + NH4NO3 + 3 H2O}\tag5$$
Therefore, your answer is correct. Answer in your textbook must be a misprint or accidently for got that one $\ce{HNO3}$ is needed to be reduced. Other $\ce{9NO3-}$ is to balance the products (as counter ions). All $\ce{10H+}$ ions needed to complete the redox reaction.


This is a uranyl ion, one of the most common forms of dissolved uranium:

Because the two U=O bonds can form a single delocalised region, there's a strong tendency towards linearity (c.f. carbon dioxide). It tends to dissolve in water by gaining equatorial water ligands (4 or 5), some of which it may then hydrolyse, depending on the pH:

The axial oxygen atoms are also valid targets for hydrogen bonding, typically gaining one hydrogen bond each.

The difficulty in dissolving uranyl formate in particular suggests that the formate ion is quite well-bound, and it seems that in this case you get a combination of water and formate ligands in the complex around the uranyl. I was expecting the formate to be a bi-dentate ligand, but according to the results of a combined computational and experimental study 1, it's uni-dentate (which incidentally allows it to form polymeric chains in solution).

Here's a picture from that study showing a computationally obtained structure of a mixed water/formate-ligated uranyl structure.

2013, Dalton Trans., Christian Lucks, André Rossberg, Satoru Tsushima, Harald Foerstendorf, Karim Fahmya, Gert Bernharda, DOI:10.1039/C3DT51711J


Drying solvents while taking care of purity

There is actually a book on preparing laboratory chemicals you can look up:

  "Purification of Laboratory Chemicals" by WLF Armarego and CLL Chai

However, given that all the chemicals you want to dry are extremely hygroscopic, I'm not sure how long they would stay super dry for calibration purposes (and I'm not sure you need super dry solvents for calibration either). If you need it to last, you may need a large dry box.


What is the product formed in the reaction of dimedone and benzaldehyde?


Does baked bread have the same amount of calories as its ingredients?

Yes, the yeast does consume sugar to produce the bubbles in the dough, but I don't think it would be more than a few grams. During the baking of the bread, some water and alcohol evaporates and proteins are denatured.

Denaturing proteins does not change the calories. Since the alcohol that was produced is lost, then the calories are reduced. Water doesn't have any calories.

So, to answer to the question in the title: there are fewer calories in the baked bread than in the ingredients. Unfortunately, it cannot be predicted, the bread needs to be tested.


What is the difference between autocatalysis and self-replication in terms of chemical terminology?

The two terms represent very distinct concepts and are everything but interchangable.

Autocatalysis means that the product of a reaction catalyses the same reaction. But there is no focus on the catalysed reaction being exactly that or only that product catalysing it. As an example, take acidic ester hydrolysis e.g. with ethyl acetate:

$$\ce{CH3COOC2H5 + H2O ->[H+] CH3COOH + C2H5OH}\tag{1}$$

This is a reaction catalysed by acids that produces an acid in its course. This acid increases the solution’s overall acidity and thus catalyses the reaction that created it. But this feature is in on way specific to the acid we have generated. For one, we might as well add sulphuric acid to the reaction mixture to generate the same catalytic effect. For two, the acid produced in this reaction would also catalyse the hydrolysis of methyl formate. So neither is a specific structure required for catalysis nor is the catalysis restricted to itself.

Self-replication again means that a product is able to catalyse its own formation but only its own formation. The catalysis must be specific to a certain (set of) reactant to produce a certain (set of) product. Similar but different compounds bearing the same functional groups should not be able to effectively catalyse the reaction in question and the reaction product should not be able to catalyse any other reaction equally efficiently.

Typically, self-replication is achieved by having the product molecule act as a template which brings two reactive moieties closer together and allows for a more efficient reaction to itself. For example, you might consider the trimeric complex shown in the figure below:

Figure 1: Self-replicating system as described by
Terford and von Kiedrowski[1]

Here, the amine and the aldehyde shown on the right-hand half of the figure can condense to form an imine. This process can, of course, be catalysed by acids or bases. However, the condensation product shown on the left-hand half of the figure is able to form hydrogen bonds to the two compounds in such a selective way that the reactive moieties are brought close together. This eases the creation of the imine because the amino group is already in the perfect position of attack.

The imine product is therefore able to catalyse its own formation by templating but not the formation of closely related systems; and there is no other compound or functional group that would have an equally good catalytic effect on the reactants. This is not just autocatalysis but also self-replication.

Reference:

[1]: A. Terford, G. von Kiedrowski, Angew Chem. Int. Ed. 1992, 31, 654–656. DOI: 10.1002/anie.199206541.


Order of reactivity of carbonyl compounds to Nucleophilic addition reaction

My instinct is that there are probably too many variables here for a definitive and accurate answer to be given strictly from first principles. When comparing molecules in which there is only one key difference, especially when that difference leads to steric and electronic effects which are essentially additive, it's fairly easy to draw accurate conclusions. This question, however, is a bit thornier.

Comparing acetone, $\ce{(CH3)2CO}$, to acetaldehyde, $\ce{CH3COH}$, is quite straightforward: acetaldehyde should be the more electrophilic of the two. Firstly, the greater steric bulk of acetone's methyl group slightly hinders nucleophilic attack. Secondly, the additional methyl group is electron-donating by hyperconjugation, comparatively reducing acetone's electrophilicity. Here, both steric effects and electronic effects are synergistic.

The other comparisons are not quite as straightforward, as they require (in the absence of empirical data) qualitative judgment calls to be made.

Consider p-nitrobenzaldehyde. Sterically, reactivity should be reduced due to the obvious bulk of the aromatic ring. Electronically, the picture seems to me to be more complex. From the perspective of valence bond theory, the $sp^2$ carbons may be electron-withdrawing by induction, particularly in concert with the electronegative nitro group. On the other hand, the carbonyl group is also highly electronegative and should remove electron density from the ring, primarily via the $\pi$ system, and I would expect that effect to be greater. Finally, from the MO theory perspective, considering the frontier orbitals, conjugation of the carbonyl to the aromatic ring should result in lowering the energy of the $\pi^*$ LUMO, which ought to enhance reactivity. That, however, may be mitigated by any overall increase in electron density around the carbonyl group. By comparison to acetaldehyde, the steric effects are clearly unfavorable, while the electronic effects are slightly murky.

Finally, for the 4,4'-dinitrobenzophenone, the analysis of electronic effects is roughly the same as for p-nitrobenzaldehyde, but here sterics is undoubtedly going to be the overriding factor, and should result in massively diminished reactivity.

Overall, I would predict the reactivity order to be acetaldehyde >> p-nitrobenzaldehyde > acetone >> 4,4'-dinitrobenzophenone. I don't claim this is in any way authoritative, so I hope others weigh in as well. If I can find some relevant literature containing, e.g., relative rates, I'll update this answer with a link.


How and why can more volatile acids be made with sulfuric acid?

An acid according to the Brønsted-Lowry definition is any substance that can release $\ce{H+}$ ions. By having done so, it becomes a Brønsted-Lowry (conjugate) base which can accept $\ce{H+}$ ions. This is shown in equation $(1)$:

$$\begin{array}{ccccc}\ce{HA} & \ce{<=>} & \ce{H+} & + & \ce{A-}\\
\text{acid} & & \text{proton} & & \text{conjugate base}\end{array}\tag{1}$$

As already hinted, a Brønsted base is any substance that can accept such an incoming $\ce{H+}$ ion (reaction $(2)$)— as such, an acid will always react with a base and vice-versa; it is not possible for only an acidic reaction to happen.

$$\begin{array}{ccccc}\ce{B} & + & \ce{H+} & \ce{<=>} & \ce{HB+}\\
\text{base} & & \text{proton} & & \text{conjugate acid}\end{array}\tag{2}$$

Let me quickly quote part of the question:

  [M]ore volatile acids can be made using sulphuric acid and their corresponding salts

The corresponding salt of a more volatile acid could also be termed its conjugate base. So the reaction we are talking about includes the conjugate base of one acid and sulphuric acid as a second acid. To exemplify things, I will use hydrochloric acid as the first acid whose conjugate base is chloride. If we thus mix chloride ions (e.g. in sodium chloride) with sulphuric acid, the following equilibrium will result:

$$\ce{Cl- + H2SO4 <=> HCl + HSO4-}\tag{3}\\[1.2em]
K = \frac{[\ce{HCl}][\ce{HSO4-}]}{[\ce{Cl-}][\ce{H2SO4}]} = \frac{K_\mathrm{a} \left ( \ce{H2SO4} \right )}{K_\mathrm{a} \left ( \ce{HCl} \right )}$$

The equilibrium constant $K$ is the ratio of the acid constants of hydrochloric and sulphuric acid. Since hydrochloric acid ($\mathrm{p}K_\mathrm{a} \approx -8$) is more acidic than sulphuric acid ($\mathrm{p}K_\mathrm{a} \approx -3$), we expect the equilibrium to be shifted towards the reactant side — which would be true for a closed, pressurised system. Once we open the system, and allow for gases to be exchanged, the picture changes. Compare equation $(\text{3'})$, wherein the aggregate states of the reaction partners are mentioned.

$$\ce{Cl- (s) + H2SO4 (l) <=> HCl (g) ^ + HSO4- (s)}\tag{3'}$$

Hydrochloric acid is a gas and can diffuse away while sulphuric acid is a liquid and thus cannot diffuse away. (And cannot drip away since we are using a proper beaker of some kind.) If we constantly remove $\ce{HCl}$ from the reaction mixture by letting it diffuse away, we are exercising pressure onto the system, and per Le Châtelier’s principle the product side (one of whose species we are removing) will now be favoured.

This general principle applies to any acid, as long as it is more volatile than sulphuric acid, i.e. has a lower boiling point under the pressure used. If the second acid happens to be liquid rather than gaseous, a simple distillation will suffice to remove it from the reaction mixture.


The only time I've ever seen that second notation is when referring to the substitution of the Helmholtz energy, as A = U-TS and therefore A = -kTlnQ. If you want an in depth explanation of how the thing is derived I suggest looking at Chapter 10 of Biomolecular Thermodynamics: From Theory to Application by Douglas E. Barrick.

Here are some photos of the key parts:


What is the meaning of the prefix "per-"?

Per- can mean a number of different things, depending on the context.
It can mean that there is a peroxi bond(oxygen-oxygen bond) in the system, persulfate is a good example of this.

More generally it tends to mean that the compound is the most completely altered product of a certain(possibly hypothetical) reaction. For example perchlorates and permanganates are the most completely oxygenated compounds of chlorine and manganese, respectively. This also describes another common use of per-, perhalogenated compounds. In that context, it usually means that all hydrogens present in the parent compound, that were bonded to a carbon, have been replaced with a halogen, for example perfluoroethylene (C2F4, also known as tetrafluoroethylene, the monomer of Teflon(PTFE))


I would solve this question rather simply.

Let's consider the general equation $\ce{ X-Y_{(s)} \rightarrow X^+_{(aq)} + Y^-_{(aq)}}$ with a $\Delta H = C$. Intuitively, if the reaction is endothermic ($\Delta H > 0$), an increase in temperature should help the dissolution (more heat to be absorbed by the salt while dissolving). You can rationalize this further using the Le Châtelier's principle:

$\ce{ X-Y_{(s)} + C \rightarrow X^+_{(aq)} + Y^-_{(aq)}}$

If you add more heat ($C$), the dissolution will be more favored.

More rigorously, you can use Van't Hoff equation (assuming a dissolution equilibrium): $\frac{d(ln(K))}{dT} = \frac{\Delta H}{RT^2}$. If $\Delta H > 0$, then $ln(K)$ increases with $T$, i.e. dissolution is more favored.


It has been already mentioned in the comments and the other answer, that fluorine is more electronegative than oxygen. Therefore it is highly unlikely that this compound exists.

To illustrate the problem at hand, take a look at similar, known compound with a fluorine oxygen bond: hypofluorous acid, $\ce{HFO}$, which is actually better described as hydroxylfluoride, $\ce{HOF}$. According to Holleman and Wiberg's[1,2] textbook the synthesis of this compound is already difficult. Low pressure fluorine gas is passed over ice at about $-40~^\circ\mathrm{C}$. Obtained is a mixture of $\ce{HF}$, $\ce{O2}$, $\ce{H2O2}$, $\ce{H2O}$ and also $\ce{HOF}$. Fractional freezing until about $-183~^\circ\mathrm{C}$ leads to a white substance, that turns pale yellow at about $-117~^\circ\mathrm{C}$, and boils at about $10-20~^\circ\mathrm{C}$. It decomposes via a radical pathway at about $25~^\circ\mathrm{C}$ and $100~\mathrm{mbar}$ with a half life of about 30 minutes to oxygen and hydrogen fluoride: $$\ce{2HOF -> 2HF + O2}.$$

The electronic structure of the molecule is most likely $$\ce{HO^{\delta+}\bond{-}F^{\delta-}}.$$

Salts, $\ce{M^{+}OF^{-}}$, that are derived from this compound are not known. Covalent compounds, however, are known, i.e. $\ce{O3ClOF}$ or $\ce{O2NOF}$.

It is easy to imagine, that a compound with an even higher oxygen load cannot be stable. Therefore we can conclude that a halogenate like $\ce{HFO3}$, let alone its salt $\ce{M^{+}[FO3]^{-}}$, cannot exist.

Notes and References

Unfortunately I only have access to the German version of this textbook. I tried to excerpt and translate the important parts to the best of my knowledge.
Holleman, A., Wiberg, N., Wiberg, E., et al. (2008). Lehrbuch der Anorganischen Chemie. Berlin, Boston: De Gruyter. Kapitel XII 4.2. Sauerstoffsäure des Fluors. S. 465.  (in german)
Translation of the 101st edition available as Egon Wiberg and Nils Wiberg (2001) Holleman, Wiberg. Inorganic Chemistry. Academic Press.


How many electrons are in the d orbitals of NiCl2·6H2O?

The hexahydrate has octahedral symmetry, and it is very probable that this is the case for the solid and dissolved state of the complex.

The compound will not separate into the ions, because the solvation of three ions vs the solvation of one neutral complex is higher in energy, especially considering the fact that opposite charges attract. So in solution state it is a $\ce{[NiCl2(OH2)4]}$ complex with octahedral symmetry.

The $d$-electron count is 8, as it starts with 10 and has a formal oxidation state of II. This is always the case for $\ce{Ni^{II}}$ in any medium but vacuum.

Whether it's a high-spin or low-spin complex depends on the splitting energy of the $e_\text{g}$ and $t_\text{2g}$ molecular orbitals that arise from the $d$ orbitals in combination with the ligand group orbitals, colloquially known as $\Delta_\text{o}$ or $\Delta_\text{oct}$. Have a look at this Wikipedia article on crystal field theory for more details.

In short, there are several factors that influence the magnitude of the splitting:

Oxidation state of central atom
Geometry
Ligand types

The octahedral geometry usually gives good energy splitting.$^\text{[citation needed]}$ In conjunction with a moderate oxidation state compared to Ni(0), and four mid-range splitting ligands (water), it is a safe bet to say that we will have a low-spin octahedral $d^8$ complex: Three pairs of electrons in the $t_\text{2g}$ orbitals and two unpaired electrons in the $e_\text{g}$ orbitals, that is.

Note at this point that the terms high-spin and low-spin are superfluous. The electrons in the $e_\text{g}$ orbitals will always be in a triplet configuration according to Hund's Rule. So it will always be in a "high-spin" configuration from an experimental standpoint.


How is this sugar classified as an aldose?

Yes its an aldose it is classified as an aldose because of the aldehyde group (cho). The compound is actually an aldotetrose called L-Threose. Try using acd Chem sketch program for the projection it is a free download for the freeware version...


HCl + NaOH: Why H prefers to move to O instead of staying with Cl?

Because hydroxide is a stronger base than chloride. Chloride is a much larger ion, and thus its negative charge is spread over more volume, stabilizing it. The negative charge on the oxygen in hydroxide is spread out over less volume, leaving it more reactive. Thus, when a $\ce{H^+}$ ion is available, it will preferentially be retained by the hydroxide rather than the chloride.


Is there way to detect surface area of nanoparticle by electrochemical technique?

Dynamic light scattering is commonly used to measure nanoparticle size: http://en.wikipedia.org/wiki/Dynamic_light_scattering

Whether you can measure this sort of thing electrochemically depends a lot on the type of nanoparticle. Conductive nanoparticles, you might be able to estimate the surface area by measuring adsorptive behaviours of various things, like CO for Pt and Pd: http://www.sciencedirect.com/science/article/pii/S1388248113000933 Non-conductive particles might be measured by how they impede diffusion to a surface. Electrochemical techniques could suffer from problems related to the shapes of the nanoparticles, as well.

The most straightforward ways of examining the surface features of nanoparticles would be electron microscopy or AFM.

Without further details of your application, it's difficult to recommend a specific technique. If the nanoparticles are being used as an electrode, the apparent electrochemical surface area may be all you care about, even though this is not necessarily the same as the geometric surface area.


Resemblance of molecular orbital formed by mixing of s and p orbitals to hybridised sp orbital

It's important to realize that both of these drawings are highly stylized renderings of the electron density that are intended to provide a qualitative understanding. They don't accurately represent the actual shape of the orbitals. For example, $p$ orbitals are much rounder and taller relative to their length than we usually draw them. They are drawn elongated so that they don't cover each other up if you are drawing more than one on an atom. So the fact that the drawings of the $sp$ hybrid and the $p+s$ $\sigma$ bond look similar should only be taken to mean that they both have a bias of density to one side. You can't really conclude much else from the drawings.

The Orbitron website is a great resource for seeing the actual shape of the electron density of different types of atomic orbitals, which you can compare to the drawings.

MolCalc produces similar images of molecular-level orbitals. The bonding orbital of HF is probably the closest you can get to the $p+s$ $\sigma$ bond orbital in your drawing.


Is there an example of a bond with negative bond enthalpy?

Bond enthalpy is like a measure that tells us how strong/stable a particular bond is.
Definition:

Bond enthalpy (also known as bond energy) is defined as the amount of energy required to break one mole of the stated bond.

To break a bond, energy must be provided to the bond. How is that energy provided?

The reactants are either heated or they take up energy from the surroundings, that is, they take in energy in the form of heat from the surroundings, overcome their bond enthalpies and break the bond.
All bonds require energy to be broken.
A negative Bond enthalpy simply means that the bond is highly unstable and would not exist in the first place.
I highly doubt if any compound has negative bond enthalpy.{Not even Dimeric Helium}
From Wikipedia:

Based on molecular orbital theory, $\ce{He2}$ should not exist, and a chemical bond cannot form between the atoms. However, the van der Waals force exists between helium atoms as shown by the existence of liquid helium, and at a certain range of distances between atoms the attraction exceeds the repulsion. So a molecule composed of two helium atoms bound by the van der Waals force can exist.The existence of this molecule was proposed as early as 1930.


Why basicity of oxides increases down a group?

As suggested in comments, major factors play a part

Atomic size

As the atomic size of the element increases, its electron affinity decreases consequently. It turns out that when moving vertically in the periodic table, the size of the atom outdoes its electronegativity with regard to basicity. The atomic radius of gallium, for example is approximately twice that of boron, so in a gallium ion, the charge is spread out over a significantly larger volume.

Electrostatic charges, whether positive or negative, are more stable when they are ‘spread out’ than when they are confined to one location.

Electron affinity

The ability of an atom “to accept an electron” decreases down the group therefore increases the tendency of the elements to be good Lewis bases (recall a Lewis base is an electron-pair donor). Thus the decrease in electron affinity down the group means the elements readily donate an electron rather than accept them.

Electronegativity

The more electronegative an atom, the better it is able to bear a negative charge. Down the group, the electronegativity of the elements decrease down the group.  B is very electronegative and doesn’t readily release electrons but instead holds on to it, additionally is able to stabilise a negative charge, and consequently has good acidic properties unlike proceeding elements down the group.


Numbering heteroatoms and groups in benzene derivatives

Quaternary carbon atoms i.e. Those with no hydrogen atoms attached and in these examples, the ones that are common to both or all rings, are not counted. Try it with 1,8-dihydroxynaphthalene. Both OH groups should be pointing in the same direction.

In your examples, as you probably know, the heteroatom is numbered first.


Why doesn't the bromide anion attack during electrophilic addition of bromine in methanol solvent?

There's a very simple explanation for this. Your reasoning regarding $\ce{Br-}$ being a better nucleophile is absolutely correct. The reason as to why methanol attacks instead of $\ce{Br-}$ is because it is the $\mathsf{\color{red}{solvent}}$. The solvent concentration is simply too high. Only one $\ce{Br-}$ is generated.

More or less the answer is this paragraph from J. Clayden - section "Unsymmetrical bromonium ions open regioselectively" in chapter 20 "Electrophilic addition to alkenes" :

  ...when a bromination is done in a nucleophilic solvent—water or methanol, for example—solvent molecules compete with the bromide to open the bromonium ion. As
  you know, alcohols are much worse nucleophiles than bromide but, because the concentration of solvent is so high (remember—the concentration of water in water is $\pu{55M}$), the solvent gets there first most of the time. This is what happens when isobutene is treated with bromine in methanol. An ether is formed by attack of methanol only at the more substituted end of the bromonium ion.

Note: Concentration of pure methanol is around $\pu{24.75 M}$, calculated from the density given on ChemSpider


How can I separate the vinegar from a 5% white vinegar to make a more concentrated solution.

If you have access to a still you can distil the vinegar solution, however this can only go so far. You may find this section of the wikipedia article on azeotropy edifying - to summarise, adding ethyl acetate to a vinegar solution generates a water:ethyl acetate azetrope with a depressed boiling point that makes distillation more effective, however this may be beyond the pale for a home experiment and would obviously render the vinegar non-food-safe.


In this simple redox reaction, why does the Hydrogen oxidize the calcium metal?

This is a simple example of the reactivity series. More reactive metals will displace less reactive metal ions, and they switch places, with the more reactive metal becoming the ion and and the less reactive ion becoming a metal. This can be very cool in many situations: for example, copper metal can displace a silver salt to plate silver metal and produce a blue copper salt. See this video! In any case, hydrogen is consider a reference "metal" in the reactivity series, and it is less reactive than calcium, so the displacement reaction you cite occurs.


How is hydration free energy of a solute measured experimentally?

Your interpretation is correct. Theoretically, you will need only an aqueous solution and a fixed volume of gas, and then be able to measure equilibrium concentrations. Although, you may not observe very much vapor pressure without using elevated temperatures (as you noted, the zwitterion is very soluble).

Derivation

When a species $x$ is in equilibrium between two states, $$x_{state1} \rightleftharpoons x_{state2}$$ the equilibrium constant is given by the ratio of concentration of $x$ in each state: $$K = \frac {[x_{state2}]} {[x_{state1}]}.$$ The equilibrium constant relates to Gibbs free energy by the following equation: $$\Delta G = -RT \ln(K)=-RT\ln\left(\frac{x_{state2}}{x_{state1}}\right)=RT\ln\left(\frac{x_{state1}}{x_{state2}}\right).$$ Substituting for an aqueous/vapor phase system yields the cited equation. The literature uses Boltzmann's constant, $k$, in place of the gas constant $R$ because they are using units of molecules instead of moles. (The extra $kT$ term within the logarithm comes from the conversion of pressure to number density concentration using the ideal gas law: $\frac{n}{V}=\frac{P}{RT}$, again using molecular units instead of moles will require $k$ instead of $R$.)

Measurement

Determining the "best" method for gas phase measurement depends on what instrumentation you have available to you and how accurate you need/want to your results. Mass spectrometry would be ideal if it employs a soft ionization technique. If the molecular ion doesn't fraction, you could use the raw counts compared to an internal standard to calculate the concentration.

You don't necessarily need to measure the vapor pressure directly, either. If you have a method for measuring the aqueous concentration, you could calculate the vapor pressure based on decrease in solution concentration: $$\Delta G=RT \ln \left(\frac{[arg_{aq}]_i-[arg_{aq}]_f}{[arg_{aq}]_f}\right)=RT \ln \left(\frac{[arg_{aq}]_i}{[arg_{aq}]_f}-1\right)$$ where $[arg_{aq}]_i$ is the initial aqueous arginine concentration (total arginine) and $[arg_{aq}]_f$ is the final concentration (equilibrium amount).  HPLC (or even UV-Vis if that is all you have available) would be a good method, but you would have to either make a calibration curve or use an internal standard (your buffer may work).

Notes

If you have access to J. Phys. Chem. and/or Biochemistry, this article cites articles (ref. 29, 82) that have experimental data for free energies of hydration for amino acids (all but arginine). They may be helpful to consult for methods.


Why is Standard heat of reaction equal to difference between standard heats of formation of products and reactants?

It is the consequence of the Hess's law and that is the consequence of the energy conservation law. Overall it says the enthalpy change does not depend on the path, but only on the start and final point.
Comparing to the real reaction path, you can formally get there also

via the path of conversion reactants to pure elements in their standard form
and then via forming products from those elements.

If the result were not the same, you could create perpetuum mobile switching both states via 2 different paths, gaining the nonzero net energy.


Using IR to detect / quantify a molecule at low concentrations

There isn't enough information to answer the question with certainty.  We'd need an IR spectrum of the culture medium (possibly dried/lyophilized medium)

I suspect that the peak at 3000 $\mathrm{cm}^{-1}$ is an alkyl CH stretch.  You're going to see similar absorption peaks in lipids and proteins, so I don't think that particular peak is going to be too promising for you.

You might look into headspace mass spectrometry.  Many terpenes are volatile and can be analyzed directly from headspace air (i.e. the air directly above your cultures).  You don't necessarily need to do gas chromatography, so the method can be quite fast.  It won't be as diagnostic as to the identity of the particular terpenes formed, however.


How to measure the electrical conductivity of a substance?

A lot of work was done in the 19th and 20th century on determining the conductivity of aqueous solutions. It was a huge field called conductometry (check Wikipedia). You experimentally measure the resistance of a solution by placing the desired liquid between two inert electrodes and pass high frequency alternating current to prevent electrolysis. In solutions, electrical current is carried by ions. If your solution has ions or if it is ionizable, it will conduct electricity. Now you can determine which solution has ions based on ionization, e.g. you can write $\ce{HCl (aq) -> H+ + Cl-}$ and can predict that it must be a conductive solution. In fact, the aqueous $\ce{H+}$ has the highest conductivity known in water. $\ce{OH-}$ is next.


Given that the fumes are scattering droplets, I would say that polar HCl acts as centre of nucleation for water vapour.

This is about the same way by which nitric acid containing small amount of nitrogen dioxide fumes white (in this case first the red-brownish oxide reacts with moisture to form the colorless acid, the latter in turn works as nucleus for water condensation and acidic droplets formation, ie white fume).

A more detailed or perhaps better answer should be under paywall:

Chang, O.-K. Why does concentrated HCl fume? J. Chem. Educ. 1985, 62 (5), 395. DOI: 10.1021/ed062p395.


I'll give you the structures. Then you'll understand everything:

Resonance structures of $\ce{SO3^{2-}}$ ion:

Structure of Sulfur Trioxide:

The absence of a double bond between two oxygen atoms in sulfite ion shows why it is bivalent.


Question about the setting of DFT (Density Function Theory)-NEB (Nudged Elastic Band) calculation

There is not a general rule for choosing the right displacement between images. For very big systems, I sometimes launch two NEB calculations: one with shallow parameters (eg: steps of 0.5 angstrom on the reaction coordinate), and later I use the almost converged geometries near the TS for another NEB calculation, with finer (less than 0.1 angstrom) steps. Of course the nature of the system might help you choose: sometimes things behave funny, and you have to choose a strategy for each system. Anticipating your last question, freezing or constraining sometime helps: while it doesn't improve convergence, it can speed up the calculation (by telling the program that some movements are forbidden, so that the NEB algorithm doesn't need to try them before discarding them). The intermediate image should not be mandatory for a NEB calculation (if not given, it is generated by interpolation and optimized at every cycle), but it can speed up a calculation, if provided. It can be taken, for instance, from a geometrical scan, or from an approximate TS geometry.
As for the core number, if you are talking about processors, then generally speaking as much as possibile is a good choice (edit: provided that the software you are using is capable of parallelizing efficiently the slowest step of the calculation: your choice could be the same you use for geometrical optimizations at the same level of theory)


It's been so long you probably won't even see this, but in case someone else is searching for an answer, here are some interesting answers I've found: (The websites are in French, so I tried to translate the intersting bits)

  Igneous rock samples can be melted in the laboratory by heating them at elevated temperatures (1000 ° C to 1300 ° C). If one operates under pressure and in the presence of volatile elements, the melting temperature of the rock and its minerals is much lower (650 ° C to 950 ° C). It is known that magmas contain varying amounts of volatile elements that lower the melting point of many silicates, thus allowing the magma to remain liquid even at temperatures below those required to melt rocks in the laboratory. Indeed, the temperatures that exist in the different levels of the earth's crust are not sufficient to melt rocks in the absence of volatile elements.
  The main volatile elements found in magmas can be commonly observed in the gaseous state at the vents of active volcanoes. It is essentially water vapor, carbon dioxide, hydrogen, chlorine, fluorine (in the form of acids) and many other gases [...].

Taken from http://www.fossiliraptor.be/rochesetphenomenesignes4.htm

  In the case of hydrated paragneiss, during the warming of this rock, the melting begins as soon as it reaches the first reaction ((4) Qz + Ab + Gold + H2O ⇄ Liquid), around 650 ° C. Indeed, this gneiss contains all the reagents necessary for this reaction to take place.
  For a dry paragneiss, on the other hand, nothing happens on the curve corresponding to reaction 4; indeed, one of the reagents, water, is lacking. However, when the temperature exceeds 850 ° C, this rock crosses the curve corresponding to reaction 5 ((5) Bt + Ab + Qz + AlS ⇄ Liquid + Cord / Gt + KF, which can be decomposed as follows: (6) Bt + AlS ⇄ Gt / Cord + KF + H2O, then 7) Qz + Ab (Pg) + Gold (KF) + H2O ⇄ Liquid) and the two coupled reactions described above take place: the biotite decomposes to liberate water; therefore, we are in a situation where all the necessary products are present: the merger can take place.
  In other words, a hydrated system will melt (cross its solidus) at a lower temperature than a system without free water: the two reactions correspond respectively to "hydrate solidus" and "dry solidus"!

Taken from: https://planet-terre.ens-lyon.fr/article/reactions-de-fusion.xml#equation4


Quantitative Analysis of Urea in aq. solution

In clinical chemistry, urea is quantitativly determined as follows:

Hydrolyze urea with urease to ammonia and carbon dioxide
react the ammonia with 2-oxo-glutarate in the presence of glutamate dehydrogenase and $\ce{NADH}$: $\ce{NADH}$ is converted to $\ce{NAD+}$
monitor the decay of the UV absorption of $\ce{NADH}$ at $\lambda$= 340 and 410 nm


Effect of different electron withdrawing/transition state stabilising groups in nucleophillic aromatic substitution

Your reasoning is correct.  The relative reaction rates would probably be quite similar if we didn't have the methyl groups on both sides of the substituent X.

In order for the substituent to stabilize the transition state through resonance, the substituent must be coplanar with the aromatic ring.  This isn't a problem for the linear $\ce{CN}$ group, but the $\ce{NO2}$ group with the two oxygens protruding out will not be able to remain planar with the aromatic ring.  The two adjacent methyl groups create steric interference with the planar nitro group; this steric interaction is minimized when the nitro group is twisted out of the aromatic plane.  When the nitro group is twisted out of plane it cannot resonance stabilize the transition state.  Hence the nitro compound reacts more slowly than the cyano compound.

While the underlying reaction is different, the thinking behind this question is identical to that found in this question from yesterday.


Does computational chemistry get the molecular orbitals of dioxygen wrong?

A triplet species, as your MO scheme shows, has two unpaired alpha spin electrons, which is why you don't have to look at the beta orbitals at all but you have to look, if the $\alpha$-HOMO and $\alpha$-HOMO-1 are equivalent or not.

After optimizing the geometry of triplet oxygen using the method you states, B3LYP 6-311G**, we end up with a bond length of $\pu{1.2058 Å}$ and are given the following energies:

 **********************************************************************

            Population analysis using the SCF density.

 **********************************************************************

 Orbital symmetries:
 Alpha Orbitals:
       Occupied  (SGG) (SGU) (SGG) (SGU) (PIU) (PIU) (SGG) (PIG)
                 (PIG)
       Virtual   (SGU) (PIU) (PIU) (SGU) (SGG) (SGG) (PIG) (PIG)
                 (SGU) (PIU) (PIU) (DLTG) (DLTG) (DLTU) (DLTU)
                 (SGG) (PIG) (PIG) (SGU) (PIU) (PIU) (SGG) (PIG)
                 (PIG) (SGU) (SGG) (SGU)
 Beta  Orbitals:
       Occupied  (SGG) (SGU) (SGG) (SGU) (SGG) (PIU) (PIU)
       Virtual   (PIG) (PIG) (SGU) (PIU) (PIU) (SGU) (SGG) (SGG)
                 (PIG) (PIG) (SGU) (PIU) (PIU) (DLTG) (DLTG) (SGG)
                 (DLTU) (DLTU) (PIG) (PIG) (SGU) (PIU) (PIU) (SGG)
                 (PIG) (PIG) (SGU) (SGG) (SGU)
 The electronic state is 3-SGG.
 Alpha  occ. eigenvalues --  -19.28009 -19.27992  -1.31462  -0.83929  -0.56862
 Alpha  occ. eigenvalues --   -0.56862  -0.55311  -0.31192  -0.31192   <------
 Alpha virt. eigenvalues --    0.19733   0.66242   0.66242   0.69937   0.71138
 Alpha virt. eigenvalues --    0.77370   0.78062   0.78062   1.23343   2.23120
 Alpha virt. eigenvalues --    2.23120   2.53472   2.53472   2.80188   2.80188
 Alpha virt. eigenvalues --    2.83167   3.40715   3.40715   3.75295   4.50288
 Alpha virt. eigenvalues --    4.50288   4.64528   4.80485   4.80485   5.97229
 Alpha virt. eigenvalues --   49.48209  49.67839
  Beta  occ. eigenvalues --  -19.24954 -19.24923  -1.25730  -0.75086  -0.51327
  Beta  occ. eigenvalues --   -0.46530  -0.46530
  Beta virt. eigenvalues --   -0.11708  -0.11708   0.24099   0.70896   0.70896
  Beta virt. eigenvalues --    0.71935   0.72389   0.79954   0.83108   0.83108
  Beta virt. eigenvalues --    1.27043   2.27711   2.27711   2.62147   2.62147
  Beta virt. eigenvalues --    2.87800   2.89802   2.89802   3.46348   3.46348
  Beta virt. eigenvalues --    3.78751   4.58054   4.58054   4.66543   4.88322
  Beta virt. eigenvalues --    4.88322   6.00448  49.51237  49.70844

You can clearly see, that the $\alpha$-HOMO and $\alpha$-HOMO-1 are degenerate with both an energy of $\pu{-0.31192 Eh}$. If you look at both MOs you can easily see why, because they are qualitatively the same combination of 2p AOs but with different orientation.

As symmetry was turned on, I guess, this needs to be the case, but the result stays the same, if it is turned off.

So, does computational chemistry get the molecular orbitals of dioxygen wrong? No, the MOs are right.

But actually you have a different question: Why are the energies of alpha and beta MOs in unrestricted open shell calculations not identical?

There are two methods to treat open shell systems. One is restricted open (RO*) and the other is unrestricted (U*). To not expand it too much, the main difference is, that as restricted closed shell methods, the restricted open shell methods force the alpha and beta orbitals to share the same MOs. Then, of course, they also would have the same energies.

But the energies that you show are from an unsrestricted open shell calculation. If you use this method, alpha and beta MOs are not forced to be in the same spatial MOs. And this freedom leads to different energies alpha and beta MOs.


Isobaric entropy relation from isothermal Gibbs free energy relation

For an ideal gas at constant temperature $\Delta H=0$, so $\Delta G=-\Delta (TS)$, and, at constant temperature, $$\Delta G=-T\Delta S$$


Bending of water and methanol towards a charged objects

As you state, in an electric field, a polar molecule such as methanol or water, having an uneven charge distribution, is attracted (or repelled) by an electric field. This orients the molecule, but is not primarily responsible for the attraction of the bulk substance.

An electric field causes a force on even a nonpolar molecule such as methane or sulfur, though. You can pick up powdered sulfur with a charged piece of plastic because of electrostatic induction, which moves electrons in the bulk material.

This induced charge is similar to the effect of moon (and sun) on tides, pulling the water underneath the moon more strongly than it pulls on the more distant core of the earth, and on the core more strongly than on the ocean on the opposite side of the planet, causing uneven distribution of the water. In a similar way, an external field causes an uneven distribution of charge.


What kind of solvents favor elimination reaction?

This website gives good rules of thumb: http://www.masterorganicchemistry.com/2012/12/04/deciding-sn1sn2e1e2-the-solvent/

And so does this website: http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch08substitutionvselimination.html

In addition, according to page 583 of "Physical Organic Chemistry" by Anslyn and Dougherty:

  Almost all bases are also nucleophiles, and hence we expect competition between elimination and substitutions. In both $\ce{S_N2}$ and $\ce{E\text{2}}$ reactions, the nucleophile or base reacts in a single rate-determining step with the reactant. In both $\ce{S_N1}$ and $\ce{E\text{1}}$ reactions, the nucleophile or base reacts in a step after the rate-determining heterolysis...

  An $\ce{E\text{2}}$ reaction requires the addition of a base, and can be performed in solvents of low or high ionizing power. Since most strong bases are also good nucleophiles, $\ce{S_N2}$ reactions are important competitors. However, elimination will dominate if the carbon with the leaving group is not susceptible to nucleophilic attack, such as a tertiary R group. Examples of this trends are given in Table 10.6.

  $\ce{E\text{1}}$  reactions involve carbenium ion intermediates, and therefore are facilitated by all the factors that stabilize carbenium ions. These are the same factors that facilitate $\ce{S_N1}$ reactions. Strongly ionizing solvents and substitution of electron donating groups on the carbon undergoing heterolysis are necessary.

  In highly ionizing solvents and with R groups that readily form carbenium ions, the ratio of substitution to elimination is typically independent of the leaving group. This evidence supports the notion that the substitution and elimination products are formed b branching from a common intermediate, and therefore the two reactions share a common rate-determining step. In contrast, in solvents of lower ionizing power, the ratio of substitution to elimination products does depend on leaving group, and indication that the two reactions do not share a common intermediate. Instead, contact ion pairs are formed, and either the leaving group or the solvent can remove the proton to give elimination products. The formation of a contact ion pair also has an influence on the stereochemistry of the elimination.

Table 10.6

Reactant            Solvent   Base      Elimination
Tert-Butyl Bromide  Ethanol   NaOEt     100.00%
Tert-Butyl Bromide  Acetone   NBu4Cl    96.00%
isopropyl Bromide   Ethanol   NaOEt     75.00%
isopropyl Bromide   Acetone   NBu4Cl    0.00%
Ethyl Bromide       Ethanol   NaOEt     9.00%
Ethyl Bromide       Acetone   NBu4Cl    0.00%


How can a Hydrogen make such a difference?

The difference in structure of these two compounds is  perhaps the presence of a more conjugated structure in biliverdin compared to bilirubin. These products (degradation products of haeme1), just as most pigments, make use of the conjugated electron systems to absorb visible light, giving rise to strong colors. The long chain conjugated system in bilverdin leads to its strong green color, the reduced biliribin (methenyl bridge) has a less conjugated system with a different absorption consenquently different color.

This is a case of effects of conjugation in pigments than merely a difference in functional groups as I see it.

1Note The degradation of haeme occurs at  the end of the red cells’  lifetime, where they are removed from the circulation and their components degraded, thus depends of many factors most importantly regulations carried out in the body, enzymes etc.

References

Voet and Voet Biochemistry Section 26-4. Amino Acids as Biosynthetic Precursors.
Conjugated system


how can carbon dating even work? Its at atomic level!

You are correct in the fact carbon-14 does not care whether or not a life form is alive; however, this is irrelevant in the case of carbon dating. You may be familiar with the carbon cycle, which is, in a nutshell, the recycling of carbon through carbon dioxide. Plants are, of course, constantly using carbon dioxide from the atmosphere to generate large bio molecules that we can eat. We are also constantly recycling the carbon, ingesting it in the form of food, and exhaling it in the form of carbon dioxide, so it is clear that living organisms are constantly replacing their "old" carbon from carbon from atmospheric carbon dioxide. The key factor to carbon dating is that atmospheric carbon dioxide has a higher percentage of Carbon-14. This is caused by high energy cosmic rays which convert nitrogen-14 (the most abundant isotope of nitrogen) into carbon-14 nuclei, which in turn result in a greater percentage of our carbon content being carbon-14. When we die, we are no longer replenishing our carbon from the atmosphere (higher carbon-14 content due to cosmic rays), so the carbon-14 will gradually decay into more stable nuclei, leaving a lower percentage of carbon-14 in our remains. By using the half-life of carbon-14 nuclei, we are able to calculate the time when a certain living organism was no longer exposed to our atmosphere's high concentration of carbon-14. As a side note, carbon-14 dating is fairly accurate because the relative concentration of carbon-14 has remained constant in our atmosphere.


Why does spontaneity of a reaction depend on temperature?

The issue here is in your definition of entropy. According to your definition:
$$\mathrm dS = \frac{\text{đ}q}{T}.\tag{1}$$
However, there is a small difference.
As you may know, the entropy of a system is a state function and so it only depends on the initial and final states. Therefore, if the entropy of the system were to be calculated for a specific path for which the parameters are simple to find, then the value would not change irrespective of the path.
The actual formula of change in entropy is:
$$\mathrm dS = \frac{\text{đ}q_\mathrm{rev}}{T}.\tag{2}$$
This value is calculated for a reversible isothermal process.
Now, let us assume that we have taken such an isothermal reversible process and discovered the value of $\Delta S$ and now used an isobaric process to discover the value of $\Delta H.$ Now, we solve for $\Delta G$ in our isothermal process (this is done so that we can assume constant temperature in our formula).
Therefore, for such a process
$$q_\mathrm{rev} = nRT\ln\frac{V_\ce{B}}{V_\ce{A}}.\tag{3}$$
Thus, we get change in entropy as
$$\Delta S = nR\ln\frac{V_\ce{B}}{V_\ce{A}}.\tag{4}$$
Now, enthalpy would be a constant value that we discerned from the isobaric process.
Therefore, since $\Delta G = \Delta H - T\Delta S,$ we get
$$\Delta G = \Delta H - nRT\ln\frac{V_\ce{B}}{V_\ce{A}}.\tag{5}$$
As you can see, there is still a temperature dependence.
This is because $\Delta S$  is temperature independent and so you would still have a temperature dependence for $\Delta G.$
Another thing that you may have forgotten to take into consideration is the fact that $q$ is dependent on the temperature.
The above example is that for an ideal gas expansion.
As Poutnik stated in the comments in case of a reaction, the calculation of $\Delta H$ becomes more complicated as we have to take into consideration the different molar capacities of the reagents and products which would $\Delta H$ would also be a function of temperature.
For standard value of entropy, we use the formula
$$\Delta S^\circ_\mathrm{rxn} = \sum nS^\circ_\mathrm{products} - \sum nS^\circ_\mathrm{reactants}.\tag{6}$$


What is the hidden meaning of electrochemical stability window?

The electrochemical stability window is most important when considering components of an electrochemical system that you do not want to be oxidized or reduced. This refers most often to the electrolyte or protective coatings.

For example, in lithium ion batteries it is highly desirable that the electrolyte does not react/change/degrade in any way as a result of battery charging/discharging or even in storage. A lithium ion battery typically has a voltage of 3.6 V. The electrolytes used are either lithium salts in organic solvents (ethylene carbonate, etc), solid ceramics, or polymerics electrolytes. If the electrolyte was water based, the electrochemical stability window would only be 1.2 V.

Why 1.2 V?

The standard reduction potential of water is 0.0 V (vs SHE)

The standard oxidation potential of water is 1.2 V (vs SHE)

You can not operate a 3.6 V lithium ion battery with an aqueous electrolyte because the electrolyte itself would react, creating dangerous hydrogen and oxygen gas. So you need an electrolyte with a larger stability window.


How can I avoid crystallisation of sugars in vinegar at low temperatures?

It is difficult to correctly measure the solubility, especially if it goes to multicomponent solutions, as your vinegar. It is possible to find tables with some components of your mixture, but you won't get the full picture, and it looks as the moment, when the hands-on experience is most important.

Nonetheless, I found some facts for you, namely:
the solubility of a-D-glucose in all the solvents increases with
increasing temperature and decreases with increasing volume fraction
of acetic acid, from Solubilities of {α-d-glucose in water + (acetic acid or propionic acid)} mixtures at atmospheric pressure and different temperatures

Another point is that the supersaturated solution can be stabilized by various means, and I think it already happens, as your vinegar contains lot of "impurities" from the grapes.

As for the second point, how to prevent it, the simplest solution will be to keep the temperature high enough, so that the crystallization does not occur. If this is not possible, the supersaturated solution will crystallize, sooner or later. If you than ask, how to slow down the process, until spring comes, filtration could help to remove the crystallization nuclei.


Is there a visible change to baking soda when it reacts with odour?

Let us assume that baking soda works by not simply absorbing, but reacting with acidic/odiferous substances via an acid-base reaction. So, for example if you had rancid butter in the fridge and it gave off the nasty-smelling butyric acid, that acid would react with the baking soda to give sodium butyrate plus carbonic acid (which breaks down to carbon dioxide and water). Let us assume that the carbon dioxide and water would evaporate. That would be a net gain of weight since the butyric acid molecule has a bigger mass that carbon dioxide and water.

Most odiferous substances will likely have a higher molecular weight than carbon dioxide plus water, so I believe over time the baking soda will increase in weight. This will also be true for those cases where there is no reaction - but simply absorption.


On the oxyacids of +III pnictogens and +IV chalcogens

Disclaimer: This isn't really the answer Jan is looking for.

Holleman, Wiberg "Lehrbuch der Anorganischen Chemie", de Gruyter, Vol. 101, notes that:

Selenium

Selenium forms a weaker, but more stable acid based on oxidation state $ +4$. No explicit mention of the structure, which suggests to me that $\ce{SeO(OH)2}$ is the structure.

Tellurium

Tellurous acid's structure is described as unknown.

Arsenic

For arsenous acid (the $+3$ oxidation state based acid) the tautomeric equilibrium is entirely on the side of $\ce{As(OH)3}$ (as opposed to $\ce{HAsO(OH)2}$). The reason given is that $\ce{As}$ does not form double bonds with $\ce{O}$, which is outdated: Current thinking is that only the first row forms double bonds, i.e. no double bonds in $\ce{H2SO4}$ or $\ce{HPO(OH)2}$. Organic esters exist that can be described as $\ce{RAsO(OH)2}$.

Antimony

The $+3$ oxidation state based acid is described as forming $\ce{H+ + [Sb(OH)4]-}$ in water. No mention of tautomerism.

The book gives sources only sparingly and on a per-section basis. Relevant items appear to be:

R. Paetzold: "Neuere Untersuchungen an Selen-Sauerstoff-Verbindungen", Fortschr. Chem. Forsch. 5 (1966) 590-630.
W.A. Dutton, W.Ch. Cooper: "The Oxides and Oxyacids of Tellurium", Chem. Rev. 66 (1966) 657-675.
M.A. Ansari, J.M. McConnachie: "Tellurometalates", Acc. Chem. Res. 26 (1993) 574-578.
J.D. Smith: "Arsenic, Antimony and Bismuth", Comprehensive Inorg. Chem. 2 (1973) 547-683.
C.A. McAuliffe: "Arsenic, Antimony and Bismuth", Comprehensive Coord. Chem. 3 (1987) 237-298.
GMELIN: "Arsenic", Syst.-Nr. 17, up to now 1 book, ULLMANN, Vol. 5: "Arsenic and Arsenic Compounds", A3 (1985) 113-141.
GMELIN: "Antimony", Syst.-Nr. 18, up to now 6 books, ULLMANN, Vol. 5: "Antimony and Antimony Compounds", A3 (1985) 55-76.

My thoughts on $\ce{As, Sb}$ are that in the MO picture, $\ce{H}$ has insufficient overlap to form a bond that would compete with the alternatives.

I have performed DFT calculations (PW6B95-D3/def2-QZVP//PBE-D3/def2-TZVP) on gas-phase $\ce{Pn(OH)3}$ and $\ce{HPnO(OH)2}$. Of course, these are gas-phase  energies on single conformers, not free enthalpies. For $\ce{Pn} = \ce{N}$, a 10 kcal/mol favorable energy difference towards $\ce{HPO(OH)2}$ was found. For $\ce{As}$, about 35 kcal/mol towards $\ce{As(OH)3}$ was found. The Mayer bond order analysis was not conclusive.


Can suspensions be separated by centrifuge or by separatory funnel?

If you are indeed dealing with a proper suspension (not a colloid and not a solution) then centrifugation should always work to separate the components. It may be that you need a very high rotation speed or a long time, but you will manage to get the solid particles to the bottom of the tube and the clear liquid above.

A separatory funnel is a bad idea for a number of reasons. For one, you don’t know how long it will take for the components to separate. For two, you will probably have difficulties removing the solid portions; they might clog the opening or they might attach together not intending to go through at all. Finally, you will always get some liquid together with the solid fraction in a separatory funnel while you should arrive at a much better dryness with centrifugation.


Acidity comparison between cyclopropene and cyclopropane

This reference does in fact indicate that cyclopropene ($pK_a=29$) has a much higher a acidity than cyclopropane ($pK_a=44$).  Deprotonation at the vinylic hydrogen would account for this.  That cyclopropene is deprotonated at a vinylic carbon is briefly summarized here with a primary reference given.


Mechanism of Hofmann Rearrangement Reaction

Water is present in large excess when compared to the amide, so it is statistically not likely that the proton would be removed from the amide rather then water. Even if it happens, proton exchange is reversible so it would be formed/unformed many times that eventually the removal of the proton by water would happen and proceed with the mechanism.

Regarding the attack to $\ce{Br2}$, I believe the reason it doesn't happen is that the ion would be surrounded by many many water molecules because: 1) the amide forms H-bonds with water, 2) the ion is solvated efficiently by water, so more molecules would be around to have their protons removed.