What if kinetic energy of ejected electron = 0

Rather than debating whether an electron with no kinetic energy is ejected consider how an experiment to measure the photoelectron effect might be conducted. A photosensitive electron emitter is irradiated and the kinetic energy of ejected electrons is measured. One way to do this experiment is described in MIT Physics Department lab course notes. The setup is described in the following schematic:

A photodiode is irradiated with the filtered output of a mercury lamp. Only one of the available narrow bands of the lamp's spectrum is selected by the filter. The light strikes the K coated surface of the cathode of the photodiode. Emitted electrons are collected at a ring-shaped anode separated from the cathode by a gap, setting up a small current. An adjustable retarding voltage can be applied between the cathode and the anode to suppress the current, and the minimal cutoff voltage V which completely suppresses the current is measured. This voltage is theoretically related to the work function $\phi$ of the material and to the energy $h\nu$ of the incident light as
$$V = \frac{h}{e}\nu - \frac{1}{e} \phi$$
Now note perhaps the most important point with regard to your question: only a limited number of light frequencies are available in the experiment, corresponding to the bands generated by the mercury lamp. In addition the motion of a collection of electrons (a current) is measured, not the velocity of a single emitted electron. While it is possible to perform single particle experiments, at no point in the MIT experiment is the kinetic energy of individual electrons directly measured.
The lab notes also provide a short history of the discovery of the photoelectric effect:

Crude though the early data were, the qualitative fact of  the  dependence  of  the  critical  cutoff  voltage  on  the wavelength of light emerged with sufficient clarity to induce the young Albert Einstein, working as a patent examiner in the Swiss Patent Office in 1905, to link the effect with the recent idea, introduced by Planck in 1900,that matter radiates its energy in quanta of energy hν.

and

It was not until 1912 that the technical problems of making  precision  measurements  of  the  photoelectric  effect were overcome [...]

In other words, 7 years passed between the time that Einstein postulated his theoretical description and a confirmation based on "clean" data.
Note also from the lab notes how inherently "messy" such an experiment can be:

It is not possible to precisely determine the work function for removing an electron, because the cathode surface interacts with the remaining gases in the photocell as a getter, so that the surface characteristics change  a  little  from  the  ideal  case.   It  is  important  to note that the electronic work function φ is a material constant which incorporates the different emission potentials of the cathode and anode.  The former is a difficult quantity to estimate due to the manufacturing process which makes  the  cathode  surface  inhomogeneous.   It  is  composed  of  a  mixture  of  potassium,  potassium  oxide  and oxidized  silver.   For  this  reason,  you  need  to  take  care that the same area is always illuminated.  The emission work function of the photoelectrons can vary locally!


Since green is the absorption of orange light, you need to supply the orange light to measure how much is absorbed. As the alcohols are oxidized to the acids, the dichromate is reduced, so you can measure how much acid has been formed. BTW, what alcohol was used? Butyl alcohol oxidation has a distinct aroma that can be detected without spectroscopy.


Does CH3CHNH exist, and what is it called?

The -CH=NH functional group is called an imine (in this case a primary imine) your molecule would be called Ethan-1-imine


Will cinnamaldehyde undergo self-aldol condensation?

Not every α-hydrogen automatically means that a ketone is enoliseable. It would need to be an α-hydrogen capable of aligning with the π system of the $\ce{C=O}$ bond such that removing the hydrogen automatically generates the required enolate anion system. In the case of cinnamaldehyde — your compound — there is conjugation across the phenyl ring, the double bond and the aldehyde meaning that this compound is entirely planar. Thus, there is no way for the α-hydrogen to align favourably for deprotonation to occur. If it did, it would generate an allene-olate $\ce{Ph-CH=C=CH-O-}$ which is not particularly favourable.

In general, α,β-unsaturated ketones and aldehydes are deprotonated vinylogously, meaning that the double bond shifts away upon deprotonation. For example crotonaldehyde $\ce{H3C-CH=CH-CHO}$ is deprotonated in the following way:

$$\ce{H3C-CH=CH-CHO ->[][- H+] H2C=CH-CH=CH-O-}\tag{1}$$

I.e. the γ-hydrogen is abstracted in place of the α-hydrogen.


Why do we call it an acidic medium instead of acidic solution?

This is just a guess, but redox reactions can take place in solution, on surfaces, in the vapor phase, underground, in pastes found in batteries etc. So to have a single word instead of listing all the different environments, we just say the redox reaction takes place in a "medium" that is acidic, neutral or basic.

There are similar collective terms for other situations in chemistry. If you want to talk about reactants, product, catalysts and spectators at the same time, you can refer to them as "species". If you want to talk about atoms, molecules and ions at the same time, you can refer to them as "particles".


If stirring negatively affects crystal purity, why is it so?

"Strong stirring" is a rather imprecise term.  There is such a thing as too much speed.  Rule of thumb for APIs, if you're working with a magnetic stirrer in a round bottomed flask in a lab (rather than in some industrial vessel or pilot plant setup) 350 rpm is too fast (and no, I can't quote a reference on that, it's one of those things you get told and then confirm by experiece - usually by being a wise-arse and bucking the "established wisdom" and ending up with a product that doesn't make medical grade).
Agitation for different crystallisation processes are generally kept in certain narrow bands to avoid inclusions as wel as a number of other reasons.
An example is described in Chow et.al [1] where when crystallizing paracetamol from water in a 500 mL round-bottom flask with p-acetoxyacetanilide present, increasing the stirring speed from 200 to 320 rpm increased product purity, but increasing further to 400 rpm lead to decline in product purity.
According to Meedara et.al. [2] stirrer-particle and interparticle collisions can facilitate crystal attrition  in suspensions with high energy agitation, leading to attrition-induced inclusions.
Urwin et.al. [3] did a reasonably thorough job testing various methods and reasons for presence of impurities in crystals, and is worth a read.
Another good reference is The Handbook of Continuous Crystallization, start with chapter 1 [4]
That said, in crystallisation especially of APIs, mechanisms can be compex and are still not completely understood. All kinds of workarounds have been developed, such as e.g. ultrasonics to help prevent agglomeration when crystallizing paracetamol from certain solutions.
If you have no literature on the process you are attempting, finding the "sweet spot" where you get the greatest purity of product, is only going to happen with a lot of experimentation.
[1] Chow, A. H. L.; Chow, P. K. K.; Zhongshan, W.; Grant, D. J. W. Modification of Acetaminophen Crystals: Influence of Growth in Aqueous Solutions Containing P-Acetoxyacetanilide on Crystal Properties Int. J. Pharm. 1985, 24 (2–3), 239– 258,  DOI: 10.1016/0378-5173(85)90024-9
[2] Meadhra, R.; Kramer, H. J. M.; Van Rosmalen, G. M. Model for Secondary Nucleation in a Suspension Crystallizer AIChE J. 1996, 42 (4), 973– 982,  DOI: 10.1002/aic.690420410
[3] Urwin, SJ, Levilain, G, Marziano, I, Merritt, JM, Houson, I and Ter Horst, JP
A Structured Approach To Cope with Impurities during Industrial Crystallization Development Organic Process Research & Development 2020 24 (8), 1443-1456,  DOI: 10.1021/acs.oprd.0c00166
[4] McGinty, J, Yazdanpanah, N, Price, C, ter Horst, JH and Sefcik, J  Nucleation and Crystal Growth in Continuous Crystallization , in The Handbook of Continuous Crystallization, 2020, pp. 1-50 DOI: 10.1039/9781788013581-00001


Why do green bell peppers become bitter when cooked?

Thousands of different chemicals, previously spatial separated, are mixed together after thermal death of cells, and kept at high temperature to react well. Final taste will depend on original system, but will develop by way unpredictable theoretically from basic principles.
The answer for a particular biological source requires detailed knowledge about particulars of green bell peppers biochemistry, which is rare.
Generally, there are basically few, possibly combined, options:

Bitter tasting compounds - possibly, but not exclusively alkaloids - are already there, but encapsulated in celular structures and released during cooking.
They are actively created by enzymes released after cell deaths ( like $\ce{\mathrm{alliin} ->[\mathrm{alliinase}] \mathrm{allicin}}$ in garlic)
Thermal decomposition of some substances creates such bitter compounds.

Further reading:
bitter-bell-peppers provides rather the opposite opinion:

Why cooked peppers are less bitter
Have you noticed that raw bell peppers taste significantly more bitter when eaten raw (for crudités or in a salad)? That's because most bitter compounds in food break down with heat. The bitterness doesn't go away completely, but it is lessened.

How to Take the Bitterness Out of Bell Peppers:

Select ripe bell peppers. Green bell peppers are harvested before they fully ripen, which causes bitterness. Choose red bell peppers; they are fully mature and slightly sweeter than orange, yellow or green varieties.
Remove the seeds from the cores. The bell pepper seeds are mostly in the cores, and they are naturally bitter. Slice the top of the bell peppers and place them upright. Use a knife to cut out the cores in a circular motion. Cut open the bell peppers and scrape off the remaining seeds.
Peel the skins. Put your bell peppers in a preheated oven until the skins blister. Let them cool for 10 minutes in a paper bag -- the steam will help loosen the skins. Run cold water over the bell peppers and peel off the skins.
Cook the bell peppers. Add soy sauce, salt or other spices to give the bell peppers flavor. The cooking process will eliminate some of the bitterness, and the seasoning will help mask the taste.

chemistry of bell peppers:


The chemical shift of protons in polar groups can vary a lot between solvents of different polarity, because the average electron density around the proton changes. Hydrogen bonding etc. A nice overview can be gotten from the article NMR Chemical Shifts of Trace Impurities: Common Laboratory Solvents, Organics, and Gases in Deuterated Solvents Relevant to the Organometallic Chemist.

For example water shifts from 4.79ppm in water($D_2O$) down to 0.4ppm in d6-benzene. Or the OH group of ethanol from 0.5 in benzene to 1.3 in chloroform to 4.6(!) in DMSO. The methyl and methene groups otoh just shift by 0.3 ppm max. between solvents.

The signals in your spectra are sharp, because there is either a fast exchange, meaning the peak comes at the average CS of both positions of the proton, or no exchange. Slow exchange (either statistically or because the reaction is slow) leads to broadening.

(Slow = exchange happens on the NMR timescale, a millisecond plus/minus a few decades, depending on your $T_{2}$ etc.)


Are the reaction rates of these equations equal?

For the reaction kinetic rate constants, there is the well known Arrhenius equation:

$$k=A \cdot \exp{\left(-\frac{E_\mathrm{a}}{RT}\right)}$$

$A$ is sometimes called frequency factor, interpreted as the rate of collisions with the proper orientation of molecules.

It has 2 terms:

The rate of general collisions, that is function of temperature and molecular masses, which determine speed of molecular motion (close relation to the kinetic theory of gases). Note that the temperature dependence is much smaller than for the exponential Boltzmann term.
The probability of the proper orientation of molecules, what depends on the molecular geometry. For molecules of  otherwise the same geometry, it depends on covalent atom radii.

The exponential term follows the Boltzmann statistical distribution, determining the probability molecules would have enough kinetic energy to overcome the reaction activation energy barrier.

All 3 terms (2 for $A$ and the exponential term) Are different for different halogen atoms.

The mass molecules increases fluorine < iodine, so collision frequency as the rate constant term is the lowest for iodine.
As the reaction mechanism I suppose SN2. The geometrical aspects of $\ce{-CH2X}$ is tricky to determine from basic principles. Bigger halogen is more sterically blocking, but is also farther from the central carbon. More polar bond of the smaller halogen should cause stronger  repelling of the other 3 bonds, so the other side is more open for sn2 reaction.
Activation energy would decrease ( and the Boltzmann term for the rate constant increase) in order F ... I

That about the principles. To compare the particular rates, it is matter of experimental data.


Separating boric acid from sodium chloride

Here is a potential solution that shouldn't even require heating. Note that I have not tried this myself.

Boric acid is soluble in lower alcohols like ethyl alcohol and isopropyl alcohol.
The solubility of sodium chloride in ethyl alcohol at room temperature is about 0.05%$^1$.

Given these properties, I suggest dissolving the dried boric acid / sodium chloride in a minimal amount of cold 95% denatured ethyl alcohol or 99% isopropyl alcohol (available at most pharmacies).

The boric acid should dissolve readily and the sodium chloride should not, so you can just filter it and let it dry to a fairly sodium chloride free boric acid solid.

To make the sodium chloride even less soluble you could cool the solution in a freezer prior to filtering.  If you notice any significant increase in solids that means boric acid is coming out of solution and you'll want to abort this step.

1) Journal of Chemical and Engineering Data, Vol. 50, No. 1, 2005


Second order Jahn-Teller effect for d0 complexes

After quite some research I finally found the original work on the problem. This comes from:
S. K. Kang, H. Tang, T. A. Albright, J. Am. Chem. Soc. 1993 115 1971 - 1981.

I won't discuss the symmetry parts, simply because I have no clue about symmetry and point groups and transformations. I copied the original description for the necessary relationships for a second order JT-effect to happen in the question, so if you know more about this you can probably confirm that.

So to my understanding what is happening for d0 compleces is the following. You really get a mixing of the lower and filled $\mathrm{t_{1u}}$ Orbital with the higher unoccupied $\mathrm{t_{2g}}$ orbital. Both of them split in a set of one orbital that is unchanged in energy, $\mathrm{a_{2}''}$ (for the original $\mathrm{t_{1u}}$) and an $\mathrm{a_{1}'}$ (for the original $\mathrm{t_{2g}}$), and two degenerate orbitals with $\mathrm{e'}$ symmetry that is stabilized and therefore lowered in energy for the original $\mathrm{t_{1u}}$ and elevated for the original $\mathrm{t_{2g}}$. According to the paper this is done by a distortion of the octahedron by a normal mode of $\mathrm{t_{2u}}$ or $\mathrm{t_{1u}}$ symmetry. The first is said to equal a rotation along a threefold axis in the octahedron to get a trigonal prism and change the $\mathrm{O_{h}}$ symmetry to $\mathrm{D_{3h}}$.

You can also see this in the appendix of the book:
T. A. Albright, J. K. Burdett, M.-H. Whangbo, Orbital interactions in chemistry, second edition, Wiley, 2013.

The next question is, why does this stop at the trigonal prism and not continue in mixing the other states as well? And the answer is, for some hypothetical hydrides this seems to happen. While in a d2 complex the new $\mathrm{a_{1}'}$ is filled and therefore an elevation in energy quite unvavorable a d0 complex should further distort. Then $\mathrm{a_{1}'}$ and $\mathrm{a_{2}''}$ mix and distort along a normal mode of $\mathrm{a_{2}''}$ symmetry that changes the trigonal prism $\mathrm{D_{3h}}$ to something they call pyramidalization with $\mathrm{C_{3v}}$ symmetry. This step can also be achieved directly from the octahedron by mixing all orbitals at once and twisting a triangular face of the octahedron.

Now the final question was, why do we observe so many trigonal prisms then? And their final remark on this (unless I skipped something) that it's about the ligands. What ligands favor a second order Jahn-Teller effect for a d0 metal center? Basically only strong σ-donors. They mustn't have any π-donating effects. Since the HOMO LUMO gap has to be quite narrow the $\mathrm{t_{1u}}$ orbital has to be high. So no electronegative ligands. But also the $\mathrm{t_{2g}}$ orbitals have to be low, so a strong ligand splitting is favored but in a way that the $\mathrm{t_{2g}}$ is lowered. And most probably a strong nephelauxetic effect is advisable as well. This leads to alkyl and hydride as possible candidates. For the hydride many of the desired complexes are just unknown but the calculations show that is could distort to $\mathrm{C_{3v}}$. For the alkyl like many prismatic organometallic alkyl complexes we know the methyl seems to be to bulky to stabilize the pyramidal structure.

In the same sentence they also mention that for example alkoxides are not forming prisms. So this leaves me with the question how Mo-S or W-O structures in solid state still show the trigonal prism. But that has probably different reasons so I'll exclude them here.


To calculate work at constant pressure, you need the following:

$$w=-P \Delta V$$

where $w$ is work, $P$ is pressure and $\Delta V$ is the change in volume.

You will need to solve some gas laws relations to get both $P$ and $\Delta V$ in terms of the single variable $T$.

Algebraically, it will simplify to $P\Delta V = nR\Delta T$, assuming ideality.

If you need to assume a non-ideal gas (which is a good assumption for water at high pressure). If we use the van der Waals equation, things get dicey and calculus is needed. I'm hoping you don't need calculus:

$$d[(P-\frac{n^2a}{V^2})(V-nb)]=d(nRT)$$

$$d(PV-\frac{n^2a}{V}-nbP+\frac{n^3ab}{V^2})=d(nRT)$$

$$PdV+VdP+\frac{n^2a}{V^2}dV-nbdP-\frac{2n^3ab}{V^3}dV=nRdT+RTdn$$

Since $n$ and $P$ are constant $dP=0$ and $dn=0$

$$PdV+\frac{n^2a}{V^2}dV-\frac{2n^3ab}{V^3}dV=nRdT$$

$$\int_{V_i}^{V_f}(P+\frac{n^2a}{V^2}-\frac{2n^3ab}{V^3})dV=\int_{T_i}^{T_f}nRdT$$


Information depth in x-ray photoelectron spectroscopy (XPS)

Do you understand the basics of XPS?  The Wikipedia article is pretty clear. In summary, a (usually monochromatic) x-ray source irradiates a sample.  The x-ray can be absorbed by electrons in the sample, and can provide enough energy for a core electron to be set free with an energy equal to the incident x-ray energy minus its binding energy. Collecting the electrons and measuring their energy gives a plot of counts vs electron energy.  Lower electron energy indicates an electron coming from deeper in the atomic shell. Given knowledge of atomic binding levels, you can associate specific peaks in the spectrum with elements (with chemical effects on binding energies giving additional information).

So - what is the "information depth"? A slightly odd term, I've generally heard/used 'detection depth'.  Two questions to ponder - (1) how far will x-rays of a few keV penetrate into a solid (if an atom doesn't see any x-rays, it won't emit any electrons), and (2) how far can a ~100eV-1000eV electron travel in a solid before being scattered (and thus losing the useful info).  Not surprisingly, (2) dominates for XPS, and is often called the 'escape depth'.

As for the 'take-off angle' - again, the interesting part is where your electron detector is placed relative to the sample, and how far the electrons have to travel through the sample to head out in the direction of the detector - the 'escape depth' will vary with exit angle (take-off angle) since the amount of material the electron has to go through is a simple trig function of the exit angle.

I hope that helps.


The absolute approach:

Note: colloidal copper is small flecks of copper mixed with purified water. Typically colloidal copper is sold in dilute quanitities (like 100 ppm).

Measure and record the density of the colloidal copper solution (in grams per liter). Measure and record the weight of an empty flask. Put a sample of the colloidal copper solution in the flask. Measure and record the total weight of the sample (and flask). Subtract the weight of the flask from the total weight of the sample and record the total weight of the sample (only). Boil off the aqueous portion of the sample. Weigh and record the total weight the container of dried copper. Subtract the weight of the flask from the total weight of the container of dried copper and record the weight the dried copper.

Use the following conversions to determine the ppm of copper: 1 ppm = 1 mg per liter; and sample weight/density = liters of sample.

Example:

sample density = 1000g/L
(total) sample weight = 250g
(dry) copper weight = 25mg

250g/(1000g/L) = 0.25L

25mg/0.25L = 100 ppm.


I have not done it with lithium but many times with sodium. Lithium is the less reactive, so it should not give more problems. We use sodium to dry methanol, so it is a common operation in the lab. The reaction is:

$\ce{2Li + 2CH3OH -> 2LiOCH3 + H2}$

Of course, hydrogen is dangerous, especially in the presence of oxygen and still more in this exothermic reaction. So, be careful.

The solution is stable, but attracts water and carbon dioxide, so it must be stored well closed.

To get dry lithium methoxide, the easiest is just to evaporate the excess of methanol.
You put it in a round bottomed flask, attach it to the rotary evaporator, open the vacuum and distill it of. With sodium it works perfectly well. A white solid remains. Again, this solid will attract water and carbon dioxide, so it must be stored properly.

Finally, if the only purpose is to have a methanolic solution of lithium, you can better use lithium chloride which is very soluble in methanol, and much safer (no reaction, no hydrogen, no heating).


Something crossed my mind - you could try titrating with Fe(III) standard using SCN- as indicator. First excess of Fe3+ would produce a sharp end-point.


What is the major product of the reaction of 2-methylbuta-1,3-diene with HBr?

The regioselectivity of the overall reaction depends on the temperature.  The carbocation intermediate can pass through the two different transition states  that lead to the 1,2- and 1,4- products, respectively. If we choose a temperature low enough, then the product distribution will reflect the difference in energy between the two activation energies Ea (1,2) and Ea (1,4). So long as the reaction is not reversible, the product with the lower energy transition state will dominate which is decided by the stability of the intermediate, so in that case 3-bromo-3-methylbut-1-ene (product-1) will be more stable product.
At higher temperatures, the reaction has the potential to be reversible, this means that both the 1,2- and 1,4- products can ionize This sets up an equilibrium. The product ratio will now reflect the relative stabilities of the 1,2- and 1,4- products, not the transition states leading to their formation. In the case of 2-methylbuta-1,3-diene, the product 1-bromo-3-methylbut-2-ene (product-2) will be more stable and dominant at higher temperatures.


How do I calculate the degree of dissociation in equilibrium?

Let us first define the terms needed here.

Degree of dissociation (DOD)
Degree of dissociation is the fraction of a mole of the reactant that underwent dissociation. It is represented by $\alpha$.
$$\alpha = \frac{\text{amount of substance of the reactant dissociated}}{\text{amount of substance of the reactant present initially}}$$
Number of moles dissociated
It is defined as the product of the initial concentration of the reactant and the degree of dissociation

Now suppose you have a reaction like this
$$\ce{A->B + C}$$

The initial state of A is always the concentration of A (should be given in the question) while initial moles of B and C are zero (if anything else is not specified).
The final state of A is always defined as $\text{(amount of substance initially present) - (amount of substance dissociated)}$ while for B and C it is just $\text{amount of substance of A dissociated)}$.

Writing our equation again,
\begin{array}{lcccc}
 & \ce{A &-> &B &+ &C}\\
\text{Initial amount of substance}& a && 0 && 0\\
\text{Final amount of substance}&   a - a\cdot(\mathrm{DOD}) && a\cdot(\mathrm{DOD}) && a\cdot(\mathrm{DOD})\\
\end{array}


The axial positions see 3 substituents at 90 degrees, while the equatorial positions only see 2.

To first order, it is generally favorable to place the smaller substituents in the axial positions. Unfortunately, when you have fluorine, higher order effects (like minimizing dipole moment) may overcome steric considerations, but I would still prefer to place the fluorines axially as my first guess.


How do you calculate the equilibrium constant with the absorbance of a substance and the absorbance constant?

I'm following the outline from the comment by user21398.

Initial concentrations after mixing

The volume of the mixture is $V_\text{mix} = \pu{10 mL}$. Using the dilution law, I get the following concentrations:

$$\ce{[Fe^3+]_\text{initial}} = \pu{2.00 mM} \cdot  \frac{\pu{5 mL}}{\pu{10 mL}} = \pu{1.00 mM}$$

$$\ce{[SCN-]_\text{initial}} = \pu{2.00 mM} \cdot  \frac{\pu{2 mL}}{\pu{10 mL}} = \pu{0.400 mM}$$

Equilibrium concentrations

The equilibrium concentration of the complex is already calculated, $\ce{[FeSCN^2+]_\text{equil}}=\pu{6.39e−5 M}.$

To get the equilibrium concentrations of the reactants, we have to consider that some reacted:

$$\ce{[Fe^3+]_\text{equil}} = \ce{[Fe^3+]_\text{initial}} - \ce{[FeSCN^2+]_\text{equil}} $$ $$ = \pu{1.00e-3 M} - \pu{6.39e−5 M} = \pu{0.94e-3 M}$$

And similar for thiocyanide:

$$\ce{[SCN-]_\text{equil}} =  \ce{[SCN-]_\text{initial}} - \ce{[FeSCN^2+]_\text{equil}} $$ $$ =\pu{0.400e-3 M} - \pu{6.39e−5 M} = \pu{0.336e-3 M}$$

Equilibrium constant

This is copied from the question, using the values calculated in this answer above and dropping the unit M because the standard state is 1 M:

$$\mathrm{K} = \frac{[\ce{FeSCN^2+}]}{[\ce{Fe^3+}][\ce{SCN^-}]}$$

$$\mathrm{K} = \frac{\pu{6.39e-5}}{\pu{} \pu{0.94e-3}\cdot \pu{0.336e-3}}$$

$$\mathrm{K} = 202$$


The general consensus is that each element is made up of unique atoms of that element's type. I address nucleosynthesis below, but this is not how elements become elements.
For example, oxygen is made of oxygen atoms. Titanium is made of titanium atoms. Neon is made of neon atoms.
The first good atomic theory, formulated by John Dalton and refined by Amadeo Avogadro and others has a the following principles (some of which are invalidated by modern experiments):

All matter consists of atoms as fundamental building blocks (we now know that subatomic particles are the real fundamental building blocks).
Atoms cannot be created or destroyed (we now have the ability to do this by nuclear fusion and fission).
Each element has its own characteristic type of atom (still valid).
All atoms of the same element element are chemically and physically indistinguishable (the discovery of isotopes invalidated this one).
Chemical reactions occur by the rearrangement of atoms (this one is still valid).

More modern atomic theory takes into account the discoveries of the subatomic particles and the development of quantum mechanics.
Atoms are now defined by their composition of the three principle subatomic particles: protons, neutrons, and electrons. Any atom can be characterized by three numbers derived from this composition:

Atomic Number ($Z$) - equal to the number of protons.
Mass Number ($A$) - equal to the sum of the numbers of protons and neutrons ($A=Z+N$).
Charge ($Q$) - equal to the difference between the number of protons and the number of electrons ($Q=Z-E$).

The atomic number, $Z$, defines to which element the atom belongs. For example, all atoms with $Z=1$ are hydrogen atoms. All atoms with $Z=6$ are carbon atoms. The periodic table is arranged by increasing atomic number. The atomic number controls the number of electrons an atom prefers (see below), and therefor controls the majority of the chemistry of an atom.
The number of neutrons (mass number) defines the isotope. For example, there are three isotopes of hydrogen, with $A=1$, $A=2$, and $A=3$: protium, deuterium, and tritium. These three isotopes have 0, 1, and 2 neutrons respectively, but they each have only one proton, so each of them is a hydrogen atom. The mass differences influence the properties of these isotopes subtly, but not in such a way that is usually noticeable to the average human.
The number of electrons (or more precisely the charge) determines whether the atom is neutral or an ion(http://en.wikipedia.org/wiki/Monatomic_ion). The number of electrons controls the behavior of an atom chemically (for example the number of covalent bonds that can be formed), and the number of electrons of a neutral atom is the same as the atomic number $Z$. For example, a carbon atom has 6 electrons (and you can go further into the quantum mechanics of electron configuration to learn that two of those electrons are less accessible than the other four, which accounts for most of carbon's chemical behavior). Carbon can have a number of electrons not equal to $Z$. If carbon were to have 7 electrons, the charge on carbon would be $Q=Z-E=6-7=-1$, making carbon an anion. Carbon behaves differently as an anion than it does as a neutral atom.
Atoms of some elements can be made by nuclear reactions. For example, in the sun hydrogen is converted into helium by the following pathway (and others). The notation below follows the generic example $^A_Z \ce{X}^Q$, where $X$ is the atomic symbol (and here $n$ means a neutron with $A=1$ and $Z=1$ and $Q=0$ and $e$ means an electron with $A=0$ and $Z=0$ and $Q=\pm1$). A proton is represented as $^1_1 \ce{H}^+$.
$$\ce{^1_1H + ^1_1H -> ^2_2He}$$
$$\ce{^2_2He -> ^2_1H + e+}$$
$$\ce{^2_1H + ^1_1H -> ^3_2He}$$
$$\ce{^3_2He + ^3_2He -> ^4_2He + ^1_1H + ^1_1H}$$
In addition to the nuclear transformations, gamma rays, neutrinos, and heat and being generated and flung around like it was nothing.
The number of possible nuclear reactions is humongous. I would go check out a book on nuclear reactions or nuclear chemistry from a library to learn more.


How must elemental symbols be ordered in the formula for a compound?

Yes there is.  The most general rule is that elements are listed in increasing order of electronegativity, and in your case the Pauling ENs are 2.19 (P), 3.04 (N) and 3.16 (Cl), so that's why the application wanted it that way.

Most of the time we approximate that rule by saying elements are listed in increasing order of column of the Periodic Table, so e.g. OF2, CO2, and BF3, and when elements appear in the same column, in decreasing order of period, so e.g. SiC, BrCl and SO3.  The column rule takes precedence over the row rule, because EN varies more strongly left-to-right than up-and-down, generally.

The place where this approximation fails most obviously is with hydrides, since H has an EN close to C but appears at the far left of the Table.  Normally we skip the approximation and mostly use the underlying rule for hydrides, so e.g. BeH2 and GeH4 even though H is above and to the left of Be and Ge  (Pauling EN is 1.57 for Be, 2.01 for Ge, 2.2 for H), but H2O and HF. However CH4 (Pauling En 2.55 for C) doesn't quite follow that rule and NH3 (Pauling EN 3.04 for N) is wildly contradictory.  These may just be historical accidents.


Mechansim of reaction between 1‐bromo‐2‐fluorobenzene and furan in the presence of Li/Hg

This is a question based on two reactions.
The first reaction is the formation of benzyne using $\ce{Li/Hg}$. After this a Diels–Alder reaction takes place between furan and benzyne. The reaction mechanism would be as follows [1]:

A more generalised form of this reaction, dealing with the conditions needed and the yield of reaction would be [2]:

Reference

Shashidher, B.; Bajjuri, R.; Guguloath, V. Formation and Trapping of Benzyne. Pharm Anal Acta 2011, 02 (07). DOI: 10.4172/2153-2435.1000137. (PDF)

Comprehensive Organic Synthesis; Wolfgang Oppolzer; Université de Genève, Switzerland 1991. DOI: https://doi.org/10.1016/B978-0-08-052349-1.00128-1


There are many reactions that do this. The simplest way is to substitute any other reactive metal for magnesium: aluminium, lithium, sodium, etc. All will burn in oxygen to give solid products / oxides.

This can be extended by using any of these metals and other reactive non metals, such as fluorine, chlorine, etc. in Group VII or sulphur, etc. in Group VI. Do not try it with $\ce{Se}$ or $\ce{Te}$ in combination with aluminium, however, as it is near-explosive and forms toxic gas on hydrolysis.

A simple experiment that does this is the reaction of sulphur with iron filings.


Figuring out the concentration of weak polyprotic acids in presence of strong acid

The first step in solving this kind of problems is to write the reactions that take place in your solution, and the equations of the acidity constants :
$$\ce{H2S + H2O <=> HS- + H3O+}$$
$$K_{\mathrm a_1}=\frac{[\ce{HS-}][\ce{H3O+}]}{[\ce{H2S}]}$$
$$\ce{HS- + H2O <=> S^2- + H3O+}$$
$$K_{\mathrm a_2}=\frac{[\ce{S^2-}][\ce{H3O+}]}{[\ce{HS-}]}$$
Since $\ce{HCl}$ is a strong acid ($\mathrm pK_\mathrm a < 0$), the dissociation equilibrium for it will be totally shifted to right:
$$\ce{HCl + H2O -> H3O+ + Cl-}$$
And the dissociation of water, occuring in any solution:
$$\ce{H2O + H2O <=> H3O+ + HO-}$$
$$K_\mathrm w = [\ce{H3O+}][\ce{HO-}]$$
*The concentration of water is not taken into account because it remains approximately constant as it is much bigger than the concentration of the $\ce{H2S}$ dissolved. It is also right to write the equations as you did above, considering the dissociation of the acid, but it's better to use the hydronium ion, $\ce{H3O+}$, as we know $\ce{H+}$ is quite unstable on its own. (In some books $[\ce{H+}]$ is used to ease writing)
As you saw, you cannot calculate the concentrations at equilibrium (written in square brackets, i.e. $[\ce{HS-}]$) considering only the first reaction (unless you consider the shift of the second reaction doesn't affect significantly the first, as Nicolau said, and so we use the analytical concentration of a compound (in this case $\ce{H2S}$ and $\ce{HCl}$) which means the specific concentration of the solute when the solution is prepared ($\pu{0.1 M}$ for both acids, if these are the concentrations in the mixture of acids solution and not the initial concentration of each acid solution). It is usually denoted $c_{\ce{H2S}}$ and $c_{\ce{HCl}}$ respectively.
Looking at the above reactions, we can write the formula for the analytical concentrations :
$$c_{\ce{H2S}} = [\ce{H2S}] + [\ce{HS-}] + [\ce{S^2-}]$$
and
$$c_{\ce{HCl}} = [\ce{Cl-}] $$
As you might notice no $\ce{H3O+}$ concentration appeared in any of these formulas, because it depends on both $\ce{HCl}$ and $\ce{H2S}$ concentrations.
However, this is not enough to solve our problem, and so the electroneutrality principle is used, stating that the concentration of the negative ions must be equal to the concentration of the positive ones, and in our case may be written as:
$$[\ce{HS-}] + 2[\ce{S^2-}] + [\ce{Cl-}] + [\ce{HO-}] = [\ce{H3O+}]$$
*The concentration of $\ce{S^2-}$ has the coefficient $2$ because of the two negative charges, unlike the other ions, which have only one negative or positive charge and don't need any.
Using the equations of the analytical concentrations in the electroneutrality formula, we get this :
$$c_{\ce{H2S}} - [\ce{H2S}] + [\ce{S^2-}] + c_{\ce{HCl}} + [\ce{HO-}] = [\ce{H3O+}]$$
And, using also the acidity constants and the water constant :
$$c_{\ce{H2S}}-\frac{[\ce{H3O+}][\ce{HS-}]}{K_{\mathrm a_1}}+\frac{K_{\mathrm a_2}[\ce{HS-}]}{[\ce{H3O+}]} + c_{\ce{HCl}} + \frac{K_\mathrm w}{[\ce{H3O+}]} = [\ce{H3O+}]$$
The equation has 2 unknown concentrations ($[\ce{H3O+}]$ and $[\ce{HS-}]$) and you may approximate the concentration of $\ce{H3O+}$ to $c_{\ce{HCl}}$ since the acidity constants for $\ce{H2S}$ are very small:
$$[\ce{H3O+}]=c_{\ce{HCl}}$$
So, the concentration of $\ce{HS-}$ can be calculated:
$$[\ce{HS-}]=\frac{c_{\ce{H2S}} + \frac{K_\mathrm w}{c_{\ce{HCl}}}}{\frac{c_{\ce{HCl}}}{K_{\mathrm a_1}} + \frac{K_{\mathrm a_2}}{c_{\ce{HCl}}}}$$
However, since $\frac{K_\mathrm w}{c_{\ce{HCl}}} << c_{\ce{H2S}}$ and $\frac{K_{\mathrm a_2}}{c_{\ce{HCl}}} << \frac{c_{\ce{HCl}}}{K_{\mathrm a_1}}$, the above equation can be approximated to:
$$[\ce{HS-}] = \frac{c_{\ce{H2S}}K_{\mathrm a_1}}{c_{\ce{HCl}}}$$
The approximations lead us to a simple formula, which could be denoted directly from the first equilibrium, but without understanding exactly why. From this the concentration of $\ce{S^{2-}}$ can be calculated easily, just like you did first. Even though some would just ignore the second equilibrium for $\ce{H2S}$, that is NOT the way to solve analytical chemistry problems. First you consider all of the equilibriums and processes occuring in a solution and then (if possible) approximate, and not the reverse.


How to calculate mass of water of crystallisation?

The mass of aluminium sulfate has to be taken along with the water of crystallization.

Now molar masses
$$
\begin{array}{c| c}
\text{compound} & \text{ mass (g)} \\
\hline
\ce{Al2(SO4)} & 342 \\
\ce{6H2O} & 108 \\
\ce{Al2(SO4)3\cdot 6H2O} & 450\\
\end{array}
$$

After that it's simply ratio and proportion find out the moles of hydrated salt then find out the moles of water finally get the mass of water.

$$\frac{13.5}{450} \times 6 \times 18 = 3.42$$


Your question is addressed by Explosive Systems Containing Liquid Oxygen: Liquid Oxygen-Liquid Methane Mixtures. J. Chem. Eng. Data, 1959, vol. 4, pp. 127–131.

If liquid methane and liquid oxgygen at 90K are mixed, they are miscible in all proportions.  Mixtures over a wide range of proportions detonated with use of a blasting cap. 100 grams of mixture punched a 10cm hole in a 3/16 inch steel plate.

For comparision, the TNT control only dented the metal plate!

Further tests on the stoichimetric liquid methane/oxgygen mixture found that it was detonated by a .22 caliber copper bullet at 40 foot range, flame and spark.

So nothing happens if the liquids are simply mixed, but a powerful explosion occurs if a stimulus is applied.

Liquid oxygen and liquid methane have also been used in rocket engines.  See Liquid Methane/Liquid Oxygen Injectors for Potential Future Mars Ascent Engines for example.


The transformation taking place in your question is a name reaction known as a Baeyer-Villiger.

The Baeyer-Villager transforms ketones into esters/lactones (cyclic esters) via an alkyl migration in which the group most able to stabilise a positive charge is observed to migrate.

A general mechanism is presented below (taken from the wonderful name reactions book cited). The key steps are (1) attack of the peroxide into the carbonyl followed by (2) the alkyl migration to form the ester grouping.

Source: Strategic Applications of Name Reactions, 2005.


When does a pinacol undergo dehydration or rearrangement?

When you treat pinacol with sulfuric acid, an E1 elimination will take place ( having a carbocation as intermediate ) and it forms pinacolone ( 3,3-dimethyl-2-butanone ). I am not aware of how concentrated the acid should be. Do you require a photo depicting the mechanism ?


Reductive amination of (protected) 3-Oxopropanenitrile: Will azetidin-2-ol be formed via intramolecular ring closure?

Your first product looks alright, $\ce{LiAlH4}$ will reduce the nitrile group to an amino group.

But I don't think that product 1 will do an intramolecular reaction to form an azetidin-2-ol. The reason is that such a reaction, being a 4-endo-trig reaction, doesn't comply with Baldwin's rules and is thus disfavoured for kinetical reasons. If you look at the orbitals that would have to interact for the 4-ring to form you see why this is so:

The "tether" by which the amino group is attached to the carbonyl group is simply too short for the molecule to get comfortably into a conformation that allows the amino group's lone pair orbital to efficiently overlap with the carbonyl group's $\pi^{*}$ orbital.

In my opinion you will get the following reactions instead (the $\ce{NaBH3CN}$ is a typical reagent for selectively reducing an iminium ion):

But closing 8-rings is often a tricky thing. Since the chain is long the activation entropy is rather high, which makes the ring-closing reaction slow (compared to 3- or 5-rings) and intermolecular reactions with other amines in the solution might compete with the intramolecular reaction I proposed.


The most acidic hydrogen among ethane, ethene, ethyne and allene

The inductive effect will always make the stronger acidic compound, especially with an sp hybridization as it will ultimately stabilize the electrophile best in solution. Though resonance will also contribute to the stability, a structure similar to allene will have more resonance as it will have more bonds focused in one area so close to the electrophilic site.
I think you over resonance, its an easy mistake, but my notes from Orgo I show that the difference in pKa from an sp2 to an sp hybridization are roughly a factor of 2.
TL:DR, Induction beats resonance.


Arsenic is toxic to humans due to the fact that it imitates phosphorous almost identically in its bonding patterns, as both elements are in the same group.

However, why would this have any affect? The most likely reason for this is due to the fact that when the phosphorous is substituted by arsenic in many of a cells vital molecules such as DNA and other phosphorous containing groups, the properties of the new substance is slightly different in terms of its polarity, reactivity, etc. Because the proteins and enzymes within a cell interacts with these phosphate containing groups in such a specific manner, these slight irregularity in chemical nature can have detrimental effects (you can think of the processes of these proteins and enzymes between the phosphate as clockwork, i.e. if the parts are slightly different, it will not work).

I am not as sure about the chemistry of antimony, as I have not researched that topic greatly, but I would guess it being in the same group adds to its toxicity.


Solubility of Organic Compound confusion

There are two regions of bond energy to look at here:

The bonds IN a molecule: a carbon-hydrogen bond has about $\pu{100
    kcal/mol}$ of stabilization. An oxygen-hydrogen bond is a little
stronger, about $\pu{110 kcal/mol}$. Stronger means HIGHER in
stabilization energy; but we draw diagrams where the stronger bonds are
designated by lines that are LOWER on the diagram. We hope no confusion
results!
The bonds BETWEEN molecules of a pure material or a mixture. These
(much weaker) bonds between molecules of a substance can be
estimated by heats of fusion or vaporization. The heat/enthalpy of
fusion of $\ce{H2O}$ is $\pu{1.435 kcal/mol}$, and for $\ce{CH4}$, $\pu{0.223
kcal/mol}$. Hydrogen bonding (not the regular bond to hydrogen, but the polar
bonding between molecules) falls in this region of energy and is
frequently encountered when examining solutions.

The bonding between water molecules is strong enough to resist separation by methane molecules because the gain could be no more than the stabilization in methane ($\pu{0.223 kcal/mol}$), but the loss would be as much as $\pu{1.435 kcal/mol}$. The first sentence in the question refers to this region of bond energies, not to the much higher energy region (~ $\pu{100 kcal/mol}$) that would be extensively covered in a chapter on bond enthalpy.


No. There are sweet, bitter, and various other salts. (Likely, there are tasteless salts too). Pure salty taste is as far as I know exclusive for table salt, though I wouldn't bet on it.

Lead and Beryllium salts are said to be sweet, though toxic. Epsom salt, $\ce{MgSO4}$, is bitter. $\ce{CuSO4}$ has an incomprehensible, persistent metallic taste. (Based on personal experience. Copper salts are slightly toxic, but not extremely, so I survived with no consequences.)

Salts with hydrolysing cation (various alums) are acidic in addition to other notes.


How do I figure out the composition of a material given specific densities?

Your mistake is that you have calculated the mass of a sample with the same volume as ours consisting entirely of sand, and a sample with the same volume as ours consisting entirely of sand.
volume = mass_total / density_total = 0.2 kg / 2300 kg/m^3 = 0.00008696 m^3
gives the initial volume of the sample before the salt is removed, however by multiplying this by the densities of sand and plastic, we calculate the mass of a 0.0000869 m^3 sample of sand and a 0.0000869 m^3 sample of plastic.
First we need to find the new volume and density of the sample after the salt has been removed.  This can be done by calculating the volume of salt removed:
volume_salt=0.03kg_salt/2100 kg/m^3=0.00001429m^3
By subtracting this volume from the initial volume you already calculated:
0.00008696m^3-0.00001429m^3=0.00007267m^3
dividing the new mass by this gives:
0.170kg/0.00007267m^3=2,339kg/m^3
Now we need to use this to calculate the proportion.  We can do this by creating and solving a system of equations.  We know that some mass of sand plus some mass of plastic gives us a 170 gram mixture, and that some volume of sand plus some volume of plastic gives us a mixture with a volume of 0.00007267m^3.
xkg_sand+ykg_plastic=0.170kg
am^3_sand+bm^3_plastic=0.00007267m^3
Now, we know that the volume and mass of each material are related by their densities, and because of this it is possible to replace the terms in one equation with the terms in the other by relating them through density, so if we decide to solve for mass, our volume equation becomes:
(xkg_sand/(2700 kg/m^3))+(ykg_plastic/(1400 kg/m^3))=0.00007267m^3
Because the x and y in these equations are the same, we can solve a system, which yields:
x=0.1418kg
y=0.0282kg
Dividing these by our starting mass gives:
proportion_sand=0.709
proportion_plastic=0.141
These can be rounded to match your answer if needed, however these proportions combined with your calculated salt proportion adds up to 1.


Fluorine is much more reactive than chlorine and would certainly cause more damage to living tissues. You can even check out https://www.youtube.com/watch?v=vtWp45Eewtw for some fun demonstrations of its oxidizing power too!
Likewise, compared to chlorine gas, I'd assume fluorine munitions would be significantly harder to manufacture, store, and handle for any side attempting to wield them in combat. (And, as Waylander mentioned in the comments, especially for "WW1-era technology" too!)
Also, due to its great reactivity, fluorine gas will readily react even with the atmospheric humidity to form hydrogen fluoride: $$\ce{2F2(g) + 2H2O(g) -> 4HF(g) + O2(g)}$$ which, while still toxic, is much lighter than chlorine gas and so may not be as good at staying "close to the ground" and/or "flowing down" into an enemy's trenches as intended. Or even worse -- potentially thus dispersing even back to your own troops and damaging them as well!


I used F'x answer as a starting point, and I found the mineral. It's called Ulexite.

  The fibers of ulexite act as optical fibers, transmitting light along their lengths by internal reflection. When a piece of ulexite is cut with flat polished faces perpendicular to the orientation of the fibers, a good-quality specimen will display an image of whatever surface is adjacent to its other side (as shown in the photograph).


What is the role of sulfuric acid and mercury(II) sulfate in the oxymercuration of alkynes?

Let's first look at the acid-catalyzed hydration of an alkyne. Just like in the analogous reaction of alkenes, the new proton goes to the least substituted position, to give the most substituted carbocation. Water attacks the carbocation, and after deprotonation, an enol is produced. Under acidic conditions, the enol tautomerizes the the corresponding ketone.

The mechanism with mercury catalysis is very similar. The main difference is that the initial intermediate is a mercurinium ion, which is opened by the nucleophilic attack of water at the more substituted site. Beyond that, you can think of the mercury atom as a giant proton. The mechanism proceeds analogously, first through a mercury substituted enol, which is protonated and then demercurated.

So why is mercury typically used? We know from oxymercuration, that using mercury avoids carbocations that are prone to rearrangement. The sp2 (vinyl) carbocation shown in the first mechanism is less stable than sp3 carbocations, so we would expect rearrangement to be an issue. Avoiding the high energy carbocation also increases the rate of reaction.


How do people check the type of defect in a crystal?

As you said there are many different kind of defects in crystals. There are basically two different methods to find the Crystallographic defect.

Mathematical classification methods
Computer simulation methods

Mathematical classification methods:

A successful mathematical classification method for physical lattice defects, which works not only with the theory of dislocations and other defects in crystals but also, e.g., for disclinations in liquid crystals and for excitations in superfluid $\ce{^{3}He}$, is the topological homotopy theory.

Computer simulation methods:

Density-functional theory, classical molecular dynamics and kinetic Monte Carlo simulations are widely used to study the properties of defects in solids with computer simulations. Simulating jamming of hard spheres of different sizes and/or in containers with non-commeasurable sizes using the Lubachevsky-Stillinger algorithm can be an effective technique for demonstrating some types of crystallographic defects.

Source: Wikipedia


Is it possible to express Gibbs free energy as following?

Yes, but only if the temperature T in the initial and final equilibrium states of the system is the same and you are able to identify a reversible path between the initial and final equilibrium states over which all the heat transfer takes place at the temperature T.


If the ferric ion is more stable than the ferrous ion, then why is it readily reduced to the latter?

As already pointed out, the notion of stability needs to be defined much more accurately.

The fact that the standard reduction potential is positive has no direct bearing on the gas-phase stabilities of the two ions. The only thing that the standard reduction potential actually reflects, is that the standard Gibbs free energy of the following reaction

$$\ce{Fe^3+ (aq) + 1/2H2 (g) -> Fe^2+ (aq) + H+ (aq)}$$

is negative.

So, for example, Curt already hinted in the comments that under an atmosphere of oxygen, $\ce{Fe^3+}$ would be more stable. This is simply because the reaction

$$\ce{4Fe^2+(aq) + O2 (g) + 4H+ (aq) -> 4Fe^3+ (aq) + H2O (l)}$$

(or some variation of it) has a negative Gibbs free energy. That is very easy to prove, since we know that $E^\circ(\ce{O2}/\ce{H2O}) = +1.23~\mathrm{V}$ and $E^\circ(\ce{Fe^3+}/\ce{Fe^2+}) = +0.77~\mathrm{V}$.

In addition if you were to carry out the reaction in base, the stability of $\ce{Fe(III)}$ relative to $\ce{Fe(II)}$ will be increased greatly, since any $\ce{Fe^3+}$ will be precipitated as $\ce{Fe(OH)3}$.

Remember that a standard reduction potential is measured against an arbitrarily determined zero. If, instead of the standard hydrogen electrode, we chose to use the $\ce{O2/H2O}$ reaction as the standard, then $E^\circ(\ce{Fe^3+}/\ce{Fe^2+})$ would have a negative value.

Let's also get this myth out of the way. $\ce{Fe^2+}$ does not, under any circumstances, have a $\mathrm{(3d)^5(4s)^1}$ configuration. Period.

Ok, with all of that settled, let's answer the question you were actually trying to ask. Yes, $\ce{Fe^3+}$ is a decently good oxidant. There are plenty of half-reactions that have $E^\circ$ below the $\pu{0.77 V}$ of $\ce{Fe^3+/Fe^2+}$. If you want to use $\ce{H+/H2O}$ as a reference point (which is perfectly fine, but you have to make it very clear that this is a relative stability, not an absolute stability), then yes, $\ce{Fe^3+}$ is a better oxidant than $\ce{H+}$.

The reason behind that has nothing to do with $\mathrm{d}^5$ stability. It is simply because $\ce{Fe^3+}$ has a pretty high charge and would quite like having another electron so that it doesn't have quite so high a charge.

More formally, you could construct a cycle for the half-reaction $\ce{Fe^3+ + e- -> Fe^2+}$, neglecting entropy for the time being.

$$\require{AMScd}
\begin{CD}
\ce{Fe^3+ (aq) + e-} @>\large\Delta H^\circ>> \ce{Fe^2+ (aq)} \\
@V\large-\Delta_\mathrm{hyd}H(\ce{Fe^3+})VV & @AA\large\Delta_\mathrm{hyd}H(\ce{Fe^2+})A \\
\ce{Fe^3+ (g) + e-} @>>\large -I_3(\ce{Fe})> \ce{Fe^2+ (g)} \\
\end{CD}
$$

The quantity $\Delta_\mathrm{hyd}H(\ce{Fe^2+}) - \Delta_\mathrm{hyd}H(\ce{Fe^3+})$ is positive since $\Delta_\mathrm{hyd}H(\ce{Fe^3+})$ is more negative than $\Delta_\mathrm{hyd}H(\ce{Fe^2+})$. So, the hydration enthalpy terms disfavour the reduction of $\ce{Fe^3+}$ to $\ce{Fe^2+}$.

However, the third ionisation energy of Fe is pretty large, and it turns out that in the 3d block, the third ionisation energy is the major factor that controls the reduction potential of $\ce{M^3+}$ to $\ce{M^2+}$. In particular, it is quite large because Fe is the 6th element in the 3d-block; you should know that $Z_\mathrm{eff}$, as well as ionisation energies, steadily increase across the 3d block due to the poor shielding provided by 3d electrons.

Put simply, the large value of $I_3$ makes it difficult to oxidise $\ce{Fe^2+}$ to $\ce{Fe^3+}$.

The $\mathrm{d^5}$ configuration of $\ce{Fe^3+}$ does have an impact. It can explain why $E^\circ(\ce{Fe^3+}/\ce{Fe^2+})$ is relatively small compared to its neighbours: $E^\circ(\ce{Mn^3+}/\ce{Mn^2+})$ and $E^\circ(\ce{Co^3+}/\ce{Co^2+})$ are both much larger than $\pu{0.77 V}$. However, this "half-filled shell stabilisation" (more properly phrased in terms of exchange energy) is not quite a large enough factor to make $E^\circ(\ce{Fe^3+}/\ce{Fe^2+})$ negative.

The question that I linked above has slightly more information.


How to calculate the energy to dissociate a bond into neutral atoms?

Your $\pu{640 kJ mol^-1}$ value is correct, so the next step is to neutralize the ions. First add an electron to a potassium ion to get a potassium atom. This releases $\pu{418 kJ mol^-1}$ because it is the reverse of the potassium atom ionization. Then take away an electron from fluoride ion to get a fluorine atom. This requires an input of $\pu{328 kJ mol^-1}$ because it is the reverse of the fluorine electron affinity. So the result is $\pu{550 kJ mol^-1}$, i.e., $\pu{640 kJ mol^-1}$ - $\pu{418 kJ mol^-1}$ + $\pu{328 kJ mol^-1}$.

This figure makes it clear:

Figure copyright information: D.W. Oxtoby, H.P. Gillis, L.J. Butler, Principles of Modern Chemistry, 8th Ed., Cengage Learning, Section 3.8, p. 87, © 2016 Cengage Learning.


What is the meaning of "charge separation" in resonance?

There are two ways to think of charge separation and both make resonance structures less stable:

Moving electrons from a bond to a single atom to generate a cation/anion pair. For example:
$$\ce{R2C=CH-CH=CR2 <-> R2C=CH-\overset{+}{C}H-\overset{-}{C}R2}$$

Moving double bonds to separate opposite charges. For example:
$$\ce{R2C=CH-\overset{+}{C}H-\overset{-}{C}R2 <-> R2\overset{+}{C}-CH=CH-\overset{-}{C}R2}$$

Usually, however, the first way is meant. The general idea behind required a minimal charge separation is that in general (as long as we do not form super-octet structures) double bonds are lower in energy than separate charges on adjacent atoms as you correctly mention. Therefore, if we minimise the number of formal charges the idea is that we are concurrently maximising the number of double bonds, thereby generating a better resonance structure which contributes more to the actual composition.
On the other hand, even in the second case it is better to keep the charges together (in the absence of other effects such as strongly electronegative or electropositive atoms). This is because the charges attract each other and they do this better if they are close.


You seem to be mixing up thermodynamic and kinetic considerations. All reactions occur spontaneously all the time, and their backward reactions too. If left alone for long enough, a system will eventually reach the point where the forward and backward reactions go at the same rate (and it's not like they both stop altogether!) - that's an equilibrium. Where that equilibrium will be is judged by $\Delta G$; if it is negative, then the equilibrium will be closer to the products.
As for ammonia, that's where the kinetics kicks in. At room temperature $\Delta G$ is in our favor, but the equilibrium would take forever to reach. We raise the temperature so as to make the reaction faster, but that also shifts the equilibrium in the wrong direction. So we raise the pressure, too. This does the trick: now the reaction goes the way we need, and at reasonable rate.


Organic solvent for cleaning autosampler syringe

Methanol is less expensive. Don’t know how well it would work in your particular situation. We do semi volatiles (EPA 624) .  Methanol works fine for the GC autosampler syringe.


Why is a buffer solution best when pH = pKa i.e. when A-/HA=1

Preliminaries

There are many ways in which you can go about defining a "Buffer Capacity". A good intuitive way to start would be to say, it is tha maximum amount of acid (or base) the solution can tolerate before it starts showing a significant change in pH. While this is not bad, it obviously begs the question what is deemed "significant"?

However, let us start at the very beginning by examining (in very basic terms) how a buffer works.

Say, you have some amount of $\ce{A^-}$ in the solution and some amount of $\ce{HA}$. You add a strong acid to the solution, and this leads to an influx of $\ce{H^+}$ ions; the buffer responds by forming more $\ce{HA}$ through the combination of $\ce{A^-}$ and $\ce{H^+}$. Similarly, you add a strong base to the solution, and you have a lot of $\ce{OH^-}$ floating around, and the soultion responds appropriately, by favouring the disassociation of $\ce{HA}$, the $\ce{H^+}$ so released combines with the $\ce{OH^-}$ and neutralises it.

The equilibrium we are interested in is $$\ce{HA <=> A^- + H^+}$$

an increase in acidity causes a shift to the left, and an increase in basicity causes a shift to the right.

So, in this way a buffer acts a proton sink, absorbing/releasing protons as appropriate to maintain the pH of the solution.

The buffer is said to be broken when we add to much acid/base, and this consumes all of the $\ce{A-}$ / $\ce{HA}$ respectively.

A solution with more weak base, $\ce{A-}$, has a higher buffer capacity for addition of strong acid. A solution with more weak acid, $\ce{HA}$, has a higher buffer capacity for addition of strong base.So although the pH of a buffer is determined by only the ratio conjugate base and acid, the ability of the buffer to absorb strong acid or base is determined by the individual concentrations of the conjugate base and acid.

So it might seem intuitive to have the optimum at equal concentrations of conjugate base and acid.

Intuitive definitions

$$\ce{HA + OH^- <=> A^- + H2O }$$
$$\ce{A^- + H3O^+ <=> HA + H2O}$$
$$pH + pOH = 14$$

The following discussion is largely inspired by this paper (free to read)

Namely, if you you have $$\mathrm{pH} = \mathrm{pK_a} + \log \left( \frac{A^-}{HA}\right)$$

Since, $\mathrm{pK_a}$ is a fixed number, an changes that can take place to the pH value are caused by the $\log \left( \frac{A^-}{HA}\right)$ term.

so, we can say $$ \mathrm{pH} = \mathrm{pK_a} + \delta \\ \text{where}\ \delta = \log \left( \frac{A^-}{HA}\right)$$

make $\delta = 0$ and you are done. You can call that your ideal case, an ideal buffer.

Now, you can consider a buffer's resistance to base and acid separately, and you can define an acid and base buffer capacity ($\beta_a$ and $\beta_b$), for very simple cases (read caveats).

The stoichiometric ratio of of conjugate base and acid is 1:1. $\Delta$ denotes an arbitrary change in $\delta$ after $\alpha$ mol of $\ce{HA}$ has reacted (after addition of a strong base)

$$\delta + \Delta = \log \left( \frac{[A^-] + \alpha}{[HA] - \alpha}\right)$$

One then notes that bases attempt to increase the pH of a solution so $\Delta = 1$, and also let us denote $\beta_b$ as the buffer base capacity. Also, note that $\alpha = \beta_b$. We are interested in the maximum amount of base a 1:1 buffer solution can tolerate before the pH increases by one unit. We are assuming a 1:1 stoichiometry between the the weak acid in our buffer and the strong base that is being added to it.

$$10^{\delta + 1 } = \left( \frac{[A^-] + \beta_b}{[HA] - \beta_b}\right) $$

We solve for beta and substitute for $\delta$ to ultimately get,

$$ \beta_b = \frac{9[\ce{HA}][\ce{A^-}]}{10[\ce{HA}] + [\ce{A^-}]}$$

Now, let us consider that we are adding a strong monoprotic acid. The approach is similar to the one described above. A few key differences is one lets $\Delta = -1$ and let $\alpha = -\beta_a$

I am omitting the algebra involved. You can attempt as an exercise it if you'd like, and ask for clarifications in the comments if you run into trouble. We obtain a similar relation

$$ \beta_a = \frac{9[\ce{HA}][\ce{A^-}]}{10[\ce{HA}] + [\ce{A^-}]}$$

$\beta_a = \beta_b$ if and only if $\delta = 0$ Otherwise, the buffer either displays stronger resistance towards acids or bases.

Caveats

This is a good exercise to develop a good basic understanding, however doesn't have a lot of practical uses. Here, only a simple buffer was considered, i.e one without multiple weak bases and acids. Additionally, polyprotic acids haven't been accounted for.

Formal defintion of buffer capacity

The formal definition of buffer is a little bit intimidating, but I will throw it in here anyway. It obviously makes little sense to have two different kinds of capacities, one for acids and one for bases in everyday life. So we need something more general, which still incorporates the intuition we have developed so far (although it might not seem obvious)

Let $n$ be number of equivalents of added strong base (per 1 L of the solution). Note that addition of $dn$ moles of acid will change pH by exactly the same value but in opposite direction. I will try to derive a formula connecting buffer capacity with pH, pKa and buffer concentration- all numbers we can obtain easily.

For simplicity, I assume that the strong base is monoprotic, and we have volume of 1 (which will allows me to use concentration and number of moles interchangeably)

$$\beta = \frac{\mathrm{d}n}{\mathrm{d}pH} \tag{1}$$

Charge balance of the solution is given by the equation
$$[\ce{A^-}]+[\ce{OH^-}] = [\ce{B^+}] + [\ce{H^+}] \tag{2}$$

$[\ce{B+}]$ is nothing but concentration of the strong base present, or the number of moles present since we assume a volume of 1 L(:D ), so it is simply $n$ in the solution.

$$C_{b} = [\ce{HA}]+[\ce{A-}] \tag{3}$$

and $$[\ce{HA}] = \frac{[\ce{A-}][\ce{H+}]}{K_a} \tag{4}$$

from 3 and 4,

$$ [\ce{A-}] = \frac{C_bK_a}{K_a+ [\ce{H+}]} \tag{5} $$

Using the definition of $K_w$, equation 2 and 5

$$n = \frac{K_w}{[\ce{H+}]} - [\ce{H+}]+ \frac{C_bK_a}{K_a+ [\ce{H+}]} \tag{6}$$

$$\beta = \frac{\mathrm{d}n}{\mathrm{d}pH} = \frac{\mathrm{d}n \mathrm{d}[\ce{H+}]}{\mathrm{d}[\ce{H+}]\mathrm{dpH}} \tag{7}$$

Now, you can evaluate the derivative given above, and after some alebraic manipulations you get
$$\beta = (2.303) \left( \frac{K_w}{[\ce{H+}]} + [\ce{H+}]+ \frac{C_bK_a}{(K_a+ [\ce{H+}])^2} \right )$$

Generalising this for solutions containing several buffers

$$\beta = (2.303) \left( \frac{K_w}{[\ce{H+}]} + [\ce{H+}]+ \sum_{}^{} \frac{C_bK_a}{(K_a+ [\ce{H+}])^2} \right )$$

I'd like to point out that the first two terms in the buffer capacity formula are not dependent on the buffer presence in the solution. Why? They just tell us solutions of high (or low) pH are resistant to pH changes. Simple.

Above plot shows how the buffer capacity changes for the 0.1M solution of acetic buffer. As expected buffer exhibits the highest resistance to acid and base addition for the equimolar solution (when pH=pKa). From the plot it is also obvious that buffer capacity has reasonably high values only for pH close to pKa value. The further from the optimal value, the lower buffer capacity of the solution. Solution containing only conjugate base (pH 8-10) has buffer capacity of zero, for the higher pH presence of the strong base starts to play an important role. In the case of pure acetic acid solution (pH below 3) pH is already low enough to be resistant to changes due to the high concentration of $\ce{H+}$.


Which complexes are stable: ammine complexes or carbonyl complexes?

That quotation from Wikipedia is bad and should be reworded (I can’t think of good ways to edit it, though). Ligands in a spectrochemical series are not ordered by bonding strength or anything, they are ordered solely by the field split they induce.

Both $\ce{CO}$ and $\ce{NH3}$ are great at forming complexes — as is iodide. The main difference is that they won’t form similarly strong complexes with the same cores. Just to give a few examples:

Carbon monoxide is neutral and a π-acid: It has an empty orbital with π symmetry which can take part in metal-to-ligand back-bonding. As such, it stabilises low oxidation states of metals. The neutral carbonyl complexes such as $\ce{[Fe(CO)5]}$ are well known, but also complexes such as $\ce{[Fe(CO)4]^2-}$ (yes, that’s iron($\mathrm{-II}$)!) are stable.
Ammonia is also neutral and a very weak π-base. However, it is a much stronger Brønsted base than carbonyl is. Therefore, it is much better at stabilising medium to high oxidation states such as in $\ce{[Cu(NH3)4(H2O)2]^2+}$.
The halides, finally, are anionic and good π bases and thus able to stabilise high oxidation states. They are classed as weak-field ligands because they generally generate a very low field split. An example of a halide complex would be the bright yellow $\ce{[FeCl6]^3-}$.

You cannot take ligand $\ce{X}$ and call it strong or weak and thus expect it to make stable or unstable complexes. You cannot even take the same ligand and expect it to make inert or labile complexes all the time. One ligand-metal combination may be stable, another may be very unstable — $\ce{[Fe(CO)6]^2+}$ can only be synthesised under a $\ce{CO}$ atmosphere in $\ce{SbF5}$ from $\ce{FeF2}$ where the antimony pentafluoride abstracts any free fluoride ions from the solution leaving a ‘naked’ iron ion which will grab the only ligands it can reach. Iron pentacarbonyl is stable. The same goes for ammonia: Some will readily accept it as a ligand, others won’t. There is no black and no white, only shades of grey.

(Although since coordination compounds are generally colourful let me rephrase that: There is no black and no white, only a wheel of colours.)


How is phosphoryl chloride attaching an aldehyde to benzene?

It is a Vilsmeier–Haack reaction.

You are actually neglecting the effect of DMF (Dimethylformamide).

The Mechanism:

1. Formation of Iminium Cation:

The reaction of the DMC i.e. Dimethyl formamide (which is basically an amide) with $\ce{POCl3}$ produces an iminium cation.

2. Aromatic Electrophilic Substitution Reaction:

The formed iminium cation now act as an electrophile and does electrophilic aromatic substitution as a result of which we get an aryl ketone or an aryl aldehyde. So that extra aldehyde group which you are speaking about came from the electrophilic substitution reaction of your initial structure with iminium ion.

Here I have drawn the whole mechanism for you.


Let's consider an aqueous solution, the concentration of which is $\pu{1 M}$ in $\ce{NH3}$ and $\pu{1 M}$ in $\ce{NaOH}$. Thus, following equilibrium would be taken place:
$$\ce{NH3 (aq) + H2O <=> NH4+ (aq) + OH- (aq)}\tag1$$
$$\ce{NaOH (aq) -> Na+ (aq) + OH- (aq)}\tag2$$
The $\mathrm{p}K_\mathrm{b}$ of equilibrium $(1)$ is 4.75, thus $K_\mathrm{b} = \pu{1.78E{-5}}$. In pure ammonia solution, from equation $(1)$:
$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}\tag3$$
If ionized amount at equilibrium is $\alpha$, then
$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{\alpha \times \alpha}{1-\alpha} = \alpha^2 \ \Rightarrow \ \therefore \ \alpha = \sqrt{\pu{1.78E{-5}}} = \pu{4.22E{-3}}$$
Assumptions: $\alpha \lt\lt 1$, and thus $1-\alpha \approx 1$, and $\alpha  \gt \gt \pu{1.00E{-7}}$ and autoionization of water can be ignored. At the end, since $[\ce{NH4+}] = [\ce{OH-}] = \alpha = \pu{4.22E{-3}}$, both of these assoumtions are correct.
Now consider, if you have $\ce{NaOH}$ in your solution. Since it is a strong base, it completely dissociate according to equation $(2)$. Thus, there is a common ion in this solution: $[\ce{OH-}] = \pu{1 M}$. Hence, from the equation $(3)$:
$$K_\mathrm{b} = \pu{1.78E{-5}} = \frac{\beta \times (1+\beta)}{1-\beta} = \beta \ \Rightarrow \ \therefore \ \beta = \pu{1.78E{-5}} $$
Hence ($\alpha \gt \beta$), the ionization amount of $\ce{NH3}$ in presence of the common ion $\ce{OH-}$ is less than that of the solution when no common ions are present.
In similar way, you can prove the ionization amount of $\ce{NH3}$ is larger in the presence of the common ion $\ce{NH4+}$ Using following equilibria:
$$\ce{NH3 (aq) + H2O <=> NH4+ (aq) + OH- (aq)}\tag1$$
$$\ce{NH4Cl (aq) -> NH4+ (aq) + Cl- (aq)}\tag4$$
$$\ce{NH4+ (aq) + H2O <=> H3O+ (aq) + NH3 (aq)}\tag5$$


Welcome to Chemistry SE!

To do this specific problem you need to use the $K_{b}$:

$$\ce{\frac{x^{2}}{.375-x}}=K_{b}$$

Finding the $K_{b}$ and solving for x will give you the concentration of $\ce{OH-}$. Remember to take the $-log$ of that and solve for the $pH$.

Also next time please show some work and take your time to format these questions so people can try to reply back to you.


How to separate two diastereomeric amines?

When you mentioned column chromatography I had to assume you meant flash chromatography or biotage, which by implication is normal phase. When people mention HPLC they are normally defaulting to reverse phase. Now, if you can get reverse phase TLC plates (C18 are available) , you could make some advances.  The issue though is then getting the silica and you also mention large quantities.

Large lab scale to me is say, 50g plus. For this, I would advocate crystallisation,  but to guarantee some differentiation,  I would make salts with organic acids. Nitrobenzenesulfonate salts are renowned for giving well defined crystals.

If you needed something to better differentiate again the two amines, I would try a camphorsulfonate salt (pick an enantiomer).  These tend not to be too expensive.  The salts then have different solubilities and the right solvent becomes a screening exercise. Otherwise, tartaric acid is very cheap and easy to salt break after crystallisation.


Can hemi-acetal rings form under basic conditions?

Using base means that the reaction will stop at the hemiacetal and acetal formation would not take place. Acidifying at the end might encourage acetal formation.


Why are the melting and boiling points of the group 1 metals lower than the group 2 metals in the same period?

In the photo, you can see that for Group $2$ metals, the cation is doubly positively charged, making the attraction stronger.

Credits

The photo is from Figure 12.20, Section 12.5, Principles of General Chemistry (v. 1.0).


Enyzmatic reduction (as would occur in yeast) could easily be a stereoselective process - probably via selective delivery of a hydride from NADH.

As the other respondent notes, there is simply no way that combining achiral reagents (pyruvic acid and NaBH4) can yield anything other than a racemic product.


Why are partial molar Gibbs energy and Gibbs energy per mol equal?

The key is that the Gibbs free energy is an extensive quantity. This means that for a single-component system it can be written as
$$ G\left(T,P,N\right) = N\cdot g\left(T,P\right) $$
where N is the number of particles and g is the Gibbs free energy per particle. In other words, if you double the number of particles and double the volume (at fixed temperature and pressure), then G will also double.

From the above expression you can see that the chemical potential is just the Gibb's free energy per particle:
$$\mu\left(T,P\right) = g\left(T,P\right)$$


Modification to improve antibacterial activity

Essentially the basis of antibacterical activity of sulfonamides is the inhibition of  the enzyme dihydropteroate synthase, an
important enzyme needed for the biosynthesis of folic
acid derivatives and, ultimately, the thymidine required
for DNA.
Now they do this by competing at the active
site with p -aminobenzoic acid (PABA), a normal structural
component of folic acid derivatives.

PABA

So how does the structure of sulfonamide affect the bacterial
activity in general?

The basis of the structural resemblance of sulfonamides to PABA is clear. The
functional group that differs in the two molecules is the
carboxyl of PABA and the sulfonamide moiety of sulfanilamide.
Now there are requirements that are vital for activity:

Para-amino group:

unsubstituted (i.e $\ce{NH2}$).
very important for activity because any modification of it other than to
make prodrugs results in a loss of activity. For example, all
of the N4-acetylated metabolites of sulfonamide are inactive.

Sulfonamide:

$\ce{N}$ must be primary or secondary
$\ce{R2}$ can be substituted

Aromatic ring:

Para substituted

Modifications
The pKa of the carboxyl group of PABA is
approximately 4.9 thus an optimised sulfonamide has a pKa somewhere close to this value.

It was observed that replacement of one of the $\ce{NH2}$ hydrogens by
an electron-withdrawing heteroaromatic ring enhanced the acidity of
the remaining hydrogen and dramatically enhanced potency.
With suitable groups in place, the pKa is reduced to the same range as
that of PABA itself. Not only did this markedly increase the
antibacterial potency of the product, but it also dramatically
increased the water solubility under physiologic conditions.

Structural variation among the clinically useful sulfonamides
is restricted primarily to installation of various
heterocyclic aromatic substituents on the sulfonamide
nitrogen.
An example  is Sulfisoxazole (pKa 5.0) which can also be formulated as a prodrug, by inclusion of N-acetyl groups. Another advantage is the acetyl derivative
is tasteless and, therefore, suitable for oral administration, especially
in liquid preparations.

Is there an alternative for C as well?

Clavulanic acid is a naturally occurring, mechanism based
inhibitor , which causes potent and
progressive inactivation of β-lactamases.
There  are alternatives e.g sulbactam (A synthetic derivative of penicillins undergone the oxidation of the sulfur atom to a sulfone which greatly
enhances the potency) or tazobactam.
References

The 12th Edition of Wilson and Gisvold’s Textbook of Organic Medicinal and Pharmaceutical Chemistry
Foye’s Principles of Medicinal Chemistry 7th ed.


One barrier is the greater tendency of lithium to form partially covalent bonds than heavier alkali metals.  This is a good thing for organic chemists, since this stabilizes organolithium compounds and enhances the solubility of lithium chloride as a chloride ion source (see this answer).  But it's not so hot for forming lithides.  Covalent bonding to more electronegative elements, meaning to practically all other elements in this case, oxidizes the intended lithide ion.  In effect the cryptand in an intended lithide is converted into one of those organolithium compounds.

Linear Christmas, in the comments to the question, hints at a way to beat this tendency.  Two of the references given there, https://doi.org/10.1021/ja056314+ and https://doi.org/10.1002/cphc.201200805, involve theoretical investigations of lithides built with pyrrole-based cryptands.  Such cryptands resist covalent bonding to lithide (or other nucleophilic) anions because this would destroy the aromatic of the ring and require electrons to overlap into a high-lying, strongly antibonding orbital.  These compounds, or counterparts with heavier alkalide ions, draw interest for their expected nonlinear optical (NLO) properties, making them synthetic targets.  So, stay tuned.


Can functional groups exist by themselves

A functional group on its own essentially has at least one non-bonded electron, thus making it a free radical, this makes them highly reactive molecules, making them react with other substances (including themselves) and can thus only exist in very low concentrations in inert medium/vacuum.

However, there are molecules such as $\ce{NO}$, $\ce{NO2}$ which do have an odd electron in their molecular state, can exist in reasonably large quantities, and can be functional groups in organic chemistry, these are in some sense functional groups that exist as molecules.


Increasing Covalent character increases Colour intensity

In molecular orbital theory, as the covalent character increases the homo-lumo gap decreases, and the excitation happens more easily. Thus the colour intensity increases.


What chemical properties that allow for colour exist in the dark?

The chemical property that creates colour is the ability to absorb light of a specific visible wavelength. There is more than one way to do this.

Mostly colour is caused by the existence of electronic transitions in substances that match the energy of some wavelength of light so when light hits the substance, some is absorbed by exciting electrons from a lower to a higher energy level. The higher energy levels will not be populated unless there is some energy input to excite the electrons. But the structure that enables this absorption to exist is present whether or not there is any light.

There are other ways a substance can create colour. For example, some compounds do it by interference: the physical structure contains features that have the same scale as wavelengths of some colour of light. When light hits the substance the reflected light creates interference patterns that can selectively eliminate some wavelengths. This is why oil films on water show rainbow colours and why the mineral opal shows beautiful iridescence. again, though, the cause of colour is present whether there is light or not.

In short, the cause of colour is sometimes a chemical thing (involving electrons being excited) and sometimes a physical thing (involving interference) but, either way, the property that creates it exists even when there is no light.


How to decide between different possible substitution/elimination mechanisms under given conditions?

Generally, your substance can react in one of the following four ways that you have already identified:

$\mathrm{S_N1}$ solvolysis on the tertiary chlorine resulting in 4-chloro-2-methylbutan-2-ol;[1]
$\mathrm{S_N2}$ solvolysis on the primary chlorine resulting in 3-chloro-3-methylbutan-1-ol;
$\mathrm{E1}$ elimination from the tertiary chlorine resulting in 4-chloro-2-methylbut-1-ene or 1-chloro-3-methylbut-2-ene;[1]
$\mathrm{E2}$ elimination from the primary chlorine resulting in 3-chloro-3-methylbut-1-ene.

It is highly unlikely that one of the $\mathrm{X1}$ mechanisms happen on the primary chlorine due to the very badly stabilised intermediate primary carbocation we would need, highly unlikely for an $\mathrm{S_N2}$ reaction to proceed on the tertiary chlorine due to steric hindrance (especially since a primary one is available) and moderately unlikely (but not impossible) for $\mathrm{E2}$ to happen on the tertiary chlorine.

We thus need to consider:

How good is the leaving group (influences everything strongly except for $\mathrm{E2}$)?[2]
How strong is a base possibly present in solution (influences which elimination reaction can happen and how well)?[3]
How good is our nucleophile (influences the probability of $\mathrm{S_N2}$ most)?[4]

Starting at the base. This is a case of rule 6:[6] There is no base present in the mixture! (Yes, technically water can act as a base, but you need a strong acid for it to happen. Water’s $\mathrm{p}K_\mathrm{b}$ is $15.7$ if I recall correctly.) This definitely rules out $\mathrm{E2}$ which requires a strong base, and mildly rules out $\mathrm{E1}$ which requires a weak base.

Let’s check out potential nucleophiles next. Well, one of our nucleophiles is water, also known as very lousy, and another possible one is acetone, also known as even lousier. There is hardly any chance that either of them will nucleophilicly attack the primary chlorine site ruling out $\mathrm{S_N2}$.

Next, what about our leaving group? Well, we have chlorine which is also a pretty lousy leaving group. It means that $\mathrm{S_N2}$ will be discouraged even more and the generated chloride will probably hang around close enough to the carbocation required for the $\mathrm{X1}$ reaction paths to disallow $\mathrm{E1}$.

Finally, there are solvent effects, too. Here, the solvent is polar protic; a combination generally preferring the $\mathrm{X1}$ pathways.[5]

All things considered, this leaves $\mathrm{S_N1}$ as the only mildly probable pathway remaining. And thus I would strongly assume the main product to be 4-chloro-2-methylbutan-2-ol.

Of course, one should always include reaction times and temperatures (and yields) to any reaction-predicting exercise. Raising the temperature should increase the likelyhood of $\mathrm{E1}$ over $\mathrm{S_N1}$ because of the entropic factor (two particles become three in $\mathrm{E1}$ resulting in a rise in entropy).[7] However, in exercises, this is usually indicated by the presence or absence of a Δ above or below the reaction arrow. Since we have no Δ, we can assume room temperature.

[1] C. A. Grob, A. Waldner, Helv. Chim. Acta 1979, 62, 1854.
[2] https://en.wikipedia.org/wiki/SN2_reaction#Leaving_group
[3] https://en.wikipedia.org/wiki/Elimination_reaction#E2_mechanism
[4] https://en.wikipedia.org/wiki/SN2_reaction#Nucleophile
[5] https://en.wikipedia.org/wiki/SN1_reaction#Solvent_effects
[6] Rule 6: There is no rule six!
[7] https://en.wikipedia.org/wiki/Elimination_reaction#E2_and_E1_elimination_final_notes


Why doesn't aromatic substitution add at multiple positions in one reaction?

The rate of Aromatic substitution depends upon the activity of the aromatic system, because when the collision happens the aromatic system has to donate electrons to an electrophile.
In the example above you have used nitro-benzene, which is very strongly deactivated due to the $\ce{-NO2}$ group and thus the ortho and para positions are completely blocked. The meta position of nitro-benzene is not good but also isn't as bad as the para and ortho positions, in terms of electron density. So, we can say that the meta position of nitro-benzene is just a smidgen less electron rich than normal benzene, and thus first bromination takes place. But, after the addition of one bromine the ring becomes even more electron deficient which just makes further bromination even harder and slower. The reaction slows down to such an extent that it practically ceases to proceed.
On the contrary, if you would use a strongly activated aromatic system like phenol or aniline. It would provide such electron density to the para and ortho positions that they would undergo complete bromination, forming a 2,4,6-tribromo product. The electron density provided by $\ce{-OH}$ and $\ce{-NH2}$ is simply so high that it overcomes the electron deficiency created by the added bromines.


Why does the reaction of the primary alcohol cyclopentylmethanol with hydrogen bromide follow SN1 mechanism?

It does not. The proposed mechanism is wrong.

The conversion of cyclopentylmethanol to 1-bromo-1-methylcyclopentane is not reported anywhere in the literature. If you could somehow generate the primary carbocation, then one would expect a rapid 1,2-hydride shift to generate the more stable tertiary carbocation, giving rise to 1-bromo-1-methylcyclopentane.

The ring-expansion product could be rationalized by a concerted 1,2-alkyl shift displacing $\ce{OH2^+}$ without violating first-principles.


What is the purpose of CaO in Oakwood reaction

Sodium hydroxide plus calcium oxide forms soda lime. Its use is mainly down to handling covenience as it is far less hygroscopic than powdered sodium hydroxide.


Green color substance forming during electrolysis

The green material is probably an iron(II) salt, corroded from what I assume is the steel composing the positive pole of the battery.  If it has a gelatinous appearance, and slowly turned orange-ish over time, my guess would be it's $\ce{Fe(OH)2}$, with any orange color deriving from slow oxidation to various iron(III) species such as the oxide-hydroxides.

Per the above link, the white flakes are possibly that same iron(II) hydroxide, prior to reaction with $\ce{O2}$ to form the green color.  Alternatively, it could be $\ce{Mg(OH)2}$, or a related magnesium compound, formed at the cathode (negative terminal) from the hydroxide ion produced by reductive water electrolysis:

$$
2\ce{H2O} + 2e^- \longrightarrow \ce{H2} + 2\ce{OH-}
$$

A photograph would be tremendously helpful in facilitating identification.


How does one separate ethanol from acetic acid using aqueous extraction?

Well, distillation is almost certainly the best method. However, since you've explicitly ruled that out, there are only a few alternatives I can think of:

Ethanol can be separated from water with reasonable efficiency (at least on the microscale) by a process of "salting out." The ethanol/water mixture is saturated with $\ce{NaCl}$. The solvation of the salt ions is a highly thermodynamically favorable process, and as that occurs the molecules of water become increasingly less able and available to form hydrogen bonds with ethanol. I've never tried this with acetic acid specifically, but I can't think of any reason why it wouldn't work.
I would imagine with the right choice of eluent, separation by column chromatography using silica or alumina should be feasible.
Fractional freezing may be possible in theory, although I have no idea as to its efficacy for this specific case.

I'm also assuming here that you're looking strictly for methods of physical separation, not chemical (so, strictly as a general example: adding something like calcium oxide to consume water in the generation of insoluble calcium hydroxide would be out of the question).


Why does Mg react vigorously in NaCl solution and less so in water?

It would still be the same reaction.

\begin{align}
\ce{Mg (s) &-> Mg^2+ + 2 e-} & E^\circ&=\pu{2.36 V}\\
\ce{2 e- + 2H2O (l) &-> H2 (g) + 2OH-} & E^\circ&=\pu{-0.828 V} \times 2
\end{align}

The combined reaction would be spontaneous ($E > 0$), but $\ce{Mg(OH)2 (s)}$ is highly insoluble,
\begin{align}
\ce{Mg(OH)2 (s) &<=> Mg^2+ (aq)  + 2OH- (aq)}, &
K_\mathrm{sp}=7.1 \times 10^{-12},
\end{align}
leaving the solution insufficient ions to conduct the electrons necessary to fast complete the reaction (despite of the escaping hydrogen gas).
$$\ce{Mg (s) + 2H2O (l) -> H2 (g) + Mg(OH)2 (s)}$$

This is like an electrochemical cell with no salt bridge (or bridge with only de-ionized water). Hence a little salt ions would accelerate the reaction forward.

In contrast, calcium solid, having comparable standard reduction potential, will undergo a similar redox reaction in water and needs no help for additional ions because $\ce{Ca(OH)2 (s)}$ is about a million times more soluble ($K_\mathrm{sp}=6.5 \times 10^{-6}$) than magnesium hydroxide, providing sufficient free ions in the solution to drive the vigorous release of hydrogen gas. (There are a couple of YouTube video demos.)


Diazomethylene is singlet therefore carbene is formed in the singlet state.
singlet carbene is highly reactive therefore it goes through reaction probably before intersystem crossing (depends on competitive speeds). this is why singlet state of carbene is usually invisible. Triplet carbenes react like radicals unlike singlet state (besides some aryl carbenes) therefore the products are different. you can look up Skell-Woodworth for reaction of different spin states of carbenes.
based on the products you can find which spin state of the carbene has been involved in the reaction.


How to calculate the number of prochiral carbon atoms in a complex organic molecule?

I consider the concept of prochirality as a property of an achiral compound (or centre), that can be turned into a chiral compound (or centre) in one step (proprochiral: in two steps) by means of substituting or addition. Hence all carbon centres with three different ligands (substituents or hydrogen) can be considered prochiral.

Prochirality is unfortunately a grown concept, which was used in different ways throughout the years and can be confusing at times. I find the definition in the IUPAC gold book quite enlightening and much better than the Wikipedia page that spawned off of it.

  prochirality
  Also contains definition of: proprochirality

  This term is used in different, sometimes contradictory ways; four are listed below.

  The geometric property of an achiral object (or spatial arrangement of points or atoms) which is capable of becoming chiral in a single desymmetrization step. An achiral molecular entity, or a part of it considered on its own, is thus called prochiral if it can be made chiral by the replacement of an existing atom (or achiral group) by a different one. An achiral object which is capable of becoming chiral in two desymmetrization steps is sometimes described as proprochiral. For example the proprochiral $\ce{CH3CO2H}$ becomes prochiral as $\ce{CH2DCO2H}$ and chiral as $\ce{CHDTCO2H}$.
  The term prochirality also applies to an achiral molecule or entity which contains a trigonal system and which can be made chiral by the addition to the trigonal system of a new atom or achiral group. For example addition of hydrogen to one of the enantiotopic faces of the prochiral ketone $\ce{CH3CH2COCH3}$ gives one of the enantiomers of the chiral alcohol $\ce{CH3CH2CHOHCH3}$; the addition of $\ce{CN−}$ to one of the diastereotopic faces of the chiral aldehyde shown below converts it into one of the diastereoisomers of the cyanohydrin. The two faces of the trigonal system may be described as Re and Si.

  The term prochiral also applies to a tetrahedral atom of an achiral or chiral molecule which is bonded to two stereoheterotopic groups. For example, the prochiral molecule $\ce{CH3CH2OH}$ can be converted into the chiral molecule $\ce{CH3CHDOH}$ by the isotopic replacement of one of the two enantiotopic hydrogen atoms of the methylene group. The carbon atom of the methylene group is called prochiral. The prochiral molecule $\ce{HO2CCH2CHOHCH2CO2H}$ can be converted into a chiral product by esterification of one of the two enantiotopic $\ce{–CH2CO2H}$ groups. The carbon atom of the $\ce{CHOH}$ group is called prochiral. The chiral molecule $\ce{CH3CHOHCH2CH3}$ can be converted into one of the diastereoisomers of $\ce{CH3CHOHCHDCH3}$ by the isotopic replacement of one of the two diastereotopic hydrogen atoms of the methylene group. The carbon atom of the methylene group is called prochiral. The stereoheterotopic groups in these cases may be described as pro-R or pro-S. Reference to the two stereoheterotopic groups themselves as prochiral, although common, is strongly discouraged.
  See: chirality centre
  The term prochirality is also applied to the enantiotopic faces of a trigonal system.

The first definition applies especially to tetrahedral carbon atoms.
You can also work off of another definition which might be a bit easier to comprehend.

  prochirality centre
  An atom of a molecule which becomes a chirality centre by replacing one of the two stereoheterotopic ligands attached to it by a different ligand, e.g. C-1 of ethanol; C-3 of butan-2-ol.

    stereoheterotopic
    Either enantiotopic or diastereotopic. (In contrast the term constitutionally heterotopic has been used to describe identical groups located in constitutionally non-equivalent positions).

As a rule of thumb you can always substitute a $\ce{-CH2\bond{-}}$ group with a $\ce{-CHD-}$ group and ask your self if it is a chiral centre now.
I personally find it easier to identify stereoheterotopic ligands as this is also important for spectroscopy (esp. NMR).
This is somewhat also mirrored in the third definition.

The second (and fourth) definition applies to doubly bonded carbon atoms especially. If there are three different ligands, then you can assign a Re- and Si-face of that centre. Hence I would also consider these carbons prochiral.

In your case that means that all carbons (except for the ones that are already chiral) are prochiral.

Marked in pink are all chiral elements (the chirality of the nitrogen depends on the inversion barrier).
Marked in dark blue are all carbon centres who contain diastereotopic protons and hence can be considered prochiral according to the first definition. This is also evident from the $\ce{^1H-NMR}$ estimation of ChemDraw.
Marked in green are all carbon centres where you can add a nucleophile to create a stereo centre and are hence prochiral according to definition two.
Marked in orange are all carbons who are part of an aromatic system. Strictly speaking they are also trigonal carbons and technically you can assign a Re/Si to them. Accordingly they fall within the second definition and can be considered prochiral centres.


How is hyperconjugation into antibonding orbitals stabilizing?

This is nothing specific to antibonding or nonbonding orbitals. There is a very general principle behind it:

  Mix an occupied and an empty orbital with each other, and the occupied orbital will be stabilised.

You’re not actually ‘transferring electron density into an antibonding orbital’. You are simply linearcombining two orbitals: one occupied one with a rather low energy and one unoccupied one with a rather high energy. With this linear combination, you receive two new orbitals, where the unoccupied one is further destabilised while the occupied one is further stabilised.

Your electron density is still in the (now more stable) occupied orbital and the antibonding/nonbonding/otherwise empty orbital (now less stable) is still empty. Only their shapes have changed a bit, because we mixed them.


Equation for salt formation for ammonium oxalate?

Where am I wrong? First of all, note that the ammonia is dissolved in water. So there is an ion that's involved in the reaction, not the ammonia itself. And second, the charge of the ion involved requires a different coefficient for ammonia. Notice that the $\ce{H+}$ cations in Hydrogen oxalate are spectator ions.

The formation of Ammonium Oxalate will be a precipitation reaction. (Or some may put it an acid-base one) Answer these in order to reach a final net ionic reaction:

What are the ions that are present in the final product (ammonium oxalate)?

   The ions involved are ammonium ($\ce{NH4+}$) and oxalate ($\ce{C2O4^{2-}}$).

Notice the charges of those ions:

   Ammonium has a 1+ charge (that's written correctly as +, but of course, I wanted to emphasize) and oxalate has a 2-.

The other ions are spectator ions. These two ions actually form the precipitate. But the final formula (where your misunderstanding rises from) is

   $\ce{(NH4)2C2O4}$

You can now write the net ionic equation and get it balanced.


How to analytically quantify a chemical's potential to intercalate DNA?

Our lab uses a thiazole orange exclusion assay to measure how well our polyintercalating peptides* bind DNA. The assay works by mixing thiazole orange with DNA so it becomes fluorescent. Then we measure fluorescence while titrating in peptide and the loss of flourescence indicates binding. This is because thiazole orange is only fluorescent when intercalated into DNA, and as the competing compound displaces it, the free thiazole no longer fluoresces. Note that this assay is not foolproof, polycationic compounds can condense DNA without intercalating, but this conformation change still forces the thiazole out and reduces the fluorescence, so loss of fluorescence implies binding, not necessarily intercalation. This method works for RNA too, but you'll need more thiazole to see signal because there are usually fewer double helices to intercalate into.

Another method is to measure viscosity, because intercalators make the DNA longer and the solution becomes more viscous, but you either need a very sensitive method to measure the change or a very concentrated DNA solution, and the viscosity can probably also be changed by condensing the DNA.

Another possible method MIGHT be fluorescent anisotropy, if your intercalator is fluorescent. It would spin rapidly outside the DNA, but slow down once it binds the double helix. However, conformational changes in the DNA can change its rotation speed and confound the results.

That paper doesn't really mention the assay, it's just to show off the peptides so you understand what I'm talking about a little more. Those peptides also bind by electrostatic interaction and condense DNA, so it's not a pure measurement of intercalation.


Can quantum entanglement affect the chemistry of molecules?

I've not really read anything on it, but a short while ago research on the role of entanglement in excitonic transfer in the photosynthetic light harvesting apparatus caused quite a stir.

Here is an ArXiV preprint of a paper on entanglement in an antenna protein from C. tepidum. It was published in Nature Physics. The authors note the extraordinary result of entanglement being observed in an inherently quite 'noisy' system.