Is it possible to make metal fulminates?

In The Preparatory Manual of Explosives  (paperback at amazon), the author, Jared Ledgard, provides a procedure for the preparation of copper fulminate that can be summarized as follows:

  To 100 mL of 70% $\ce{HNO3}$, 30 g of anhydrous $\ce{Cu(NO3)2}$ is added under stirring. If the copper nitrate doesn't dissolve properly, add water dropwise until it does. (NOTE: Adding water to concentrated acids usually isn't the best idea!) The solution is stirred for 24 hrs at room temperature and then heated to roughly 80 °C. Under stirring, 120 g of 95% ethanol are added. After addition of the ethanol, the heat source is removed and the mixture is alllowed to reach room temperature. The precipitate is rapidly filtered off, washed with several hundred mL od cold water and dried in a desiccator over magnesium sulfate. The product, copper fulminate, should be stored in a desiccator over sodium sulfate in a refrigerator.

NOTE

To my opinion, the procedure leaves out significant information, such as the exact temperature of the aqueous solution while adding the ethanol. It is not clear whether and how much ethanol is supposed to evaporate. While the author suggests to carry out the reaction in a beaker, a setup with a two-neck flask, reflux condenser and a dropping funnel might be a better idea.
Handling fairly concentrated nitric acid requires safety precautions: A fume hood, eye protection, gloves and a lab coat are mandatory!
The reaction product is a primary explosive! While less shock sensitive than mercury fulminate, it still will go off on impact (or upon heating above ~200 °C).

In the case of these or similar sensitive materials, there is a good rule of thumb: If you have to ask on a Q&A site - Don't touch it!


Why do nanoparticles have a different color than their macro counterparts?

There is a fairly simple explanation of why small-enough particles are different from bulk materials. Once the particle becomes similar in size to the wavelengths of light involved then quantum effects start to matter for how the particle behaves.

The actual mechanisms that give specific colour may vary, but the main point is that the size of the particle comes to dominate the bulk properties of the material. In a bulk semiconductor, for example, the electronic properties depend on the band gap between occupied electron levels and unoccupied electron levels. When the particle is small enough, however, the electrons are more constrained and become more like an electron confined to a small box. This can mean that the possible energy levels of the electrons are determined by the physical size of the box not the bulk properties of the material. Hence any emission or absorption of light will be different to the bulk material. The additional physical constraints impose a different set of energy levels on the electrons.

The actual details can be quite complicated but the basic idea is that when particles are small compared to the wavelengths of light, you have to take into account the extra constraints on the system when calculating possible energy levels involved in electron transitions which are what causes colour.


NMR chemical shift range of different elements

The are a number of important factors that contribute to the shielding of a nucleus. Chemical shifts arise due to differences in the local magnetic field in the different environments within a molecule, driven by overall electron density about that nucleus. That is to say, the effective magnetic field experienced by a nucleus is a function of the shielding constant, σ. There are a number of components of σ.

$$B_\mathrm{eff} = B_\mathrm{o}(1- \sigma)$$

σdia is the local diamagnetic contribution
σpara is the local paramagnetic contribution
σm is the neighbour anisotropy effect
σrc is the ring current contribution
σef is the electric field effect
σef is the solvent effect

The two most important factors for the chemical shift range are the diamagnetic and paramagnetic contribution. All other contributions are quite small and operate within the effects of these fist two.

The diamagnetic contribution comes predominantly from electrons in the s-orbital; for most nuclei this is almost constant for all arrangements. Its effect diminishes with increasing distance from the nucleus, which is why s-orbitals play a more important role than p-orbitals. Only for those nuclei that have their outer orbitals as s-orbitals ($\ce{^1H, ^2H, ^7Li, ^{23}Na}$ for example) will the diamagnetic contribution be the dominant contributing factor, and this effect is generally small, providing only modest chemical shift ranges of up to tens of ppm.

The paramagnetic contribution is the dominating factor for all nuclei that have electrons in outer shell p, d and f orbitals. Paramagnetic shielding comes from the imbalance of electron density of the valence electrons in the outer orbitals. The way I get my brain around this is by thinking that the larger the non-spherical orbitals, the harder it is to get uniform distribution. The effect of paramagnetic contribution is large; up to many thousands of ppm.

To address your question, there is a general relationship that describes the expected chemical shift range being proportional to the inverse cube of the valence shell radius and also to the populations of the p and d orbitals. So, you are correct in your observation of chemical shift range being basically proportional to atomic number.

A useful reference for you is:
W. von Philipsborn, Chem. Soc. Rev. 1999, 28, 95. DOI: 10.1039/A706424A


Why does acetonitrile have a larger dipole moment and boiling point than acetaldehyde?

Using molcalc.org to get a qualitative idea about what is going on, here are the charge distributions and the partial charge assigned to the electronegative N (-0.19) or O (-0.40):

Even with the quick cheap calculation done in the educational software, the calculated dipole moment of acetonitrile (4.05 D) is larger than that or acetaldehyde (3.16 D).

How do we rationalize a smaller dipole for acetaldehyde given the larger negative partial charge on the oxygen atom? You could consider the dipole of the methyl group separately from the dipole of the two functional groups. In acetonitrile, these dipoles line up, so they are additive. In acetaldehyde, they are at an angle, so the (vector) sum will be less.

If you look at all the calculated partial charges in acetonitrile, the hydrogen atoms are assigned most of the positive charge (3 x 0.08), and the nitrogen atom most of the negative charge. This results in a large charge separation, which influences the dipole moment (charge difference x distance). Surprisingly (to me), the carbonyl carbon is assigned a charge of -0.06.

If you look at all the calculated partial charges in acetaldehyde, the largest positive partial charge is assigned to the carbonyl carbon (+0.33), making for a shorter distance of the charge separation.

To get a more quantitative picture, one would have to perform a more serious quantum-chemical calculation. This would also be an opportunity to look at the orbital shapes and their relation to the charge distribution. This quick calculation, however, already points out some of the salient differences.


Could the difference between absorption time of Diclofenac Sodium and Diclofenac Potassium be related to the cation?

Under physiological conditions, the counter-ion is almost entirely dissociated from the drug, meaning that the identity of the counter-ion isn't directly responsible for how well a drug performs. This has been looked at extensively by medicinal/formulation chemists, who often screen many salt forms looking for desirable properties.

From the graph below, you can see that the plasma concentration curves are essentially identical for sodium, potassium, and calcium salts (slight variation in Cmax and Tmax but within experimental error):

Source: Drug Like Properties: concepts, structure, design and methods, Chapter 7

One case in which the identify of the cation does matter is where there is a common ion in the body. This 'common ion' effect is most common when amine drugs are presented as their hydrochloride salts. The stomach is acidic (hydrochloric acid), and therefore the presence of a high concentration of chloride ions can prevent the salt from turning into the free drug.

What is likely to be different between different cations the form that the salt takes. Different salt forms have different packings (some might be amorphous, some might be highly crystalline and ordered). The formulation may also differ (powder coated tables vs gelatine capsules), which will change the time it takes for the drug to be released/absorbed.


I'm going to continue to resist calling C the heat capacity because, as used here, it is not path-independent.  Instead, I will take it to merely represent the derivative of the heat added per mole of ideal gas with respect to temperature for a polytropic process path.

With this caveat, here is my interpretation of what is happening mathematically when C takes on negative values:  If the gas is expanding, a negative value of C means that we added heat to the gas during the expansion, but not enough to offset the expansion cooling that would occur in an adiabatic reversible expansion, so its temperature decreases while we add heat.  If the gas is being compressed, a negative value of C means that we removed heat from the gas during the compression, but not enough to offset the compression heating that would occur in an adiabatic reversible compression, so its temperature increases while we remove heat.  It's as simple as that.


Is Gibbs free energy the sole determiner for spontaneity?

For a spontaneous process at constant temperature and pressure, $\Delta G$ must (emphasis on must) be negative.

Spontaneity of a process , explicitly, depends on the enthalpy change and the entropy change. The Gibbs free energy change includes both these factors to predict the spontaneity of a process.

So, yes it can be said that Gibbs free energy change is the main factor that decides the spontaneity of a process.

Putting into better words, $\Delta G$ being negative is a consequence of a process being spontaneous.


Cathodic protection of tubes in Shell & Tube Heat Exchanger

Cathodic protection requires an electrolyte to work. You could just pass current through the tubes but how is it going to return to the power supply? If there is no electrolyte to conduct ions it cannot provide any protection to your system. Have you considered replacing your heat exchanger with glass or glass-coated metal? If the capital costs of this are prohibitive, I would think your best bet would be to try drying the gas some more, as you mentioned. Do you happen to know the water content of the gas going into the exchanger now?


Difference between chemical energy and potential energy

There exist many different potential energies. Maybe you would prefer an analogy. For example let's consider an object on a table (a ball, a book or a piece of chalk). It has a potential energy which is not chemical for the moment. It is not visible. It is a hidden energy. It can only be shown if you give the object an opportunity to fall down and to go to a place where the potential energy is lower. During the fall, the potential energy is transformed into kinetics energy (its speed increases). When it touches the ground, the kinetics energy is transformed into heat which is another sort of energy. This gravitational potential energy is equal to mgH, where m is the mass of the falling object, H is the beginning height of the table, and g is a universal constant (9.81 m/s2 on Earth). Before the fall, the object has a sort of hidden energy, which we call gravitational potential energy (or mechanical energy), and this energy could only be expressed if you give the object an opportunity to go to a lower place. The object has a higher gravitational potential energy on the table than on the ground.

It is the same in chemistry. H2 and O2 have a similar hidden energy, which is a chemical potential energy. If you break the bonds H-H and O-O, you need energy to do it. You give energy to the atoms. It is similar to the operation of picking up the object on the ground and put it higher, on the table. If you imagine going back to H2 and O2, you would get the energy you had consumed for breaking the bonds. But now, if you prefer recombining the H and O atoms into H2O molecules, you will get much more energy (much more heat).

The object on the ground is similar to H2 and O2. When you give it some energy, you can lift it up to the table, or better to a state of separated H and O atoms. But now if the table is just at the edge of an abyss, you can get more energy by letting the object fall down the cliff. In this case you obtain H2O and not H2 and O2. The chemical potential has decreased to a very low level if the final state is H2O. The fall has produced much more heat than the recombination H2 + O2.

Chemists often use this analogy between height and chemical potential energy (which is usually called enthalpy). They draw short horizontal lines above one another to simulate  different states, as if one line represents the table (H and O atoms), and another one (lower) the level of the ground (H2 and O2). Of course the level of H2O is another horizontal line : much much lower.

Have you followed me ? Is it what you needed to understand the potential energy ?


Finding percentage of the gas in a binary gaseous mixture of the given density

Well, I can see all the relations you require knowledge of in the question itself!

Calculate the effective molecular mass from

$$M = \frac{dRT}{P}$$

$R$ is known, $d$ given. $P$ and $T$ are available from the fact that it is at STP.

$M$ comes out to be ${20.176 u}$.

Effective molar mass is easily calculated below:

$$M_\mathrm{eq} = M_1x_1 + M_2x_2$$

where $x_i$ is the mole fraction of each gas. You can replace $x_2$ by $1-x_1$.

Plugging in the values and solving for $x_1$, argon mole fraction comes out to be $0.4493$.

I believe you can now calculate the demanded percentage, be it by mass or by moles.


I think a really full answer will require delving into the theory of NUS, which (just like the Fourier transform) goes way beyond NMR as a field: the idea of undersampling a signal and then reconstructing it is of great interest in a variety of engineering areas. I can't give that level of detail, so I don't pretend that this is really an answer, but maybe having some of this written down will help.
I'm really not an expert in this, but I can backup Mad Scientist's comment that for 2D data, TopSpin's reconstruction mode (the MDD_mod processing parameter) generally should be set to cs and not mdd. I suspect this may require a license from Bruker, so it might be best to check with the relevant NMR manager. Mnova, on the other hand, uses a different reconstruction algorithm entirely, so it's not overly surprising that there might be some differences in performance.
The amount of undersampling you want to do will also affect the quality of your results. I think a good rule of thumb is to make sure the ratio of points measured in $t_1$ to number of peaks in each column of the spectrum is somewhere around 2 to 5. The reasoning is simple: if you need to detect $k$ peaks in the frequency domain, then you need to measure some multiple of $k$ signals in the time domain. In other words, the more data you want to pick up, the more signals you need to measure. For a HSQC where each column has very few frequencies, you can get away with very sparse sampling, and 25% is likely to be fine (as long as your TD1 is not too small). For other spectra like COSY, TOCSY, HMBC where there is less empty space in the spectrum, this may not work quite as well.
As far as I'm aware, increasing the number of scans should have a similar effect for both uniform and non-uniform sampling. The main changes are the increase in S/N, as well as possibly better artefact suppression achieved via phase cycling. However, with modern gradient-selected experiments (especially the standard experiments like HSQC, COSY, etc.) you should not really need extremely long phase cycles to achieve good artefact suppression. In my experience the difference in spectral quality between NS=2 and greater is already pretty minimal. Bumping up the number of scans is mainly useful for increasing S/N. [Obviously, if the reason why the NUS reconstruction is failing is because of too poor S/N, then increasing the number of scans will help.]
Lastly, this is a trivial point, but I'd also suggest double-checking the other processing parameters (window functions, zero-filling etc.) after the reconstruction is done. It's quite easy to miss something out (for example, changing QSINE to SINE) and inadvertently get ugly peak shapes. I learnt (and am still learning) it the hard way.
Having said all this, what should you do if you find your NUS data is not good enough? I don't think there are really many answers to this, although someone more knowledgeable will be able to make better suggestions. One is to play around with the reconstruction algorithm, but unless you are very well-versed with the theory, this is essentially limited to trial-and-error with a dropdown box. The other is to just reacquire the spectrum with less undersampling, especially if you suspect you may have pushed it a bit too much.


Sr(NO3)2 + Na2SO4 = SrSO4 + 2NaNO3

So, here we have, 1 mol Sr(NO3)2 reacts with 1 mol of Na2SO4 by unitary method, you can say, 0.003 mol Sr(NO3)2 reacts with 0.003 mol of Na2SO4.
But checking the reactants we have 0.003 moles Na2SO4 and 0.014 moles Sr(NO3)2

So only 0.003 moles of each reacts with the other to leave out any excess in either of the reactants, as in this case, Sr(NO3)2.

When 0.003 mol reacts of Sr(NO3)2, we have $$0.014 (initial) - 0.003 (used) = 0.011mol$$ of Sr(NO3)2 left.


How can I determine if a chemical will dissolve aluminum oxide, on paper?

Normally you would learn about properties of chemicals from general chemistry textbook. Al2O3 is pretty stable and is hard to dissolve in anything at room temperature.

It also depends a lot on what is the physical state of Al2O3. Fresh fine Al2O3 is much more reactive than aged (or coming in big chunks) Al2O3.

Searching in general chemistry literature is the best way to go, but Al2O3 is really robust.


IUPAC: Would I prioritize low numbering to highest-priority group, OR try to assign lowest numbers overall?

Very loosely speaking, if your OH group is being named as a prefix, i.e. "hydroxy-something", then it's no longer "higher priority" than the methyl group, which is itself also a prefix. The "priority" only matters when you're deciding on the suffix, and the aldehydes already occupy that spot.
Yet, that rather informal and unsubstantiated claim probably won't satisfy some readers, and it shouldn't satisfy you, as I could be any useless person on the Internet feeding you false information (which is sadly common). Let's just say that if you had to speak to your teacher about it, you would probably be embarrassed to tell them that "somebody on the Internet wrote a comment that said this".
So, I will quote a few rules in the IUPAC Blue Book (Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013) which need to be looked at to assign the name. The first step is to decide on the  "principal characteristic group". The seniority order is listed in Table 4.1, P-41:

1 Radicals
2 Radical anions
[...]
15 Aldehydes and chalcogen analogues
16 Ketones [...]
17 Hydroxy compounds and chalcogen analogues (includes alcohols and phenols) [...]

The principal characteristic groups here are obviously the aldehydes. Notably, P-33.1 states that "only one [suffix] can be placed at the end of a name to indicate the principal characteristic group", so the alcohol is now relegated to being a prefix, i.e. "hydroxy-". The next step is to decide on the "parent structure":

P-44.1.1
The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

Again, it is quite clear that the parent structure here is the chain containing both aldehydes. Lastly, we come to the numbering:

P-14.4
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
[...]
(c) principal characteristic groups and free valences (suffixes);
[...]
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
[...]

(c) doesn't help us resolve the ambiguity at hand, because the aldehydes are at both ends of the principal chain whichever way round we count. So (f) is what matters, and note that it says "all considered together": it no longer distinguishes between any kind of "priority" between different prefixes. The locant set (4, 21) is lower than the locant set (19, 36), and is consequently the preferred numbering.
If you search Chem.SE for "P-14.4" there will no doubt be similar questions and answers, which may help to clarify the situation with similar examples.
Note that the "lower" locant set is obtained by point-by-point comparison i.e. comparing the first locant of both at the same time, then the second locant of both, etc. until any ties are broken (P-14.3.5). In this case, (4, 21) is lower than (19, 36) because 4 is lower than 19; there is no need to compare the second locants. A common and baseless misconception is that the smallest "sum of locants" wins. This is not true and is not written anywhere in the IUPAC Blue Book.


How could a rechargeable battery recharge itself?

As the comment indicated, it is probably equilibrating. When you are discharging the battery, you are causing the materials in the electrodes and the electrolyte to rearrange. The surface of the electrodes are where the reactions are happening, so the products of the discharge reactions tend to accumulate at or near the surfaces of electrodes. This causes the measured potential to go lower.

Once the discharge stops, the contents start to homogenize within the cell bringing the discharge products away from the electrodes and bringing the material from the charged state closer to the electrodes. This in turn increases the measured potential.


Optical Isomerism in Coordination Compounds of type MA₂B₂(CC)

I came to the same conclusion as you, total 4 stereoisomers (2 of them create one enantiomeric pair).

In the 3rd party solution, they depict the same isomer at bottom, i.e. only 3. The missing one is that with two $\ce{NH3}$ ligands in the same plane with $en$, and $\ce{Cl-}$ ligands in straight line (trans), perpendicular to that plane (not that horizontal, which is depicted)

Here is my simplified schematic representation of the octahedral complex with ethylenediamine ligand bound by back-pointing bonds in horizontal plane, hidden; the rest is vertical axis and two front-pointing horizontal bonds (seesaw geometry, not to be confused with Fischer projection for tetrahedrals!) Dotted line is mirror plane, equal sign means identity (by 180° rotation around observer-to-model-line axis).

has B ligands in the (en)-defined plane, A ligands perpendicular (axis)
(it's their bottom left isomer, idenitical with bottom right, for A=$\ce{NH3}$, B=$\ce{Cl}$; note that they have en oriented vertically, everywhere, with one end in dotted horizontal plane, other one in upper vertical position, I have it somewhat simpler, rotated for en to be horizontal);
has the opposite, A's and B's swapped (and it's not mirror image to 1.)
(it's their missing isomer);

If A and B are in the plane then there are two possibilities how to relatively place A and B in the perpendicular axis (resulting in mirror images),

(it's their top right);
(it's their top left).


Understanding the CIP configuration for cyclic substituents with hetero-oxygen atoms

CIP rules for cyclic substituents are different. See
this question for the rules. (The rules are given in the question itself).

This will be the hierarchical digraph for the right side of the molecule.

The atoms encircled are the ghost atoms. From the digraph one can see why the compound is Z.


Why did HF disolve this lightbulb at the surface rather than more submerged points?

Near surface of the liquid the glass is a bit hotter. All glass of the bulb gains comparable amount of heat energy from tungsten wire, but parts above the liquid are cooled only by air, so they are much hotter. Part of this heat is transferred into slightly submerged part, so it is the hottest part that is under liquid, and all reactions are quickened by increase of temperature.

However, this exact experiment is a bit ambiguous, as glass here is under extreme thermal stress, and shattering near surface of the liquid may be addressed to that.


Potassium and sodium nirate can be used to make black powder (gun powder), but they are not explosives by themselves. It is safe to store bags of salpetre in the garage.

If there's salpetre forming on the brickworks, I'd rather be concerned to keep the bags closed and away from that wall: sodium nitrate is rather hygroscopic and salpetre crystals on and between the bricks indicate that the wall is wet.


Calculating molar mass of an oxidant by iodometry

The molar mass of a substance, as you said earlier, is:

  simply the mass of the substance divided by the number of moles in the same amount of substance

In this question, you can see that as per part b), we found that there were $7.553\times10^{-4} \text{ moles}$ of $\ce{M^{n+}}$ present.

Hence, the salt, i.e., $\ce{M_xA_y.ZH2O}$ contains $\ce{M_x}$ = $7.553\times 10^{-4} \text{ moles}$ $\ce{M^{n+}}$.

=> $\ce{M}$ = $ \frac{7.553\times 10^{-4}}{x} \text{ moles}$ of $\ce{M^{n+}}$

$\therefore$ taken moles of $\ce{M_xA_y.ZH2O}$ = $\frac{7.553\times 10^{-4}}{x} \text{ moles}$

But we know that the taken moles have a mass of $0.2642 g$.

Now, divide $0.2642 g$ by $\frac{7.553\times 10^{-4}}{x} \text{ moles}$ to get the molar mass, in terms of x, which is $349.8x$


What is Hydrogen bond Auto-correlation function

In eqn 1 and 2 the correlation functions are written in the form $<A B >/<A>$ , the denominator  is the average of $A$ and the numerator the correlation function. This is the the integral product $U(t_d)=\int A(t) B(t-t_d)dt$ where $t_d$ is some time delay. When $A$ and $B$ are the same thing then this is the auto-correlation. It measure how similar the signal A is to B at different times. If $A$ and $B$ are random signals (such as white noise, or from a random number generator) then $U$ is 1 at zero time delay (both signals exactly superimposed) and zero at all other delays and this is because the correlation has no 'memory'; any point in a random signal is not related to any other. If $A$ and $B$ are constant then $U$ is one at all times, i.e perfect 'memory'. In your example the H bonds have a certain lifetime because are being broken due to molecular motion (as explained in the paragraph under eqns 1 & 2) and so the correlation function decays away. A plot of the correlation gives the lifetime of the H bonds.


Relationship between Productivity and Yield

Depending on your application and the specifics, to keep it simple. . .

Yield: The amount of product obtained in a chemical reaction
Productivity: The rate of speed a product can be obtained in a chemical reaction

For the compromise, as vague example, perhaps the more of the product you want to yield, the more time it may take thus a longer productivity you'll have.


Are you asking about catalytic hydrogenation or non-catalytic hydrogenation (e.g. diimide)?  For the time being, let's assume that your question is about the catalytic process.  All the catalyst does is lower the activation energy required to reach the transition state (TS) and the TS will look about the same independent of which isomer you start with.  Therefor, ground state stability differences between the two olefins will determine which isomer reacts faster.  Since trans-olefins are generally more stable than their corresponding cis isomers, it will take more energy (the difference in stability between the cis and trans isomers) for the more stable (lower energy) trans isomer to reach the TS, and less energy for the cis isomer to reach the TS.  Consequently cis isomers react somewhat faster than their trans-counterparts during catalytic hydrogenation.  I suspect that the same considerations would apply to non-catalytic processes.

As an aside, remember that in the catalytic process, depending upon catalyst, temperature, solvent, pressure, etc.; many other things can be going on, such as cis-trans isomerization of the starting material, positional isomerization, etc.  So if the reaction conditions are not chosen carefully and 1) if cis-trans isomerization is faster than hydrogenation and 2) let's say the cis isomer reacts 100 times faster than the trans; then both isomers could appear to give the same rate of hydrogenation because all hydrogenation would proceed through the cis isomer.


How to convert Excel comments to ChemDraw .cdx?

Because you find these entries in a Excel workbook, it is likely that this spreadsheet was created with Chemdraw's add-in for Excel (company's demo) to embed chemical formulae:

(source).
to provide a bridge between the two worlds to display chemical libraries (promotional example, or manual chapter 13).  To use it, you need both access to Excel as well as ChemDraw.
The (part of the) string shown by you does not look like the typical content of the binary .cdx, nor the human readable .cdxml file which would be understood by openbabel, a program written in the intent to interconvert structure files used in chemistry (it may read .cdx; but read and write .cdxml) on a one-by-one basis (e.g., in a GUI), or programmatically (e.g., C, Python).
Maybe you may identify the author of the spreadsheet.  Or find support by the software's authors (for some years now by PerkinElmer) / add-in maintainers (e.g., on GitHub).  It is not obvious their format were readable by similar products (e.g., ChemAxon's JChem).


Racemization isn't "exact," but rather very very close to equality. It is just simple probability.

Think of flipping a coin, p=probability for heads, and q=probability of tails. Now for a fair flip p=q=0.5. From binomial theory  the standard deviation is $\sqrt{n\cdot p \cdot q}$ where n is the number of flips. Now let's assume 2 standard deviations difference, which is roughly at the 95% confidence interval.

If you flip 10 pennies then a two standard deviation difference is $2\times \sqrt{0.5^2\times 10} \approx 3$ in the number of heads.

Now flip $6.022\times10^{23}$ dimes then a two standard deviation difference is $2\times \sqrt{0.5^2\times 6.022\times10^{23}} \approx 7.8\times10^{11} $ in the number of heads.

But now think of the % difference.

$3$ heads in 10 tries for the pennies is $30\%$.
$7.8\times10^{11}$ more heads when flipping $6.022\times10^{23}$ dimes is only a difference of $1.3\times 10^{-10}\%$ which is an insignificantly small difference.


What does the osmotic pressure do in an electrolytic solution?

Actually the ions cause what is called osmotic pressure, not the pressure cause the ions to flow. As positive charge accumulates on the electrode, it has a tendency to attract extra negative charge towards itself, as alike attract each other. It starts behaving as the cathode and this phenomenon starts to develop what is called osmotic pressure.


Why and how carboxylate ion has greater stability than carboxylic acid?

First of all, it would have been clearer if it's a misprint or not if you had attached a screenshot of the wording in your textbook (you still can).

At any rate, the subject of stability of carboxylic acids and carboxylate anions usually arises when discussing the level of acidity or in other words, the strength of the acid. It is pointed out in most organic chemistry books (where electron delocalization is explained) that electron delocalization or resonance is a concept that contributes to a certain structure's stability. That is to say, when a negative charge can be delocalized over several atoms (as is the case with a carboxylate group), instead of being localized on a single atom (as is the case with an alkoxide group), it is considered more stable energetically. (see here, section B - Resonance).

The relative stability of the carboxylate anion means that a carboxylic acid will favor (again, due to energetic considerations) the release of a proton in a basic environment (this is important, since in an acidic environment, the equilibrium will shift towards the acid, and much less carboxylate will exist in the solution), since the conjugate base (the carboxylate anion) is relatively stable, and therefore can exist in the solution in this form.

 (Taken from Organic Chemistry, by Paula Yurkanis Bruice, any edition really...)

To sum up, Carboxylic acids and their conjugate carboxylate anions are not (and cannot be) compared in terms of stability. Under the right conditions, i.e. in a basic solution, a carboxylate anion will be favored over a carboxylic acid because:

a. There is a base to take a proton (be it a hydroxide ion, an aminic group or any other H+ acceptor).

b. The carboxylic group will release the proton rather easily, because what's left (the carboxylate anion) is relatively stable due to electron delocalization.


What is the need for so many theories of acids and bases?

First came the Arrhenius theory. It defines an acid as a species that dissociates to produce $\ce{H+}$ in solution, and a base as a species that dissociates to produce $\ce{OH-}$ in solution. (Examples: $\ce{HCl}$, $\ce{NaOH}$)

Next came the Brønsted-Lowry theory, which expands the notion of acid and base. It defines an acid as a species that loses $\ce{H+}$ in a reaction, and a base as a species that gains $\ce{H+}$ in a reaction. All Arrhenius acids and bases are Brønsted-Lowry acids and bases; hence the Brønsted-Lowry theory generalizes the concepts of Arrhenius acids and bases and allows the application of these concepts to a larger range of species. (Examples of "new" acids and bases under this theory: $\ce{NH4+}$, $\ce{CH3COO-}$)

Finally came the Lewis theory, which further expands the notion of acid and base. It defines an acid as a species that accepts an electron pair, and a base as a species that donates an electron pair. All Brønsted-Lowry acids and bases are Lewis acids and bases; hence the Lewis theory generalizes the concepts of Brønsted-Lowry acids and bases and allows the application of these concepts to a larger range of species. (Examples of "new" acids and bases under this theory: $\ce{Li+}$, $\ce{CO}$)

Remark: Hopefully the repetition and structure within my answer has been suggestive. The benefits of generalization are obvious: the theory can explain more, and more things can be explained; this is the natural progression of science.


Show that this mechanism is consistent with the rate law - how to?

You have a three step mechanism where only one step is slow and the other steps are fast.

Since the first two steps (which are, in fact equilibria) are fast, it can be assumed that each of the equilibria holds throughout the reaction. Therefore, the expression of the equilibrium constant for each step can be used:
$$\ce{K_1=\frac{[I]^2}{[I2]} \,\,\, and \,\,\, K_2=\frac{[H2I]}{[H2][I]}}$$
So, $\ce{[I]=\sqrt{\ce{K_1[I_2]}}\,\,\,\,and,\,\,[H2I]=K_2[H2][I]=K2[H2]\sqrt{\ce{K1[I2]}}}$

As only the last step is slow, the rate of formation of the product (which is the rate of the overall reaction) only depends on the last step.

$$\ce{rate=\frac{1}{2}\frac{d[HI]}{dt}=k_s[H2I][I]}$$
Here, $\ce{k_s}$ is the rate constant of the last, slow step. Substituting the expressions for $\ce{[H2I]}$ and $\ce{[I]}$ in the rate law should give you an expression of the form $\ce{rate=constant\times[H2][I2]}$, which is what the given rate law is.


Calculate the number of moles produced per hour per liter of air?

That is the correct way to go about it. In your calculations the rate of the reaction should include L because M is defined as moles/L. (1.27×10^-3 moles/Ls)(3600s/Hour) = 4.572 moles/Lh NO₂ produced.


What is the basic concept of positive and negative charge of an atom?

Sodium's proton number doesn't change else it'd become a different element.

In basic terms, sodium donates an electron to chlorine's outer electron shell; there is no electron 'stealing'. As much as both ions' nuclei are surrounded by electrons, the charges of the ions are based on the ratio of protons to electrons. For example, in an atom, the number of protons and electrons are equal to give no overall charge.

However, in an ion it's different; the sodium ion has 11 protons but only 10 electrons, so has a relative charge of 1+, as there's one more positive charge (proton) than there are negative charges (electrons). On the flip side, the chloride ion has 17 protons and 18 electrons, so has a relative charge of 1-, as there's one more negative charge than there are positive charges.

This then causes the electrostatic attraction between the two ions, which we call ionic bonding.


How to distinguish between a coordinate bond and a covalent bond?

Coordinate bonds can be distinguished from (other or ‘normal’) covalent ones experimentally.

Let’s look at the bond creation first as Newbie suggested. Typically, one would use a Lewis base and a Lewis acid to create a coordinate bond between them.

$$\ce{NH3 + BH3 -> H3N\bond{->}BH3}\tag{1}$$

In contrast, one would usually consider a normal covalent bond to be created from two radicals:

$$\ce{H3C^{.} + {}^{.}CH3 -> H3C-CH3}\tag{2}$$

But we can also create covalent $\ce{C-C}$ bonds by using a carbanion source and a carbocation source (slightly different example, because both ions are actually known and react with each other):

$$\ce{(H3C)3C+ + {}^{-}CN -> (H3C)3C-CN}\tag{3}$$

Thus, we can’t really tell a difference here. The difference in fact lies in the way a bond dissociates if the bond dissociation enthalpy $D = \Delta_\mathrm{D} H^0$ is provided. Standard covalent bonds will dissociate homolytically by the following equation:

$$\ce{H3C-CH3 -> H3C^. + ^.CH3}\tag{4}$$

A coordinate bond will, however, dissociate heterolytically thus liberation the original Lewis acid and base:

$$\ce{H3N\bond{->}BH3 -> NH3 + BH3}\tag{5}$$

In cases such as the ammonium ion ($\ce{NH4+}$) or say $\ce{[AlCl4]-}$ this means we have a certain number of normal bonds and another certain number of coordinate bonds (three and one, respectively in both examples). Although they are all between the same atoms, we can distinguish if we remember that neither ion can exist by itself, there will always be a corresponding counterion. Thus, if we consider $\ce{NH4Cl}$ or $\ce{Na[AlCl4]}$, we realise how the first bond is broken differently:

$$\begin{align}\ce{NH4Cl &-> NH3 + HCl}\tag{6}\\[0.6em]
\ce{Na[AlCl4] &-> NaCl + AlCl3}\tag{7}\end{align}$$

And all further bonds are cleaved homolytically. We gain two different bond dissociation energies, one corresponding to the coordinate bond and one corresponding to the remaining bonds. (Note that salts such as $\ce{AlCl3}$ or $\ce{NaCl}$ are cleaved homolytically by heating in the gas phase — flame-colouring would not happen if we had $\ce{Na+}$ in the flame.)


The very question you pose is addressed thoroughly in this open access work:
https://www.nature.com/articles/ncomms9287 .
The short answer to your question is that the electron density can be mapped using a technique akin to diffraction as described in their work.  You mentioned a distinction between the electron density and the orbitals, and the orbital is not itself observable; but as you suggested, the electron density can be mapped.


Work can be understood as a variation of energy during which the transporters (may they be molecules) are spatially aligned or bonded, meanwhile heat is a variation of energy in which the transporters aren't aligned (therefore heat dissipates energy).
With this in mind, any machine that uses an ordered energy transfer (like your hand waving) can induce heat in a fluidic means, as particles in fluids are initially moving randomly and are not likely to move far on the orientation of your hand.
In this example, the induced work is little and restricted to the analysis on the surface of the moving hand, but if the analysis is made from further on a wider system, we can assume you're doing work to generate heat alone.


The gases in the distillation column boil off according to their boiling point.

   Gas          Boiling point
   Nitrogen     77.355 K (−195.795 °C)
   Argon        87.302 K (−185.848 °C)
   Oxygen       90.188 K (−182.962 °C)

The distillation column shown is setup for continuous distillation. But imagine that a pot distillation was being done. Nitrogen would boil off first, then argon, then oxygen. So the various trays in the distillation column represent different temperatures. Nitrogen boils off at the top since it has the lowest boiling point, and oxygen collects in the "pot" since it has the highest boiling point.


How to convert chemical rate of reaction into time?

Well, there isn't usually a specific time that a reaction reaches completion or equilibrium. You need to choose an arbitrary limit, such as "when 90% of the reactant has been converted to product".

You can calculate your time from an integrated rate law. This will require you to know what order your reaction is (perhaps CSE can help you there).

For example, for a first-order reaction, the integrated rate law is:

$C_t = C_0e^{-kt}$

Or, in a form more convenient for you:

$t = \frac{-1}{k}ln(\frac{C_t}{C_0})$

where $k$ is the rate constant of the reaction, $C_0$ is the initial concentration, and $C_t$ is the concentration at the time $t$ you are interested in.

If you choose the arbitrary limit I mentioned earlier (convenient for demonstration, but you'd probably want more product formed that 90%), at your "completion" time there would be 10% reactant left. The math would look like:

$\frac{C_t}{C_0} = \frac{10%}{100%} = \frac{1}{10}$

$t = \frac{-1}{k}ln(\frac{1}{10}) = \frac{1}{k}ln(10) \approx \frac{2.303}{k}$

where you have now found the time it takes for the reaction to proceed to your chosen point. Note that if you choose 100% conversion, the remaining reactant concentration would be 0. This gives you $ln(0)$, which is not defined for real numbers. To get a more meaningful understanding, you can take the limit as the reactant concentration approaches zero (the reaction approaching completion), and you will find:

$t_{completion} = \lim_{C_t \to 0}\frac{-1}{k}ln(\frac{C_t}{C_0})$
$= \frac{-1}{k}\lim_{u \to 0}ln(u) = \frac{-1}{k}(-\infty) = \infty$

And thus by our most useful models, it takes literally forever for every single molecule to react. This is true for any reaction order. This is why we cannot easily say when a reaction is "done", and we must pick an arbitrary point.

Note that all the math examples I presented only apply to first-order reactions.

To find the time your reaction "takes" you will have to: 1) choose an arbitrary point to call completion, 2) know the order of the reaction, and 3) know the rate constant of the reaction.

P.S. If you wanted, you could set $\frac{C_t}{C_0} = \frac{1}{2}$ in the equation solved for $t$, and you will derive the half-life formula $t_{1/2} = \frac{ln2}{k}$ mentioned elsewhere.


As above, your labor all by itself (even if you are grad student) is worth more than the sulfuric acid.  As a pedagogic exercise, run the solution down a strong anion exchange resin washed OH-.  Only glucose comes out.  Recover the sulfuric acid by displacing it with phosphoric acid.


Can tin(IV) sulfide be synthesized from tin(II) chloride via oxidation with sulfite in acidic media?

Yes, the reaction

$$\ce{2 Na2SO3 + 6 SnCl2 + 12 HCl → 4 NaCl + 5 SnCl4 + SnS2 + 6 H2O}$$

is described in chapter 35.6.4 Reactions of chalcogen oxo-compounds of the Russian translation [1, p. 526] of Blumenthal's Anorganikum 10th Ed. (in German).
Unfortunately, I don't posses original German edition, so the following is my English translation of the procedure:

  To the 0.5 M solution of $\ce{SnCl2}$ in 12 M $\ce{HCl}$ in three test tubes 0.2 M solutions containing $\ce{SO3^2-}$, $\ce{SeO3^2-}$ and $\ce{TeO3^2-}$ are added. In all three tubes precipitations of $\ce{SnS2}$, $\ce{Se}$, and $\ce{Te}$, respectively, are formed.

Original Russian text:

  К 0.5 M раствору $\ce{SnCl2}$ в 12 M $\ce{HCl}$, находящемуся в трех пробирках,
  приливают 0.2 M растворы, содержащие $\ce{SO3^2-}$, $\ce{SeO3^2-}$ и $\ce{TeO3^2-}$. Во всех трех пробирках наблюдается выпадение осадков, которые представляют собой $\ce{SnS2}$, $\ce{Se}$ и $\ce{Te}$ соответственно.

References

G. Blumenthal, S. Engels, I. Fitz, W. Haberditzl, K.-H. Heckner, G. Henrion, R. Landsberg, W. Schmidt, G. Scholz, P. Starke, I. Wilke, K.-Th. Wilke Anorganikum: Lehr- und Praktikumsbuch der anorganischen Chemie mit einer Einführung in die physikalische Chemie; Ed. of Russian translation: L. Kolditz; Mir: Moscow, 1984; Vol. 1.


Why do Rydberg atoms have electrons that can transition above the first ionization energy?

A useful concept in Rydberg atoms is the notion of a channel. This concept is borrowed from scattering theory and describes all solutions of the Schrodinger equation for the scattering of the Rydberg electron of specified angular momentum with the ionic core in a certain energetic state. The bound Rydberg states correspond to scattering at negative energy and the scattering at positive energy corresponds to the ion-electron continuum of the channel.
If the ion has multiple electrons, it can have several excited states. Each of these excited states can form a channel. The energy of a Rydberg electron is given by
$E=E_\text{IP} + E' - \frac{hcR}{(n-\delta_\ell)^2}$
Where $E_\text{IP}$ is the first ionization energy, $E'$ is the excitation energy of the ionic core, $R$ is the mass-corrected Rydberg constant, $n$ is the principal quantum number and $\delta_\ell$ is the quantum defect for a Rydberg electron with orbital angular momentum $\ell$.
Each excited state of the ionic core thus gives a new Rydberg series with each own set of principal quantum numbers. So, Rydberg states do exist above the first ionization energy, but their principal quantum number is finite; it just belongs to a different channel.

Image source


What's the most appropriate name for this reaction?

As Koch's reaction (Gatterman-Koch Formylation) adds a carbonyl group, and not a carboxylic acid group, I decide to go with Kolbe's reaction.

You started right, and then took the wrong turn ;-)

Under acidic conditions

the proton adds to the $\ce{C=C}$ double bond and gives a secondary cation
the cation reacts with carbon monoxide and yields an acylium cation
the latter reacts with water to result in the formation of a carboxylic acid

The whole process is known as the Koch reaction (without the Gattermann).


Why does hand sanitizer turn into a liquid after coming out of the bottle?

A little research and perhaps a small experiment might help you figure something like this out on your own. First, have you tried wikipedia? A small mention of research can go a long way. Describing what product you are interested in never hurts either. Is the polyacrylic acid, a dispersant in this substance? That material can absorb water effectively, so here is what I would do:

Take a very small amount of water and smear it on the same surface that it does not "turn into a free flowing liquid" and see if it does with the addition of water. If it works, you might conclude that your alcohol-based rubs use a suspension material that when contacted with a little water, such as from your palms, will cause the gel to have just enough water absorbed inside to decrease its viscosity.

Perhaps you will like to share the results of your experiment.

EDIT: OP seems to be observing a gel demonstrating Thixotropy, this I have determined from a quick google search


Is it relevant to pursue a master degree or a doctorate when applying to a job in the petroleum field?

Professional careers in the petroleum industry can be very competitive, so a higher qualification such as masters degree could make you a more attractive candidate for a scientific or engineering position. Many companies nowadays specify post-graduate qualifications as necessary or desirable.

A PhD is generally preferred for research positions.


What are the units of 85000 Molecular Weight of a polymer

First of all $M_w$ is not the number of monomers in a chain of PVA. The number of monomers in a chain is called the degree of polymerization.
$\overline {M_w}$ is the weight average molecular weight. It is a way of determining the molecular mass of a polymer. As polymer molecules have different chain lengths, the average molecular mass will depend on the method of averaging. For example, we can use the number average molecular weight which is based on different type of averaging.

As for the unit, it's as you mentioned: g/mol. In biochemistry, they use the dalton (Da) as a mass unit for biopolymers (1 $N_a.$Da= g/mol, where $N_a.$ is the Avogadro number). So, when we say a protein has a molecular weight of 64000 g·mol$^{−1}$, it has  also a mass of 64 kDa.

P.S. To be rigorous, we have to say ** molecular mass** instead of ** molecular weight**. But this was an old way of naming things.


Graph of compressibility factor vs pressure when real gas is assigned Z=1

For a real gas, $(P+a/V^2)(V-b) = RT$

Now, since we are considering $Z=1$ for real gas, all the real gas equations will be reversed for the ideal gas. Meaning, we had to subtract $b$ for volume correction of real gas, for correction of ideal gas we need to add $b$. Same, with pressure.

The equation then comes out to be: $(P-a/V^2)(V+b) = RT$

Now, for the graph, we can calculate the approximate values of $Z$ (relative to unity) for very low pressure, low pressure and high pressure.

Case 1: Very low pressure: $P \downarrow  \downarrow \implies V  \uparrow  \uparrow $

$(P-a/V^2) \approx P$ and $(V-b)\approx V \implies Z \approx 1$

Case 2: Low pressure $P \downarrow \implies V \uparrow$

$(V+b)\approx V$. On solving, we get $Z= 1+a/(RTV)\implies Z>1$.

Case 3: Very high pressure $P \uparrow  \uparrow \implies V^2\downarrow  \downarrow$

$(P - a/V^2) \approx P$. On solving, we get $Z= 1-Pb/(RT)\implies Z<1$.

The graph would thus be (A), because it is the only one satisfying all these conditions.


Mechanism with hypohalite in haloform reaction

The mechanism of haloform reaction will also be almost similar in the case if we use hypohalites (here, it is $\ce{NaOCl}$).

In the first step, instead of $\ce{OH-}$, the acidic $\alpha$ proton will be taken up by $\ce{OCl-}$ (Remember, $\ce{HOCl}$ is a weak acid (but not as weak as water), so it's conjugate base i.e $\ce{OCl-}$ is a moderately strong base to take up a proton), and $\ce{HOCl }$ will be formed, which will be attacked by the enolate formed in the previous step, and $\ce{OH-}$ will leave to form the $\alpha$-haloketone. The reaction will proceed further two steps to form the tri-halo substituted ketone, and correspondingly $\ce{OH-}$ will be generated. These generated $\ce{OH-}$ can then attack the electrophilic carbonyl carbon give the haloform and the salt of the organic acid .

The mechanism is shown below,


What exactly is a solubility product, and what is its use?

If we take small particles of a sparingly soluble salt, say AgCl, then one or two particles may dissolve in water and a saturated solution is obtained pretty soon. Upon adding extra AgCl to the solution precipitation of AgCl starts. At this stage, you'll find an equilibrium is established between the dissolved and the undissolved AgCl.

$$\ce{AgCl(s) <=> Ag+ (aq.) + Cl- (aq.)}$$

And for this equilibrium if you're to find the equilibrium constant, let's call it  $K_{sp}$, you get:

$$ \ce{K_{sp} = [Ag+] \cdot [Cl-]} $$

(Both the concentrations are the ones taken at equilibrium.)

This $K_{sp}$ is called solubility product. It's useful to predict whether or not more solute can be dissolved in your solution or not. We do this by comparing it to something which is called the ionic product.


Unable to find sources for uncommon specific heat

For potassium sulfate see here.

For potassium carbonate see here.

Generally, search in the NIST Chemistry WebBook and when you find the compound of interest, either click on "gas phase thermochemistry data" or "condensed phase thermochemistry data" as appropriate.


Rearrangements of α-aminoketones in Clemmensen reduction

Regarding this question, there are two things that should be stated beforehand.

Firstly, the mechanism of the Clemmensen reduction is not well-understood. There are two plausible pathways that have been proposed:1

so, the mechanism for this rearrangement is equally unclear.

Secondly, this compound does not actually undergo rearrangement in the Clemmensen reduction, according to studies by Leonard and Ruyle.2 Curiously, the authors could only isolate the alcohol product. No explanation was offered as to why the alkane was not obtained.

However, α-aminoketones are known to undergo rearrangement when the nitrogen and the carbonyl carbon are in the same ring. Two examples which Leonard and Ruyle studied are offered below:2,3

Their proposed mechanism occurs via migration of the C–N bond, following protonation of the carbonyl oxygen. The carbocation is captured by what appears to be hydride, although the existence of hydride is inconsistent with the proposed mechanisms we have seen above. It might just be their shorthand for sequential electron + proton transfer; I can't really tell.

The second half of the mechanism is not explained, but it is plausible to protonate the hydroxyl group and use the nitrogen lone pair to expel water.

The authors posit that the tendency to undergo rearrangement here is related to a "geometrical directing effect": in the case where the nitrogen and the carbonyl carbon are in the same ring, the nitrogen lone pair has a larger opportunity to approach the carbonyl carbon.2 In modern-day terms, this might be considered to be a stereoelectronic effect.

How can this be reconciled with the proposed mechanisms in Kürti-Czakó? Considering that the mechanism of the reduction itself is unclear, I do not want to make any assertions about what the exact mechanism of this rearrangement is. However, I think it would be plausible that C–N bond migration can occur at some stage in the above mechanism - especially in the intermediates where the carbon bears a positive charge.

References

Kürti, L.; Czakó, B. Strategic Applications of Named Reactions in Organic Synthesis; Elsevier: Amsterdam, 2005; p 92.
Leonard, N. J.; Ruyle, W. V. Rearrangement of α-Aminoketones During Clemmensen Reduction. II. Contraction of a Six-membered Ring in the Monocyclic Series. J. Am. Chem. Soc. 1949, 71 (9), 3094–3098. DOI: 10.1021/ja01177a043.
Leonard, N. J.; Wildman, W. C. Rearrangement of α-Aminoketones During Clemmensen Reduction. I. Bicyclic Compounds Containing a Bridge-head Nitrogen. J. Am. Chem. Soc. 1949, 71 (9), 3089–3093. DOI: 10.1021/ja01177a042.


On the meaning of distinguishability, and wavefunctions for 3 electron atoms

The only case in which electrons can be considered distinguishable is when they are so well separated that their wave functions do not overlap. Any electrons on the same atom or within the same molecule are considered indistinguishable.

In chemistry we like to talk about electrons being in specific orbitals, but the indisinguishable nature of them means that they are all in all orbitals simultaneously. This is why a mathematically rigorous description of a molecule using hybrid orbitals is just as valid as one using canonical orbitals. Neither is true in the sense of describing where each electron is, and both are true in the sense of describing the collective electron density.


How can I undo the tarnish on my aluminium headphones?

I would try any sort of weak acid that wouldn't cause damage to other parts of the headphones.

Citric Acid (found in lemon juice), Acetic Acid (found in vinegar), and Phosphoric Acid (found in cola) are probably all safe bets for sticking on a cloth and wiping off the casing. The primary difficulty you have here is the issue that comes from the fact submerging your headphones in anything would likely damage the electronics inside. You can rig up a shallow flat pan full of acid and resting the headphones on it's side in there, though I'd take caution.


The effect of "reverse shielding", as I call it, is on the order of one volt of potential difference in light elements where the $n=2$ electrons are the valence electrons.

Let's look at lithium.  From this data page the effective nuclear charge on the $1s$ electrons in the neutral atom is 2.691.  If we assume that the effective nuclear charge is related to the electronic energy level by

Energy$=(-13.6 eV)(Z_{eff}^2)$

then we infer that the orbital energy level is -98.5 eV for each electron.

Now what is that energy level if we take away the 2s electron? For this we need the electronic energy level of the $1s$ orbital in the lithium ion, which is obtained by averaging the second and third ionization energies of lithium given here.  That leads to -99.1 eV per electron.  The 2s electron has effectively increased the energy level of each 1s electron by 0.6 eV.

A similar calculation for beryllium gives a difference of about 1.1 eV per $1s$ electron (-184.7 eV from the effective charge on $1s$ in neutral Be vs -185.8 eV  from the third and fourth ionization energies after forming the di-cation).  The second 2s electron has basically doubled the shielding.

The electron volt differences determined above have some impact on chemical reactions, whose energies are on the order of one electron volt per electron transferred in a redox reaction.


What exactly happened when I added KOH to hydrogen peroxide?

According to THE HOMOGENEOUS BASE-CATALYZED DECOMPOSITION OF HYDROGEN PEROXIDE J. Phys. Chem., 1961, vol. 65 , pages 304–306:

$\ce{H2O2 + OH- <=> HO2- + H2O}$

$\ce{H2O2 + HO2- ->}$ [6-membered ring with 2 "H"s bridging 2 "O-O"s, with another H bonded to one of the "O"s] $\ce{-> H2O + O2 + OH-}$

  It is clear that the rate will go through a maximum
  when the peroxide is 50% ionized, that is
  $[\ce{H2O2}] = [\ce{HO2-}]$.


Finding chair and boat transition states during ring-forming reactions

As you have figured out, this is a non-trivial problem.

It's okay to find a boat conformation that might be a bit higher in energy and see if it can map to a TS. Keep in mind that for 6-membered rings, there are many conformations beyond chair and boat. A half-chair, for example, might be a useful test.

In general, for 6-membered rings, using ring-puckering searches are useful. Calculate the plane of best-fit for the 6-membered ring and displace atoms up or down -- this is similar to sampling the out-of-plane vibrational modes of the ring.

So the ideal situation for the TS would be to consider a set of conformations, that might be higher in energy than the reactant and product. After all, it's a transition state. Then use a nudged elastic band or similar method (your bond-lengthening) to look for the real TS.

Good luck!


Your misconception is directly at the beginning of your question and turns up multiple times. There are no $\mathbf{sp^3}$ nitrogens in guanidine.

Guanidine is isolobal to urea, where the carbonyl oxygen has been replaced by an imine $\ce{NH}$. However, in principle it is still the same flat, resonance-stabilised molecule. The main difference is that there is no ‘preferred’ site for the double bond — it could point towards any of the three nitrogens in theory; you could say the diamide resonance is enhanced.

This is also why the $\ce{-NH2}$ nitrogens are not basic. Their lone pair is not sitting in an orbital all alone waiting for a proton. It is actively resonating with the π system of the other two. The only available lone pair guanidine has is that of the $\ce{=NH}$ nitrogen sitting in an $\mathrm{sp^2}$ hybrid orbital. Thus, a second protonation requires somewhat superacidic media.


Best way to prepare reference crystal structure with a given force field

I'm not entirely sure what you're asking. Are you attempting to match the geometry of a molecule from a crystal structure or the entire crystal packing?

Computational prediction of crystal structures is very much an unsolved problem. There have been multiple "blind tests" from the Cambridge Crystal Database, with varying levels of success. In general, the most accurate results come from dispersion-corrected DFT, not force field methods.

In some cases, to speed results and screening, teams will prepare "tailored force fields" by deriving optimized parameters for a particular compound on-the-fly from the DFT calculations. In no way are these intended for general purpose force field methods.

Also, most methods typically have to optimize both the unit cell parameters and the geometry of the compounds in them. Your note about expanding the volume is important - if you are working in this field, you may wish to use such heuristics to optimize the cell parameters, but typical minimizers will focus on either atomic coordinates or unit cells but not both.


Is this galvanic cell representation right?

The chapter 1 of the reference 1 describes the Line Notation for Galvanic Cells as follows:

  A shorthand way of completely describing an electrochemical cell such as Galvanic cell is called line notation. It is a shorthand way of expressing the redox reaction progressing in the electrochemical cell. In this notation:

  A slash (/) or a vertical line (|) represents a phase boundary, and a comma separates two components in the same phase. A double slash (//) or a double vertical line (||) represents a phase boundary whose potential is regarded as a negligible component of the overall cell potential (e.g., salt bridge).
  When a gaseous phase is involved, it is written adjacent to its corresponding conducting element.
  The constituents of the cathode on the right and the constituents of the anode on the left.
  The phases of all reactive species are listed and their concentrations or pressures are given if those species are not in their standard states (i.e. $\pu{1 bar}$ for gasses and $\pu{1M}$ for solutions).

Thus, in the given cathode representation, both $\ce{Fe^3+ (aq)}$ and $\ce{Fe^2+ (aq)}$ are separated by a vertical line (|), which is an incorrect representation, according to the description (1) above. They should be actually separated by a comma (as $\ce{Fe^3+ (aq)}$, $\ce{Fe^2+ (aq)}$) because there is no phase separation between them (both are in same phase).

Also, according to the description (2) above, $\ce{I2 (g)}$ cannot be introduced into $\ce{Pt (s)}$ anode (as indicated in the representation) because $\ce{I2 (g)}$ is the product of the reaction, not a reactant.

Based on these, given answers (1) and (3) should be correct. Also, OP's representation of 'anode half' is also correct since there is phase separation.

References:

Allen J. Bard, Larry R. Faulkner, In Electrochemical Methods: Fundamentals and Applications; Second Edition, John Wiley & Sons, Inc.: New York, NY, 2001 (ISBN 0-471-04372-9).


Does a polyethene shrinking signify negative or positive entropy?

When stretched, the polymer chains elongate and are held taut. This means there are more crosslinks and fewer opportunities for internal degrees of freedom (movement). A relaxed polymer is not held taut and therefore can freely move, having more degrees of freedom, and an associated increase in entropy.


Benchtop NMR spectroscopy parameter explanation; tau and number of points

AS Karl said read the instrument's manual and add the meaning of the symbols. However there is one important point in discrete Fourier transform (DFT).

I think I kind of understand how np works; it's the discrete points of analogue FID that are being converted to digital points along with the FID signal envelope. Therefore it's associated with the resolution of the spectra. *Let me know if I interpreted this wrongly.

I have been recently discussing this exact point with signal processing engineers and spectroscopist friend with a lot of good constructive discussions. No, the number of points cannot enhance the intrinsic resolution of the signals whether it is NMR or FTIR or any other FT spectroscopic technique. What you increase, when you collect more points, is the frequency resolution on the x-axis, and consequently more points per peak. That is it.
For an N-point DFT, the frequency step is
$$\frac{1}{N\Delta t}$$
For example if your sampling frequency is fixed to 1 s (an arbitrary number), in one experiment you collect 1000 points, the frequency step on the axis is
$$\frac{1}{1000 . 1 s}=0.001 Hz $$
In another experiment, you only collect 250 points with a time step of 1 s
$$\frac{1}{250 . 1 s}= 0.004Hz$$
Note the time to sample the signal in each case is different, because it is given by $N\Delta t$. So by chance, if your original experimental spectrum was "limited" by the number of points per peak by increasing the number of points, you might see an improvement in the peak shape and apparent resolution when you increase the number of points in an FID.


How to reduce Fe(III) impurity in Iron(II) sulfate?

I recommend you dissolve your iron sulfate in hot water, dissolving as much sulfate as possible in as little hot water as possible. While keeping it hot, dip some coiled iron wire in it as well as some steel wool, and stir gently. The metallic iron should reduce your iron III back to iron II.

Once the solution is clear green, remove the iron wool, but keep the iron wire in it, and let it cool down as slowly as possible, while protecting it from air the best you can. Finish the cooling by putting it in a fridge

You should be left with beautiful large, green crystals of highly pure iron(II) sulfate heptahydrate, and a clear solution of relatively dilute iron(II) sulfate.

Since you have 50 kg of it, you will likely have to proceed in several batches, so I recommend you keep that solution for the next batch to minimize product losses.

Once you are done recrystallizing your iron sulfate, you can heat up that solution to a boil, until some crystals start forming on its surface. Then, immediately stop the heating and cool down that solution again. You'll have some lower grade iron sulfate, but that should still be perfectly usable for gardening.

And if you want to keep some of the prettiest crystals you made in the process, I recommend you keep them in a sealed container with a small piece of moist cloth. Otherwise, they may tend to slowly dehydrate and turn into powder.


In the book by Clayden et al. you can find a very nice description of the mechanism of the halogenation of enols and the haloform reaction on pp. 533-536. Especially on p. 536 the authors give the mechanistic details of the acid catalysed halogenation of enols and some reasons why multiple halogenation steps on the same carbon atom (as happens in the haloform reaction) occur easily under base catalysis but do usually not occur under acidic conditions. Since I lack the time to write it up myself (and since the authors already did a marvelous job at describing it) I simply post the relevant passage here:


Determining reaction formula from optical absorption graphs

Let's see. You are right in that it can't be (a). Now consider (c): A and B are consumed, while F and G accumulate. Where is that intermediate that goes up, then down?

As for (d), I believe there is no way we can tell that from (b) based on the picture alone, without looking at the numbers. See, suppose $\lambda_1$ actually measures B, and A is in huge excess; what would be the qualitative difference between the two? I can't see any.


It has already been pointed out that your text is incorrect. Saccharose (commonly known as sugar) is not an ionic compound yet it dissolves very well in water. It is not alone although I am too lazy to think up a non-carbohydrate example.
I’ll also point out that ionic compounds exist that are insoluble or poorly soluble in water, although some may rightly argue that their bonds tend to be more covalent than ionic.
As for two different liquids mixed together, the answer is: You can’t say a priori. (Or, in a highly logical reading of your question: no, the result will not always be a solution.) Many polar liquids will form a solution with many other polar liquids but in some cases the pairing will not. Conversely, a polar liquid such as ethyl acetate or acetone can form a solution with a practically unpolar liquid such as hexane.


What allows all 3 states of H₂O to be present at the same time?

At the triple-point of a substance, all three phases can be present. Since it takes additional energy to go from ice at 0 C to liquid water at 0 C, there can be a considerable overlap for the two substances, e.g. 1 g ice with 99 g water or 99 g ice with 1 g water, but for thermodynamic equilibrium, the triple point is very narrow.

BTW, the first well-know American chemist, Josiah Willard Gibbs, pioneered in defining the relationship of temperature, pressure and state. See  Gibbs' phase rule for more information.


Is it true that all molecules of a given point group such as, $C_{2v}$, will have the same MO diagram, the only thing that will change is the number of electrons?

No. There are so many different possibilities of how molecules with $C_{2v}$ may look, that they cannot possibly all have the same energy level diagram. Take water and acetone for mild examples.

What you can do, however, is estimate how different molecular orbitals will look based on the atomic orbitals you put into the estimation. Let’s take formaldehyde as it is a nice and simple example.

We have two hydrogen atoms, one carbon and one oxygen. Carbon and oxygen both lie on the $C_2$ axis so we can consider them separately, but we should use group orbitals for the hydrogens (I am defining the $yz$ plane to be the plane of the molecule). Luckily, they only have s orbitals, so it doesn’t get too complicated. Check out how the s-orbitals transform with the different symmetry operations to find their irreducible representations: $\mathrm{A_1 + B_2}$. Carbon and oxygen each have an s orbital and three p orbitals which will transform in the same way: the s orbital as $\mathrm{A_1}$ (naturally), the p orbitals transform as $\mathrm{A_1 + B_1 + B_2}$ (for $\mathrm{p}_z, \mathrm{p}_x$ and $\mathrm{p}_y$, respectively).

Thus, we have five atomic (or group) orbitals of the $\mathrm{A_1}$ type which will combine to form most of the σ bonds, three of the $\mathrm{B_2}$ which will also combine to σ bonds and two of the $\mathrm{B_1}$ type which will combine to form π bonds. Had we chosen water, we would only have had six orbitals altogether transforming as $\mathrm{3 A_1 + B_1 + 2 B_2}$ — a clear distinction showing that the diagrams cannot always be the same.

We now have to estimate the energy levels the orbitals initially have. It is a good guess to put the highest-energy atomic orbital at the respective atom’s ionisation energy; then you would just have to esimate a lower level for the s orbitals of carbon and oxygen. With these levels in hand, you can educatedly guess the different energy levels of the resulting molecular orbitals and educatedly guess what the orbital would look like: The resulting MO must transform in the same way as the AOs used to build it. So any $\mathrm{A_1}$ orbital will be symmetric to any symmetry operation while $\mathrm{B_2}$ will always have a nodal plane in the $\ce{C=O}$ bond axis, etc. I’m sure you can now come up with a well-educated guess of the formaldehyde orbitals; you can check out what they actually look like on the homepage of Professor Zipse. (You can ignore the accompanying German text. The higher-level antibonding MOs LUMO+2 and LUMO+3 that are required by theory are not shown; one must be $\mathrm{B_2}$, the other $\mathrm{A_1}$.)


Units of the rate constant for reactions whose order is greater than 1

As an example if the reaction is $\ce{A + B -> C}$ the rate expression for the appearance of $\ce{C}$ is

$$\frac{\mathrm d[\ce{C}]}{\mathrm dt} = +k[\ce{A}][\ce{B}].$$

The units have to be the same on both sides of this equation, these are concentration/time on the left and to be the same on the right the rate constant $k$ has to have units $(1/\text{concentration})(1/\text{time})$ usually expressed as $\mathrm{dm^3\,mol^{-1}\,s^{-1}}.$

It is true that the numerical value of the rate constant usually depends of several things such as temperature, but the units remain the same for each type of rate constant, 0th,  1st, 2nd, 3rd simply because the rate equation units have to balance.


How to determine the empirical formula for an unknown iron oxide?

First assume the mass of an oxide of iron to be $\pu{100 g}$ which implies mass of iron would be $\pu{69.9 g}$ and mass of oxygen would be $\pu{30.1 g}$. Now calculate amount of substance of both the elements by dividing the assumed mass by their atomic number.
Then calculate relative amount of substance of each element by dividing the obtained amount of substance of each element by the smallest of them. (E.g. suppose you've got $\pu{5 mol}$ of iron and $\pu{2 mol}$ of oxygen, divide 5 by 2 and 2 by 2 to get relative amounts of substance of both.)
Now just write relative amount of substances in the form of a ratio and bring the obtained ratio to least simplified form. (Remember the ratio should come in whole numbers.)
After obtaining ratio, write the empirical formula as I did in this example.
(eg. suppose I got this ratio $2:5$ from last example. So the empirical formula will be $\ce{Fe2O5}$ .)

Actually, you can skip the relative amount of substance step when you've got 2 elements in empirical formula, but you will need to perform all the steps when there are more than 2 elements as the answer comes to be different in some cases.


Is there a difference between a chelate ligand and a polydentate ligand?

A chelating ligand should bind more than once to the same metal atom. Therefore, it must be polydentate. However, a regular polydentate ligand does not have to ligate to the same metal, and therefore, it is not necessarily chelating.


The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is $38$.

This means, the reaction this value is describing is:
$$\ce{NH3 <=> NH2- + H+}$$

The $\mathrm{p}K_\mathrm{b}$ of $\ce{NH3}$ is $4.74$.

This means, the reaction this value is describing is:
$$\ce{NH3 + H+ <=> NH4+}$$

There is a lot of misuse going on with the term $\mathrm{p}K_\mathrm{a}$ since many people are too lazy to say

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$’s conjugate acid is $9.26$
$$\ce{NH4+ <=> NH3 + H+}$$

rather than

The $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is $9.26$.
$$\ce{NH3 <=> NH2- + H+}$$

but that is really what they should be doing. Thankfully, when looking up values in a table the person compiling the table (hopefully) knew what they were doing and used the correct terminology.
Only for a $\mathrm{p}K_\mathrm{a}$ and a $\mathrm{p}K_\mathrm{b}$ that describe the same reaction (from opposite viewpoints) is it true that $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = 14$.


Critical separation at which molecule breaks

... As far as I know, in the absence of an external influence, the molecule never actually breaks. I mean, most people wouldn't call two atoms a metre apart a molecule as such, but there's no negative curvature stationary point in a potential like this at which you could say "if $r > x$, this molecule is broken and will not reform".


You are correct.  There is a theoretical ratio of air-to-fuel that provides exactly enough air to completely burn the fuel, converting it to $\ce{CO2}$ and $\ce{H2O}$.  This theoretical ratio can be calculated knowing (a) the exact composition of the fuel and (b) how much oxygen is in the air (generally ~21%), and (c) the formula weight(s) of the molecules that are in the fuel.  For fuel that are pure octane, the combustion reaction is:

$$\ce{\frac{25}{2}O2 + C8H18 -> 8CO2 + 9H2O}$$

Which means 12.5 moles of oxygen are required for every mole of octane.  For 12.5 moles of oxygen you need about 60 moles of air because air is only 21% oxygen.  That works out to 1740 grams of air (~29 grams per mole).  One mole of octane is 114 grams.  So the theoretical, or stoichiometric ratio of air to fuel is 1740/114 or about 15.

Most motor fuels are not pure octane.  The additives complicate the math a bit, but the stoichiometric ratio is around 14 for many fuels.

The equivalence ratio is just the actual air-to-fuel ratio divided by the theoretical.

  My understanding of the equivalence ratio is that if it is below 1, then there is too little air (compared to the stoichometric mix). If it is above 1, that means that there is too much fuel, and not all of it can be used leading to what is called incomplete combustion.

You said the second part backwards.  If the ratio is below one, there is too little air (i.e. too much fuel), and if it above one, there is too much air (not enough fuel).  So incomplete combustion should happen more at lower equivalence ratios.

Also, the exhaust temperature will be maximized for equivalence ratios close to 1.  Otherwise, there are extra reactants, either fuel or air, which remain after combustion and absorb some of the combustion energy, resulting in a lower temperature increase.  Since temperature is important for NOx formation, I figured it was worth mentioning.


There exist a physical theory called quantum mechanics which describes the Universe at atomic and subatomic length scales. Chemical systems, such as, for instance, molecules, are one types of systems one encounter at this level, and as such, they can in principle be completely described by quantum mechanics. What is meant by a "complete description" is that all the properties of a system (say, a molecule) can be theoretically predicted by the application of the principles of quantum mechanics.

However, the direct application of the principles of quantum mechanics to chemical systems is practically impossible in the absolute majority of cases due to the complexity of the resulting description. To simplify matters a series of approximation has to be introduced. But even the approximate description is usually too complex to be handled by a human being analytically (on a piece of paper), and thus, numerical methods has to be developed to shift this burden to our digital mates (computers). All these questions (approximations plus numerical treatments) are addressed by a branch of chemistry known as quantum chemistry.

GAMESS[1] and Gaussian are indeed programs; programs that implement different methods of quantum chemistry. And just for your information there are a lot of other such programs. These programs allows us to calculate the properties of molecules, to predict their behaviour in different environments, etc. And all this can be done theoretically using computers without actually synthesising a molecule in question and then doing some real experiments on it. To understand what these programs do exactly, you have to spent a lot of years by learning some advanced mathematics, quantum physics, quantum chemistry, programming, numerical methods, ... At this point I can only tell you that they are doing some magic. The quantum magic.

[1] There are in fact 3 different programs: PC GAMESS (now Firefly), GAMESS (UK), and GAMESS (US).


Calculation of equilibrium constant for dissolution of iron(II) fluoride in acid

Let’s approach the entire problem from the other direction. Consider your final reaction:

$$\ce{FeF2(s) + 2H+ (aq) <=> Fe^2+ (aq) + 2 HF (aq)}\\
K_\mathrm{tot} = \frac{[\ce{Fe^2+}][\ce{HF}]^2}{[\ce{FeF2}][\ce{H+}]^2}$$

We have our initial dissociation with the following equation:

$$\ce{FeF2(s) <=> Fe^2+ (aq) + 2 F- (aq)}\\
K_1 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]}$$

And of course we have the second equilibrium which equates to:

$$\ce{F- (aq) + H+ (aq) <=> HF (aq)}\\
K_2 = \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]}$$

Now we want to express $K_\mathrm{tot}$ in terms of $K_1$ and $K_2$. Since we are dealing with products all the way through, it will somehow be a multiplication. So let’s try $K_1 \times K_2$:

$$K_1 \times K_2 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]} \times \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]} = \frac{[\ce{Fe^2+}][\ce{F-}][\ce{HF}]}{[\ce{FeF2}][\ce{H+}]}$$

We’re not quite there yet. We still have an extraneous fluoride concentration in the numerator and we are missing an $\ce{HF}$ concentration and a proton concentration. Luckily, that is exactly what an additional $K_2$ will give us:

$$K_1 \times \left( K_2 \right )^2 = \frac{[\ce{Fe^2+}][\ce{F-}]^2}{[\ce{FeF2}]} \times \left ( \frac{[\ce{HF}]}{[\ce{F-}][\ce{H+}]} \right )^2 = \frac{[\ce{Fe^2+}][\ce{HF}]^2}{[\ce{FeF2}][\ce{H+}]^2} = K_\mathrm{tot}$$

This is the mathematical way of saying ‘we need the second equation twice’. Due to the way equilibrium constants work, you always have to multiply or divide, so ‘doing something twice’ is equivalent to squaring.

Numerically, this means the answer is:

$$K_\mathrm{tot} = K_1 \times \left ( K_2 \right )^2 = 2 \times 10^{-6} \times \left ( 1 \times 10^3 \right )^2 = 2 \times 10^{-6} \times 1 \times 10^6 = 2$$

Yes, $\ce{FeF2}$ is a solid, so it has an activity of $1$ so we could ignore it when calculating the equilibrium constant. But we don’t actually use its value, so its being there doesn’t hurt either.


Mass percent and mole percent of a composition

Your approach in general looks OK except for

terminology (e.g. "total grams" should be total mass);
units (e.g. values provided for molar masses should have dimensions $\pu{g mol-1},$ not $\pu{g});$
excessive precision (you are given fractions with two significant figures, but there are mass values reported in your intermediate calculations having five significant digits; also, there is no need to calculate molar masses up to six significant figures — two would suffice);
math notations ("textual formulas" including chunks such as "molar_mass" or "1.5 mol CO" are not acceptable; use standardized symbols instead, e.g. $M$ or $n(\ce{CO}) = \pu{1.5 mol},$ correspondingly).

I was, however, taught to solve problems differently.
Solve the problem algebraically first, then plug in the physical quantities.
This way your solution becomes modular, generalized, and errors are easier to notice.
No imaginary "set" or "fixed" masses, volumes or other quantities introduced out of nowhere.
There is no point in intermediate calculations and they make tracking significant symbols harder than it should be.
Finally, assigning numerical indices to the compounds in the mixture saves you some time and paper, so let's do it now:
$$\underset{1}{\ce{H2}}\quad\underset{2}{\ce{CO}_\vphantom{}}\quad\underset{3}{\ce{CO2}}$$
and move on to the questions:

a) Calculate the mass% hydrogen if the mixture contains 15 mol% $\ce{CO}$ and no $\ce{CO2}.$

By definition, mass fraction of $i$-th component in the mixture is
$$ω_i = \frac{m_i}{\sum_j m_j}.\tag{1.1}$$
For a binary system with given mole fraction of the second component $x_2:$
$$ω_1 = \frac{m_1}{m_1 + m_2} = \frac{1}{1 + \frac{m_2}{m_1}} = \frac{1}{1 + \frac{x_2M_2}{(1 - x_2)M_1}} = \frac{1}{1 + \frac{0.15 × \pu{44 g mol-1}}{(1 - 0.15) × \pu{2.0 g mol-1}}} = 0.29.\tag{1.2}$$

b) Calculate the molar composition if the mass fraction of $\ce{CO}$ and $\ce{CO2}$ are both $0.35.$

By definition, mole fraction of $i$-th component in the mixture is
$$x_i = \frac{n_i}{\sum_j n_j} = \frac{m_i}{M_i\sum_j\frac{m_j}{M_j}} = \frac{ω_im}{M_i\sum_j\frac{ω_jm}{M_j}} = \frac{ω_i}{M_i\sum_j\frac{ω_j}{M_j}}.\tag{2.1}$$
Plugging in the corresponding quantities:
$$x_1 = \frac{0.30}{\pu{2.0 g mol-1} × \left(\frac{0.30}{\pu{2.0 g mol-1}} + \frac{0.35}{\pu{28 g mol-1}} + \frac{0.35}{\pu{44 g mol-1}}\right)} = 0.88 \tag{2.2.1};$$
$$x_2 = \frac{0.35}{\pu{28 g mol-1} × \left(\frac{0.30}{\pu{2.0 g mol-1}} + \frac{0.35}{\pu{28 g mol-1}} + \frac{0.35}{\pu{44 g mol-1}}\right)} = 0.07 \tag{2.2.2};$$
$$x_3 = \frac{0.35}{\pu{44 g mol-1} × \left(\frac{0.30}{\pu{2.0 g mol-1}} + \frac{0.35}{\pu{28 g mol-1}} + \frac{0.35}{\pu{44 g mol-1}}\right)} = 0.05 \tag{2.2.3}.$$
As you can see, using this approach takes less space and in my opinion is much more presentable.
Given you know elementary algebra and use standardized notations, you can greatly improve literacy of your answers and make them more concise.


Will bridged compounds the undergo SN1 reaction?

Very interesting question… I would give two answers, at different levels:

At undergrad level, I think the carbenium ion formed at the bridge head is less stabilized that “regular” ternary carbenium ions, because it cannot adopt the planar trigonal geometry which would be ideal for it.
At higher level, this is much more complicated: the norboryl cation is an example of non-classical ion…


Would solar panels produce energy if uniform temperature in the universe?

Solar panels work on the principle of photovoltaic effect. The only criteria for a voltage difference to be affected is that the light be incident on the solar panel. The energy of the incident light needs to be around that of the band-gap.
According to the question if both sun and the solar panel were of the same temperature, a solar panel on which light rays are incident would indeed generate a voltage. The band gap on the other hand depends on the temperature of the panel. Higher the temperature, smaller the band gap. Therefore, if the light incident on the panel is energetically close enough to the band gap (which depends on the temperature) the panel will generate a voltage difference.


How does a Lewis acid differ from an oxidizing agent; Lewis bases from reducing agents?

An acid and an oxidising agent are very different concepts that apply to different phenomena — although a single compound can be both at the same time, e.g. $\ce{H2SO4}$. However, a compound can also be an acid and a reducing agent, e.g. $\ce{HI}$.

The Brønsted acid/base theory is clearly distinct from any type of redox reaction, so I suppose that the confusion lies with the extension of said theory that the Lewis theory provides. But first a short word on the Brønsted theory: therein, an acid is a compound that can transfer a $\ce{H+}$ ion onto another compound.

The Lewis theory realises, that the acidic transfer reagent need not be hydrogen. Instead, anything with an unoccupied low-energy orbital can function as a Lewis acid; most notably boron compounds and metal cations. Upon creation of a Lewis acid-base pair, a coordinate bond is formed. Basically, this means that the energy of the lone pair of the Lewis base is lowered because it interacts with the empty orbital of the Lewis acid whose energy is raised. This can be depicted as $\ce{LB\bond{->}LA}$. The key concept is that the interaction is fully reversible and that upon dissociation the Lewis base keeps its lone pair.

When discussing oxidising or reducing agents, however, the electrons are irreversibly transferred. They actually move from orbitals of the reducing agent (which will be oxidised) into orbitals of the oxidising agent (which will be reduced). This transfer is thermodynamically favourable and thus the electrons won’t just ‘go back’.

To exemplify this, let’s consider two reactions of $\ce{H2SO4}$; one where it is an oxidising agent and one where it is an acid (Brønsted type; the proton of $\ce{H2SO4}$ is actually the Lewis acid).

$$\text{Acid-base:}\\
\ce{H2SO4 (l) + NH3 (g) <=>> NH4+ HSO4- (s)}$$

Note that ammonia, our base, is a gaseous compound at room temperature. If you employ appropriate conditions (i.e. temperature and low pressure), the equilibrium will be drawn to the left, releasing ammonia as ammonia ($\ce{NH3}$ molecules) including its lone pair. $\ce{H2SO4}$ will stay behind.

$$\text{Oxidising agent:}\\
\ce{2H2SO4 + Cu -> CuSO4 + SO2 ^ + 2 H2O}$$

$\ce{SO2}$, being a gas, will diffuse away. However, even if you perform this in a pressurised vessel, there will be no significant back-reaction because the electrons have been fully transferred from copper to sulfur.

The argument for Lewis bases/reducing agents is the same in reverse.


How do EM waves cause molecules to rotate, move, and cause electrons to move up energy levels?

But how do EM waves cause atoms to rotate, and vibrate?

In no way do EM waves cause atoms to rotate and vibrate. Atoms (more precisely, nuclei) in a molecule are always moving, so EM radiation can only change the way they move.

The thing is that within the Born–Oppenheimer approximation an energy  state of a molecule is the sum of the electronic, vibrational, and rotational components,
$$
  E = E_{\rm{el}} + E_{\rm{vib}} + E_{\rm{rot}} \, ,
$$
and there are accordingly three types of states and energy levels: electronic, vibrational, and rotational. For each and every electronic state there are few vibrational states associated with it and for each and every vibrational state there are few rotational states associated with it.

A transition can happen not only between different electronic states, but also between different vibrational states associated with the same electronic state or between different rotational states associated with the same vibrational state. Different vibrational and rotational states correspond to different ways atoms of a molecule move in space, so vibrational and rotational transitions change the way atoms move.

And, of course, any transition emits or absorbs a photon, the frequency of which is proportional to the difference in energy levels. Vibrational and rotational transitions spectra are studied by rotational–vibrational spectroscopy.


What is the difference between "reaction pathway" and "reaction mechanism"?

Usually with the reaction mechanism the sequence of events (electron transfers) is indicated, while the term reaction pathway usually refers to the reaction coordinate diagram associated with the reaction, i.e. the change in energy with an extra notion on the transition states.


For a solution of $\ce{Na2SO4}$, the sodium sulfate will ionize as you indicated:

    $\ce{Na2SO4(s) ->[aq] 2Na+(aq) + SO4^{2-}(aq)}$

This reaction is essentially irreversible in that the ions in an unsaturated solution won't suddenly form solid $\ce{Na2SO4}$ again.

$\ce{HSO4-}$ is a not a weak acid, but has a $\text{pK}_\text{a2} = 1.99$ which makes a moderately strong acid. So for  pH > 3 essentially all the $\ce{HSO4-}$ will ionize:

    $\ce{HSO4-(aq) + H2O ->[pH > 3] H3O+ (aq) + SO4^2- (aq)}$

Thus $\ce{SO4^{2-}}$ is such a weak base that reverse reaction doesn't happen when you dissolve $\ce{Na2SO4}$, and a solution of $\ce{Na2SO4}$ in distilled water stays neutral rather than becoming basic.

    $\ce{\require{cancel}H2O + SO4^2- \cancel{->} HSO4- + OH-}$

To think about a different example acetic acid is a weaker acid, $\text{pK}_\text{a} = 4.76$, so the acetate anion is a stronger base than the sulfate anion. A solution of sodium acetate would become weakly basic.

One last point - the statement should be "sodium sulfate a neutral salt in aqueous solution." In other solvents sodium sulfate could act as a base.


Painting metal, does it matter if all rust is removed before?

How exactly is the water broken down into hydrogen and oxygen?

As the acid is formed and the iron dissolved, some of the water will begin to break down into its component pieces -- hydrogen and oxygen.

That's extreme oversimplification. The reaction common for active metals (not including iron) is
$$\ce{Mg + 2 H2O ->  Mg(OH)2 + H2 ^ }$$
Iron does not react this way. However, it does react with acids, say, with hydrochloric acids
$$\ce{Fe + 2 HCl(aq) -> FeCl2(aq) + H2 ^}$$
The reaction is quite slow even for concentrated nonoxidizing acids. However, in presence of oxygen and in diluted acids two subsequent reactions occur, the one noted above is followed by
$$\ce{4FeCl2 + 10H2O + O2 = 4Fe(OH)3 v + 8HCl }$$
The second reaction effectively removes iron from solution, significantly fastening the overall reaction.  However, even without acid present, even in slightly basic conditions iron slowly reacts with water in presence of free oxygen, forming rust. The process is hindered only at pH above 10 or at absence of free oxygen. Since natural water is usually slightly acidic due to presence of dissolved carbon dioxide, iron always rust on contact with water and atmosphere. This, actually, is a reason why concrete layer over steel frame is regulated: concrete always contains water, and surface layers quickly looses basicity due to reaction with atmospheric carbon dioxide, so steel near surface of concrete quickly rusts, increasing in volume and tearing the concrete from inside.

In the context of automobile repair work, assume a metal surface has
rust over it. Does it matter if all of the rust is removed before the
metal surface is painted? If it does matter, then why?

Yes, it does. Rust is weak, so any paint over it will hold weak. As long as the paint film over rusty part is broken, water and atmosphere contacts the metal and start to erode it, resulting in more paint film be destroyed. So it is critical to remove all rust from the surface being painted. Abrasive blasting, if you have the equipment, is probably the best way. While some paints may be applied over rust, it should be a very thin layer, and even then I would prefer traditional paint over cleaned surface anyway.


Largest alkane having a given alkane as its base name

Any alkyl substituent of butane in position 2 or 3 cannot be longer than $\ce{CH3}$ since that would lead to a longer parent chain. And obviously, there cannot be any alkyl substituent at all in the first or the last position of the butane chain. Therefore, the largest structure based on a butane parent chain is 2,2,3,3-tetramethylbutane.

This principle can be expanded to a heptane parent chain. The maximum length for alkyl substituent chains are 0 for position 1 and 7, 1 for position 2 and 6, 2 for position 3 and 5, and 3 for position 4. Therefore, the largest theoretical structure based on a heptane parent chain is 3,3,5,5-tetra-tert-butyl-4,4-bis[3-(tert-butyl)-2,2,4,4-tetramethylpentan-3-yl]-2,2,6,6-tetramethylheptane ($\ce{C53H108}$).


What is the physical significance of molecular partition function?

The partition function $q=\sum_i\exp(-E_i/k_BT)$ in your question can be regarded as the effective number of levels accessible to the molecule at a given temperature. It also means that in the equilibrium distribution the partition functions tells us how the systems are partitioned or divided up among the different energy levels.

In determining the form of the Boltzmann distribution the total number of 'systems' is N and these are divided up as $n_1$ systems have energy $E_1$, $n_2$ have energy $E_2$ etc. subject to conditions $\sum_i n_i=N$ and $\sum_in_iE_i=E$ where E is the total energy. It is found overwhelmingly that the chance that $n_i$ states have energy $E_i$ is given by $n_i = N\exp(-E_i/k_BT)/q$

In practice the partition function acts as a normalisation term in the calculation all sorts of thermodynamic properties, for example the average (internal) energy of N molecules with quantum energy levels $E_i$ is

$$ U=N\frac{\sum E_i e^{-E_i/k_BT}}{\sum e^{-E_i/k_BT} }$$

and the calculating the entropy $S=-Nk_BT\sum p_i \ln(p_i)$ where $p_i=n_i/N =\exp(-E_i/k_BT)/q$  is the probability that the ith state is occupied.

[ If the states are degenerate, with degeneracy $g_i$, then the partition function is $q=\sum_i g_i\exp(-E_i/k_BT)$.]


Why is enthalpy of atomization of Mn exceptionally low?

According to "Atomic Structure and Periodicity" by Jack Barrett, the reason is a combination of the $d^5$ electron configuration (half filled subshell) in the gas state and, relative to chromium, "the extra electron in the metal occupying an anti-bonding region of a band, s or d".


What are the degrees of freedom (from Gibbs phase rule) for an azeotropic mixture?

Gibbs' phase rule simply counts the number of variables required to specify the system and the number of relationships between these variables (constraints). Any failure of Gibbs' phase rule is symptomatic of additional variables or constraints that have not been taken into account.

This system is generally specified by four variables, the mole fraction of one species in the liquid phase, $x$, the same in the gas phase, $y$, the pressure $P$, and the temperature $T$. Equilibrium of the liquid and gas phases in both species provides two constraints. The specification of an azeotrope, for which $x = y$, provides a third constraint, as has been mentioned in the comments. One degree of freedom remains.


Does the salt bridge in a galvanic cell deplete over time?

Consider a salt bridge as a part of the cell electrolyte, that is not allowed to mix with the rest of electrolyte.

For a particular ion, there are 2 cases:

The ion is shared with the electrolyte part, from which the ion migrates toward the bridge. Then the ion gets depleted from electrolyte and not from the bridge.
The ion is not shared with the electrolyte part, from which the ion migrates within the bridge toward the other electrolyte part. Then the ion gets depleted from the bridge.

But there is needed to add the bridge maintains ion neutrality, so ions that leave the bridge are replaced at the other bridge end by the same or other ions of the same total charge.

Bridges are not usually used in power cells and cells where significant portion of chemicals gets depleted or react. If they are used, they usually use at least one of ions shared.

Power cells often use a single electrolyte and a diaphragm, soaked with this electrolytes, so ion depletion does not apply.

Bridges are used mostly in cells where accurate potentials matter, to minimize phase potentials, where the current is minimal. Like calomel ( $\ce{Hg2Cl2}$ ) or $\ce{AgCl}$ reference cells, or in laboratory test cells.