How do I approach drawing all of the possible constitutional isomers?

I approach questions like this by considering all possibilities of arrangements of the non-hydrogen atoms. Obviously, there's C-C-O, C-O-C, and then the cyclic ethylene oxide.

Then I make sure I've matched bond orders to get the hydrogen count.

So I get vinyl alcohol, acetaldehyde, and ethylene oxide. H2C-O-CH2 would only work for a biradical, which is probably not expected.

The systematic fashion then is to arrange the "heavy" atoms, first in all linear forms, then all cyclic forms, then all two-ring forms, etc. Of course with large formulas, the number of isomers goes up exponentially.


Comparing explosive properties of mercury(II) cyanate and mercury(II) fulminate

This can be attributed to the formal charges of fulminate ion and cyanate ion. Structural stability of a compound sometimes depend on formal charge.

We observe that formal charges of fulminate ion is higher than cyanate ion for which fulminate ions are said to have a less favorable molecular configurations which leads to facile decomposition of the fulminate ion.
Fulminate vs cyanate stability has been demonstrated by using their corresponding silver compounds. Here is a video demonstration.


What is the molecular structure of polymetatelluric acid?

It would be ideal to check out x-ray analysis data, but it seems like polymetatelluric acid is a rather amorphous substance and does not form suitable crystals. As of now, I haven't discovered anything relevant neither in ICSD, nor in CCDC.

Another method of structural analysis would be vibrational spectroscopy. There is an article "Ultrarotspektren von Tellursäuren, Telluraten und Antimonaten" [1, p. 165] (in German) containing an extensive research of various $\ce{Te}$-containing substances performed with infrared and Raman-spectroscopy (I reduced the table to the frequencies for polymetatelluric acid exclusively):

  \begin{array}{ll}
\hline
\ce{(H2TeO4)_x} &\text{Zuordnung} \\
\hline
450~\mathrm{(m, b)} & \delta~(\ce{TeO}) \\
\hline
600~\mathrm{(st, b, Sch)} & \nu~(\ce{TeO}) \\
720~\mathrm{(sst, b)} & \\
600~\mathrm{(sst, b, Sch)} & \\
\hline
1085~\mathrm{(mst, b)} & \delta~(\ce{TeOH}) \\
\hline
1618~\mathrm{(s, b)} & \delta~(\ce{H2O}) \\
\hline
2360~\mathrm{(m)} & 2\delta~(\ce{TeOH}) \\
\hline
3200~\mathrm{(sst, b)} & \nu~(\ce{TeOH}) \\
3360~\mathrm{(sst, b, Sch)} & \nu~(\ce{H2O}) \\
\hline
\end{array}

  Polymetatellursäure $\ce{(H2TeO4)_x}$ ist eine feste amorphe Substanz. In ihrem UR-Spektrum sind alle Banden sehr breit, wie man es bei amorphen Substanzen meist antrifft. Das Tellur ist in dieser Verbindung wahrscheinlich wie im $\ce{Te(OH)6}$ sechsfach koordiniert, wie die annähernd gleiche Lage der $\ce{TeO}$-Valenz- und Deformationsschwingungen in beiden Verbindungen zeigt. Hieraus folgt, daß $\ce{TeOTe}$-Bindungen vorhanden sein müssen. Bei Viererkoordination würden die Valenzschwingungen höher liegen, wie es in den vergleichbaren Molekeln $\ce{IO}$, und $\ce{H5IO6}$ der Fall ist.[...]

  Im Gegensatz zum $\ce{Te(OH)6}$ tritt im Spektrum des $\ce{(H2TeO4)_x}$ eine Frequenz bei $\pu{1618 cm-1}$ auf, die nur von freiem Wasser herrühren kann. Ebenso beobachtet man neben der $\ce{OH}$-Valenzschwingung der $\ce{TeOH}$-Gruppen ($\pu{3200 cm-1}$) die Bande $\pu{3360 cm-1}$ des freien Wassers. Aus den Intensitäten dieser Banden ist zu entnehmen, daß $\ce{H2O}$ und $\ce{TeOH}$-Gruppen in größenordnungsmäßig gleicher Menge vorhanden sind. Eine stöchiometrische Konstitutionsformel wird man dieser Substanz wohl nicht zuordnen können.

Briefly, the IR-spectra bands of $\ce{[H2TeO4}]_x$ bands are very broad, suggesting amorphous non-stoichiometric compound (as Mithoron suggested) with polymeric structure where $\ce{Te}$ atom is 6-coordinated. Bands were only assigned for $\ce{Te-O}$, $\ce{Te-OH}$ bonds and $\ce{H2O}$. IR-spectra do not show the presence of $\ce{Te-Te}$ bonds. Most likely there are $\ce{[TeO]6}$ units with edge- or corner-sharing, linked together in a chain-alike fashion.

Reference

Siebert, H. Z. anorg. allg. Chem. 1959, 301 (3–4), 161–170. DOI: 10.1002/zaac.19593010305 (in German).


Is center of symmetry only valid for symmetric substituents?

You assume too much, starting with the words "The given compound". There is no given compound. The picture corresponds to quite a few different compounds, and here is why.

As I told you before on a remotely similar question, the biphenyl fragment totally can't be planar. The xylene substituents must be rotated so their ortho methyl group points either up or down relative to the plane of the central ring. Since there are two of them, we have three options: "both up", "both down", and "one up, one down". The first two compounds are indeed chiral, and enantiomers of each other. The third one is centrosymmetric, just as you've reasoned.


Lets see this step-by-step.

Bromocyclohexane is reacted with alcoholic KOH. This, as you would know, results in the formation of an alkene by elimination reaction. Product P would be Cyclohexene.

Cyclohexene is heated with KMnO4. You might think a cis-1,2-cyclohexanediol would form, and indeed it does. But vigorous oxidation provided by the heat cleaves the C-C bond between the two hydroxy groups and essentially converts into 1,6-hexanedioic acid (product Q, a.k.a adipic acid).

Calcium Oxide, being a pretty strong base,  reacts with the dicarboxylic acid and forms a saltcalcium adipate (product R)

Heating a calcium salt of any carboxylic acid results in the elimination of a $\ce{CaCO3}$ molecule, and forms a ketone. If you take calcium adipate, upon heating would give cyclopentanone(product S).

Now, cyclopentanone, just like any other carbonyl compound with α-hydrogens, undergoes an aldol reaction, heating which would result in the formation of the compound given in option (B).

Here is the complete reaction mechanisms:


What elements in Tomato and Spinach make them cook quickly when cooked together?

The table salt is probably irrelevant to why the veges cook down quickly, although it does cause veges to sweat, it may leach the moisture out of the tomato and spinach faster than if you didn't add it but I think heat would be the main variable. Also do you stir the food more when you add the salt than if you didn't?

Most organisms have a ratio of about 《80% water, spinach and tomatoe may have a high water content and also contain oils and other volatile components that are released when heated so when you heat / cook them the water and volatile components evaporate and the solid mass(other 20% of initial weight) is left behind.

They are soft veges too so the material can be sweated easier than say a carrot and will lose its structure and water content at a faster rate.

Some oxidation could be occurring along with other chemical reactions between components such as decarboxylation, hydrolysis, hydration, dehydration and many more depending on what compounds are present in the process, the kind of environment (eg. aqueous) and the techniques used (eg. frying, baking).


Yes, we can definitely see atoms in modern microscopes.  I'll try not to get deep into the subject of "seeing", but I will note that most people say they are seeing when they use their eyes to observe objects.  That is, they are using an optical system (eye) consisting of a lens and a detector (retina) and some image processing carried out by a computer (brain).  Thus, any instrument which uses some sort of detector and a computer to turn the signals of that detector into an image we feel comfortable looking at qualifies to me as "seeing in a microscope."  If you're comfortable with that then read on.

My favorite microscope which is capable of seeing atoms is the transmission electron microscope (TEM).  Using a high quality TEM, we can easily resolve objects around an angstrom in size, which is the distance between atoms.  To image a single atom, you also need to have a sample which is only a single atom thick at some point.  This is tough to accomplish but can be done.  Today, images of graphene provide arguably the best "single atom" TEM images.  A seminal paper is here (paywall).  You can also search the internet for graphene images such as this.

In the links above, the images show the shapes of atoms as little blobs, and to a large extent the blobs are more a description of the microscope's ability to focus rather than the shape of the atom itself.  However, the atomic force microscope (AFM) excels at seeing the electrons at the surface of an atom.  With AFM, researchers are even able to image the chemical bonds of molecules sitting on a substrate.  In this case, the image shows the bonds as rods and the atom centers are not really visible.  Chemistry textbooks often show the "shapes" of atoms by drawing the electron orbitals, so in a sense these rods are the shapes of these atoms too.

This just points up that the "shape" of atoms is a non-trivial subject, which is very dependent on how you look at them.


Is it possible to use diatomic carbon to utilize a new type of fuel that takes away the waste product of carbon dioxide being released into the air?

No doubt, diatomic carbon does exist - in blue flames and in comets.

The characteristic uv absorption pattern (Swan bands) has already been described almost 100 years ago.

At low temperature (77 K) it reacts with hydrocarbons by H-abstraction to form $\ce{HC#CH}$, and the addition to $\ce{C=C}$ double bonds has been reported too.

But I can't figure out any reasonable way for a reaction with carbon dioxide.


Why are nitro compounds with at least one alpha-H soluble in NaOH?

The compounds containing nitro groups when dissolved in basic medium, the following things can happen.The base can take the $\ce{$\alpha$-H}$ of the nitro-group, which will create a negative charge on that $\ce{$\alpha$-C}$, which can further delocalise with the Nitro-group (similar to Nitro-Acinitro tautomerism).
So, the whole compound will be in the state of a Resonance Hybrid where there will be partial double bond character between $\ce{$\alpha$-C, N }$ and two $\ce O$ atoms. and also there will be a $\delta$- charge on the $\ce{$\alpha$-C }$ atom as well as on the $O$ atoms. , and $\delta$+ charge on the $N$ atoms.Due to such salt like compound formation, it becomes soluble. This $\delta$- charge created on the Carbon atom is stable if the side chain is short, but starts to get destabilised as the side chain grows longer. So, I think the solubilities of the Nitro-compounds decrease as the side chain grows longer, and the compound with $10-12$ C atoms may not be at all soluble in base because of such large Hydrophobic part and destabilisation of the $\delta$- charge.


I went through the derivation myself and I can reproduce the equation you got. But after using $C = [\ce{A}]_0$ and some rearrangements I get the following formula

\begin{equation}
\ce{[A]} = \frac{k_{-1}}{k_{1} + k_{-1}} \left( \ce{[A]_{0}} + \ce{[B]_{0}} \right) + \frac{\mathrm{e}^{-(k_{1} + k_{-1})t} }{k_{1} + k_{-1}} \left( \ce{[A]_{0}} k_{1} - \ce{[B]_{0}} k_{-1} \right)
\end{equation}

and not the one your professor got. But your professor's formula might be incorrect because for $t=0$ his formula yields the incorrect result $\ce{[A]}_{t = 0} = \ce{[A]_{0}} + \ce{[B]_{0}}$ instead of the correct result $\ce{[A]}_{t =  0} = \ce{[A]_{0}}$, so it doesn't satisfy the given boundary conditions.


Protons and neutrons are both baryons, subatomic particles composed of three quarks. Quarks are smaller fermionic particles that make up protons, neutrons, and other hadrons.

Electrons are leptons, fermions that are not composed of quarks.

Quarks have mass, spin, charge, and (confusingly) color charge. Color charge is a quantum property that quarks have, and all hadrons need to have a net zero color charge.

The proton is composed of three quarks: two up quarks and one down quark. Each up quark has a spin of $+\frac{1}{2}$, a mass of $2.3~\text{MeV/c}^2$, and a charge of $+\frac{2}{3}$. Each down quark has a spin of $-\frac{1}{2}$, a mass of $4.8~\text{MeV/c}^2$, and a charge of $-\frac{1}{3}$. The total charge of the proton is $2(+\frac{2}{3})+(-\frac{1}{3})=+1$. The total spin of the proton is $2(+\frac{1}{2})+(-\frac{1}{2})=+\frac{1}{2}$. The masses of the quarks do not add up because most of the proton's mass is in its confinement energy.

We can do the same analysis for a neutron, which is composed of one up quark and two down quarks. The charge of the neutron is $(+\frac{2}{3})+2(-\frac{1}{3})=0$. The total spin of a neutron is $(\frac{1}{2})+2(-\frac{1}{2})=-\frac{1}{2}$. Again the masses do not add up.

Protons and neutrons are different because the are composed of different quarks. However... enough physics. Now for the nuclear chemistry.

  Interconversion of protons, neutrons, and electrons

The proton is a stable particle with a lifetime at least $2.1\times 10^{29}$ years.

However, protons can be converted into neutrons (and electron neutrinos) by electron capture:

$$\ce{p+ + e- -> n +\nu}_e$$

This requires an investment of energy because the rest mass of the neutron is higher than the rest mass of a proton by $1.29~\text{MeV/c}^2$.

Free neutrons are not stable, undergoing beta decay to a proton, electron, and electron antineutrino with a halflife of about 10 minutes.

$$\ce{n -> p+ + e- +\bar{\nu}}_e$$

Most importantly, and fortunately, most neutrons bound in nuclei are stable. The protons and neutrons in a nucleus can be treated quantum mechanically analogously to electrons from several atoms bound in a molecule. There are not a collection of independent particles that do not interact. The nucleons occupy quantum mechanical energy levels. Neutrons can only decay if the proton it would become can occupy a lower energy level, which is usually not possible except for those radioisotopes which tend to undergo beta decay, thanks to the Pauli Exclusion Principle! Thus, the bound neutron is stabilized.


How significant is air decomposition on formation of nitrogen oxides?

In situations where NOx originiates form $\ce{N2}$, such as lightning or a high temperature furnace, the Zeldovich mechanism dominates:

$$\ce{O2 -> O +O}$$

$$\ce{O + N2 -> NO +N}$$

$$\ce{O2 + N-> NO +O}$$

If a combusting fuel actual contains nitrogen, other mechanisms can be involved.


I suspect that the ethanoic acid contains a residual amount of water and that the concentrated sulphuric acid is reacting with the water. Since this is a very exothermic reaction the test tube gets heated up which evaporates some of the ethanoic acid, causing the bung to pop off. If it's just slight evaporation this would also explain why you couldn't see any effervescence.


Quasi-static implies that T aprox equals Tsurr?

In my judgment, when they are talking about irreversible cycles, they are referring to idealized cases in which (a) the process is internally reversible with respect to the working fluid but (b) irreversible with regard to the region outside the working fluid where there is a finite thermal resistance (e.g., a cylinder wall with finite thermal conductivity) between an (external) ideal constant-temperature reservoir and the working fluid.  So the rate of heat transfer to the working fluid is very low (quasi static), and all the entropy generation takes place within the (higher) thermal resistance wall.  At least, this is how the irreversibility is often modeled.

To illustrate all this, see the following thread from Physics Stack Exchange:  https://physics.stackexchange.com/questions/443527/entropy-generation-in-an-endoreversible-engine


Using electrolysis of water as a proton generator

In a chemical sense, "anode" of an electrolytic cell is the most powerful oxidizing reagent known, so much so that it oxidize F(-) to elemental fluorine. Don't forget that the electron balance in an electrolytic cell is always maintained so the gram-equivalents of x reduced at the cathode must equal gram equivalents of y oxidized at the anode. As a result, half cell reactions cannot occur alone. So something is being oxidized at the anode, this is your nickel electrode. Platinum is the best one but it is too expensive. Graphite electrode might be an economical substitute. Secondly, you have to add an electrolyte to the anode chamber as well because tap water resistance is pretty high. You can add a non-corrosive electrolyte such as sodium sulfate. With these changes, hopefully you will see the desired trend.


Comparing equilibrium constants between acetone and benzaldehyde in nucleophilic addition reaction

The information given by your teacher is incorrect. The $K_\mathrm{eq}$ value of acetone for cyanohydrin formation (a nucleophilic addition reaction) is less than the $K_\mathrm{eq}$ shown by benzaldehyde
According to Organic Chemistry: Structure, Mechanism, Synthesis [1 , p. 595], the values of $K_\mathrm{eq}$ for cyanohydrin formation are as follows:
$$
\begin{array}{lr} \hline
\mathrm{Compound} & K_\mathrm{eq} \\ \hline
\text{Acetone} & 20 \\
\text{Ethanal} & 10^5 \\
\text{Benzaldehyde} & 210 \\
\textit{p}\text{-Methoxybenzaldehyde} & 30 \\
\text{Acetophenone} & 0.8 \\ \hline
\end{array}
$$
Hence the statement that the $K_\mathrm{eq}$ is greater for acetone is incorrect. This seems accurate considering that acetone is more sterically hindered than benzaldehyde, which is a dominating factor for nucleophilic addition reaction since they are kinetically stable reactions.
This document has a much larger cases of compounds that are compared for multiple reactions, in all of which benzaldehyde is more reactive than acetone.
However, there is one minor issue with your reasoning given in the question.

Benzaldehyde does not have a −M group in it. However if you add another functional group into the ring, then we say that the $\ce{-CHO}$ group acts as a −M group. Actually, $\ce{-CHO}$ has a +R effect from the phenyl ring attached to it which makes the ring in benzaldehyde deactivated. So, the carbon in benzaldehyde is less electrophilic than formaldehyde which has no such effects.
Reference

Ouellette, R. J.; Rawn, J. D. Organic Chemistry: Structure, Mechanism, and Synthesis, 1st ed.; Elsevier, 2014. ISBN: 978-0-12-800780-8 (Google Books Preview)


Difference between coordination entity and coordination sphere

Coordination entity is the part of complex compound having central atom surrounded with the number of secondary linkage.
But,if this entity is written along with charge on it ,then it is called coordination sphere
For example  [Fe(CN)6]4- is a coordination sphere while Fe(CN)6 is the coordination entity


Chemical Biology - synthesizing/designing siRNA

At this time, it it possible to completely chemically synthesize oligonucleotides (DNA/RNA) between 30-50 base pairs. siRNAs are a class of RNAs that, as the name suggests, bind to other RNAs in a sequence specific manner. Specifically, siRNAs preferentially target mRNAs, which blocks translation of those mRNAs into proteins. This regulation of gene expression has absolutely nothing to do with "creating genes for a human being".


Does the amplitude of the RF pulse affect the NMR signal?

V. approx explanation. The rf pulse is at 90$^\text{o}$ to the static field $B_0$ and has the effect of moving the magnetisation (sum of spin vectors) away from alignment along $B_0$ (z-axis normally) into the xy plane where it rotates and cuts through detector coils and induces FID signal. The FID can only be detected with a magnetisation component in the xy plane. The size of the rf pulse's magnetic field $B_I$ is proprotional to angle $\theta$ moved away from z-axis as $B_I\sim \tan(\theta) $ and so via magnetisation's component in xy plane (arrow on axis in sketch) to the size of the FID.


I am not entirely sure what purpose the lower sigma bonding orbital in the MO-diagram serves, because an ethane molecular orbital would, as you have surmised correctly, have six degenerate $\sigma_{\ce{C-H}}$ MOs and a $\sigma_{\ce{C-C}}$ MO, lower in energy.

My best guess would be the position of the $\sigma_{\ce{C-H}}$ on the right hand is quite arbitrary and the basic purpose of the diagram is to show the possible overlap between the $\sigma_{\ce{C-H}}$ of one $\ce{C-H}$ bond and the $\sigma^{*}_{\ce{C-H}}$ of $\ce{C-H}$ bond anti to it - you don't really need to be worried about the $\sigma_{\ce{C-H}}$ of that anti bond because it cannot overlap with the other $\sigma^{}_{\ce{C-H}}$.

That brings us to the second part of your query. A $\sigma_{\ce{C-H}}$ can donate electron density to an empty $\sigma^{*}_{\ce{C-H}}$ anti to it, because the latter is oriented correctly (anti) to accept the electron density from it. It's a question of orbital symmetry and not energy difference.


What is meaning of unit (mg-Pt/cm^2) in relation to fuel cells?

The unit in question means milligrams of platinum per square centimeter.

In your application - fuel cells, and specifically the membranes used to fabricate them - it describes how much of the metal is present per square centimeter of membrane.


How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

The IUPAC recommendations on this matter have been changed. In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed:

  (…), the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second criterion.

For example, a preferred IUPAC name is 3-methylidenehexane (not 2-ethylpent-1-ene) since the longer chain now supersedes a shorter unsaturated chain. (This example is given in the Blue Book.)

Besides, the prefix ‘methylene’ is no longer recommended in IUPAC nomenclature to designate the $\ce{=CH2}$ group. The preferred prefix is ‘methylidene’.

Therefore, the correct name for your example is 2-bromo-4-methylidenehexane.


There are two issues here:

First, you will be oxidizing water. You would end up making oxygen in most of those cases, before oxidizing your metal and making the cation you need.

Second, even in cases where your metal will oxidize before water, the concentration of the metal cation in the solution will be low and variable. This will create problems with controlling rates of deposition and uniformity of deposition.

That is why the electroplating baths rely on the already dissolved cation and not the metal as the source. This way you have lots of the cation you need in the solution to work with and you can put some inert counter electrode.


The solubility can be thought of as an equilibrium. The Van 't Hoff Equation describes the relationship between any equilibrium constant and temperature.

$lnK_2 - lnK_1 = \frac{\Delta H }{R} (\frac{1}{T_1}-\frac{1}{T_2})$ . Thus, the graph is not linear.


Gravimetric analysis of barium or sulfate

I think $\ce{BaCl2}$ should be added later. This is because $\ce{BaCl2}$ is soluble in water, but the metal sulphate you will be given may not be soluble. So if you add $\ce{M2SO4}$ later, you may get a wrong result as you have to make sure that the whole of $\ce{BaCl2}$ is used up. On the other hand all chloride salts are soluble. So while filtering the $\ce{MCl}$ will be removed easily along with the solvent and you will get $\ce{BaSO4}$.


Acid (Brønsted–Lowry Acid) is a substance that gives away its $\ce{H+}$, but the environment should be able to accept that $\ce{H+}$. After it accepted it - the acid (which is now its conjugate base) can accept it back. There are 2 extremes:

The environment already has plenty of $\ce{H+}$ already. So even if acid gives its $\ce{H+}$ to the environment, it immediately receives $\ce{H+}$ back. Hence the acid always has its $\ce{H}$ and you see $\ce{HA}$ on the picture.
The environment is basic - it can take lots of $\ce{H+}$ and won't give them back to the acid. Hence acid stops being acid - it gives away $\ce{H+}$ and becomes $\ce{A-}$.

$\mathrm{pH}$ is a measure of how many $\ce{H+}$ there are in the solution. Low $\mathrm{pH}$ means there's plenty of $\ce{H+}$ (our 1st case), high $\mathrm{pH}$ means there's a "shortage" of $\ce{H+}$ (2nd case).
In the middle (where $\ce{H+}$ equals acid's $\mathrm{p}K_\mathrm{a}$ which in your case is 6.5) you'll have a case when 50% of $\ce{H+}$ is taken by the environment and the other 50% of acid molecules are $\ce{HA}$.
When you titrate you do exactly that - you change the $\mathrm{pH}$ of your environment which then changes how many molecules are ionized. And then at some point your indicator changes the color which means you reached some $\mathrm{pH}$ you were looking for.
Titration curves that I've seen have the volume of solvent added (your graph probably shouldn't be called Titration Curve?) - not $\mathrm{pH}$. But these are related since if you're adding a basic titrant to acidic solution you're changing its $\mathrm{pH}$.


Does the shielding effect affect atomic bonding?

Yes, it does. It is due to the poor shielding effect of f-orbital electrons that we see the inert pair effect; that is, $\ce{Pb^2+}$ is more stable than $\ce{Pb^4+}$. Lower oxidation states are more stable as we go down the 13, 14 & 15 group. This affects the number of bonds that an atom can form.


Conformation of six-membered ring constituting a bridged compound

According to structure determined by electron diffraction, norbornane is present in bridged boat conformation. http://pubs.acs.org/doi/abs/10.1021/ja01014a032, see Figure.

Norbornane and its derivatives are often used for studying effect of equatorial and axial substitution of cyclohexane and is considered rigid.
Therefore, I don't expect significant population of the second conformer you sketched.

Edit:
For the [2.2.2]bicyclooctane (not the lactone mentioned in question), both conformers are known in crystalographic database.

Bear in mind, that bicyclooctane constist of three hypothetical cyclohexanes, possibly equivalent by symmetry.

For any compound of interest, you should find enough data in crystallographic database, but I'd expect all reasonable conformations to exist.


Stereochemistry of epoxide hydrolysis under acidic conditions

Standard epoxide opening in acidic conditions would happen at the more
  substituted end

This premise is false. Usually $\ce{S_{N}2}$ is faster than the cation-stabilized pathway (most textbooks don't even bother to explain the subtleties involved), yielding mainly the less substituted (less sterically hindered) product. It is not clear cut which of the two mechanisms will be predominant in an acid catalyzed epoxide cleavage given the reactants and reaction conditions.

See also Regioselectivity of acid-catalyzed ring-opening of epoxides (especially the second answer and then the excerpt from Clayden et al.).


Nickel plating with nickel acetate, nickel chloride, sulfur, saccharin and coumarin

The formula calls for nickel sulfate; is it possible to simply add the powdered sulfur to the solution, or do I have to use a different process to synthesize the nickel sulfate?

This will not work. Sulfur ($\ce{S}$) and sulfate ($\ce{SO4^{2-}}$) are two pairs of shoes. The oxidation state is different and there is no way that sulfur powder will turn into sulfate during the electroplating.

  If so, can I do so with the materials that I have?

It might be possible to heat another nickel salt in sulfuric acid, cool down and wait for the nickel sulfate to precipitate, but I wouldn't do it!

Given that the nickel salts are highly toxic, might cause cancer on inhalation and are harmful for aquatic organisms, I wouldn't mess around with crystallisation and the remaining solution for which you would have to find a safe way for disposal.

It might be much easier to aquire nickel sulfate and set up the bath for electroplating according to the given specifications.


How to acurately measure pH in heterogeneous solution?

The $\mathrm{pK}_{a3}$ of phosphoric acid is 12.319 according to Wikipedia. So at a pH of 11 most of the phosphate species would be in the form of $\ce{HPO4^{2-}}$. Thus as $\ce{Ca3(PO4)2}$ precipitates the solution will become more acidic. So an unanswered question here is how fast would equilibrium be established in the solution. Normally once precipitation starts, then equilibrium is established fairly quickly.

Thus it seems likely that equilibrium would depend a lot on how well the solution was stirred, thus swirling the ppt in the solution rather than having it sit at the bottom...

Also the solution is basic enough that $\ce{CO2}$ from the atmosphere will be forming carbonate species which would also slowly make the solution more acidic. (not sure how much drift in pH that you are observing...).


Do the particles of thermometer's bulb attain the same translational, rotational and vibrational energy (vibration of atoms within a particle) as of the particles of the liquid?

No, but they will have the same temperature. Solids don't have any translational energy, the atoms time-averaged positions are constant. If you want to link the temperature to the energy of the particle, it is easiest to start out with a monoatomic ideal gas because here, there is a linear relationship.

Relationship between change in temperature and energy

Although under some ideal circumstances, particles have an energy of $\frac{1}{2} k T$ per degree of freedom, how much energy is needed to raise the temperature of a sample depends on the sample (and the temperature). The heat capacity is a measure of change in energy per change in temperature. So the heat capacity tells you the total energy going into all degrees of freedom when the temperature increases. The splitting up into translational, rotational and vibrational energy depends on the nature of the sample (for example, the physical state) and the temperature.


How does copper rich "weld thru" primer protect iron / steel?

Well there are also many different kinds of steels with different galvanic characteristics are well. For example, all of these metals have a higher reduction potential than copper:
1. Red Brass
2. Stainless steel 347 (active)
3. Molybdenum, Commercial pure
4. Copper-nickel 715
5. Admiralty brass
6. Stainless steel 202 (active)
7. Bronze, Phosphor 534 (B-1)
8. Monel 400
9. Stainless steel 201 (active)
10. Carpenter 20 (active)
11. Stainless steel 321 (active)
12. Stainless steel 316 (active)
13. Stainless steel 309 (active)
14. Stainless steel 17-7PH (passive)
15. Silicone Bronze 655
16. Stainless steel 304 (passive)
17. Stainless steel 301 (passive)
18. Stainless steel 321 (passive)
19. Stainless steel 201 (passive)
20. Stainless steel 286 (passive)
21. Stainless steel 316L (passive)
22. AM355 (active)
23. Stainless steel 202 (passive)
24. Carpenter 20 (passive)
25. AM355 (passive)
26. A286 (passive)
27. Titanium 5A1, 2.5 Sn
28. Titanium 13V, 11Cr, 3Al (annealed)
29. Titanium 6Al, 4V (solution treated and aged)
30. Titanium 6Al, 4V (anneal)
31. Titanium 8Mn
32. Titanium 13V, 11Cr 3Al (solution heat treated and aged)
33. Titanium 75A
34. AM350 (passive)
35. Silver
36. Gold

(source: Corrosion Doctors)
So, as long as you're welding any of those under copper's boiling point of 2,562°C (or 4,644°F, at which point it could just float away from your intended target), then it should be fine for your application.


The periodic table actually looks like this:

...you can see that the first 14 elements in either series don't belong to any group from the main body of the table.

You could say that the lanthanides and actinides form fourteen new 2-element groups, with Lu and Lr belonging to group 3. But IUPAC does not number these groups:


What would happen if someone consumed a Silica Gel Packet?

Essentially you need to understand that the silica gel packet is just $\ce{SiO2}$. These packets are used to absorb moisture (water vapor) to protect goods from spoiling.

To be honest with you here, nothing ever happens when you eat $\ce{SiO2}$. You consume it all the time from food and water. But, the reason these packets are so dangerous is because they might contain impurities such as $\ce{CoCl2}$ (blue color) and $methyl~violet$. These colored products are used to indicate moisture. These products are carcinogenic and are mutatic poisons. In most cases, it will not be fatal immediately, but given time, yes.

Generally, if the silica gel is colored, it is highly likely that they might be very dangerous. So don't eat them. :)

On top of this the silica beads tend to fracture so you have essentially consumed broken glass which will be quite painful as it tears up your digestive tract causing you internal bleeding.

Just to add a couple references: 1, 2, 3.
I don't know how far the the second and third ones are true but the first one is the most accurate.


Does Br2/H2O oxidize all aldehydes including carbohydrates?

Aldehydes, including aldoses, are oxidized to their respective carboxylic acids in the presence of $\ce{Br2}$ in $\ce{H2O}$. The reason this reaction is often discussed with carbohydrates is that it is useful for differentiating aldoses from from ketoses, which cannot be further oxidized. A solution of $\ce{Br2}$ and $\ce{H2O}$ will lose its characteristic brown color in the presence of aldoses, but will not in the presence of ketoses.

The general mechanism for the oxidation of an aldehyde under these conditions is:

Though ketones will be able to form a hydrate like aldehydes, no deprotonation can occur to eliminate the bromide ion as we see in aldehydes. Hence no further oxidation can occur.


Conductivity measurements were done in late 1800s by Friedrich Kohlrausch, who painstakingly measured the conductivities of solutions so accurately that they are still valid today. Most of his work is in German was published in relatively unknown journals. In his time, conductivity theory was not advanced enough to predict the conductivity of simple salt solutions in water, so there was no a priori theoretical prediction of conductivity, from the first principles. I don't think it is doable today after 100 years (may well be wrong(.
However, the accurate measurement of resistance was well established in physics. See below.
A paper in the J. Am. Chem. Soc. 1933, 55, 5, 1780 describes what Kohlrausch and his meticulous colleagues they did. Note the level of details which is remarkably missing from papers today.

After Kohlrausch had developed his method of measuring the electrical
conductance of solutions by means of an alternating current bridge, he
undertook with the aid of Holborn and Diesselhorst’ the determination
of the specific conductance of seven standard reference solutions in
absolute units in order that the results might be available to himself
and to other experimenters for the calibration of conductance cells.
For this purpose they used two cells whose cross sectional areas, a,
and length, l, were determined mechanically. The cell constant,
l/a, could then be computed in absolute units (cm$^{-1}$) from the measured dimensions.
The cell constants of a considerable number of other cells with
sealed-in electrodes of the usual type were determined electrically by
comparison with the primary cells by measurements of resistance when
filled with a common solution. These cells, whose cell constants were
thus determined indirectly, were then used for the determination of
the specific conductance of the seven standard reference solutions at
approximately 0, 9, 18, 27 and 36 C. The specific conductance of each
of these solutions for every degree between 0 and 36 C was obtained by
interpolation. No actual measurements were made at either 20 or at
25’. The water-bath used to fix the temperature of the cells was not
provided with an automatic thermostatic control. The majority of the
data on conductance in the literature are based on the values ascribed
to these standard reference solutions by Kohlrausch, especially the 1
N and 0.1 N potassium chloride solutions.
In the original paper by Kohlrausch, Holborn and Diesselhorst it is
stated that the normal potassium chloride solution was prepared by
dissolving 74.555 g. of potassium chloride weighed in air and making
up to 1 liter at 18 C. The 0.1, 0.02, and 0.01 normal solutions were
prepared by dilution by volume. The book “Das Leitvermogen der
Elektrolyte” by Kohlrausch and Holborn, which was published in the
same year as the paper referred to, contains alternative directions
for the preparation of these solutions by weight. This book has
probably been used by other experimenters more than the original paper
because it is more generally available.


The 1-specification for the ketone function is superfluous, there is no information obscured by omitting the '1'.  It is the most important functional group in molecule, and therefore establishes the 1-position.

Also, cyclopentone is not a proper name.  You are missing the 'ane' suffix to indicate saturation.  It would be a cyclopentanone.

Final name:

2-fluorocyclopentanone.


Does cooling a potato change the nature of its carbohydrates?

Yes, retrogradation is a reaction that takes place when the amylose and amylopectin chains in cooked, gelatinized starch realign themselves as the cooked starch cools.

Resistant Starch:

  There is weak evidence that resistant starch can improve fasting glucose, fasting insulin, insulin resistance and sensitivity, especially in individuals who are diabetic, overweight or obese. In 2016, the U.S. FDA approved a qualified health claim stating that resistant starch might reduce the risk of type 2 diabetes, but with qualifying language for product labels that limited scientific evidence exists to support this claim. Because qualified health claims are issued when the science evidence is weak or not consistent, the FDA requires specific labeling language, such as the guideline concerning resistant starch: "High-amylose maize resistant starch may reduce the risk of Type 2 diabetes. FDA has concluded that there is limited scientific evidence for this claim."

U.S. National Library of Medicine - "Studies on effect of multiple heating/cooling cycles on the resistant starch formation in cereals, legumes and tubers" (non-paywall source), (July 2009), by Yadav BS, Sharma A, and Yadav RB:

  Abstract
  'Resistant starch' (RS) is defined as starch and starch degradation products that resist the action of amylolytic enzymes. The effect of multiple heating/cooling treatments on the RS content of legumes, cereals and tubers was studied. The mean RS contents of the freshly cooked legumes, cereals and tubers (4.18%, 1.86% and 1.51% dry matter basis, respectively) increased to 8.16%, 3.25% and 2.51%, respectively, after three heating/cooling cycles (P< or =0.05) with a maximum increase of 114.8% in pea and a minimum of 62.1% in sweet potato (P< or =0.05). Significant positive correlations were observed between the RS content and amylose (y=0.443x-5.993, r=0.829, P< or =0.05, n=9) as well as between the percentage increase in RS and insoluble dietary fiber content (y=2.149x-24.787, r=0.962, P< or =0.05, n=9). A differential scanning calorimeter study showed an increase in the T(0), T(p), T(c) and DeltaH values of the repeatedly autoclaved/cooled starches. The intact granular structure was also observed disappear, as studied using scanning electron microscopy.


You simply can't make a transoid molecule of cyclohexene because there is not enough "length" in the carbon chain.

I don't know if it is "strain" per se--but a putative transoid cyclohexene would have extraordinarily exotic bonding arrangement--thus being very high in energy.


Which material would work best as "sacrificial material" for melting aluminum powder in a kiln?

You need some oxygen scavengers before you start heating up the system. However, please note that you are going to need a lot of scavengers. To get rid of all the oxygen, you will have to put in a large excess of these scavengers into the chamber.

There is some information on oxygen scavengers on Wikipedia


Carbon dioxide is not stable in water in its original form, so it undergoes an acid-base reaction to give the carbon dioxide.

On the other hand, oxygen is stable and soluble in water in its $\ce{O_2}$ form, so it will stay like that, no reaction other than solubilization. Besides, it could react with organic matter or other compounds in water, oxidizing them, and getting reduced to $\ce{OH^-}$


What determines the atom or molecule that receives electrons during oxidation?

When a metal like iron is oxidized, each iron atom loses 3 electrons. These electrons are kept by oxygen atoms from the air. The iron atom becomes an ion charged 3+, and the oxygen atom becomes an ion charged 2-. The positive ions still attract negative ions, and they will never separate. Two iron ions and three oxygen ions are necessary to make a new compound which must be not charged. This assembly is a new substance called rust, which has a brown color. Once such a transformation has happened, the next iron atoms can repeat this operation : the amount of rust will slowly increase.

All metals can be oxidized by more or less the same way. But it could be quicker or longer.  For copper, this reaction is not so frequent. Copper makes a copper oxide which is black, but you may wait for months to see it.


How long will it take for a measurable change in mass (corrosion) in an iron copper corrosion cell to happen?

In order to get any measurements, you would probably need to leave the coupon in the solution for about a week or so.

To see the difference due to the salt concentration differences, I would recommend going even longer, probably at least 2-3 weeks.


Should chemical compounds and elements be capitalized?

The names of chemical compounds and elements should be capitalized if they appear at the beginning of a sentence or in a title - that is, they are treated just like any other common noun.
For example, a title:

Why I Don't Like Zinc

or a sentence:

Boron is my favorite element.

Within a sentence:

We used boron and zinc in the experiment.
Vinegar contains acetic acid.

The symbols for chemical elements are always capitalized, no matter what:

We combined $\ce{As}$ and $\ce{W}$ to make a new alloy.


Model of London forces and Hydrogen Bonding; the question is carried on forward, partially, as a personal question, from another question

If i underatabd correctly,
Now a days when in molecular dybamics simulation or docking studies, to calculate everything requires precise knowledge of polar, hydrophobic, ionic and non-bonding interactions. We have models to deal with each, and each has limitations. With polar groups the issues are not so much the models, which have distance dependance and orientation dependance, but computer power. This is really what limits the models. Often solvent effects and hydrophobic effects are ignored or estimated and simplified.

If there is a model missing is a hydrophobicity model. There isn't anything that is super.


Wittig reaction in presence of carboxylic acids

This works because the acidic starting material is going to be deprotonated as there is excess bicarbonate present. Because of the adjacent carboxylate group, the pKa of the phosphonium species is going to be around 6-7  pKa Table here so within the range where bicarbonate can deprotonate it. The reaction is between the anion of the starting material and the anion of the phosponium species in aqueous phase. This will only work for readily deprotonated phosphonium species and aqueous soluble aldehydes in the presence of excess base.


The primary reason for this and other seemingly implausible features of element distribution is this: some elements are concentrated, and some are scattered. Strontium is much like calcium (which is way more abundant) and can hide as a tiny admixture in its minerals, while sulfur is pretty much on its own.

Just think of the rare earth elements. Their name is in fact a misnomer; they are not rare at all, just very scattered and thus difficult to come by, hence the relatively late discovery. On the other hand, silver and gold, which are thousands times less abundant, were known since antiquity (or probably before that).

Also, you see yourself that there are several columns with the same name "Abundance in crust", and the data for the same element vary quite significantly between the columns. This is due to the fact that finding an element's content in the crust is not easy at all. Wait, don't they say explicitly that the numbers are but estimates?


Is there a reaction to get hydrogen from hydrocarbons?

Hydrocarbons are in fact the main source of hydrogen gas. According to Wikipedia, 95% of all hydrogen is produced by the steam reforming reaction:

$$\ce{CH4(g) + H2O(g) -> 3H2(g) + CO(g)}$$

This is a very important industrial process which has been studied in depth. It is done at very high temperatures (~1000 °C) and under pressure (~20 atm), and can be "up to 75%
efficient" (though efficiency here may not be defined as solely reaction yield).

As an aside, the fact that so much hydrogen is obtained from methane (itself generally sourced from natural gas, a fossil fuel) is one reason why hydrogen fuel cells currently have limited environmental benefit, though there is a huge amount of research into more efficient techniques for obtaining hydrogen from water splitting (especially using light) which could be carbon-neutral.


Comparing van der Waals constant for gases

The reason that in your sample set water seems to have a low value of $a$ despite having hydrogen bonding is because most of the molecules that you chose were of much higher molecular mass. You should compare water with some gas of approximately equal molecular mass, then you'll see that value of $a$ in water is quite high. Your other observations are generally correct.


Wikipedia, in a short article, indicates that there are actually two types of azirenes:

1H-azirenes, where there is one ligand on each ring atom as shown in the question.  Since all the ring atoms have ligands the lone pair contributed by nitrogen enters the conjugated pi system, which then has four conjugated electrons and is antiaromatic.  Such a molecule is thereby unstable and prone to rearrange into the alternative structure that follows.
2H-azirenes have two ligands on one carbon, saturating that atom so we no longer have a conjugated ring.  The other ring carbon has one ligand and is still unsaturated.
The nitrogen atom is left without a ligand, and its lone pair now replaces the missing ligand in an outward pointing in-plane orbital.  The only pi bonding is an isolated bond between the nitrogen and the remaining unsaturated carbon.  These compounds are non- (not anti-)aromatic and may be isolated.


Why is ordinary soap and water adequate to decontaminate possible nerve agents?

Nerve agents are usually susceptible to hydrolysis leading to non-toxic degradation products

Most modern nerve agents are based on fluorophosphonate cores. That is they contain a $\ce{PFO(OR)(X)}$ core where R is some organic radical (-iPr in sarin) and X can be a range of radicals (Me- in sarin, OR or N=R in novichok agents).

Most are not very stable in storage as they are subject to degradation from hydrolysis so have to made not long before use or have to be made ultra pure to avoid any contaminants catalysing their degradation by acid or base hydrolysis. As a result any were deployed as binary agents (where two tables components are mixed inside the weapon to create the active ingredient).

The key to their reactivity is the P-F bond which is easily hydrolysed to give much less harmful products. This hydrolysis is not fast enough to degrade the compounds inside the body before they kill you but is fast enough that they don't hang about for long periods in the environment after use (it is worth bearing in mind that they are extremely powerful inhibitors of acetylcholinesterase, without which your nervous system can't function: it doesn't take a big dose or a long time for them to work). It is also something that can be accelerated by washing. It seems likely, for example, that mildly contaminated clothing can be decontaminated by a standard domestic clothes washing cycle as this would dramatically accelerate any hydrolysis reactions.

It is also worth noting that, if the contamination were large enough, you would notice it very rapidly. So any residual contamination in people not already ill would be of a very low level and easily rendered insignificant by washing.-


What results from the reaction of pent-3-en-2-ol with hydrogen bromide?

It is very rare for a reaction to have no side products, and this reaction is certainly not exceptional in that regard. I'm actually a bit surprised your book chose this reaction to illustrate the regiochemistry of $\ce{HBr}$ addition to an alkene, as this would compete with the formation of an allylic cation, which are generally more stable than alkyl carbocations. Being more stable, these cations will form more quickly:

giving the product you correctly assumed would also be formed. You probably wouldn't obtain this though, as it would continue to react, forming a dibromoalkane (as would the products listed by your book):


Have you checked Wikipedia?

  Upon thermal decomposition, $\ce{LiNO3}$ gives lithium oxide ($\ce{Li2O}$),
  nitrogen dioxide, and oxygen:

  $$\ce{4 LiNO3 → 2 Li2O + 4 NO2 + O2}$$

  Other group I nitrates decompose differently, forming the nitrite salt and oxygen. Because of its relatively small size, the lithium
  cation is very polarizing, which favors the formation of the oxide.


Let me explain you in the simplest manner.

For a two object system,

If they repel each other when you move them towards each other, that means it takes work to do this, and the potential energy is positive.

If they attract each other when you move them towards each other, then potential energy of the two object system goes on decreasing.

Potential energy of a two charged object system at a particular orientation is defined as work done by external agent to bring the second body from infinity to desired point in presence of the field created by the first body. Now think the first body to be nucleus and the second body to be electron.  Now to find the potential energy, according to potential energy defination I just said above, we need to being the electron from infinity to any desired location, lets say to a distance of "x" from the nucleus. As we have to bring electron from infinite that means initially electron is infinitely far away from nucleus and hence at zero potential.

Potential Energy of electron:

When we are bring the electron towards the nucleus, offcourse they are going to attract each other and as said earlier if two objects attract each other, then potential energy of the system goes on decreasing. So as we move electron towards the nucleus, energy goes on decreasing. Note that initially as electrons and nucleus were at infinite distance from one another, hence initially at zero potential and any further decrease in potential energy implies negative potential energy.
So potential energy of the electron is negative inside any atom.

Kinetic Energy of electron:

The electron also has kinetic energy. Kinetic energy can never be negative (because in K.E expression we have squared of velocity which is always positive) and it is smaller in magnitude than the potential energy.

Total Energy of electron:

So the total energy i.e kinetic energy  + potential energy is negative.


How to synthesize Boron tri-fluoride using Boron Oxide

The reaction of boron trioxide with sulfuric acid and calcium fluoride to obtain boron fluoride is given here.

$$\ce{B2O3 + 3CaF2 + 3H2SO4 ->[\Delta] 2BF3 + 3CaSO4 + 3H2O}$$

Conditions

  Boron trioxide react with calcium fluoride and sulfuric acid to produce boron trifluoride, calcium sulfate and water. Sulfuric acid - concentrated solution.

The reaction will be easy to understand if you break the reaction into two reaction.

Babounet says:-

  $\ce{CaF2 + H2SO4}$  will give you $\ce{HF}$ so why bother if you add directly the $\ce{HF}$ instead ?

Indeed $\ce{CaF2 + H2SO4}$  will indeed yield  $\ce{HF}$ along with calcium sulfate as indicated here.

$$\ce{CaF2 + H2SO4 ->[\Delta] 2HF + CaSO4}$$

This $\ce{HF}$ will now react with boron trioxide to yield boron trifluoride as indicated here.

$$\ce{B2O3 + 6HF → 2BF3 + 3H2O}$$

Thus the  products are boron trifluoride, calcium sulfate and water and not calcium bisulfate.


Chemical reaction that can produce lots of heat from 2-3 simple liquid ingredients?

The first thing that comes to mind, if you have access to a stock-room, is mixing two solutions: (1) NaOH; (2) HCl. This can release a lot of heat if your solutions are concentrated enough, and it forms salt-water if your NaOH and HCl are of the same molarity.

Edit: I'm on my lunch break, so I did some of the math...

The chemical reaction of interest in this case is:

$$\ce{OH- + H+ -> H2O}$$

The standard enthalpies of formation ($\Delta H_\mathrm f^\circ$) of these species are:

$$\begin{array}{lr}
\hline
\text{Species} & ΔH_\mathrm f^\circ/\pu{kJ mol-1} \\
\hline
\ce{OH-} & -229.99 \\
\ce{H+}  & 0.00       \\
\ce{H2O} & -288.83 \\
\hline
\end{array}$$

Thus, the change in enthalpy for the reaction is:

$$\Delta H^\circ = -288.83\ \mathrm{kJ/mol} - (-229.99\ \mathrm{kJ/mol}+0\ \mathrm{kJ/mol}) = -58.84\ \mathrm{kJ/mol}$$

Therefore, for $1\ \mathrm{mol}$ of $\ce{NaOH}$ + $1\ \mathrm{mol}$ of $\ce{HCl}$, you get $58.84\ \mathrm{kJ}$ of heat. Say you want to release enough heat to get the net solution up to $100\ \mathrm{^\circ C}$.

Water has a heat capacity of $4.18\ \mathrm{J/(g\ ^\circ C)}$. Say you have $1\ \mathrm L$ of $\ce{NaOH}$ + $1\ \mathrm L$ of $\ce{HCl}$, you'll need enough heat to raise the temperature of $2\ \mathrm L$ of water to $100\ \mathrm{^\circ C}$. I'll assume the water starts off at $25\ \mathrm{^\circ C}$, so you have $2\,000\ \mathrm g$ and $75\ \mathrm{^\circ C}$ to go.

$$4.18\ \mathrm{J/(g\ ^\circ C)} \cdot 2\,000\ \mathrm g \cdot 75\ \mathrm{^\circ C} = 62\,700\ \mathrm J = 62.7\ \mathrm{kJ}$$

How many moles of $\ce{NaOH}$ + $\ce{HCl}$ do you need for that much heat?

$$\frac{q}{\Delta H^\circ} = \frac{62.7\ \mathrm{kJ}}{58.84\ \mathrm{kJ/mol}} = 1.065\ \mathrm{mol}$$

That would mean that you can mix $1\ \mathrm L$ of $1.065\ \mathrm M$ $\ce{NaOH}$ + $1\ \mathrm L$ of $1.065\ \mathrm M$ of $\ce{HCl}$, and would theoretically expect to get a temperature close to $100\ \mathrm{^\circ C}$.

This might be off a bit because I've made some assumptions:

$\Delta H^\circ$ is constant with respect to temperature from $25\ \mathrm{^\circ C}$ to $100\ \mathrm{^\circ C}$. This may not be true.
the dissolved salts in water dont significantly affect its heat capacity
You have $2\,000\ \mathrm g$ of water in $1.065\ \mathrm M$ $\ce{NaOH}$ + $1.065\ \mathrm M$ $\ce{HCl}$

I think this could get you close though? I'm a bit surprised that the molarities aren't higher... It's a starting point at least.

Disclaimer of course: be careful with the $\ce{NaOH}$ + $\ce{HCl}$ solutions, they can be dangerous. Use proper chemistry hygeine protocols. The mixture should be benign, but you should confirm this with pH paper.


I am trying to understand ppm, too. As far as I have understood there are different kinds of ppm, which is basically a ratio: it can be a ratio of amount of substance, masses, or volumes.
Assuming your ppm is a molar ratio I did this reasoning:
Indicating with $n$ the amount of substance, with $M$ the molar mass and with $V$ the volume, the concentration of your gas is:
$$c = \frac{n_\mathrm{gas} \cdot M}{V},$$
and defining the $\mathrm{ppm}$ as:
$$\mathrm{ppm} =\frac{n_\mathrm{gas}}{n_\mathrm{total}}\cdot 10^6.$$
Using the gas law:
$$n_\mathrm{tot} = \frac{p\cdot V}{R \cdot T}, $$
where $T$ is the temperature in kelvin and $p$ the pressure in pascal, and substituting, you get:
$$c = \frac{\mathrm{ppm} \cdot M \cdot p}{R\cdot T}\left[\frac{\mu \pu{g}}{\pu{m^3}}\right].$$


How can nucleophilic substitution happen at the sp²-hybridised carbon of 4-chloropyridine?

The mechanism is indeed a nucleophilic substitution but it does not follow the bimolecular substitution paradigm ($\mathrm{S_N2}$) or the unimolecular one ($\mathrm{S_N1}$) that one learns about in basic organic chemistry classes — and it also does not follow the general electrophilic aromatic substitution that is also often presented in introductory courses.

$\mathrm{S_N2}$ can be ruled out because a rear-side attack would require an attack through the ring — a path which is blocked by the nitrogen atom (and the other bits of the ring).
$\mathrm{S_N1}$ is highly unlikely because we would need a phenyl cation — a highly unstable intermediate.

The aromatic ring of 4-chloropyridine can be understood as electron-deficient since pyridine’s nitrogen exercises a strong $-I$ and $-M$ effect. You can draw the $-M$ effect by pushing the electrons of the $\ce{C=N-C}$ double bond to nitrogen to give $\ce{\overset{+}{C}-\overset{-}{N}-C}$ — and this cation can be moved around the ring, most notably also to the 4-position where chlorine is located. See the figure below.

Figure 1: $-M$ effect of the pyridine nitrogen.

The nucleophilic attack is that of the amino nitrogen’s lone pair into this π system; the attack flips electrons so that the pyridine nitrogen has an additional lone pair, two $\ce{C=C}$ bonds are adjacent to this and a tetrahedral carbon bonded to both $\ce{Cl}$ and $\ce{\overset{+}{N}H2R}$ completes the ring.

This zwitterion can break down by reverse electron flipping which displace the chloride to give the skeleton of the reaction product. Additionally removing a proton from the amino nitrogen gives the neutral product. Considering that we have an acid (2-aminobenzoic acid) in the reaction mixture, I have drawn the pyridine nitrogen (the more basic one) to be protonated throughout the mechanism. This assists in the nucleophilic addition step by further reducing the ring’s electron density. See figure 2 for the mechanism.

Figure 2: Mechanism of the nucleophilic aromatic substitution reaction. Nitrogen being protonated due to the acidic effect of 2-aminobenzoic acid facilitates the reaction.

This mechanism is called nucleophilic aromatic substitution $\mathrm{S_NAr}$ and can only occur on very electron-deficient aromatic compounds: typically pyridines or pyrimidines with additional electron-withdrawing potential leaving groups such as chlorine.


Why using smaller portions are better for extractions

Suppose you have 1 liter of water solution of something you want to extract, 1 liter of solvent, and a partition coefficient of 1 (not very practical, but this is just to make the math simpler). You mix them all together, the compound distributes between water and solvent at 1:1 ratio, then you separate the solvent and end up with 50% of your compound. Can we do better than that?

OK, now let's divide the solvent in two equal portions. The first portion of 500 ml takes $1\over3$ of our compound, then the second portion takes $1\over3$ of the rest, or $2\over9$ of the original amount, thus making the total of ${5\over9}\approx55\%$, which is somewhat better than $50\%$.

Believe it or check it yourself, splitting the solvent into more portions will get us even more of our compound. You may call 5% a marginal gain, but with more favorable partition coefficients, the figures may grow quite a while.

And finally, if we split the original solution as well, and apply the portions in the right order, we may get pretty close to the industrial process.


The concentration of dopamine is low
The concentration of other protons is very high
The magnetic field is not homogeneous.
The brain cannot be rapidly rotated the way a laboratory solution state NMR sample is to average magnetic field inhomogenity
Compare to laboratory solution state NMR, the viscosity in the brain would be high causing further line broadening.

If you look at the spectra in 1H NMR Probe for in Situ Monitoring of Dopamine Metabolism and Its Application to Inhibitor Screening J. Am. Chem. Soc., 2012, 134 (30), pp 12398–12401, you'll see that even in tissue samples, there are too many other stronger signals for conventional 1H NMR to be successful, so C-13 labeling is used in combination with 1H NMR.


What physical/chemical reasons could have caused Dorian Gray's picture to degrade?

Preface: So far, I do not know the novel you refer to, and assume the portrait is a classical oil painting.

Such a painting consists of several layers of products applied.  Similar to a wall in your living room, the canvas is prepared to accomodate the pigments dispersed in linseed oil, what in other fields of application would be called a "primer".  Then there are multiple layers of the painting itself, topped by the varnish.  To mention a few of the potential issues,

Pigments may change their colour upon exposure to air. Pigments like lead white turned over time black, for example.
Pigments may interact with the oil, forming complexes, of different mechanical and complexation strength:

(source)

Components may migrate across the picture and locally, and reactions may occur.  The formation of soaps of lead and zinc are the most prominently known ones, for example based on palmitic, stearic, or azelaic acid.  These processes may alter colors, even rendering them transparent enough to recognise the support of the paint, and they may eventually protrude the surface of the paintings as shown from a topview:

(source, p. 14)

or cross section:

(source)

These may not simply be "washed off" as a normal cleansing would do, as water and humidity in general amplify the problem.  If left untreated, the painting however will age and degrade.

Further reading may be found, for example, here, here, here, here/here.


How does the state of iodine affect its HOMO–LUMO gap?

I will broadly mention several mechanisms that may influence color of a solid vs. gas, not just the HOMO-LUMO transition.
Without going into the literature and/or performing (extensive) calculations, I cannot confirm or reject your teacher explanation.

In the solid state, you may find intermolecular MO interactions that will affect orbital energies. Depending on the strength, you may even have to abandon the concept of molecular orbitals in favor of crystal orbitals. Quantum dots and their dependence of color on size are somewhat of an example.
Molecules may alter their shape in solid state compared to gaseous state. This is less of an issue in $\ce{I2}$.
In the solid state, intermolecular excitations are more likely (i.e. going beyond HOMO-LUMO transitions). This is especially true if you do not have a pure substance, think doped solids.
Note that supermicroscopic spatial distribution of a substance will have influence on the color as well: finely distributed silver or mercury (as in calomel) may appear as black rather than shiny.


Producing methane, reactants and products

The answer is:

$$\ce{2CH3COONa(s) + 2NaOH(s) + Ca(OH)2(s) <=> 2CH4(g) + Na2CO3(s) + CaCO3(s) + H2O}$$

You have to get same number of equal atoms in each side of the reaction (this technique is called mass balance). If there were charges, they must be balanced also.

Example

There are $2\times 3 + 2\times 1 + 2\,$ hydrogen atoms in left side, and there are $2\times 4 + 2$ $\ce{H}$ in the right side. So both sides give $10$ hydrogen atoms.

I hope it helps.


Why does oxygen in OH have an oxidation state of 2-?

First step: draw the Lewis structure of $\ce{OH-}$ with all lone pairs. You should be getting three lone pairs on oxygen and one $\ce{O-H}$ bond.

Second step: ‘[cleave] [all bonds] heterolytically in such a way that the electronegative partner gets both electrons.’[1] You should now have eight electrons around the oxygen and zero on hydrogen.

Third step: count the electrons on both (or all) atoms and check what the difference is to their elemental valence electron count. Oxygen formally has eight electrons on it which is two more than the six it would in its elemental state. The two extraneous electrons give an oxidation state of $\mathrm{-II}$. Likewise, hydrogen has lost its electron for an oxidation state of $\mathrm{+I}$.

Control step: add up the oxidation states of all atoms. If you did everything correctly, they will add up to the entire charge of the ion. Here, this works out:

$$(-2) + (+1) = -1$$

[1]: https://chemistry.stackexchange.com/a/66242


Standards for drawing chemical molecules

As it happens, there are a set of quite involved documents from the "IUPAC Chemical Nomenclature and Structure Representation Division".

There's a list here, but to avoid later link-rot, they are currently:

Jonathan Brecher Graphical representation of stereochemical configuration (IUPAC Recommendations 2006) 2006, Vol. 78, Issue 10, pp. 1897-1970
Richard M. Hartshorn, Evamarie Hey-Hawkins, René Kalio and G. Jeffery Leigh: Representation of configuration in coordination polyhedra and the extension of current methodology to coordination numbers greater than six (IUPAC Technical Report)
2007, Vol. 79, Issue 10, pp. 1779-1799
W. Mormann and K.-H. Hellwich: Structure-based nomenclature for cyclic organic macromolecules (IUPAC Recommendations 2008) 2008, Vol. 80, Issue 2, pp. 201-232
Jonathan Brecher: Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008) 2008, Vol. 80, Issue 2, pp. 277-410
Andrey Yerin, Edward S. Wilks, Gerard P. Moss and Akira Harada: Nomenclature for rotaxanes and pseudorotaxanes (IUPAC Recommendations 2008) 2008, Vol. 80, Issue 9, pp. 2041-2068


Are precipitations exothermic and/or endothermic?

No, absolutely not.
Precipitation reactions can be either endothermic and exothermic.

Table 1. Thermodynamic data of precipitation for some salts
\begin{array}{cccccc}
\hline
\text{Salt} & \Delta G_\text{ppt}^\circ & \Delta H_\mathrm{ppt}^\circ & -T\Delta S_\mathrm{ppt}^\circ(\pu{25 °C}) \\
\hline
\ce{Be(OH)2} & -121 & -31 & -90 \\
\ce{Mg(OH)2} & -63 & -3 & -61 \\
\ce{Ca(OH)2} & -28 & 16 & -44 \\
\ce{Li2CO3} & -17 & 18 & -34 \\
\ce{MgCO3} & -45 & 28 & -74 \\
\ce{CaCO3} & -48 & 10 & -57 \\
\ce{SrCO3} & -52 & 3 & -56 \\
\ce{BaCO3} & -47 & -4 & -43 \\
\ce{FePO4} & -102 & 78 & -180 \\
\hline
\end{array}

It is more than clear that all precipitations are not exothermic.

Also definitely noticable is the fact that $\Delta S$ is a positive quantity.

But that is unexpected, as an increase in the disorder of the system upon precipitation may seem counterintuitive to many.

What is often overlooked is that several other factors are at work.

Paving the way

Ions can be of two types:

Electrostatic structure breakers
Electrostatic structure makers

This classification is based on how ions interact with water.

Electrostatic structure breakers

Sometimes referred to as chaotropic ions, these disrupt the hydrogen bonding interactions between water molecules. Examples include $\ce{NH4+}$, $\ce{Cs+}$, $\ce{Br-}$, $\ce{I-}$.

It is tempting to assume that large cations with higher possible coordination numbers would attach and order water molecules more effectively than small cations, but this is not experimentally observed.

Smaller the radius and/or greater the charge (the more acidic or basic the ion), the greater its structure-making properties in aqueous solution. It is hence clear that the above mentioned ions are structure breakers rather than makers, evidently having very low values of $Z/r^2$.

It is easy to deduce that upon their introduction into aqua, they will correspond to an increase in entropy.

Electrostatic structure makers

AKA kosmotropic ions, these increase the ordering of water. $\ce{Fe^2+}$, $\ce{Al^3+}$, $\ce{Mg^2+}$, are members of this club.

These ions have large charges and relatively smaller radii, resulting in a higher $Z/r^2$ value.

Via hydrogen bonding these ions create relatively larger shells of hydration than their low $Z/r^2$ counterparts.

The differences in hydration for the two types of ions is visible in the following table.

Table 2. Hydration data for some cations

Uh..What does this have to do with the question?

Everything.

The reason that most acidic cations and basic anions combine to give precipitates is due to the disorder that results from the release of large number of water molecules from the well structured hydration spheres.

The second law favours this.

For an example, let us consider the precipitation of $\ce{MgCO3}$

$$\ce{Mg(H2O)36^2+ + CO3(H2O)28^2- <=> MgCO3(s) + 64 H2O}$$

The number of water molecules are derived from Table 2.
The release of $64$ molecules of water per molecule of magnesium carbonate precipitated is very indicative of the positive entropy of the process.

Now for the enthalpy.
Rather than the Gibbs–Helmholtz, let's return to the good old Hess law.

According to Hess' law,

$$\Delta H_\mathrm{ppt}(\ce{M_yX_m)}) = -y\Delta H_\mathrm{hyd}(\ce{M^m+})- m\Delta H_\mathrm{hyd}(\ce{X^y-}) + U(\ce{M_yX_m})$$

The electrostatic structure makers (type 1) would have higher negative

$$y\Delta H_\mathrm{hyd}(\ce{M^m+})+ m\Delta H_\mathrm{hyd}(\ce{X^y-})$$

leading to an overall positive heat of precipitation. e.g. $\ce{FePO4}$, $\Delta H_\mathrm{ppt} = \pu{+78 kJ/mol}$

Vice-versa for type 1: the structure breakers.

I hope I managed to enlighten you on this matter.

P. S. This was a great question. Pardon the short conclusions, I am currently travelling. Members of the community may feel free to edit and make a more dramatic exit.

References:

Hanzlik, R. , Inorganic Aspects of Biological and Organic Chemistry 1976, II, 26-30

https://www.wou.edu/las/physci/ch412/ppt2.htm

https://www.wou.edu/las/physci/ch412/pptrxn.html


Finding out the number of atoms in the sun.

The problem is that you did not include units with your numbers. If you had, you would have easily seen that you are multiplying $m^3$ and $cm^{−3}$ and coming up with 1 instead of 1000000.


Strongest B/L Acid Which Can Exist (Leveling Effect)

Your reasoning is not entirely correct, as you consider that the charge is only located on the atom which bears the proton upon protonation. But nearby atoms, bonds and molecules will also participate in the stabilization of the positive charge, through several factors.

In the gas phase (so considering no solution phase effects), you should at least consider that, in addition to the instrinsic capability to bear a proton of the atom (which goes as the inverse of electronegativity, as you state in your discussion), one has also to consider the following factors:

Resonance effects (delocalization of the charge on other atoms, that may be less electronegative than the atom directly bearing the proton). In the molecule given above, this would lead to a stabilization of the $sp$ and $sp^{2}$ hybridized nitrogens, as there would be a partial positive charge on the neighbouring carbon atom, as: $$\ce{RCH=N+H-R' <-> RC+H-NH-R'} $$ or $$\ce{RC#N+H <-> RC^{+}=NH}$$
Inductive effects (or if you prefer availability of electrons in nearby atoms that can be displaced towards the charged site). In your constrained textbook example, this would be hard to indicate: there are a lot of heteroatoms (nitrogens and oxygens) and electron withdrawing groups (CN) which are likely to remove electrons from the carbon atoms, and in the central part of the molecule, these are likely to contribute to destabilize the molecule.
Intramolecular hydrogen bonds. Although sometimes neglected, these can contribute quite a lot to the stabilization of the charge when mutiple protonation sites involving heteroatoms are present. Just as a reminder, in the gas phase, a hydrogen bond in a protonated molecule can reach up to 120 kJ/mol. In your example they would be hard to predict without modelling the complete structure, but one could consider a possible H-bond between the protonated imine ($sp^{2}$ nitrogen) and the alcohol oxygen.

Then if we were to consider solution phase, the interactions between neighbouring molecules and the propensity to form hydrogen bonds with the protonated molecule will also play a role on the possible protonation sites.

Based on the above considerations, one should consider that the most likely stable forms would be protonation on either the nitrile group ($sp$ nitrogen) or the imine group ($sp^{2}$ nitrogen), as both allow to put a partial positive charge on the neighbouting carbon atom. The protonated nitrile group is a protonation on an $sp$ nitrogen, with charge delocalization on an $sp$ carbon which means that the charge stabilization will be less efficient than for an imonnium ion.

If one considers purely gas phase consideration, one can use the proton affinity scale to rank these various contributions.

If we look for proton affinities in the NIST database we have:

for a secondary amine $\ce{H3C-NH-CH3}$ 929 kJ/mol
for a nitrile $\ce{H3C-C#N}$ 779 kJ/mol

Unfortunately little experimental data is available for imines. But some theoretical data is available in this article by Hammerum et al.. If one considers the methylimine $\ce{Z-CH3-CH=N-CH3}$ proton affinity is calculated at 943 kJ/mol, higher than dimethylamine. But on the other hand if one considers $\ce{H2C=N-CH3}$, its proton affinity is of 880 (exp) or 902 (theo), lower than dimethylamine.

So inductive effects clearly play a very important role in the order between amine protonation and imine protonation in the gas phase. In a really constrained molecule as above, they are likely to play a major role, as is the possibility to form hydrogen bonds in the solution.

If one now puts in the possible formation of intermolecular hydrogen bonds, protonated secondary amines can make two hydrogen bonds with two hydrogen bond acceptors, whereas protonated secondary imines can only make one hydrogen bond. This will likely tip the balance towards amines and would make answer (4) most likely in solution, whereas I would be quite at loss to give an answer in the gas phase.


In regard to collision theory, why do molecules have to have the correct orientation in order to react?

Suppose we have a simple reaction of a diatomic and a single atom $\ce{AB + C ->A +BC}$. So the reaction is breaking the bond between $\ce{A}$ and $\ce{B}$, which then bonds to $\ce{C}$. So we might expect the transition state to look like $\ce{A--B--C}$. This suggests the reaction is highly orientation dependent.

Looking at the process as a function of the angle formed by ABC we would expect that a 0 degree angle would vary likely lead to a reaction, since everything is lined up as we would expect, while a 180 degree angle would have almost no chance of forming the desired product as it would aligned more like this $\ce{C--A--B}$. While its not a perfectly linear relationship, we should expect that the reactivity will decrease monotonically as it goes from 0 to 180 degrees.

For more complex structures, the orientation becomes even more important. A large molecule may only have a very small reactive site, and if the other reactant collides with a completely different section of the molecule, there is virtually no chance the desired reaction will occur.


What typically happens to H2 molecules released into atmosphere (with no spark source)?

The most common, reactive stuff in the atmosphere are $\ce{-OH}$ radicals (Wikipedia). So I guess they're the natural antagonist for $\ce{H2}$ molecules.
Btw., as soon as it's diluted in the surrounding air, it will no longer rise. Only bulk masses have buoyancy. The g gradient in the atmosphere is much too small to separate gases.
In the end, hydrogen however does diffuse upwards and is ultimately lost into space. Hydrogen and helium are light enough to be able to reach escape velocity in the upper atmosphere.


I believe the problem was that the LDA+C energies quoted in the paper were extra, additive energies which were supposed to be added to the main band gap energy of the bulk material. In this case that was something like $\pu{2.7 eV}$.

This is because the quantum dots will have higher band gaps the smaller they get. As the quantum dots get larger, in a sense, they approach a bulk material size, hence their band gap energy gets smaller and smaller and approaches the bulk material band gap energy limit.

Hope this helps anyone who happens upon a similar problem.


Why are hydrogenation reactions exothermic?

Let's take a look at an example of a hydrogenation reaction, namely the hydrogenation of ethene to give ethane

$$\ce{CH2\bond{=}CH2\ (g) + H2 (g)-> CH3\bond{-}CH3 (g)}$$

And now let's determine the enthalpy change of this reaction. By Hess's Law, the enthalpy change of the reaction is essentially the same as the enthalpy change of breaking all the bonds of the reactants and forming all the bonds in the molecules of the product. We can see that

1 $\ce{H\bond{-}H}$ bond has been broken
1 $\ce{C\bond{=}C}$ bond has been broken
1 $\ce{C\bond{-}C}$ bond has been formed
2 $\ce{C\bond{-}H}$ bonds have been formed

The other bonds remain unchanged by the end of the reaction

The following is the list of bond enthalpies of the aforementioned bonds, obtained from the Internet

$\ce{H\bond{-}H}$ bond: +432 kJ/mol
$\ce{C\bond{=}C}$ bond: +614 kJ/mol
$\ce{C\bond{-}C}$ bond: +347 kJ/mol
$\ce{C\bond{-}H}$ bond: +413 kJ/mol

Breaking bonds gives a positive enthalpy change since it is an endothermic process, and vice versa for bond formation. We can now estimate the enthalpy change of the reaction

$\Delta$$H$ = 2(-413) + (-347) + 614 + 432 = -127 kJ/mol

As indicated by the negative value of the enthalpy change, the reaction is exothermic.

In fact, the enthalpy changes of hydrogenation reactions of other alkenes are also similar (though more substituted alkenes may have a lower magnitude of heat of hydrogenation, but still the reaction is exothermic). You may notice that for other alkenes, the way you calculate the enthalpy change is the same since they involve the same bond breaking and bond forming processes aforementioned (any substituents bonded to the carbons with the $\ce{C\bond{=}C}$ bond remain unaffected throughout the reaction and will still be present in the product being bonded to the carbon).

To simply put, the energy released from bond forming is larger than the energy required (absorbed) for bond breaking, and that is why you get net energy released from the reaction.


1H splitting pattern of benzyl CH2 protons

This is a bit speculative. I'm still thinking about it, and maybe this shouldn't even been an answer. I drew out the left structure and the enantiomer of the right structure because I want the configuration at the carbon with the OMOM group to be the same.

I think you're calling these the 1,2-anti-1,4-anti and 1,2-syn-1,4-syn, respectively. The main focus here was on the conformation about the 1,2-carbon-carbon axis. I put the two hydrogens anti to each other. In the anti-anti case, the alkyl chain extends off away from the benzyl moiety while in the syn-syn case, the alkyl chain is pointed more towrads the benzyl moiety.

I'm not in love with this explanation. It's a bit speculative, but I used the tools I have and at least, it's consistent with your observation about splitting.


Is there any molecule visualizing software for Python?

In case you really need it has to be written in python probably you want to try Atomic Simulation Environment (ASE). It has a GUI that "allows users to visualize, manipulate, and render molecular systems and atoms objects".

Let me add (even when you are asking explicitly for codes written in Python), just in case that what you need is to create images from .xyz files (regardless of the language), Jmol probably will suits you, as it reads .xyz files straightforwardly. For saving your image once it is loaded, you will need to open a console in Jmol and type:

write image 1200 720 jpg 90 "filename.jpg"

where:

1200 and 720 are the width and height of your image (choose the ones you want to)
90 is the quality of the compression


Directionality of ring current-induced magnetic fields in purines

This is an interesting problem, I think it is best for a moment before I get stuck in that we considered a slightly different compound. I will consider it with an external magnetic field applied to it. Napthalene, now napthalene has two aromatic hydrogen environments in its NMR spectrum. These are very similar in chloroform to benzene in terms of chemical shift.

This makes me think that the ring currents go in the same direction in this compound.

It may be better to consider the ring current as flowing along the perimeter of the aromatic system rather than the loops formed by the six membered rings.

It is important when considering aromatic rings to escape being trapped by six membered rings. A lot of aromatic molecules have six membered rings but there are plenty of five and seven membered aromatic rings. It is more important to understand that a pi system which is cyclic and has 4n+2 pi electrons is the defining thing which makes a ring system aromatic.

If napthalene had ring currents going in different directions I would expect that if the interconversion of the sites (by ring current reversal or flipping of the molecule) was slow then I would expect to see four different chemical shifts in the proton spectrum.

If this exchange of sites was fast then I would expect to see two sites which are each the average of the two sites. I would also expect the chemical shifts of the napthelene to be very different to that of benzene.

If we allow ourselves a little diversion for a moment, the cyclopropanes have a rather rare proton chemical shift, this is becuase the C-C bonding in cyclopropane has some pi character and the hydrogen nuclei are above the ring in such a place that the ring current has the opposite effect on the NMR shift to the magnetic field due to the ring current in benzene.

Now if we move onto purine, in case anyone is not sure what the shape of it is then here it is.

If we look at the literature (S2-75) in the supporting literature of Thomas H. Graham, Wensheng Liu, and Dong-Ming Shen, Org. Lett., 2011, 13 (23), pp 6232–6235
DOI: 10.1021/ol2026813. This compound in DMSO has three very similar proton NMR shifts. This suggests that the ring current goes the same way around in both rings. These NMR shifts are all in the aromatic area and are further left (higher ppm) than benzene.


Where can I find a collection of density functional parameterizations?

Unless you are doing it just for fun and curiosity how the think works, you are better with some existing library. Most prominent is the libxc, offering the lowest level interface, i.e. given density (and possibly gradient) at series of points, it returns the energy. What is left for you is the construction of the grid and optimization of density, ie. you can go for the real Hohenberg-Kohn DFT, not just Kohn-Sham. For the usage examples (of the really nice and simple interface) see http://www.tddft.org/programs/octopus/wiki/index.php/Libxc:manual

And most importantly, you can find definitions of all the functionals in the list of implemented functionals


How is dissolving a physical change and not a chemical change?

This is indeed a bit of a grey area. Consider the following: chemical or physical?

$$\ce{Na+ Cl- (aq)  + Ag+ NO3- (aq)   ->  Ag+Cl- (s) + Na+ NO3- (aq)}$$

There is a tendency in more advanced chemistry to view the formation / breaking of ionic lattices as a physical change, because there are no differences in the electron shells of the atoms. Ions are considered as independent species. As such we could consider the sodium and nitrate irrelevant and reduce the above to simply

$$\ce{Ag+ (aq) + Cl- (aq) -> Ag+ Cl- (s)}$$

Note that I have shown the charges all the way through, to indicate that there are no covalent, dative, or metallic bonds made or broken.

Reactions such as these are definitely "chemical":

$$\ce{Na + \frac{1}{2}Cl2 -> Na+ Cl-}$$
(metallic bonds in $\ce{Na}$ and covalent bonds in $\ce{Cl2}$ broken, electron transferred from $\ce{Na}$ to $\ce{Cl}$.)

$$\ce{C2H4 + H2 -> C2H6}$$
(double bond of $\ce{C2H4}$ opened up, bond of $\ce{H2}$ broken, new covalent bonds formed.)

$$\ce{SOCl2 + 2H2O -> H2SO3 + 2HCl}$$ (covalent bonds broken, new bonds made.)

This last reaction sets up another point. It is usually done with excess water, so the products (which are themselves acids) will then dissolve in water, thereby protonating it:

$$\ce{H2SO3 + H2O -> H3O+ + SO3H-}$$
$$\ce{HCl + H2O -> H3O+ + Cl-}$$

These particular dissolutions are more likely to be seen as a chemical change than the dissolution of $\ce{Na+Cl-}$, because there is a rearrangement of covalent bonds (even though it's only a $\ce{H+}$ that moves from one molecule to the other.)

Finally, $\ce{H2SO3}$ has another equilibrium, again involving covalent bonds being broken and made, and therefore a chemical change:
$$\ce{H2SO3 -> H2O + SO2 (g)}$$

Conclusion

Making or breaking of metallic, covalent or dative bonds, or the formation/destruction of new ions or molecules, is always considered a chemical change.

The simple dissolution of an ionic salt, where none of the above mentioned bonds are made or broken, and no new ions or molecules are formed/destroyed, is frequently considered a physical change.

Note the care taken to always write $\ce{Na+Cl-}$ (and never $\ce{NaCl}$) in this answer.


Work done when P,V, and T are changing simultaneously

As far as I am aware, there is no generic formula for when all three variables are changing at once.

In this case, the easiest way is to simply use the area of a trapezium.

If you insist on integrating, then you need to know what $P$ is as a function of $V$. This derives from basic principles of integration - if you have a function $f(x)$, then $\int_a^b f(x)\,\mathrm{d}x$ gives the area under the curve of $f(x)$ from $x = a$ to $x = b$.

The line $\mathrm{C \rightarrow A}$ can be easily found since the beginning and end points are given. The gradient of the line is

$$m = \frac{2P_0 - P_0}{2V_0 - V_0} = \frac{P_0}{V_0}$$

and the the intercept is clearly 0 from the diagram (you can work it out anyway). In general the equation of a line $y = y(x)$ is given by $y - y_i = m(x - x_i)$ where $(x_i,y_i)$ is any point on the line.

That was perhaps an unnecessarily long explanation but the point is that you now know the relation

$$P = mV$$

holds (where $m$ was determined earlier). Therefore

$$\begin{align}
w &= -\int_{2V_0}^{V_0} P\,\mathrm{d}V \\
&= -\int_{2V_0}^{V_0} -mV\,\mathrm{d}V \\
&= m\left[\frac{1}{2}V^2\right]_{2V_0}^{V_0} \\
&= m\left(\frac{1}{2}V_0^2 - 2V_0^2\right) \\
&= \frac{P_0}{V_0}\left(-\frac{3}{2}V_0^2\right) \\
&= -\frac{3}{2}P_0V_0
\end{align}$$

which is the same as you found via the first method (you also forgot to put the negative sign in the right-hand side of the equation, but I assume that is just a typo).


Is equilibrium potential for K+ always ~-90 mV?

Most ions don't pass freely through the membrane. This is prevented by the hydrophobic (water "fearing") fatty acid tails of the phosholipids that provide the basis of the membrane. The regions outside and inside the membrane are aqueous and attract ions much more than does the hydrophobic fatty region in the interior of the membrane bilayer.

Under usual physilogical conditions, the only way ions can cross the membrane is through protein channels. But these channels are gated and only open in response to certain stimuli. Stimuli can include a chemical substance binding to the gate and opening it, this is how neurotransmitters work. Many gates respond to changes in the overall potential across the membrane. These are termed voltage regulated gates and, among other finctions, explain how action potentials (impulses) travel down neurons and muscle cells.

Regardless of how these gates open, the effect is temporary. Neurotransmitters break down, the membrane potential continues to change in a manner that voltage regulated gates once again close.

During these events only small amounts of ions cross the membrane compared to the vast supplies on whichever side has the surplus. That would be the interior of the cell in the case of potassium. Yes, this might eventually deplete the concentration if allowed to continue without some mediation.

That mediation is the sodium-potassium pump. The Na-K pump, using the energy in ATP, pumps Na+ out and K+ into the cell. But it pumps out 3 Na+ for every 2 K+ it brings in which makes the overall resting membrane potential -70 mV. This pump is busy and important. It is responsible for using about 20% of the ATP your body produces. Even more if you're inactive.

So, after some of the K+ ions are allowed out through gated channels, the potential changes very locally. Then other K+ ions migrate into this area, down their concentration gradient, restoring mmost (but not all) of the normal K potential. Operation of the Na-K pump provides the additional K+ ions to restore it completely