Indeed, this effect is observed in certain select cases. The effect of spin-spin cross over has been mentioned in this regard. There can be some other effects at play, namely: phase transitions, change in ligand geometry, equilibria between different molecular structure, and change in the number of solvent molecules in the coordination sphere.
Thermochromic behavior can be divided into two main classes: continuous and discontinuous. Some color changes are gradual; as the temperature increases over time, the color slowly changes. This can be due to the breaking or rearrangement of the lattice. This type of rearrangement is also known as continuous thermochromism.
On the other hand, a dramatic change in color can occur at a specific temperature or over a very small temperature range. This is known as discontinuous thermochromism.
I shall attempt to supplement what others have said by providing some illustrative examples.
I. $\ce{[Fe(NH2trz)3]Br2·H2O}$
This compound exhibits a spectacular reversible purple-white thermochromic transition upon heating−cooling with a large hysteresis centered at room temperature. The thermochromic transition is related to the spin crossover (SCO) from the low-spin state (LS, purple, $S = 0$, diamagnetic) to the high-spin state (HS, white, $S = 2$, paramagnetic).

(A) after cooling at $77~\mathrm{K}$ and (B) after heating at $353~\mathrm{K}$. The white form (B) is attracted to the magnet, whereas the purple form (A) is slightly repelled by the magnet.
Spin crossover (SCO) complexes are essentially $\mathrm{d}^4$−$\mathrm{d}^7$ first-row transition-metal octahedral complexes.
The occurrence of ST in coordination compounds of transition-metal ions is governed by the relationship between the strength of the ligand field (the electrostatic field acting at the central metal ion) and the mean spin-pairing energy. Octahedral complexes of $\mathrm{d}^4$–$\mathrm{d}^7$ ions may be either HS or LS, depending on whether the ligand field strength is weaker or stronger, respectively, than the spin pairing energy.
In order for thermally induced ST to occur the difference in energy must be of the order of thermal energy, i.e $k_\mathrm{B}T$. An increase in temperature favours the HS state, while lowering the temperature favours the LS state.
Thermal spin transition occurs nearly exclusively with coordination complexes of $\mathrm{3d}$ metal ions, because for $\mathrm{4d}$ and $\mathrm{5d}$ transition element compounds on the basis of ligand field theory, because the strength of the ligand field increases considerably, relative to analogous $\mathrm{3d}$ compounds and is generally much greater than the spin pairing energy (hence, these mostly show LS behaviour).
II. $\ce{Cu2HgI4}$
$\ce{Cu2HgI4}$, (and the related $\ce{Ag2HgI4}$) undergo a temperature-dependent order-disorder phase change at ca. $67~^\circ\mathrm{C}$, reversibly changing from a bright red solid material at low temperature to a dark brown solid at high temperature, with intermediate red-purple states. the colours are probably due to $\ce{Cu(I)}$-$\ce{Hg(II)}$ charge-transfer complexes.

All the metals involved are $\mathrm{d}^{10}$ so one can rule out $\mathrm{d}$-$\mathrm{d}$ transitions, and spin-spin crossover. An electron transfer of some kind might be occurring, especially since the colours observed are so intense. Based on the pi-donating characteristics of iodide, one can also rule out metal to ligand charge transfer. The colour could be ligand to metal or metal to metal charge transfer, although the electron would have to enter something other than the d orbitals.
As the temperature increases and the lattice is rearranged, the $\ce{Cu(I)}$ and $\ce{Hg(II)}$ ions remain in the same environment relative to the iodide ions, but their environment in relation to the other metal ions is changing, and this supports a metal-metal charge transfer explanation for the colour.
III. Bis(diethylammonium)tetrachlorocuprate(II)
At room temperature, this compound is a bright green colour and changes
discontinuously to bright yellow at the phase transition temperature.
The thermochromic phase transition in this compound is quite complex Structurally, the salt transforms between two monoclinic unit cells, with three independent $\ce{CuCl4^2-}$ anions in the low-temperature phase and two independent $\ce{CuCl4^2-}$ ions in the high-temperature phase. The phase transition is first order due to the lack of any crystallographic
relationship between the two phases.
From reference 5:

No simple structural transformation pathway can be envisioned for converting one phase to the other. This accounts for the stability of the supercooled high-temperature phase as crystals of that phase, grown from the melt, can frequently be kept at room temperature for periods of
several weeks.The crystallographic study of the high-temperature phase shows the presence of disorder for both the cationic and anionic species
in the structure.

At temperatures below ca. $50~^\circ\mathrm{C}$, the four coordinate compound has a square planar geometry even though it is not a $\mathrm{d}^8$ compound, owing to the short hydrogen bonds between the hydrogens of the counter ion, diethylammonium, and the chlorines on the metal compound (this favours the square planar geometry).  When the temperature passes the phase transition temperature, the colour changes to yellow because the hydrogen bonds lengthen to (because of the phase transition between the two crystalline phases) and the geometry changes to a distorted tetrahedral  The change in geometry changes the arrangement of the d orbitals.

Square planar is pictured on the left, and tetrahedral is pictured on the right.
There is no stable intermediate between the square planar and the distorted tetrahedral geometry, so the change is an example of discontinuous thermochromism. Individual clumps of crystals may appear as either green or yellow depending on their exact temperature.
References

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3596041/

http://pubs.acs.org/doi/abs/10.1021/ed4000487

http://people.umass.edu/mjknapp/Chem242/2004%20Experiments/CuHg2I4Lab.pdf

http://www3.amherst.edu/~thoughts/contents/amberger-thermochromism.html

http://pubs.acs.org/doi/abs/10.1021/ja00230a020


Why does the stability of hydrides decrease down the group in nitrogen family?

The increasing size of the elements play a role here along with the overlapping factor.

Bismuth is humongous as compared to hydrogen so it's hydride is unstable. However nitrogen being small in size can form strong bonds with hydrogen. This also accounts for the fact that the reducing nature of the hydrides increase down the group.

As far as overlapping is concerned the bonds in the hydrides are - $\ce{2p-2p}$ up to $\ce{5p-2p}$.

Clearly the upper hydrides will be more stable as they have more effective overlapping between the orbitals.


How do transmembrane protein alpha helixes bind to the membrane?

Well, it binds via hydrophobic interacions - non-polar R groups are attracted to hydrophobic interior of the membrane via dispersion forces, and also don't disrupt the structure of water as if they would outside of membrane. It's the same thing which holds molecules of the membrane together.


The answer to this depends on how the reagents are mixed. For one, you can have the reaction without water, as pointed out by @Klaus in the comments. If you blow dry HCl on dry NaOH, they will react without involving water in the reactants.

$$\ce{HCl + NaOH -> H2O + NaCl}$$

This is probably not the case though, and if you're dealing with solutions, then yes, water will react. If you start by adding the HCl to pure water, you'll have

$$\ce{HCl + H2O -> H3O+ + Cl-}$$

This representation is prefered over the simple dissociation $\ce{HCl -> H+ + Cl-}$ because it shows that you don't have "free" protons in solution. You could actually even say that this reaction has already happened since, considering you're using hydrochloric acid it's already dissolved, and this step would be just a dilution.

If you start with the NaOH solution, you could argue that water does not react since

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

but that only happens in the liquid state (usually in solution). Also, you can't ignore the ions solvation (which is one, if no the most important factor which affects the solubility of the sodium hydroxide salt) and equilibrium

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

And what this means is that you don't have fixed hydroxil ions and water molecules, so you can't simply ignore the solvent and assume it's just a spectator molecule waiting around.

To write the two step reaction, it's not a good idea to do it in one line, because as you can see on the first step you have NaOH on both sides, which is something that doesn't make much sense. It's better to do it like this:

$$\ce{HCl + H2O -> H3O+ + Cl-}$$

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

It's usually a good idea to represent salts in their ionic form when dissolved, which would be

$$\ce{H3O+ + Cl- + NaOH -> 2H2O + Na+_{(aq)} + Cl^{-}_{(aq})}$$


Test for chlorine and chloramine in city water supply

Test strips for chlorine should work for both chlorine- and chloramine-treated water. See Here-to-learn's comments to this answer for the chemical kinetic reason why chlorine test strips will also work for chloramine-treated water.

Notice that your city supplies the water with chloramine. Depending on how much chloramine is present in your tap water, leaving water to sit for days on the countertop would be necessary to remove a significant amount of chloramine.

Please see this research paper: http://hbd.org/ajdelange/Brewing_articles/BT_Chlorine.pdf

Page 17 estimates the half-lives of chlorine vs. chloramine.


Why does enthalpy change differ in a chemical reaction because of temperature?

From Hess' law, the heat of reaction at temperatures $T_1$ and $T_2$ are related by $$\Delta H_{T_1}=C_{p,\mathrm r}(T_2-T_1)+\Delta H_{T_2}+C_{p,\mathrm p}(T_1-T_2),$$where $C_{p,\mathrm r}$ is the heat capacity of a stoichiometric molar mixture of reactants and $C_{p,\mathrm p}$ is the heat capacity of a stoichiometric molar mixture of products.


How polarization effects from cations affect H-bond strength

Okay, I realized what was wrong with my picture above.

It is not the hydrogen bond that is weakened, but the OH bond within a water molecule that is weakened, and then stretches at a lower frequency.
When the cation is introduced, and is near the oxygen atom, the electron cloud in the water molecule is moved toward the cation (polarised), leading to electron density in the OH bond being reduced => weaker bond => lower frequency vibration.

The hydrogen bond is not really of interest here, although the same effect takes place when two water molecules are H-bonded. Hence, the OH stretch of an isolated water molecule will vibrate a bit faster than the OH stretch in a H-bonded water molecule.


If you want a quick and dirty expressions in terms of hydrogen atom wave functions, you can try this:

For ground state you can get a quick approximation by applying effective nuclear charge.  Then you get a hydrogen-like wave function with a modified nuclear charge between 1 and 2;  solve that and get orbital.  It will be doubly occupied, spin singlet.
The lowest excited states of He are all Rydberg states.  The roughest approximation would be to consider it He+ plus a Rydberg electron, and you can even use hydrogen-like atom formula to get the Rydberg orbital.  The total wave function would be antisymmetrized product of these two orbitals, coupled to be either singlet or triplet.

I am not sure you want transition density matrix or transition dipole matrix.  Either way, its just quite straightforward integration from there.  I think there might be ways to do this analytically, but frankly I do not remember and you can probably try that on Mathematica.


IMO, the only reaction that you are studying is the first reaction:  the conversion of $I^-$ to $I_2$ by hydrogen peroxide.  The thiosulphate reaction is simply a dramatic way to indicate when a particular amount of iodine has been produced.

The reaction was not slower when you doubled the amount of thiosulphate.  Twice as much iodine needed to be produced (and reconverted to iodide ion) before the thiosulphate was consumed and the first permanent iodine changed the starch indicator blue.


Reaction of ethylene glycol with acidified potassium permangnate

I think the final oxidation product should be carbon dioxide ($\ce{CO_2)}$ if acidified potassium permanganate is present in excess amount.

Firstly, ethylene glycol will be oxidised to oxalic acid. But oxalic acid can still be oxidised. In fact, it is the concept of titration of oxalates with potassium permanganate ($\ce{KMnO_4}$). Under acidic conditions, oxalic acid ($\ce{H_2C_2O_4}$) will be oxidised to $\ce{CO_2}$ by $\ce{KMnO_4}$ as:

$$\ce{2MnO_4^- + 5C_2O_4^{2-} + 16H^+-> 2Mn^2+ + 10CO_2(aq)  + 8H_2O}$$

Thus the final product should be $\ce{CO_2}$

Also, note that, formic acid is an oxidisable acid because it gives silver mirror in reaction with Tollen's reagent $(\ce{[Ag(NH_3)_2]OH})$, and thus, formic acid can not be the end product, as it is still oxidisable and hence it will definitely react with acidfied $\ce{KMnO_4}$ if both are present in  the reaction medium.


Do noble gasses besides Helium form diatomic molecules at low temperatures?

Yeah, dimers of noble gas atoms are stable due to dispersion aka London forces. More generally, neutral, non-polar molecules can still form very weak bound states arising entirely from the induction and interaction of instantaneous dipole moments (that is, fluctuations in the electron density around an atom arising from the motion of electrons). My understanding is that physicists would explain this interaction as a special case of the Casimir-Polder interaction. This interaction means that even noble gases are nonideal.

Correctly predicting the negative interaction energy of a noble gas dimer is a pathological test of dispersion-corrected DFT functionals, as F'x's reference1 and the references therein indicate, and many methods fail qualitatively. I recall that Zhao and Truhlar used noble gas dimers to test the efficacy of their various metahybrid functionals2,3. One of their papers2 makes reference to a table of binding energies and equilibrium distances derived4 by Wang and Ogilvy, giving for instance a $\ce{Ne\cdots{}Ne}$ binding energy of 0.35 kJ mol-1 (that's tiny) and less than half that for $\ce{He\cdots{}Ne}$. The $\ce{Ne\cdots{}Ne}$ equilibrium distance is given as 309 pm, which is extreme. As the paper5 you refer to indicates, however, the interaction energy for the $\ce{^4He\cdots{}^4He}$ dimer is much, much smaller than either of these and has an equilibrium distance of 5.2 nm (!!!), which is something else entirely.

(1) Ruzsinszky, A.; Perdew, J.P.; Csonka, G.I.; Binding Energy Curves from Nonempirical Density Functionals II. van der Waals Bonds in Rare-Gas and Alkaline-Earth Diatomics; J. Phys. Chem. A; 2005, 109, pp. 11015-11021

(2) Zhao, Y.; Truhlar, D.G.; Hybrid Meta Density Functional Theory Methods for Thermochemistry, Thermochemical Kinetics, and Noncovalent Interactions: The MPW1B95 and MPWB1K Models and Comparative Assessments for Hydrogen Bonding and van der Waals Interactions; J. Phys. Chem. A; 2004, 108, 6908-6918

(3) Zhao, Y.; Truhlar, D.G.; Comparative DFT Study of van der Waals Complexes: Rare-Gas Dimers, Alkaline-Earth
Dimers, Zinc Dimer, and Zinc-Rare-Gas Dimers; J. Phys. Chem. A; 2006, 110, pp. 5121-5129

(4) Ogilvie, J.F.; Wang, F.Y.H.; Potential-energy functions of diatomic molecules of the noble gases: II. Unlike nuclear species; J. Mol. Struct.; 1993, 291, pp. 313-322

(5) Grisenti, R.E.; Schöllkopf, W.; Toennies,  J.P.; Hegerfeldt, G.C.; Köhler, T.; Stoll, M.; Determination of the Bond Length and Binding Energy of the Helium Dimer by Diffraction from a Transmission Grating; Phys. Rev. Lett.; 2000, 85, pp. 2284-2287


How do you determine the pH of a solution at different volumes?

With regard to your understanding of the correlation between concentration and volume, you should know that concentration $c$ is defined as
$$c=\frac nV$$
where $n$ is amount of substance and $V$ is the volume of the solution.

When you dilute a sample, you change the volume $V$, but you do not change the amount of substance $n$ in your sample. Therefore, if you know the initial concentration $c_1$ in the initial volume $V_1$
$$c_1=\frac n{V_1}$$
you can calculate the new concentration $c_2$ after dilution to the new volume $V_2$:
$$c_2=\frac n{V_2}$$
You can rearrange both equations to calculate the amount of substance $n$:
$$\begin{align}
c_1\cdot V_1&=n\\[6pt]
c_2\cdot V_2&=n
\end{align}$$
Since the amount of substance $n$ is constant, you get another useful equation:
$$c_1\cdot V_1=c_2\cdot V_2$$

In your exercise, after addition of all solutions, the new volume may be estimated as
$$V_2\approx1\ \mathrm{ml}+0.05\ \mathrm{ml}+6.95\ \mathrm{ml}=8\ \mathrm{ml}$$

The initial concentration of the base (sodium acetate) is $c_{\text{base},1}=0.10\ \mathrm{mol\ l^{-1}}$ and the initial volume is $V_{\text{base},1}=1\ \mathrm{ml}$.

Therefore, the new concentration of the base is

$$\begin{align}
c_{\text{base},1}\cdot V_{\text{base},1}&=c_{\text{base},2}\cdot V_2\\[6pt]
c_{\text{base},2}&=\frac{c_{\text{base},1}\cdot V_{\text{base},1}}{V_2}\\[6pt]
&=\frac{0.10\ \mathrm{mol\ l^{-1}}\times1\ \mathrm{ml}}{8\ \mathrm{ml}}\\[6pt]
&=0.0125\ \mathrm{mol\ l^{-1}}
\end{align}$$

The initial concentration of the acid (acetic acid) is $c_{\text{acid},1}=0.10\ \mathrm{mol\ l^{-1}}$ and the initial volume is $V_{\text{acid},1}=0.05\ \mathrm{ml}$.

Therefore, the new concentration of the acid is

$$\begin{align}
c_{\text{acid},1}\cdot V_{\text{acid},1}&=c_{\text{acid},2}\cdot V_2\\[6pt]
c_{\text{acid},2}&=\frac{c_{\text{acid},1}\cdot V_{\text{acid},1}}{V_2}\\[6pt]
&=\frac{0.10\ \mathrm{mol\ l^{-1}}\times0.05\ \mathrm{ml}}{8\ \mathrm{ml}}\\[6pt]
&=0.000625\ \mathrm{mol\ l^{-1}}
\end{align}$$

Finally, you can use the Henderson–Hasselbalch equation to calculate the $\mathrm{pH}$ of the acetate buffer from the new concentrations $c_{\text{acid},2}$ and $c_{\text{base},2}$.


Significance of shapes of Conical flask and Kjeldahl flask

Question: What is the practical significance of the shape of conical flask (in US: Erlenmeyer flask)?

An Erlenmeyer flask (also known as a conical flask or a titration flask) is a type of laboratory flask, which consists of a flat bottom, a conical body, and a cylindrical neck. The flask is created by the German chemist, Richard August Carl Emil Erlenmeyer (1825–1909) in 1860, and hence its name. With its wide flat bottom, it was perfect for heating solutions (large surface area). Hot vapor condenses on the upper section of slanted sides of the Erlenmeyer flask, reducing solvent loss. Also, these slanted sides and small mouth of the flask cause the hot solutions in the flask to cool slowly, keeping the empty air space hot. This feature greatly aids certain techniques in chemistry, especially appropriate for recrystallization. For example, during the hot filtration step, when the hot solution is filtered through a fluted filter paper into the receiving Erlenmeyer flask, hot vapors from the boiling solvent would keep the filter funnel warm, avoiding the premature crystallization on the funnel and its stem.

Equally, the sloping sides prevent risk of spillage of solutions in it, therefore ideal to hold content, which needs to be mixed by swirling. This nature also makes them suitable for use in titrations. Again, the flat bottom of the flask has the advantage of giving its content a uniform color across the entire flask, which makes it easier to determine sharp end points.

  Question: What is the practical significance of the shape of Kjeldahl flask?

In 1883, Danish chemist Johan Kjeldahl (1849–1900) of the Carlsberg Laboratory published a method to determine nitrogen content in organic matter. This is the first accurate, simple, and speedy method to determine nitrogen content in organic matter including protein content in food, which bares discoverer's name, the Kjeldahl method. The first step of the method is the digestion of sample in concentrated sulfuric acid at $\pu{360–410 ^\circ C}$, which decomposes the sample by oxidation to liberate the reduced nitrogen as ammonium sulfate. This heating would cause splash-back, which is a treat because of conc. $\ce{H2SO4}$. Thus, along with his famous method, Kjeldahl had also developed a piece of laboratory equipment he developed in 1888, which bares his name as well, Kjeldahl's flask. The long-necked, round-bottomed (actually pear shaped) flask was ideal for avoiding splash-back when heating solutions. The long-neck also function as air condenser.


Reason for enthalpy being non-zero at constant temperature

The only way that an exothermic reaction can occur with "constant temperature" is if the heat generated is constantly removed in some way. Even if the heat removal is essentially instantaneous, it is incorrect to say that there is no temperature change. Rather there is an infinite number of infinitely small temperature changes as the system heats up by an amount dT and then is returned to the set T. If you were able to measure these infinitely small changes and also to know the instantaneous Cp (which changes as the reaction composition changes), you could sum them up and get a non-zero value for the enthalpy change. Obviously, that's not a practical approach.
Instead, to determine the enthalpy change experimentally, it is easiest to perform the reaction within an insulated apparatus such that the temperature is not constant and the change in temperature can be used to determine the enthalpy change.


Balancing ionic equations and determining number of ions

Actually $Mg(OH)_2$ has (low) solubility ($K_{sp}=1.5\cdot 10^{-11}$). Since it is asking for the net ionic equation, the only ions reacting (since $NO_3^-$ stays in solution) are $H^+$ and $OH^-$.

You forgot that each molecule of ammonium carbonate produces 3 ions, so multiply your answer by 3.


Making 1,3,5-Tribromobenzene from Phenol

First react phenol with ammonia in presence of zinc chloride to obtain aniline.

Then react aniline with bromine water to form 1,3,5-tribromoaniline

Then react 1,3,5-tribromoaniline  with sodium nitrite and hydrochloric acid at 273 K to form 1,3,5 tribromobenzenediazonium chloride.

Then react the diazonium salt with ethanol/phosphorous acid and water in presence of copper to form  1,3,5 tribromonbenzene.

Detail explanation about the reaction, conditions  and the mechanism is given here by actually conducting the experiment.


Has IUPAC been inaccurate in their 1994 definition of kinetic stationary state?

IUPAC should say they are defining the "steady state approximation" rather than the steady state.
There are two approximations in this technique.
First, that final reaction step is not an equilibrium (reverse reaction rate is zero).
Second, that the intermediate is so unstable that its concentration (and therefore the rate of change of the concentration) is approximately zero.
Compare the IUPAC definition to the following 1954 U.S. National Research Council explanation:

STEADY STATE APPROXIMATION
The most generally applicable of these simple methods is the so-called "steady-state approximation." A steady state may be defined as a condition in which the rates of change of the concentrations of the several intermediates are very small compared to the rates of change of the concentrations of the reactants and products. This condition is realizable whenever the ratio of the concentrations of the intermediates to the concentrations of the reactants is very much less than unity. When this condition is not attained, the method is not applicable; however, it should not then be necessary since the (larger) concentrations of the intermediates could be measured by experimental means. ... The steady-state approximation consists in setting the rates of change of each of the intermediates equal to zero and in solving simultaneously the resulting algebraic equations.

Overall, if the situation was truly, not just approximately, steady state, yes the concentration of the intermediate would be constant, but the point is that the "stead state approximation" is useful even when the concentration of the intermediate, though always near zero in absolute terms, decreases to say half of its original concentration over the course of an experiment (as the starting material, A, decreases to half its original concentration).
Additionally, a very small $\ce{[X]}$ does not purely-mathematically imply a small rate of change of $\ce{[X]}$, because $\ce{[X]}$ could rapidly oscillate within a small, near-zero, range, resulting in brief instances of large rate of change.  Instead, $\ce{[X]}$ being small over a given period of time really just places a limit on how long of a time the rate of change can exceed a certain value.  For example, if $\ce{[X]} < 0.001\ \mathrm{M}$,
$|\mathrm{d}[\ce{X}]/\mathrm{d}t|$ must not exceed $1\ \mathrm{M/s}$ for more than $0.001\ \mathrm{s}$.


How do I lookup a polymer by its structure?

That looks like bilirubin, from hemoglobin catabolized in the liver  Monomer, dimer..tetramer...oligomer...polymer...high polymer.  It is a tetrapyrrole, not even an oligomer.


This video shows how a black flame is achieved. If you illuminate the fire with a monochromatic light source (sodium vapor lamp) and introduce a species in the fire that absorbs that wavelength (sodium ions) then the fire will in fact appear black under the illumination. A screen shot is shown below:

Click image for video.


How dried up detergent water led to precipitation of snowflake like shapes?

There are many water soluble molecules in soap. I believe that as the water evaporated, the concentration of these molecules became great enough for them to start crystallizing. This process is used in chemistry as a means of purification and is known as recrystallization.

Are you considering the X type shape of the crystals to be snowflake like? I am not quite sure why they all seem to have similar geometries, I want to guess that they are crystals of citric acid.


Increase in bond length of BF₄⁻ compared to BF₃

There seems to be a bit of confusion between bond length and bond angle here.  It is true that VSEPR theory predicts that bonding pairs of electrons (and lone pairs, for that matter) will repel one another as much as possible, thereby increasing bond angles, but this has little to do with bond length.

One way to explain the different bond lengths in $\ce{BF_3}$ and $\ce{BF_4^-}$ is by looking at the molecular orbital diagram of $\ce{BF_3}$:

The linked diagram isn't great since it does not show the filled orbitals: the $1a_2'$ $1e''$ and $4e'$ orbitals are all filled, making the lowest unoccupied orbital (LUMO) the $2a_2''$.  If you haven't wrapped your head around molecular orbital diagrams of this complexity, don't sweat it.  The bottom line is, the $2a_2''$ orbital has antibodning interactions with the $2p_z$ orbitals of boron and the $2p_z$ orbitals of the surrounding fluorines.  Thus, adding electrons to this orbital makes the $\ce{B-F}$ bonds weaker, and therefore longer.


Recognising a rate-determining step in E1

Anindya Prithvi has said everything you needed to hear in two sentences. It is true that rate determining step for E1 mechanism is the carbocation formation. Thus, I present here the detail energy diagram for reaction progress of acid-catalyzed dehydration:

The first step is the protonation of starting compound (SM), which is shown in the right hand top corner in blue box. The next step is the rate determining step, which has the biggest activation energy and progress through the transition state 2. The resultant intermediate is the relatively higher energy 2°-carbocation, which is quickly rearrange to lower energy 3°-carbocation by 1,2-hydride shift through the transition state 3. It is worth noting that this 1,2-hydride shift does not involve net bond-breaking or bond-forming (it is believed to be undergoing concomitant bond breaking-making mechanism involving orbital overlapping). Thus its transition state (transition state 3) would not be higher in energy than that of transition state of RDS. The 3°-carbocation would finally give away the expected product through the transition state 4. It doesn't matter if all other steps happen at once or separately. All those steps have to wait to happen until the formation of the 2°-carbocation. That's why it is the rate determining.


Estimation of dipole moments in aromatic compounds

I read up in my text book that dipole moment direction of phenol is towards OH group so I guess O being highly electronegative pulls the electrons more and participates less in resonance hence it has deta negative charge on it thus acc to convention dipole moment direction is from positive to negative hence direction is towards OH group.


What causes transition from smoldering combustion to flame combustion?

Burning in open air isn't a steady state process, but a very dynamic one. Likely when the fuel is smouldering there isn't enough of a convection current in the air to bring in fresh oxygen. As the smouldering continues something about the geometry of the area smouldering changes and fresh oxygen is brought in by a strong convection current. At that point the fuel bursts into open flame.


Size of target substance in IR and Raman Spectroscopy

This all depends on the exact geometry of your spectrometer light path, if you measure in transmission or reflection, using ATR, etc. AND of course on your sample concentration.

If you want to ask about the minimal size: As small as you can get your focal point or laser spot. A few dozen microns, maybe less. Local sample heating can be a problem. You can buy scanning IR and FT Raman microscopes.

In transmission, you can measure IR on a cm sized sample, given that is is translucent. With Raman, that would be tricky, because you should focus the scattered light again.

Minimum number of molecules for IR is very low, but that also depends on how strong the absorption line in question is. Raman used to be very insensitive, that has become a lot better with FT Raman.


How to calculate the magnitude of magnetic field of an orbital?

Hyperphysics means that the energy difference between the 2p J=3/2 state and the 2p J=1/2 state is as if the magnetic moment of the electron were aligned with and against a magnetic field of 0.4T.

The energy of the two states, relative to the ground state, is (according to NIST)

2p J=3/2 : 10.19885142904 eV
2p J=1/2 : 10.19880606470 eV

The difference is 0.00004538204 eV or $7.271004 \times 10^{-24}$ J

The electron magnetic moment is $−9.28476462 \times 10^{-24}$ J/T

So the energies are as if they are raised and lowered by being in a field of 0.3915556 T.

For the ground state, 1s, there is no orbital angular momentum.  The 1s state is only split by interaction with the proton.  This is sometimes referred to as hyperfine structure. The energy difference is $5.87433 \times 10^{-6}$ eV.


How does a catalyst accelerate both forward/reverse reactions simultaneously?

The idea is that the activation energies is lowered for both the forward and backward reactions.

Though the activation energy for the backward reaction is higher than the activation energy for the forward reaction, it is nevertheless lowered.

To answer your second question, reactions aren't always reversible, but a lot of the time they are, and are subject to the principle of microscopic reversibility. That means when reactions are performed as a series of steps, at equilibrium, each individual reaction occurs in such a way that the forward and reverse rates of reaction are equal, and these equilibriums can be affected to drive the reaction one way or the other.


The primary compounds of potential theraputic interest are all far less volatile than ethanol.  Ethanol based tinctures are sold and even have   available MSDS, so at least some of the oils of interest are soluble.  In looking at the chemical structures of the predominant oils of nigella sativa I would guess that they should all be fairly soluble in ethanol.  In some ways it seems that pure acetone (available at pharmacy) might make a better solvent and is even much more volatile even than ethanol.  If you used acetone, be certain to dry it via heating and time such that there is no scent of acetone remaining.  This shouldn't be difficult.

Anyway, to answer your question: ethanol should be a good solvent and it should not be difficult to remove the majority of the ethanol from the far less volatile oils just by drying in air for sufficient time (until you don't smell alcohol).


Does Mie Scattering occur in Liquid or is Rayleigh scattering?

So does Rayleigh or Mie scattering occur in liquid or bulk water? So why does Rayleigh scattering (?) occur in liquid/bulk water when the particles are much larger than the wavelength of light?

What is the approximate size of water molecule? It is on the order of 2.7 Angstroms or 0.27 nanometers. When we are talking about visible light, the wavelength is 500 nm/0.27 times larger than the scatterer! In case of pure water and with visible light, Rayleigh scattering is observed along with Raman effect. If you disperse larger "particles" in water such as milk, one sees Mie scattering.


The answer depends on which teaching you follow.

Traditional teaching, dating back to Linus Pauling, would prefer three double bonds to minimise charge separation, requiring an expanded octet.

Even older traditional teaching would favour the validity of the octet rule and take charge separation as a tolerable by-product.

Both sides have their advantage and disadvantage:

The Pauling method

correctly shows that all three bonds are equal without resorting to mesomery
incorrectly assumes a bond order of $2$ rather than the actual $1.4$

The charge-separation method

predicts a bond order of $1.33$ which is much closer to $1.4$
requires mesomery of three structures to explain the three equal bond lengths.

I for my part prefer the mesomeric charge-separation description, because we have no trouble drawing benzene in the same way and because I like the octet rule. Others will disagree and prefer the three double bonds.

I wouldn’t know anybody who prefers a description with two double bonds and one single bond. It is just between the two models making it all the less correct.

Please also see this excellent answer to a related question by Martin which includes molecular orbitals for $\ce{SO3}$


For a grignard reaction to work, the ether must be free from water. This is usually achieved via distillation over sodium/potassium using benzophenone as an indicator. As you are in a school lab, the solvent provided to you is likely already dried/dry enough for purpose.

If the solvent is a little wet, two things will happen:

Firstly, it is more difficult to get the grignard formation to 'initiate' as the water effectively quenches the reagent as it forms. However, with persistence it is usually possible to get them to initiate when making simple grignards such as PhMgBr.

The other issue however, is that as the water is quenching any grignard reagent formed until all the water is used up, you will make less reagent than you think you are (in this sense the reaction is self-drying) meaning if you do think your ether is wet, you should aim to make a greater quantity of reagent than required.

It is possible to titrate a grignard reagent to identify the actual concentration of solution made, although in an undergraduate practical it is likely that you will use the reagent in excess anyway.

In terms of identifying your product, you should carry out the reaction, work it up, purify, and use a technique such as NMR to see if your product has formed. If this is unavailable, simple IR spectroscopy can help you determine if the functional groups in your starting materials have changed (e.g. Carbonyl to alcohol).

Good luck with your reactions


You also may want to consider Hard-Soft Acid-Base Theory. The carbanion is generally said to be a better nucleophile than hydroxide, presumably as measured in a highly controlled setting using a standard electrophile like methyl iodide. I'm under the impression nucleophilicity is very nearly defined as how fast an electron donor reacts with $\ce{CH3I}$. Note that there are no hydrogens available for elimination on this electrophile, which precludes the possibility of the nucleophile acting as a base instead.

In practical reaction conditions (protic solvents, beta-hydrogens on electrophile, etc), we may need to worry about our nucleophile acting as a base. This is where HSAB Theorycomes in. Hydroxide is considered a "hard base", which means it may be more inclined to react with "hard acids" like $\ce{H+}$ rather than "soft acids" like methyl iodide. As a "soft base", a carbanion will probably prefer to bond with methyl iodide (as a classic nucleophile) than with any available protons (as a Brønsted-Lowry base). So in practice, the difference between a carbanion's nucleophilicity and hydroxide's might be larger than expected due to competing "hard base" reactions (Brønsted-Lowry rather than Lewis). Just food for thought.

Qualitative nucleophilicty ratiomale


An acid chloride is a covalent molecule.  It has no pH.  If you put a pH electrode into a reactive acid chloride you will dehydrate and acylate its hydrated glass surface ($\ce{#Si-OH}$) and damage its future response when properly used.  pH electrodes are stored sitting in buffer or saline solution.

An acid chloride is an oxo-acid whose acidic $\ce{-OH}$ has been replaced with $\ce{-Cl}$.  Hydrolysis restores the oxo-acid, displacement makes derivatives. Hydrochloric acid is not an acid chloride.


What does the neutralisation of acids and bases mean?

There are many theories put forward to explain acid-base reaction.
According to Arrhenius acid-base theory. All acid liberate protons (H+ ions) when dissolved in water. All bases liberate Hydroxyl ions (OH- ions) in water.
When the reaction takes place H+ and OH- combine to form water, and the remainder of the ions (what are left after liberating H+ and OH- ions) form what is called a salt.
In such reactions, Water is always formed.

What you have shown is an example of a Bronsted acid-base reaction.
Bronsted acids are proton donors (have ionisable hydrogen ion).
Bronsted bases have the capability to accept them (either via hydroxyl ion liberation or by their lone-pair as in NH3)

Clearly, the formation of water is not a necessity in a Bronsted acid-base reaction.


Noyori hydrogenation and the Curtin–Hammett principle

There's no way to predict this.

Mnemonics are just rules, and rules are just rules. There will be exceptions, and you won't know when you should apply them a priori. If you could, it would be part of the rule by definition.

The Curtin-Hammett conditions apply when the rate of interconversion between intermediates is significantly faster than the rate of reaction from the intermediates onto their respective products. If the rate of interconversion is slow, then the ratio of products is simply governed by the ratio of intermediates.

The problem comes about from determining if the interconversion is fast. If you're looking at proton transfers or some kinds of conformational changes, it's usually pretty straightforward. Otherwise, only intuition can guide you ahead of time. Once you have experimental results, you might be able to rationalize this fact and determine that the reaction is governed by Curtin-Hammett kinetics, but there's no way to know for sure. This is just due to the complexity of binding entropy, solvent effects, etc, coupled with the intrinsic barrier for the forwards reaction.

Chemistry is hard.


Explain the difference in stability of permanganate ions in acidic/ alkaline solutions?

Actually, yes.

As we know, $\ce{H2O = H+ +OH-}$ . Chemical equation can be threated as algebraic equations as long as the final result has chemical meaning (i.e. no negative coeffecients ... unless you understand their meaning), thus we can add $\ce{4H+}$ to both sides of the second equation. This gives us equation

$\ce{MnO4- +4H+ + 2H2O +3 e- \rightarrow MnO2 + 4H+ +4OH-}$

Obviously, we can collapse four pairs of $\ce{H+}$ and $\ce{OH-}$ into $\ce{4H2O}$

$\ce{MnO4- +4H+ + 2H2O  +3 e- \rightarrow MnO2 + 4H2O}$

and now we can drop two $\ce{H2O}$ from both sides

$\ce{MnO4- +4H+   +3 e- \rightarrow MnO2 + 2H2O}$

which is exactly as your first equation. So, in fact it is exactly the same reaction. However, WTF they are written differently and the corresponding potentials are different? This is because standard potentials are implied to be measured at concentration of all ions and molecules (except water) in corresponding half-reaction  1 mol/l. However, what happend when one concentration deviates from 1 mol/l ? The corresponding potential deviates from its value as well, in accordance with Le Chatelier's principle.

When we set up concentration of $\ce{OH-}$ at 1 mol/l, we set $\ce{H+}$ at roughly $10^{-14}$, since $\ce{[H+][OH^{-}] \approx 10^{-14}}$ (water autoprotolysis constant). According to Le Chatelier's principle the equilibrium should move to the left, which is observable as a change in electrode potential.

The quantitative expression of this kind of dependence is known as Nernst equation.


Periodic boundary conditions in DFT for cylindrical system

Let's dissect this:

  As I understand it, a DFT calculation uses periodic boundary conditions, whenever you want to simulate a bulk material. This is convenient because all information can then be extracted from the fundamental unit cell.

One important point is that you employ periodic boundary conditions (PBC) only when you are treating a periodic system, i.e. something with translational symmetry. And in those cases you use plane waves as a "basis set".

In the ordinary molecular case you use atom centred basis functions, which fulfil the basic requirements of a wave function (in short): orthonormality, continuous, continuous differentiable, vanishing at infinity.
Therefore a quantum chemical calculation electronic energy is always in vacuum at "0 K".

  Currently I am faced with the problem of calculating the band structure of a carbon nanotube, which has a cylindrical geometry, i.e. only bulk in only one direction, say z, and circular in the x and y directions. As my unit cell I am using a ring of C-atoms surrounded by vacuum and I then employ periodic in the z-direction. Obviously, for large diameters, this approach involves a rather large unit cell, which makes the calculation slow.

This approach actually only makes sense if you are treating a very, very long nanotube. I'd believe a molecular approach using appropriate symmetry, Dnh would be far superior.

  My question is: Is there an alternative way to do this, which exploits the cylindrical symmetry of the system? Specifically what I have in mind is to also apply periodic boundary conditions in the other directions, but where the Schrödinger equation is expressed in cylindrical coordinates with the middle of the nanotube as the origin.

A nanotube is discrete in two directions, i.e. the wave function has to vanish in these directions, egro apart from applying vacuum as PBC, there is no other choice in this framework.

An alternative approach is to treat your nanotube as a discrete molecule and cap it off at the end with hydrogen, then increase the length of the tube until you have sufficient convergence, obviously using Gaussian or Slater type (atom centred) orbitals.

  Another, less important, question is also: Do you ever NOT use periodic boundary conditions in DFT? For example, when you do a calculation for an atom, it always seems like you have to add vacuum around it, which is essentially, because you are applying pbc, and do not want it to interact with other unit cells. Is there another way to solve this problem, where you assume the density to vanish far away?

For discrete molecules: always. For periodic systems: only sometimes not (cluster approach).
Prior to the golden 90s, almost every calculation was a molecular calculation. In the early stages of quantum theory nobody ever dared to dream that we would eventually able to treat a huge stone quantum chemically (a priori) with a computer.

A nanotube is typically right in the middle, between a bulk material and a molecule, and you should either rely on the experience of others, maybe you have an advisor, through the literature, or built it yourself.
There are quite a few groups (I suppose) who work with nanotubes, and some of them have surely tried computations. Do a thorough literature review and apply common practices.

I fear that you are getting your knowledge about DFT calculations from working with a calculation suite designed for bulk materials. You should probably brush up on some of the fundamental theory. I recommend Frank Jensen's book for a rough overview, and Attila Szabo & Neil Ostlund's book for some better maths  (See resources for learning chemistry for more details).
A fairly comprehensive "review" of DFT can be found here: DFT Functional Selection Criteria and in the links within.


The equation of the reaction provides you with the composition of the gas liberated.

You assume ideal gas, which means that the gas particles are indistinguishable. This means the reaction equation could look like
$$ \ce{2Pb(NO3)2 -> 5X + 4PbO} $$
where 5 moles of gas are liberated for every 2 moles of lead nitrate.

Reforming the ideal gas equation to $n=pV/(RT)$ will lead you to the numerical value. The amount of lead nitrate will then be $$n_{\ce{Pb(NO3)2}} = \frac{2n}{5}$$

From that calculating the mass is a trivial task.


You could just as well have wondered is there any number which is neither negative nor positive. Yes, there is such number, and it is called zero. It is rarely encountered among the real-life physical values, though, unless enforced by some law (which is not the case here). After all, there are immensely many positive numbers, and equally many negative numbers, and only one zero. You can hardly expect to hit it by choosing numbers at random.

Then again, reaction enthalpies are subject to variation with temperature, so if we pick any reaction with low enthalpy, it is quite probable that there is a certain temperature at which the said enthalpy is exactly 0. That would be the answer to your question. It is for entropy to decide whether the reaction will occur spontaneously at this temperature. Chances are it will.

On a side note, there is a huge number of "void reactions" like $\ce{H2 + H -> H + H2}$ with guaranteed zero enthalpy at any temperature, but many would object against calling these things reactions in the first place.

So it goes.


The final solid mass would be a powder of sort, right? If you grind it / mix it will be homogeneous irregardless of the selectivity of their precipitation in the prior solution.


Are Isopropyl Alcohol (2-Propanol) and 1-Propanol tautomers?

Tautomers need to be able to interconvert readily and quickly. This is done via proton transfer. For example in keto-enol tautomerism (https://en.wikipedia.org/wiki/Keto%E2%80%93enol_tautomerism#/media/File:Keto-Enol-Tautomerie.png) the proton on the alpha carbon donates its electrons to form a pi bond between the 2 carbons. As a result the pi bond between the carbonyl carbon and oxygen breaks and the now negatively charged oxygen picks up the proton which was originally lost.

In propan-1-ol and propan-2-ol, the oxygens are on different atoms and there is no way they can interconvert, so they are not tautomers.


Why are neutralisations involving weak acids and bases less exothermic than those involving only strong acids and bases.?

I think the answer is closer to what was stated in the original question: "some energy is needed to cause the weak acid (acetic acid) to completely ionise". Stated differently, less energy is released from making acetate ion from acetic acid than from making chloride ion from hydrochloric acid (or water from hydronium ion).

Despite the lesser tendency of acetic acid to ionize, the overall stoichiometry of the two reactions is the same--as pointed out in a comment by the OP.


How to find if the unpaired electron is placed in hybrid orbital or pure orbital in odd electron species?

When discussing hybrid orbitals, they are justified by the experimentally observed or calculated/simulated molecular geometry. For instance, the $\ce{^.C(CH3)3}$ radical was found to be planar by ab initio calculations (http://www.tandfonline.com/doi/abs/10.1080/00268977600102711). Thus, the molecular shape is best approximated by $\ce{sp^2}$ hybridization and the unpaired electron is most likely to be found in a $\ce{p}$ orbital.

As a simple heuristic under the arguably outdated VSEPR theory, electronegative atoms want to be the furthest away from each other and other electrons as possible due to electric interactions. When considering the $\ce{^.CF3}$ radical, the fluorine atoms are highly electronegative and want to be the furthest they can be from the unpaired electron that holds negative charge. A planar $(\ce{sp^2})$ conformation would make the distance between the unpaired electron and each fluorine atom $90^\circ$. While a pyramidal $(\ce{sp^3})$ conformation would separate the two groups by $109^\circ$, which is much more favorable according to VSEPR theory. The later is what is observed under calculations (http://www.chem.mun.ca/courseinfo/c4420/handouts/handout-03a.pdf) with a $\ce{F-C-F}$ bond angle of $112^\circ$.

So, when considering hybridization problems, you must remember that hybridization fixes geometry. Thus if any plausible argument for a hybridization must be made, there should be experimental or computed evidence of the geometry. One cannot simply look at a molecule and know what the geometry could be with absolute certainty, but we can make a good guess based on simple heuristics, like those of VSEPR, molecular orbital diagrams, previous experience/similar molecules, etc.


Why methyl group is 2,4-directing?

When toluene undergoes electrophilic aromatic substitution the products are primarily the ortho and para isomers; usually only a small amount of the meta isomer is produced.  In order to explain this observation there are two effects to consider, inductive and resonance effects.

Inductively the methyl group releases electron density into the benzene ring.  This is because the methyl group, being $\ce{sp^3}$ hybridized, is less electronegative than the $\ce{sp^2}$ hybridized aromatic carbon.  The carbon with more s-character is more electronegative, electron density likes to flow from low to high s-character orbitals since the more s-character in an orbital, the lower its energy.  If inductive arguments were most important we would expect the ortho isomer to predominate followed by the meta and para isomers in that order.  This is not observed, the meta isomer is only a minor isomer.

Resonance arguments are based on the concept of hyperconjugation involving the $\ce{C-H}$ bonds in the methyl group.  The following figure shows some of the possible hyperconjugative resonance structures for toluene.

Additional resonance structures can be drawn using the other methyl $\ce{C-H}$ bonds as well.  As the resonance structures show, electron density is increased at the ortho and para positions, not the meta.  This hyperconjugative resonance effect makes the ortho and para carbons more susceptible to electrophilic attack.

Hyperconjugation explains how alkyl groups can be electron releasing in a resonance sense.  Hyperconjugation explains the mild activating effect of an alkyl group in electrophilic aromatic substitution and its ortho-para directing effect.


Iron can oxidize to "rust" (which is actually a mixture of hydrated iron(iii) oxide $\ce{Fe2O3*nH2O}$ and iron(iii) oxide hydroxide $\ce{FeO(OH)}/\ce{Fe(OH)3}$). The iron hydroxide species look like they could come from basic solutions.
Even if consider only the formation of $\ce{Fe2O3}$, this can occur by several pathways.
Using Wikipedia's Standard electrode potential data page as a source of known redox half reactions, the process may be something like this:
In acidic medium
The net reaction in acid is between the iron and the protons in the acid to produce $\ce{Fe^{3+}}$, which reacts with water in a redox neutral (but generates more acid) process to form the insoluble oxides/hydroxides. The initial redox reaction is spontaneous. The water is the net source of oxygen.
$$\ce{2Fe(s) + 6H+(aq) - 2Fe^{3+}(aq) + 3H2(g)} \ \ E^\circ=+0.04\ \text{V}$$
$$\ce{Fe^{3+}(aq) +3H2O -> Fe(OH)3(s) + 3H+(aq)}$$
$$\ce{Fe(OH)3(s) -> FeO(OH)(s) + H2O}$$
$$\ce{2FeO(OH) -> Fe2O3 + H2O}$$
In basic medium
The net reaction in base is between oxygen (in the air, which is not in contact with your submerged nail) and the iron in the nail. The iron initially reacts with the hydroxide ions, which are regenerated by the reduction of oxygen at the surface of the water.
$$\begin{align}
\ce{4Fe(s) + 8OH- &-> 4Fe(OH)2 + 8e-} \ \ \ &&E^\circ=\pu{+0.89 V}\\
\ce{4Fe(OH)2 + 4OH- &-> 2Fe2O3 +4e-}\ \ \ &&E^\circ=\pu{+0.86 V}\\
\ce{3O2 + 6H2O + 12e- & -> 12OH-}\ \ \ &&E^\circ=\pu{+0.40 V}\\ \hline
\ce{4Fe(s) +3O2(g)& -> 2Fe2O3(s)}\ \ \ &&E^\circ=\pu{+1.95 V}\\
\end{align}$$


What are the concentration changes when diluting an equimolar acetic acid: acetate buffer?

Consider an acid buffer with $pK_a=5$. Ignore the hydroxyl concentration, assuming it is negligible.

Dilution leads to an identical (proportional) reduction in the concentration of all solutes: undissociated acid, complementary base (anion), and protium. Because the concentration of anion and protium is reduced, encounters between the two will occur less frequently. On the other hand, dissociation of the acid into component ions continues unabated, since this process is (we are ignoring activities) a first order reaction. The result is an increase in the concentrations of anion and protium, thus reestablishing a value of the protium concentration close to that prior dilution.

Other  important factors should be remembered in the context of ideal buffers and the Henderson-Hasselbach equation: (1) the concentrations of $\ce{A^-}$ and $\ce{HA}$ are set equivalent in an ideal buffer (2) the $\ce{H^+}$  concentration is set equal to the $K_a$ and (3) the concentrations of $\ce{A^-}$ and $\ce{HA}$ are much higher than those of $\ce{H^+}$ . That means that in both absolute and relative amounts, the concentrations of $\ce{A^-}$ and $\ce{HA}$ need to change by only a small amount in order to bring the pH back near the desired set point. This is very clear from the following plots (concentrations are molar and $pK_a = 5$). Every 0.1 time units the concentration is diluted by 17%. The pH spikes up but then settles back again as the acid dissociates. You practically don't see the changes in the undissociated acid and anion on the plot during the recovery because they are tiny.

You can compare this to the case when your buffer is not as good (same $pK_a$, but lower capacity, 100 $\mu M$ concentration). I altered the kinetics too so the pH would stabilize after each dilution:

I should add: ignoring $[OH^-]$ is admittedly bad near neutral pH/low buffer capacity, but then I was attempting to provide a visualization. Adding terms to cope with the hydroxyl equilibrium etc would complicate things without adding much value or accuracy (since the kinetics are not meant to be accurate anyway, just illustrative).

Addendum: Here's the code:

K = 1e-5;  % M
kd=10;     % arbitrary, set to alter convergence speed, numerical stability
dt = 1e-7; % arbitrary, alter to change convergence speed, numerical stability
v0 = [0.2; 0.2 ; K]; % M [HA], [A], [H],

%% for dilute conditions
%kd=80;
%dt = 1e-5;
%v0 = [0.0001; 0.0001 ; K]; % [HA], [A], [H]

Nstep = 10000;
t = [0:dt:(Nstep-1)*dt];
tdilute = t(round(Nstep*[0.1:0.1:0.9]));

dv_dt = @(kd,K,v) [ -kd*v(1)+kd/K*v(2)*v(3); kd*v(1)-kd/K*v(2)*v(3)] ;
dv_ = @(v,dt) dv_dt(kd,K,v)*dt;
dv = @(v) dv_(v,dt);
v = v0;
conc = v0;
for ii=2:Nstep
    vstep = dv(v);
    v = v + [vstep(1); vstep(2); vstep(2)];
    conc(:,ii) = v;
    if any(t(ii) == tdilute)
        v = v/1.2;  % <-- change factor to dilute by here
    end
end

figure
subplot(2,1,1)
plot(t,conc(1,:),'r')
hold on
plot(t,conc(2,:),'g--')
plot(t,conc(3,:),'b--')
ylabel('[c]')
legend('[HA]','[A^-]','[H^+]')
subplot(2,1,2)
plot(t,-log10(conc(3,:)),'b--')
ylabel('pH')
xlabel('time (au)')


What's the difference between aqueous hydrogen chloride and hydrochloric acid?

There is no difference.

Actually, it would be more accurate to refer to gaseous (and dry) $\ce{HCl}$ as "hydrogen chloride", and to aqueous $\ce{HCl}$ as "hydrochloric acid", but often the terms are used interchangeably.

The difference is due to the fact that, as you might guess, an acid is a proton donor (it means that it tends to give off $\ce{H+}$ in a solution). On the other side, gaseous and dry $\ce{HCl}$ is simply a diatomic molecule. The dissociation in its ions (and the showdown of its acidic properties) only happen in a polar medium (typically water).

In the paper - Zotikov, V.S. Russ J Appl Chem (2009) 82: 1733. https://doi.org/10.1134/S1070427209090389 - you can find a cool example of the difference between a dry hydrogen halide and a wet hydrogen halide (here, the "terrible" HF is involved):

  In  dry  hydrogen fluoride,  carbon  steel  is  subjected to  minor  corrosion  up  to  $\pu{250^\circ C}$.  In  aqueous  solutions the  carbon  steel  is  stable  only  at  the  concentration  of hydrogen fluoride  above  75  wt  %  and  temperature not exceeding $\pu{50^\circ C}$


How does hybridisation affect an otherwise chiral centre?

I really wouldn't put too much stock into YouTube comments. Often, it's just the blind leading the blind. The comments on this video have all sorts of incorrect statements, it's really not where you want to be learning from...

Firstly, chiral centres are not restricted to only carbon. Source: IUPAC Gold Book. With that out of the way, we can enumerate the three main possibilities for nitrogen.

$\mathrm{sp^3}$ nitrogen with a lone pair is usually not considered to be a chiral centre, as the rate of inversion is very rapid (google "nitrogen inversion").

There are exceptions, such as if the amine geometry is constrained by a bicyclic system (something similar to quinuclidine), but that's a discussion for another day.
$\mathrm{sp^3}$ nitrogen without a lone pair e.g. in ammonium cations, can be chiral if there are four different groups attached. So these two molecules are enantiomers of each other.

In fact, it is this context in which Wikipedia says that nitrogen can be a chiral centre.

  Racemization by Walden inversion may be restricted (such as ammonium cations), or slow, which allows the existence of chirality.

The implication is that nitrogen inversion is fast in the case where there is a lone pair. However, it is very unclear, so I don't blame you.
$\mathrm{sp^2}$ nitrogen, e.g. amide nitrogen, is planar so it cannot be chiral. So, this video you watched is incorrect in stating that the amide nitrogen is a chiral centre.

And, that also means the comment you quoted is correct. But, I would caution that it is not a direct cause and effect relationship between "$\mathrm{sp^2}$ hybridisation" and "not a chiral centre". It is more like: because it is $\mathrm{sp^2}$, therefore the geometry around it is planar; and because it is planar, it cannot be chiral.

As was mentioned in the comments, phosphorus and sulfur behave differently. This has been covered multiple times here; here are some links. 1 2 3 4


How do I name this alkane with a branch that has bromine?

3-(bromomethyl)hexane

You can't have a 2-position on a methyl group, there is only a 1-position and we omit redundant 1-enumerations of position.


Are there limits to waters energy absorbtion capacity

The heat transfer is limited by the flask material heat transfer rate, which is proportional to the thermal gradient within the flask. The external flask temperature will follow.
On the water side, liquid can jump out of flask in form of the thick foam, powered by the vapour. This is the major limiting factor.
On external side, at high enough power, the eventual plastic holders/features may start to melt/decompose. At higher power, the flask may start shining red. Or start to melt if metallic.


Finding the pH of a mixture of weak acid and strong base

Buffer equation

The Henderson equation for buffers is:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$

$\mathrm{p}K_\mathrm{a}$ and $\mathrm{p}K_\mathrm{b}$ add up to 14, as do $\mathrm{pH}$ and $\mathrm{pOH}$. So the expression for $\mathrm{pOH}$ is:

$$\mathrm{14 - pOH = 14} - \mathrm{p}K_\mathrm{b}  + \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$

or

$$\mathrm{pOH} = \mathrm{p}K_\mathrm{b}  - \log{\frac{[\ce{A-}]}{[\ce{AH}]}}$$

Amounts of weak acid and weak base

  Also, shouldn't the buffer solution have a common ion?

You start with $\pu{10 mmol}$ of ammonium salt (weak acid), to which you add $\pu{7.5 mmol}$ of $\ce{NaOH}$. The result is $\pu{7.5 mmol}$ of ammonia (weak base) with $\pu{2.5 mmol}$ ammonium salt remaining. So the ratio of weak base to weak acid is 1:3.

If you plug this into one or the other buffer equation, you get the answer.


Does 99.8% acetic acid cause severe skin burns like formic acid?

Yes it is corrosive to skin.  The Safety Data Sheet identifies it as Category 1A for skin corrosion, which means:

A substance that, according to animal data acquired from a scientifically validated method, produces irreversible damage to the skin after an exposure of three minutes or less, and within one hour of observation, in at least one of three animals.

Don't let the "organic" part fool you.  In terms of handling hazards glacial acetic acid rivals concentrated hydrochloric acid, and actually beats it in skin corrosion classification.


How to calculate the pH of a sodium bicarbonate-sodium hydroxide buffer solution?

Your idea is good, but you have thought of the wrong reaction.

You have written that $\ce{HCO_3^-}$ hydrolyses to give $\ce{H_2CO_3}$ and $\ce{ OH^-}$, but think carefully when you add base, $\ce{OH^-}$ is in excess in this medium. So, shouldn't your proposed reaction proceed in backward direction instead? Moreover, your reaction is actually not going to occur.

When you are adding strong base, $\ce{HCO_3^-}$ acts an acid to give $\ce{CO_3^2-}$ like:

$$\ce{NaHCO_3 + NaOH -> Na_2CO_3 +H_2O}$$
 or, more simply
$$\ce{HCO_3^- + OH^- -> CO_3^2- + H_2O}$$

Initially, number of millimoles of $\ce{HCO_3^-}$ were $=50\times0.05 = 2.5$ and number of millimoles of base added $=5\times0.1 = 0.5$

So, $\ce{CO_3^2-}$ produced will be also 0.5 millimoles as the base added was a limiting reagent, and $\ce{HCO_3^-}$ left $= 2.5-0.5 = 2$ millimoles.

As, you can see now, the solution acts an acid buffer as, $\ce{HCO_3^-}$ is a weak acid and $\ce{CO_3^2-}$ is the salt after reacting with a strong base. According to Henderson-Haselbach equation,

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{[\text{salt}]}/[\text{acid}]$$
$\mathrm{p}K_\mathrm{a}$ of $\ce{HCO_3^-}=\mathrm{p}K_{\mathrm{a2}}$ of $\ce{H_2CO_3^}=10.3$ . Now [$\ce{HCO_3^-}$] = $\pu{0.02 M}$, and [$\ce{CO_3^2-}$] = $\pu{0.005 M}$. Now if you calculate $$\mathrm{pH}= 10.3 +\log\frac{0.5}2 = 10.3- \log(4) = 9.69$$ So, the final $\ce{pH}$ of the medium has come out to be $= 9.69$.

The difference in digits after decimal may be due to some given $\mathrm{p}K_\mathrm{a2}$ which is experimental and you should do calculation according to that.


Which has higher priority according to Cahn-Ingold-Prelog priority rules, vinyl or secondary butyl?

The assignment of CIP priorities has some intricate details when it comes to multiple bonds.

First, the general procedure (Wikipedia):

  If there is a tie, we must consider the atoms at distance 2 from the stereocenter — as a list is made for each group of the atoms bonded
  to the one directly attached to the stereocenter. Each list is
  arranged in order of decreasing atomic number. Then the lists are
  compared atom by atom; at the earliest difference, the group
  containing the atom of higher atomic number receives higher priority.
  If there is still a tie, each atom in each of the two lists is replaced with a sub-list of the other atoms bonded to it (at distance
  3 from the stereocenter), the sub-lists are arranged in decreasing
  order of atomic number, and the entire structure is again compared
  atom by atom. This process is repeated, each time with atoms one bond
  farther from the stereocenter, until the tie is broken.

Now, concerning double bonds in particular:

  If an atom A is double-bonded to an atom B, A is treated as being
  singly bonded to two atoms: B and a ghost atom that has the same
  atomic number as B but is not attached to anything except A. In turn,
  when B is replaced with a list of attached atoms, A itself is excluded
  in accordance with the general principle of not doubling back along a
  bond that has just been followed, but a ghost atom for A is included
  so that the double bond is properly represented from both ends.

I have constructed a table of these atom lists for the vinyl, isopropyl and sec-butyl rests:
$$\small
\begin{array}{llll}
\hline
\text{Dist} & \text{Vinyl} & \text{Isopropyl} & sec\text{-Butyl} \\
\hline
1 & \text{C} & \text{C} & \text{C} \\
2 & \text{C} & \text{C} & \text{C} \\
  & \text{(C)} & \text{C} & \text{C} \\
  & \text{H} & \text{H} & \text{H} \\
3 & \text{(C)} & \text{H} & \text{C} \\
  & \text{H} & \text{H} & \text{H} \\
  & \text{H} & \text{H} & \text{H} \\
  & & \text{H} & \text{H} \\
  & & \vdots & \vdots \\
\hline
\end{array}
$$

The sec-butyl rest has thus the highest priority of the three, followed by vinyl and then isopropyl.

The ghost atom at distance three is the ghost carbon at distance one, as seen from the ghost carbon that was doubly bonded to it. If the distance 4 was to be considered, no more ghost atoms would appear, as the ghost atom at position 1 does not have any more ghost atoms bonded to it (atom 1 however is bonded to the ghost atom at position 2).


Is it correct to say that the pKa is the pH at equilibrium?

$\mathrm{p}K_\mathrm{a}$ and $\mathrm{pH}$ are apples and oranges.* We can't equate them on a graph. The correct wording should rather be that the $\mathrm{pH}$ at which the concentration ratio of an acid and its conjugate base is equal to $1,$ then the $\mathrm{pH}$ is numerically equal to $\mathrm{p}K_\mathrm{a}.$

* It will be another painful story if one begins to argue on the units of $\mathrm{pH}$ and $\mathrm{p}K_\mathrm{a}.$ They are dimensionless. Just a subtle point $\mathrm{p}K_\mathrm{a}$ is constant as long as the temperature/pressure is constant. It is not always a constant.


Shape of Weak-Strong Acid-Base Titration

Suppose $\ce{NH3(aq)}$ is being titrated with $\ce{HCl(aq)}$. The products are $\ce{NH4+ (aq)}$ and $\ce{Cl- (aq)}$. As the titration proceeds [$\ce{NH3}$] decreases and [$\ce{NH4+}$] increases. The presence of $\ce{NH3}$ and $\ce{NH4+}$ together constitutes a buffer system, in this case a weak base and its conjugate acid. When molarities of both species are relatively large, the buffering capacity is significant.

Halfway to the equivalence point, the molarities of both species are equal, which allows for the greatest possible buffering action. In the very early part of the titration, the molarity of $\ce{NH4+}$ is not great enough to allow much of a buffering effect, so the pH falls rapidly with the addition of $\ce{HCl}$. But more toward the point where [$\ce{NH3}$] becomes equal to the [$\ce{NH4+}$], the buffering action is increasingly able to "resist" changes in pH so the slope of the titration curve becomes more shallow. Past this point, the [$\ce{NH3}$] steadily diminishes to the point that buffering is no longer possible and the pH dives through the equivalence point as we expect.

Past the equivalence point, only $\ce{NH4+}$ is present so no buffering can occur and there are no inflections like the one that occurs at the start of the titration. The importance of the "midpoint" as it is labeled in the illustration is that pH vs. titrant volume analyses are sometimes done specifically to determine the K$_a$ or K$_b$ of a weak acid or weak base, respectively. Continuing with the $\ce{NH3}$ and $\ce{HCl}$ example, the K$_b$ for $\ce{NH3}$ is $[\ce{NH4+}][\ce{OH-}]/[\ce{NH3}]$. Halfway to the equivalence point, [$\ce{NH4+}$]=[$\ce{NH3}$] and we have K$_b = [\ce{OH-}$] or pK$_b$ = pOH.


Laboratory and commercial preparations of aniline other than reduction of nitroarenes

It is true that until recently, the syntheses of primary anilines mostly rely on the reduction of corresponding nitroarenes. However, recent publications allow the synthesis of primary anilines or their derivatives in laboratory.
One such method is Buchwald–Hartwig amination:

The palladium-catalyzed cross-coupling of amines and aryl (pseudo)halides, now commonly known as the Buchwald–Hartwig amination, was first reported 25 years ago. Since the simultaneous breakthrough reports of Buchwald and Hartwig in 1995, this reaction has transformed the way synthetic chemists think about synthesizing aromatic amines... (Ref.1):

A continuous process for Buchwald–Hartwig amination in different scale synthesis has also been introduced recently (Ref.2). Further, a version of Buchwald–Hartwig amination for the preparation of primary arylamines has been introduced recently as a general method (Ref.3):

The importance of the method is reflected by the development of a synthetic route to manufacture the drug candidate AMG 925 on kilogram scale using Buchwald–Hartwig amination (Ref.4):

Another amine synthetic method not following nitro reduction is Ullmann amine synthesis and its variations (Ref.5 and references therein):

In addition, secondary or tertiary amine can be prepared using Chan-Lam coupling reaction. For example see following reaction (Ref.6):

Other than using aryl halides, there is an another method to convert phenol derivatives to anilines. One such example describes a direct conversion of phenols into primary anilines with hydrazine catalyzed by palladium (Ref.7):

References:

Paola A. Forero-Cortés, Alexander M. Haydl, "The 25th Anniversary of the Buchwald–Hartwig Amination: Development, Applications, and Outlook," Org. Process Res. Dev. 2019, 23(8), 1478–1483 (https://doi.org/10.1021/acs.oprd.9b00161).
Sebastian Falß, Giovanna Tomaiuolo, Antonio Perazzo, Paul Hodgson, Polina Yaseneva, Jacek Zakrzewski, Stefano Guido, Alexei Lapkin, Robert Woodward, Rebecca E. Meadows, "A Continuous Process for Buchwald–Hartwig Amination at Micro-, Lab-, and Mesoscale Using a Novel Reactor Concept," Org. Process Res. Dev. 2016, 20(2), 558–567 (https://doi.org/10.1021/acs.oprd.5b00350).
Giang D. Vo, John F. Hartwig, "Palladium-Catalyzed Coupling of Ammonia with Aryl Chlorides, Bromides, Iodides, and Sulfonates: A General Method for the Preparation of Primary Arylamines," J. Am. Chem. Soc. 2009, 131(31), 11049–11061 (https://doi.org/10.1021/ja903049z).
Caroline Affouard, Richard D. Crockett, Khalid Diker, Robert P. Farrell, Gilles Gorins, John R. Huckins, Seb Caille, "Multi-Kilo Delivery of AMG 925 Featuring a Buchwald–Hartwig Amination and Processing with Insoluble Synthetic Intermediates," Org. Process Res. Dev. 2015, 19(3), 476–485 (https://doi.org/10.1021/op500367p).
Andrea Francesca Quivelli, Paola Vitale, Filippo Maria Perna, Vito Capriati, "Reshaping Ullmann Amine Synthesis in Deep Eutectic Solvents: A Mild Approach for Cu-Catalyzed C–N Coupling Reactions With No Additional Ligands," Frontiers in Chemistry 2019, 7, Article 723 (14 pages) (https://doi.org/10.3389/fchem.2019.00723)(Full Paper).
Carolin Fischer, Burkhard Koenig, "Palladium- and copper-mediated N-aryl bond formation reactions for the synthesis of biological active compounds," Beilstein J. Org. Chem. 2011, 7, 59-74 (https://doi.org/10.3762/bjoc.7.10)(Full Paper).
Zihang Qiu, Leiyang Lv, Jianbin Li, Chen-Chen Li, Chao-Jun Li, "Direct conversion of phenols into primary anilines with hydrazine catalyzed by palladium," Chem. Sci. 2019, 10(18), 4775-4781 (https://doi.org/10.1039/C9SC00595A)(Full Paper).


What is the correct product of the Diels-Alder reaction between 2-ethoxybuta-1,3-diene and (Z)-acrylaldehyde-3-d?

These resonance structures and the polarization shown within explain the regioselectivity (1,4 isomer instead of 1,3) in the product.


To start with, I recommend to reformulate Beer-Lambert's law in a way more frequently met in (analytical) chemistry, i.e.

$\mbox{ABS} = c \cdot{} l \cdot{} \varepsilon = \log_{10} \frac{I_0}{I}$

because then it more obvious that the (dimensionless) absorption (ABS) recorded is the product of analyte concentration ($c$), optical path length ($l$) across the sample investigated, and the molecular extinction coefficient ($\varepsilon{}$).  Absorption equally is defined as the logarithm (to base of ten) of the fraction of incoming light intensity ($I_0$) and light intensity that is transmitted across the sample ($I$).

Now to your question.  While concentration $c$ and optical path length $l$ may be changed (the later, for example, by exchanging a measurement cell of 1 cm path length to one of 2 mm or 5 cm -- depending on the setup of the spectrometer), the molecular extinction coefficient $\varepsilon$ is indeed more valuable than just assistance to solve the equation.  It relates to the ease the analyte may absorb irradiation.

UV-Vis spectroscopy investigates electronic transitions, prominently of $\pi{}$-systems (like of alkene double, and aromatic bonds), yet investigation of electronic transitions of $n$- (non-bonded, like of carbonyl-O electrons), $d$- (for example in transition metal complexes) and $\sigma{}$-electrons (in the short wavelength UV-range).  In principle, these transitions are discrete, i.e. only photons of this attuned energy ($h\nu{}$) that fits the energy gap between the electronic ground state and the excited state are absorbed.

An analyte molecule may possess more than one allowed electronic transitions.  They may occur simultaneously, with different efficiency.  For example, many organic dyes have a strong absorption in the visible range, like indigo has one around 600 nm.  Yet if you tune the wavelength of investigation to short wavelengths, below 250 nm, for example, absorption may still occur, too, yet to a lesser extent.  (This eventually contributes to processes that indigo stained cloths fade over time, for example.)

Side note:  The molecular extinction coefficient $\varepsilon$ depends on other, additional parameters than the "solvent" and "wavelength of investigation".  Varying the pH-value, ionic strength of the solution, presence/absence of other dissolved molecules than of your analyte and solvent may alter $\varepsilon{}$, too.  Do not forget, the concentration $c$ of your analyte may influence $\varepsilon{}$, too -- pointing to (concentration dependent)  potential dimerisation / aggregation of your analyte (for example charge transfer complexes).


How to desalinate seawater at freezing temperature or high pressure?

If there is a problem raising the freshwater to the top, why not make ice
cubes from the freezing ocean depth and have them float to the top via a vertical pipe? The high ocean depth pressure would make membrane technology
less energy intensive.


rock ($\ce{SiO2}$)?

Rock is not strictly $\ce{SiO2}$. Rocks are composed mostly of silicates: minerals such as feldspars $\ce{CaAl2Si2O8}$ or $\ce{(Na,K)AlSi3O8}$, olivine $\ce{(Mg,Fe)2SiO4}$, pyroxene $\ce{CaMgSi2O6}$, etc.

There are two ways in which rocks can absorb carbonates, one being a more direct form of the other.

Weathering of rocks by carbonic acid. $\ce{CO2}$ reacts with $\ce{H2O}$ to form acid and bicarbonate: $\ce{CO2 + H2O -> H+ + HCO3^-}$. This acid would then trigger: $\ce{4KAlSi3O8 + 4H+ + 2H2O -> 4K+ Al4Si4O10(OH)8 + 8SiO2}$. This is the reaction of orthoclase (a type of feldspar, common in granites) to form kaolinite (a clay, aka dust or mud). All of the resulting cations ($\ce{Ca2+, Mg2+, K+, Na+}$, etc. and the bicarbonate end up washing to the oceans, where they precipitate carbonates: solid rocks composed of mostly calcite ($\ce{CaCO3}$) and dolomite ($\ce{CaMgCO3}$), thus overall removing  $\ce{CO2}$ from the atmosphere.
Directly reacting of $\ce{CO2}$ with solid rock. All of the above basically happens in-situ, directly transforming solid silicate rocks to solid carbonate rock. When this happens, minerals such as magnesite ($\ce{MgCO3}$) or siderite ($\ce{FeCO3}$) also occur. This is actually one of the way to do industrial carbon sequestration: instead of putting $\ce{CO2}$ in the atmosphere, for example in power stations, you inject it down to the rock to lock it up in solid minerals.


Calculating collision diameter with viscosity, but not having the mass given

Answering my own question here (just received the answer to the problem.)

the value for $(\mathrm{m})$ was supposed to be in $\mathrm{kg}$, and could be found by taking the Molar mass and converting:

$\mathrm{M} \cdot 1.661 x10^{-27}\mathrm{kg\cdot amu^{-1}}$

The solution came out looking like

$\overset{-}{v} = \sqrt{\frac{8x8.314 \mathrm{J \cdot K^{-1} \cdot mol^{-1}} x 288\mathrm{K}}{3.14x28.05x10^{-3}kg \cdot mol^{-1}}}$

$d=\sqrt{\frac{m \overset{-}{v}}{3 \sqrt{2} \pi \eta}}$

$d=\sqrt{\frac{(28.05 \mathrm{amu} \space x \space 1.661x10^{-27} \mathrm{kg} \cdot \mathrm{amu}^{-1}) \space x \space 466.23 \mathrm{m} \cdot \mathrm{s}^{-1}}{3 \sqrt{2} \pi \space x \space 99.8x10^{-7} \mathrm{N} \cdot \mathrm{s} \cdot \mathrm{m}^{-2}}}$

$=4.04x10^{-10} \mathrm{m} = 4.04 \overset {\circ}{A}$


Grouping of symmetry classes in chemistry character tables

The $\sigma_\mathrm v$ mirror planes in the $C_\mathrm{3v}$ point group are themselves related by symmetry: note that they can be interchanged via rotation by 120 degrees about the preexisting $C_3$ axis. To be technically precise, they belong to the same conjugacy class, in the sense that applying $C_3$, then one mirror plane, and then the inverse of $C_3$ is the same as applying a different mirror plane:

$$C_3\sigma_{\mathrm v, i}C_3^{-1} = \sigma_{\mathrm v, j}$$

As a result of this, they have to be treated together and not separately, in the sense that the characters in the character table represent how each irrep transforms under all the three mirror planes collectively. Unfortunately, this is hard to explain properly without delving into the actual matrices. But it should be possible to see that the two mirror planes in $C_\mathrm{2v}$ are not related to each other in the same way.


Most of the original question has been solved by discussion in the comments, but a pertinent question remains:

  [W]hy you do not simply take the HF expression for the exchange part and insert it into the DFT equations [?]

The answer is that it has been tried and does not work well. In a sense, the DFT quantities "exchange energy" ($x$) and "correlation energy" ($c$) do not quite match the HF quantities "exchange energy" and "correlation energy".$^1$ In fact, even within DFT, the quantities do not match between functionals, as can be seen in some $x$- and $c$-functionals working together well and some not so well. In particular, those that have been fitted in combination with each other often only work in that combination. This suggests to me an illusion of a clear-cut way to differentiate the two in the DFT framework. It is also true that in some functionals, the $x$-term appears to account for some static correlation (which, broadly speaking, often becomes important when a single Lewis structure does not describe the molecule well: transition states, resonance structures). Entirely replacing it with Fock exchange forgoes this advantage.

Some programs, such as NWChem, and to some extent Gaussian, allow you to combine functionals and Fock exchange (often with arbitrary weights) to conduct your own experiment. A small benchmark set for main-group thermochemistry would probably give you an indication.

I wish I could you references for this, but this has been related to me verbally by my advisor back in the day and I no longer have access to most of the literature. If anybody has such references, I would be happy to see them included.

$^1$ The notation is a bit sloppy, but we're not striving for rigor in this answer.


Is the ammonium substituent (-NH3+) really meta-directing in electrophilic substitution?

how is anilinium ion meta directing for electrophiles?

Actually, anilinium is not meta directing (I know it is often taught that way), but rather it inductively deactivates the entire aromatic ring.

To explain the electrophilic substitution pattern observed with any aromatic system we must consider both resonance and inductive effects.

Resonance Effects: You are correct that you can't delocalize the positive charge on the anilinium nitrogen into the benzene ring through resonance.  As you point out, resonance structures with 5 bonds to nitrogen aren't allowed.

Therefore resonance effects are not involved in explaining the reactivity of the anilinium ion towards electrophilic aromatic substitution.

Inductive Effects: Inductively we expect the anilinium nitrogen to strongly withdraw electron density from the ortho position since it is closest to the positively charged nitrogen, this makes the ortho position the most deactivated.  Inductive effects die off quickly with distance, so we would expect the meta position to also be deactivated, but less so than the ortho position.  Finally, we would still expect some deactivation at the distant para position, but slightly less than we saw at the meta position. Overall, in terms of reactivity we would expect para > meta > ortho, with the difference between para and meta being much smaller than the difference between meta and ortho due to the effect distance has on inductive electron withdrawal.

If we examine the nitration of aniline in strong acid ($\ce{HNO3}$, $\ce{H2SO4}$), under these conditions the strong acid will completely protonate the aniline and reaction will proceed through the anilinium ion.  We see that, as expected, there is very little ortho product and slightly more para product than meta product.  This is pretty much as we expected and is consistent with the electron withdrawing inductive effect of the positive anilinium nitrogen.


FeS2 has an odd structure; the iron atom has a +2 oxidation number and each of the sulfurs has a -1 oxidation number.

This can be balanced by inspection.

4 FeS2 + 11 O2 --> 2 Fe2O3 + 8 SO2

Just to check, using oxidation numbers we get:

Sulfur goes from -1 to +4.  (Total change from sulfur = +40)

Iron goes from +2 to +3.    (Total change from iron = +4)

(I'm not sure where you got 4 for the oxidation number of iron in FeS2.)

Oxygen goes from 0 to -2.   (Total change from oxygen = -44)

ETA:
Half reactions are messy here, since both iron and sulfur are oxidized.  Similarly, using oxidation numbers is problematic (except to check the solution) because there are three substances changing oxidation states.  You could write a system of equations to describe it, but that's a lot more trouble than it's worth.

You know from the structure of iron(II) disulfide that there are twice as many sulfur atoms as iron atoms.  That means that the number of sulfur dioxide molecules must be four times the number of iron(III) oxide molecules.  With that relationship in mind, the smallest ratio of molecules that fits the pattern is the one I wrote above.

If you really want to use oxidation numbers, here's what I've come up with:

Let a equal the number of iron atoms, b equal the number of sulfur atoms, and c equal the number of oxygen atoms.

Iron increases its oxidation state by 1, sulfur by 5, and oxygen decreases by 2. So:

1a + 5b - 2c = 0

We also know that iron and sulfur are in a 1:2 ratio because they come from pyrite.

2 a = b

Substitution yields:

1a + 5(2a) - 2c = 0

11a - 2c = 0      Iron and oxygen are in a 2:11 ratio.

Oxygen must be even since it comes as O2, so its smallest possible value is 22 atoms.  Using the ratios listed above, we get 22 O: 4 Fe: 8 S.

ETA #2: Half Reactions

Fe+2 --> Fe+3 + e-

S-1 --> S+4 + 5 e-

O20 + 4 e- --> 2 O-2

We know from FeS2 that there must be twice as many sulfur atoms as iron, so the second equation has to be multiplied by two.

2 S-1 --> 2 S+4 + 10 e-

If we add together all of the species that are being oxidized (Fe and S) then we get an oxidation half reaction of:

Fe+2 + 2 S-1 --> Fe+3 + 2 S+4 + 11 e-

This must be multiplied by 4 and the reduction equation by 11 to balance the number of electrons.

4 Fe+2 + 8 S-1 --> 4 Fe+3 + 8 S+4 + 44 e-

11 O20 + 44 e- --> 22 O-2

Adding this mess together gives:

4 Fe+2 + 8 S-1 + 11 O20 --> 4 Fe+3 + 8 S+4 + 22 O-2

Since we know the structure of the molecules, putting it all back together is fairly straightforward, and it gives the same result as listed above.


Stronger ESI signal for deuterated substances

There are a couple reasons this could be happening.
The most interesting possibility: isotope effects during ion fragmentation
You are performing LC-MS/MS, i.e., your mass spectrometer is chemically reacting the estradiol and estradiol-$d_5$ with gas molecules in the collision cell of the instrument.  This process, known as CID or collisionally induced dissociation, will break chemical bonds in the parent molecule and form fragment ions.
On triple-quad and qTOF instruments, the CID is of the "beam" or non-resonant type.  This means a fairly large amount of energy is imparted to parent ions.  In general, this means that a wide variety of fragment ions are formed.  This is true of estradiol; publicly available fragmentation spectra show many fragment ions for this compound (see several GNPS spectra for example).
Deuteration is likely to change the relative propensity of the different fragmentation pathways.  For example, these reactions could represent the fragmentation of a parent ion $\ce{P}$ into two different fragments $\ce{F1}$ and $\ce{F2}$.
$$\ce{P -> F1}$$
$$\ce{P -> F2}$$
Isotopically normal $\ce{P}$ might give an intensity ratio of say 1 to 1 for $\ce{F1}$ and $\ce{F2}$.  But due to deuterium isotope effects, the pathway to $\ce{F1}$ may be much less favored in deuterium-labeled $\ce{P}$.  This could lead to an enhanced signal for $\ce{F2}$ as a much higher fraction of the $\ce{P}$ ion remains available to proceed down that fragmentation pathway.
However, a three-fold increase is pretty large.  I was able to find good measurements of various deuterated tocopherols, and judging from Table 1 in that paper, the response ratio for $d_0$ (unlabeled) tocopherol vs. $d_9$ is 11.8/0.967/10, or 1.22, not 3.  For measurement of trimethylamine N-oxide, authors note "the intensity ratios of the two product ions are different between TMAO and $d_9$-TMAO due to the difference in bond energy among C and H and C and D (deuterium)" but did not quantify this difference.  Eyeballing Figure 1, the intensity ratio of the light and heavy fragment is about 2.5:1 for $d_0$ but 1.4:1 for $d_9$.  None of those values are close to 3-fold.  But maybe estradiol and its site of deuteration and its fragmentation pathway differ and that a value of 3-fold is possible.
Errors in sample preparation
Maybe the concentration of one of your samples is off.  How many times have you independtly prepared the standards and repeated the measurement?
Ways to tell
If you can see a signal with a single quadrupole (i.e. do MS1 instead of MS/MS), do you see the same three-fold difference?  If so, error in sample prep is likely to blame.  If not, then isotope effects during ion fragmentation are to blame.


How to construct NMR spectra from chemical shift tensors?

Yes, it is possible. It is done often. The solution depends on the dynamics and orientation(s) of the spins in the ensemble (as will be shown bellow). I will assume you want to calculate the spectrum in frequency space.

$$
\omega_{cs} = \sigma_{zz}^{LF}(-\gamma \vec{B}_{0})
$$

We can hide a few variables and rewrite this in a much more useful way as:
$$
\omega_{CS} = (\vec{b}^t_0 \overset {\leftrightarrow}{\omega}\vec{b}_0)^{LF} = (\vec{b}^t_0 \overset {\leftrightarrow}{\omega}\vec{b}_0)^{PAS} =(\vec{b}^t_0 \overset {\leftrightarrow}{\omega}\vec{b}_0)^{RF}
$$

where,

$$
\vec{b_0} = \frac{\vec{B_0}}{B_0}.
$$

The above allows us to calculate the chemical shift in many reference frames. The frame where $\omega$ is diagonal (the PAS) is often convenient.

$$
\omega_{CS} = \omega_{xx}^{PAS}(\vec{b}^{PAS}_{0,x})^2 + \omega_{yy}^{PAS}(\vec{b}^{PAS}_{0,x})^2 + \omega_{zz}^{PAS}(\vec{b}^{PAS}_{0,x})^2
$$

We can define $\theta$ as the angle between $Z^{PAS}$ and $b_0$ and $\phi$ as the angle between $X^{PAS}$ and the x-component of the $b_0$ field. This lets us write:

$$
\omega_{CS} = \omega_{xx}^{PAS} \cos^2\phi \sin^2\theta + \omega_{yy}\sin^2\phi \sin^2\theta+ \omega_{zz}^{PAS}\cos^2\theta
$$

The above is the main equation you will need to calculate a chemical shift spectrum in frequency space. Numerically you will sum over all the angles in your ensemble. If your sample is a powder you could use angles sampled from a Gaussian distribution on a sphere for example. Typically people then apply convolution with a Lorentzian function (to account for relaxation and smooth the numerical artifacts). If your sample is a solution and the molecular reorientation time is fast compared to the inverse of the anisotropy, everything but the isotropic component will be averaged away. To put this in context (and to get to some possibly familiar equations) it will be useful to introduce some convention. In the Haeberlen-convention:

$$|\delta_{zz} - \delta_{iso}| \ge |\delta_{xx} - \delta_{iso}| \ge |\delta_{yy} - \delta_{iso}|$$
$$\delta_{iso} = \frac{1}{3}(\delta_{xx} + \delta_{yy}+ \delta_{zz})$$
$$\delta = \delta_{zz} =\delta_{iso}$$
$$\eta = (\delta_{yy}-\delta_{xx})/\delta$$

Where $\delta$ is called the anisotropy and $\eta$ is called asymmetry. Graphically:

Using this convention and some geometry you can get to:
$$
\omega _{CS} = \frac{\delta}{2} (3\cos ^2\theta -1 + \eta \sin^2 \theta \cos 2\phi) + \delta_{iso}
$$

Also, ab initio chemical shift calculations normally are reported in shielding values, you will probably want to convert to shift values using a sutable reference.

Reference:
Multidimensional Solid-State NMR and Polymers, by Klaus Schmidt-Rohr and Hans Wolfgang Spiess, chapter 2


Reason for protonation of amino group in amino acids at near neutral pH

Prerequisite knowledge

A proton is deprotonated when the $\mathrm{pH}$ exceeds the $\mathrm pK_\mathrm a$ of the proton.

Analysis
There are 3 different forms that an amino acid can exist in. For the sake of simplicity of analysis, I am taking the simplest amino acid - glycine.
$$\underset{\mathrm{glycine}}{\ce{HOOC-CH2-NH2}}$$
Continuing to state the conditions of analysis, let's take a highly acidic medium to begin with and observe the changes that take place on increasing $\ce{pH}$.
Case 1: pH of solution < 2.34
In such a scenario, the $\mathrm{pH}$ is lower than the $\mathrm p K_\mathrm a$ of the acid group and lower than the $\mathrm p K_\mathrm a$ of the amino group. This means that both the amino group and acid groups are protonated. The amino acid is positively charged at this $\mathrm {pH}$.
$$\underset{\mathrm{glycine\,(pH \,<\,2.34)}}{\ce{HOOC-CH2-NH3+}}$$
Now, we increase the $\ce{pH}$ to a value greater than $2.34$ but less than 9.6  Specialty of these numbers? We'll see at the end.
Case 2: pH of solution greater than 2.34 but less than 9.6
In such a scenario, the $\mathrm{pH}$ is higher than the $\mathrm p K_\mathrm a$ of the acid group but lower than the $\mathrm p K_\mathrm a$ of the amino group. This means that the amino group remains protonated and the acid group is now deprotonated.
$$\underset{\mathrm{glycine\,(2.34\,<\,pH \,<\,9.6)}}{\ce{^-OOC-CH2-NH3+}}$$
Case 3: pH of solution greater than 9.6
In this last case, the $\mathrm{pH}$ is higher than both the $\mathrm pK_\mathrm a$s of the amino group and the acid group. This means that both groups are now deprotonated. The amino acid is now negatively charged.
$$\underset{\mathrm{glycine\,(pH \,>\,9.6)}}{\ce{^-OOC-CH2-NH2}}$$

Observations and Conclusion
As you noticed, there are some specific numbers that are so important they are mentioned. They are the $\mathrm p K_\mathrm a$ values of the acid group ($\mathrm pK_{\mathrm a_1}$) and the amino group ($\mathrm pK_\mathrm {a_2}$). There is one more important number and that is the exact point of net average neutrality. It is known as the isoelectric point.
$$\mathrm {pI} = \frac{\mathrm pK_\mathrm {a_1} + \mathrm pK_\mathrm {a_2}}{2}$$


Can the amount of shielding in proton NMR caused by ring currents be predicted?

Johnson and Bovey first proposed a theory describing the ring current effect,[1] and in 1972 Haigh and Mallion published their often-cited paper on ring current shielding.[2] A later review by these authors in 1979 describes the mathematical prediction of ring currents very well.[3] You may often see this graphic describing ring current contributions to chemical shifts in textbooks.

So, in answer to your question, it is possible in theory to make chemical shift predictions for nuclei within the shielded regions of ring currents, however in practice, it is very difficult and prone to large errors. Calculations require a very thorough knowledge of distances and angles between the aromatic plane and the nucleus of interest. And a reasonable computer.

Very few chemical shift prediction tables in texts or literature give examples of upfield-shifted aromatics, and none that I'm aware of give enough tabulated data to extend an even ballpark prediction. This makes it very, very difficult to make a prediction unless you have experience with the particular type of molecules you are dealing with. For example, I work with one research group who makes quite good estimations of chemical shifts for the inner ring proton within porphyrin systems, but this has evolved over many years of experience of dealing with this particular class of molecule. They would really be at a loss trying to predict any of the shifts for the 4 compounds you have shown.

Most desktop chemical shift predictors use an additive model based on combined chemical shift contributions, and do not do a full calculation of ring current properties. In fact, many desktop predictors give very poor estimations for these types of protons for this very reason, as they do not distinguish shielded and deshielded regions arising from ring current. For small organic molecules the variables dictating the ring current contribution are difficult to predict, especially for those signals experiencing increased shielding, such as overhanging methyl groups. They require knowledge of distance from the centre of the aromatic ring and angle from the plane. This information is often used qualitatively in spectral analysis to construct a 3-dimensional structure of a molecule, in explaining upfielded shifts of certain nuclei.

In proteins, NMR shift prediction models are more inclusive, as the molecule conformations are often very highly constrained and well predicted. Many shift predictors for protein NMR includes a ring current contribution.

References

Johnson, C. E.; Bovey, F. A. Calculation of Nuclear Magnetic Resonance Spectra of Aromatic Hydrocarbons. J. Chem. Phys. 1958, 29 (5), 1012–1014. DOI: 10.1063/1.1744645.
Haigh, C. W.; Mallion, R. B. New tables of ‘ring current’ shielding in proton magnetic resonance. Org. Magn. Reson. 1972, 4 (2), 203–228. DOI: 10.1002/mrc.1270040203.
Haigh, C.; Mallion, R. Ring current theories in nuclear magnetic resonance. Prog. Nucl. Magn. Reson. Spectrosc. 1979, 13 (4), 303–344. DOI: 10.1016/0079-6565(79)80010-2.


Are there any general rules for choosing solvents for recrystallization?

Generally speaking, the best solvent will be dependent on the impurity that you are trying to remove.

The solvent must dissolve both the desired compound and the impurity at a high temperature, but only the desired compound at lower temperatures.  The solubility product of the impurity, as well as the common ion effect should both be taken into consideration.


Why do only chiral molecules rotate the plane of polarized light and how do they rotate it?

See also: How do Optically Active Compounds Rotate Plane Polarized Light?

This is because optical rotation is a chiral phenomenon.

Take a molecule, and draw arrows depicting the polarization of incoming and outgoing light on it. Even if the molecule is achiral, the molecule with the arrows is chiral.

Chirality can't spring out of nowhere, it can only arise from a chiral process.

Also, see my answer here. Since molecules will exist in all rotations in a given fluid, for an achiral molecule the mirror image molecules cancel each other out.


How does a diatomic molecule vibrate in response to heat?

The diatomic can vibrate only by extending and contracting the bond. The more energy that is added the bigger the bond excursion. (More vibrational levels are excited and have bigger excursions). The bond only extends by a few % of its length in lower vibrational levels so its pretty hard to see. If heaps & heaps of energy are added (temp say $1000$ K) then in a real molecule the bond can break as it is actually anharmonic in its motion. When this happens the atoms fly apart.

Additionally the molecule can also rotate while it is vibrating. Rotation is a slower motion than vibration so many vibrational periods occur within a $360$ degree rotation.

At any given constant temperature in a sample of molecules the energy moves between vibrations, rotations and translations so that each type of energy is in equilibrium.

Some of the energy can  also be transferred as  radiation, in the infra-red (for vibrations hundreds to a few thousand $\pu{cm^{-1}}$) and microwave region ( in the range few $\pu{cm^{-1}}$) for rotational transitions but only if the diatomic molecule has a permanent dipole, i.e. is not homonuclear. If these molecules are in equilibrium with surrounding then these surroundings must also emit/absorb radiation to keep the total energy constant. If you were to suddenly isolate your block of (heteronuclear diatomic) molecules, say in space, then the radiation emitted would over a few hours remove energy so that the molecules will end up in their zero point levels.

The figure shows the potential energy profile of HI. The the average length for the for the lowest level (quantum number $0$) is approx $106.4 $ pm and the chnage due to vibration is about $\pm 10 $ pm and clearly greater for higher vibrational levels. The vibrational period is $\approx 14 $ fs.