I believe that when you say "a solution of enol acetate" you are refering to a solution of the corresponding enol-acetal of hetanal? You should use full names of the compounds used, if not, it might be confusing.

My opinions

  (a) Do you think that this reaction should be done using a drying tube, under nitrogen or open to the air? Explain your answer.
  It can be done in drying tube, as the reactant have to be prevented from coming into contact with water vapor from the air as the water vapor could potentially react with the reactant.

I don't see a reason you would need any inert conditions for this experiment. I think you will be fine if you perform it open to the air.

  (b) Without doing TLC or any other form of spectroscopy or chromatography, how would you be able to tell if sufficient Bromine had been added?
  (b) By observing the colour of the reactant mixture. Since bromine is brown in colour, Once sufficient bromine has been added, there will then be excess bromine if bromine was continued to be added, hence the reactant will turn brown once enough bromine is added.

Indeed. The end point of the reaction is reached when the calculated amount is absorbed and the bromine is no longer decolorized.

  (c) Will the organic layer be the upper or the lower layer? How can you quickly check this experimentally?
  The organic layer will be at the bottom. Since it is how that the density of the organic layer is more that that of the ether.The density of the organic layer can be measured and compared to that of the ether.

This is wrong. The organic layer IS the one which is will be a mixture formed by chloroform, unreacted enol acetate and ether. The other phase will be the one composed mainly by water and methanol.
Methanol+water solution will probably be less dense than chloroform+ether+enolacetate solution, and therefore, organic layer will be the upper one. I cannot tell for sure, since you have mixtures in high proportions of compounds with quite different densities, but and easy way to tell is:
- Separate both phases using an extraction funnel. Put some tap water on the separated phase you suspect is the aqueous phase. It dissolves? That is the aqueous phase. It does not dissolve and forms a two-phased system? It was the organic phase, extract it again.
You can double check adding water to the other phase you extracted, and see that the opposite effect should be observed.

  (d) Sodium carbonate is added just before the final distillation. Why?
  To remove any excess acid that might be present in the mixture.

Yeah. The acetal must be free of acid; otherwise decomposition takes place during distillation.

  (e) The procedure calls for removal of the organic solvents by “distillation under reduced pressure”. Which one piece of laboratory apparatus would you use for this?
  Rotatory evaporator

Correct. Rotovap is your guy.

  (f) Suggest two serious and different hazards associated with this experiment.
  Bromine is toxic if inhaled, Methanol is a carcinogen

Bromine will definitely be your top priority concern. Even more if you have to prepare the solution from pure bromine. Also, it has a very strong unpleasant smell, which gives it the name it has (from Greek: βρῶμος, brómos, meaning "strong-smelling" or "stench")
It is corrosive very toxic, volatile (fumes a lot, MUST be handled on a fumehood).
Chloroform is also toxic; Heptanal is volatile and flammable.

Methanol, even though you should not drink it wouldn't be of an important concern.

To sum up, some of the reactants are of unpleasant odor or have lachrymatory effects it is best to use a good hood.

And also a reference: Organic Syntheses, Coll. Vol. 3, p.127 (1955); Vol. 29, p.14 (1949).


Why does electron configuration in the periodic table is different when you follow the aufbau principle?

As a general rule, full electron shells are more stable than unfilled shells. In silver, promoting an electron from the 5s to the 4d shell completes the d orbital, which increases the stability of the 4d shell more than it destabilizes the 5s shell.
This is also explained here:

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=21&t=2807


Which atom goes in the middle of a lewis dot structure?

ETA: I'm unclear about your question.  Are you asking why nitrogen isn't the central atom or why sodium isn't?  If it's the former, the explanation is below.  If it is the latter, sodium, then it's because this is an ionic compound, and we build the lewis structures of the cation and anion separately.

First and foremost, carbon (EN=2.5) is less electronegative than nitrogen (EN=3.1), so by your rule it should be in the center.

A more complete explanation of why that is:

Your molecule is sodium cyanide, an ionic compound.  The cyanide group, CN-, carries a negative charge.  If we consider where that charge lies in the ion, it is a bit confusing.  The basic structure of the cyanide is  ${\ce :C:::N:^-}$  .

At first glance we might think that the nitrogen would be the more negative end of the ion, since it is the more electronegative atom.  However, if we consider the formal charges of the atoms in the ion, we see that nitrogen has a formal charge of 0 and carbon a formal charge of -1.  Formal charge is not the same as actual charge, but it does tell us that the carbon will be the negative end of the ion, and that is what we see in its bonding.

Here, the sodium ion, Na+, bonds with the carbon in the cyanide ion.  Even if it weren't a sodium ion, but something like hydrogen cyanide (with covalent bonding) the bond would be formed at the carbon end.  This is because that structure would allow carbon to minimize its electron excess and reach a lower energy state than if the hydrogen stuck to the nitrogen.  Again, formal charges predict the HCN structure over the HNC structure.


Will a Grignard reagent react with an epoxide or episulfide first?

There are quite a few reasons why the episulfide might open before the epoxide:

Sulphur is less electronegative than oxygen. Therefore, by Bent's rule , more atomic s-character will be directed by the ring towards sulphur in the episulfide as compared to the oxygen in the epoxide. An increased s-character will lead to a bigger increase in the $\ce{C-S-C}$ bond angle as compared to the $\ce{C-O-C}$ angle, and thereby increasing the angle strain in the episulfide relative to the epoxide
Sulphur will also tend to be a better leaving group, as basicity decreases down the group(in this case, group 16)
The carbon-sulphur linkage is weaker than the carbon-oxygen one

So, the mechanism should look like this:

Some points to note:

Sulphur compounds also tend to be pretty good nucleophiles due to their softness as per HSAB as elaborated here. Therefore, the $\ce{RS-}$ formed after ring opening will again attack on the epoxide to open the latter.
This will also lead to the formation of an  ethereal linkage in the final product, which will be quite favorable due to the high stability of $\ce{C-O}$ ether bonds


How to interpret a luminescence intensity vs wavelength graph?

This is a partial spectrum of the luminescence from a hydrogen atom, the graph presents only the Lyman series of emissions.

When a hydrogen atom is excited (absorbs energy), the electron can be promoted to the n=2, 3, 4, 5...infinite level.  This excited electron can eventually emit its excess energy and return to the n=1, 2, 3, 4...etc. state.  All returns to the n=1 state (ground state) are termed the Lyman series,  all returns to the n=2 state are termed the Balmer series and so on.

For the hydrogen atom, the wavelength of these various emmissions is given by the Rydberg formula

$${1 \over \lambda} = R \left( {1 \over (n^\prime)^2} - {1 \over n^2} \right) \qquad \left( R = 1.097373 \times 10^7 \ \mathrm{m}^{-1} \right)$$

for the Lyman series this equation becomes

$${1 \over \lambda} = R \left( {1} - {1 \over n^2} \right) \qquad$$

and six lines can be seen in the ultraviolet.  After the 5th line (n=6 -> n=1), all other transitions are so closely spaced that they appear as one line (the sixth line).

\begin{array}{|c|c|}\hline
n & λ (nm) \\ \hline
2 & 122 \\ \hline
3 & 103 \\ \hline
4 & 97.3 \\ \hline
5 & 95.0 \\ \hline
6 & 93.8 \\ \hline
\infty & 91.2 \\ \hline
\end{array}

As to the relative intensities, I'm not sure why they are all the same in your graph.  The intensities do depend upon how many electrons are excited to the various excited states.  If for some reason all excited states were created in a manner such that they were equally populated, then something like your spectrum would result.  Usually the relative intensities differ and are used much like a fingerprint.  For example, the relative intensities of the Lyman series from the hydrogen atoms in one galaxy differ from those in another galaxy due to the difference in temperature of the two galaxies.  My guess is that your data was artificially created (e.g. the title in your graph says "Model").


Why does sodium require energy to ionize?

There are two very different questions here so let's break things down a bit
Why does sodium have an ionization energy?
Since electrons are negatively charged and nuclei (owing to the presence of protons) are positively charged, electrons will typically be drawn to the nucleus.  The situation gets a bit more complicated by (a) the quantum nature of energy and (b) the presence of other electrons.  Complication (a) is the reason why electrons do not behave like planets orbiting the sun because if they did the electron would eventually be pulled in to the nucleus.  (More details probably requires a different question.)  The second complication is more relevant to your question, because additional electrons will both repel the electron we are trying to eject through ionization and also shield some of the positive charge of the nucleus.  (Think about wearing sunglasses; they don't make the sun less bright, they just make less light reach your eyes.)
Your comparison of Na to that of an object on a cliff is not correct; the sunglasses analogy I used above is better (although it's flawed as well).  The reason why sodium, or any other atom/ion for that matter, has an ionization energy is because it requires energy to separate two oppositely charged particles.  The attraction may be weak, as it is in sodium, but it is still non-zero.
Why does $\ce{FeCl3}$ bond the way it does?
First, it is a good idea to look at the wikipedia page for iron(III) chloride, which shows that the molecule actually binds in an octahedral fashion, where two additional chlorides from neighboring molecules bind in a bridging fashion.  This is a type of bonding that is typically not discussed until one's third year of chemistry education.  Valence shell electron counting works really well for identifying "stable" electron configurations of ions and covalent molecules made out of lighter elements (C, N, O, for example).  The presence of d-orbitals, plus the ionic nature of the interaction between Fe(III) and chloride ions, plus the Lewis acid/base concepts that do a better job at describing and rationalizing bonding in transition metal complexes, makes the concept of valence electrons not the best model to explain bonding in $\ce{FeCl3}$.
Without access to molecular orbital theory, it is perhaps best to explain the bonding in iron(III) chloride from a strictly ionic bond nature.  Fe(III) is a +3 cation and therefore requires three -1 anions to balance the charge.


I would say it's worth investigating. My suspicion is that the answer won't change much, but the way to find out is to try it both ways and see how the assumptions affect the answer under various conditions.

Instead of a CSTR, for this particular reaction a batch reactor might be more appropriate, since that is the easiest setup for a lab. In that case, to model the reaction all we have to do is not assume that some of the rates of change are zero, and see how that changes the reaction.

Fortunately for us, this site has already done the heavy lifting and even has a solution in a mathcad file.

I haven't looked at the solution closely enough to determine if they are making the approximations you mentioned (I don't have mathcad installed), or if they are just directly integrating all four equations. They only show graphs for the concentrations of X and Y. However, it probably wouldn't be too hard to modify in either direction to compare the results under different sets of assumptions.


Is there an easy way to find number of valence electrons"

You can get the valence electrons in an atom's electronic arrangement by consulting the periodic table:

The Group 1 atoms have 1 valence electron.
The Group 2 atoms have 2 valence electrons.
The Group 3 atoms have 3 valence electrons.
The Group 4 atoms have 4 valence electrons.
The Group 5 atoms have 5 valence electrons.
The Group 6 atoms have 6 valence electrons.
The Group 7 atoms have 7 valence electrons.

The atom with electric configuration 2, 8, 3 is Aluminium and you can find it in Group 3 of the periodic table.

The periodic table is usually given in exams so...

The first two groups of the periodic table are the groups concerned with the filling of the $s$ orbital. So, all Group 1 atoms will have an outermost $s$ orbital filled with a single electron.

Similarly, Group 3 - Group 8 (Group 8 is also referred to as Group 0) are where the $p$ orbital is being filled up. All the Group 5 atoms thus all have an outermost $p$ orbital filled with 3 electrons.

The Periods in the periodic table can be used as an indication of the number of shells the atom has. Sulfur for instance is in Period 3 and thus has 3 shells.


Can I determine the rate law order from stoichiometric coefficients?

There are multiple issues here:

You should not assume that there is a simple rate law. Chemistry is complicated, and even simple reactions may proceed through multiple pathways in parallel.
More generally, you should not assume that you are given enough information to determine the solution. Just because the stoichiometric coefficients are there, does not mean that you can use them to determine the answer. So, if you just have a balanced reaction, you know nothing. If you want to become a scientist, you should learn to be comfortable with not having the answer.

So, I would say that your bottom line is incorrect.

The experimental rate law should be determined experimentally.

Given a putative mechanism with elementary steps and knowledge of rate constants for those steps, you can propose a rate law that for that mechanism. Do not assume it has any bearing on reality, since your putative mechanism may be wrong.

Frequently, the approach is to take the experimental rate law and see if the derived rate law for a proposed mechanism is consistent with the experimental result.


Measuring Colloidal copper PPM & how to ignore the salt PPM in soluton

First method is to look for Faraday's law of electrolysis and put in the equation your data.

Second method is to add to TDS reading the initial reading of the salt solution before the electrolysis.


What happens to the crystal structure of ice as it gets colder and colder

Well, the easiest way to address this question is take a look at the phase diagram for ice. Ordinary ice as we know consists of lots of hexagonal rings of water molecules. First off, as temperature is decreased well below $0~^\circ$C, there will be a slight contraction of the lattice due to decreased thermal contribution to vibrational frequencies. This will cause the average $\ce{OH}$ bond length to decrease which will cause a subsequent contraction of the average $\ce{OO}$ distance.

It is not obvious in any way that there should be another phase transition at low temperatures and pressure for ice Ih, but it turns out there is. One can understand why there is a phase transition by observing that in the low temperature limit, entropic effects will make a diminishing contribution to the free energy of any state, and the ground state will in the low temperature, low pressure limit will simply be the global minimum of the potential energy surface.

Now, experiments demonstrate that while the oxygen atoms in ice Ih are all arranged in these hexagonal rings, the hydrogen atoms are essentially distributed randomly subject to the conditions that two hydrogen atoms are covalently bound to each oxygen atom, each oxygen receives two hydrogen bonds, and no two hydrogen atoms point towards each other. These are known as the ice rules (or sometimes Bernal-Fowler rules). This, however, is not the most stable structure that one can imagine as the protons are disordered, and hence there is not optimal cooperativity in the direction of the hydrogen bonds.

For this reason, people thought there might be a low-temperature phase of ice which is proton-ordered. That is, you have the same lattice as in regular ice Ih, but the protons all point in the same direction when forming hydrogen bonds. This is easier to see, so I have put a picture of the lattice below.

This phase of ice is called ice XI. I think they are numbered based on the order they were discovered, which is very annoying.

In practice, even though this is essentially the ground state of bulk water, it is very uncommon to find in nature, and very difficult to synthesize in a lab. This is because at such low temperatures, the rates of proton rearrangements are very slow. These rearrangements occur by proton transfer in ice, or by rotating to rearrange which oxygen the hydrogen bonds are donated to. These processes individually have high barriers, and hence have low rates even at relatively high temperatures and pressures. So, going from ice Ih to ice XI occurs much more slowly than most phase transitions.

You can read about how ice XI is prepared in practice on the wikipedia page for ice XI, but basically you introduce defects which act as catalysts for the rearrangements for the hydrogen atoms.

TL;DR:
To be really pedantic, the answer to your question depends on how long you are willing to wait as you keep cooling down the block of ice. That is, if you have an ice cube and cool to e.g. 5K, then nothing will happen besides a slight contraction of the lattice. If you keep it at that temperature for, literally something like 1,000 years (evidence from Antarctic ice cores), then your ice cube will look exactly the same, but microscopically, the protons will be organized much more neatly. How pleasant.


Why can't Al3+ be reduced to solid, pure aluminum?

It is aluminium’s standard electrode potential that prevents it from easily being reduced. For the half-cell $\ce{Al^3+ + 3e- <=> Al(s)}$, the standard potential is $E^0 = -1.66~\mathrm{V}$. For comparison a few other values:

$\ce{Fe2O3 + 3 H2O + 2e- <=> 2 Fe(OH)2(s) + 2 OH-}$ and $\ce{Fe(OH)2 + 2 e- <=> Fe(s) + 2 OH-}$ have potentials of $E^0 = -0.86~\mathrm{V}$ and $-0,89~\mathrm{V}$, respectively;
$\ce{Zn^2+ + 2 e- <=> Zn(s)}$ has a potential of $E^0 = -0.76~\mathrm{V}$;
$\ce{Cr^3+ + 3 e- <=> Cr(s)}$ has a potential of $E^0 = -0.74~\mathrm{V}$;
$\ce{Co^2+ + 2 e- <=> Co(s)}$ has a potential of $E^0 = -0.28~\mathrm{V}$;
$\ce{Cu^2+ + 2 e- <=> Cu(s)}$ has a potential of $E^0 = +0.34~\mathrm{V}$.

(All values taken from Wikipedia)

The higher the standard potentials are, the easier the ion can be reduced. Hydrogen, of course, has a potential of $E^0 = 0~\mathrm{V}$ at pH 0 by definition. So in acidic media, hydrogen is reduced first. So let’s lower the pH to impede the formation of hydrogen. According to the Nernst equation:

$$E = E^0 + \frac{0.059~\mathrm{V}}{z} \lg \frac{a_\mathrm{Ox}}{a_\mathrm{red}}$$

Where $z$ is the number of electrons transferred in the half-cell, and $a$ is the activity of the reduced/oxidised species. $a_\mathrm{red}$ is $1$ by definition and $z = 2$ for the hydrogen half-cell. So we need to solve the following simplified equation to determine the acid concentration that will allow reduction of aluminium:

$$-1.66~\mathrm{V} = 0~\mathrm{V} + \frac{0.059~\mathrm{V}}{2} \lg \frac{[\ce{H+}]}{1}\\
\frac{0.059}{2} \mathrm{pH} = 1.66 \\
0.059 \times \mathrm{pH} = 3.32 \\
\mathrm{pH} = 3.32 / 0.059 = 56$$

Whoops, so we would need a pH of $56$ to allow the reduction of aluminium in the presence of water. This calculation was highly theoretical, because:

A pH of 56 is impossible in anything that once could have been water. Wikipedia gives the $\mathrm{p}K_\mathrm{b}(\ce{O^2-}) \approx -38$; because $\mathrm{p}K_\mathrm{a}(\ce{HA}) + \mathrm{p}K_\mathrm{b}(\ce{A-}) = 14$ in water that must mean that $\mathrm{p}K_\mathrm{a}(\ce{OH-}) \approx 52$ — thus we would be dealing with an ‘oxide solution’ rather than water.
Although it is nice that both equations (pH and Nernst) use acitivities, we cannot really say anything about the activities at this large a concentration range.
The redox system of both species would be totally different in these environments.

Note that you can reduce most air-stable metals in thermite reactions from their oxides thereby creating aluminium oxide. An example reaction would be:

$$\ce{Fe2O3 + 2Al -> 2Fe + Al2O3}$$

That alone should show the unwillingness of aluminium to be reduced in aquaeous solutions; think about the heat that the thermite reactions generate.


Electron flow: How does osmium tetroxide react with alkenes?

I have a suggestion that could explain the attacK of electron pair of alkene pi bond to the oxygen of osmium tetroxide : The electron pair of alkene pi bond repels the electron density of the oxygen when the osmium tetroxide approaches the alkene, creating a temporary δ+ on the oxygen that is closer to the pi bond of alkene. The electron-rich alkene then attacks the electron-poor oxygen.

Dimitrios Kariotoglou


From Calculation of Coefficients in the Power Series Expansion of the Long‐Range Dispersion Force between Atoms J. Chem. Phys. 56, 2801 (1972)

  Highly accurate calculations
  of $C_6$, the coefficient of the $R^{-6}$ term, resulting from
  an induced dipole-induced dipole interaction, have
  been made for many atomic systems, while values of
  $C_8$ (dipole-quadrupole interaction) and $C_{10}$ (dipoleoctopole
  plus quadrupole-quadrupole interactions) are...

In other words, in London dispersion interaction is usually represented as induced dipole - induced dipole interaction, the energy of which is proportional to $R^{-6}$, but there are other components of the interaction.


How does beta decay increases the stabilization of the nuclei?

While this is nuclear physics, the relative number and relative stability of isotopes is of some relevance to chemistry. Further, I spend more time than I'd like perhaps at the intersection of materials science, chemistry, and nuclear physics, so...

There are several good on-line resources to get information on nuclear energy levels and the relative stability of nuclei. These include:
The Nuclear Data Evaluation Project at Triangle Universities Nuclear Laboratory (TUNL), and the Evaluated Nuclear Data Structure File data base at the National Nuclear Data Center hosted at Brookhaven National Laboratory. From these you can get more information than you might want with regard to what is known about nuclear energy levels and stability of the elements.

One cut at looking at stability is the so-called isobar diagram. This compares the relative nuclear energy levels of all nuclei with the same $A$, where $A = P + N$, that is, the sum of the number of protons and neutrons in the nucleus. Since $\beta \pm$ converts between protons and neutrons, keeping $A$ constant, this is the best vantage point for the purpose at hand.

Borrowing from TUNL one finds for $A=14$:

So, we see 4 different nuclei shown on the isobar diagram. The lowest energy (most tightly bound) nuclei is $^{14}$N. To the left is $^{14}$C, which you likely know is unstable and $\beta -$ decays to $^{14}$N, used for carbon dating. To the right is $^{14}$O and you will note that it $\beta +$ decays to $^{14}$N. Neither $^{14}$C or $^{14}$O has as much binding energy as $^{14}$N due to the fact that protons and neutrons are not actually identical in the nucleus. So, the ground state of $^{14}$C or $^{14}$O is about 2.4 MeV higher than $^{14}$N.

High and off to the left you can barely see $^{14}$B, 24.71 MeV above $^{14}$N. It desperately would like to be $^{14}$N, but can't get there directly through the available $\alpha$, $\beta$ or $\gamma$ decay channels. So, it happily $\beta -$ decays to $^{14}$C with a half-life of 16 ms, which will then ultimately decay down to $^{14}$N.

Now, your misconception is that you don't know nuclear physics (and I don't mean that negatively - you applied a fairly simple model, did not get the expected answer, so you asked a question here - GREAT!). But, in the nucleus you have Coulomb repulsion, the strong force, and the weak force, and it all has to be treated quantum mechanically. Taken all together one can determine the binding energy (total or per nucleon) of nuclei, the excited nuclear energy levels, and possible decay paths. It is great fun, but is a different world to explore.


Yes there is and it has a very simple structure and is the strongest acid known: HeH$^+$. You might find it interesting to know that HeH$^+$ is also most likely the first molecule to have formed after the Big Bang.


Can rearrangement in carbocations cause inversion?

Originally this was a comment, but I figure it's good enough for an answer.

As far as I can tell, you seem to be confusing the notion of suprafacial/antarafacial with retention/inversion.

You're asking whether the migrating group can loop around to the other side of the carbocation, right? That's known as an antarafacial process, not an inversion.

Retention, or inversion, refer to the stereochemistry at the migrating centre. If it is a hydride shift, then the migrating centre is a hydrogen, and there is no stereochemistry to speak of. The terms retention or inversion are therefore not applicable to hydride shifts.

In any case, the crux of the matter is that this migration is a concerted process (it occurs in one step), and an antarafacial migration cannot occur due to geometric reasons. The lobe of the p orbital on the other side is way too far away, and the $\ce{C-R}$ $\sigma$ orbital (which contains the electrons that you are trying to migrate) is not going to be able to reach it.

[1,2]-alkyl (or hydride) shifts are always going to be suprafacial!

(As a side note, the suprafacial [1,2] shift has to proceed with retention, as per Woodward-Hoffmann rules.)


Are there any magnetic particles that are safe to inhale?

$\ce{O_2}$ is paramagnetic.

source: http://www.chem1.com/acad/webtext/chembond/CB-images/liquidoxygeninmagnet.jpg

The stable Xenon isotope Xe-129 has been used in MRI studies:

Hyperpolarized Xenon MR Imaging of the Brain

  A radically new magnetic resonance imaging (MRI) technique, using hyperpolarized 3He and 129Xe, is being developed to produce high-contrast images of important body tissues that have resisted conventional MRI techniques. Hyperpolarization of the noble gas nuclei is achieved by spin exchange with rubidium atoms that have been optically pumped into one electronic spin state with the use of circularly polarized light from a high-power diode laser system. Hyperpolarization can increase the detectability of 3He and 129Xe up to a hundred thousand times. This technology is showing dramatic results for diagnostic imaging of the lungs, brain, and other parts of the body. Most notably, we have achieved high-resolution dynamic images of the lungs and airways using 3He, and measured cerebral perfusion in the brain using 129Xe.

See video of lung being imaged: http://www.spl.harvard.edu/archive/HypX/about.html


Potassium carbonate and water/alcohol/toluene?

The synthesis you are referring to is outlined here. Potassium Carbonate acts as a drying agent as you said as the later part of the distillation requires a specific temperature to remove the toluene and alcohol.

The water comes from the concentrated sulphuric acid solution added with apidic acid. The residue is also rinsed with alcohol which is a strong indicator that water needs to be removed.


Basic concept of the first rule of thermodynamics

Chemical energy of system has been converted to thermal energy of both  the system and the surrounding water.

Additionally, water did work on system, as the system shrinked down.

$$\Delta E_\mathrm{system}=-C_\mathrm{water} \cdot \Delta T_\mathrm{water}-p \cdot \Delta V_\mathrm{system}$$

For $H=E + p \cdot V$,
$\Delta H=\Delta  E + p \cdot \Delta V + V \cdot \Delta p$

As p is constant,

$\Delta H=\Delta  E + p \cdot \Delta V $

$\Delta H_\mathrm{system}=-C_\mathrm{water} \cdot \Delta T_\mathrm{water}$

where $H$ means enthalpy, energy at the constant pressure.

If we considered other thermodynamic quantities like Gibbs energy or entropy, we would need to examine the underwent reaction and system composition. But if the system energy is considered, its change is related just to exchanged thermal energy and work.

$q$ in the equation $\Delta E = q+w$ means heat transfers to the system, so it is negative. $w$ is work done on system and is positive.  $\Delta E$ is very probably negative, as work value would be much less than transferred heat.


To answer the question in the title:

Yes, ATP can be synthesised, isolated and you can even eat it. It would be very expensive to do so but considering others put gold on most of their food that’s not a reason in itself not to. While eating too high a dose of ATP is not beneficial as per Paracelsus’ law, small amounts are certainly not harmful.

However, ATP is rarely shuffled into or out of cells. I’m not sure whether the human body provides mechanisms to shuffle ATP across cell membranes. I highly doubt it, since ATP is typically produced cell-internally by mitochondria and what gets shuffled around via the blood stream is primarily glucose. Most transport mechanisms are adopted to that.

Furthermore, even assuming that ATP does get shuffled across cell membranes in significant amounts, eating a crumb of it would first place it into the stomach. There, in the strongly acidic environment there is practically no way that ATP won’t get hydrolysed into first adenotine and diphosphate and then further broken down. Even if some of it magically manages to get through the stomach unharmed, the slightly alkline pH of the intestines are no better and the digestive tract is swarming with enzymes designed to break down practically everything. Thus, you can be sure that ATP has been fully metabolised before the body has a chance of taking it up into the blood stream.

Some of these problems can be overcome by injecting ATP subcutanously or intravenously but the problem with getting it into the cell remains.


Calculate the molar fraction of a mixture component given only its pressure

Your mole fraction is $$x_\ce{Ar}=\frac{n_\ce{Ar}}{n_\ce{He}+n_\ce{Ne}+n_\ce{Ar}}$$ and you somehow want to insert your pressures. You can try to convert all moles that you don't know into pressures that you know by using the ideal gas law $$p~V=n~R~T$$ which will make you find that everything you have (1.5, 2.0, 4.9 atm) is all what you need to come up with $x_\ce{Ar}=0.583$.

As your question was answered by midi-chlorian without further explanation, here is why you don't need more than those pressures given to you.

I said that you can use the ideal gas law to find out the amount of substance of each gas. Therefor we need to rearrange it to give $$n=\frac{p~V}{R~T}$$
We know the gas constant is constant ... that's where the name comes from ... and we can assume the temperature to be constant at whatever temperature we want ... it's not important to know the temperature here. On top of that the missing volume of each gas is equal for all three ideal gases, as Avogadro's law says.

This means that not only R and T are constant but that the volume will also not change, which in turn simplifies everything a lot.
$$n=p~\frac{V_m}{R~T}=p~k$$
Let's insert it into the mole fraction (as I told you to do) and see what happens.
$$\begin{align}
x_\ce{Ar}=&~\frac{n_\ce{Ar}}{n_\ce{He}+n_\ce{Ne}+n_\ce{Ar}}\\
=&~\frac{p_\ce{Ar}~k}{p_\ce{He}~k+p_\ce{Ne}~k+p_\ce{Ar}~k}\\=&~\frac{p_\ce{Ar}}{p_\ce{He}+p_\ce{Ne}+p_\ce{Ar}}\cdot\frac{k}{k}\\=&~\frac{p_\ce{Ar}}{p_\ce{He}+p_\ce{Ne}+p_\ce{Ar}}\end{align}$$

And this is the simple reason why you do not need more than your pressures.


Why zinc sulphate solubility is higher in ethylene glycol than water?

Zinc is an amphoteric metal which means it can dissolve both in acidic and basic conditions. However, neutral pH is the transition point. Please see this article Recovery of Rare Earths from Waste Cathode Ray Tube (CRT) Phosphor Powder by Selective Sulfation Roasting and Water Leaching on its behavior in water and eg. I believe the reason is that in eg there is no mechanism for zinc to form insoluble hydroxide complexes which will decrease the solubility of zinc. But in ethylene glycol you simply cannot form hydroxides..


Does any known substance ignite on cooling?

Actually... yes! Iron(II) oxide is thermodynamically unstable below $848~\mathrm K$. As it cools down to room temperature (it has to do it slowly) it disproportionates to iron(II,III) oxide and iron:

$$
\ce{4FeO -> Fe + Fe3O4}.
$$

The iron is in a form of a fine powder, which is pyrophoric (it may catch a fire when exposed to air). You can see it in action here.


When is it necessary to include solvent effects?

Well, first of all, explicit is more expensive than implicit. So, most people do not go for the former if they can help it. So, in my experience, you want to use explicit solvent only when you have to, like in the formation of $\ce{H2SO4}$ from $\ce{SO3}$ and $\ce{H2O}$. Here, extra water molecules act as catalyzers [1]. As for the case keto-enol tautomerization, I guess the results are in good agreement with the experiments using implicit solvent.

The bifunctional catalytic role of water clusters in the formation of acid rain. Chem. Commun. 2017, 53, 3516-3519.


Why is CO a major species only in fuel-rich combustion?

I think it is quite simple: the higher the proportion of the compound that will burn in the mixture, the more oxygen will be required to a complete combustion. If you have a constant amount of oxygen (in the reactor), and more is requires while increasing the amount of fuel, the complete combustion will gradually decrease, and the ratio $CO/CO_2$ produced will increase as a result of that.

As it can be seen on the graphic, from the point that the equivalent ratio of fuel and oxygen is larger than one, the $CO$ starts being formed in a considerable proportion, and over a ratio of 1:3, starts being the major product.


Your question is about immobilized metal affinity chromatography (IMAC), often used to separate a protein expressed in the E. coli bacterium from other proteins made by it. The protein of interest is expressed with a histidine-containing affinity tag which binds to immobilized nickel or cobalt at neutral or basic pH. Immobilization is achieved by complexing to NTA (nitrilotriacetic acid) or IDA (iminodiacetic acid). For a discussion of the two ligands, see e.g. here.

There are a few proteins from E. coli (listed in this paper) that are known to bind along with the protein of interest. According to a post written by Damien Soghoian in 2004, nickel-NTA tends to bind stronger to histidine-tags while cobalt-NTA tends to bind more weakly, but this does not affect specificity (e.g. it would be across the board, not making the separation easier or harder).


Does fumaric acid form stronger hydrogen bonds than maleic acid?

Intramolecular hydrogen bonds do not have to be weak, but the strength of a hydrogen bond depends a lot on the $\ce{O-H\bond{....}O}$ angle. With intramolecular hydrogen bonds, it is often hard to achieve the perfect $180^\circ$ angle while the same is not a real exercise for intermolecular hydrogen bonds which have full degrees of freedom of movement on both sides.

In fact, I want to argue that maleic acid only forms intramolecular hydrogen bonds if there is no solvent (or crystal water) that can otherwise supply hydrogen bonds. If there is, all hydrogen-bonding sites — both donating and accepting — will be fully saturated with external hydrogen bonds instead; they are slightly stronger due to the better angle.

The question remains why fumaric acid — which is definitely unable to build up intramolecular hydrogen bonds — has a higher boiling point, because in theory maleic acid should also be able to form as many intermolecular hydrogen bonds. However, fumaric acid’s structure already dictates that it practically must form a network of coplanar molecules interacting with each other via hydrogen bond dimers from carboxylate to carboxylate. This is a very nice structure to keep. Maleic acid cannot form such an extensive network no matter how much it distorts. It will be restricted to much less favourable crystal structures meaning weaker intermolecular attraction.

In the liquid phase, fumaric acid will keep the near-sphere ordering somewhat intact, i.e. conserve a high number of hydrogen bonds to its neighbours. However maleic acid will probably stick with one fully intramolecular hydrogen bond that does not contribute to intermolecular attractions, thereby lowering its boiling point.


Will the carboxylate groups of the citrate anion undergo hydrogen bonding?

As the comments indicate, you don't need a formal charge on oxygen at all.  "Neutral" oxygen atoms are usually negatively charged anyway just from bond polarity, as in water.  That and the lone pairs on oxygen are quite enough to enable hydrogen bonds to protic hydrogen.


Why does aluminum have a lower first ionization energy than magnesium?

The easiest way to explain it is that $\ce{Al}$ has one unpaired electron in it's highest energy orbital ($\mathrm{3p}$), and $\ce{Mg}$'s highest energy orbital ($\mathrm{3s}$) the electrons are paired. It is energetically favorable for all the electrons in an orbital to be paired, which means that breaking up this pair would require more energy.

Here's what their orbital pairings look like:

and if you look at the general trend you can see that this occurs whenever all the valence electrons are paired:

Sources:
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch7/ie_ea.html
http://en.wikipedia.org/wiki/Electron_pair


Enantioexcess for a non-chiral molecule?

The Ph.D. thesis that is cited is probably near-impossible to obtain and may very well be in Japanese, which certainly doesn't help.

2-Phenylethanol is not a chiral compound, so as Ivan pointed out, the concept of enantiomeric excess (ee) makes no sense. The fact that the direct reduction product of acetophenone (e.g. via hydride transfer) is 1-phenylethanol strongly suggests that this is nothing more than a typo. However, I thought that it might be fun to go to great lengths to verify the applicability of Occam's razor to this situation.

A SciFinder search for "acetophenone Euglena gracilis" did throw up one Chinese patent, CN 103757061. There is a translation on Google Patents. The patent describes the reduction of various prochiral ketones with various microphytes (also known as microalgae) to form chiral alcohols, with near-perfect enantioselectivity.

I managed to figure out the technical terms in the patent. E. gracilis in Chinese appears to be 纤细裸藻 (xian1 xi4 luo3 zao3). Acetophenone is 苯乙酮 (ben3 yi3 tong2). The patent does not, in fact, describe the reduction of acetophenone by E. gracilis, and the text in Table 2 is so small that I cannot quite make out what compounds they tested (they seem to be aliphatic). Perhaps they simply did not include this combination in the patent if the enantioselectivity they obtained was very low.

However, there are experiments with different species that lead to the reduction of acetophenone to "S-苯乙醇" (ben3 yi3 chun2). This translates to (S)-phenylethanol, and leaves open the possibility of it being either 2-phenylethanol or 1-phenylethanol. Since only the 1-phenyl isomer is chiral, I assume the reduction product is (S)-1-phenylethanol.

This is supported by Figure 1 in the patent. I have scribbled some loose translations. In hindsight I think "Photosystem" should be "Light-harvesting complex". The text translates to "Light-harvesting system" which is a bit of a mix and match of the two, so I wasn't sure what to go for.

Having $\ce{R_1 = Me}$ and $\ce{R_2 = Ph}$ would mean acetophenone is reduced to 1-phenylethanol.

In conclusion, I would therefore assume with confidence that the handbook contains a typo and the authors meant 1-phenylethanol.

(On a side note, the Chinese also got >95% ee for all their acetophenone reductions, go figure...)


Is enthalpy of a reaction meaningful at variable pressure?

Why should the enthalpy of a reaction be defined at constant pressure? Why can't it be defined at variable pressure?

Not sure "should" is the right choice of words: It can be defined for a reaction either at constant volume (variable pressure) or constant pressure (variable volume). You might choose one measurement over the other, but you'll get two different quantities.

At constant volume, there's no work done by the system on the surroundings, so you're just getting the change in internal energy of the system (this is typical of a bomb calorimeter lab setup); in the other case, you get the internal energy change as well as the work, and that is how enthalpy is defined (mathematically). See this Wikipedia article for a little more detailed discussion.


Why does HCO₂⁻ exhibit delocalized bonding while CO₂ does not?

Here is an orbital diagram for $\ce{CO2}$.

Note how the red pi orbitals on the terminal oxygens are rotated 90 degrees with respect to one another; they are said to be "orthogonal" to one another.  This means that the p-orbitals on the terminal oxygens can only interact with the p-orbitals with which they are aligned on the central carbon; there is no way for an electron on one oxygen to interact with the orbitals on the other oxygen.

Here is the analogous drawing for the formate anion ($\ce{HCO2^-}$).  Note how the p-orbitals on the terminal oxygens are now both aligned parallel with the p-orbital on the central carbon atom so that continuous overlap (delocalization) is now possible.

For completeness, we can restate the drawings from above in terms of resonance structures.  In the top row of the following picture I've drawn the possible resonance forms for $\ce{CO2}$.  The only charge-separated structures we can draw involve charge separation on each of the two orthogonal $\ce{C=O}$ groups.

In the bottom row of the picture I've drawn 4 resonance structures for the carboxylate anion (when R=H this is the formate structure in the title of your question).  Structure II corresponds to the charge separated structure I drew above for $\ce{CO2}$.  But notice that I have drawn 2 new resonance (III and IV) where the negative charge is delocalized over the O-C-O 3 atom array.  This is because the p-orbitals on the terminal oxygens are now parallel and can overlap with the p-orbital on the central carbon.  Just like the "particle in a box" problem you may have studied in physics, it turns out the longer the path the electron has to roam over, the more stable the electron.  So resonance structures III and IV, where the electron can roam (delocalize) over 3 atoms are very important contributors to the description and stability of the carboxylate anion.  Carbon dioxide has no comparable resonance structures due to the orthogonal alignment (90 degree rotation) of the orbitals on the terminal oxygens


Periodic trends: why is effect of protons greater than electrons?

This is because the attraction is not between each proton and it's corresponding electron, but between each electron and the nucleus. Since the number of protons increases the charge of the nucleus, but the charge of the electron stays constant, the attraction is greater. The attraction of the electron to the nucleus depends on the charge of the nucleus, and it is independent of the number of electrons in "orbit"

A great analogy for this is the solar system. (Gravity works very similarly to electrostatic force.) It won't matter if suddenly there is a mother earth orbiting in the outskirts of the solar system, Earth would be unaffected by that change, and we would still orbit the sun at the sae distance. However, if the sun suddenly becomes noticeably heavier, the planets will each be pulled in with a stronger force, which reduces the orbit radius. This will happen equally to each individual planet. It doesn't matter how many planets we have, The pull of the earth is only affected by the mass of the sun and the mass of earth.

This applies to the atom if you think of the nucleus as the sun and each electron as a planet. The force attracting the electron to the nucleus is only affected by the charge of the nucleus as each electron carries the same charge, and the charge is not the collective  charge of all electrons, but the individual charge. Hence all each electron perceives is the increase in the charge of the nucleus that is increasing its attraction to the nucleus.


Conversion process of lanolin to 7-Dehydrocholestrol

The conversion of lanolin to 7-dehydrocholestrol is an important procedure in the production of vitamin D.
Lanolin is converted to cholesterol by a method which comprises of following steps:

dissociating cholesterol from lanolin by transesterification
reaction;

performing short-path distillation to obtain cholesterol crude
product

purifying by silica gel column chromatography

crystallizing to obtain refined cholesterol with purity higher than
95% and yield higher than 70%.

The extracted cholesterol is converted to 7-dehydrocholesterol and then treated with UV light to produce vitamin D3. The method has such advantages as high product consumption, little pollution, simple process and easy operation, thus is suitable for mass production.
References

https://patents.google.com/patent/CN1958596A/en
https://patents.google.com/patent/CN1263769C/en
https://www.sigmaaldrich.com/technical-documents/articles/biofiles/nutrition-research.html


What makes an X-ray machine's white photographic film go black?

X-ray film is very similar to standard black and white negative film. It's basically a transparent plastic base coated with an emulsion of silver halide particles (a salt of $\ce{Ag+}$ with a halide, usually $\ce{Cl-}$ or $\ce{Br-}$). As shown below, when light hits one of these particles, it reduces a small amount of the halide to elemental silver (1). When this is exposed to a photographic developer (essentially a mild reducing agent), more elemental silver is produced around the initial sites (2).$$\ce{Ag^+ +e- -> Ag}$$

Finally, a fixing agent is added which solubilizes the unexposed silver halide so it can be removed from the emulsion, preventing further light exposure from fogging the film.

Thus, the regions that are exposed to light appear dark when developed as the tiny silver particles ($\ce{Ag}$) block light passing through the film and the unexposed regions are transparent, taking on the colour of the plastic base (often blue or grey). X-ray films are typically viewed on an illuminated white background, making the unexposed areas appear white (or off-white, coloured by the plastic base).

Normal B&W film is actually slightly sensitive to x-rays but, the emulsion of dedicated x-ray film contains other components to facilitate the silver reduction from x-ray exposure.

Source


Electrons are not actually separated in shells. That is a simplified model to explain the atom's structure when people first start approaching Chemistry or Physics.

In reality, electrons' movements around the nucleus are governed by complex equations (some of them not even resolvable so approximations are needed) to form what is called an electron cloud. This cloud is confined in proximity of the nucleus by the electromagnetic force. Every electron is defined by its wave equation, which tells us the probability that the electron appears to be at a particular location when its position is measured. These equations allows us to define regions in space where the electron spend the most part of its time around the nucleus, and they can be visualised as atomic orbitals, as shown below. Orbitals can have one or more ring or node structures, and they differ from each other in size, shape and orientation.

Each orbital (which is defined by a different equation) has different associated parameters, called quantum numbers, and a particular energy. These properties allow us to divide them into levels and sublevels, to form a structure that resembles multiple shells one inside each other. But as you can see from what I explained the structure is in fact far more complex.

This knowledge has been gained over time both from the mathematical and physical analysis of different equations (like Schröndiger equation) and by experimental data collected by the means of different techniques. Starting from Thomson and Rutherford the model of the atomic structure has seen many changes (Bohr, Moseley, Van der Broek, de Broglie, etc.) and it has come closer to a realistic representation of the actual atom. Take a look at their experiments to understand how it has been studied. However, they often involve particle physics and are not really straightforward to understand.


Equilibrium is the state at which the rates of the forward and reverse reactions are equal.

$$\mathrm{rate_{forward}=rate_{reverse}}$$

Let's take a generic reaction:

$$\ce{A + B <=> C + D}$$

For simplicity's sake let's have both the forward and reverse reactions be concerts fundamental reactions. Then we can write the rate laws for the forward and reverse reactions:

$$\begin{aligned}
\mathrm{rate_{forward}}&=k_1 [\ce{A}][\ce{B}]\\
\mathrm{rate_{reverse}}&=k_{-1} [\ce{C}][\ce{D}]
\end{aligned}$$

If we set them equal to each other, we can rearrange the equation to get the rate constants on one side and the concentrations on the other.

$$\begin{aligned}
k_1 [\ce{A}][\ce{B}]&=k_{-1} [\ce{C}][\ce{D}]\\
\dfrac{k_1}{k_{-1}}&=\dfrac{[\ce{C}][\ce{D}]}{ [\ce{A}][\ce{B}]}
\end{aligned}$$

The ratio of rate constants is the equilibrium constant. Thus is derived the law of mass action.


Is it dangerous to inhale the steam coming from electrolysis of salt-water solution

So if I understand the question correctly, you heated up water with some salt by performing electrolysis.

If so, you would also generate chlorine gas. However, the heat capacity of water is really large, so I doubt that you would even be capable of bringing water to a boil by just jolting current through it.

To answer your question: Yes, when performing electrolyzation of any solution you should apply the usual precautionary measures, such as working in an open space or under a fume hood. Under no circumstances should it be possible for you to inhale the gases produced.


Generalizations regarding functional groups on benzene ring

Generalisation while often is a very useful tool at the zeroth approximation level, – that is crude, very crude, barely resembling the truth, – is a temptingly dangerous game to play. It often boils down complicated concepts beyond the point of usefulness and leads to wrong conclusions.
Generalisations are oversimplifications.
An example for this is the ever so annoying inclusion of d orbitals into hybridisation schemes. This was a result from the post rationalisation of VSEPR theory with molecular orbitals. It has been disproved many times, yet it still is used as a common explanation in textbooks. It is a myth that continues to live, and live, and live...
Another example for generalisation taken too far is resonance. The concept most often employed, yet understood so little about it; especially why it is neccessary in the first place. As a result we hear phrase like "most stable resonance structure", which are complete garbage.

As such, while making generalisations you might not yet realise what damage that may cause later. Some concept simply are complicated and should not be broken down further.
It is quite common to draw reaction mechanisms with Lewis structures and electron pushing arrows. The problem with these is the extremely simplified nature of these drawings. Molecules are (usually) not planar, and yet we ignore this for the purpose of representation and depiction. A picture may sometimes be worth a thousand words. However, another, probably worse, problem with these schemes is the depiction of discrete electron pairs; a contributing factor to the resonance problem mentioned above.
When using these mechanisms it is always implied, that these are proposals, simplified to highlight certain aspects of a reaction.

Let's analyse your statements a little and see where you might derail.

  I claim that negative charge on phenoxide ion first ALWAYS delocalises onto the ring and then attacks through carbon atom (like in Reimer Tienman reaction).

That statement has already one big problem, it's wrong. The charge in the phenoxide ion is never localised. The electronic structure of a molecule will always adapt much more easily and fast than any given change in the molecular geometry. When phenoles are deprotonated, while the hydron leaves, the electronic structure instantaneously adapts.
You are thinking in terms of discrete resonance structures, which simply do not exist. The negative charge is delocalised. That's it. The corresponding observable is the electron density, which is concentrated at certain positions due to the excess electron. In the approximation of molecular orbital theory, this can be partially explained with the HOMO and lower orbitals. In the context of valence bond theory we need to employ resonance, i.e. the superposition of various electronic structures that feature localised electron pairs. Both representations are approximations, but currently our best way to understand complicated mechanisms.

The way how molecules react with each other depends on many things. The product which is formed depends on a fragile system of well balanced energy levels. You need to understand, that molecules react in all possible ways, but only the most stable products will eventually be retained.

In that regard you statement is also wrong, the phenoxide ion is a nucleophile, it will act through every position with increase electron density, most likely through the oxygen though.

  This happens because negatively charged carbon atom is less electronegative and hence a better nucleophile than a negatively charged oxygen atom.

The pure notion of better or worse nucleophiles is highly dependent on the context. Just because formed carbon-carbon bonds tend to be a little more stable than the carbon-oxygen bonds, it is not useful to generalise that. I also highly doubt the mechanism presented on Wikipedia for the Reimer–Tiemann reaction is complete enough. As has been stated before, this is not a clean reaction, hinting at a variety of semi-stable intermediates.

  But a normal lone pair like that on aniline or phenol attacks by itself without delocalisation (as in Schotten Bauman reaction). This is because delocalised resonance structures have a formal positive charge on the more electronegative oxygen/nitrogen and are thus less stable.

I don't quite understand where you are going with that, but the lone pair of nitrogen in aniline certainly has overlap with the aromatic system. In general amines and phenols are quite alright nucleophiles.  There are other factors determining why they react the way they do.
Your statement suffers from the most-stable-resonance-structure-syndrome, i.e. you take resonance structures much too seriously. Please see: What is resonance, and are resonance structures real? If there is resonance, then this effect is always stabilising, because otherwise it would adopt a molecular structure where resonance would not be possible (compare cycloocta-1,3,5,7-tetraene).
Additionally formal charges are exactly what they are called: formal. There is little to no truth to them other than keeping track of all the electrons.

One of the most important things is also to realise that stability always is a relative concept. More often than not it highly depends on the chemical environment, if an intermediate is "stable" and if it would be more "stable" than another intermediate.

In conclusion,
as a high school level student, you will learn a lot of useless or even wrong stuff. Some concepts are still taught even though they are known to be wrong. Some concepts are often not correctly understood by teachers themselves. Even at the university level you will encounter hardliners who teach chemistry like it was still the 20th century. Unfortunately that can't be changed (in the immediate future).
However, critical thinking is the way to gain knowledge. Unfortunately that means letting go of generalisations. Many things are much more complicated that they appear at first sight, only a detailed look will yield a better understanding. And once you have obtained a deep and fundamental knowledge, you won't need these generalisations or oversimplifications.


How to calculate the specific heat of ethanol?

Each gram of water has an specific heat of 4 J/(K.g)*. So heating $\pu{100 g}$ of water from $\pu{15 °C}$ to $\pu{70 °C}$ would take:

Heat = (70°C - 15°C) * 100g * 4 J/(K.g) = 22,000 joules = 22 kJ

$\pu{20 g}$ of ethanol would be:

Ethanol mol = (20g / 46,07 g/mol) = 0.43 mol

So this means that:

0.43 mol = 22 kJ
     mol = 22 / 0.43
     mol = 50677 J = 50.67 kJ

So ethanol has 50.67 kJ/mol vs 53.72 kJ/mol said by you. I would say that you have made fine your homework, there is a very little difference (I have used some decimal to calculate, maybe you not. (Below you can find a more accurate answer)).

* Water specific heat:  water changes it's specific heat when it changes it's heat. For example: (Wikipedia)

Water at  100 °C (steam)  2.080  J/(K.g)
Water at   25 °C          4.1813 J/(K.g)
Water at  100 °C          4.1813 J/(K.g)
Water at  −10 °C (ice)    2.05   J/(K.g)

Normaly it's used:        4.1813 J/(K.g)

A better table can be found in a french page.

Using this table we can do a better aproximation:

Water heat = (70°C - 15°C)        * 100g
Water heat = (293.03 J - 63.04 J) * 100g
Water heat =    229.99 J          * 100g
Water heat = 22,999 J = 22.999 kJ ≈ 22.9 kJ

Ethanol mol = (20g / 46.07 g/mol) = 0.43 mol
Exactly: 0,4341219882787063164749294551769 mol

0.43 mol = 22999 J
     mol = 22999 / 0.43
     mol = 52978.19 J = 52.97819 kJ ≈ 52.97 kJ
Exactly: 52978,196500000000000000000000001 J

Answer: Ethanol has 52.97 kJ/mol. A value closer to your own value.

PD: I am only a normal person who like chemistry, maybe I am wrong. For example in the theory I have no idea what means $Q = c \cdot m \cdot \Delta{}T = \pu{4.2 \frac{J}{g \cdot K}} \cdot \pu{100 g} \cdot \pu{55 K} = \pu{23100 J}$ Well, I have some ideas but I haven't learned it in the school.

Edit:

I notice that you have your own mol of ethanol (46 g/mol instead of my 46.07 g/mol) and your own specific heat of water (4.2 J/(K.g)). You have to do:

Water heat = (70°C - 15°C) * 100g * 4.2 J/(K.g)
Water heat =      55°C     * 100g * 4.2 J/(K.g)
Water heat =
Water heat = 23100 J = 23.1 kJ

Ethanol mol = (20g / 46 g/mol) = 0.43 mol
Exactly: 0.43478260869565217391304347826087 mol

0.43 mol = 23100 J
     mol = 23100 / 0.43
Using mol decimals (0.43478260869565217391304347826087)
     mol = 53130 k = 53.13 kJ
Using only 2 decimals (0.43)
     mol = 53720,93 J ≈ 53.72 kJ
     Exactly: 53720,930232558139534883720930233 J

Answer: Ethanol has 53.13 kJ/mol (With decimals) or 53.72 kJ (with 2 decimal). So yes! You have done right!


Concentrations are a geometric series

First, let's focus on the reactions where one of the reactants is a monomer $\ce{X_1}$:

$$\ce{X_1 + X_n <=> X_{n+1}}$$

and call the equilibrium constant for this reaction $K_{\mathrm{extend}}$.

From this, we can figure out the equilibrium constant for making a polymer from monomers:

$$\ce{n X_1 <=> X_n}$$

$$K_n = (K_{\mathrm{extend}})^{n-1} = \frac{[X_n]}{[X_1]^n}$$

So we can express the concentration of a polymer $X_n$ dependent on the monomer concentration:

$$[X_n] =  (K_{\mathrm{extend}})^{n-1} [X_1]^n = \frac{1}{K_{\mathrm{extend}}}(K_{\mathrm{extend}} \cdot [X_1])^n$$

This is a geometric series (if you take the concentration of polymers as a series, the ratio of neighbors in the series is constant). Because the amount of matter is finite, $K_{\mathrm{extend}}\cdot [X_1] < 1$ for an equilibrium scenario (as opposed to ending up with a single chain containing all the monomers). So the concentration of $[X_1]$ is the highest, and concentrations drop with increase of $n$.

Expression for $[X_1]$

Let $[X]_\mathrm{tot}$ be the total concentration of monomer, no matter whether in a polymer or not.

$$ [X]_\mathrm{tot} = \sum_{j=1}^{\infty} j[X_j]$$  $$ = \frac{1}{K_{\mathrm{extend}}} \sum_{j=1}^{\infty} j (K_{\mathrm{extend}}\cdot [X_1])^j $$

Using $$\sum_{j=1}^{\infty} j \beta^j = \frac{\beta}{(\beta - 1)^2}$$

we get

$$ [X]_\mathrm{tot} = \frac{[X_1]}{(K_{\mathrm{extend}}\cdot [X_1] - 1)^2}$$

which you can solve for the monomer concentration.

Relationship to real polymerization reactions

The polymerization reactions in chemistry and biochemistry that I am aware of are kinetically controlled. There are fibers (such as actin) in biochemistry that are dynamic (i.e. there is polymerization and depolymerization), but the processes are affected by nucleotide binding, complicating matters. I would be interested to hear whether the OP has an example that behaves like this.


It looks like a isochroman-1,3-dione to me.

For the reference there is a similar compound (published in French): Kakou-Yao, R.; Saba, A.; Ebby, N.; Pierrot, M.; Aycard, J. P. Tautomérie de la 4-(hydroxyphénylméthylène) isochroman-1,3 dione à l'état solide, Acta Cryst C 1999, 55 (9), 1591–1593. DOI: 10.1107/S0108270199006782


Question about the Law of Definite Proportions

In reality, there is not a sudden changeover of $\ce{CO}$ to $\ce{CO2}$ production when the oxygen supply is increased. This is the reason that all dwellings should have carbon monoxide detectors.

As an example, along with $\ce{CO2}$, gasoline (petrol) engines commonly produce 3-7% $\ce{CO}$ before additional air is injected.

The amounts you cite are how much oxygen is in pure samples of each gas, but producing either gas often requires filtering out traces of other gases that were also made in side reactions.

Producer gas is a fuel made by the incomplete oxidation of carbon (from coal or other source, sometimes with the addition of water) and contains both flammable $\ce{CO}$ and nonflammable $\ce{CO2}$ (and $\ce{N2}$ from the air supply). During fuel shortages, gasifiers were cobbled onto vehicles to enable them to run on this type of fuel.


Does the order of protonation/deprotonation in the formation of hemiacetals matter?

The first (productive) step of this mechanism in acidic medium MUST be protonation of the carbonyl.  Alkoxides and hemiacetal-alkoxides have no appreciable concentration in acidic solutions, and the carbonyl is not a strong enough electrophile to be attacked by alcohol without protonation.


Relation between enthalpy and internal energy for chemical reaction at constant volume

why is dH=dE at constant volume?

It isn't, and nowhere in the problem or answer is this implied.
First of all some definitions.
For a combustion reaction, the enthalpy change can be equated with the heat of combustion at constant pressure, whereas the internal energy is the heat of combustion at constant volume:
$$\Delta U = q_V ~~~~~\text{constant volume} \\
\Delta H = q_p ~~~~~ \text{constant pressure}$$
There is a simple integral form for the relation between internal energy and enthalpy that follows from the definition of the latter:
$$\begin{align} \Delta H &= \Delta (U+pV) \\ &= \Delta U + \Delta (pV) \end{align}$$
This relation always holds (it is a definition).
For this particular problem, the key to the next step is to use the ideal gas law, $pV=n_gRT$. This is allowed only if that equation of state holds for the substances in question (here carbon dioxide and oxygen). Inserting the ideal gas relation gives
$$\begin{align} \Delta H &=  \Delta U + \Delta (n_gRT) \\ &= \Delta U +  RT\Delta n_g\end{align}$$
where T is assumed constant. This is the expression used to find the solution. Clearly it shows that the change in enthalpy and internal energy are different.
You could derive this last expression slightly differently to emphasize the differences in the energy and enthalpy. The enthalpy change includes the heat input required to perform work against a constant pressure atmosphere and isothermally. In the case of an ideal gas, this heat input is
$$q_p-q_V = \Delta q = -w_{pV} = p\Delta V = p \Delta (n_g RT/p) = RT \Delta n_g$$
Note that for an ideal gas the enthalpy is a function of T only, so it's true then that a change in enthalpy is the same under constant V or p, provided you converted equal amounts of substance at same T. The same holds for the energy (since for an ideal gas it is also a function of T only).


What is the common denominator of all the lipids?

As you note, most biomolecules — or more precisely primary metabolites — exist as polymers that can theoretically be infinitely long and consist of well-defined monomers. This is true for nucleic acids and the building block nucleotides, for proteins made up of amino acids and for carbohydrates made up of monosaccharides.

Lipids, on the other hand, do not have that kind of polymeric possibility. In essence, they are trimers or monomers by definition, depending on how you look at them. Furthermore, all lipids contain the same (with same I mean identical) connective molecule namely glycerin or propan-1,2,3-triol $\ce{CH2OH-CHOH-CH2OH}$. Glycerin has three ‘sockets’, if you wish, namely its three hydroxy groups.

The ‘plugs’ that connect to glycerin always contain an acid functionality: typically this is either a carboxy group (fatty acids) or a phosphate (e.g. phophatidyl choline). The bond between ‘plug’ and ‘socket’ is an ester bond.

Both the fatty acids and the substituted phosphates can be structurally diverse. If you wanted, you could label the different fatty acids ‘monomers’ and the overall lipid molecule a ‘trimer’. However, the most typical difference between fatty acid residues is simply their length ($\ce{C16}$, $\ce{C18}$ or $\ce{C20}$ — typically multiples of 2) and their degree of unsaturation (e.g. stearate, oleate, linoleate and linolenate have increasing degrees of unsaturation). Attempting to teach students more than a few exemplary fatty acid residues or phophate residues is rather pointless since the choice of fatty acid hardly has any effect on the properties of the overall molecule, and aside from generally knowing what constitutes cellular membranes the same can in principle be said for the phosphates. Hence, students are hardly ever bothered.


Dipole moment - calculation of percentage ionic character

The calculated dipole moment is charge on electron * radius of
  molecule

That's off a bit, the calculated dipole moment (calculated assuming one electron has been completely transferred from hydrogen to bromine) is dependent on the bond length (interatomic spacing), not the radius.
\begin{aligned}
\mu_{100\%} &= \mathrm{electron ~charge \times interatomic ~spacing}\\
\mu_{100\%} &= (1.6 \times 10^{-19})\ \mathrm C \times (1.41 \times 10^{-10})\ \mathrm m\\
\mu_{100\%} &= 2.26 \times 10^{-29}\ \mathrm{C\ m}\\
\end{aligned}

As you point out,
$$\mathrm{\% ~ionic~ character} = \frac{[\mu_\mathrm{obs}]}{[\mu_{100\ \%}]} = \frac{[2.6 \times 10^{-30}]}{[2.26 \times 10^{-29}]} = 0.12 = 12\ \%$$


Molar Absorptivity of Copper(II) Sulfate

These are the absorbance spectra of copper sulfate in water (open markers) and heavy water (filled markers) (1):

The absorbance in water is maximum at 780 nm, as stated in the reference. From the spectrum, the molal absorption coefficient, at 780 nm, appears to be approximately 12.5 $\mathrm{(mol/55.51 mol \ of \ water)^{-1} cm^{-1}}$ with aquamolality equal to 0.5.

Since only ordinary water, i.e., $\ce{H2^16O}$, is of interest in the OP's question, the aquamolality is the same as the molality, so the solution is simply 0.5 molal. The measurements were reported to have been made at room temperature, with no specific temperature given. Hence, the solution molarity is approximately 0.5 M.

Reference

G. Jancso, "Effect of D and $^{18}O$ isotope substitution on the absorption spectra of aqueous copper sulfate solutions", Radiation Physics and Chemistry, 74 (2005) 168-171.


How does one measure the change in enthalpy of hydration?

Usually, one would measure these changes in enthalpy calorimetricly — or at least it is a nice student excersize in physical chemistry lab courses. I was once tasked to determine the solvation enthalpy of copper(II) sulphate pentahydrate.

The calorimeter consisted of an isolated thermos in which we filled a defined amount of water and a concealed inner container in which we filled a sample of the salt; the apparatus contained a stirrer inside the isolated area. We then let the water temperature equilibrate itself, opened in the inner container from the outside (mechanically) and measured the change in temperature. From $\Delta T$ we could calculate $\Delta q$ and from that $\Delta H_\mathrm{solv}$. All corresponded to the equation

$$\ce{CuSO4 . 5 H2O + H2O -> Cu^2+ (aq) + SO4^2- (aq)} ~~~~~~ \Delta H_\mathrm{solv}(\ce{CuSO4 . 5 H2O})$$

One could then perform a second experiment using anhydrous copper(II) sulphate. According to Hess’ law, the following is true:

$$\ce{CuSO4 + H2O -> Cu^2+ (aq) + SO4^2- (aq)} ~~~~~~ \Delta H_\mathrm{solv}(\ce{CuSO4})$$
$$\ce{CuSO4 + 5 H2O -> CuSO4 . 5 H2O} ~~~~~~ \Delta H_\mathrm{hyd}(\ce{CuSO4})$$
$$\Delta H_\mathrm{solv}(\ce{CuSO4}) = \Delta H_\mathrm{hyd}(\ce{CuSO4}) + \Delta H_\mathrm{solv}(\ce{CuSO4 . 5 H2O})$$

The two $\Delta H_\mathrm{solv}$ can be determined experimentally, thus the $\Delta H_\mathrm{hyd}$ can be calculated from these two values.


Why is ΔE described as Q-W and Q+W in different contexts?

The difference in sign in the two versions of the first law of thermodynamics is to handle the two ways in which work can be defined.

The work done (assuming only pressure-volume work) can be defined as

$$w=P\Delta V$$

This is the definition often used in in scenarios when we care about the fate of the work. This definition is about the surroundings. Thus, when the system expands, it does positive work on the surroundings, which is a good thing. Think about an internal combustion engine. When the reaction occurs in a cylinder, gas is produced causing the volume to expand and move the piston, turning the drive shaft, etc. Since we care about the work after it leaves the cylinder, it is positive because the surroundings are gaining energy.

The work done by a system can also be defined as:

$$w=-P\Delta V$$

We use this definition often in chemistry and in general when we care more about how much energy is left in the system than we care about what the energy that left the system is doing. This definition is about the system. In the case of the engine above, energy has left the system, so the work done is negative to reflect the fact that the system has less energy than it used to.

The sign convention in the definition of the change in internal energy (and I prefer $\Delta U$ to $\Delta E$) reflects that we can define work done with respect to the system or the surroundings. The sign in front of $w$ guarantees that we always get

$$\Delta U=q-P\Delta V$$

With respect to the surroundings:

$$w=P\Delta V$$
$$\Delta U = q-w=q-P\Delta V$$

With respect to the system:

$$w=-P\Delta V$$
$$\Delta U = q+w=q-P\Delta V$$

A better question might be

  Why then do we not change the sign of $q$ when we switch reference frames from system to surroundings?


There is not such as general/ic rule but water treatment has used, waste water as inflow. It is not as we are discussing the oxidation of X to obtain Y, if you get the point.  The target is to get an outflow of water that can be reused or discharged in nature without significant impact. The lines below do not pertain to the treatment of industrial wastes nor the removal of a specific pollutant.
Normally no toxic compounds go to the sink, and even in the lab we are not supposed to flush chemicals down.

Oxidative processes in waste water treatment mainly consists of biological aerobic ones, basically the demolition of a wide range of chemicals as it would naturally occur in a natural environment by means of bacterial activity.

What is peculiar here is that the waste water is kept in a regulated physical and chemical environment able to sustains and ease the process (for instance the oxygen required by the microorganisms is guaranteed by injecting air, or the temperature is regulated, ecc.).

This aerobic biological oxidation is targeting biodegradable pollutants is not targeting any specific chemical. Again, keep in mind that we are speaking of waste water not of special industrial waste. Even a chemical toxic to the bacteria, or in other words interfering with the bacterial  metabolism, does not pose a limitation as far is getting degraded in the while.

Most of the organic pollutants are decomposed into carbon dioxide and water leaving a water with high content of nitrites, nitrates, phosphates that must be lowered in subsequent steps and is full of particulate and biomass that must be separated as well.

Let assume all the steps above have resulted in a sufficiently clear and relatively pure water, then yes a chemical oxidation is performed. This step might use chlorinated chemicals and other oxidizers, but it is just targeting the residual microorganisms and tailored just to do that, like in the case of the disinfection of a swimming pool. It is performed when virtually there is nothing to oxidize but the residual organisms that were at the start of the whole treatment.


How to find the volume of the products in methane combustion?

One mole $\ce{CH4}$ reacts with 2 moles $\ce{O2}$, producing 1 mole $\ce{CO2}$ and 2 moles $\ce{H2O}$.
So 1 liter $\ce{CH4}$ produces 1 liter $\ce{CO2}$ and 2 liter $\ce{H2O}$, if $\ce{H2O}$ is taken in the gaseous phase.


Molecules having higher enthalpy of vaporization and boiling point than another but with lower entropy of vaporization?

The enthalpy of vaporization is greater for hydrogen-bonding molecules than for plain alkanes.  For low-molecular weight alcohols, this effect is pronounced.  The longer the alkane chain becomes, the more the compound behaves like a pure alkane.  For example, using data obtained from the NIST Webbook:

Table 1. Enthalpy of vaporization in kj/mol:
$\ce{Methanol\ \ \ \ \ 38}$
$\ce{Ethanol\ \ \ \ \ \ \  42}$
$\ce{n-propanol\ \  47}$
$\ce{n-butanol\ \ \ \  52}$
$\ce{n-pentanol\ \  57}$
$\ce{n-hexanol\ \ \ \  61}$
$\ce{n-heptanol\ \  67}$
$\ce{n-octanol\ \ \ \  71}$
$\ce{n-nonanol\ \ \  77}$
$\ce{n-decanol\ \ \ \  82}$

Keeping in mind the relative molecular weights of the compounds, you can see there is a decreasing effect of the hydrogen bonding (and other) effects on the n-alcohol series as we move to larger chains and become less alcohol-like and more alkane-like.  This is much more obvious in the next table, which is simply Table 1 normalized to a per-carbon basis:

Table 2. Enthalpy of vaporization on a per-carbon basis in kj/mol:
$\ce{Methanol\ \ \ \ \ 38}$
$\ce{Ethanol\ \ \ \ \ \ \  21}$
$\ce{n-propanol\ \  16}$
$\ce{n-butanol\ \ \ \  13}$
$\ce{n-pentanol\ \  11}$
$\ce{n-hexanol\ \ \ \  10}$
$\ce{n-heptanol\ \  10}$
$\ce{n-octanol\ \ \ \  9}$
$\ce{n-nonanol\ \ \  9}$
$\ce{n-decanol\ \ \ \  8}$

Table 2 even more drastically illustrates the deviation in $\mathrm{\Delta}H$ changes as we move from alcohol-like compounds to alkane-like compounds. So, given that the enthalpy of vaporization is greater for hydrogen-bonding molecules than for plain alkanes, this should explain your findings.


Is there a reaction happening between Coconut Oil & Hydrogen Peroxide?

Did any chemical reaction occur between the coconut oil and H2O2 to make a new compound?

Most likely, no chemical reaction occurred.

Whether any reaction occurred depends on the conditions you used when mixing the oil and hydrogen peroxide. How long did they mix for? What temperature was the mixture at? Did you add anything else to the mixture?

Hydrogen peroxide is used in the production of epoxidized vegetable oils. However, the reaction is performed for prolonged times at elevated temperatures with concentrated hydrogen peroxide in the presence of a catalyst.

I am guessing that you just poured the two together at room temp, then soon after used the mixture to wash your hands. Compared to the commercial reaction, you are missing time, heat, and a catalyst, so it is very unlikely that any reaction occurred.

  My guess would be, since Coconut Oil is mainly lauric acid, small quantities of peroxylauric acid may have been produced?

Oils are composed of triglyceride esters of fatty acids, not fatty acids themselves. So you would not see any peroxylauric acid.

  It definitely felt less greasy than pure coconut oil, left my hands feeling clean and soft :)

It felt less greasy because it was less greasy: you added water to it (3% hydrogen peroxide is $\ce{H2O2}$ in water). You probably made an emulsion of oil and water and used that to wash your hands (kind of like using soap).


The available energy levels are arranged in the following order, going from the lowest level to the highest :
$1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p,$  etc. So the $4s$ level is filled before the $3d$.


What do you mean by "periodic" in the periodic table?

First you need to understand what a periodic function is in mathematics. For example, if you are standing at any point P on a circle, you start walking on the circumference and count the degrees you completed. Once you complete 360 degrees, and you will find yourself on exactly the same point P. Sines, cosines and many other functions are periodic. This term existed before the chemists started using the term periodic law in the early 1800s or perhaps well before that. If $f(x)$ is a function with period $T$, then $f(x+T)= f(x)$ for a periodic function in a defined range of $[a,b]$. The period $T$ may change to $T'$, when x is in a different range say, $[c,d]$. This is  an example of a discontinuous periodic function. As you will see, periodic properties of the elements form a discontinuous periodic pattern in a loose sense.

Now come to the periodic law: "The law that the chemical elements, when listed in order of their atomic numbers (originally, atomic weights), fall into recurring groups, so that elements with similar properties occur at regular intervals." [unabridged OED].

Instead of $x$ for a mathematical function, the independent variable is the atomic numbers Z (Zahl for a number in German) for the chemists. So this periodic law has a very loose connection with mathematical periodic function, because none of the values are actually repeated. Each element is unique, but in a loose way, their properties or even a better word is trends "repeat" themselves after a certain number of elements. See this chart for example. Atomic radii vs. Z, do you see a loose periodic pattern?

Once you have got the idea of periodic function in a loose way for the chemical elements, get ready to the see the physicists style of periodic table. It explains why some properties loosely repeat themselves.

Each element in the column, corresponds to filling of a certain orbital, for example look at the column $s^1$. All elements falling in that column correspond to the filling of $s^1$ electrons. Then move to the next column $s^2$, all elements listed there corresponding to filling of $s^2$. Can you see a pattern? In this sense, one calls it periodic function in a modern way, that certain elements repeatedly fall in $s^1$ column, some elements repeatedly fall in $s^2$ column.


SN2 on compound containing halogen and carbonyl

If it attacks the carbonyl carbon, then you form an unstable tetrahedral intermediate with a negative formal charge on the oxygen. Unless you treat the compound with water so that the oxygen can remove it's formal charge, what will happen is that the reaction goes backwards where the CN group is kicked off.
However, if the CN attacks the Cl group, then you can have a substitution where you get a stable compound finally.


We sometimes call this type of complex 'pseudotetrahedral' since there is an isomerism from a tetrahedral to a square planar complex possible. I was unable to find the original work here but this link gives some information. As you already mentioned there are two strong and two weak ligands so it's hard to tell how strong the ligand field splitting will be. For your particular complex it seems to be right on the spot where it would change from one to the other so depending on what you do you can influence the equilibrium. From what I read this may depend on the ability of the solvent to coordinate to the complex as well, the temperature, etc.

This is also mentioned in Earnshaw's Chemistry of the elements

  Planar-tetrahedral equilibria. Compounds
  such as $\ce{[NiBr2(PEtPh2)2]}$ mentioned above as
  well as a number of sec-alkylsalicylaldiminato
  derivatives (i.e. Me in Fig. 27.6b replaced by
  a sec-alkyl group) dissolve in non-coordinating
  solvents such as chloroform or toluene to give
  solutions whose spectra and magnetic properties
  are temperature-dependent and indicate the presence
  of an equilibrium mixture of diamagnetic
  planar and paramagnetic tetrahedral molecules.


Predicting changes in pH of water when chlorinating using two different methods

In aqueous solution, chlorine can undergo disproportionation: one atom in $\ce{Cl2}$ (oxidation state $0$) is reduced to $\ce{Cl-}$ (oxidation state $-1),$ the other is oxidized to a species in which the chlorine atom has the oxidation state $+1.$

$$\ce{Cl2 + 3H2O -> 2H3O+ + Cl- + OCl-}$$

Ask yourself how that influences the pH of the solution.

What about adding calcium hypochlorite to water? If $\ce{CaCl2}$ is the calcium salt of hydrochloric acid, what about $\ce{Ca(OCl)2}?$

 You may have guessed it: $\ce{Ca(OCl)2}$ is the calcium salt of chloric(I) acid.

You may remember that adding the soluble salt $\ce{MA}$ of a strong acid $\ce{HA}$ to water will not influence the pH of the solution.

 If $\ce{HA}$ is a strong acid, it will fully dissociate in water:
 $$\ce{HA + H2O <=>> A- + H3O+}$$
 This means in return that the corresponding base $\ce{A-}$ is very weak:
 $$\ce{A- + H2O <<=> HA +OH-}$$

Now, is $\ce{OCl-}$ a strong or a weak base? In order to decide, you might want to have a look up the $\mathrm{p}K_\mathrm{a}$ of the corresponding acid and keep in mind that for a corresponding acid-base pair in water $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14.$


Why is the 2nd ionization energy of chloride similar to that of sulfur?

The second ionization of chlorine leaves $[\text{Ne}]3s^23p^3$ where the $3p$ subshell is half-filled with one electron per orbital and gives a favorable exchange energy.  With sulfur this relatively favorable configuration is achieved instead after the first ionization, the second ionization of sulfur breaks the half-subshell and is thus less favorable.

The effect of the half-filled subshell can be seen elsewhere, as well.  The chart below (reference) shows trends in first ionization energies.  Counting backwards from the third peak (argon) we see that phosphorus exceeds sulfur violating the usual rules based on effective nuclear charge.  The first ionization of sulfur gives the half-filled $3p$ subshell after removing the electron, whereas the first ionization of phosphorus would have to break that configuration.  The half-filled shell effect beats effective nuclear charge here.


Carbon has four spare electrons. In graphite structure, two of them make two single (sigma) bonds while the other pair form a double bond (consists of one sigma and one pi bond). Electron that forms pi bond is freer to travel between atoms compared to the others and is theoretically associated with all adjacent atoms (imagine the 3d structure of graphite) rather than a single atom. Because of that, it is called delocalized electron. In other words, electrons that form pi bonds continously replace each other causing temporary, local crystallizations within the structure. According to Wikipedia "It is increasingly appreciated that electrons in sigma bonding levels are also delocalized". But it is not the subject we are discussing here.

Electric current is basically the flow of charged particles (ions and electrons, but electric currents in everyday life can simply be defined as flow of electrons). If graphite is exposed to an electric current in a circuit, "the pi electrons" which are already traveling between carbon atoms that forms the graphite structure begin to flow through the circuit and leave the structure from one point of connection between graphite and the circuit. At the same time, other electrons come via the other connection point to replace them as new delocalized electrons. So, the new delocalized electrons simply come from the current.


How are Grignard reagents stable in an ethereal solution?

We tend to assume that acid-base reactions are instantaneous. But they need to go through an activated complex of higher energy, like every other reaction.

These reactions are usually done at - 78°C. As long as the energy provided to the system is below that which is required to reach the activated complex there should be no reaction.

I know for sure that tetrahydrofuran is fully deprononated by nBuLi at room temperature. The color turns deep yellow, if I recall correctly.


Mechanism for synthesis of Meldrum's Acid

The first step (reaction with acetic anhydride) creates a mixed anhydride, this is a good electrophile with acetate as a leaving group. This is nucleophilically attacked by the oxygen of acetone (drawn as its enol in the scheme, but it is easier to visualise it as acetone) to give a cationic intermediate which is immediately attacked by the -OH of the neighbouring carboxyl group to give the product.


Is the reaction thermodynamically favored or not?

Your reasoning is correct but there is more to it.

Look at it from the point of view of Gibbs Free Energy,

$\Delta G = \Delta H - T \Delta S$

For a given reaction if:

$\Delta G >0$, the reaction is nonspontaneous in the forward direction, not thermodynamically favourable

$\Delta G < 0 $, the reaction is spontaneous in the forward direction, thermodynamically favourable

If we substitute the values for $\Delta H$ and $\Delta S$:

$\Delta G=(-128\ \mathrm{kJ/mol}) - T\left(-0.409\ \mathrm{kJ/(mol \cdot K)}\right)$

Mathematically, $\Delta G$ will become positive only when $T$ is greater than 313K. Because of this,  the reaction is thermodynamically favorable at any temperature less than 313K (since $\Delta G$ will be negative), but is not thermodynamically favorable at any temperature greater than 313K (since $\Delta G$ will be positive).


Enzymes are very, very large in comparison to the typical molecules you encounter in chemistry (excluding any polymers). A 10 kD protein is considered to be small, and there are proteins of several hundred kD.

If you remove all the water from a typical protein you end up with a solid. But for most enzymes this would be an irreversible step, and they would lose their activity.

Usually enzymes are kept in aqueous solution, often with some glycerol or other cryoprotectants, and frozen.

Strictly speaking, they would probably be solid, but likely inactive in that form so I'm not sure if it counts. The relevant form of an enzyme is dissolved in water.


Converting sulfate ion to sulfur trioxide?

A good analogy to your proposed reaction is the formation of carbon dioxide above a solution of carbonate salts.  Like sulfur trioxide, carbon dioxide is an acid anhydride.  The acid for sulfur trioxide is sulfuric acid, $\ce{H2SO4}$.  The acid for carbon dioxide is carbonic acid, $\ce{H2CO3}$.

Dry carbonate salts like sodium carbonate can be made to release $\ce{CO2}$ by heating to high temperatures.  For example, calcium carbonate decomposes to calcium oxide and $\ce{CO2}$ at around 800 °C to 900 °C.  Since such high temperatures are required, this technique does not work with aqueous solutions at ambient temperatures.   (The water would boil off way before gas is released.)

Thus, sulfur trioxide evaporation from sulfate salts would be even more disfavored at room temperature than carbon dioxide evaporation from carbonate salt solutions, because of sulfur trioxide's affinity for water is much higher than $\ce{CO2}$'s.

Also, the reaction mechanism you have drawn is not a dehydration, but a reduction of the sulfate.  That would not form sulfur trioxide, but a reduced sulfur compound, such as sulfite.  Sulfur dioxide gas evaporation from above sulfite salts is probably more favored than sulfur trioxide evaporation, but that isn't saying much. Either one is still highly disfavored under neutral or alkaline pH.


How much electricity could be generated by a person's stomach acid?

Well a battery works on the principle of dissolving one of the electrodes with the electrolyte. This means that you would have a solution of the metal and electrolyte in your stomach. If consumed in large quantities or in quick succession, then you most probably would notice some physiological effects, however, I would assume they'd be very minor given the mass of the tablet in question. Other than under those circumstances, I would expect the physiological effect to be non-existant. Given this has been approved by the FDA, you can be pretty confident that it's safe.


Vapour density dependence on temperature

For ideal gas vapour behaviour, the vapour density is the ratio of its density and of the density of hydrogen at the same conditions
$$\frac { \frac {pM}{RT}}{ \frac {pM_{\ce{H2}}}{RT}} \simeq \frac M2 $$.
There for assumption of vapour ideal gas begaviour, "vapour density" does not depend either on temperature either on pressure.

I see a clash of definitions, or rather their terminology for the relative definition of vapour density versus the absolute density definition with the unit kg/m3 in sense of "density of vapour" or substance density.
It is easy not to be aware of "vapour density" is not "density of vapour". In fact, lack of context gives the reader the confusing choice if vapour is an adjective to density or a part of the term.


The "strength" of a bond can't be evaluated just based on its enthalpy.

According to this relation

$$ \Delta H = \sum \Delta H_\text{bonds broken} - \sum \Delta H_\text{bonds created} $$
the strength of your bond will depend on the reactive nearby the bond. If the molecule is likely to create bonds with higher enthalpy after the bond's breakdown, then the enthalpy variation will be negative. The process will then be spontaneous and the initial bond will breaks "easily".
There is also some steric consideration, the hydrogen atom is smaller, the bond is then more easy to be "attacked".
Finally, for a polar bond, the electronic cloud is delocalised closer to one atom rather than the other (according to their electronegativity) so you have "less perturbation to add" to fully remove the covalent electron from the less electronegative atom.


There is a misconception here. A p orbital is a 3D-function, and these functions don't have shapes, they have values at any point in space. If you describe an electron distribution with one of these functions, you can plot contours at a chosen value, and these contours have shapes. For example, you could choose a contour level such that an electron has a 90% chance of being inside the shape defined by the contour level.

[OP...] What confuses me is that there are 2 shapes of p orbitals.

There are actually four common graphs depicting p orbitals, and a lot of conceptual drawings that don't have an exact mathematical shape. The graphs differ in whether they show the wave function or the square of it (the latter is proportional to the probability of finding the electron in that location). They also differ in whether they use a cartesian coordinate system, or plot the angular dependence in a polar plot.
The one you probably want is a contour plot of the probability density (you could also say the electron density). Here is an example from an article in the Journal of Chemical Education (center panel):

The shape of the contour lines of the wavefunction are the same, but you can have positive and negative values (and because of the squaring involved, the contour lines also have different values). Your "shape 2" is a contour plot of the wavefunction, with two contours of equal magnitude and opposite sign.
Polar plots
The other type of depiction is a polar plot, where the distance from the origin is the value of a function. This is different from a Cartesian coordinate system, where the nucleus is at the origin, and the distance from the origin corresponds to the distance from the nucleus.
If you plot the angular component of the wavefunction (you can think of this as keeping the distance from the nucleus constant, and sampling different angles), you get what looks like two spheres. A polar plot has no good way of dealing with negative values, so often one part of the curve is shown in a different color.

If you plot the square of the angular component, you get a shape closer to your "shape 1". For technical reasons, the plots are rotated by 90 degrees (the angle zero is defined along the x-axis in Wolfram Alpha, which I used to plot these).

So what does your "shape 1" represent? The left-most panel looks like it wants to show an electron density, but it is very inaccurate. There are other artists' interpretations floating around, some that look like they are actual graphs of functions, such as this one:

It is not clear which function, if any, this shows a graph of.

[OP...] The true shape of p orbitals

There is none. You have a choice of representations, and it depends on the context which one, if any, makes sense.


Electrolysis or electrolytic refining is a technique used for extraction as well as purification of metals obtained by other refining methods. In the electrolytic refining process, a block or a strip of impure metal is used as the anode, and a thin sheet of pure metal is used as the cathode. Both the cathode and anode are dipped in an electrolytic cell containing an aqueous solution of the metal salt. Upon the application of direct electric current of a suitable voltage, pure metal is deposited at the cathode (reduction) by the dissolution of impure metal at the anode (oxidation).

Since electrolytic refining id done in aqueous medium, the key to successful refining is the oxidation of the metal should be more facile than the oxidation of water (see the standard oxidation potentials below). Hence the solid metal would dissolve into the solution as cataions, leaving behind many of the impurities, which are inactive to electrolysis. The possible anodic oxidation reactions would be:

$$\ce{Ag <=> Ag+ + e-} \quad E_\mathrm{ox}^\circ = \pu{-0.7996 V} \tag1$$
$$\ce{4H2O <=> O2 + 4H+ + 4e- } \quad E_\mathrm{ox}^\circ = \pu{-1.229 V} \tag2$$

When value of $E_\mathrm{ox}^\circ \ (\ce{Ag})$ is compared with that of $E_\mathrm{ox}^\circ \ (\ce{H2O})$, it is safe to say that $\ce{Ag}$ oxidation at anode is more spontaneous (less negative value) than that of $\ce{H2O}$ (greater negative value) at anode. As a result, impure $\ce{Ag}$ anode will be slowly dissolved into solution as $\ce{Ag+}$ ins as direct current applied.

Similarly, the possible cathodic reduction reactions would be:

$$\ce{Ag+ + e- <=> Ag}  \quad E_\mathrm{red}^\circ = \pu{0.7996 V} \tag3$$
$$\ce{ 2H2O + 2e- <=> H2 + 2OH-} \quad E_\mathrm{red}^\circ = \pu{-0.8277 V} \tag4$$

Since the refining solution is dilute $\ce{HNO3}$ to provide the counter-ion for $\ce{Ag+}$ cation, we should consider possibility of  another reduction reaction at cathode:

$$\ce{NO3- + 3H+ + 2e- <=> HNO2 + H2O} \quad E_\mathrm{red}^\circ = \pu{0.934 V} \tag5$$
$$\text{And, } \ce{NO3- + 4H+ + 3e- <=> NO + 2H2O} \quad E_\mathrm{red}^\circ = \pu{0.957 V} \tag6$$

When the value of $E_\mathrm{red}^\circ \ (\ce{Ag})$ is compared with that of $E_\mathrm{red}^\circ \ (\ce{H2O})$, it is clear that $\ce{Ag}$ reduction at cathode is much more prominent (high positive value) than that of $\ce{H2O}$ (greater negative value) at cathode. However, the competition reduction would happen as well when comparing the value of $E_\mathrm{red}^\circ \ (\ce{Ag})$ with those of $E_\mathrm{red}^\circ \ (\ce{NO3-})$, both of which  are slightly higher positive values. However, if you use appropriately dilute nitric acid to begin with (c.f., Nernst equation), you will be able to avoid nitrate reduction at cathode.


If there is more than one branch in a hydrocarbon, which branch do you name first?

Branches must be numbered in alphabetical order, along with their positions.
Suppose you have a chloro group on the first carbon, a bromo group on the second carbon and an amino group on the fourth carbon of your carbon chain (butane).
One must therefore go in alphabetical order which is amino, then bromo and then chloro.
Therefore you naming will be something like this:-

$$\ce{3-Bromo-4-chloro-butanamine}$$

The naming is in alphabetical order but the numbering starts according to the following rule:

$$\ce{RCOOH>RCOOR>RCOCOOR>RCOCl>RCONH2>RCHO>CO>OH>NH2>double bond>triple bond>halogen>NO2>ROR>alkyl}$$


Expansion work by a gas when pressure is not explicitly constant?

You can calculate the work done and Heat exchange if you assume that the process is quassi-statically performed and it is reversible and most importantly done in a single step.

Assume that the process through which the $\ce{He}$ is expanded obeys the equation, $PV^x = K$(constant). This equation can also be alternatively stated as, $TV^{x-1} = K$. Now, suppose the gas is transformed to the state $(T_2,V_2)$ from the state $(T_1,V_1)$. So, we can thus write, $$T_2V_2^{x-1} = T_1V_1^{x-1}$$ so, putting the given conditions, we can find $x$ as, $$x= 1+ \frac{\log\left(\frac{T_2}{T_1}\right)}{\log\left(\frac{V_1}{V_2}\right)} \approx 0.676$$ Thus, after knowing $x$ ,you can calculate the work done by the gas as,  $$W= \int_{V_1}^{V_2} P\mathrm{d}V = \int_{V_1}^{V_2} \frac{K}{V^x}\mathrm{d}V = \frac{P_2V_2 - P_1V_1}{1-x} = \frac{R(T_2-T_1)}{1-x}$$ We know all the value of $T_1, T_2 $ & $x$, and work done can be easily calculated. and we already know the change in Internal energy  and thus heat exchange can be easily calculated and also $\Delta H $ can be calculated   ($Q = (C_V + \frac{R}{1-x}) \Delta T$ and $\Delta H = C_P \Delta T $) and thus all the quantitites can be known.