I am assuming that you want the alcohol to react with the $\ce{OH}$ group and the alkyl halides to react with the $\ce{Cl}$ group.

Both groups like to withdraw electrons, so they tend to be negative, and if we take them away, the resulting ions would be positive so they cannot react.

This leaves us to the possibility of the $\ce{H+}$ being cleaved off from the $\ce{OH}$, producing an alkoxide anion, which can react with the cation resulting from the cleaving off of the halide.

However, to effectively cleave off an $\ce{H+}$, we would need an acidic alcohol, such as phenol, which does react with alkyl halides (taken from here):

  In the presence of a base, such as potassium carbonate ($\ce{K2CO3}$), and heat, phenol reacts with an alkyl halide to form an ether.

For example:

$$\ce{\underset{phenol}{PhOH} + \underset{1-chloropropane}{CH3-CH2-CH2Cl} ->[K2CO3] \underset{phenyl propyl ether}{PhOCH2-CH2-CH3} + HCl}$$

where $\ce{Ph}$ denotes the phenyl group.

The reaction proceeds through the ionization of phenol:

$$\ce{\underset{phenol}{PhOH} ->[K2CO3] \underset{phenoxide}{PhO-}}$$

The next step is an SN$2$ step:

$$\ce{\underset{phenoxide}{PhO-} + \underset{1-chloropropane}{CH3-CH2-CH2Cl} ->[K2CO3] \underset{phenyl propyl ether}{PhOCH2-CH2-CH3} + Cl-}$$

For other alcohols, they are generally not acidic, so you would need the corresponding alkoxide by reacting with sodium:

$$\ce{\underset{alcohol}{R1-OH} ->[Na] \underset{alkoxide}{R1-O-} ->[R2Cl] \underset{ether}{R1-O-R2}}$$

This, as the comments have pointed out, is called the Williamson ether synthesis.


how to use di-tert-butyl dicarbonate (diBOc) as a solvent?

Di-tert-butyl dicarbonate melts around room temperature (m.p.=23°C).  It does not decompose at this or even slightly higher temperatures.  For example, it is typically purified by distillation under reduced pressure at temperatures up to around 65°C.  At higher temperatures it will decompose to isobutene, t-butyl alcohol and carbon dioxide.

In any case it is OK to melt di-tert-butyl dicarbonate and use it between room temperature and up to around 40-60°C.  Just make sure that when you open the bottle and use it that you do so in a dry environment as water will decompose di-tert-butyl dicarbonate.


Technically, even simple water can cause rust, so nothing surprising here.
However, spoilage of milk most probably produced a lot of organic acids (lactic acid and similar) which can speed up any corrosion process. Lactic fermentation is a natural fermentation process in milk, when bacteria start converting the sugar content of milk to lactic acid. It is not the only possible product, but a very common one. Link: https://en.wikipedia.org/wiki/Lactic_acid_fermentation

There are two important factors that help corrosion: First, acidic environment (low pH) itself generally help corrosion. Second, the organic acids produced in fermentation can often form metal complexes, therefore they help remove any protecting oxide layer from the surface of metal objects as well as shift the  redox equilibrium towards an oxidized complex form.


What are point groups and the benefits of using them?

Point groups are a very valuable tool for analysing molecules without knowing much about them. They rely on analysing a molecule’s symmetry and deducing both physical parameters and the shape of molecular orbitals from the information gathered. The downside is that one must know a molecule’s geometry before one can apply group theory to it, but for most molecules the geometry is known anyway.
Picture a molecule, for example ammonia $\ce{NH3}$. Remember that it has a trigonal pyramidal shape with $\angle(\ce{H-N-H}) \approx 107^\circ$ Consider all atoms of the same type (only applicable to hydrogen here) identical. Look at this molecule naively and think what kind of symmetry it has.

Figure 1: Ammonia (image taken from Wikipedia).
One of the easiest symmetry elements that you can find is probably a mirror plane that transects the molecule along one $\ce{N-H}$ bond and places the image of the second hydrogen onto the third hydrogen. They are shown in figure two by the blue-ish ovals. A plane of symmetry is typically called $\sigma$.

Figure 2: Planes of symmetry in ammonia (image taken from Researchgate)
You may also see, that if you stick an axis into the nitrogen atom through its lone pair (the thick, blue axis in figure 2), you can rotate the molecule $120^\circ$ around this axis and it will again look the same. Three rotations along this axis make $360^\circ$, so due to some formalism (which is really helpful if you think about it), the rotation is called $C_3$.
All we have gotten so far is a simple set of symmetry operations but they don’t help much. We need to put them together. We can define the connection of two symmetry operations by saying ‘we’ll perform one first, the other after it’; since neither change the molecule the final molecule will remain unchanged. Now to prove that the symmetry elements present in a molecule constitute a mathematical group, we need to prove the group axioms:

Closure The connection $A \circ B$ of two elements of a group must always be an element of the group.
I just proved this before. A symmetry operation will transform all hydrogens onto a hydrogen position, no two hydrogens on the same and nitrogen will always remain unchanged. We will always gain the same molecule. We just need to make sure we find all possibilities correctly when searching.

Associativity $(A \circ B) \circ C = A \circ (B \circ C)$.
If you think about this practically, it makes sense. Why should it matter whether I rotate first and apply a mirror plane later or the other way around?

Existence of an identity element One element must exist so that $E \circ A = A$ and $A \circ E = A$.
Well, this is easily done; we just include the identity, i.e. no transformation as one of our symmetry operations. It is labelled $E$.

Inverse elements For each element $A$ there must exist an element $A^{-1}$ so that $A \circ A^{-1} = E$.
This requires a bit of working but in the end we can prove this. For $\sigma$, the same $\sigma$ is always inverse element. For $C_3$, $C_3^2$, the rotation around $240^\circ = -120^\circ$ is the inverse element.

The good thing is that if we didn’t identify all the possible operations initially, we can use the connection $\circ$ to indentify the ones we missed.
Now because there is only a certain number of inherently different groups around, they have been given names for easier identification. Our ammonia example would be the point group $\bf\it C_\mathrm{3v}$. $C$, because we have a rotation axis, $_3$ because rotation is $\frac{1}{3}$ of the full circle and $_\mathrm{v}$ because the planes of symmetry are vertical to the rotation axis (rather than horizontal).
(Naming point groups is rather simple. Find the most important axis — the rotation with the highest index. If that is the only rotation, it is a $C$ group. If there are rotations perpendicular to that one, the group is called $D$. If there is an improper rotation, it is labelled $S$. The highest index of rotation is also the group’s index. $_\mathrm{v}, _\mathrm{h}$ and $_\mathrm{d}$ denote planes of symmetry vertical, horizontal or diagonal to this axis. Finally, $C_s$ is the group that contains only one plane of symmetry, $C_i$ contins only the inversion and finally, tetrahedra ($T_\mathrm{d}$), octahedra ($O_\mathrm{h}$) and icosahedra ($I_\mathrm{h}$) have their own names due to their high symmetry.)
Part of the beauty of point groups is already apparant here. If we know what symmetry operations transform the atoms onto each other, this must also be true for some physical properties — those that can be expressed as vectors. Most notably, a dipole moment must transform onto itself during any symmetry operation, so ammonia’s dipole moment must be along the rotation axis. Molecules that display inversion symmetry cannot be a dipole. Molecules can only be chiral (but are always chiral) if their point groups do not contain inversion, planes of symmetry or improper rotation (a rotation followed by a mirror image perpendicular to the rotation axis). There are more examples, but these are probably the most important.
Finally, for each point group a character table exists. This groups sets of similar symmetry operations together according to a set of specific rules (e.g. the two rotations $C_3$ and $C_3^2$ will be grouped together) whose details I won’t go into, and also includes irreducible representations. Any spec of the molecule (and all its symmetry-equivalent specs) must transform as one of these irreducible representations. The $\mathrm{p}_z$ orbital of nitrogen in ammonia (the one along the $C_3$ axis) transforms according to the irreducible representation $\mathrm{a_1}$ — essentially, $\mathrm{a_1}$ means that something will always be the same. The $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals together will transform as $\mathrm{e}$. Knowing the irreducible representations of atomic orbitals helps to predict molecular orbitals, because only those of the same irreducible representation will mix and combine (for others, the constructive overlap is always $0$).
You can google character tables to find lists of them, and thereby also which point groups actually exist.

Point groups are even more powerful than shown here; but I tried to keep the application parts low-level. There are a number of questions here on Stack Exchange tagged group-theory that go more into depth.


How would one go about making a cheap freeze spray?

The reason the 'canned air' dusters and other aerosol-based things like anti-perspirant cans produce a cryogenic effect is from the decompression involved. They are stored at room temperature, but the can is pressurised.

There are two effects that happen when you spray out some aerosol.

Firstly, by the ideal gas law, $PV=nRT$, when you release the pressure (i.e. lower $P$), the temperature ($T$) also drops. (The increase in volume $V$ mostly compensates for this, though.)

Secondly, the contents of the can are stored under enough pressure to keep at least one of the parts of the mixture in a liquid state. When you release the pressure by spraying it out, the liquid boils and evaporates, and takes energy in the form of heat to do so. You'll notice that the contents of a canned air duster, if you hold them upside-down when you spray, include more than just air -- there's also a liquid in there which is designed to be mostly inert to electronics and to quickly evaporate just as described above when the pressure inside the can drops, maintaining the pressure needed to produce a gas jet. If you spray for an extended period of time, you'll normally notice a boiling sound coming from inside the can, and the can getting cold, as that liquid evaporates. (As a safety note, these substances are often flammable, so don't use these air dusters near naked flames or ignition sources unless you want fire.)

Most people don't have the equipment needed to produce this sort of thing at home.

None of the cooling baths in that list are designed to be left as-is for months at a time -- they're designed to be made up as and when needed, with occasional addition of liquid nitrogen or dry ice to keep them cool. While you could produce a cold pressurised spray on the spot using dry ice and ethanol, there's a high probability that any sort of robust switched-release mechanism could make the vessel you're using explode under the pressure of sublimating carbon dioxide. Also, unless you relied on a pressure source other than the carbon dioxide itself, your spray probably wouldn't have time to get very cold.

So, I don't recommend it.


How do I analyse this data? (optimization experiment)

Some materials (e.g. ionic salts) in certain solvents (e.g. water) actually do decrease volume of the solution in certain concentrations (obviously at some point it inevitably increases).

So you certainly can end up with non-linear functions of displacement from $\Delta_{conc}$.

That said, a negative quadratic may fit your data well, e.g. $ax^2 + bx + c$, with $a < 0$. Maybe even a damped exponential, e.g. $ae^{-bx}$. You can use Microsoft Excel etc. for a quadratic fit or fancy math skills for other functional forms.

Another Ben describes in more detail here but the image he used is worth repeating. Here the concentration of an added substance actually shrinks/increases volume displacement "oddly".


How to name an alkyne with cyclic substituent?

As far as I am concerned, your attempt was incorrect.
You must use a parent chain with the highest number of carbon molecules. If I am not mistaken, if a cyclic carbon chain has the same number of carbon molecules as a straight carbon chain, the cyclic carbon chain should be the parent chain.
In addition to what I said above, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), rings/cyclic carbon chains are always selected as the parent chain.

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).

(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

Thus, the name should be: 1-methyl-1-(3-methylpent-1-yn-1-yl)cyclopentane.
Note that an alkyl substituent that is also an alkyne (i.e. 3-methylpent-1-yn-1-yl in your question above) is called an alkynyl group. It is also worth noting that rings/cyclic carbon chains are always selected as the parent chain for the preferred IUPAC name as I mentioned previously.


How much KMnO4 is required for the complete oxidation of ferric/ferrous oxalate in acidic medium?

I'd like the question to be tackled in two parts. In each part, we're going to deal with one of ferric oxalate and ferrous oxalate.

Since the mixture ($\pu{5 mol}$) given to us is equimolar, we have $\pu{2.5 mol}$ of ferric oxalate and $\pu{2.5 mol}$ of ferrous oxalate. I second your understanding that $\ce{MnO4-}$ will be oxidised to $\ce{Mn^{2+}}$, because the medium is acidic, and ferrous ions to ferric ions.

Ferric oxalate doesn't contain any $\ce{Fe^3+}$ ions, so our main focus here is the conversion of oxalate ($\ce{C2O4^2-}$) into carbon dioxide. The oxidation state of carbon in oxalate is $+3$, while it's $+4$ in carbon dioxide. So there's a net change of $+1$ per carbon atom. Oxalate contains two carbon atoms, so that gives us a valency factor of $2$ per mole of oxalate ion. However, for $1$ mole of $\ce{Fe2(C2O4)3}$, you'll have $3$ moles of oxalate ions that are undergoing the aforementioned change. This gives us an overall valency factor for $\ce{Fe2(C2O4)3}$ as  $2 \times 3 = 6$. The valency factor for the $\ce{MnO4- -> Mn^2+}$ change is $5$. Now, we're going to equate the equivalents as per the Law of Equivalence.

$$ \begin{align*}
n_{\ce{KMnO_4}} \times V_{f_1} &= n_{\ce{Fe2(C2O4)3}} \times V_{f_1'}\\
x \times 5 &= 2.5 \times 6 \\
x &= 3 \end{align*}
$$

For the part of the mixture that contains $\ce{FeC2O4}$, we need to change our valency factor. This is because here we're taking into account the conversion of ferrous ions into ferric ions; this conversion has a valency factor of $1$. So the overall valency factor, in this case, would be $2+1=3$. Now, since this has been done, we proceed to apply the same equation as in the previous case.

$$ \begin{align*}
n_{\ce{KMnO_4}} \times V_{f_2} &= n_{\ce{FeC2O4}} \times V_{f_2'}\\
x \times 5 &= 2.5 \times 3 \\
x &= 1.5 \end{align*}
$$

Thus, overall, it can be concluded that you'll need $\pu{3+1.5=4.5 mol}$ of $\ce{KMnO4}$ for the complete oxidation. The beauty of this? I didn't use any reaction-equilibrium equations.

This is the reason we use equivalent concept, to save ourselves from the hassle of writing (and perhaps balancing) chemical reactions.


Is it true that todays crude oil is getting more 'dirty'?

We've already extracted most of the easy-to-get, easy-to-use, relatively pure, oil, and are now digging up oil from other sources, like the tar sands of Alberta, which contain a lot more contaminants than most other sources we've drawn from in the past.


In the context of photochemistry, radiation time or half life period are not used, as far as I know.

The only relevant technical term i'm aware of -in that context- is half-life (time), which describes the time needed for the concentration of an excited species to decrease to 50% of the initial value, cf. S. E. Braslavsky, Glossary of terms used in photochemistry,
Pure Appl. Chem., 2007, 79, 293-465
[DOI].

For longer-lived excited triplet states in the gas phase, radiationless deactivation upon collision with a ground-state molecules will play a role.

Thus, the lifetime might be correlated with the mean free path of the molecules.


What is the conjugate acid of NaOH using the Brønsted-Lowry definition of acids?

Your first equation is more properly written as

$$\ce{NaOH + H3O+ -> Na+ + 2H2O}$$

in aqueous media. However, we can do better if we explicitly show the dissociation of $\ce{NaOH}$ as

$$\ce{NaOH -> Na+ + OH-}$$

and substitute that into the first expression (note that I write $\ce{2H2O}$ as $\ce{H2O + H2O}$) to get

$$\ce{Na+ + \underbrace{OH^{-}}_{base} + \underbrace{H3O^{+}}_{acid} -> Na+ + \underbrace{H2O}_{conjugate\;acid} + \underbrace{H2O}_{conjugate\;base}}$$

where we see that $\ce{H2O}$ is the conjugate acid of $\ce{OH-}$ as well as the conjugate base of $\ce{H3O+}$. The last bit - where water plays 2 roles - is due to water being amphoteric, or able to act as an acid or a base.


Let's say that the amount of $\ce{CH4}$ and $\ce{O2}$ at the start is $x$.

After the reaction is complete, let's say that $y$ moles of $\ce{O2}$ reacted.

The amount of moles of $\ce{CH4}$ left is $x-2y$;

The amount of moles of $\ce{O2}$ left is $x-y$;

The amount of moles of $\ce{CO}$ formed is $2y$;

The amount of moles of $\ce{H2}$ formed is $4y$;

That should make one of the choices clearly false, one clearly true and one might not be so obvious, unless you assume that the forward reaction path is less favourable ($\ce{CO}$ is not very stable)


Solubility determination of Ca(OH)₂ in water

The self ionization of $\ce{OH}$ is $10^{-7}$. If the concentration of $\ce{Ca(OH)2}$ is greater by a few magnitudes, then the self ionization should not matter much. In fact, $\ce{Ca(OH)2}$ is considered soluble, so if not in miniscule amounts, the self ionization should not matter much


Electronic configuration and the periodic table

In the 4th period of the periodic table there is 1 s orbital, 3 p orbitals, and 5 d orbitals. That's a total of 9 orbitals. $9 \times 3=27$

That's 27 elements.


How does ring size affect SN2 reactions?

TL;DR - Bond angle strain gets a lot worse in going from cyclpropyl chloride to the transition state for the 3-membered ring example, than for cyclopentyl chloride going to the corresponding transition state.

The rate for each reaction will be determined by the free energy barrier in each case. Cyclopropyl chloride is strained because the carbons "want" to be tetrahedral, with sp3-hybridized orbitals (109° bond angles), but are forced to have much more acute (60°) bond angles within the ring. However, at the transition state, the carbon undergoing nucleophilic attack "wants" to be trigonal planar, with sp2-hybridized orbitals (120°), but is still geometrically-constrained to 60° bond angles within the ring. This makes the free energy barrier for SN2 very large.

Cyclopentyl chloride has bond angles close to the ideal 109°. There will be some bond angle strain at the transition state, but the effect won't be nearly as bad as for the three-membered ring.

Now, can the effect be generalised? Mostly. As expected, cyclobutyl chloride reacts faster than cyclopropyl, but slower than 5 or 6-membered rings. Five vs six is a tricky one. Cyclopentyl chloride is actually slightly more reactive. If you look at cyclohexane, it is very stable in the chair conformation. All nice bond angles, no eclipsing interactions. At the transition state for SN2 with cyclohexyl chloride, bond angle strain is introduced and there is steric interaction between the incoming nucleophile and the adjacent axial hydrogens. Cyclopentyl chloride doesn't start off quite so perfect - there are some eclipsing interactions. Also, the transition state may actually even relieve some of these eclipsing interactions.

Beyond six-membered rings, things get very complicated quickly. You could apply the same approach, but the number of possible conformations quickly rises from 8-membered and above. It becomes difficult to take all the possible factors and conformations into account, so without high-level calculations, it is difficult to make predictions.

If you have access, the reactivity of 3-6 member cycloalkyl chlorides was included in a publication by Rablen et. al..[1] It includes calculated energy barriers for SN2 reactions of these cycloalkyl chlorides with cyanide ion as nucleophile.

Reference:

Rablen, P. R.; Mclarney, B. D.; Karlow, B. J.; Schneider, J. E. How Alkyl Halide Structure Affects E2 and SN2 Reaction Barriers: E2 Reactions Are as Sensitive as SN2 Reactions. J. Org. Chem. 2014, 79 (3), 867–879 DOI: 10.1021/jo4026644.


The only way to get a "100% pure" sample of an element is by mass spectroscopy, or something similar, which acts on each atom (or ion, actually) one at a time.

So you could get a sample of pure rubidium of a few dozen atoms. Caveat: "100 % pure" applies only to those atoms actually observed... there could be other elements in the vicinity that have not been observed.

Practically, though, there is no way to prevent a single atom of hydrogen, for example, from eventually contaminating your "100 %" pure sample.


How can one calculate the pH of a solution?

I'm tring to write the methodology
If anything is wrong please point it out to me i'll be happy to correct!

To find $\small\ce{pH} $ of a solution you can:

Find the conjugated base of your acid  (for example $\small\ce{HCOOH}$ becomes $\small\ce{HCOO-}$ (correct me if i'm wrong here)
Write the equation of the solution
Now you want to prepare an aqueous solution.
Find the concentration of $\small\ce{[CH_3COOH]}$
you use the formula to determine the concentration in solution of acid

$$
   \ce{[CH_3COOH]\ =\ C\ -\ [CH3COO- ]}
   $$

($\ce{C}$ is the concentration of brought acid in $\ce{mol/l}$)
Then you write that:

$$\qquad\ce{[H+ ]\ =\ [CH3COO- ]}$$
as both dissociate from the same acid as

$$\ce{CH3COOH\ ->\ CH3COO- \ +\ H+}$$
the $\small\ce{pH}$ concentration is given by:
$$\ce{pH\ =\ -\log(\small[H_3O+ ])}$$

$pH$ here you are! (without any unit)


What is the basis for naming the p-orbitals with reference to the Cartesian coordinate system as x, y, and z?

The choice of axes is arbitrary in this case (isolated atom with no external influence).
You could choose about any set of co-ordinate axes, and that wouldn't really affect the behavior of any of the electrons there ;)
However, in the presence of external influences, such as a magnetic field, it's customary to take the positive axis in the direction of the field as the positive z axis. Though ultimately, the choice of notation is left to you.

EDIT: Since you seem to have an issue here, I'll mention this as well:
Quoting from chemguide.co.uk
A)  $\mathrm{d_{xy}}$,  $\mathrm{d_{yz}}$,  $\mathrm{d_{zx}}$

The names tell you that these orbitals lie in the x-y plane, the x-z plane and the y-z plane respectively.

Each orbital has four lobes. Notice that each of the lobes is pointing between two of the axes - not along them.

B) $\mathrm{d_{x^2 - y^2}}$, $\mathrm{d_{z^2}}$

Although these two orbitals look totally different, what they have in common is that their lobes point along the various axes. That's different from the first three where the lobes pointed in between the axes
The $\mathrm{d_{x^2 - y^2}}$ orbital looks exactly like the first group - apart, of course, from the fact that the lobes are pointing along the x and y axes, not between them.

The $\mathrm{d_{z^2}}$ orbital looks like a p orbital wearing a collar! The main lobes point along the z axis.


Perturbation Theory in Orbital Interactions in Chemistry

Rather than focusing on the individual matrix entries, it may be more intuitive to look at the matrix $\mathbf{T}$ formed by putting together all entries $T_{ij}$. In this case, $\mathbf{T}=\mathbf{C}^T\mathbf{S}^0\mathbf{C}$, where $\mathbf{C}$ is the coefficient matrix with column $k$ denoting how atomic orbitals mix together into molecular orbital $k$.

Now the point is that molecular orbitals, or the columns of $\mathbf{C}$, are orthonormal and are eigenfunctions of $H_{eff}$ (see page 34). This orthonormality makes sense if you think of the Hamiltonian matrix, which is certainly self-adjoint, and solving the eigenproblem is equivalent to diagonalizing the matrix with an orthonormal basis of eigenvectors.

Hence, $\mathbf{C}$ is unitary, and $\mathbf{T}=\mathbf{C}^T\mathbf{S}^0\mathbf{C}$ can be simply seen as a change-of-basis transformation done on the overlap matrix from atomic basis to the molecular orbital basis.

I hope this rather mathematical perspective may be of help to you. Thanks!


Rates of solvolysis of p-dimethylamino benzyl halides

In the second compound, steric repulsions (denoted by the curly lines) between the N-methyl groups and the methyl groups on the ring occur:

This forces the $\ce{-NMe2}$ group to rotate such that the nitrogen lone pair is no longer parallel to the ring's $\pi$ system. As a result, the nitrogen lone pair can no longer contribute to the mesomeric stabilisation of the second carbocation.

The rate of solvolysis therefore increases in the order: $2 < 1 < 3$. The effect described above is called "steric inhibition of resonance".

A similar effect is observed in azo coupling:

In the second reaction, the nitrogen lone pair is not delocalised into the ring and the ring is therefore not nucleophilic enough to attack the diazonium ion.


A simple and efficient method for the analysis and comparison of inks is thin layer chromatography (TLC).

In order to separate the various carboxylic and phenolic compounds that may be contained in iron gall ink, the mobile phase should be a polar solvent (e.g. ethyl acetate) or solvent mixture and made acidic (e.g. by adding formic acid).

The coloured spots can be compared under normal lighting; other spots may be detected under UV light. The phenolic compounds may be made visible by spraying with $\ce{FeCl3}$ solution.


If gases assume the volume of the container that they are in, then wouldn't a gas that isn't enclosed in a container have an infinite volume?

I give you a case. Suppose you opened an airtight jar containing an ideal gas. Suppose by some mysterious power, you are able to observe the molecules of the gas. What do you think you'll see on opening the jar?
Obviously you'll see the molecules dissipating in space(the correct word would be diffusing). Initially they suppose on opening occupied 100ml, then they occupied 200, then 300... Thus at every point of time the molecules have a finite volume, the value may be very large, but still its finite. Thus we say a gas "will occupy infinite volume" but only after "infinite time".


Why Freundlich adsorption isotherms fails at high pressure conditions?

There is a saturation level for the adsorption of any adsorbate on the adsorbent due to limited surface area of the adsorbent. So once that saturation level is reached , more pressure won't increase the number of particles adsorbed .


The main substances that cause the yellowish color of the milk are carotenoids [1]. The main carotene involved is the beta-carotene coming from the feed that cows eat.

Some studies have been carried on and it has been noticed that the milk with a more yellow tinge was collected during late spring and early summer when carotene levels are at a maximum [2] unfortunately for this reason in many products as mozzarella they add titanium oxide.
The amount of Riboflavin is too little compared to that of beta-carotene and generally gives a slightly green-ish tinge to the milk as reported here and here [3].

P. F. Fox, Fundamentals of Cheese Science,p. 13
Kosikowski, Frank V., and David P. Brown. "Application of titanium dioxide to whiten mozzarella cheese." Journal of Dairy Science 52.7 (1969): 968-970.
Marion Eugene Ensminger,Audrey H. Ensminger,Foods & Nutrition Encyclopedia, 2nd Edition, Volume 1,p. 798


Why do we scratch the walls of test tube in test for magnesium or calcium basic radicals?

Scratching causes microscopic shards of glass to chip off the scratched surface which can break the surface tension of the liquid, allowing the solute to precipitate/crystallize from of the supersaturated solution.


Bromination of but-3-en-1-ol in basic medium and Baldwin's rules

Your reasoning is totally correct according to Baldwin's Rule - the 4-exo-product would be more stable than the 5-endo-product ( see : This for more details).
But your method may be wrong in the given question the product A was first formed and then a ring closure step was performed using $\ce{KOH}$ as shown below :

here the 5-exo-trig ring closure is favoured according to Baldwin's rule .
Let me also address something that may be a doubt ( It was mine ) i.e. why doesn't the oxygen atom attack the bromonium Ion in the first step ? this is because even if it will attack you will also form a acid $\ce{HBr}$ which will cause the cleavage in the cyclic ester we just formed returning us to the original reagent hence it doesn't attack.


Precipitation titration with Volhard method

Yes filtering is necessary as thiocyanate can complex the solid silver chloride per:

$$\ce{AgX_{(s)} + SCN-_{(aq)} -> Ag[SCN]X-_{(aq)}\tag{X = Cl, Br, I}}$$

This additional reaction requires more thiocyanate for back titration and thus gives a lower chloride concentration than is true.


How much pressure does Gallium exert when it freezes?

Enough to break a glass container.

Gallium is one of the few substances that expands on freezing (like water). The solid is about 3% less dense than the liquid. This is enough to shatter a glass bottle full of gallium on freezing (see this video).

Solids that expand on freezing can exert a lot of pressure (clearly more than enough to break glass) but the exact amount of pressure probably depends on the vessel and the circumstances of freezing.


Decomposition of total Interaction energy with SAPT or EDA-NOCV

SAPT(O) is an amazing tool for partitioning the interaction energies, it yields meaningful values and data. ETS-NOCV is also very good for partitioning the interaction energy, but it does it in a much different way, it yields values that correspond more to the orbital interactions, and more so the sigma and pi orbital mixing (although orca will not partition more than just the overall orbital mixing). ETS-NOCV is really good for giving qualitative orbital pictures for electron donation and back-donation. For hydrogen bonding, i'd suggest SAPT


Mathematical explanation of blank titration in Mohr method for argentometry

$\ce{AgCl}$ ppts before $\ce{Ag2CrO4}$. But in order to visually detect the $\ce{Ag2CrO4}$ you must add a slight excess of $\ce{Ag^+}$. Hence you use a "blank" with no $\ce{Cl^-}$ to account for the amount of $\ce{Ag^+}$ needed to get enough of the $\ce{Ag2CrO4}$ ppt to form so that it can be detected visually.


These are examples of enantiomerism or optical isomerism. If you imagine a mirror placed between the two images you will see that they are mirror images of each other. If you try to rotate them to match up (easiest to do with a model, either physical or virtual) you will find that they are non-superimposable and this is what gives them their isomeric properties. The two isomers rotate plane polarised light in opposite directions (this is where the name, optical isomerism, comes from).

The $\Delta$ and $\Lambda$ labels distinguish the two isomers by their absolute stereochemical configuration.

An decent explanation of this type of isomerism can be found here.


Chemistry underlying blacks spots in banana bread

There are, at least, two reactions taking place and causing the brown colour of your bread. One is the oxidation of monophenols by an enzyme called PPO and the other is Maillard reaction, which is the same reaction that makes your steak looks deliciously brown.

Reaction 1. Oxidation of monophenols.

Bananas have a lot of polyphenol oxidase (also known as phenolase or PPO), an enzyme that catalyzes the oxidation of polyphenols naturally present in the banana fruit. The product is a colored ortho-quinone, which is brown or dark reddish. When you smash bananas to make a banana bread you end up releasing a lot of PPO which will catalyze the polymerization of the phenols and, therefore, make your bred looks darker then others. Below is a representation of the oxidation of monophenols into the colored ortho-quinone.

Reaction 2. Maillard reaction

The Maillard reaction is one of the most important reactions in food chemistry. Its products can add sensory features to many preparations, from steaks to coffee. Basically, sugars from the ingredients react with an aminoacid's amine group to form a glycosylamine. This compound is then isomerized to a ketosamine, which breaks into smaller substances. These substances are the Maillard products, and one class is called Melanoidins, responsible for the dark spots in your bread.
The reason why it is formed when you add sodium bicarbonate to the mixture is because alkaline media speed up the fragmentation of the ketosamine, yielding more Maillard products. 2

I have included an infographic from CompoundChem with other information and the Maillard reaction in chemical notation.

2Owusu-Apenten R., Introduction to Food Chemistry, CRC Press, 2004


How to calculate wavenumbers of normal modes from the eigenvalues of the Cartesian force constant matrix?

Short answer: (1) mass weight each force constant (see below). (2) Diagonalize the matrix. (3) Convert the eigenvalues ($\{ k_1^{\prime\prime} \}$) to wavenumbers $\tilde{\nu}=\sqrt{k_i^{\prime\prime}}/(2\pi \times 137 \times 5.29\times 10^{-9})$.  137 is the speed of light in atomic units and $5.29\times 10^{-9}$ converts bohr to cm.

More detail

Harmonic vibrational frequencies come from solving $F=ma$ assuming $E(x)=\frac{1}{2}k(x-x_0)^2 = \frac{1}{2}k^\prime l^2$ where $l=\sqrt{m}(x-x_0)$ and $k^\prime = k/m$. $F=-(\partial E/\partial x)$ so $F=ma$ becomes
$$ \frac{\partial^2 l}{\partial t^2}+k^\prime l = 0 $$
The solution to this differential equation is
$$ l = K \cos(\sqrt{k^\prime}t) +\varepsilon$$
where $K$ and $\varepsilon$ are arbitrary constants.

The solution tells us that the displacement coordinate will increase and decrease in a repeating fashion every $\sqrt{k^\prime}/2\pi$ times, which is the definition of frequency ($\nu$). The units of $k^\prime$ are energy/(mass length$^2)$ or 1/time$^2$, because energy = (mass)(length/time)$^2$.  So the units of $\nu$ is 1/time.  This can be converted to a wavenumber ($\tilde{\nu}$, 1/length) by dividing $\nu$ by the speed of light.

Diatomic molecule

You have 6 Cartesian coordinates $\{q_i\}$ and 6x6 Cartesian force constants (some have the same value),
$$ k_{ij} = \left( \frac{\partial^2 E}{\partial q_i \partial q_j} \right) $$
and it's not clear what to do with the cross terms.
So you diagonalize the matrix to get rid of the cross terms, but you need to mass-weight first for the reasons described above.
$$k^\prime_{ij} = \frac{1}{\sqrt{m_i m_j}}k_{ij}$$
After diagonalization of $\mathbf{k^\prime}$ you have 3N-5(6) non-zero force constants $\{ k_i^{\prime \prime}\}$ , which you can convert to wavenumbers by $\sqrt{k_i^{\prime \prime}}/2\pi c$


You might want to read about the pH of a neutralized solution. But in case you don't, one of my answers here on chem.se covers the calculation of titration curves with and without increasing volume. It is quite easy to create a titration curve using an adapted (see exact solution for triprotic acids here) version of equation 8.

  $$V(B)=\frac{V(H_3A)~\left(\frac{k_w}{x}-x+C(H_3A)\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}\right)}{\left(x-\frac{k_w}{x}+\frac{C(B) \cdot k_b}{\frac{k_w}{x}+k_b}\right)}$$

I did that for you, so you can see, why it is not wrong, that you only needed 12-15 ml to reach pH 7 but also that that pH 7 is not what you searched for.

What you probably actually wanted, was to reach one of the equivalence points at 10 or 20 ml. You will also not have a chance to reach the third EP in an aqueous solution, which is why 30 ml is not a good goal to go for.


Why do diacidic and triacidic alkalis dissociate in one step?

Polyprotic acids

In most acids, even the strong acids, the bond to the acidic hydrogen atom is covalent (though frequently very polar). Thus, for acids, the more correct description of what happens in solution is ionization - a covalent bond is breaking heterolytically to form ions. The solvent is involved in solvating the proton, which is why we usually write $\ce{H3O+}$ for protons in water and not $\ce{H+}$.

Polyprotic acids ionize in separate steps because after the first ionization, the new species has a different strength covalent bond to break. Consider your example of sulfuric acid. Sulfuric acid is a strong acid that ionizes readily and (nearly) completely in water:

$$\ce{H2SO4 + H2O -> HSO4- + H3O+}$$

The resulting hydrogen sulfate anion is a different chemical species. Its bonds have different strengths. The second proton is not as easily ionized. In fact, the hydrogen sulfate anion is considered a weak acid. It does not completely ionize in water. An equilibrium is established with an equilibrium constant of $K_a = 1.02 \times 10^{-2}$. This means that at $\mathrm{pH = 2}$, the hydrogen sulfate anion is 50% ionized: half is hydrogen sulfate and half has ben converted to sulfate.

$$\ce{HSO4- + H2O <=> SO4^2- + H3O+}$$

What evidence do we have for this step-wise ionization? We have titrations! If we add small volumes of a solution of a stong base (say $\ce{NaOH}$) of known concentration to a solution of sulfuric acid and monitor the pH, we can see two distinct equivalence points (the inflection points on the graph where the pH change is very large over a small volume of base added) corresponding to the two ionization events. However, because the first ionization of sulfuric acid is complete before the base is added, a titration  curve of sulfuric acid will only show one equivalence point when two equivalents of base are added (though there is a blip at 1 equivalent).

If we consider maleic acid $(\ce{H2C4H2O4})$, for which both ionizations are "weak" $(K_1=1.26\times 10^{-2},\ K_2=8.51\times 10^{-7})$, we get a different looking titration curve with inflections at 1 and 2 equivalents of base. If the ionization of both protons occurred simultaneously, we would not see a curve like this.

$$\ce{H2C4H2O4 +H2O <=> HC4H2O4- + H3O+}\\
\ce{HC4H2O4- + H2O <=> C4H2O4^2- + H3O+}$$

  Bases

Bases that are the hydroxide salts of metals are ionic compounds. Though many of them have poor solubility in water, what is dissolved is completely ionized. Thus, for copper (II) hydroxide in water at equilibrium, there is only solid copper (II) hydroxide ad solubilized copper(II) and hydroxide ions:

$$\ce{Cu(OH)2 (S) <=> Cu^2+ (aq) + 2OH- (aq)}$$

A titration curve for copper (II) hydroxide with a strong acid like hydrochloric acid would show only one equivalence point at two equivalents of acid:

All ionic bases will behave similarly.

However, some bases are covalent and ionize by reacting with water:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

If there was to be a base of this type that had two ionizable groups, we would see two equivalence points.


Relation between rate of adsorption and the critical temperature of a gas

You can argue that the effect of the solid is to "seed" formation of the condensed phase, as a set of interactions that lower the chemical potential below that of the gas, leading to "condensation" (well, adsorption). This is all very approximate as the details matter of how (as in, through what interactions) the adsorbent interacts with the gas molecule.

In addition, the magnitude of the critical temperature may be considered as a proxy for the strength of intermolecular interactions. For instance, if you assume the gas follows the van der Waals equation of state, then the critical temperature $$T_\mathrm{c} = \frac{8a}{27bR}$$ where $a$ is a measure of the strength of attractive interactions. You can find other ad hoc justifications for associating higher $T_\mathrm{c}$ with stronger strength of adsorption. For instance, larger, more polarizable molecules will tend to have higher boiling points and critical points. Such molecules will also adsorb more strongly. However the argument that the rate of adsorption increases should be considered carefully, since steric effects may be associated with larger molecules.

Regarding the point that "we want gases to be adsorbed", well, the point is that you want to predict behavior. The book merely arms you with guidelines that allow you to make predictions. So that, for instance, if you have a number of compounds in a gaseous mixture, you can estimate which adsorb more. Or perhaps you can use that ability to select new substances with desirable properties based on the adsorption behavior of other compounds.


Why does ATP has a lower entropy than ADP

If you treat the entropy as a "level of disorder" the overall entropy of ADP and phosphate seems to be higher than the entropy of ATP.


How to calculate the solubility of CaF2 at different temperatures?

It depends what you have access to… but if we suppose that you have access to decent thermodynamic tables, here's a hint:

You know how solubility is linked to dissolution reaction (Gibbs) free energy. Now, how does the free energy depend on temperature? In other words, what is $\displaystyle\frac{\partial G}{\partial T}$?


Are food calorie values really integers?

They're not exact numbers. These numbers aren't exact for three reasons:

Each type of carb, protein, and fat has a different caloric value.  These are overall averages for each class.

Even if you were dealing with a single pure compound, the value couldn't be exact because there is individual variation in how much of that compound is metabolized based on digestion, absorption, etc.

Even if #1 and #2 were not issues, the number still couldn't be exact, because it is a measured value, and measured values are never exact.  Only counted numbers, and values that are exact by defintion (e.g., the speed of light) are exact.

To get a better idea of how these values are determined, and the issues and uncertainties associated with measuring them, I'd recommend reading through:
Food and Agriculture Organization of the United Nations. FAO food and nutrition paper. Chapter 3:  Calculation of the energy content of foods - energy conversion factors. Food and Agriculture Organization of the United Nations, 1979.
Here is a direct link to chapter 3:
http://www.fao.org/3/Y5022E/y5022e04.htm


Not exceptions, but see, theories say that going $L→R$ electron affinity increases up to halogens.
For the noble gases, $\Delta_\mathrm{eg}H$ is positive (energy required). While electron affinity is $0$. Which implies no tendency to lose electrons.
For $\ce{Be}$, the small size and nuclear charge are responsible, which is related to $\ce{Mg}$ due to a diagonal relationship.


Why do we not using concentrated HCl and zinc reaction to prepare hydrogen?

If metallic zinc $\ce{Zn}$ is in contact with concentrated $\ce{HCl}$ solution, $\ce{H2}$ is produced, but the reaction is exothermic  : the solution gets hot. As $\ce{HCl}$ is not so highly soluble in hot water, a fraction of the dissolved  $\ce{HCl}$ will be vaporized. So the gas produced will be a mixture of $\ce{H2}$ and $\ce{HCl}$, which is not wanted.
Anyhow, the reaction between pure $\ce{Zn}$ and pure concentrated $\ce{HCl}$ is not really rapid. And, after some time, when the solution contains a certain amount of $\ce{Zn^{2+}}$, the reaction rate decreases in such a manner that the reaction looks finished. It is rather surprising, and even deceiving.
However, if some metallic impurities are present in solution, like $\ce{Co^{2+}}$ or $\ce{Cu^{2+}}$, the zinc metal reduces these ions according to $$\ce{Zn + Co^{2+} -> Co  +  Zn^{2+}}$$or $$\ce{Zn + Cu^{2+} -> Cu  +  Zn^{2+}}$$ As a result, some small amounts of metallic cobalt or copper are deposited on the zinc metal. The contact of the two metals $\ce{Zn + Co}$ or $\ce{Cu}$ produces a galvanic cell, and the reaction rate increases a lot. $\ce{Zn}$ gets dissolved much quicker, producing electrons that go onto the copper or cobalt spots, where they reduce $\ce{H+}$ to produce a lot of $\ce{H2}$.
So the best way of producing H2 by the reaction $\ce{Zn + HCl}$ is to use not too concentrated solutions of $\ce{HCl}$ solution, with small amounts of copper or cobalt salts added into the acidic solution.
The same phenomena of galvanic cell happens if  $\ce{HCl}$ is replaced by $\ce{H2SO4}$


Influence of neighboring fluorine and hydroxyl on carbocation

The answer is actually quite straightforward. Fluorine is more electronegative than oxygen, thus it is less willing to share its lone pair with the carbon bearing the positive formal charge. Thus, the $\ce {-OH}$ group provides greater stabilisation. In fact, the donation of the lone pair by the $\ce {O}$ generates a familiar moiety, an epoxide. The donation by $\ce {F}$ is almost negligible since the positive formal charge on $\ce {F}$ would be highly unfavoured.


The colour of molecules come from electronic transitions that are very sensitive to the nature of the molecule
To simplify a little, molecules absorb light when the energy difference between electrons in an occupied orbital and an unoccupied orbital match the energy of a photon of light (and that light will have a specific colour). This much you already understand.
But the energy levels of the electrons that matter for colour are usually the bonds in the molecule that join one atom to another, not the other electrons in the atom that don't get involved in the bonding. So atoms are rarely thought of as having an inherent colour that is independent of what they are attached to (some isolated atoms do have colour: sodium vapour has narrow absorption bands in the yellow, for example, but this isn't important in sodium compounds).
The assumption you make that is false is that "If we look at the molecules, it seems to me that they are basically the same". If this were true there would be very little interesting in the whole subject of chemistry. When bromine reacts with benzene a Br-Br bond is replaced by a Br-C bond and a Br-H bond. Those are very, very different bonds. A Br-Br molecule has a readily accessible electron transition that results in a dark red colour. That this particular transition is very sensitive to the exact nature of the molecule is obvious as other dihalogens (with very similar electronic structures) have very different colours (iodine is dark purple, chlorine is light yellow). But the orbitals available in other Br-X bonds will be very different as will their energy. Often the corresponding colour of light will be in the UV and there will be no visible colour.
The essential point is that colour comes from electronic transitions that vary a great deal with the atoms involved in the bond. And in many colourful compounds those transitions involve molecular orbitals including multiple atoms across the whole molecule. There is no simple relationship between atoms and the colours of molecules. Every bond is (very) different.


How does increasing temperature reduce the efficiency of solar cells?

While perhaps better suited to Physics SE, I will quote from Sze's Physics of Semiconductor Devices:

  As the temperature increases, the diffusion lengths in Si and GaAs will increase ... and the minority lifetime increases with temperature. The increase in minority-carrier diffusion length causes an increase in $J_{L}$. However, $V_{oc}$ will rapidly decrease because of the exponential dependence of the saturation current on temperature. The increase in the "softness" (roundness) in the knee of the $I-V$ curve as temperature increases will also degrade the fill factor. Therefore, the overall effect causes a reduction of efficiency as the temperature increases.

Here, $J_{L}$ is the current caused by light generation of electron-hole pairs. So, it goes up. But, the device is still a diode and must follow the standard diode equation, so the open circuit voltage ($V_{oc}$) drops. This then also impacts the power output.

In brief, neither of your proposed reasons is the basis for it. Instead, you have to probe deeper into the physics of semiconductor devices.


What is the electrophile in Perkin's condensation reaction?

The reaction mechanism shown on the Wikipedia page for the Perkin Reaction is perfectly clear.

The first step is deprotonation of acetic anhydride by acetate anion.
The second step is nucleophilic attack by the anion of acetic anhydride on the aromatic aldehyde; thus the aldehyde is the electrophile.


Is quite reductive think that pH is the onlyfactor that determinate the health of your teeth, the content of carbohydrate have a strong effect too.
However in fact to my knowledge clear vodaka have one of the highest pH and very few carbohydrates so indeed is one of the most "teeth friendly alcohol drink".
These is take from Victor R. Preedy, Ronald Ross Watson Academic Press, 24/dic/2004 Comprehensive Handbook of Alcohol Related Pathology with the adding of Vodka and Coca Cola.

\begin{array}{|c|c|c|c|}\hline
Alcolic \space beverage & Carboydrate \space content(\%) & pH \\ \hline
Beer               & 3.0-5.0   &  4.1-4.5    \\ \hline
Wine & 0.0-12.9 &  2.8-3.8     \\ \hline
Sweet \space liquor           & 30.0-31.0 & 3.3-3.9 \\ \hline
Strong \space alcohol & 0.0-1.2 &   6.5-6.9\\ \hline
Vodka                & very \space few       & 6.0-7.0        \\ \hline
Coca Cola & 10.6(?) & 2.8\\ \hline
\end{array}

However keep out  from Vodka lemon!


Can a base like OH- attack aldehyde and ketone to create hydrate instead of enol?

Ester hydrolysis and the soap-making process

In aqueous solution, esters are subject to hydrolysis to the corresponding carboxylic acid and alcohol components:

Hydrolysis is,  the reverse of the esterification reaction. The addition of
base accelerates ester hydrolysis because the nucleophile is a hydroxide ion, which is much more reactive than a water molecule.

Enol formation

The a-protons of carbonyls, are somewhat acidic, because the negative
charge on the conjugate base is stabilized by resonance with the carbonyl oxygen. In the formation of an enol, a base abstracts an a-proton from a carbonyl compound, and that same proton (or a proton on a nearby acid group) is
transferred to the carbonyl oxygen. The deprotonated form of an enol is an enolate.

However in this scenario  I don't think an enolate will form in this case.

enol formation is favored in acidic environments rather than basic

The preffered route would be the first one above.

Edit:

To avoid a bit confusion:

To avoid generalisation, the R1 group given on the ester can be anything from -C(CH2)3,  in this case 2 alpha carbons and no alph hydrogen. (No enol possible)
an alpha hydrogen may be there but reaction conditions won't favour enol formation.

Hope this clears doubts


The three functions you mentioned are applicable only to thermodynamic equilibrium states of the system, for which the pressures and temperatures of the system and its surroundings match one another to within insignificantly small differences.  In the integral to get the entropy change (Q2), the same matching applies; that is, a reversible path consists of a continuous sequence of thermodynamic equilibrium states.

Note that, at the interface between the system and the surroundings, the temperature and pressure of the system always matches the temperature and pressure of the surroundings (irrespective of whether the system is at equilibrium).  It's just that, along an irreversible path of the system (for which the system passes through a sequence of non-equilibrium states), the pressure and temperature within the system vary with spatial position, and their average values (averaged over the volume of the system) do not match those of the surroundings at the interface.


One key problem with astatine is that it's incredibly unstable. There are no known stable isotopes, and the longest-lived has a half-life of ~8 hours. So no one has been able to (yet) prepare enough to make real measurements. Thus, we don't know for sure whether solid At is diatomic (like the other halogens) or monatomic.

On the other hand, high-level first-principals (quantum) calculations have been performed that suggest astatine has some unusual properties. The full paper by Roald Hoffmann and Neil Ashcroft at Cornell was published in 2013 in Phys. Rev. Lett.: "Condensed Astatine: Monatomic and Metallic" and also covered in multiple science news sites, including Chemistry World.

The calculations on At:

  predict that it will be a metallic solid. That’s perhaps not so surprising given that iodine, the halogen above it in the periodic table, is itself a dark silvery solid (although not truly metallic itself) and that astatine’s melting point is 302°C. But the real surprise of the new results is that the solid wouldn’t be composed of diatomic molecules, like all the other halogens, but would be monatomic.

Why would At be different than solid $\ce{I2}$?

  ... The monatomic form appeared only with spin-orbit coupling included. "These effects influence the atoms’ propensity to form certain bonds and crystal structures, usually by reducing molecular binding energies," says Hermann. Earlier calculations have indicated that in the gas phase astatine would still form a weakly bound diatomic molecule.

In other words, it's the spin orbit coupling contribution of relativistic effects that would prevent the formation of diatomic molecules in solid At.


Melting and boiling points of transition elements

I don't really like parts of this explanation (was hoping to find a better one here actually), but it's the best I know. I'll build on suggestions by @michielim and @vrtcl1dvoshun.

(Note: I think most of this argument can be transposed to the orbital overlap picture of bonding, but it might be slightly trickier to describe.)

For a backdrop to this answer, I have plotted the melting points and boiling points of the transition metals, so make sure to stare at these for a while before reading on.

From the physicists' "electron sea" point of view of metal bonding, the higher the ionic charge the metal atom can support, the higher the element's melting and boiling points. This explains why group 1 metals such as sodium have quite low melting/boiling points, since the metal would be composed of electrons delocalized in a $\ce{M}^+$ lattice. Going towards group 2 and group 3 elements, one can expect to find a $\ce{M}^{2+}$ and $\ce{M}^{3+}$ lattice, and so on. However, this does not mean that that, for example, metallic osmium is expected to be made out of $\ce{Os}^{8+}$ cations surrounded by a dense electron sea; as each successive "ionization" of the metal atom is performed, it becomes more unlikely that the next electron can become delocalized, since it has to fight against a higher effective nuclear charge relative to the electron before it. Therefore, we can expect some "optimal average charge" for the metal ions in the pure metal. Since melting and boiling points tend to increase towards groups 7 and 8, we can safely assume that this optimal average charge is at least greater than $+3$, and it is likely higher for the heavier elements since they have significantly higher melting/boiling points.

If we can somehow figure out how easily electrons can be removed from their parent atoms to create the electron sea, we may have an argument to explain the trends and apparent discontinuities. First, notice that the $4^{\ce{th}}$ period metals are in general a little bit underwhelming compared their $5^{\ce{th}}$ and 6$^{\ce{th}}$ period counterparts when looking at the boiling points. This can be attributed to the fact that the $3d$ orbitals are anomalously compact relative to higher $nd$ orbitals. This happens because all first occurrences of a given sublevel (i.e. $1s$, $2p$, $3d$, $4f$, $5g$, etc.) are composed of orbitals whose wavefunctions have no radial node. Since the wavefunction's value is monotonically increasing up to its maximum value, it tapers off quickly. Thus, when a 4$^{\ce{th}}$ period metal atom tries to delocalize its $3d$ electrons, it meets a slight increase in resistance due to its stronger interaction with the nucleus, meaning that the optimal average charge of the ions is slightly lower and leading to lesser interatomic interactions holding the metal together.

Now, for the sudden dips, most visible for $\ce{Cr,\, Mn}$ and $\ce{Tc}.$ As has been suggested, one is tempted to take into consideration the fact that they all have $d^5$ configurations, which represent a half-filled $d$ subshell. It can be argued that they are particularly stable due to exchange energy, though Cann suggests it is not the most adequate explanation and instead forwards the "parallel spin avoidance factor".

In half-filled subshell configurations, there is a maximum in the effective nuclear charge felt by the electrons (compared to the previous elements with no doubly-occupied orbitals) combined with relatively low interorbital repulsions due to the Pauli exclusion principle. The element that follows a half-filled subshell must now put an electron in a previously occupied orbital, creating stronger, intraorbital electron repulsion. These effects combine to suggest that removing an electron from or forcing an electron into half-filled configurations is particularly difficult (notice that $\ce{Cr}$ doesn't follow $\ce{Mn}$ and $\ce{Tc},$ more on that soon). If this is the case, then the delocalization of the third electron (which would create $\ce{Mn}^{3+}$ and $\ce{Tc}^{3+}$ ions) and all those beyond it would be somewhat suppressed. This would effectively reduce the optimal average charge of the ions in the metal, and decrease the strength of interactions.

Finally, $\ce{Cr}$ doesn't show the same electronic behaviour as $\ce{Mn}$ and $\ce{Tc},$ even though it has a $d^5$ configuration, and yet it is still anomalous. What gives? Well, according to Greenwood and Earnshaw (Chemistry of the Elements, 1997), the low boiling point of $\ce{Cr}$ is due to a combination of increased effective nuclear charge and the anomalously small size of $3d$ orbitals. When going from $\ce{V}$ to $\ce{Cr},$ the increase in nuclear charge is enough to pull the already small $3d$ orbitals quite close to the nucleus, disfavouring electron delocalization to the extent that a sudden drop appears. Mo has the same electron configuration as $\ce{Cr},$ yet it does not show nearly as large a dip because its $4d$ orbitals are already significantly more diffuse, and aren't so easily suppressed from delocalizing.


Joh,

Please let me know if you referred to this compound:

Also, if this is your compound, please find the name.


Hybridization of Cu2+ in tetraamminecopper(II) sulfate complex

I feel a bit like a broken record. Do not use hybridisation to describe coordination compounds; it is not helpful.

The cation you called tetraamminecopper(II) is more accurately described as tetraamminediaquacopper(II) $\ce{[Cu(NH3)4(H2O)2]^2+}$, because two aqua ligands are there, just further away. The four ammine ligands donate two electrons each, as do both the aqua ligands, totalling 21 electrons overall. Due to the strongly Jahn-Teller distorted geometry and other reasons, the bonding interactions are overall stronger than antibonding ones making the complex favourable.

The interaction between the metal and ligand involves $\mathrm{e_g, a_{1g}}$ and $\mathrm{t_{2u}}$ orbitals (when viewed as an undistorted octahedral complex). That would include two d-orbitals, three p and one s. If your life depended on it and you are the last person on Earth, you may attempt to extract a hybridisation from that. But don’t do it.


How to identify which of two given electrodes with corresponding solutions is reducing more?

The solution is to immerse each rod in both of the solutions (separately of course). Let's imagine that $\ce{A}$ is the more reducing metal. We can consider what happens in each case:

If we immerse $\ce{A}$ in a solution of $\ce{A^m+}$ nothing will happen.
Similarly, if we immerse $\ce{B}$ in a solution of $\ce{B^n+}$ nothing will happen.
If we immerse $\ce{B}$ in a solution of $\ce{A^m+}$ nothing will happen because the electrode potential for $\ce{A|A^m+}$ is more negative (more negative means its a stronger reducing agent) than for $\ce{B|B^n+}$ so the reaction potential is negative and the reaction is not feasible.
However, if we immerse $\ce{A}$ in a solution of $\ce{B^n+}$, $\ce{B}$ will precipitate out on the surface of $\ce{A}$. This is because $\ce{A|A^m+}$ has a more negative electrode potential than $\ce{B|B^n+}$ and so the reaction potential will be positive. $\ce{A}$ will reduce $\ce{B^n+}$ ions to form $\ce{B}$ metal and $\ce{A}$ will be oxidised to form $\ce{A^m+}$:
$$\ce{A (s) + $\frac{m}{n}$B^n+ (aq) -> A^m+ (aq) + $\frac{m}{n}$B (s)}$$

In short, the most reducing metal is the one which causes precipitation of the other metal out of its solution.


About undocumented conventions(?) in PDB files

First, your PDB data seems to have some formatting issues, in that the columns are in the wrong places.  I'm not sure if it's actually messed up, or if it simply got messed up when you posted it.

"In every amino acid there are two O atoms":  True for a single amino acid; not true for an amino acid residue.  I presume you're referring to the two oxygens on the carboxyl group.  However, one of these is lost when the amino acid forms an amide linkage with the next amino acid.

"What's this entry's nr.31 supposed to describe? ":  I opened the PDB in PyMol (after I corrected the format), and it seems to be the amide hydrogen

Regarding the branch designator -- I've never heard of such a thing in a PDB file, and I don't see any mention of it in the format description.  Could you clarify?

ETA:  I've tried to post the correctly formatted PDB data, but it seems to be messing up the formatting.  I'll try to figure out how to do that.
Here's an image of the residue


How would you go about solving this pH problem?

The $K_\mathrm{b}$ value of benzylamine, the conjugated base of benzylmmoniumchloride, is given. The acid constant of benzylammoniumchloride can be inferred via the identity
$$ \mathrm{p}K_a + \mathrm{p}K_b = 14. $$
Thus, we have
$$ \mathrm{p}K_\text{b,benzylamin} =  4.65$$
$$ \mathrm{p}K_\text{a,benzylammoniumchloride} = 9.35 $$
$$ \Rightarrow K_\text{a,benzylammoniumchloride} = 10^{-9.35}$$
As furthermore
$$ K_\text{a,benzylammoniumchloride} = \frac{[\mathrm{H}^\oplus] \cdot [\mathrm{PhCH_2NH_2}]}{[\mathrm{PhCH_2NH_3^\oplus]}} = \frac{[\mathrm{H}^\oplus]^2}{[\mathrm{PhCH_2NH_3^\oplus]}}$$
and
we see that
$$ K_\text{a,benzylammoniumchloride} = \frac{[\mathrm{H}^\oplus]^2}{[\mathrm{PhCH_2NH_3^\oplus]}}$$
This leaves us with an equation with only one unknown variable and is easy to solve as $[\mathrm{PhCH_2NH_3^\oplus}] = 0.24~\mathrm{\frac{mol}{L}}$ is given:
$$
[\mathrm{H^\oplus}] = \sqrt{[\mathrm{PhCH_2NH_3^\oplus]}\cdot K_\text{a,benzylammoniumchloride}} \approx 1.0354 \cdot 10^{-5} ~\mathrm{mol/L}
$$
And as the pH is the negative of the logarithm with base ten we have:
$p\mathrm{H} \approx 4.985$


Physical gelation result in a single massive polymer chain?

The physical alignments can occur anywhere. Keep in mind that these regions are both inter- and intra- molecular. (CH4 can't align with itself, but CH3(CH2)1000CH3 can and does.)


Meaning of the Term "Heavy Metals" in CofA?

You are right heavy metals (as Pb) is a vague term. It is being discouraged and individual analysis should be reported. When someone says "I tried to find a clear standard describing this to no avail", it is a good illustration that not all information is on the webpages. It is worthwhile to search books :-)
I used to think this was an old-school analytical terminology and it turns out to be correct because it is based on classical methods. From the Food Chemicals Codex (ISBN 0-309-08866-6), which is a 1000 paged book, has the following information. Later they describe the tests as well.
Heavy Metals (as Pb)

This test is designed to limit the content of common metallic
impurities colored by sulfide ion (Ag, As, Bi, Cd, Cu, Hg, Pb, Sb, Sn)
by comparing the color with a standard containing lead
(Pb) ion under the specified test conditions. It demonstrates that the
test substance is not grossly contaminated by such heavy metals, and
within the precision of the test, that it does not exceed the Heavy
Metals limit given as determined by concomitant visual comparison with
a control solution. In the specified pH range, the optimum
concentration of lead (Pb) ion for matching purposes by this method is
20 micrograms in 50 mL of solution. The most common limitation of the
Heavy Metals Test is that the color the sulfide ion produces in the
Sample Solution depends on the metals present and may not match the
color in the Lead Solution used for matching purposes. Lead sulfide is
brown, as are Ag, Bi, Cu, Hg, and Sn sulfides. While it is possible
that ions not mentioned here may also yield nonmatching colors, among
the nine common metallic impurities listed above, the sulfides with
different colors are those of As and Cd, which are yellow, and that of
Sb, which is orange.
If a yellow or orange color is observed, the following action is
indicated: Because this monograph does not include an arsenic
requirement, As should be determined. Any As found should not exceed 3
mg/kg. If these criteria are met, Cd may be a contributor to the
yellow color, so the Cd content should be determined. If an orange
color is observed, the Sb content should be determined. These
additional tests are in accord with the section on Trace Impurities in
the General Provisions of this book, as follows: ‘‘if other possible
impurities may be present, additional tests may be required, and
should be applied, as necessary, by the manufacturer, vendor, or user
to demonstrate that the substance is suitable for its intended
application.’’


The most important simplified criteria for the numbering of simple compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Therefore, if no decision can be made about the locants for an ethyl and a methyl group using the previous rules, the lowest locant is given to the ethyl group according to Rule (g) because ‘ethyl’ is cited first in the name (since alphabetical order is used to establish the order of citation of substituent prefixes in the name).

For example, a correct systematic name is 4-ethyl-5-methyloctane rather than 5-ethyl-4-methyloctane since ‘ethyl’ is cited first in the name.


Thermodynamic model of inclusion process of alpha-cyclodextrin

This is a variation on the application of Hess's Law (or looks like it, not having read the article). The idea is that the individual steps add up to the complete reaction

cyclodextrin with included water + hydrated guest $\rightarrow$ hydrated cyclodextrin with included guest

According to Hess's Law you can break up this process into a sum of intermediate steps. One of those steps is dehydrating the guest molecules and shoving the waters into the bulk solvent. That last step is what is meant by collapse of the cavity  with redistribution of the water molecules..


Why are the reactants in the solubility product constant multiplied by their coefficients and raised to the power of their coefficients?

The reason why in the equilibrium constant expression, the concentration is to the power of coefficient is because the above equation can be written like this:
$$\ce{MgF_2(s) -> Mg^{2+} (aq) + F^- + F^- (aq)}$$
Hence when you plug this into the equilibrium constant expression, you get this:
$$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}][\ce{F-}]$$ which becomes the equivalent to:$$K_{sp}=[\ce{Mg^{2+}}][\ce{F-}]^{2}$$ So really the equilibrium constant is proportional to the square of the concentration of fluoride. So that is the reason for the square. However, the concentration of fluoride is twice as much as the concentration of Mg, hence that is the reason for the multiplying the concentration of fluoride by 2. So that is why fluoride is denoted as $(2X)^2$.


How do you Synthesise a Fatty Acid Amide?

You are perfectly right on that you need to convert the carboxylic acid to a more reactive species to eventually obtain the amide!

Playing with the pH will not help:

under acidic conditions, the amine is protonated and therefore unreactive
under alkaline conditions, the acid is deprotonated and the carboxylate won't react at all

The Schotten-Baumann reaction of fatty acids with $\ce{SOCl2}$ will probably work fine:

reflux 1 mol of the acid with 1.5 mol of thionyl chloride
dIstill the remaining thionyl chloride off
purify the fatty acid chloride by distillation in vacuum. Note that the bp of stearoyl chloride is about 175 °C at 2 mm Hg!

This is quite some work to obtain a fatty acid derivative that eventually reacts with an amine!

Wouldn't it be much more convenient to form the amide in a one-pot reaction, possibly at room temperature, by activating the carboxylic acid  in situ?

(In situ means: at the same spot, without isolating the reactive intermediate.)

There's a large number of coupling reagents available, such as

carbodiimides like DCC (dicyclohexylcarbodiimide)
$1H$-benzotriazoles
chloro-substituted triazines

Eric Valeur and Mark Bradley published a critical review titled Amide bond formation: beyond the myth of coupling reagents, in Chem. Soc. Rev., 2009, 38, 606-631 (DOI) on the topic.


How does Resonance stabilize a molecule?

With the concept Resonance you can describe delocalized electrons in some molecules. To go with your example:

(source)

Imagine the ozone molecule to be frozen in one of those two states. Positive and negative charges are localized, which is generally connected to a higher energy, i.e. a more reactive species. This would be an extreme case for ozone - in reality those electrons will be delocalized all over the molecule. So, what is the reason that molecules with localized electrons/charges have a higher total energy and are, hence, less stable?

Maybe you remember the Particle in a box. I wont go into the quantum chemistry here, but one of the take-away-messages is, that the energy for particles in a infinite square well, which is a very rough approximation of the potential in which the outer electrons reside in, does decrease with $ L^{-2} $. $L$ is the size of the box in which the particle can exist. So, the more delocalized the electrons are, the less total energy in a system.

With this, to correct you question a little bit: Resonance is not the reason why molecules are more stable, it is a concept which describes the bonding in a molecule, in which electrons are delocalized, better than a single Lewis Structure.

I would like to have a feedback if this answer is somewhat understandable.


How to determine the angle between non-bonding electron pairs?

Yes, there are ways one could claim to calculate an angle between two non-bonding electron pairs.

BUT: As Mithoron points out, this Chem.SE question illustrates how quantum chemical calculations and photoelectron spectroscopy both demonstrate the non-equivalence of the lone pairs of $\ce{H2O}$, an analysis which presumably applies equally well to the analogous $\ce{H2S}$. Thus, the methods used for calculating such an angle will be controversial, and the results may or may not be of any particular practical value given that they're at odds with PES data. That being said, I'll show here one way that a putative 'non-bonding electron pair angle' can be calculated.

The calculation below is based on application of the quantum theory of atoms in molecules (QTAIM) to the quantity called the electron localization function (ELF), which is a scalar function in $\mathbb R^3$.  QTAIM is useful for identifying intrinsic features of various three-dimensional fields that arise in quantum chemical calculations, with a significant focus placed on the 'critical points,' where the field gradient is zero. It was originally developed for analysis of the electron density distribution, but has been extended to the ELF and other quantities. One of my favorite papers illustrating coupled QTAIM/ELF analysis is a 2009 review by Matito and Solà (Coord Chem Rev 253: 647); it also discusses the localization and delocalization indices (see here and here), which I won't go into in any detail here.

I'm going to start out by emphasizing the distinction between electron density and electron localization. The electron density at a point is fairly straightforward: it's a measure of, on average, how many electrons (or fraction thereof) are located at that point, per unit volume. Due to its waveform nature, each electron's position is a continuous distribution function, and the electron density of the system represents the collective distribution function of all the electrons present.

The electron localization at a point can be harder to get a grasp of: it's a measure of how "spread out" the "location distribution" is, of the electrons that contribute to the density at that point. Regardless of the magnitude of the electron density, the electron localization at a point is high when the electrons contributing to the density at that point overall contribute relatively little to the electron density in other parts of the system.  That is to say, these electrons associated with the point of interest don't "travel around very far" from that point, in their quantum-mechanical meanderings about the system.  Conversely, the electron localization is low when the electrons associated with the point do contribute significantly ("wander") to "distant" regions of the system.

My answer here provides a more concrete example of how the electron localization can differ significantly among systems whose electron density distributions are otherwise relatively similar.  It also anticipates the discussion below, noting features of the three-dimensional ELF field that suggest the locations of electron lone pairs on N, O and F atoms. It is this ELF field that I will be focusing on in the following analysis.

I performed DFT calculations of $\ce{H2O}$, $\ce{H2S}$, $\ce{H2Se}$, $\ce{H2Te}$ and $\ce{H2Po}$ for this analysis using ORCA v3.0.3, and I carried out the subsequent QTAIM/ELF analysis in Multiwfn v3.3.7.  The ORCA input for $\ce{H2S}$ was

! RKS PBE0 def2-TZVP def2-TZVP/J RIJCOSX
! OPT GRID4 GRIDX5 PRINTBASIS

* xyzfile 0 1 H2S.xyz

H2S.xyz was an initial geometry file generated by Avogadro, overwritten with the optimized geometry at the end of the ORCA run.  The inputs for $\ce{H2O}$ and $\ce{H2Se}$ were identical except for the identity of the central atom; I added relativistic effects for $\ce{H2Te}$ via the ! ZORA simple keyword, and for $\ce{H2Po}$ via the use of a 60-electron effective core potential by adding ! ECP{def2-TZVP,def2-TZVP/J}. (While there is an argument to be made that I perhaps should've included relativistic effects for $\ce{H2Se}$ also, I expect them to minimally affect the geometry and overall valence-electronic structure of that system, which is what is of interest here.)

I generated a MOLDEN wavefunction file for each computation, to serve as inputs to Multiwfn. Using $\ce{H2S}$ as an example, the shell command was:

$ orca_2mkl H2S -molden

I then renamed the resulting H2S.molden.input to H2S.molden, which is necessary for Multiwfn to correctly identify it as an ORCA-generated MOLDEN file.

After loading H2S.molden into Multiwfn, I used the following series of commands to locate the 3-D maxima ("attractors") of the ELF distribution:

Main menu:  [17] Basin analysis
[1] Generate basins
[9] ELF
[3] High quality grid

Multiwfn's search routine found six ELF attractors, one of which was identified as a cluster of 'degenerate attractors'. The figure below plots these attractors (via Multiwfn sub-command [0] and some MS Paint manipulation) atop two different views of the $\ce{H2S}$ molecule (click to enlarge):

As can be seen, there are two attractors located right where chemical intuition would expect lone pairs to sit, along with two more attractors located at the hydrogen atoms.  The final non-degenerate attractor is right in the center of the sulfur atom, and could be interpreted as representing the $n=1$ core electron shell. The degenerate attractor is distributed around the $n=1$ attractor, and could similarly be interpreted as the $n=2$ core electron shell. Interestingly, the two $n=3$ valence electrons of the sulfur that are involved in bonding to the $\ce{H}$ atoms do not exhibit independent attractors---each shares an attractor with the electron originating from its respective hydrogen atom.

The QTAIM method provides a way to subdivide the space occupied by a molecule based solely on the properties of the ELF distribution, and associate portions of the electron density to each of these attractors. Integrating these subdivisions of the electron density then provides the number of electrons associated with each attractor:

$$
\begin{array}{cccc}
\hline
\text{lp} & \text{H} & \text{S 1s} & \text{S 2sp} \\
2.114 & 1.855 & 2.143 & 7.865 \\
\hline
\end{array}
$$

All of these values seem pretty reasonable---the attractors corresponding to single orbitals $(\ce{lp}$, $\ce{H}$, $\ce{S 1s})$ have about $\ce{2 e-}$ associated with them, and the $\ce{S 2sp}$ attractor has about $\ce{8 e-}$. In my experience, the signs and magnitudes of these deviations from $2.0$ and $8.0$ are typical for ELF basins.

So, leaving the question of actual physical significance aside, I would argue that these ELF attractors provide a reasonable representation of the (non-)bonding structure of the molecule. Thus, to the main question: what is the structure?  Conveniently, angles and distances among the attractors and atoms can be calculated via Multiwfn sub-command [-2]:

$$
\begin{array}{cccc}
\hline
\angle\,\ce{H-S-H} & 92.3^\circ & r_\ce{S-H} & 2.535\,\mathrm{Bohr} \\
\angle\,\ce{lp-S-lp} & 127.8^\circ & r_\ce{S-lp} & 1.838\,\mathrm{Bohr} \\
\angle\,\ce{H-S-lp} & 108.5^\circ \\
\hline
\end{array}
$$

(In the above, the actual nuclear positions of the atoms were used where relevant, not the positions of the associated attractors. While each such attractor generally falls close to its associated nucleus, it's usually displaced by a tiny amount, $<0.1\,\mathrm{Bohr}$. I've always attributed this to numerical precision limits of the calculation methods, but I don't know for sure if that's what causes it.)

Therefore: By this QTAIM/ELF method, the angle between the two lone pairs of $\ce{H2S}$ is calculated to be $127.8^\circ$. This is considerably greater than the $\ce{H-S-H}$ angle, consistent with the intro-chem conception of the increased 'steric bulk' of a lone pair.

For comparison, I used the same procedure to generate results for $\ce{H2O}$, $\ce{H2Se}$, $\ce{H2Te}$ and $\ce{H2Po}$:

$$
\begin{array}{c|c|ccccc}
\text{Quantity} & \text{Units}  & \ce{H2O} & \ce{H2S} & \ce{H2Se} & \ce{H2Te} & \ce{H2Po} \\
\hline
\int_\text{lp}{\rho} & \ce{e-} & 2.265 & 2.114 & 2.206 & 2.265 & 2.390 \\
\int_{\ce{x}\ce{-H}}{\rho} & \ce{e-} & 1.667 & 1.855 & 1.861 & 1.807 & 1.806 \\
\int_\text{core}{\rho} & \ce{e-} & 2.129 & 10.008 & 27.273 & 44.567 & 77.609 \\
\hline
r_{\ce{x}\ce{-H}} & \mathrm{Bohr} & 1.813 & 2.535 & 2.772 & 3.129 & 3.289 \\
r_{\ce{x}\ce{-lp}} & \mathrm{Bohr} & 1.103 & 1.838 & 2.526 & 3.216 & 3.399 \\
\hline
\angle \ce{H-x}\ce{-H} & ^\circ & 105.1 & 92.3 & 91.2 & 90.9 & 89.6 \\
\angle \ce{lp-x}\ce{-lp} & ^\circ & 114.9 & 127.8 & 139.5 & 156.8 & 156.8 \\
\angle \ce{H-x}\ce{-lp} & ^\circ & 109.1 & 107.8 & 104.0 & 98.1 & 98.2 \\
\hline
\end{array}
$$

Due (presumably) to numerical precision limitations, the calculated values of the four different $\ce{H-x}\ce{-lp}$ angles for each system varied by $1^\circ$ or so; I've reported the means of the obtained values above.

Those at home who are doing their math carefully will note that there is some electron density missing in the systems with the heavier central atoms.  This is because at the 'high' grid quality used, the deep core electrons are poorly captured by the numerical integration involved.  (Indeed, the deep core electrons may not be accurately captured at just about any computationally reasonable grid size!) I think the $\int_\text{lp}{\rho}$ and $\int_{\ce{x}\ce{-H}}{\rho}$ values should be pretty reliable, though.

Naturally, $\ce{H2O}$ does not exhibit a degenerate attractor for the $n=2$ valence electron shell, though it does have a non-degenerate attractor for the $n=1$ core shell.  The two heaviest chalcogenides examined possess degenerate attractors representing their $n\geq 2$ core electrons, as with sulfur.

In general, the $\ce{x}\ce{-H}$ and "$\ce{x}\ce{-lp}$" bond lengths increase as the central atom is varied down the group, which is unsurprising ("bigger atoms are bigger"). Interestingly, while the $\ce{x}\ce{-lp}$ distance starts out considerably shorter than the $\ce{x}\ce{-H}$ bond length for $\ce{x}=\ce{O}$, by the time one reaches $\ce{x}=\ce{Te}$ the relative magnitudes are reversed.  I don't have a particularly good explanation for this trend; presumably it's due to some sort of counterbalance between the nucleus-nucleus repulsion and nucleus-electron attraction forces.

Where these results are particularly dramatic is in the clear trends in the bond angles: as the central atom is made heavier, the $\ce{lp-x}\ce{-lp}$ bond angle increases markedly while the $\ce{H-x}\ce{-H}$ and $\ce{lp-x}\ce{-H}$ angles steadily decrease. The lone pair 'steric bulk' apparently increases uniformly down the group for this series of analogous systems, at least as determined by this QTAIM/ELF approach.  The answers to this question provide more detail into the physical reasons for this decreasing trend in $\ce{H-x}\ce{-H}$ bond angles.


Lewis dot structure of beryllium fluoride

You are correct. $\ce{BeF2}$ is octet deficient.

To solve this problem, they would form polymers by themselves, but that would be out of syllabus.

(I cannot find the polymer of $\ce{BeF2}$, so here is the polymer of $\ce{BeCl2}$:)


Provided that mellitic anhydride ($\ce{C12O9}$, empirical formula $\ce{C4O3}$) is considered one of carbon oxides, then e.g. hexaiodobenzene ($\ce{C6I6}$, empirical formula $\ce{CI}$) could be considered a form of carbon(I) iodide, formally. But please rather don't teach that in regular lectures.


Why can't we oxidise allylic alcohol to aldehyde using pyridinium chlorochromate?

There are indeed examples of $\mathrm{PCC}$ oxidation of allylic alcohols to $\alpha, \beta$-unsaturated aldehydes. The first example is from the original Corey-Suggs paper while the other two cases are from the work of Dauben and Michno.

References:

E. J. Corey, J. William Suggs, “Pyridinium chlorochromate. An efficient reagent for oxidation of primary and secondary alcohols to carbonyl compounds,” Tetrahedron Letters 1975, 16(31), 2647-2650 (https://doi.org/10.1016/S0040-4039(00)75204-X).
William G. Dauben, Drake M. Michno, “Direct oxidation of tertiary allylic alcohols. A simple and effective method for alkylative carbonyl transposition,” J. Org. Chem. 1977, 42(4), 682–685 (https://doi.org/10.1021/jo00424a023).


What is the lowest temperature tungsten starts sublimating at?

Well, you see, searching online for "vapor pressure at such-and-such temperature" (or conversely, "temperature to get such-and-such pressure") is often fruitless, and unnecessary too, because pretty decent approximations are known. Here is one:
$$\ln\mathfrak p=-{\Delta H\over R}\left({1\over T}-{1\over T_0}\right)$$
where $\mathfrak p$ is your vapor pressure in atmospheres, $\Delta H$ is the enthalpy of vaporization and $T_0$ is the boiling temperature at normal pressure. For tungsten, the latter two are reported as 774 kJ/mol and 6203 K, respectively. Now you may plug in your desired pressure and solve for temperature.

That's a relatively crude approximation, for the enthalpy will hardly remain constant over such a wide range of temperatures. This page may contain better approximations designed for specific ranges.


Why are heterogeneous catalysts calcined?

For heterogeneous catalysis calcining of the catalyst allows for the following:

mechanically and thermodynamically stable macrostructure (matrix) with defined geometry, surface and pore size;
predictable activity based on calcining regime (generally, there is a temperature peak for maximum activity, and further extreme temperature treatment often lowers catalyst's activity);
pure composition without traces of organics and majority of other undesired impurities;
production convenience and scalability (for example, metal oxide catalyst manufacturing process can be easily implemented within mining and processing plants);
when distributed over support, more uniform distribution of metal oxide is achieved, and upon slow treatment crystallinity of the catalyst can also be tweaked.


Lennard Jones liquid vapor phase transition

I think your transition, if present, may be drowned out by noise, as your RDF is only produced by 108 particles. As the Lennard-Jones interaction is quite simple and you are using cutoffs (good idea!), is it possible to increase the number of particles in your simulation somewhat, or else take the mean of some large number of trials?

This said, it does not take much imagination to see a continuous PV plot from about 0.7 onwards, which might mean you have something analogous to a supercritical fluid. Maybe try a number of trials at different pressures?


Yes. $$n = \frac{m}{M}~~~~~~~~~~n = cV$$
So $$c = \frac{m}{VM}$$

Given $\frac{m}{V}$ you can work out $c$ (in $\mathrm{mol~dm^{-3}}$).


Why does the crystal sublime more quickly at the corners where there is a higher vapour gradient?

The vapor gradient reflects the fact there is much more space at the corner than in the middle of a face. The vapor builds up near the surface but because there is much more space near a corner, the vapor pressure drops off, where as the vapor pressure near a point on the middle of a face is reinforced by sublimation at points near it.


How do I find the ratio in which the residue of a reaction combines with HCl?

In the comments I mentioned one way to solve the problem, but it turns out that was the roundabout way! There is a simpler solution, as follows.

From the neutralization reaction, it is possible to calculate that the mixture reacted with $0.06\ \mathrm{mol}$ of $\ce{H+}$. The carbonate and the oxide react in a proportion of 1:2 with the hydrogen ions, so we know that $n_{\ce{BaO}}+n_{\ce{XO}}=0.03\ \mathrm{mol}$. The mass balance from the thermal decomposition then suggests that $m_{\ce{XCO3}}-m_{\ce{XO}}=0.44\ \mathrm g$. These two equations are already enough.

Notice that the for the molar masses of the unknown oxide and carbonate, we have $M_{\ce{XCO3}}-M_{\ce{XO}}=44\ \mathrm{g/mol}$. If one converts the masses in the second equation into expressions in $M$ and $n$, the result is $n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XO}}M_{\ce{XO}}=0.44\ \mathrm g$, which can be turned into $n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XO}}\left(M_{\ce{XCO3}}-44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$. It is easy to see that the thermal decomposition does not alter the number of moles of compound containing $\ce{X}$, so that $n_{\ce{XCO3}}=n_{\ce{XO}}$. Thus:

$$n_{\ce{XCO3}}M_{\ce{XCO3}}-n_{\ce{XCO3}}\left(M_{\ce{XCO3}}-44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$$

$$n_{\ce{XCO3}}\left(M_{\ce{XCO3}}-M_{\ce{XCO3}}+44\ \mathrm{g/mol}\right)=0.44\ \mathrm g$$

$$n_{\ce{XCO3}}\times 44\ \mathrm{g/mol}=0.44\ \mathrm g\quad\Rightarrow\quad n_{\ce{XCO3}}=0.01\ \mathrm{mol}$$

Now the problem unravels easily. We get that  $n_{\ce{BaO}}=0.02\ \mathrm{mol}$, which implies $m_{\ce{BaO}}=3.08\ \mathrm g$ and $m_{\ce{XCO3}}=1.00\ \mathrm g$. Hence, $M_{\ce{XCO3}}=100\ \mathrm{g/mol}$, and from there $M_{\ce{X}}=40\ \mathrm{g/mol}$. A quick look on a periodic table yields calcium as the most likely candidate, and indeed its carbonate has the formula $\ce{CaCO3}$ and a molar mass of $100\ \mathrm{g/mol}$.


Are all degenerate d-orbitals identical?

Yes, they are identical.
One thing that we don't really teach well with orbitals is thinking about the symmetry of the orbital with respect to the name of the orbital.
$p_{x}$ has the same symmetry as the function $f(x,y,z)=x$.
Likewise, $d_{x^{2}-y^{2}}$ has the same symmetry as $f(x,y,z)=x^{2}-y^{2}$.
What about $d_{z^{2}}$? You should note that $d_{z^{2}}$ is really $d_{2z^{2}-x^{2}-y^{2}}$.
Imagine taking $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$, summing them up, and renormalizing. So the symmetry is like: $z^{2}-x^{2} + z^{2} - y^{2}=2z^{2}-x^{2}-y^{2}$.
What you get is a big lobe of the same phase along the $z$-axis. And around it in the $x$-$y$-plane (but smaller), you see a smear of the opposite phase that's shaped kind of like a torus. That's why it looks different. It's really two of the other looking orbitals put together.
Why don't we just use $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$ instead? Because they're not linearly independent with the 4 other "normal"-looking orbitals. $d_{x^{2}-y^{2}}$ summed with $d_{z^{2}-x^{2}}$ is $d_{z^{2}-y^{2}}$. The 5 conventional orbitals are just the nice linear combinations of the 5 spherical harmonic solutions to the angular part of the Schrodinger equation. See here.
You don't seem to have a problem with $d_{xy}$, $d_{zx}$, $d_{yz}$, and $d_{x^{2}-y^{2}}$ all being degenerate. So hopefully, you can see that $d_{z^{2}}$ is not really different and therefore also of the same energy as the other 4.
EDIT:
Based on comments, it's also important to point out that the pictures of orbitals with smooth surfaces are not real pictures of orbitals. These are boundary surface diagrams where we've drawn the surface that represents a constant probability of finding the electron for that orbital. This basically means that electron densities for $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$ will smear together in the $x,y$-plane (the orbitals are the same phase) to create something that is symmetrical and shaped like a torus.
This analysis extends fully to $f$ orbitals. Just identify the full functional name for each orbital and apply the same analysis here.


How does the Freezing Point fall in a solution?

Focus more on free energies rather than on vapor pressures (which derive, ultimately, from free energies after all).  For a mixture of B (solute) in A (solvent), the entropy of mixing is $RT(x_A ln(x_A) + x_B ln(x_B))$, and the enthalpy of mixing will go approximately as $x_A x_B \Omega$, with $\Omega$ as a measure of the interaction of A and B. The entropy term will always result in a reduction of free energy at small $x_B$ regardless of the sign of $\Omega$, but in the case of salt in water $\Omega$ is negative, driving further solubility.

Now, about those temperatures of phase transitions (which as you should recall occur when the free energies of the phases are equal, more fundamental than vapor pressures).  On the boiling end, water with salt in it has a lower free energy than water without salt - the boiling point of the salt water has to be higher than for pure water.  The presence of the solute makes it happy to stay liquid.

So, why isn't the freezing point raised in an analogous way? For pure A, of course, the freezing point remains that of pure water.  The entropy of mixing is equivalent for solid and liquid in this case, and it is not clear how different the excess enthalpy is going to be - so why does liquid continue to be stable?  Well, it only does while it has a higher concentration of salt in it than is in the solid.  It forms a classic eutectic point on the binary phase diagram (see, for example http://antoine.frostburg.edu/chem/senese/101/solutions/images/saltwater-phase-diagram.gif).  The only requirement of melting point depression is that water at a high concentration of salt has a lower free energy than the solid at a lower concentration of salt.  Given the outline of free energy above, this is likely to hold for some range of temperatures.

Again, remember that what is being lowered is the freezing point of water with salt in it - what starts freezing out is ice with a lower salt content in it.


This is where your logic fails:

since the concentration of reactants and products remains the same since B is different from A

As soon as you mix two components you change the concentration of both, you effectively dilute all components.
Otherwise your intuition is correct. The gases can often be assumed to act independently. This is true if the solution is ideal (that is, assuming interactions between all molecules in solution are roughly the same). Then each component equilibrates independently with its vapor. However, the vapor pressure of each component is in general not the same as it would be if it were pure. Instead, for an ideal solution the partial pressure of each component follows Raoult's law $p_i=p^*_i\chi_i$, where $p^*_i$ is the partial pressure of the pure component and $\chi_i$ is the mole fraction in solution.
The total pressure above the solution is then the sum of the partial pressures of the components (following Dalton's law, where $\chi_i^g$ is the mole fraction in the gas phase):
$$ \begin{align} p &= \sum_i p_i = \sum_i \chi_i^g p \\ &= \sum_i \chi_i p^*_i \end{align} $$


A theoretical framework for the formalization of chemical interaction and configuration based properties

Technically, one could argue that the entirety of chemistry is trying to formalize how properties of molecules change under a given perturbation. From a computational perspective, some work is being done in the context of machine learning towards developing a continuous mathematical representation of molecules, essentially mapping out chemical space. In this way, you could determine how properties change based on how close different configurations of a compound are in this chemical space.

For details/an example, see 1.
For more on this topic, search "chemical space" or, for some of the commonly used mathematical tools used, "principal component analysis" and "variational auto encoder".

Overall, I think a chemical Theory of Everything would have to necessarily be much more qualitative than that of Physics, just due to the emergent complexity that comes from the vast number of combinatorial possibilities of different compounds.

Aspuru-Guzik, A. et al. Automatic chemical design using a data-driven continuous representation of molecules. ACS Central Science. 2018, 4.


It's not advisable to consider these types of reaction schemes in a slavishly literal fashion. In reality, there are multiple proton transfer equilibria occurring simultaneously, and the reactions will generally be quite rapid. While it's logical and technically correct to depict the most basic of the molecules (among those present in sufficiently large concentration) performing the proton abstraction, it's also of literal practical interest or importance in many cases. I suspect the reason that $\ce{HSO4-}$ was chosen is to emphasize that the sulfuric acid is actually catalytic, so it should be regenerated in the course of the depicted reaction mechanism. Of course, water is produced as a byproduct, so the sulfuric acid will not exist primarily in its neutral molecular form.