So I am currently reading the book: Physical Inorganic Chemistry A Coordination Chemistry Approach, S. F. A. Kettle

One of the chapters deals with Lanthanides and Actinides and their spectra. After a long introduction on different energetic levels and spin-orbit coupling there are a few remarks on $\ce{f->f}$ transitions although, as the author chooses to explain different ideas and phenomena during his explanation I am not sure if I fully understood his final conclusion. I guess it's probably a sum of many contributions. To summarize it quickly the aspects he is discussing are:

We can see that the intensities for $\ce{f->f}$ transitions in coordination polyhedra with high symmetry tend to be weaker than in low symmetries. That means that there are probably mechanisms such as vibrations and mixing with other orbitals, presumably d-orbitals (like in $\ce{d->d}$ transitions).
For the vibronic coupling (in order to compensate for a parity forbidden transition) he gives $\ce{NaCs2[YCl6]:M^3+}$ as an example, where at low temperatures both electronic and vibrational transitions seem to be observable for the trivalent lanthanide dopant.
Then there are also 'bands' under high-resolution conditions that cannot be explained by vibronic coupling. This is where he introduces magnetic dipole allowed transitions. In his explanation he compares how an electric dipole will invert at an inversion center while, much like in the detectors for mass spectrometers, electrons will be forced to move in a spiral along the magnetic vector and a spiral or circle does not invert here. According to his explanation the magnetic dipole is an even function and as f-orbitals are odd an $\ce{f->f}$ transition is $$u * u = g$$
and therefore magnetic dipole allowed.
Then he also refers to the fact that even though light may not be absorbed in matter it will still travel slower through a medium than through vacuum since it causes oscillations that have to propargate through the medium first.
I'll just quote the book at this part:

  It has been suggested that the light wave used in absorption studies
  polarizes the ligands and that the lanthanide ion feels the dipoles created.
  That is, it is suggested that the ligands provide a non-centric field at the
  lanthanide ion not by virtue of any vibrational movement of their nuclei but
  because of the instantaneous polarization of their electron density caused by
  the light wave itself.

Then there was a remark on the environment as well. Since the f-orbitals don't really take place in any bonding for the lanthanides, do they actually feel their environment? And for this they considered weak nephelauxetic effects.
And then there are so called hypersentisitve-transitions (transitions with environment sensitive intensities). And those seem to be related to quadrupole-allowed transitions. The two dipoles in a quadrupole may arise from a polarization by the electric vector as mentioned above. So one dipole is the light vector itself (with the energy corresponding to the hypersentisitve-transitions and the other one being the induced dipole).
The difference in environments would arise from different polarizabilities of the ligands in different environments.

The other mechanism seems a bit more complicated. It considers low-symmetry crystal fields with an inherent dipole. So high-symmetry crystal fields distort via a vibration to a low symmetry. The ligand environment is distorted to a point where it become dipolar.
And then the light itself is the other dipole.
The final remark was on the bonding in f-orbitals. Here the complex $\ce{[UF8]^3-}$ was considered. By repulsion a square antiprism would be favored if there was no interaction at all with the ligands. What we find however is a cubic arrangemenent.
And the $\ce{f_{xyz}}$ orbital can contribute to a bond only in the cubic case.


Why is the ground state electronic configuration of Group 10 (Ni group) elements so anomalous?

I think the relevant comparison should be with the earlier transition metal groups where $s^2$ configurations are more predominant.

In multielectron atoms, electrons interact in such a way that for a given shell quantum number $n$, the orbitals with higher total angular momentum $l$ are raised, thus for example $3d$ ($n=3, l=2$) becomes higher than $3p$ ($n=3, l=1$).  This angular momentum effect is most important when the orbitals are well shielded with a low effective nuclear charge, where the electron-electron interactions don't get swamped by electron-nucleus interactions.

In earlier transition groups the valence shells have this low effective nuclear charge and the angular momentum effect is so strong by comparison that the $d$ valence subshell is raised above the $s$ subshell even with one higher $n$ quantum number.  So, for instance, in the fourth period we see $3d$ orbitals filled only after $4s$ in neutral atoms.  Not so much in the ions, however, where ionization leaves the remaining electrons with more effective nuclear charge, hence we often see $d$ instead if $s$ subshell electrons remaining in the early transition ions.

But by the time we get to the later transition metals the effective nuclear charge has increased even in the neutral atoms, the lower-$n$ $d$ subshell drops relative to the higher-$n$ $s$ subshell, and we begin to see a greater preference for the $d$ subshell being filled first.  By the time we get to Group 12, this transition is complete and only the $s$ electrons remain valence electrons except, possibly, in extreme circumstances.


The Wikipedia page on "Cute Poison" episode says that it wasn't anhydrous copper (II) phosphate, but Gypsum.

I haven't seen the episode, and the Wikipedia page's reaction seems to be the backwards reaction of yours. But I imagine yours is the one the pro/antagonist is gonna use to escape the prison; as

Gypsum is commonly found and accessible:

  Gypsum is a soft sulfate mineral composed of calcium sulfate dihydrate, with the chemical formula $\ce{CaSO4·2H2O}$. It can be used as a fertilizer, is the main constituent in many forms of plaster and in blackboard chalk, and is widely mined. A massive fine-grained white or lightly tinted variety of gypsum, called alabaster, has been used for sculpture by many cultures including Ancient Egypt, Mesopotamia, Ancient Rome, Byzantine empire and the Nottingham alabasters of medieval England.

Calcium phosphate's solubility will decrease with temperature increase and it will precipitate, leaving you with a solution of sulfuric acid.

It's possible for a double displacement reaction to occur in aqueous medium (with a spark):
$$\ce{2H3PO4(aq) + 3CaSO4(aq)·2H2O(l) \leftrightharpoons 3H2SO4(aq) + Ca3(PO4)2(aq) + 6H2O(l)}$$

$\ce{H2SO4}$ is sulfuric acid. It's a very strong acid in water, a diprotic acid. Its $p {\rm K_a}$s are −3 and 1.99 according to Wikipedia.

  Its corrosiveness on other materials, like metals, living tissues or even stones, can be mainly ascribed to its strong acidic nature and, if concentrated, strong dehydrating and oxidizing properties.

Later in the same article, you can read how it reacts with different material. It can weaken metal pretty fast and easily, as 3 moles of it will be present in any 7 moles of products, which is high above the dilute concentration (arbitrarily) defined in the Wikipedia page.

Copper sulfate also reacts with phosphoric acid and forms sulfuric acid, but it's hardly accessible in prison, I reckon.


Potential Energy Curve from Hartree-Fock Calculations

The energy calculated by a quantum chemistry program is essential always the total potential energy $U(R)$.  For example Fig 1 and Fig 4 of Dutta and Sherrill referenced in the question both plot the total sum of electronic and nuclear energies $U(R)=E(R)+V_{NN}(R)$.
For an example, have a look at a PES I have generated for $\ce{H2}$, using Restricted Hartree-Fock and the STO-3G basis set using PySCF.  The characteristic PES, with a single minimum is actually the total sum of energies $U(R)$, labelled the "Potential Energy Surface"

N.B. The zero bond-length electronic energy is calculated using a helium nucleus as the program doesn't allow superposition of atoms.


Relation between number of conjugated double bonds and absorbance?

When 2 p orbitals mix, a pi and a pi* orbital is formed, and the electrons occupy the lower energy pi orbital.  The UV-absorbance is the energy difference between these two orbitals such that on electron may be 'promoted' to the p* orbital.

When 3 p orbitals mix, three pi-type orbitals are formed.  One is obviously bonding and one is obviously antibonding.  There is a third one 'in the middle', frequently called non-bonding.  Because of this third orbital, there is a smaller difference between any two states, and as such absorbing a photon to promote an electron requires less energy, redshifting the absorbance.

Add a fourth p orbital and the energy differences get smaller, but with two bonding orbitals and two anti-bonding orbitals.  Add a 5th p orbital and now you have two bonding orbitals and two anti-bonding orbitals and a non-bonding orbital, with even smaller differences.  Repeat ad nauseum.

Smaller energy differences means longer wavelengths.

Check out this picture for a graphical summary:


In a strict sense, all quantities in the equilibrium constant are not molar concentrations but activities. Activity is a thermodynamic quantity which is proportional to concentration. Now by IUPAC definitions of standard states, the activity of pure liquids and solids is taken is unity (by convention and for reference purposes. Therefore in the following expressions, $\{\ce{H2O}\}$ or $\{\ce{H2O}\}^2$ are all unity, no matter what the power is (real number). Elementary textbooks tend to say a lot of wrong or oversimplified stuff.

$$K_\mathrm{eq} = \frac{\{\ce{H+}\}\{\ce{OH−}\}}{\{\ce{H2O}\}}$$

and

$$K_\mathrm{eq} = \frac{\{\ce{H3O+}\}\{\ce{OH−}\}}{\{\ce{H2O}\}^2}$$

Here the curly bracket indicate activities. The IUPAC notation is $a$ for activity. The species of interest is written as a subscript.
A more detailed answer as to why activity is taken as unity is already provided:
Why are activities of solids and liquids taken to be unity?


Why is phenol soluble in sodium carbonate but not in sodium bicarbonate?

The liberation of $\ce{CO2}$ from $\ce{Na2CO3}$ requires two protonation steps.

$$\begin{align}
\ce{CO3^2- + HA &<=> HCO3- + A-}\\
\ce{HCO3- + HA &<=> CO2 ^ + H2O + A-}
\end{align}$$

Neglecting the concentrations, the position of these equilibria will depend on the acid strength of $\ce{HA}$. If $\ce{HA}$ is a stronger acid than $\ce{HCO3-}$, for example, the first equilibrium will favour the products.

This is the case for phenol, which has a $\mathrm{p}K_\mathrm{a}$ of $9.95$, whereas $\ce{HCO3-}$ has a $\mathrm{p}K_\mathrm{a}$ of $10.32$. Therefore, as long as you don't have too much phenol, it will exist in the soluble phenolate form:

$$\ce{PhOH + CO3^2- <=>> PhO- + HCO3-}$$

The solubility of phenol does not necessitate that it releases carbon dioxide.

Now, the second protonation. If you want to quantitatively determine whether $\ce{CO2}$ will actually be liberated, you need more data than the $\mathrm{p}K_\mathrm{a}$ values, since quite a significant amount of $\ce{CO2}$ can exist in the aqueous form. However, one thing is certain: if you want to produce $\ce{CO2 (g)}$ you first have to protonate $\ce{HCO3-}$ to form $\ce{H2CO3}$ (which dissociates to give $\ce{CO2 (aq)}$).

This is not possible with phenol, since $\ce{H2CO3}$ has a $\mathrm{p}K_\mathrm{a}$ of $6.3$ (even after correcting for dissolved $\ce{CO2}$).


Friedel-Crafts synthesis - How is HCl produced?

You could perform the electrophilic aromatic substitution with acetyl chloride as well instead of acetic anhydride - in the first case it is evident where $\ce{HCl}$ comes from. However, the mechanism is similar if you use acetic anhydride. The first step consists in the electrophile activation by $\ce{AlCl3}$, which generates in situ the "acylium" ion ($\ce{H3C-C#O+}$) - the actual acylating agent. The second step is the electrophilic aromatic substitution, in which the electrophile is preferentially "attacked" by the aromatic ring at the ortho- or para-position (ortho- and para-orienting effect of halides). The driving force for the reaction is the last step, which leads to restore the aromaticity and it is assisted by $\ce{(AcO)(Al)Cl2}$ with liberation of $\ce{HCl}$.

Here I have drawn the mechanism:


Attack at ketone vs carboxylic acid carbonyls

The keto group is more reactive compared to the ester group due to the mesomeric effect (+M effect) caused by the hydroxyl group in the ester functionality, as already pointed out by orthocresol. See also  https://en.wikipedia.org/wiki/Carbonyl_reduction. Generally:
$$
 \mathrm{acylchloride} > \mathrm{anhydride} > \mathrm{aldehyde} > \mathrm{ketone} > \mathrm{ester} \approx \mathrm{carboxyl} > \mathrm{amide}
$$

A very detailed discussion on the subject is given for example in Warren Organic Synthesis: Strategy and Control or in Francis A. Carey Advanced Organic Chemistry - Part A.


Rust is a hydrated iron oxide that has a volume greater than the iron it contains. When iron rusts, the oxide grows in layers and curls away from the iron, exposing it to more corrosion.

Phosphoric acid is a moderately strong acid, but it forms an iron phosphate that is insoluble but adherent to iron (and steel). In that sense, phosphoric acid deactivates corrosion on iron (not perfectly, but pretty good), and there are commercial products (like Naval Jelly) which are essentially gelled phosphoric acid, designed to rub on iron and steel to give some corrosion protection. It is a sort of passivation.

Phosphate ions adhere to rust, making rust particles negative and promoting dispersion in the liquid and disadhesion from the iron metal. Some elbow grease is generally needed.

Overall, the extra corrosion from Coke or other phosphoric acid is minimal, and the dispersive effect is helpful, and the iron is better protected after the treatment than before.


What is the noble gas configuration of copper?

$\ce{[Ar] 4s^1 3d^10}$  is the correct configuration for copper. Why?

In simplest terms, some elements do not follow the Aufbau principle, so there are some alternate ways that electrons can arrange themselves that give these elements better stability.

Copper is a transition element and transition elements like either half filled orbitals or fully-filled orbitals. This makes them more stable.

For example:

Using the Aufbau principle, you would write the following electron configurations:

$\ce{Cr = [Ar] 4s^2 3d^4}$
$\ce{Cu = [Ar] 4s^2 3d^9}$

However, the actual electron configurations are:

$\ce{Cr = [Ar] 4s^1 3d^5}$
$\ce{Cu = [Ar] 4s^1 3d^10}$


Relationship between electric dipole moment and polarization?

In fact, the first equation is applicable on a microscopic scale  (for an atom or a molecule):
  $\space$ $\vec{p}_\text{ind}=\alpha \vec{E}_\text{local} $

While the second one is applicable on a macroscopic scale (for the bulk):
$\space$ $\vec{P}(\omega) \propto \chi^{(1)}(\omega) \vec{E}_\text{ext}(\omega) $
So, the polarisability is a microscopic quantity, while the susceptibility is a macroscopic quantity.

You have to notice that the electric field in the first equation is the local field, while it's the external field in the second one.
For more details, please see this page.


As pointed out in my earlier comment, you won't find tetraalkyl lead compounds in current petrol used for cars. In most countries, it was banned decades ago and replaced by MTBE.

However, the situation seems to be different for fuels used in aviation,   such as AvGas 100 where lead concentrations up to $\mathrm{1\,g\,L^{-1}}$ can be found.

Even the low lead variant, AvGas 100LL contains $\mathrm{0.56\,g\,L^{-1}}$ lead, see product specifications from TOTAL.

In an older technical note by R. E. Mansell and T. A. Hiller, published in Anal. Chem., 1973, 45, 975-975 (DOI), a possible approach for the decomposition of tetraethyl lead and subsequent detection of $\ce{Pb^{2+}}$ is outlined, referring to the ASTM D526-56 method:

  […] recommends refluxing 50 ml of gasoline with 50 ml of $\ce{HCl}$ for 30 minutes to decompose the tetraethyllead.[…]

In the aqueous layer, $\ce{Pb^{2+}}$ can be detected by complexometric titration with EDTA, either directly, or by back titration with a solution of a $\ce{Mg^{2+}}$ salt.


Compounds forming between silicon gallium and nitrogen

The ASM Alloy Phase Diagram Database (ASM Database if your institution has access) shows GaN and Si$_{3}$N$_{4}$ as the only compounds in that system. Standard binary alloy references agree. No ternary compounds are listed. Given the difference in crystal structures between all end members and the two nitride phases, mutual solubility of any two phases seems unlikely (little to none exists in the binaries, even Ga in Si has a pretty low solubility).


Can the total amount of solution be found as a ratio between molar mass of a component and total mass of solution?

To sum up the comments, only the following relation for the total amount of solution $n_\mathrm{tot}$ is universally true:
$$n_\mathrm{tot} = n_\mathrm{solvent} + n_\mathrm{solute} = \frac{m_\mathrm{solvent}}{M_\mathrm{solvent}} + \frac{m_\mathrm{solute}}{M_\mathrm{solute}}\tag{1}$$
The best you can do is to assume that $n_\mathrm{tot}\approx n_\mathrm{solvent}$ for the diluted solutions of small molecules. Also, if the molar masses are similar $(M_\mathrm{solvent}\approx M_\mathrm{solute}\approx \bar{M}),$ the expression can be lead to a common denominator:
$$n_\mathrm{tot} \approx \frac{m_\mathrm{solvent} + m_\mathrm{solute}}{\bar{M}}\tag{2}$$
This can be the case, for example, for the solution of ammonium nitrate $(M(\ce{NH4NO3}) = \pu{80.043 g mol^-1})$ in dimethyl sulfoxide $(M(\text{DMSO}) = \pu{78.13 g mol^-1}).$
Algebra aside, your mistake was also neglecting $M_\mathrm{solvent}.$
Keep in mind you can always check dimensions.
For your proposed formula

$$n_\mathrm{solvent} + n_\mathrm{solute} = \frac{M}{m_\mathrm{solvent} + m_\mathrm{solute}}$$

it doesn't work out well:
$$\mathrm{dim}~n_\mathrm{tot} = \mathsf{M}\cdot\mathsf{N}^{-1}\cdot\mathsf{M}^{-1} = \mathsf{N}^{-1} \neq \mathsf{N}\tag{3}$$
Illustrating with common units used in chemistry:
$$[n_\mathrm{tot}] = \frac{[M]}{[m_\mathrm{solvent}] + [m_\mathrm{solute}]} = \frac{\pu{g mol^-1}}{\pu{g}} = \pu{mol^-1} \neq \pu{mol}\tag{4}$$


Find a function of length depending on mass

Conversion factors:

$2.54 ~\mathrm{\frac{cm}{in}}$
$454 ~\mathrm{\frac{g}{lb}}$

Constants:

Wire diameter $d = 0.129~\mathrm{cm}$
Density of copper $\rho = 8.92~\mathrm{g\, cm^{-3}}$

Thus, our function $l(m)$ for the wire length from the wire mass in pounds reads:
$$ l(m) = m \cdot 454~\mathrm{\frac{g}{lb}} \cdot \frac{1}{\pi d^2 \rho}$$

Evaluate the function

Since I'm too lazy to evaluate something like this by hand (and since I somehow had to record this thought process and did not want to waste the processed remnants of a dead tree) there is a small python script available to do this for you, which you can find here and which, if you run it using the command python3 wires.py, will give you the following output:

  Wire length of 1.00 lb of 16-gauge wire: 9.71 m

(Bonus: You can run the command as python3 -i wires.py and will then be dropped into an interactive python shell, where you can call the wirelength function as many times as you wish using different masses for the wire.)


Did Berzelius unify the names of chemical elements using the Latin language?

What I have found from the literature is that Berzelius's proposal helped unifying the symbols of chemical elements, but the names remain diverse in language sources (not all are Latin) and some remain disputed (not all unified). Thus, I gather that Berzelius's proposal helps, but it is far from the statement that he unified all the names of chemical elements using Latin.

This is essentially true. Berzelius was the first to suggest using symbols abbreviated from the names of the elements and this system, with some modifications, is still used today. Most of Berzelius' symbols were abbreviated from Latin names but a large number of the elements discovered since then have had symbols drawn from their common names and some of Berzelius' original symbols have been dropped in favour of more modern versions. Dalton had earlier used a system of drawn symbols to represent the elements but this was not widely taken up as the symbols were difficult to draw and memorise. (source)

Looking though this wikipedia article on the etymology of chemical elements, the most common language of origin appears to be Greek (argon, chlorine etc.), with Latin coming in second (mercury, rhenium etc.), including quite a large number of names which come originally from another language (often Greek) via Latin.

Additionally there are a large number of elements which are named after places (Swedish - ytterbium, yttrium, etc. ; English - americium, californium, etc. to name just a few) or people (mendelevium, einsteinium, fermium etc.) and have been given the Latinate endings -ium or occasionally -um (see this question for more discussion). Also note that according to IUPAC recommendations:

  For linguistic consistency, the names of all new elements should end in “-ium”

The common, and IUPAC recognised, names of gold and gilver both have old Germanic origins and are recognised as official names because of their widespread usage. I have never seen the Latin names for gold, silver or lead (aurum, argentum and plumbum) used in a serious chemical context, although of course the elements' symbols (Au, Ag and Pb) are derived from them.

As such I would not recommend using the Latin names as the main recognized name in a Wikipedia page, although a side note about them might be justified.

EDIT: Thanks to @KlausWarzecha for pointing this out. There is one exception where the latin names (aurum, etc.) come into play: in anionic complexes, the latin version of the name is officially used. $\ce{[Cu(CN)4]^{3-}}$ is tetracyanocuprate(I), $\ce{[Ag(Cl)2]^-}$ is dichloroargentate(I), $\ce{[Au(Cl)4]^- }$ is tetrachloroaurate(III). However, this does not change the viewpoint that the Latin names should not be used when commonly discussing the element.


First you find the number of moles of $\ce{Ca3(PO4)2}$ reacting. You have 75% of 1 kg, i.e. 750 grams, of calcium phosphate. The rest of reagents are assumed to be in excess as we have to find the maximum amount of $\ce{P4}$ obtained, hence reaction should not have other limiting reagents.


Why is the HOMO-LUMO gap in N2 and CO of different sizes?

As correctly noted in the question, $\ce{CO}$ and $\ce{N2}$ are iso-electronic, so they should be comparable. But a difference remains: $\ce{CO}$ is hetero-nuclear, $\ce{N2}$ is homo-nuclear. The latter case typically means that there is better overlap between the orbitals (because they are of similar extent/energy). This translates into a greater gap between the bonding and the anti-bonding combination of the orbitals. This will definitely change the HOMO-LUMO gap. However, without performing a detailed quantum chemical calculation, it is not clear to me whether the HOMO-LUMO gap should be bigger or smaller for the homo-nuclear case, because one has to take the entire molecular orbital diagram into account.


How to drain engine oil by understanding oil viscocity issues

If you drain the oil when the engine is hot or at least warm, it is much less viscous and will drain pretty efficiently. Oils of different viscosity or even different density will surely mix well while the engine is running, so if you still want to do a flush, be sure to run the engine for a while after you add the new oil and before you do your second oil change.


The given graph indicates that $\ce{[NH3]}$ remains constant till the 15 minute mark which implies that the equilibrium was established till then. Now, after the 15 minute mark, $\ce{[NH3]}$ decreases and reaches a constant value which means, the equilibrium was disturbed and re-established.
$$\ce{N2 + 3H2 <=> 2NH3 + heat}$$
Since we observe a decrease in $\ce{[NH3]}$, the reaction must have shifted in the backward direction according to the given equilibrium.
Out of the five options, only match for this is (C) removing some amount of $\ce{N2}$. Let us analyse the other four options,
(A) Decrease in temperature: Note that the above reaction is exothermic and hence according to Le Chatlier’s principle, the reaction shifts in the forward direction for a decrease in temperature.
(B) Decrease in volume: Since we are dealing with a gaseous(ideal) equilibrium, a decrease in volume implies an increase in pressure and hence, according to Le Chatlier’s principle, the reaction shifts in the direction where there is less no. of moles ie., forward direction here.
(D) Removal of $\ce{NH3}$: As Gaurang pointed out in the comments, removal of $\ce{NH3}$ will cause a sharp fall along a vertical line, but the graph indicates a gradual fall in $\ce{[NH3]}$. Hence this is also not valid.
(E) Increase in volume: Assuming a sudden increase in volume, say n times, $\ce{[NH3]}$ should also fall by n-times. Clearly, the graph doesn’t match with our condition. Hence this is also incorrect.

So (C) must be the correct answer, as removing some $\ce{N2}$, shifts the reaction in the backward direction, thereby decreasing $\ce{NH3}$ gradually till equilibrium is re-attained.


Products of Lead(II) nitrate decomposition

Check if the green condensate is $\ce{N2O3}$. It has been reported to be blue-green.
In the presence of water vapor and oxygen, the nitrogen dioxide that you are trying to obtain is in equilibrium with nitric acid and nitrogen trioxide, as explained here.
As reported in the Journal of Industrial and Engineering Chemistry, volume 12, number 6, page 531-538 (1920), condensation of gaseous nitrogen dioxide in a spiral condensor at -5 °C in the presence of oxygen and water vapor resulted in liquid with distinct upper and lower layers, the upper layer being a concentrated nitric acid layer (essentially no water) and the lower layer being a diluted nitric acid layer (27 %) water.
So the reason you are getting side products is likely that water vapor and oxygen are also present and causing equilibrium reactions that yield nitric acid, nitrous acid and nitrogen trioxide.
Reference

CCXIV.—The dissociation of gaseous nitrogen trioxide, Bernard Mouat Jones,  J. Chem. Soc., Trans., 1914, 105, 2310-2322, DOI: 10.1039/CT9140502310


Why does a protein crystallize with a specific salt but not with another?

It would be a lie to postulate we knew much about the formation of protein crystals. Yes, we know the basic theory — that we need a supersaturated solution in which the protein is not denatured and which would then slowly evaporate until hopefully a single crystal starts growing — but we don’t have the slightest of clues of what exactly helps and what doesn’t help.

This is why protein crystallography is still trial and error (and nothing else). You try thousands of different conditions in an automated manner, select those with promising results for further examination and finally — if and only if you are lucky — end up with crystals suitable for X-ray diffraction.

Therefore, we don’t know and nobody knows why magnesium chloride might help the crystallisation of your protein but sodium chloride, calcium chloride, aluminium chloride or magnesium bromide don’t.
We don’t know and nobody knows why a $\pu{1.3M}$ solution of magnesium chloride might help the crystallisation but a $\pu{1.4M}$ or a $\pu{1.5M}$ solution don’t.
We don’t know and nobody knows why $\pu{23^\circ C}$ might be a good temperature for crystallisation but $\pu{21^\circ C}$ and $\pu{25^\circ C}$ aren’t.
And so on and so forth.


How to balance the disproportionation reaction of manganate anion in basic conditions?

You are referring to the disproportionation reaction of $\ce{Mn(VI)}$ to $\ce{Mn(IV)}$ and $\ce{Mn(VII)}$, which is:

$\ce{MnO4^{2-}(aq) -> MnO2 (s) + MnO4^{-}(aq)}$

Reduction: $\ce{MnO4^{2-} + 2e^{-} + 2H2O -> MnO2 + 4OH-}$

Oxidation: $\ce{2(MnO4^{2-} -> MnO4^{-} + 1e^{-})}$ = $\ce{2MnO4^{2-} -> 2MnO4^{-} + 2e^{-} }$

Combine both, and the final balanced equation is:

$\ce{3MnO4^{2-} + 2H2O -> MnO2 + 2MnO4^{-} + 4OH-}$


How did Scientists Capture The First Images of Molecules Before and After Reaction?

There are a ton of papers on how AFM works, but I'll have a go at explaining it anyhow.

Atomic Force Microscopy (AFM) is a venerable technology for imaging nanostructures down to individual molecules by physically contacting atoms on surfaces. The whole process is very similar to a finger reading Braille.

What does it mean to physically contact an atom? It means that the AFM works by detecting some kind of repulsion between an atom and a very, very sharp probe due to their respective electron clouds repelling. The forces between surfaces on the nanoscale are a complicated mixture of attraction due to Casimir-Polder/Van der Waals interactions, dipole-dipole interactions, electrostatic repulsion, Pauli exclusion, etc. etc., but the key point is that if two objects get too close together they will experience a rapidly increasing repulsive interaction, which the AFM picks up.

This repulsion is detected through the deflection of a cantilever (a thin, long, elastic beam of material). The probe is mounted on the end of this cantilever, and the cantilever bends when the probe is pushed up against a substrate. The deflection of this cantilever is measured by bouncing a laser beam off the end, the change in position of the reflected beam indicating the force.

The AFM probe has an atomically sharp tip that can localise single atoms on a surface, and the tip is moved around using exquisitely precise actuators so that the tip can be scanned back and forth over a surface, picking up the bumps and valleys of the sample that indicate the positions of atoms. I've seen ads for AFM probes that actually have flared tips - these can obtain even more 3D information about nanostructures as they can hook into nanostructures from the side and measure the texture and relief parallel to the surface.


Have a look at the MSDS of the hazardous materials, which you are planning to store together. At least in Germany, these materials are assigned to storage classes (in German: Lagerklasse (LGK)).

The German regulations are to be found in the document TRGS 510, TRGS stands for Technische Regeln für Gefahrstoffe (Technical Regulations for Hazardous Materials). Document 510 of this series specifies Lagerung von Gefahrstoffen in ortsbeweglichen Behältern (Storage of hazardous materials in non-stationary containers). Within TRGS 510, you'll find a "compatibility table" that specifies, which storage classes may (not) be stored together.

Sodium azide belongs to the LGK 6.1 A (flammable , acute toxic hazardous materials of category 1 and 2).

Materials from LGK 6.1 A may be stored together with other materials of the same LGK, but not, for example, with explosives (LGK 1), strong oxidants (*LGK 5.1 A), organic peroxides (*LGK 5.2), pyrophororic materials (*LGK 4.2), etc.


How to calculate the dissociation constant of a weak acid from the titration with a strong base?

This is a strong base weak acid titration. If the weak acid is of the form $\ce{HA}$, then at equivalence point, all the acid has been converted to its conjugate base.

Total volume of base required for equivalence point is $\mathrm{16~ml}$. Let the molarity of base used be $M$. Then, number of milli moles of base for equivalence point is $M\times 16$. This is also the number of milli moles of acid taken.

If only $\mathrm{2~ml}$ of base was used, there will be some acid and some salt of acid and base (A buffer, since the acid is weak and the base is strong).
$$pH=pk_a+\log\frac{M\times 2}{M\times 16-M\times 2}=pk_a+\log\frac{2}{14}$$
From this you will get $$6.912=pk_a-0.845$$
$$pk_a=7.757$$


What does *oxidizing* have to do with *hydrogen*?

An oxidizing acid is a Brønsted acid that is a strong oxidizing agent.
All Brønsted acids can act as oxidizing agents, because the acidic
proton can be reduced to hydrogen gas."

The whole sentence is written in a very convoluted fashion, and it is partially wrong as well. Someone should improve this Wikipedia section.
All they are saying is that strong acids do react with many metals and oxidize them. In turn their acidic proton is reduced to hydrogen.
For example, metallic Zn reacts with dilute sulfuric acid to form zinc sulfate and hydrogen gas. Zinc (0) has been oxidized to Zn(II). Consequently, Zn(0) also reduced the hydrogen ions to hydrogen gas.
Why this statement is wrong? The reason is that it says ALL Bronsted acids can do this, which is simply an over-generalization. For example, benzoic acid would not produce hydrogen with zinc. It is sparingly soluble in water.
Independent of this point, sulfuric and nitric acid are strong oxidizers in concentrated form. Here it is not behaving like an acid. Sulfuric acid, if concentrated will oxidize copper to Cu(II), and itself get reduced to sulfur dioxide.


Can protection group be used in a Wolff–Kishner reaction?

@Halake: The protection of adipaldehyde as the monoacetal was solved by Schreiber and Claus some years ago [Also see, Schreiber, S.L.; Claus, R.E.; Reagan, J. Tetrahedron Letters, 1982, 23, 3867.] Ozonolysis of cyclohexene (1) in the presence of methanol leads to the molozonide 2, that opens to the zwitterion 3 which is trapped as the methoxy hydroperoxide 4. Addition of p-toluenesulfonic acid catalyzes the formation of the acetal 5. After neutralization with sodium bicarbonate, reduction of the hydroperoxide with dimethyl sulfide affords the monodimethyl acetal of adipaldehyde 6. Since acetals are base stable, Wolff-Kishner reduction of the aldehyde can be accomplished followed by aqueous acid hydrolysis of the acetal.


Change in Gibbs free energy vs. change in enthalpy

To start off, $\Delta G$ is concentration-dependent and $\Delta H$ is not. You already showed the concentration-dependence of $\Delta G$. There is no equivalent relationship for $\Delta H$ because it does not depend on concentration. $\Delta H$ reflects how the bond energies change from reactants to products. Bond energies don't depend on concentration. (It also reflects other more subtle things like solvation and intermolecular interactions, but these don't change in an ideal solution).  In contrast, $\Delta G$ has an entropy term, which has to do with the entropy of mixing, resulting in a significant concentration-dependence even in ideal solutions.

  Is this understanding correct? Or does ΔH change throughout a reaction like ΔG?

The enthalpy H of the system changes throughout the reaction. Because the enthalpy is not concentration-dependent, it changes in a linear way with the amount that reacts. There are two versions of $\Delta H$ with different dimensions and units. One does what you say: compare H between two states. It has dimensions of energy (typical units: kJ). The other asks what the change of enthalpy is per amount reacted. It has dimensions of energy per amount (typical units: kJ/mol) and is called molar change in enthalpy.

An example is the molar enthalpy of formation. If you want to know how big the enthalpy change is to make a certain amount of a compound, you multiply the amount by the molar enthalpy of formation. The dimensions of the result will be energy again (units: kJ / mol * mol = kJ). Introductory textbooks sometimes gloss over the difference between molar (intensive) and extensive enthalpies.

  If ΔG gives information about a specific point in the reaction, then why is there a "Δ" at all?

The Gibbs energy G of the system changes throughout the reaction. The change per amount of product formed decreases as the reaction approaches equilibrium. Again, ΔG can mean two things, either the difference in Gibbs energy between an initial and a final state, or the molar change in Gibbs energy with given concentrations.

You might say concentrations change while the reaction is going on, so it is impossible to make a statement about specific concentrations. You solve this problem by either looking at a very small amount of change (infinitesimal, if you want) or by imagining very large volumes of reaction mix, where concentrations change very little even if - say - one mole of reactants turn into one mole of products. If you want to communicate that this is what you are talking about, you can either use $\Delta_r G$ or $\frac{dG}{d\xi}$. Both have dimensions of energy per amount of substance (typical units kJ/mol).

Again, introductory textbooks tend to gloss over this difference, not giving $\Delta G$ and $\Delta_r G$ different symbols even though they have different dimensions (one is extensive and the other intensive) and mean different things.


What is the name for the relationship between two elements in the same group?

The correct term is congener. Congeners refer to elements in the same group in periodic table.
For example, congeners of the Group 1 elements are Li , Na and K etc.
Therefore , X is a congener of Y is the right statement.


How does zinc become oxidized when it's not in direct contact with copper ions in a galvanic cell?

Zinc electrode is anode and copper electrode is cathode. The two electrodes are connected by a electrically conducting wire. At anode, $\ce{Zn}$ atoms are oxidised to $\ce{Zn^2+}$ which go into the solution, and the electrons are left on the metal strip.

The electrons produced in the oxidation half-reaction at anode are transferred through the wire to $\ce{Cu}$ electrode where they are consumed. The $\ce{Cu^2+}$ ions are reduced to metallic $\ce{Cu}$ which coats on the $\ce{Cu}$ strip.

Thus, oxidation and reduction half reactions occur at separate electrodes and the current flows through the wire. A salt bridge maintains neutrality in both the compartments by a flow of ions.

The overall cell reaction can be given as:

$$\ce{Zn (s) + Cu^2+ -> Zn^2+ + Cu (s)}$$


Light can be ‘felt’ because light is transmitted by photons. Depending on how you choose to observe light, it will either manifest itself as waves — for example in interference experiments — or as particles — for example when describing the photoelectric effect.

When described as a wave, the energy of a certain source of light will be determined by its wavelength by equation $(1)$ in which $h$ is the Planck constant and $c$ is the speed of light.
$$E = \frac{hc}\lambda\tag{1}$$
It may not immediately be obvious how this translates to the energy of a photon, but the key is the de Broglie wavelength as given in equation $(2)$.

$$\lambda = \frac hp\tag{2}$$

This equation connects all particle and wave properties of types of matter. (Indeed, due to the wave-particle dualism all particles can theoretically behave as waves and vice-versa; the wave nature of particles as large as entire molecules has been shown in interference experiments.)

When a photon with a certain energy travels through an atom or molecule, it may interact with the electrons present; often this interaction causes the photon to be absorbed and the electron to be excited to a higher energy level. As chemists, we describe this as an electronic transition from one orbital to another. The entire process can also be reversed, i.e. if an electron is excited to a higher energy level by any means, it can relax following certain quantum-mechanical selection rules back down to a lower energy level and emit a photon of a definite wavelength. The latter process can be observed in the flame colour given by sodium atoms: their 3s electron is excited to a 4p level thermally due to the heat of the flame. Upon relaxation, the electron emits a photon of $\pu{589nm}$ wavelength which we perceive as yellow. We could write that in an equation:
$$\ce{4p ->[][$\lambda = \pu{589nm}$] 3s}\tag{3}$$

If the target of a photon happens to be a molecule rather than an atom, we need to consider what type of orbitals we are exciting from and exciting into. Typically, the occupied orbitals will be bonding orbitals while unoccupied orbitals will be antibonding. Thus with the correct photon wavelength, an electron can be excited from a bonding into an antibonding orbital which we can, in layman’s terms, call ‘breaking a bond’. Whether or not the bond actually breaks depends on whether or not the atoms can dissociate apart.

A beautiful reaction to exemplify this is the reaction of chlorine and hydrogen in equimolar mixtures. Under ambient conditions, these two are metastable. If the activation energy is supplied, they will react to form hydrogen chloride.

$$\ce{H2 + Cl2 -> 2 HCl}\tag{4}$$

Activation energy can be supplied not only by a flame but also by light as long as its wavelength is short enough. A flash of blue light will work, a flash of red light will not. What happens? First, a photon excites an electron in a chlorine molecule. It will reach an antibonding, unoccupied orbital, the $4\sigma$ orbital. This causes the $\ce{Cl-Cl}$ bond to break liberating two chlorine radicals.

$$\ce{Cl2 + $h\nu$ -> [Cl2^*]^\ddagger -> 2 Cl^.}\tag{5}$$

These chlorine radicals are then energetic enough to react with hydrogen molecules to give hydrogen radicals and so on — a radical chain reaction.

$$\begin{align}\ce{Cl^. + H2 &-> HCl + H^.}\tag{6}\\
\ce{H^. + Cl2 &-> HCl + Cl^.}\tag{7}\end{align}$$

And this goes on until all the reactive species have been consumed and the radicals recombine. The reaction releases $\pu{92.3kJ/mol}$ and is therefore strongly exothermic, even explosive.

This mechanism of course also works for infrared light which we perceive as heat. However, since the energy of infrared light is very low, few reactions can be initiated by it.


How do catalysts provide alternative routes for reactions?

So catalysts provide an alternative pathway by allowing a new intermediate to be formed.
For example consider the reaction of $A$ and $B$ to form $C$. this means that
$$\ce{A + B -> C}$$
Say this reaction has  an activation energy of $x$ $kJmol^{-1}$.

This reaction could also be possible with a catalyst $X$. This means that a new possible route for the reaction could be (there a multiple routes the new reaction could take this is just a simple example):
$$\ce{A + X->AX}$$
$$\ce{AX + B->C +X}$$
Now you have ended up in the same position using a catalyst and when not using a catalyst, seen as you add both reaction and cancel out common species then you get back to the equation of:
$$\ce{A + B -> C}$$
These two reactions with the catalyst will have their own activation energies. Say these two reactions have activation energies $\pu{p}$ and $\pu{q}$ $\pu{kJmol^{-1}}$ and that $\pu{p>q}$ (doesn't matter if $\pu{p}$ applies to the first or second reaction) then if $x>p$ a particle needs less energy to proceed via the route with the catalyst compared to the route without the catalyst.

As a particle now needs less energy for the product $\pu{C}$ to be formed, more particles will now be able to take part in the reaction thus increasing the amount of successful collisions per second which is the same in an increase in rate. This can be seen nicely in a general Maxwell-Boltzmann distribution of molecular energies as the new route will mean there's a lower activation energy overall meaning a lot more particles can react.


How much does butter shrink when frozen?

Given that fat (consisting mainly of assorted triglycerides of varying saturation) is the major constituent by a wide margin, it's probably reasonable to compare the change in density w/r/t temperature of butter to the change measured in various vegetable oils, since they have roughly comparable chemical composition (and because extensive data on them actually exists). I found a study in which the densities of various vegetable oils were measured over a range of temperatures. On average, the increase in density resulting from cooling from 110.0°C down to 23.9°C was roughly 6-7%. The data points graphing density against temperature arrange very linearly, and the regression lines generated fit them exceedingly well. I see no reason to suppose that the trend wouldn't continue as temperature decreases further. Assuming the comparison is reasonably valid, you could extrapolate that the increase in density you'd observe over the temperature range you specified should be on the order of approximately 2% at best.

As for the amount of shrinkage, strictly considering volume, the ratio of densities (assuming the 2% approximation is reasonable), would be:

$\frac{D_2}{D_1} = 1.02$

That is, the density after cooling, $D_2$, is 2% greater than $D_1$. That implies, holding mass constant:

$\frac{V_1}{1.02} = V_2$

Hence, given that volume is in units of cubic length, you can get a rough idea of how much the length of each side/edge should shrink by taking the cube root of both sides of the equation (obviously, this is only completely accurate for a perfect cube, but it emphasizes that the shrinkage in each dimension is smaller than the shrinkage in volume):

$\sqrt[3]{V_2} \approx 0.993\sqrt[3]{V_1}$

Unless your stick of butter is really gargantuan, a decrease of 0.7% on an edge is not going to account for 1-2mm.

Edit: Additionally, I've read that, in general, the magnitude of relative thermal expansion/contraction that liquids experience is usually greater than that experienced by solids, meaning that liquid density varies more over a given temperature range than solid density. Of course, I'll add the caveat: that's a very broad statement that admits of various exceptions. Nevertheless, this suggests that the estimate of 2% I made above by comparison to data on liquid vegetable oils is, if anything, more likely to be too large than too small.


Why does a pickled gherkin glow only at one end when a current is passed through it?

It looks like an electric arc to me, more precisely, a single-phase arc -- it seems like AC current was applied, as "ground" and "neutral" ends were used. Hence the light emission, high observed current density, and high temperature. As you noticed, the yellow color is most likely due to the presence of $\ce{Na+}$ ions.

Any conducting (or not so, if the voltage is high enough) object put in contact with a couple of separated electrodes and applied current (DC, single-/three-phase AC) can undergo arc discharge. For example, with the current from the standard household outlet one can do the same with lemons (it's going to smell better though than the gherkin, at least at the beginning due to evaporating essential oils); one can also light a tip of a cigarette moistened with saliva; or put a piece of paper with a pencil drawing on fire within seconds (starting the ignition from the graphite layer) and so on. Even a piece of wood can also  glow when arc discharge paves the way in a form of graphite path after dielectric breakdown  in between electrodes of HV-generator ($\pu{5 .. 10 kV}$ depending on wood and moisture).

It is not the particular end of the gherkin that glows, it is an area where arc discharge takes place. One can also observe arc(s) hopping along between two elongated parallel electrodes in the very same gherkin:

Video courtesy of YouTube channel "ПРОСТАЯ НАУКА" (Russian; English "SIMPLE SCIENCE")

Predicting the path of the arc is a rather non-trivial physics problem. To put it simply, it partially depends on the contact area, which constantly changes mainly due to vigorous boiling substance, as well as the local pressure and composition of formed gases, which results in equilibrium or non-equilibrium plasma formation in positive column (luminous region in the environment between the electrodes).

The process of forming an electric arc can be simplified down to three distinguished stages:

Arc ignition, when, as a result of shock ionization and emission of electrons from the cathode, an arc discharge begins and the ionization intensity is higher than deionization.  As the electrodes are diverged from each other, the contact pressure and the contact surface are decreased due to increased transient resistance in addition to the resistance of the media (mostly vinegar/salt solution in this case), overheating  starts. High temperature subsequently contributes to thermionic emission, when the electron velocity increases so they can be eliminated from the electrode surface. The voltage on the contact gap is quickly restored, and since the distance between the contacts is small, an electric field of high intensity arises, under the influence of which more electrons get accelerated and escaped from the electrode surface. When they struck a neutral atom or molecule, they give away their kinetic energy. If this energy is sufficient to remove at least one electron from the shell of a neutral atom, then the ionization process takes place.
Stable arc combustion maintained by thermal ionization in the arc itself, when the ionization and deionization intensities are approximately equal. Free electrons and ions form plasma arc content, that can be represented as an ionized channel, in which the burning act takes place and continuous motion of the particles is ensured. Negatively charged particles (primarily electrons) move toward the anode, and positively charged particles (deprived of one or several electrons atoms and molecules of gases) in the opposite direction (toward the cathode). This contineous flow makes the conductivity of plasma comparable with the conductivity of metals. So the voltage drop along the length of the arc is small due to the intensive thermal ionization.
Arc extinction, when the deionization intensity is higher than the ionization. In addition to ionization, there is also deionization of atoms and molecules. The latter occurs mainly through diffusion of charged particles to the environment, and the recombination of electrons and positively charged ions, which are reunited into neutral particles releasing energy in a form of heat.

Video bloggers all over the internet popularized the heck out of this effect, which is, strictly speaking, has more to do with physics, rather than with chemistry.


Calculating enthalpy change for a real gas

You started out correctly by getting the relationship for $\mathrm{d}H$ for your van der Waals gas algebraically. The usual approach to evaluating a change like this for a real gas is to break the change into three steps (since you will typically know $C_p$ only in the limit of ideal gas behavior at low pressures):

Step 1: Calculate $\Delta H$ for the change from $\pu{10^5 Pa}$ and $\pu{300 K}$ to $\pu{0 Pa}$ at $\pu{300 K}$.

Step 2: Calculate $\Delta H$ for the change from $\pu{0 Pa}$ and $\pu{300 K}$ to $\pu{0 Pa}$ and $\pu{600 K}$ (i.e., using the ideal gas heat capacity).

Step 3: Calculate $\Delta H$ for the change from $\pu{0 Pa}$ and $\pu{600 K}$ to $\pu{2e5 Pa}$ and $\pu{600 K}$.

Based on Hess' law, add the three changes together.

For this particular problem, the 3 step procedure is not necessary because of the following:

$$\left(\frac{\partial H}{\partial P}\right)_T=b$$

So,

$$\mathrm{d}H = C_p\mathrm{d}T + b\mathrm{d}P$$

Since $b$ is independent of temperature, $C_p$ is independent of pressure. So for this system, the equation integrates immediately to

$$\Delta H = C_p\Delta T + b\Delta P$$


How can I obtain oxygen gas from plagioclase

Yes, I collect oxygen isotope data from plagioclase using the laser flourination method. To liberate the oxygen from the plagioclase, crystals are reacted with purified $\ce{BrF5}$. Then the plagioclase is hit with a $\ce{CO2}$ laser. Gasses are sent through a couple of cryogenic (liquid nitrogen) traps to filter, and then through a mercury diffusion pump to remove that nasty flourine.


Why do tosylation and mesylation of alcohols follow different mechanisms?

The only real difference between an alkanesulfonyl chlorides and arylsulfonyl chlorides is the $\alpha$-hydrogen. In a mechanistic study of the hydrolysis of methanesulfonyl chloride, the reaction was found to have a very small secondary kinetic isotope effect for $\mathrm{pH}<6.7$, indicating no bond breakage to the isotopically labelled hydrogen and thereby an $\mathrm{S_N2}$ reaction mechanism.

For $\mathrm{pH}>6.7$ there is a large primary kinetic isotope effect, indicating bond breakage to the isotopically labelled hydrogen, corresponding to the formation of the sulfene intermediate.$^{[1]}$ This shows that though methanesulfonyl chloride isn't incapable of direct attack by alcohols, in the presence of basic amines, which serve the additional purpose of driving the reaction to completion by accepting the alcohol proton, it is more quickly deprotonated to form the sulfene.

While pyridine isn't that great of a base ($\mathrm{p}K_\mathrm{b}=5.21$) and won't form the sulfene to any great extent, triethylamine ($\mathrm{p}K_\mathrm{b}=11.01$) and trimethylamine ($\mathrm{p}K_\mathrm{b}=9.81$) are much stronger bases and will deprotonate methanesulfonyl chloride to give the sulfene to a greater extent. The mechanism is just as dependent on the base as it is the sulfonating agent, so all these considerations must be taken into account.

$^{[1]}$ King, J. F.; Lam, J. Y. L.; Skonieczny, S. Journal of the American Chemical Society 1992, 114 (5), 1743–1749.


What are compounds with the same mass called?

In mass spectrometry, ions with the same nominal mass (but possibly differing exact masses) are called isobaric.


More acidic - protonated carbonyl or protonated hydroxyl?

You have outlined the correct reasoning but drawn the wrong conclusion. The protonated carbonyl structure (top left) is stabilized by resonance (bottom left), giving two equivalent resonance structures. Although the protonated hydroxyl (top right) has a resonance structure that can be drawn (bottom right), it should be considered a negligible contributor since there is a double positive charge on a very electronegative atom that would be sp2 hybridized (also toss in the charge separation). Since the protonated hydroxyl is not stabilized by resonance, it is more acidic, i.e., it is more willing to give away the proton (and keep the electrons).


How can I calculate the molar conductivity of a solution with multiple solutes?

If you are interested in the modelling of electrical conductivity of aqueous solutions containing more than a single salt, then there are already some references in literature, for example for the salinity of natural waters (Limnology and Oceanography Methods, doi 10.4319/lom.2008.6.489) or in systems relevant to chemical engineering (Ind. Eng. Chem. Res., doi 10.1021/ie9605903).

(from the latter)


What happens to iron(III) sulfate (ferric sulfate) in aqueous solution?

In aqueous solution $\ce{Fe2(SO4)3}$ dissociates into $\ce{Fe^{3+} (aq)}$ and $\ce{SO4^{2-} (aq)}$ ions. The sulphate ions will be solvated by hydrogen bonding with the water and the iron ions will form the hexaaquairon(III) complex, $\ce{[Fe(H2O)6]^{3+}}$.

As @Mithoron mentioned, the $\ce{Fe^{3+}}$ ions cannot easily be further oxidised so no additional reaction takes place.


Has a carbon compound ever been found having an ionic bond?

Why unlikely? Ionic compounds of carbon have been known for ages. There are ionic carbides ($\ce{Al4C3}$, $\ce{CaC2}$, etc.), graphite intercalation compounds like $\ce{KC8}$, ionic derivatives of fullerenes and more.

Come to think of it, common $\ce{CaCO3}$ is certainly ionic and at the same time a compound of carbon, but this is most likely not what you want.


Reaction rates are in most cases influenced by concentration for liquids/dissolved solids and surface area for solids.

Surface area

The surface area of the calcium carbonate is one factor in the rate of the reaction. The more surface area, the faster the reaction. In this case, certainly, the five $\pu{1kg}$ cubes will react much faster. The reason large cubes are used to demonstrate this instead of powder is because the surface area is more or less constant throughout the reaction.

The rate law
$$\text {Rate} = k[\text A]^\text m[\text B]^\text n $$ where $\text A$ and $\text {B}$ are the reactant concentration, $k$ is the rate constant (does not change in the same reaction), and $m$ and $n$ are the rate orders (which also does not change in a reaction)

This means that unless m and n are 0 making it a zero order reaction which is independent of concentration, in most cases concentration does have an effect on rate.

In this case, with $k$, m and n constant and either  m or n non-zero, and increase in $[\text A] \text{ or } [\text B]$ would increase the rate.

If $[\text A]$ is calcium carbonate and $[\text B]$ is hydrochloric acid, it shows that most likely an increase in concentration of $\text B$ will increase rate. However, calcium carbonate is a solid and has no concentration. We can imagine $[\text A]$ as the surface area of $\ce{CaCO3}$ instead, and m as 1.

We could also imagine the $\ce{CaCO3}$ dissolved in a solvent, like sugar in water. Now, the reaction rate will be dependent on the amount of
$\ce{CaCO3}$
(Note: the rate orders can only be determined experimentally)


Valence Bond theory, VSEPR theory and predicting the shapes of the molecules

VSEPR theory is a purely observational model, based on empirical knowledge. Many of the "rules" were added to fit the theory to what has been observed. It has quite a lot limitations and may only provide an initial guess for the structure of a molecule.

According to VSEPR, the structure of water is bent. While we would expect $\angle^\mathrm{theo}_\mathrm{bond}(\ce{HOH})=109.5^\circ$, it is indeed smaller $\angle^\mathrm{expt}_\mathrm{bond}(\ce{HOH})=104.5^\circ$. This discrepancy is explained with the reasoning that lone pairs require more space. This is wrong, and a caveat added to make the theory fit.
It would also predict the lone pairs of water to be equivalent. A statement which has experimentally been disproved. Quoting myself:

  One popular example is water. In many texts the central oxygen is described as having (approximately) four sp3 orbitals, which essentially makes the lone pairs equivalent. That view, however, is not in agreement with the photoelectron spectrum, which clearly shows that the lone pairs are not equivalent. At this point I'd like to refer you to Michael Laing's article: "No rabbit ears on water. The structure of the water molecule: What should we tell the students?" (J. Chem. Educ. 1987, 64 (2), 124.).

The theory sometimes predicts correct structures, but that has no theoretical foundation. It is basically educated guessing.
When it comes to more complicated molecules, for example hyper-coordinated compounds like $\ce{IF7}$, most explanations (the involvement of d orbitals) break completely down and have been disproved many times. However, just by minimising the inter-ligand distances, it yield approximately the correct shape. This is a nice coincidence.

Valence Bond theory is on the other hand a highly sophisticated and complex approach, deeply rooted in theory. If applied at a modern, computational level, it will always yield the correct molecular geometry for the correct reasons (within its other approximations).

While we can approximately use a piece of paper and a pencil to do these calculations for simple molecules, it really is an approach at formulating a wave function and approximately solving the electronic Schrödinger equation. That requires a lot of maths and integral solving. It is not so much a theory that can be done on-the-go.
As it is such an extensive and accurate theory, it is complementary to molecular orbital theory. That means that they will yield the same results at their respective accurate (infinite) expansions.

For the example of water, VB theory would predict a bent structure. It would also predict non-equivalent lone-pairs, which agrees nicely with the photoelectron spectrum.

In conclusion; A sad fact is that both theories, VSEPR and VB, are usually (more than just often) taught completely wrong at below university levels (sometimes even there) and will introduce many artefacts into young scientists brains. Your statement, that you "used" valence bond theory is a reminder of that.
These theories could not be more different from each other. In most cases they predict structure on completely different grounds. It is basically a coincidence if they predict the same thing.

Take home message: VSEPR theory is a rule of thumb, VB theory is an accurate description of physics (within its approximations).


What is an example of a biodegradable polyethylene?

In a short research I found little to no evidence, that a there is  polypropylene/polyethylene, that can actually be called biodegradeable. I have found a couple of articles that have this as a topic. I don't have the time to actually read all of them and since this is going to be to long for a comment, I'll share it as an answer.

S. Bonhomme, A. Cuer, A-M. Delort, J. Lemaire, M. Sancelme, G. Scott, Polymer Degradation and Stability 2003, 81, 441–452.
This is a general overview of the mechanisms of biodegradation of such materials.
Ignacy Jakubowicz, Polymer Degradation and Stability 2003, 80, 39–43.
This article investigates the degradation of polymers, which contain pro-oxidant additives. It mainly focusses on the kinetics of the thermo-oxidative degradation. I am certain you can learn from this article which kind of additives are used to make braking long polymers into oligomers easier and hence complete degradation faster.
Prasun K. Roy, Minna Hakkarainen, Indra K. Varma, Ann-Christine Albertsson, Environ. Sci. Technol. 2011, 45 (10), 4217–4227.
This article is very promising as it also deals with some long-term effects of polymers with pro-oxidative additives. The abstract already lays out the main problem, when talking about biodegradation: entering the eco-cycle harmlessly.

  However, the “degradable polyethylene” which is presently being promoted as an environmentally friendly alternative to the nondegradable counterpart, does not seem to meet this criterion. [...] It appears from the existing literature that our search for biodegradable polyethylene has not yet been realized.
  (Emphasis by me.)

All in all I think it is very important to carefully outline, what biodegradation is or should be. I hope these few articles can be a starting point for your further research.


How are solubilities of hydrate salts determined when competing equilibria are unavoidable?

Let there be three species: a hydrated solid $H$, an anhydrous solid $A$, and a complete solvated (liquid) aqueous solution, $S$.

Your question is about differences in the two equilibria $\ce{H <=> S}$ and $\ce{A <=> S}$.  Can $H$ really have a different solubility than $A$?

I think the answer is yes, but only if the equilibrium $\ce{H<=>A}$ cannot and is not established.  In the presence of an "infinite" kinetic barrier that blocks conversion of the hydrate to the anhydrous form and vice versa, it makes sense to treat $\ce{H<=>S}$ and $\ce{H<=>A}$ and independent equilibria which might have different equilibrium constants.

In a comment to your question, I made it seem like "any" precipitation from a (super)saturated solution of $S$ would be the hydrate.  On further reflection, I think that is true, unless (i) the equilibrium $\ce{A<=>H}$ cannot be established due to kinetic barriers, and (ii) there is solid phase $A$ present in the solution.  In that case, the supersaturation could just result in "precipitation" of solid onto the existing crystal surface of already-present $A$, a process that would increase the mass of $A$ (solid) in the system but which would not require formation of any crystallites or nucleation intermediates.

  How, then, was the solubility of the hydrate ever experimentally determined?

I think the answer is that the hydrate solubility can differ from the anhydrous solid solubility only in cases when there is no mechanism by which equilibrium can be established between the hydrate and the anhydrous solid.


What type of bond forms between an ionic compound solute and a polar solvent

Firstly, There is no such bond existing inbetween the polar solvent molecule and the solute ions. It is ion-dipole interaction which exists between the polar solvent molecules and the ions of the solute.

I hope you got your answer.

I wanted to add something more which I think you got wrong.

I agree that if hydration enthalpy is greater than lattice enthalpy then solute will dissolve but it is not compulsory for hydration enthalpy to be greater than lattice enthalpy for dissolving. What i mean is: For dissolving its important to break the ionic bonds between the ions of the solute molecule but that energy required to break the  ionic bond is not necessarily from hydration only. That is sometimes hydration enthalpy couldn't meet the energy required to break the ionic bonds (lattice energy). In those cases the extra amount of energy can be from heat of the solvent. In the cases where hydration enthalpy is larger than lattice energy, ΔH is negative. In cases where hydration enthalpy is lesser than lattice energy, in those cases the more energy needed is driven from the heat of the solvent and in this case ΔH becomes positive. Negative ΔH is preferred to be better for dissolving because it helps in making ΔG more negative. But it doesn't mean solute with positive ΔH doesn't dissolve. If ΔS is highly negative, then even though ΔH is positive still it can make ΔG negative and hence dissolving possible. Dissolving of any solute depends on ΔG. For dissolving, ΔH doesn't matter as long as ΔG is negative. Say for example in case of Nacl, net solvation enthalpy is positive, but still it dissolves in water.

Nacl hydration enthalpy is not greater than enthalpy required to break the bonds in between Na+ and Cl-. As hydration enthalpy is not sufficient to release energy which is required for bond breaking incase of Nacl, hence the extra amount of energy required comes from the heat of the water ,which in addition to the hydration enthalpy is able to break the bonds in between Nacl. As energy released is less than net energy absorbed, hence net enthalpy (solvation enthalpy) is positive.The reason behind the dissolving of the salt is ΔS being highly positive which in turn ends up making ΔG negative as ΔG= ΔH −TΔS.

So for a well dissolving ionic solute, hydration enthalpy is always not greater than the energy required to break the ions apart in the solute molecule. An ionic solute with positive ΔH would still dissolve if ΔG is negative.


This is due to the fact that we also consider +M effect . In fluorine it’s 2p orbital that participated in overlapping but in chlorine it’s 3 p orbital. 3p of chlorine does not overlap efficiently with 2p of benzene carbon so +m decreases in chlorine thus acidic nature increases


Explosives that use ammonium nitrate and a fuel are mostly referred to as ANFO explosives, you might find more detailled information by using that term.

Most explosives used in military (especially bombs) are based on TNT, mostly in the form of Tritonal, a mixture of TNT and aluminium powder. Generally, only outdated bombs use ANFO explosives (due to their low price and ease-of-use). (See BLU-82, which does indeed use ammonium nitrate)

ANFO explosives are rather used in an industrial context, especially mining.

In the common mixture of ammonium nitrate with nitromethane (ANNM), no oxygen from the air is needed to sustain the explosion:

$\ce{3NH4NO3 + 2CH3NO2 -> 4N2 + 2CO2 + 9H2O}$

Since ANFO explosives carry their oxygen with them by definition (the ammonium nitrate molecule contains 3 oxygen atoms), they do not need oxygen from the surroundings and are therefore not thermobaric or fuel.air weapons.


How do I convert this ion exchange capacity into water softening capacity?

In the H form there are one plus charged ions ($\ce{H+}$) being exchanged in the water softening process. But the species being exchanged for the $\ce{H+}$ ions are doubly charged ($\ce{Mg^2+}$ and $\ce{Ca^2+}$), so two $\ce{H+}$ ions must be exchanged per $\ce{Mg^2+}$ or $\ce{Ca^2+}$ ion.

So, "1.0 mEq/mL min of 99% in H form" could equivalently be read as:

The exchange resin has the ability to exchange $\pu{0.5mmol}$ of $\ce{Mg^2+}$ or $\ce{Ca^2+}$ ions per $\pu{mL}$ of resin.


In water it should dissociate into $\ce{K+}$ and $\ce{[Fe(CN)6]^3-}$. This is because the $\ce{Fe}$ forms an octahedral complex with $\ce{Fe}$ at the centre and six $\ce{CN-}$ ions distributed around this each at 90 degrees to one another. This ion is sufficiently stable that it will not dissociate in water. The bonding is due to $\mathrm d$ orbitals on the $\ce{Fe}$. All undergraduate inorganic text books will give details of this. The Wikipedia page on ferricyanide gives pictures of the complex's geometry.


First of all, you are extremely lucky in asking this question for cyclohexane: cyclohexane serves as reference material for checking your intensity calibration, so it is one of the few substances for which integrated band intensity ratios are available in the literature.

Integral intensity 2567 - 3068 cm$^{-1}$ is higher than integral intensity 700 - 900 (i.e. 801 cm$^{-1}$ band). At 532 nm excitation, the ratio is roughly 10.3 : 1.

You always have the fundamental $\nu^4$-dependence of Raman scattered intensity. If you look closely, that's $\nu_0 (\nu_0 - \nu_j)^3$ for the band with Raman shift $\nu_j$ for photon count and  $(\nu_0 - \nu_j)^4$ for power. Thus, the relative band area differs depending on excitation wavelength.
For dispersive Raman instruments, the sensitivity is highly wavelength-dependent and this varies from instrument to instrument and for instruments with moving grating also depending on the grating position.

Thus, for intensity comparisons across different instruments, you'll need to calibrate the intensity axis for the spectrometers (and possibly fold the higher-resolved spectrum with the difference in line shape if that isn't negligible due to integration of broad spectral ranges)
However, when training chemometric models (calibration, quantitation, classification, ...) for one instrument (or maybe very similar instruments: say, same manufacturer & model), as long as that instrument is stable intensity calibration is not necessary as it will automatically become part of the model.
Which is why many Raman spectra are shown as raw (uncalibrated) intensity.

This doesn't have much to do with "highly sophisticated" Raman spectrometer rather than doing sophisticated calibration or not. And among the highly sophisticated Raman spectrometers, you have tradeoffs that can make such a calibration more or less stable. In fact, different instruments are highly sophisticated in different aspects (e.g. spatial resolution, spectral resolution, signal stability/drift, versatility/possibility to adapt to different experimental setups, ...) some of these influence how easy intensity calibration is and/or how often it needs to be re-done.

Literature:

McCreery: Photometric Standards for Raman Spectroscopy, in: Handbook of Vibrational Spectroscopy, Wiley 2006
McCreery: Raman Spectroscopy for Chemical Analysis, Wiley 2000


What kind of energies are present in an electron?

From your question I assume you are talking about an electron bound to a nucleus, probably hydrogen. Since the mass of the nucleus is ~1836 times larger than the mass of the electron, we may assume the nucleus to be stationary. In a classical model of the problem, we thus have an electron with a certain velocity $v_n$ and kinetic energy $\frac{1}{2}m_ev_n^2$ at a position $r_n$ from the nucleus where it is bound by a Coulombic potential $-Ze^2/(4\pi\epsilon_0r_n)$ (where for hydrogen $Z=1$). The total energy is given by the sum of the two

$$
E_n=\frac{1}{2}m_ev_n^2-\frac{Ze^2}{4\pi\epsilon_0r_n}
$$

In order for the electron to make a (classical) stable trajectory around the nucleus, the electrostatic force should be equal but opposite to the centripedal force, hence

$$
\frac{m_ev_n^2}{r_n}=\frac{Ze^2}{4\pi\epsilon_0r_n^2}
$$
from which it follows that $\frac{1}{2}m_ev_n^2=Ze^2/(8\pi\epsilon_0r_n)$ and the total energy is given by
$$
E_n=\frac{Ze^2}{8\pi\epsilon_0r_n}-\frac{Ze^2}{4\pi\epsilon_0r_n}=-\frac{Ze^2}{8\pi\epsilon_0r_n}
$$
This equation shows that the total (average) kinetic energy in the hydrogen atom is half the size and opposite to the (avarage) potential energy, i.e., $\langle E_\text{kin}\rangle=-\frac{1}{2}\langle E_\text{pot}\rangle$, a result that is also predicted by the virial theorem.

Using Bohr's postulate concerning the quantization of angular momentum, the last expression can be recast in the familiar expression for the hydrogen energies.


$\mathbf{c}_n$ is a vector with $N$ entries (I switched to boldface to indicate - hopefully slightly more clearly - that it is a vector):

$$\mathbf{c}_n = \begin{pmatrix}c_{1n} \\ c_{2n} \\ \vdots \\ c_{Nn}\end{pmatrix}$$

This effectively means the same thing as writing

$$\psi_n = c_{1n}\phi_1 + c_{2n}\phi_2 + \cdots + c_{Nn}\phi_N = \sum_j c_{jn}\phi_j$$

I'll try to contextualise it. The columns of the eigenvector matrix are simply the eigenvectors themselves, so if (for example) the eigenvectors of an operator $\mathbf{A}$ are

$$\mathbf{c}_1 = \begin{pmatrix}p \\ q\end{pmatrix}; \qquad \mathbf{c}_2 = \begin{pmatrix}r \\ s\end{pmatrix}$$

then the corresponding eigenvector matrix $\mathbf{C}$ is

$$\mathbf{C} = \begin{pmatrix} \mathbf{c}_1 & \mathbf{c}_2 \end{pmatrix} = \begin{pmatrix} p & r \\ q & s \end{pmatrix}$$

[As an addendum: If the eigenvectors $\mathbf{c}_i$ are normalised, then you will find that

$$\mathbf{C}^{-1}\mathbf{A}\mathbf{C} = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$$

where $\lambda_i$ is the eigenvalue of $\mathbf{c}_i$, i.e. $\mathbf{Ac}_i = \lambda_i\mathbf{c}_i$.]


Naturally occuring amino acid with basic sidechain - L-Arginine:

  Oral supplementation with L-arginine at doses up to 15 grams daily are generally well tolerated.

source here


What causes electrons to move from zinc to copper?

There is no reaction since the electrons will only flow one way in the wire. When you connect the zinc metal to the copper metal with a wire, there is a voltage potential generated between them, just like in a thermocouple.

From Wikipedia:

  Any junction of dissimilar metals will produce an electric potential related to temperature.

In this case, we don't care about the temperature effects.

When a salt bridge is placed between the solution, then the electrons are carried by the anions in the solution from one electrode to the other electrode.

So the zinc metal releases electrons when it dissolves, which travel through the wire easily, to the copper. The $\ce {Cu^{2+}}$ in solution grabs the electrons from the copper electrode, adding copper metal to the electrode. Since there is now more negative ions in this solution, the anions move across the bridge to the zinc electrode. The anions attack the zinc metal, causing the zinc to loose electrons when it dissolves.

I'm not sure what starts the reaction, nor why the zinc dissolves easily.


I don't know about that particular GitHub archive, although I'll point out that the smiles.csv contains several kinds of SMILES - the final column containing the correct SMILES.

IMHO, this is an unfortunate result of people mis-using SMILES. As you note, it's perfectly possible to take a chiral molecular structure and output a SMILES with no stereochemical information.

Take alanine - to a chemist, if you simply say 'alanine,' they immediately think 'amino acid' and translate to L-alanine, or C[C@@H](C(=O)O)N as the so-called "isomeric SMILES":

Unfortunately, if you go to PubChem for L-Alanine, you'll see a "canonical SMILES" which sounds official and is CC(C(=O)O)N - no stereochemistry!

The official version of QM9 can be found on FigShare based on the corresponding publication.

While the official files are in a somewhat non-standard XYZ format, they do have the stereochemically-correct SMILES at the end of the file, e.g. C1N2[C@H]3[C@@H]4[C@@H]5O[C@]13[C@H]2[C@H]45 Just watch that some numbers are formatted like -8.0386*^-6 (i.e., $-8 \times 10^{-6}$) and can sometimes be misinterpreted by programs like Open Babel or RDKit .. practically everything I've seen.


Molecular orbital theory and the conservation of molecular orbitals

The two statements are not contradictory. There are indeed two ways to combine the two atomic orbitals to form new molecular orbitals. It is not either one way or the other--in fact both occur - one way creates a bonding molecular orbital and the other creates an antibonding molecular orbital.

So for example, if you have two ground state hydrogen atoms, each with an electron in the 1s orbital, you end up with a hydrogen molecule, with 2 electrons in the lower energy (sigma) bonding molecular orbital and 0 electrons in the (sigma*) antibonding molecular orbital. The same thing is true for formation of pi bonds from p orbitals.


Making carbon powder using fireplace and weak acid

The result of your first pyrolysis must be charcoal plus the mineral constituants that are present in wood ashes. Now the wood ashes are usually a mixture of : $40 - 70$ % $\ce{CaCO3}$, $5 - 10$ % $\ce{Na2CO3}$,  $5 - 10$ % $\ce{K2CO3}$, $2 - 3$ % $\ce{K2SO4}$, $2 - 5$ % $\ce{Ca3(PO4)2}$, $5 - 10$ % $\ce{MgCO3}$, $2 - 10$ % $\ce{SiO2}$, < $1$ % $\ce{FePO4}$. Washing your charcoal with water and hydrochloric acid must remove all these substances, except $\ce{SiO2}$. This impurity cannot be removed.


What does "dye content 80%" mean on Sigma Aldrich's product page for "Allura Red AC" (aka Red 40)

Many dyes are sold as powders with 80-90 % purity. It is because some dyes are difficult to purify (e.g. they are difficult to crystallize). The remaining 10-20 % are impurities, such as isomers or closely chemically related side-products, but mostly they are "volatile impurities" (i.e. water). For most purposes these impurities do not interfere, so nobody bothers to remove them. Here is MSDS of Allura Red:
http://www.vidhifoodcolour.com/uploads/gallery/14067976084allura_red.pdf


Dehydration of secondary alcohol via E1 mechanism

Your proposed mechanism looks reasonable. I think that acetic acid primarily serves as an acidic solvent for the reaction. However, in the presence of concentrated sulfuric acid, $\ce{CH3CO2H}$ can be protonated by $\ce{H2SO4}$, like in the first step of this esterification. The resulting resonance-stabilized cation is strongly acidic ($\mathrm{p}K_\mathrm{a} = -6.1$ according to this source), and should readily protonate the alcohol. This would be a possibility to include acetic acid into the mechanism.

$$\ce{CH3CO2H + H2SO4 <=> CH3C(OH)2^{+} + HSO4-}$$

(Image source: ChemGuide)

So in the first step, the alcohol can either be protonated by $\ce{H2SO4}$ or $\ce{CH3C(OH)2^{+}}$, and in the end, the eliminated proton can either be picked up by $\ce{HSO4-}$ or $\ce{H2O}$.


Which argon compounds exist, supported by experiment or computation?

ArBeO and ArBeS are experimentally known.

ArAuF, ArAgF and ArCuF are also experimentally known.

CUOAr$_n$ and related complexes are known.

ArBeNCN and ArBeNBO are predicted by theory.


Whats the Balanced chemical equation for heating sodium chloride and what type of a reaction is it?

My guess would be that this was rather the dissociation of Sodium Carbonate or Sodium Hydrogen Carbonate.

The equations for the dissociation of Sodium Carbonate and Sodium Hydrogen Carbonate, respectively, are as follows:

  $$\ce{Na2CO3 -> Na2O + CO2}$$

$$\ce{and}$$

  $$\ce{2 NaHCO3 -> Na2O + H2O + 2 CO2}$$


The Crisscross method for finding the chemical formula

What is the logic behind this method, how to prove this method?

The logic of the method is in fact trivial. Suppose you have to write a compound knowing his oxidation numbers :
$$\ce{xA^{a} + yB^{b} -> A_{x}B_{y}}$$

Where $a$ and $b$ are oxidation numbers  that have opposite sings,  $A$, $B$ chemical species and $x$, $y$ stoichiometric coefficient $ \in \mathbb{N}$.
We can start from this assumption: if a compound is neutral the algebraic sum of the oxidation numbers of his atoms must be zero ( point .4 of current IUPAC definition of the oxidation state) so we have to find when:

$$ ax+by=0 \space [1] $$

But we assumed that $a$ and $b$ have opposite sings so we can write:

$$ ax - by=0$$

This is a single equation with two variables so you can't solve it easily because it can have several solutions. However when $\textbf{x=b}$ and $\textbf{y=a}$ the equation is always solved, because:

$$ ab - ab=0$$
And so you can always write a compound formula that satisfy equation 1 using this formula:
$$\ce{xA^{a} + yB^{b} -> A_{b}B_{a}}$$

This works for $\ce{Al^{+3} , O^{-2}}$:
$$\ce{xAl^{+3} + yO^{-2} -> Al_{2}O_{3}}$$
 but what's about $\ce{Ti^{+4} , O^{-2}}$?

This turns out $\ce{Ti2O4}$ and not $\ce{TiO2}$ we need another step, we need to divide the subscripts by their greatest common factor gcf(a,b) because we have found another solution of the equation 1 with higher coefficient. In this case the gcf(2,4) is 2 and dividing the stoichiometric coefficient by two we get $\ce{TiO2}$.

Who, how and when developed this method?

I've found it the first time in Chemistry and Human Affairs by William Evans Price, George Howard Bruce, World Book Company, 1949. I don't know who is the inventor.
I think you can easily deduce the crisscross method simply looking at many compounds and their oxidation number! Is not so hard to imagine!

Is this method only applicable to ionic compounds?

No, it works for every compounds that can be describe with oxidation numbers. Oxidation numbers are the real charge in ionic compounds and a formal (or imaginary) charge in compounds with covalent bonds, look at $TiO_2$ example.

Why compounds like $GaAs$ do not follow this rule?

Is not right if we assume that we are dealing with $Ga(+3)$ and $As(-3)$ the equation is this:
$$\ce{xGa^{+3} + yAs^{-3} -> Ga_{3}As_{3}}$$
if we divide the subscripts by the gcf(3,3)  that is 3, we get $GaAs$. So this method will be always valid until you can describe the compound effectively with oxidation numbers, unfortunately not all the oxidation numbers are so easy to determine.

In fact I don't find the criss cross method useful I think is better to calculate the least common multiple of $a$ and $b$ and then divide it by $a$ for $x$ and $b$ for $y$.
I think this photo, from J. Chem. Educ., 2012, 89 (11), pp 1436–1438, summarize the best way to do it:

Edit:

The general solution of |a|x=|b|y  s.t.  x,y∈ℕ is x=n|a| and y=n|b| where n∈ℕ, that is we can use x=2|a|,3|a|...n|a| and similarly y=2|b|,3|b|...n|b| then why use x=|a| and y=|b| only?

(rearranged after the advices in the comment of mannaia)
In fact you can write $Na_{2}Cl_{2}$ without breaking any charge neutrality or valence rule. NaCl is a crystal with a lattice and you  can't consider $NaCl$ like a stand alone unit, what matter is the charge balance, the proportion between positive and negative ions. Maybe if you are dealing with a crystal made of 23456 atoms of $Na^{+}$ and 23456 atoms of $Cl^{-}$ you could  write $Na_{23456}Cl_{23456}$ this is in fact a solution of the equation and is not theoretically wrong however if you are talking about a generic crystal of NaCl you should write the ratio between $Na^{+}$ and $Cl^{-}$ ions it has no sense to choose randomly one of the solutions because has stated mannaia: "there's no need to choose a larger unit, whereas with a smaller one you can fully explain how the crystal is formed". Mathematically speaking we write the ratio to Lowest Terms.

Why always take the "gcf"?

In fact we don't take the "gcf" we divide for the "gcf" because this is the best way to get the ratio in the lowest terms.

Furthermore as explained in the $TiO_{2}$ example.
$$
A=\{(b,a),(2b,2a),(3b,3a),\ldots\}.
$$
Is a set of solutions of the equation but not all the solutions. If you have a=6 b=-4 you have the solution (2,3) too, here is the plot of the solutions in $\mathbb{N}^{2}$.

For compounds with covalent bond is different, because you are dealing with different stand alone units called molecule. For example if you have $C^{-4}$ and $H^{+1}$ the solutions not divide by gcf of the equation have no sense. For example $C_{2}H_{8}$ doesn't exist due the valence of the atoms involved.


The definition of n-factor, what was it when it was introduced?

Instead of asking what n-factor really was, a better question would have been what an equivalent was (But that would have been a duplicate)
According to the above question, the formal definition of an equivalent is as follows:

An equivalent (symbol: officially equiv; unofficially but often Eq) is the amount of a substance that reacts with (or is equivalent to) an arbitrary amount of another substance in a given chemical reaction.
In a more formal definition, the equivalent is the amount of a substance needed to do one of the following:

react with or supply one mole of hydrogen ions ($\ce{H+}$) in an acid-base reaction.
react with or supply one mole of electrons in a redox reaction.

Therefore given any chemical reaction, one equivalent of each reactant reacts to form one equivalent of each product, irrespective of the stoichiometric coefficients
$$\ce{aA + bB -> cC + dD}$$
In all such scenarios, the n-factor of the compound is defined as,
$$\textit{n}\text{-factor} = \frac{M_\mathrm{molar}}{M_\mathrm{equiv}}$$
However, why are there so many different definitions for calculating the same thing. This is because an equivalent doesn't have one fixed definition, rather depending upon the situation, the definition of an equivalent changes and hence the n-factor definition changes as well.
Therefore there is no formal definition per se for the n-factor, simply because it was never formally introduced. As M.Farooq pointed out in the comments,

I have only seen n-factor in South Asian textbooks. This is predominantly a South Asian college term for calculating chemical equivalents in current curricula. Here n stands for normality. You will not find this usage in British/US books. The n-factor is simply a coversion factor for molarity to normality concentration. Since the definition of an equivalent is dependent on the type of reaction we need to be aware of several scenarios which you have listed viz. acid-base, salts, redox reactions.


Reaction of haloalkane with hydroxide ions

You are on the right track with the hydroxide substituting the chloride, so you can predict the alcohol product in each reaction. In order to determine whether the reactions will be SN1 or SN2, we must look at the mechanism of each reaction:
SN1

SN2

Wikipedia:Nucleophilic_Substitution
So we can see that in an SN1 reaction, a stable carbocation intermediate must be formed, while in an SN2 reaction there is no intermediate. We can determine the stability of our carbocation intermediate by looking at the following stability diagram:

Wikipedia:Carbocation
From the diagram, you can see that carbocation stability increases with the number of alkyl groups attached to the ionized carbon. If the carbocation is relatively stable, the reaction will likely be SN1, if the carbocation is relatively unstable, the reaction will likely be SN2.
Another important idea, which you have mentioned, is that a central carbon surrounded by bulky groups will be more difficult to attack due to steric hindrance (thus not likely an SN2 reaction), and becomes more easily attacked when dissociated to a carbocation.
You can use these concepts to find/explain the answers to your question.


If I understand you correctly you are asking why osmosis happens the way it does. And obviously you want a more "real-life-approach" to this than the pure mathematically description via potentials.

Think about a cell filled with water that is divided by a permeable membrane that will only let water molecules pass through it. In this initial state the water molecules on both side will occasionally pass through the membrane switching sides. They do so by pure chance and as the particle-concentration on both sides are equal (there are 100% water particles on both sides) the pass through on both sides with an equal rate leaving the overall particle-distribution unchanged.
Now we go ahead and dilute salt in one half of the box. Leaving aside effects like hydration-hulls forming around the ions (which will magnify the effect described below) the effect of this dilution is that we now have a lower water-particle-concentration on that side. Therefore every now and then when a water particle would have been "thrown" at the membrane to pass through it it's not a water particle but an ion instead that can't pass the membrane. Therefore the rate at which the water-particles switch to the non-dilution-side is lower than the (unchanged) rate in which the water-particles flow into the dilution (by pure statistics). This leads to an increase in the overall particle count on the dilution-side.
This increase in particles also increase the pressure and therefore the rate at which random particles (on the dilution side) are being "thrown" at the membrane until this higher rate ultimately compensates the lower water-particle-concentration. At this point both sides are at an equilibrium again. However this equilibrium is in a state in which there are more water-particles on the dilution-side than there have been in the beginning.

And this is what we call osmosis.


The mole is a base unit as specified in the Système international d’unités (SI) by the bureau international des poids et mesures. Its decisive definition is that published in French:

  La mole est la quantité de matière d’un système contenant autant d’entités élémentaires qu’il y a d’atomes dans 0,012 kilogramme de carbone 12 ; son symbole est « mol ».
  Lorsqu’on emploie la mole, les entités élémentaires doivent être spécifiées et peuvent être des atomes, des molécules, des ions, des électrons, d’autres particules ou des groupements spécifiés de telles particules.
  La mole est une unité de base du Système international d’unités.

The proposal was brought forth by the International Union of Pure and Applied Physics (IUPAP), the International Union of Pure and Applied Chemistry (IUPAC) and the International Organisation for Standardization (ISO). As with all SI texts, the decisive French version has a semi-official English translation:

  The mole is the amount of substance of a system which contains as many
  elementary entities as there are atoms in $0.012$ kilogram of carbon 12; its
  symbol is “mol”.
  When the mole is used, the elementary entities must be specified and may be
  atoms, molecules, ions, electrons, other particles, or specified groups of
  such particles.

(The third point is not translated.)

This definition is more or less identical with the one in your first paragraph.

In practice and well within experimental error, this means that your later definition will hold true for any substance. I.e. take $1~\mathrm{mol}$ of an entity and the combined mass of that mole will be the same numerical value in grams as a single entity has in atomic mass units ($\mathrm{u}$). It is not the correct definition, and for any entity that is not carbon-12 the masses will differ slightly (but well within margin of your macroscopic experimental error) but it is good enough for most contexts.

A redefinition of the SI units is being discussed and will likely be adopted at the 26th General Conference of Weights and Measures in autumn 2018. This would redefine the mole in a way that the Avogadro constant is defined to be numerically exactly $6.02214 \cdot 10^{23}~\mathrm{mol^{-1}}$ (with a few further digits appended to the end of the number that yet need agreement). This would mean that the new definition of the mole would be along the lines of:

  The mole, mol, is the unit of amount of substance of a specified elementary entity, which may be an atom, molecule, ion, electron, any other particle or a specified group of such particles; its magnitude is set by fixing the numerical value of the Avogadro constant to be equal to exactly $6.02214X \cdot 10^{23}$ when it is expressed in the unit $\mathrm{mol^{-1}}$.

Currently, the Avogadro constant must be measured experimentally giving a value of $6.022140857(74)~\mathrm{mol^{-1}}$; the digits in brackets express the numerical uncertainty.

This will mean that $1~\mathrm{mol}$ of carbon-12 atoms will no longer have the mass of exactly $12~\mathrm{g}$ (but again, it will be well within experimental error for everybody not practising theoretical physics).

To answer the follow-up question you asked in the comments: The coefficients in chemical equations such as

$$\ce{Zn + 2 HCl -> ZnCl2 + H2}$$

are always and exclusively to be understood as ratio coefficients. Thus, instead of thinking one atom or one mole of zinc, think amount $n$ of zinc and amount $2n$ of $\ce{HCl}$.


Would sodium metal react differently with aqueous solutions than with pure water?

It helps to know what the reaction actually is. When sodium metal reacts with water, it is oxidised and the hydrogen in water is reduced according to the following equation:

$$\ce{2Na (s) + 2 H2O (l) -> 2 Na+ (aq) + 2 OH- (aq) + H2 ^ (g)}\tag{1}$$

Note that no other components play a role; the only thing that does is the $\mathrm{pH}$ value of the water (lower $\mathrm{pH}$ would mean a more violent reaction) — this is a spoiler and later shown in equation $(8)$. Since this is a redox reaction, we can use standard half-cell theory to analyse it. Most importantly, thanks to equation $(2)$, only the effective cell potential influences $\Delta G$ ($z$ is the number of electrons transferred while $F$ is Faraday’s constant) which in turn can be separated into two half-cells according to equation $(3)$ which can then be analysed by the Nernst equation $(4)$.

$$\Delta G_\text{cell} = - n_{\ce{e-}}FE_\text{cell}\tag{2}$$
$$E_\text{cell} = E_\text{cathode} - E_\text{anode}\tag{3}$$
$$E = E^0 + \frac{RT}{zF} \ln \frac{a_\text{Ox}}{a_\text{red}} = E^0 + \frac{0.059~\mathrm{V}}{z} \lg \frac{a_\text{Ox}}{a_\text{Red}}\tag{4}$$

The half-cell reactions are shown in equations $(5)$ (oxidation) and $(6)$ (reduction) respectively, along with their standard potentials $E^0$ (under standard conditions, i.e. most importantly $\mathrm{pH\ 0}$).

$$\ce{Na+ + e- <=> Na (s)} \qquad E^0 = -2.71~\mathrm{V}\tag{5}$$
$$\ce{2 H+ + 2 e- <=> H2 (g)} \qquad E^0 = 0.000~\mathrm{V}\tag{6}$$

Let’s take a look at the anodic oxidation first. The initial concentration of sodium ions could be, say, $3~\mathrm{M}$. Remember that the activity is defined as $1$ for solids. With the assumption that $a \approx c$, that would mean our oxidation potential is $-2.68~\mathrm{V}$ as shown in equation $(7)$.

$$E_\mathrm{anode} = -2.71~\mathrm{V} + \frac{0.059~\mathrm{V}}{1} \lg \frac{3}{1} = -2.71~\mathrm{V} + 2.82 \cdot 10^{-2}~\mathrm{V} = -2.68~\mathrm{V}\tag{7}$$

The corresponding calculation for the reductive side, equation $(8)$, is a little harder since we have no real starting value for the partial pressure (and thus activity) of hydrogen gas.

$$E_\text{cathode} = \frac{0.059~\mathrm{V}}{2} \lg \frac{a(\ce{H3O+})}{p(\ce{H2})} = -0.0295~\mathrm{V} \times \mathrm{pH} - 0.0295~\mathrm{V} \times \lg p(\ce{H2})\tag{8}$$

But we do know that a reaction will happen if $E_\text{cell} > 0~\mathrm{V}$ and we have a good idea of the anode’s potential. Thus, if we want no reaction we must set the $E_\text{cell} < 0$ (equation $(9)$):

$$\begin{align}0 &> E_\text{cathode} - E_\text{anode}\\
- 2.68~\mathrm{V} &> E_\text{cathode}\\
- 2.68~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH} - 0.0295~\mathrm{V} \times \lg p(\ce{H2})\end{align}\tag{9}$$

The second summand is negative if $p(\ce{H2}) > 1~\mathrm{bar}$. (The equation itself requires a dimensionless pressure which is typically defined to have a the same value as if it were in $\mathrm{bar}$). However, that basically requires performing the reaction under a hydrogen atmosphere. This setup is rather hard and not part of the question. Instead, let’s take the natural concentration of hydrogen which is $1~\mathrm{ppm}$ according to Wikipedia. That results in $0.177~\mathrm{V}$ for the second summand (equation $(10)$).

$$-0.0295~\mathrm{V} \times \lg 10^{-6} = (-6) \times -0.0295~\mathrm{V} = 0.177~\mathrm{V}\tag{10}$$

At least we didn’t lose too much. But there is still the first summand, which is also negative by sign. We just need a high enough $\mathrm{pH}$ value to compensate. The calculation is in equation $(11)$.

$$\begin{align}-2.68~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH} + 0.177~\mathrm{V}\\
-2.86~\mathrm{V} &> -0.0295~\mathrm{V} \times \mathrm{pH}\\
96.9 &< \mathrm{pH}\end{align}\tag{11}$$

So we would need a $\mathrm{pH}$ value of almost $97$ to prevent the reaction from happening due to too basic conditions. Of course, this value is unobtainable in water or practically any solvent known to us. Even tert-butyllithium, one of the most basic species known has a $\mathrm{p}K_\mathrm{a}$ value of only $\approx 50$.

You could now try to argue that you would just need to increase the hydrogen gas partial pressure appropriately. So how does the value change if the reaction is performed under hydrogen atmosphere? See equation $(12)$ for the answer.

$$\begin{align}-2.86~\mathrm{V} &> -0.0295~\mathrm{V} \mathrm{pH}\\
90.8 &< \mathrm{pH}\end{align}\tag{12}$$

As you may or may not have guessed, a logarithmic difference of $6$ for the partial pressure results in a lowering of the $\mathrm{pH}$ by $6$ units. So to arrive at achieveable $\mathrm{pH}$ values, your reaction would need to happen at a hydrogen pressure of $10^{75}~\mathrm{bar}$ — go figure.

All things considered, the reaction is so violently endergonic that there is basically no way to inhibit it whatsoever. If metallic sodium sees a sufficient amount of water, it will start reacting, end of discussion.


Why does sulfonation of aniline occur para rather than ortho?

The sulfonyl group is kind of bulky due to the size of the sulfur, so steric hindrance is probably making it energetically disfavored for the group to be placed ortho. It's unlikely you can see this easily from a model or a drawing. If you have access to computational resources, you might be able to see the issue.