What are possible side reactions for the generation of hydrogen via electrolysis?

First of all, check again your electrodes to see if this 'black precipitate' is formed at the negative electrode as you state; I predict it's actually happening at the positive electrode for the following reason:

Since you're using two graphite electrodes, it isn't a black precipitate that's forming, but is indeed a side reaction as you state. I suggest what you're actually making is oxygen gas at the positive terminal (as one would predict) which then reacts with the graphite electrode surface to generate $\ce{CO2}$. To verify this,take a look at your electrode—it should have started to break up, now having a non-uniform surface. Perhaps even a bit crumbly.

As LDC3 points out in his comment, you're forming $\ce{Cl2}$ as well as $\ce{O2}$ which is the characteristic smell you're noticing. To understand this we need to look at the standard electrode potential for each of these reactions. Forming $\ce{O2}$ from water at constant $\mathrm{pH}$ (compared to the standard hydrogen electrode) requires a potential difference of $-1.23~\mathrm{V}$ (this value is negative since this is the reverse of the tabulated version). Chlorine evolution on the other hand requires a p.d. of $-1.36~\mathrm{V}$. Remember that the larger (the less negative in this case) the reduction potential, the higher the tendency for reduction. So since $\ce{Cl-}$ reduction potential is more negative, it's less likely to undergo reduction and hence more likely to be oxidised (lose electrons) than water, which agrees with our chemical intuition.

So what we say here is that $\ce{Cl-}$ is preferentially oxidised over water, and as a consequence $\ce{Cl2}$ is formed preferentially over $\ce{O2}$. The reason why you have any $\ce{O2}$ forming at all is due to the electrolyte being the limiting reagent. Your supporting solvent (water) is in such abundance, essentially completely surrounding the electrode, that the system simply has to oxidise some water molecules some of the time.

Of course, experiment should rule the day, so revise your experiment to provide evidence for/refute my theoretical framework. Replace your positive terminal with a high purity metal electrode (copper should work and be readily available at high purity). As for the choice of electrolyte, test a couple of salts whose reduction potentials are less negative than water. Hint: perhaps look at sulphate salts and try and find a further reason why these are very useful in electrochemistry.


It is due to that zinc is more reactive than copper which is less reactive metal so, zinc can easily displaces copper from its solution. We can understand it as zinc ion, $\ce{Zn^2+}$ ion form more strong electrostatic attraction with sulphate ion, $\ce{SO4-}$ ion.

Also, we can also consider the electrochemical series or activity series to give a better reason for the same. In activity series, zinc has greater value of standard reduction potential than copper's standard reduction potential value. It means, zinc can easily reduce Copper and it get oxidised.

$$\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}$$


And in what conditions can $\ce{H^{−}}$ occur?

In any phase. The hydride ion however doesn't stick around for long in water since it's such a strong base. First, hydrogen isn't that electronegative to start with, so tacking on an extra electron isn't helping with its stability. Second, combination with an hydrogen proton, $\ce{H+}$, is highly entropically favorable because the product, $\ce{H2}$, is a gas.

The $pK_a$ of $\ce{H2}$ has been estimated to be ~35 so that speaks a lot to the strength of the hydride ion as a base.


how many calories per unit of sweetness between table sugar and aspartame?

Aspartame 4 calories per gram source
Sucrose 3.9 calories per gram source
But you need far less aspartame than sucrose to get the same level of perceived sweetness


What do graphene "armchair(5,5)" and "zigzag(5,0)" refer to?

The third picture, subpic (d) here shows you the geometric armchair(red) and zigzag(blue) structure:

It's derived from  structure description of Carbon nanotubes similar to surface reconstruction/overlayer structures:

  The integers n and m denote the number of unit vectors along two directions in the honeycomb crystal lattice of graphene. If m = 0, the nanotubes are called zigzag nanotubes, and if n = m, the nanotubes are called armchair nanotubes. Otherwise, they are called chiral. The diameter of an ideal nanotube can be calculated from its (n,m) indices

I hope you have some background, otherwise here is a nice visualisation, or look the lecture linked above. It's important to understand what the unit cell of a 3D crystal or 2D surface is.


There is not many simple tests. One of possible tests would be passing gas through neutral solution of suitable acidobasic colour indicator, e.g bromthymol blue, turning green to yellow while getting slightly acidic.

But it would not be specific, any acidic gas would do that.

You have not mentioned context, why do you need to know ? Or is it just homework-like question ?


There are two ways that I interpret the question you're asking, and I'll try to address each in turn.

If your teacher boiled a container of water and held a piece of glass above it to collect drops of liquid water, he or she may have been trying to show that water doesn't disappear when it boils, that instead it turns to a gas that can move through the air and can be condensed back into a liquid.  Essentially, $\ce{H2O}$ is $\ce{H2O}$, regardless of the state of matter it occupies.
If instead the teacher wanted you to observe the bubbles of gas forming in a container of hot or boiling liquid, he or she might have been on the right track, but either failed to convey the correct message to you or had the wrong idea in the first place.
If I take water from the tap and heat it strongly, bubbles will form on the inside of the vessel I'm heating it in prior to it boiling.  These are bubbles of gas that are no longer soluble because the temperature of the liquid is too high (and its pressure has decreased as compared to the tap.)  The gas is dissolved air (mostly carbon dioxide, I suspect) with a large portion of water vapor.  In any case, they are bubbles of gas that were dissolved but are no longer.  You can see this happen any time you boil a pot of water on the stove.
Once the pot reaches boiling, any bubbles are almost exclusively gaseous water formed via vaporization.
An easier, if slower, way to see this is to run a cold tap until the water is as cold as it will get, then pour a glass.  As it sits (perhaps for an hour or two), bubbles will eventually form on the walls of the glass.  These eventually will dissolve again and the gas will escape through the top of the glass, so it's not something you can set up at night and come back to the next morning.  The bubbles are air that was able to dissolve in the high pressure/low temperature water in the pipes but not in the low pressure/high temperature water in the glass.


Could one synthesise unsubstituted "cyclohexa-1,3,5-triene"?

You cannot make (= synthesise) isolated, nonaromatic cyclohexa-1,3,5-triene[1] much like you cannot make (= synthesise) pseudoaromatic, biradicallic cyclobuta-1,3-diene. Hypothetical unsubstituted cyclohexa-1,3,5-triene will immediately give benzene, as this answer demonstrates.

The only way to actually generate nonaromatic cyclohexa-1,3,5-triene is to apply an intramolecular constraint that disallows the aromatization. Since cyclobuta-1,3-diene is the most obvious antiaromatic, planar system, that will attempt to always adopt a localised double bond structure, structure 1 below immediately suggests itself as a model system for the localised cyclotriene.

Scheme 1: Diercks’ and Vollhardt’s synthesis of a cyclohexa-1,3,5-triene system 1.

Indeed, that was the approach that Diercks and Vollhardt used to synthesise a model of cyclohexa-1,3,5-triene.[2] They report a crystal structure of 4 ($\ce{R}= \ce{TMS}$) wherein the alternating single and double bonds are clearly seen. The cyclobutyl sub-species adopts a cyclobutene-structure, meaning that the central cyclohexatriene ring’s double bonds are localised between the cyclobutane cores as indicated by the structure drawn. The outer benzyl rings are relatively aromatic (but not perfect hexagons).

Note and reference:

[1]: IUPAC nomenclature would allow systematically naming benzene as cyclohexa-1,3,5-triene. This is because IUPAC nomenclature assumes no knowledge of the chemistry of a given structure and requires names to be applicable ab initio and a priori. However, henceforth in this answer I shall make a distinction of (nonaromatic, localised, $D_\mathrm{3h}$-symmetric) cyclohexa-1,3,5-triene and (aromatic, $D_\mathrm{6h}$-symmetric) benzene.

[2]: R. Diercks, K. P. C. Vollhardt, J. Am. Chem. Soc. 1986, 108, 3150. DOI: 10.1021/ja00271a080.


How to reconcile conflict between assumed oxidation state and the most stable multiplicity from electronic structure calculations?

First reflex: As long as the calculations are otherwise reliable, I don't think you can ignore the possibility of a broken-symmetry/multiconfigurational solution. If such a solution ends up even lower in energy than your closed-shell singlet, I think you have your answer right there.

Alternatively, if the closed-shell singlet ends up exhibiting a lower energy than a BS solution, perhaps this is some sort of cationic Zn analogue of a peroxo species? If this were the case, then the $+1$ oxidation state wouldn't be nearly so surprising.  As you note, the non-$\ce{Zn}$ portion of the system would be $\ce{[AlO3H]^{2-}}$, and thus the species as a whole might be better thought of as $\ce{[Zn2][AlO3H]}$. I have no feel for how realistic this hypothesis is, though Wikipedia notes that organozinc(I) compounds have been prepared, with decamethyldizincocene being the first such compound reported (public domain image; click to enlarge):


The Mayr group at LMU Munich (Germany) spent a great deal determining rate constants for organic reactions between nucleophiles and electophiles.  They established a public, searchable database (explained here), and grouped their findings on posters (C-, N-, P-, O-nucleophiles; entries relevant to organometallic chemistry), too.

(picture source)


Many commercial detergents are composed of pH-neutral surfactants like alkyl polyethylene oxides plus alkaline additives like sodium silicate or carbonate or borax, which help dissolve greasy materials better than acids.

There are some acidic detergents, like citric acid cleaners for stainless steel and other metals and toilet bowl cleaners. The acidic function is paramount, and the reduced surface tension is secondary, useful in providing better wetting.

It is instructive to consider what cleaning really means, and the individual steps that go on. First, water is the primary cleaning agent. It is the universal solvent, but not quite good enough or fast enough. So we add surface tension lowering agents - big improvement: oily materials are wetted and can be suspended better as the first step in removal. Alkaline additives then combine with polar surfaces to further displace oily materials, improving cleaning action. Acids don't help with oily materials much but can remove some contaminants better.

Dispersants for solids, like mud or clay, are typically anionic polymers, usually alkaline because the anion contains groups that will adsorb on polar surfaces and disperse the particles by electrostatic repulsion, assisting their removal.

There are classes of surfactants where the oleophilic group is anionic, like sulfonates or phosphonates, and these can be acidic if the cation is hydrogen or alkaline if an alkali metal, or nearly neutral if the cations are mixed. Another class of surfactants, called cationic, has the oleophilic group attached to nitrogen, and is typically somewhat acidic because the anion is the anion of a strong acid like HCl.

So you could optimize your cleaner by selecting a pH of water that cleans best; then add a specific surfactant to improve removal of your specific dirt - or the reverse. And consider interactions and whatever other additives might help.


What is the optical nature of the tris(ethylenediamine)cobalt(III) cation?

Ethylenediammine or $en$ is an symmetrical didentate ligand. You can remember a rule of thumb for this if you like:

  In the case of $3$ symmetrical didentate ligands, i.e. $[\ce{M(AA)3}]$, the compound exists in only one form, which is optically active

So the complex you've mentioned will be optically active, no doubt.

Also, as a reference, I'm quoting the wikipedia article at https://en.wikipedia.org/wiki/Tris(ethylenediamine)cobalt(III)_chloride#Stereochemistry

  The complex can be resolved into enantiomers that are described as $\Delta$ and $\Lambda$.

The only symmetry elements that can be found in $\ce{[Co(en)3]^3+}$ are the three $C_2$ axes of rotation, one of which is shown in this image (colored in red) on wikipedia: https://upload.wikimedia.org/wikipedia/commons/e/e5/Coen3lel3.png

As a side note, here's another rule of thumb:

  In the case of $3$ unsymmetrical didentate ligands, i.e. $[\ce{M(AB)3}]$, the compound exists in two forms, namely fac (stands for facial) and mer (stands for meridional), and both of these forms are optically active.

Important note: There's a minute error in this answer; and that's the distinction between optically activity and chirality: I used the word "optically active" while it should've been "chiral". It's bad I missed it, but I'd like you to read Ben Norris' comment on this below. I'm leaving the answer untouched, editing it rip the comment of its value. I encourage you to learn from my mistake.


Rate law of reactions, what about the proportions?

[OP, in a comment] oh ok, so if we double the amount of only one substance, that substance will get left?

Yes. The rate law tells you how fast product is being made. The change in the different species is still given by the chemical reaction equation (which tells you in which proportion the amount of species will change, not in which proportion they are present at the beginning or at the end of the reaction).

You can see that when you look at the definition of rate:

$$\mathrm{rate} = \frac{d[X_i]}{\nu_i dt}$$

where $\nu_x$ is the (signed, i.e. negative for reactants) stoichiometric factor of the $i$-th species. Dividing by the stoichiometric factor ensures that the law of definite proportions is followed.

  [OP in another comment] we cant simply put more of a substance to get more of the product, the reactants must have a fixed ratio.

The law of definite proportions is not about the amounts present, but about the change in those amounts. I can set up a reaction with 1000 times more elemental oxygen than hydrogen, but they will still react in a 1:2 ratio if the reaction yields water.


Why are some smells so hard to get rid of?

Food flavors and odors are comprised of complex mixtures of hundreds of chemicals, which act as stimuli to the human sensory organs, for example, human nose. The odor of food may be attributed to the odor of a single chemical or an integrated response by nose to a mixture of chemicals. This reference to a single chemical or a mixture of chemicals allows trained analysts to develop a common terminology in describing odor, one of which called odor thresholds. Threshold is defined as the degree of stimulation of a nerve or nerve center that just produces a response. If the concentration of the odorous chemical (usually given in parts per million or ppm) is below this point, it will be judged not odorous. Hence, accepted definition of threshold is the minimum
identifiable odor (MIO) or recognition threshold. MIO of some chemicals is very high while some is extremely low. The latter chemicals are the ones you said,  "hard to get rid of." For example, the oder of dimethylformamide is described as 'fishy' and pungent, but you won't smell it if its concentration in air (or on your skin) is below $\pu{100 ppm}$. On the other hand, tar-like, pungent smell of p-cresol can be detected if its concentration in air (or on your skin) is greater than $\pu{0.001 ppm}$.
Now considering the fish smell, it was described short and sweet in following website from American Society for Nutrition: Read
Accordingly, it says:

Water in the open ocean is about 3% salt by weight, but the optimal levels of dissolved minerals inside an animal cell is less than 1%. In order to maintain fluid balance, ocean creatures must fill their cells with amino acids and amines to counter the saltiness of seawater. Ocean fish tend to rely on trimethylamine oxide (TMAO) for this purpose.
The problem is that when fish are killed, bacteria and fish enzymes convert TMAO into trimethylamine (TMA), which gives off the characteristic “fishy” odor.

Now, it is nice to know MIO of trimethylamine (TMA) is $\pu{0.00021 ppm}$ or $\pu{0.21 ppb}$ (parts per billion). That's why it is so "hard to get rid of" fishy smell whenever you touch or cook dead fish. But the good news is TMA is a organic base, which would react with any acid to form water soluble ammonium salts. Citric acid from lemon (or lime) or acetic acid from vinegar would do the job.


The term metamer isn't terminology that is widely used today, and indeed doesn't appear in the IUPAC Gold Book (the International Union of Pure and Applied Chemistry publishes guidelines for standardised nomenclature in chemistry and is widely adopted).

The description you're looking for is structural isomerism, with the compounds you show being examples of positional isomers (isomers which differ by the position of the functional groups, for instance in this case the relative positions of the methyl and methoxy moieties).


Formating a chemical equation - Physical states

Overall science is not affected by whether you write subscript or a superscript or in line. Afterall all notation was invented by humans. When professor quibble over these things my suggestion for them is to read the problem of "How many angels can stand on the point of a pin?" Apparently when Constantinople was falling, allegedly scholars were discussing these problems. Anyway, the lesson is that science is more important, notation comes later.

The International Union of Pure and Applied Chemistry (IUPAC), makes recommendations to systematize notation. Show the book "Quantities, Units and Symbols in Physical Chemistry" Third
Edition, https://doi.org/10.1039/9781847557889. It is available for free.

There was a similar question What is the standard way to denote physical states in a chemical reaction?

but the reference is old. The above reference is recent but there is no difference. Just for your interest, to see how old that reference is, it looks like type-written text!

They suggest to write the state of aggregation as they call it, just next to your symbol like $\ce{Na(s)}$, not as a subscript.


Citric acid chelates calcium carbonate more than ethanoic acid, why does this result in a faster rate of reaction?

Acetic acid has a $\mathrm{p}K_\mathrm{a}$ value of $4.76$ according to Wikipedia.

Citric acid has a $\mathrm{p}K_\mathrm{a,1}$ value of $3.13$ and a $\mathrm{p}K_\mathrm{a,2}$ value of $4.76$, also according to Wikipedia. It is about ten times as acidic as acetic acid is.

The greater acidity translates to a lower $\mathrm{pH}$ of the resulting solution. The lower $\mathrm{pH}$ translates to bases being protonated more and quicker. Considering that the decomposition reaction of carbonate is actually:

$$\ce{HCO3- + H+ <=> H2CO3 <=>> H2O + CO2 ^ }$$

The $\mathrm{pH}$ effect alone should be enough to explain a faster reactivity.

However, you can also invoke the following equilibrium which will be right-side shifted with citric acid (but obviously not so much with acetic acid):

$$\ce{CaCO3 (s) + cit^3- <=> Ca^2+ + CO3^2- + cit^3- <=>> [Ca(cit)]- + CO3^2-}$$

Where $\ce{cit}$ is citrate or the anion of citric acid. Complexation of calcium will remove free calcium ions from solution, and the solubility product only considers free ions in solution. Thus by complexing the free ions, more calcium can be dissolved.

I would still perform control experiments to make sure it is actually the complexation effect of citrate and not its acidity. For example, try reaction $\ce{CaCO3}$ with acetic acid in the presence of $\ce{Na3cit}$ or $\ce{Na2Hcit}$. These should be equally good complexing agents.


Alloys/metallurgy : promising elements for for Sn-based solders

Lithium might be tricky to alloy in an oxidizing atmosphere (e.g. air), and Pt might take some effort to form an alloy, because of its high m.p., which, however, is still less than than the boiling points of Bi or Sn. Other than that, there should be little trouble forming alloys.

You can search on line for "phase diagram alloy components" to find the melting point for at least some of the alloys. E.G. an online search for

phase diagram "copper-indium" alloy

turned up this link at Springer-Verlag GmbH, with the diagram below.

That said, I would expect the "10-20 % copper" requirement precludes melting point less than 300 C, even with low-melting In, Hg, Ga etc.

BTW, I would not use "dopant" for a component of an alloy that will not appreciably change its electronic properties; it is just another component, be it minor, of a ternary, quaternary etc. alloy.


Deriving an alternative expression for the Van der Waals equation using given parameters

$\def\p{\partial} \require{enclose}$Start with the van der Waals equation $(1)$.

$$\left(P + a\frac{n^2}{V^2}\right)\left(V - bn\right) = nRT\tag1$$

Note that you have a slightly different form of the equation, namely you have used molar volume $V_m$. Divide both sides of $(1)$ with $n$, use $V_m = V/n$, and $n^2/V^2 = 1/V_m^2$ to reach your variant.

Values for critical parameters in terms of $a, b, n, R$

The critical point $(P_c, T_c, V_c)$ is a critical point and an inflection point of the isotherm by definition.$^{[a]}$ Therefore we arrive at two equations;

$$\left(\frac{\p P}{\p V}\right)_{T\ =\ T_c} = 0,\tag2$$
$$\left(\frac{\p^2 P}{\p V^2}\right)_{T\ =\ T_c} = 0.\tag{2'}$$

Equation $(2)$ is a sufficient condition for a critical point (mathematically, a derivative's non-existance works too). Equation $(2')$ gives a necessary condition for an inflection point.

So our first step should be to find the equation $P = P(T, V)$. Just use equation $(1)$.

$$P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\tag{3}.$$

Apply conditions $(2)$ and $(2')$ on $(3)$.

$$\left(\frac{\p P}{\p V}\right)_T = -\frac{nRT}{(V - bn)^2} + 2a\frac{n^2}{V^3} \overset{(2)}{=} 0\tag4$$

$$\left(\frac{\p^2 P}{\p V^2}\right)_T = \frac{2nRT}{(V-bn)^3} - 6a\frac{n^2}{V^4} \overset{(2')}{=} 0\tag{4'}$$

Both $(4)$ and $(4')$ are equal to zero. Thus we may multiply either equation by non-matching real numbers, and still maintain equality: $(4) = (4')$. We will use this to our advantage by multiplying $(4)$ with $2/(V-bn)$ and $(4')$ by $-1$.

$$-\frac{2nRT}{(V - nb)^3} + 4a\frac{n^2}{V^3(V- nb)} \overbrace{=}^{(4)\ =\ (4')} -\frac{2nRT}{(V - nb)^3} + 6a\frac{n^2}{V^4}\tag5$$

$$\frac{2}{V - nb} = \frac{3}{V} \implies 3V - 3nb = 2V \implies \enclose{box}[mathcolor="green"]{V_c = 3nb}\tag{5'}$$

Plug result $(5')$ into equation $(4')$.

$$-\frac{nRT}{(3nb - nb)^2)} + 2a\frac{n^2}{27n^3b^3} = 0 \implies -\frac{RT}{4} + \frac{2a}{27b} = 0 \implies \enclose{box}[mathcolor="red"]{T_c = \frac{8a}{27bR}}\tag6$$

The outcomes $(5')$ and $(6)$ are to be placed into equation $(3)$.

$$P_c = nR\frac{8a}{27bR} : 2nb - a\frac{n^2}{9n^2b^2} \implies \enclose{box}[mathcolor="blue"]{P_c = \frac{a}{27b^2}}\tag7$$

Values for $a, b, R$ in terms of critical parameters

$$b \overset{(5')}{=} \frac{1}{3n}V_c\tag{8a}$$
$$a \overset{(7)}{=} 27b^2P_c \overset{(8a)}{=} \frac{3P_cV_c^2}{n^2}\tag{8b}$$
$$R \overset{(6)}{=} \frac{8a}{27bT_c} \overset{(8a, 8b)}{=} \frac{8P_cV_c}{3nT_c}\tag{8c}$$

Reduced van der Waals equation

Replace the values of $a, b, R$ from equations $(8a)$$-$$(8c)$ into the original equation $(1)$. Immediately after this substitution multiply both sides by $3/(P_cV_c)$. You should have the result

$$\left[\frac{P}{P_c} + 3\left(\frac{V_c}{V}\right)^2\right]\left(3\frac{V}{V_c} - 1\right) = 8\frac{T}{T_c}.\tag9$$

Define $P/P_c = \pi, V/V_c = \varphi, T/T_c = \tau$. Finally, enjoy the reduced van der Waals equation

$$\enclose{box}[mathcolor="orange"]{\left(\pi + \frac{3}{\varphi^2}\right)(3\varphi - 1) = 8\tau}.\tag{10}$$

What does equation $(10)$ imply by there being only reduced quantities $\pi, \varphi, \tau$?

   Hint: theorem of corresponding states.

$[a]$ While I believe this to be technically true, I also think it deserves a question of its own. The fact that a phase critical point physically always has to (?) be a inflection point confuses me a bit, so is best left to another answerer.


Strictly speaking, once we have anything more than a single cycle without conjugated side chains, the Hückel's rule does not really apply, and the very definition of aromaticity becomes somewhat muddy. But that's not an answer they expect from you, anyway. So just point to the first and third rings and call them aromatic, and that would be it.

Why not the second? Well, because the quinone fragment is not aromatic. Its bond length are different from each other. These $\ce{C=O}$ bonds are always double, and the $\ce{C-C}$ bonds next to them are always single; you can't draw a resonance structure where that would be otherwise.


You are missing an important part in the quote (the first sentence):

  Fixing the point of inversion at the origin implies that one of the symmetry operators is -x,-y,-z. What are the symmetry operators for the glide plane and screw axis? (The answer is clearly listed, but suppose it was not given here!) The symmetry operator for the c-glide plane is of the form x,T-y,1/2+z, while the symmetry operator for the two-one screw axis is of the form T′-x,1/2+y,T′′-z. It is possible to show (using matrices) that the only values for T, T′, and T′′, which permit the symmetry operators to form a closed group, are 1/2, 0, and 1/2, respectively. This produces the symmetry operators listed on the right-hand side of the space-group diagram shown above.

Once you fix the origin, T, T' and T'' are indeed fixed. Otherwise, they are not. So to show that they are, you have to combine a glide plane or screw axis with the inversion operation for the chosen origin, and then make an argument.

These arguments usually go as in this example: Two symmetry operations combined give a pure translation. This translation has to be compatible with the crystal lattice, otherwise you have a contradiction.


The reaction in question involves the addition of water across a double bond. The proton from the water is added to one carbon in the double bond - this carbon is reduced.  The hydroxyl group is added to the other end of the double bond - this carbon is oxidized. Even though the oxidation number of one olefinic carbon (the one that winds up with 2 hydrogens attached) changes from-1 to -2, and the other from -1 to 0, overall, for the entire molecule, there is no net oxidation or reduction.


Why is a singlet state called singlet and a triplet state called triplet?

The terms arose back in the early days of quantum physics when spectral lines that were expected to be singlets were actually observed to be more complex (doublets, triplets, etc.).

An electron can have a spin quantum number of $+\frac12$ or $-\frac12$.  For a system that exists as a singlet, all spins are paired and the total spin for the system is $S=0$.  If we have a single electron, $S=\frac12$.  If we have a triplet system with 2 unpaired electrons, $S=1$.  Also associated with these electrons/ systems is a spin-angular momentum vector, $L$.  Quantum mechanics tells us that $L$ is allowed to have $2S+1$ distinct values.  So

for a singlet, $L = 2(0) + 1 = 1$     only one state exists (singlet)

for a doublet, $L = 2\left(\frac12\right) + 1 = 2$   two states exist (doublet)

for a triplet, $L = 2(1) + 1 = 3$     three states exist (triplet)


Adhesion of polymerized fats to cast iron

Thank you for asking this question. As a French, I am 200+% obsessed with food, cooking and every dish my grandma was cooking for the family. I am also an organic chemist so I don't buy the "everyday bullshit" (if that makes sense).

My first (personal) remark would be that if there were a way to quickly season a cast iron pan, every manufacture would use it and build its reputation on that.

This brings me to a more physical/chemical remark: (1) seasoning cast iron takes time ; (2) many different fats are recommended, from pork fat (low in unsaturated fat) to olive oil (high in unsaturated fat). But only unsaturated fat can polymerize.

Overall, that points to something else, which is that burnt fat would play a major role in seasoning. Burning fat will create carbon (which iron cast contains) and aromatic compounds (from Maillard's reactions) which have affinity for carbon.

As a whole, I would suggest you to cook in this/those pan(s) as often as you can and deglaze it to make some sauce as we do in French cuisine.


In some cases, a primary cell can be recharged, e.g. Mn-alkaline cells can take a partial charge. However, there are a number of reasons primary cells cannot be fully recharged, and that secondary cells have to be specifically designed to allow recharging.

Gas evolution: once most of a cell is charged (though all material has not been converted to its original state), the electrolyte is electrolyzed (apologies for the pleonastic redundancy), usually evolving hydrogen and oxygen, causing the cell to rupture. Nickel-metal hydride cells, for example, get around that issue by incorporating a catalyst to recombine them, $\ce{2 H2 + O2 -> 2 H2O + heat}$.
Physical change in the reactants: the standard "dry cell" (Leclanché cell) is made with an outer shell of zinc. As the cell discharges, parts of the shell are dissolved, eventually leaving holes. Reversing the current flow will cause zinc to redeposit, but not necessarily filling in the weakened areas. Even lead-acid cells designed for storage have limited lifetime because of the buildup of large, insoluble crystals of lead sulfate (sulfation, aka sulphation in GB), which may break off from the lead plates, preventing reversal of the chemical reaction during charging.
Physical damage to the insulator: metal dendrite may grow during charge-discharge cycling, piercing the dielectric between cells, even in well-designed secondary cells.


How many total absorptions would appear on proton -NMR spectrum for this molecule?

To answer this question properly one must understand that the field of the instrument is important. At 400 MHz, ibuprofen displays 7 distinct carbon-bound proton signals. Below is the spectrum from the SDBS website. Although the two methyls of the isobutyl chain are recorded [G] as a "singlet" at δ 0.889, they are diastereotopic and inherently different no matter how far they are removed from the asymmetry on the other side of the ring. If this spectrum were recorded on a 1,020 MHz 1H NMR, perhaps separation of the two methyl signals would occur. That the two methyls are different, imagine that one of them is a perdeutero methyl group. This compound is a diastereomer of the compound having the other methyl group as the perdeutero group. So there are actually 9 total distinct types of hydrogens (including the carboxyl hydrogen) that, given the right spectrometer, could appear in a spectrum of ibuprofen. It is a matter of resolution. [As a case in point, see the 1H NMR of cholesterol at the SDBS website and note that the two methyl groups at the end of the side chain have different chemical shifts even though the are separated from the nearest center of asymmetry by three methylene groups.]
Acknowledgment: http://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi


It says fully: TRUE OR FALSE: Boat cyclohexane is a slightly higher
  energy conformer than chair cyclohexane because: (a).... (a sub
  question) T OR F (b).... (a sub question) T OR F (c)chair cyclohexane
  has more axial hydrogens. T OR F

The answer to part (c) of that question is "False."  It turns out that the chair and boat conformations of cyclohexane both have 6 axial hydrogens - so the chair does not have more axial hydrogens.  Here's a picture of the chair (on the left) and boat (in the middle), you should be able to see the 6 axial hydrogens in both the chair and the boat.


How does the crystal know where to stop stacking the unit cells?

The ideal crystal is terminated by planes, which are described by
Miller Index. Some of these planes are more favorable than others, and as a result, the crystals do get macroscopic shape which is different from the shape of unit cell. For some discussion on reasons for different stability, I have found this article: iopscience.iop.org/0022-3719/12/22/036/pdf/jcv12i22p4977.pdf


Consequences of using reagent preserved with sodium azide on gold film

http://www.ehs.neu.edu/hazardous_waste/fact_sheets/sodium_azide/

"Sodium azide reacts with heavy metals (such as silver, gold, lead, copper, brass, or solder in plumbing systems); and metal salts to form an accumulation of the highly explosive compounds such as lead azide and copper azide."

Assume reaction is fast.  How many micromoles of gold/cm^2 constitute your film?  How many micromoles of azide/volume are in your application?  Will the gold be corroded through?

Is the gold there as a convenience or as a necessary surface?  A brief soak in dilute decanethiol solution plus rinse will cover the surface with a monomolecular layer of hydrocarbon, each chain being anchored by its sulfur.  The surface will be passivated and probably hydrophobic. 1-hydroxy-10-decanethiol will give a hydrophilic surface.  Is it worth the bother and expense to diddle with the gold surface?

Gold is selling for $(USD)42.15/g today.  How much gold is in your annual discard pile? When my school dumped a load of ancient undergrad combustion bombs I went through them.  They originally shipped each with a platinum crucible.  One remained.  Half a Troy ounce of platinum was a good hour's tax-free wage.


Can two precipitates form in a precipitation reaction that starts with only two compounds in water?

The number of examples is considerably great.
Consider the following:

$$\ce{Ba(OH)2(aq) + MgSO4(aq)  -> BaSO4(s) + Mg(OH)2(s)} $$

Above reactants are reasonably soluble in water.
From Wikipedia, $\ce{Ba(OH)2} $ has a solubility of 1.67 g / 100 mL at 20 °C and $\ce{MgSO4} $ a solubility of 25.5 g / 100 mL at 20 °C


What determines the coefficients in the wavefunction of a hybrid orbital?

For the $sp$ case, the other wavefunctions you listed would be equally valid. However, these are just $sp_+$ and $sp_-$ with the signs flipped. Since we square wavefunctions to get the probability density, these would give the exact same physical observables.

As for $sp^2$, the reason for the particular choice of $\alpha,\beta,\text{ and }\gamma$ comes precisely from the name of this hybridization: we want an orbital that is 2 parts p orbital for every 1 part s orbital. In other words, we want a mixture that is 1/3 s and 2/3 p which happens with exactly the normalization coefficients you listed. This is a second constraint. Now this still will lead to an infinite number of different solutions, as we can consider rotating these orbitals by any amount in the xy plane, which would give another valid solution. One of the nice things about the choice you listed is that one of the MOs is aligned with the y-axis and the other two are mirrored across the y-axis, so it is easy to consider the effect of axis symmetry operations on this choice of orbitals. If we include this third constraint of having one of the MOs aligned with an axis, we get a unique solution (besides possible sign flips and that we could align the orbital along the x axis instead) for $sp^2$.


Directional nature of the acetamido group in electrophilic substitution reaction

First of all, your second diagram concerning canonical structures are INCORRECT. There $\ce{N}$ has formed five bonds. $\ce{N}$ can't form five bonds because of unavailability of valance $\ce{d}$ subshells. It can only have the maximum covalency upto $4$ and how do you even think of keeping $\ce N$ atom neutral after forming five bonds which is anyway wrong. So, please make that correction.

There will be only one resonance structure:

(source)

Due to that delocalisation with carbonyl carbon, $\ce N$ will acquire a $\delta$+ charge which will increase its electron withdrawing effect (similar to $\ce{-I}$ effect) to some extent. Still, due to large no. of canonical structures of delocalisation in the benzene ring from the $\ce N$ atom, its $\ce{+R}$ effect is much stronger (strong $\ce{2p$\pi$-2p$\pi$} $ overlap) than its $\ce{-I}$ effect and thus the $\ce{-NHCOCH3}$ group still acts as an ortho-para directing group. But, its activating ability is slightly decreased from $\ce{-NH_2}$ group due to that $\ce{-I}$ effect.


What am I doing wrong with the fuel and packing I am using for a cannon I made?

Flash powder and gun powder are two different mixtures, designed to do two different things. Gun powder is designed to produce a lot of gas very quickly (Carbon dioxide, Sulfur dioxide, Nitrogen) which propels the projectile out of the barrel.

Flash powder - typically potassium chlorate/perchlorate and aluminium powder, examples here and here, produce a lot of heat and light very quickly but not much gas - so no propulsion.


electrolysis - selection of what is going to be on anode and cathode

It depends on its oxidation/reduction potentials. For example, in the electrolysis of a NaCl solution (in water), the reduction potential of water to H2 is -0.83 V while the reduction potential for Na+ to Na is -2.71 V. As the potential for water is greater, water will be reduced instead Na+.
Another example, a solution of CuSO4 in water, the potential reduction for Cu2+ to Cu is +0.34 V, greater than the -0.83 V for water, so the Cu2+ will be reduced.


Why does KMnO4 oxidize iodide to iodine in acidic medium, whereas iodide to iodate in alkaline medium?

A simple way to explain this is to illustrate with the following Latimer diagrams:

If you calculate the standard electrode potential of the highlighted portions it will be positive, so the reaction is spontaneous under standard conditions.

For acidic conditions, iodine is not converted to iodate because the electrode potential of the reaction would become negative, whereas for basic conditions it is possible.

Note that both the manganate and the iodide ions are both affected by acidic and basic conditions; thus it is important to consider their respective electrode potentials.

(Data was taken from Inorganic Chemistry, Shriver et al., Sixth Edition, Oxford University Press)


Why is it that protons and electrons undergo the same amount of deflection in an electric field if they have the same energy?

Now that's a mildly non-trivial observation. Why would they be equal, really?
Let's say a particle with mass $m$, charge $q$, and initial velocity $v$ enters an area of length $L$ where an electric field $E$ starts to deflect it sideways. This is a clear example of uniformly accelerated motion, and its laws are well known: $x=vt,\;y={at^2\over2}$, where the acceleration $a={qE\over m}$. By the time the particle reaches the right side of the picture, it gets deflected by $y={at^2\over2}={1\over2}{qE\over m}\left({L\over v}\right)^2={1\over2}{qEL^2\over mv^2}$.
That pretty much sums it up. $E$ and $L$ are the same for everyone, $q$ is the same in absolute value for electron and proton, so the only difference is $mv^2$. Now, considering your two situations:

Same speed: same $v$, but a proton is slightly (that is, a thousand-and-something times) heavier, hence lower deflection.
Same energy: same $mv^2$, same deflection (just to the opposite sides).

Things get hairy in relativity, but that's another story.
So it goes.


Jahn Teller Effect for d4 low spin configurations

First off, I don't think all 4 electrons will be occupying the $d_{xz}$ and $d_{yz}$ orbitals in z-out distortion. If you do that, you'll be changing spin multiplicity on distortion. That doesn't happen. Rather, all the (originally) $t_{2g}$ levels would fill up, leaving you with a paired up electronic configuration in either $d_{xz}$ or $d_{yz}$ for z-out and a spin paired $d_{xy}$ for z-in.

Now, calculate the relative stabilization such that:

a. z-out:  $d_{xz}$ and $d_{yz}$ are stabilized by 1/3$\delta$ and $d_{xy}$ is destabilized by 2/3$\delta$.

b. z-in:  $d_{xz}$ and $d_{yz}$ are destabilized by 1/3$\delta$ and $d_{xy}$ is stabilized by 2/3$\delta$.

Your electronic configurations would then read ($e_g)^3$($b_{2g})^1$ for z-out, ($b_{2g})^2$($e_g)^2$ for z-in. Now calculate the relative energies. You will find that z-in is favoured by a factor of 1/3$\delta$.


In general, deprotonation of a $\ce{C-D}$ bond would require more energy than that for $\ce{C-H}$ bond because the zero point vibrational energy of the bond is lowered. By referring to our harmonic oscillator zero point energy equation,
$E_0 = \frac{1}{2}h \nu$
where frequency is,
$\nu = \frac{1}{2 \pi} \sqrt{\frac{k}{\mu}}$
and the reduced mass ($\mu$) being,
$\mu = \frac{m_1m_2}{m_1 + m_2}$
we see that as the mass for either atom increases, the zero point energy decreases, therefore, the bond is relatively more stable than its lighter isotopologue. However, swapping out isotopes in a molecule does not change the potential energy surface for the reaction [1]
1.) Sulphonation
Changes in the rate determining step may arise from the type of sulfuric acid used. Since fuming sulfuric acid has a higher concentration of $\ce{SO3}$ (the electrophile) then more dilute forms of $\ce{H2SO4}$, the rate limiting step is no longer forming the bond between benzene and $\ce{SO3}$, rather, it is re-forming the aromatic ring through deprotonation. Ergo, if you did a KIE study with different concentrations of sulfuric acid, you may see that only fuming $\ce{H2SO4}$ shows significant values of $\frac{K_\ce{H}}{K_\ce{D}}$.
2.) Iodination
This article [2] studied the iodination of benzene with the source of iodine being generated through the "oxidation of 0.5 g of $\ce{I2}$ in acetonitrile-tetraethylammonium perchlorate to 2.4 faradays/mol."

The article reported that the secondary kinetic isotope effect for $k_1$ and $k_{-1}$ is negligible, however, $\frac{K_\ce{H}}{K_\ce{D}} = 2.25$ for $k_2$.
Here is a thesis I found that discusses the iodination of aromatic rings [3]
The thesis, in short, suggests that whether the deprotonation is rate limiting has to do with the concentration of the reactants, along with the transition state of the molecule. The transition state is affected by the strength of the base, since a higher $\mathrm{p}K_\mathrm{b}$ may give a transition state where the hydrogen is partially bound to both the ring and the base, making this step in the reaction faster (therefore, not rate limiting). However, it did only investigate substituted aromatic rings, where substituents can play a significant role in the transition state, making the reaction with benzene a little less comparable.
3.) Nitration
This article reports no KIE observed in nitration of benzene [4].


Why the pH of xanthan gum in aqueous solution is 7 (neutral) when it has glucuronic and pyruvic acids in its chemical structure?

Pyruvic and glucuronic acids are both pretty strong acids (pK's of respectively 2.5 and 3.28) but that does not mean that the polymer of xanthan gum will be. Those protons are still there but they may be more tightly bound in the polymer than in the free acids. To find out try titrating a solution with strong base. Eventually you should start prying some of those protons off and you should see an inflection point. OTOH as there are so many carboxylic protons in the polymer the inflection points would probably smear and not be visible but you should see  some buffering above some pH if my hypothesis is correct.

I should make clear that this is purely speculation on my part but it sounds like an interesting experiment.


Is it more energy efficient to desalinate seawater by flash boiling or by flash freezing?

The cheapest way is probably to use the sun, as it is done in Tunisia. In Tunisia, they dig long channels on the beach along the sea. The channels must be long and narrow, and they are filled by sea water. Now the channel is covered by an inclined glass roof. The sun heats the water through the roof, and the steam  is slowly evaporated and condensed on the roof. It makes mist that streams and flow in a drainpipe placed at the lower part of the glass, just above but nearly on a level with the sea. It produce pure water for free.


Why does clearance increase after phenytoin overdose?

Phenytoin is primarily eliminated by hepatic metabolism (>95%).
  Hepatic metabolism is mainly via the CYP2C9 enzyme system with a
  smaller amount metabolized by CYP2C19. About 5% of a phenytoin dose is
  recovered in the urine as unchanged drug. Phenytoin follows
  Michaelis-Menten or saturable pharmacokinetics.

This is the type of nonlinear pharmacokinetics that occurs when the number of drug molecules overwhelms or saturates the enzyme’s ability to metabolize the drug. When this occurs, steady-state drug serum concentrations increase in a disproportionate manner after a dosage increase:

In this case the rate of drug removal is described by the classic Michaelis-Menten relationship that is used for all enzyme systems: rate of metabolism =$\ce{(Vmax ⋅ C) / (Km + C)}$, where Vmax is the maximum rate of metabolism in mg/d, C is the phenytoin concentration in mg/L, $\ce{Km}$ is the substrate concentration in mg/L, and where the rate of metabolism = $\ce{Vmax /2}$.

The clinical implication of Michaelis-Menten pharmacokinetics is that the clearance of phenytoin is not a constant as it is with linear pharmacokinetics, but is concentration- or dose-dependent. As the dose or concentration of phenytoin increases, the clearance rate (Cl) decreases as the enzyme approaches saturable conditions: $\ce{Cl = Vmax / (Km + C)}$. This is the reason concentrations increase disproportionately after a phenytoin dosage increase.

However, since clearance is dose or concentration-dependent, half-life also changes with phenytoin dosage or concentration changes. As doses or concentrations increase for a drug that follows Michaelis-Menten pharmacokinetics, clearance decreases and half-life becomes longer for the drug.

  For drugs such as phenytoin with a low hepatic extraction ratio
  (≤30%), the numeric value of liver blood flow is much greater than the
  product of unbound fraction of drug in the blood and the intrinsic
  clearance of the compound (LBF >> fB)

  Plasma protein binding
  displacement drug interactions cause major pharmacokinetic alterations
  but are not clinically significant because the pharmacologic effect of
  the drug does not change.

Because the clearance of the drug is

  dependent on the fraction of unbound drug in the blood and intrinsic clearance for a low hepatic extraction ratio agent, a
  decrease in plasma protein binding and increase in unbound fraction
  will increase clearance.

Currently there is no substantive evidence of phenytoin auto-inducing its own metabolism (the theory of phenytoin displacing itself from plasma proteins really isn’t convincing at all but perhaps in a drug interaction with a drug which is predominantly plasma bound). The reason being that the information regarding phenytoin auto-induction is sparse and conflicting. The results of this study aren’t convincing enough when taken in perspective of its ability to auto-induce itself after an overdose.

References

Applied Clinical Pharmacokinetics: A. Bauer, Phenytoin: Basic Clinical Pharmacokinetic Parameters


Which of the following carbocations is more stable?

The second one will be more stable.
We usually first check for resonance as it is more effective.
But here it has to be explained on basis of hyper conjugation because the resonance due to the ring is same in both. But the second one will have more hyper conjugating structures as greater number of alpha hydrogens are present.
More about hyper conjugation concept:
https://en.wikipedia.org/wiki/Hyperconjugation

Edit :  No the double bond wont participate in resonance.


Systematic IUPAC names for beta-glucose and lactic acid

A compound that can be isolated from a natural source is commonly given a trivial name. These trivial names are usually related to the biological origin of the material. When the full structure is known, a systematic name may be generated in accordance with the IUPAC recommendations for nomenclature of organic chemistry; however, this name may be too cumbersome (for example for repeated use in the text of scientific publication).

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) for lactic acid is the systematic name 2-hydroxypropanoic acid. Nevertheless, the name lactic acid is retained for use in general nomenclature.

  P-65.1.1.2.3 The names citric acid, lactic acid, glyceric acid, pyruvic acid, and tartaric acid, related to natural products, are also retained; no substitution is recommended, but the formation of salts and esters is allowed.

For carbohydrates such as glucose, the current IUPAC recommendations do not identify a PIN. The systematic name for the β-pyranose form of ᴅ-glucose is (2R,3R,4S,5S,6R)-6-(hydroxymethyl)oxane-2,3,4,5-tetrol. Since this name may be too cumbersome, a semisystematic name may be formed based on the retained name ᴅ-glucose as follows.

  P-102.3.4 Cyclic forms of monosaccharides

  Most monosaccharides exist as cyclic hemiacetals or hemiketals. Two aspects of the internal cyclisation must be examined: first, the size of the ring, and secondly, the conformation of the newly created chirality center.

  P-102.3.4.1 Ring size

  Out of the various possible heterocyclic ring sizes resulting from hemiacetal or hemiketal formation, those with five and six members, including an oxygen atom, prevail and are discussed in this Section. Their names are based on those of the parent heterocycles furan and pyran, respectively. Names are formed by including the terms ‘furan’ and ‘pyran’ before the ending ‘ose’ in the name of a sugar. (…)

Thus, the retained name ᴅ-glucose is changed to ᴅ-glucopyranose to indicate the cyclic form having a six-membered ring.

  P-102.3.4.2 Anomeric forms; the stereodescriptors ‘α’ and ‘β’

  P-102.3.4.2.1 In the cyclic form, the configuration of the newly created chirality center ‘C-1’ must be expressed. This center is called the ‘anomeric center’. The two stereosiomers are called ‘anomers’; they are designated by the stereodescriptors ‘α’ and ‘β’ according to the configurational relationship between the anomeric center and the so called ‘reference center’.

  P-102.3.4.2.2 Configurations ‘α’ and ‘β’ for monosaccharides

  The anomeric reference center in a monosaccharide having a retained name is the configurational atom as defined in P-102.3.3. (…)
  The anomeric stereodescriptor ‘α’ or ‘β’, followed by a hyphen, is placed immediately before the configurational stereodescriptor ‘ᴅ’ or ‘ʟ’ of the carbohydrate name.

Therefore, the resulting name for the β-pyranose form of ᴅ-glucose is β-ᴅ-glucopyranose.


Nomenclature - Lanthanoids vs Lanthanides (and Actinides vs Actinoids)

According to p 52, Nomenclature of Inorganic Chemistry: IUPAC Recommendations 2005 (‘Red Book’):

  The ending ‘ide’
  normally indicates a negative ion, and therefore lanthanoid and actinoid are preferred to
  lanthanide and actinide.

In practice this is almost never followed, and consequently both terms are (for all intents and purposes) interchangeable, unless the context demands a differentiation.


Is Heisenberg uncertainty principle really a demerit of Bohr model?

The text is actually describing the observer effect, which is very commonly confused with the Heisenberg uncertainty principle. The latter is a far more fundamental property of the Universe, and has strong theoretical backing (the non-commutativity of certain mathematical operators in quantum mechanics), regardless of experimental setups. The uncertainty relations thus establish a fundamental "fuzz" to any object, precluding its complete and precise characterization even in principle.

The lack of the uncertainty principle is indeed a severe conceptual limitation to the Bohr model (or rather it is better to say that the lack of the groundwork which results in the Heisenberg uncertainty principle is a limitation); whereas the Bohr model still makes predictions in terms of the classical concept of orbits, a more fully-fledged theory of quantum mechanics applied to an atom necessarily predicts the formation of orbitals, where the position and momentum of the electrons is smeared out.

Indeed, the uncertainty principle is ultimately why electrons have stable configurations around atomic nuclei that stay on average around 100 picometers away from the nucleus. The nucleus itself is only around 1-2 femtometers (~0.001 picometers) in diameter, so electrons could have their potential energy greatly reduced if they closely hugged the nucleus at a similar distance. So what's stopping them? If the electrons did scrunch up so tightly against the nucleus, they would essentially be localizing themselves in space very accurately, and thus the fundamental uncertainty in their positions would be very small. However, because of the Heisenberg uncertainty principle, this would cause a very large increase in the uncertainty of the momentum, which in turns causes a very large increase of the average momentum and from there a large increase in their kinetic energy, so that the electrons would ultimately need a higher total energy to stay so close to the nucleus. Again, this behaviour is completely independent of whether the system is being measured or not.

Furthermore, Bohr's model is actually very empirical. In particular, one of its most important considerations is that the angular momentum of the electrons around an atom obey the formula $m_evr=n\hbar$, where $n$ is the positive integer responsible for introducing quantization to his model. However, this is just a postulated equation, with no rigorous derivation. It was chosen because it happened to describe the emission spectrum of hydrogen exceedingly well, so he knew it was right in at least some regime, but had no proof for it.


Is the ideal gas constant not a constant if Celsius is used instead of Kelvin?

Charles' Law certainly DOES work with temperature in degrees Centigrade!

If we express it (relative to $T=0$ absolute) as $V/T = k$ and if we convert $T$ (K) to $t$ (°C) as $t = T-273$ then obviously Charles' Law would have the form $V/(t+273) = k$.

More generally if, instead of $T$ we use some linear function of $T$, say $Y= mT + b$, then $$V/T = Vm/(Y-b) = k$$ or $$V/(Y-b) = k/m,$$ where $k$, $b$, and $m$ are constants. We can substitute $k'$ for $k/m$ and get $V/(Y-b) = k'$, but we can't get rid of the $-b$ that way.

The claim that a law (even if it is an approximate one like Charles') "doesn't work" with a different unit of measure confuses a formula with a law. Formulas are great for solving high school level problems, but a deeper understanding is expected in college and beyond. Any strictly increasing function can be used as a temperature scale, for instance $X  = π^T$ could be a temperature scale.

The formula we'd use to express Charles' Law would look very different, but the relationship between volume and $\log(X)/\log(π)$ would be just as constant (and since $\log(π)$ is a constant, we'd have $V/\log(X) = k''$.)


Consider the $pK_b$ of the base $\ce{NH3}$ and the $pK_a$ of the acid $\ce{NH4+}$.

$pK_b$ = $4.8$

$pK_a$ = $9.2$

$\ce{NH3}$ is a stronger base than what $\ce{NH4+}$ is as an acid, being $4.8$ smaller than $9.2$: let's compare $pH$ for equal concentrations, say $0.1 M$.

For $\ce{NH3}$, you get $11.1$, while for $\ce{NH4+}$, you get $5.1$.

$11.1$ is about 100 times more "distant" from neutral $pH$ than $5.1$, if we move from the $pH$ scale to the concentration scale. That's what also stems from the comparison of the $pK$ values.

To simplify the problem, take the $pK_a$ for the acid/base couple, e.g. $\ce{NH4+/NH3}$, that is equal to $9.2$. If that value is larger than $7$, then, within that couple, the base will be stronger than its conjugated acid. The other way round, if $pK_a$ is lower than $7$.


Mechanism for conversion of ketone to dichloride with phosphorus pentachloride

I would encourage you to think about how these molecules are likely to interact with each other and try outlining a possible mechanism.

Step 1: The carbonyl oxygen carries a partial negative charge due to its higher electroneegativity. It also has a high  affinity towards bonding with the $\text{III}^{\text{rd}}$ row atoms which have $\ce{d}$-orbitals, such as $\ce{Si}$, $\ce{P}$ and $\ce{S}$.

$\ce{PCl5}$ exists as $\ce{[PCl4+][PCl6-]}$. The lone pairs on the oxygen attack the central phosphorus atom.

Step 2: The negatively charged phosphonate complex has high energy (nucleophilic) $\ce{P-Cl}$ bonds. They attack the electrophilic carbonyl carbon. Note that these bonds are more nucleophilic than basic and would prefer substitution over elimination.

Step 3: The lone pairs on the oxygen are next to the $\ce{d}$-orbitals of the phosphorus. They can push out one of the chlorines attached to the phosphorus.

Step 4: The chloride pushes the $\ce{-OPCl3+}$ leaving group out giving the dichloride. As orthocresol pointed out, this step is likely to be an SN1 type of substitution owing to the excellent leaving group and the rather weak nucleophile $\ce{Cl-}$.

Note: Steps do not represent elementary reaction steps. The $\ce{PCl5}$ formed after $\ce{PCl6-}$ gives its $\ce{Cl-}$ is ignored. I have seen/heard Step 3 and Step 4 being described as concerted, but I have shown them here separately as I am not sure of those "rumours".


Coming up with a Mathematical Model to Predict how long a redox reaction can last

In addition to the redox reactions the transport of material towards the electrodes also needs to be considered. This involves using Fick's equations for diffusion to calculate the concentration of species at distances away from the electrodes as reaction starts and reaches constant conditions. The steady state equations are of the form

$$ D_{ox} \frac{\partial C_{ox}}{\partial x}= k_cC_{ox}-k_aC_{rd} \\
   D_{rd} \frac{\partial C_{rd}}{\partial x}= -k_cC_{ox}+k_aC_{rd}$$

These lead to the Cottrell equation for the limiting flux (current density)  $j=-nFC^o_{ox}\sqrt{D_{ox}/\pi t}$. To get what you want you will probably need to consult some specialist text books on electrochemistry.


Confused about identifying delocalized electron pairs in Isoniazid

The reason why the amine lone pair is delocalized is that to make the last conformer you proposed, the nitrogen closest to the oxygen would have to form five bonds. Nitrogen is a second-row element that cannot readily possess 10 electrons from those 5 bonds, even if nitrogen has to possess those 10 electrons for the very short time required for the electrons to switch conformations between different resonance conformers.


Is there a chemical compound can transform heat into light at 70 °C?

It might be possible in a few ways:

In the process of thermoluminescence, a previously-activated material gives off light when warmed. This is used to date materials such as ceramics, which accumulate energy from terrestrial radiation and cosmic rays. Though ceramics have to be heated quite a bit to release this stored energy, some substances exhibit thermoluminescence at room temperature or lower. $\ce{BeO:Zn}$ has peak emission at 307K, ~34oC.

N.B thermoluminescence is the release of energy previously stored in a material, not the direct transformation of heat energy into electromagnetic.

In an anti-Stokes process, a substance absorbs two or more photons of lower energy to release one of higher energy. This is used in IR beam visualization cards. However, it is unlikely that thermal energy would be sufficient to produce visible light.
Non-blackbody radiators can emit more energy at short wavelengths than would be expected according to Planck's law. "Limelight" owes its intense white light to candoluminescence, or non-blackbody radiation. Using a "heat mirror", the efficiency of incandescent lamps can be greatly improved. That said, though theoretically a material might exist that could prevent emission of almost all wavelengths longer than visible light, it is doubtful that a detectable amount of visible light would be emitted at 70oC.


How to synthesize a material at 2000atmospheres?

I will try to give some more information knowing what you added in the comments. As you didn't precise which temperature I can't answer you in a very precise way.

However as I already said you can try to figure out how a reactor for polyethylene's production is made. If you're able to have access to some engineering literature, then you should have a look to tubular and autoclave reactors. If not you will have to contact someone or a company to explain what you want and get maybe an answer. In any case I doubt that having a reactor able to handle both ultra high pressure and (ultra) high temperature could be cheaper. The main reason is the savety. I bet that if you need such high pressure and temperature, you might use some dangerous chemicals and if not you have the savety and security problems of high pressure and temperature.

I found something that may be interesting for you here. You can achieve a pressure of 2.000 bars at a maximum temperature of 427 °C. On these reactors you can go up to 10.000 bars but with a lower temperature.

I would advise you to contact them they might have some solutions for your problem.


Atoms and molecules, as you surely know, are not tiny hard pebbles that you may measure using a slide gauge, unambiguously and with great precision. Instead, they are clouds of electron density with rather fuzzy edges, like things on an out-of-focus photograph. Their apparent size depends on how you define what "edge" is, or how hard you squeeze them.
This is one of the reasons why we don't have a single definitive list of atomic radii, but a whole variety thereof.

As to the point of your question, yes, $\sigma$ is a good measure of molecular size. But there are also others, equally good and all subtly different. Which one to use depends on your purpose, and may be a hard problem by itself.

Also, remember that (1) the Lennard-Jones potential is defined for spherically symmetrical objects, which your molecules are not, and (2) even for these objects (atoms) it is but an approximation.


What is a "hydrogen-like" or "hydrogenic" atom?

A hydrogen-like atom (or ion) is simply any particle with a nucleus and one electron.

That should be sufficient to answer the question at hand, but I thought I should say a bit more, as some of these answers are potentially confusing.

The historical reason why the Rydberg formula only works for hydrogen-like atoms is because it was originally formulated to explain the spectral lines of hydrogen. It was never intended to explain the spectra of multi-electron atoms.

The physical reason, though, is because the Rydberg formula uses energy levels that only depend on the principal quantum number $n$, which has to be a positive integer:

$$\bar{\nu} = Z^2\mathcal{R}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \qquad n_1,n_2 \in \mathbb{Z}^+$$

and nowadays we know that this is only true for hydrogen-like atoms;$^*$ the energy levels of multi-electron atoms depend on both $n$ and $l$.$^\dagger$

The $n$-dependency was later successfully rationalised by the Bohr model, but saying that "the Rydberg formula only works for hydrogen-like atoms because the Bohr model only works for them" is misleading and misses the point, as:

This implies that the Rydberg formula was derived from the Bohr model, which is not true; it was merely empirically determined, and the formula predated the Bohr model by 25 years.
The Bohr model simply does not work for hydrogen-like atoms. The fact that it reproduces the Rydberg formula should merely be considered a serendipity; Bohr arrived at the correct result by the wrong method.
It does not lend any real insight into the proper reason why the Rydberg formula does not apply for helium, etc. (which I briefly mentioned above).

$^*$ In fact, the energy levels of hydrogen are not only dependent on $n$ (due to various small effects such as – but not limited to – spin-orbit coupling, and hyperfine splitting). Wikipedia has a good overview of the topic here and most QM textbooks have a chapter on the hydrogen atom, where they discuss these perturbations to the Hamiltonian and their effects on the energies. Not surprisingly, the inability to explain this was one of the failures of the Bohr model.

$^\dagger$ Of course, there is also a series of approximations here. The energy levels of multi-electron atoms are only approximately described by sums of orbital energies, so the transition energies are only approximately equal to a difference in energy between two orbitals.


Why should I add lithium bromide to my GPC eluent?

The answer to my question appears to be in a 1977 paper by Nancy D Hann (DOI: 10.1002/pol.1977.170150604; full citation below). In it, she compares GPC of polyurethane in DMF with and without added LiBr. The chromatograms with no added LiBr show multiple peaks. The chromatagrams with added LiBr show just a single peak. The addition of LiBr also shifts the peak to later elution volume.

Hann explains these results by the polymer having a strong polar interaction with the pure solvent. This causes a higher effective hydrodynamic radius, resulting in the earlier elution time. The polar interactions also case aggregation of the polymer, giving multiple peaks. Adding a salt breaks up the polar interactions, giving just a single, later, peak.

Hann, N. D. (1977). "Effects of lithium bromide on the gel-permeation chromatography of polyester-based polyurethanes in dimethylformamide." Journal of Polymer Science: Polymer Chemistry Edition 15(6): 1331-1339.


What is the difference between convection and conduction?

Convection is a mechanism that enhances conductive heat transport, rather than being a separate physical phenomenon.  It operates by bringing hotter parcels of fluid and colder parcels of fluid into closer physical proximity with one another so that conduction can take place more easily and rapidly.  One way this is accomplished is by shearing the fluid near a heated wall in such a way that the colder faster-moving layers further from the wall flow over the hotter slower-moving layers closer to the wall to physically cause the temperature gradient normal to the wall to increase.  This enhances the rate of conductive heat transfer.  Another situation is where the fluid is being mixed by a stirrer to create thin striations of hotter fluid and colder fluid throughout the fluid.  Conduction between these hotter and colder striations increases, again because of the much larger temperature gradients that have been created.  This is typical of how convection works.

What I'm saying is that the deformations experienced by the fluid while flowing are such that they cause the temperature gradients in the fluid to get higher, and this increases the rate of heat transfer.

Imagine a deck of cards split in half, with all the red cards (hot) in one half and all the black cards (cold) in the other half. If you bring these two slabs of hot and cold material together and wait, heat transfer will occur very slowly between the hot and cold regions. However, if you shuffle the deck so that there is a hot (red) card sandwiched in between every pair of cold (black) cards, and a cold (black) card sandwiched in between every pair of hot (red) cards, the heat transfer between the hot and cold layers will be much more rapid. This shuffling effect is what convection does.


How do you calculate a unit cell's dimensions from an XRD spectra?

You should not really have assumed that the compound was tetragonal. If it was indeed tetragonal, then the peaks corresponding to the $(200)$ and $(020)$ planes would overlap. Instead, you can see that one occurs at $2\theta = 46.7^\circ$ and the other at $2\theta = 47.6^\circ$. The question may be considered a little misleading, but it does hint at this; it says that the compound is "nearly tetragonal", and it gives you the general formula for an orthorhombic crystal, which is one with $a \neq b \neq c \,(\neq a)$ and $\alpha = \beta = \gamma = 90^\circ$.

The next problem is that you are given values of $2\theta$ (which happens to be the convention), but the formula reads $\sin^2{\theta}$. As such, you need to make the appropriate adjustments.

This question is relatively easy because the peaks have already been indexed for you (i.e. the values of $h$, $k$, and $l$ are already found out). Therefore, all that is left is to plug the numbers into the formula you are given. Which peaks you use is entirely up to you, although it is obviously easier to use $(h00)$, $(0k0)$ and $(00l)$ peaks.

The reflections of the $(200)$ plane occur at $2\theta = 47.6^\circ$. So:

$$\begin{align}
\theta &= 23.8^\circ \\
\sin^2{(23.8^\circ)} &= \frac{(1.54~Å)^2}{4} \cdot \frac{2^2}{a^2} \\
a &= 3.81~Å
\end{align}$$

Similarly, you can show using the $(020)$ and $(003)$ peaks that $b = 3.89~Å$ and $c = 11.64~Å$. This is actually pretty good. The reported values of $a$, $b$ and $c$ are $3.8231~Å$, $3.8864~Å$, and $11.6807~Å$ respectively.

Regarding units, if you make sure that you leave your units inside the intermediate workings (as they should be), you should never run into confusion about "what unit should the answer be". In your question, you converted $\lambda$ into units of $\mathrm{nm}$. $\sin^2\theta$, $h$, $k$, and $l$ are all dimensionless. Therefore, your calculated values of $a$, $b$, and $c$ must match the units of $\lambda$ you used, i.e. $\mathrm{nm}$.

In my answer, I used ångströms throughout. It is not wrong to use $\mathrm{nm}$, but as Babounet noted in the comment section, ångströms are more commonly used in crystallography.


What is the binding mode of the ligand in complexes with quadruple metal-metal bonds?

To determine what orbitals do, it is often a good idea to look at the symmetry group as it can provide a rough overview over what can (symmetry-allowed) and cannot (symmetry-forbidden) happen. In the case of a quadruple bond such as $\ce{[Re2Cl4]^2-}$, we can approximate the rhenium(III) centres as being close to $C_\mathrm{4,v}$. A quick glance at the corresponding group theory table (copied below to avoid page loading issues in the large linked page) will show us which irreducible representations we may need to consider.

$$\begin{array}{c|ccccc|cc} \hline
C_\mathrm{4v} & E & 2C_4 & C_2 & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_1} & 1 & 1 & 1 & 1 & 1 & z & x^2+y^2, z^2 \\
\mathrm{A_2} & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_1} & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_2} & 1 & -1 & 1 & -1 & 1 & & xy \\
\mathrm{E} & 2 & 0 & -2 & 0 & 0 & (x,y),(R_x,R_y) & (xz,yz) \\ \hline
\end{array}$$

The final two columns show the relevant orbitals. The s orbital always transforms as $\mathrm{a_1}$, the p orbitals as $\mathrm{a_1+e}$ and the d orbitals as $\mathrm{a_1+b_1+b_2+e}$. The four chlorine atoms will bond using p orbitals (but the orbital type does not matter as we will only be considering σ bonds). They transform as $\mathrm{a_1+b_1+e}$ (calculation not shown). Immediately, we notice that $\mathrm d_{xy}$ does not contribute to bonding to the ligands but only to the $\ce{Re-Re}$ quadruple bond. Furthermore, we notice that $\mathrm d_{x^2-y^2}$ is practically required to bond to the chlorines as there is no other rhenium orbitals of that symmetry.

To determine what is left, it may be easiest to draw out the potential chlorine orbitals. $\mathrm{a_1}$ means that all phases pointing to rhenium are the same. This can bond either with rhenium’s 6s or $\mathrm{5d}_{z^2}$ orbital or a linear combination of both. The $\mathrm p_z$ orbital does not have any contribution in the $xy$ plane and thus cannot participate.

The chlorine’s $\mathrm e$ orbitals correspond to two in which two opposite chlorines have opposite phases and the other two have zero contribution (in one case, the $x$ chlorines have contribution, in the other the $y$ ones). Thus, to bond with these we will use $\mathrm{6p}_x$ and $\mathrm{6p}_y$. Again, this is because the corresponding d orbitals do not have any contribution in this $xy$ plane.

Thus, to sum things up:

$\mathrm{6p}_x,\mathrm{6p}_y$ and $\mathrm{5d}_{x^2-y^2}$ can bond to chlorines
$\mathrm{6p}_z, \mathrm{5d}_{xz}, \mathrm{5d}_{yz}$ and $\mathrm{5d}_{xy}$ can bond to the other rhenium
the $\mathrm{6s}$ and $\mathrm{5d}_{z^2}$ orbitals can bond to either; they probably linear combine with $\mathrm{6p}_z$ to form two bonds and keep an empty orbital.

If the symmetry around a rhenium atom is assumed to be $C_4$ instead of $C_\mathrm{4,v}$, that doesn’t change much of the analysis; only the distinction between $\mathrm d_{x^2-y^2}$ and $\mathrm d_{xy}$ becomes irrelevant.


Determining sulfate content in Mohr's salt

If there is $\ce{0.00501 mol}$ of $\ce{BaSO4}$ in the precipitate, this means that there was $\pu{0.00501 mol}$ of sulfate in the Mohr's salt solution and $\pu{0.00501 mol}$ of sulfate in $\pu{0.982 g}$ of Mohr's salt. And there was one half, that is $\pu{0.0025 mol}$ of $\ce{Fe(NH4)2(SO4)2.xH2O}$ in $\pu{0.982 g}$ of Mohr's salt. This allows you to calculate the molar mass of the salt, and to deduce $x$.


Product of the reaction between hexane‐2,5‐dione and ammonium carbonate

Heating $\ce{(NH4)2CO3}$ will generate $\ce{NH3 + CO2 + H2O}.$ Out of these, $\ce{NH3}$ will act as a nucleophile as it's the most nucleophilic of the products. Addition of $\ce{NH3}$ to the two ketones, followed by elimination of water molecules (due to the high temperature provided by ∆) will generate the final product as follows:

The product is a five-membered aromatic ring called 2,5-dimethylpyrrole.


Would cerium(III) in solution be oxidized to cerium(IV) in ESI mass spectroscopy?

I found one study that examined cerium(III) polyoxometalate salts and concluded that at least under the conditions of their study, $\ce{Ce(III)}$ did not oxidize under electrospray conditions.  The study is Bray et al. 2011 in the International Journal of Mass Spectrometry.

They say:

  The presence of the redox active $\ce{Ce}$ metal ion introduces additional complicating effects as it would be difficult to differentiate such ions as $\ce{HCe(IV)P2Mo22O75^3−}$ from $\ce{H2Ce(III)P2Mo22O75^3−}$. However, successive degradation of $\ce{H_{g}Li_{2−g}Ce(III)P2Mo22O75^3−}$ to eventually form $\ce{HCe(III)PMo7O26−}$ yields an ion that can be directly compared to $\ce{LiCe(III)PMo7O26−}$. In particular, the two ions $\ce{HCe(III)PMo7O26−}$ and $\ce{LiCe(III)PMo7O26−}$, are separated by ∼6 m/z (Fig. 6), supporting the proposition of substitution of H+ by Li+ and, therefore, the identity of $\ce{LiCe(III)PMo7O26−}$. These results also aid in assigning the proton count on $\ce{H2Ce(III)P2Mo22O75^3−}$, $\ce{HCe(III)P2Mo18O62^2−}$, and $\ce{HCe(III)PMo13O44−}$, thereby indicating the trivalent oxidation state of cerium throughout. Although Ce(III) would not necessarily expected to be oxidized to Ce(IV) under the conditions of these experiments, confirming that cerium remains in the Ce(III) oxidation state from solution throughout the electrospray and CID processes is a useful methodology for oxidation state determination of redox active metal-complexes with f-block elements such as Ce(IV) and Pu(III/IV).

So they didn't expect Ce to be oxidized by electrospray in general, although the basis for this expectation wasn't clear.  They also report hard data showing that at least for polyoxometalate salts, no oxidation was apparent under electrospray conditions.

In general, the thousands of volts applied during the electrospray process does not mean that individual ions are exposed to fields of thousands of volts, because electrosprayed solutions are conductive and field strengths in solution are lowered because of Debye shielding.


Both, Bent about Oxygen and tetrahedral about chromium:

$\hskip2.5in$


IUPAC basic terminology of stereochemistry defines anomers and epimers:

Anomers
Diastereomers of glycosides, hemiacetals or related cyclic forms of sugars, or related molecules differing
in configuration only at C-1 of an aldose, C-2 of a 2-ketose, etc.
Epimers
Diastereomers that have opposite configuration at only one of the two or more stereogenic centers present in the respective molecular entities.

So yes, anomers are a subset of epimers.
(Interestingly, on the contrary, ‘anomer’ term is exceptionally used in much broader meaning than ‘epimer’, in the term ‘anomeric effect’, which became generalized to e.g. conformations of molecules completely unrelated to saccharides.)


Whenever some says they are just starting out with electronic structure theory, I always recommend that they read Szabo and Ostlund's Modern Quantum Chemistry. The book is great introduction to HF and post-HF methods (some of the discussion of coupled cluster is a bit outdated) and its very affordable. It is nice to have around as a reference.
Getting to your actual question, I think you are confusing yourself by trying to hybridize the orbitals. During the HF procedure, you will solve for a set of MOs that is are linear combinations of the AOs, so there isn't a need to form these $S_n$ functions beforehand.
There is also the issue of actually forming the Slater determinant; while the solution to HF will be a Slater determinant, you generally don't explicitly write it out this way. Generally, you assume that each MO in the Slater determinant is a linear combination of the AOs and you solve for the coefficients of each AO, such that the energy is minimized. So the Slater determinant for methane will have 10 (initially unknown) MOs in it to account for the 10 electrons and you will then need to build these MOs using the Roothaan-Hall form of the Hartree-Fock equations.


Geometry of the pentachloridoindate(III) anion

TL;DR Initially published crystal structure of $\ce{[NEt4]2[InCl5]}$ [1] according to the further investigations [3], is not valid. The $\ce{InCl5^2-}$ ion does not have $C_\mathrm{4v}$ symmetry, and VSEPR theory pretty much explains formation of numerous slightly distorted trigonal bipyramidal $\ce{InCl5^2-}$-containing complexes according to the most recent single-crystal structure experiments.

From the other side, the bond lengths and geometry index refer to the dominating character of square pyramidal environment, which is in this case can be dictated by the crystal packing, towards which $\ce{[InCl5]^2-}$ is more sensible than $\ce{[SnCl5]-}$ due to the noticeable size difference ($\ce{[InCl5]^2-}$ is bulkier). Either way, it looks like another crystallographic experiment (preferably at lower temperatures to decrease the size of thermal ellipsoids) is needed to determine the angles more precisely.

As a general advice, always pay attention to the $R_1$ and $R_2$ values of the crystal structure. A rule of thumb: a good structure refinement of small molecules should lead to $R_1 < 0.05$ and $R_2 < 0.12$.

Crystal structure of $\ce{[NEt4]2[InCl5]}$, determined in 1969 [1], has been subsequently criticized in several publications, mainly involving additional symmetry vibration analysis of Raman spectra for $\ce{MX5^2-}$ anions [2]. It's been established that $\ce{InCl5^2-}$ ion does not have $C_\mathrm{4v}$ symmetry (but retains $C_\mathrm{2v}$), and that the crystal may not be centrosymmetric.

The results are summarized in [3]:

  Deconvolution of the single-crystal data was again performed with the aid
  of the low temperature data, revealing $11$ $\ce{InCl5^2-}$ bands between $300$ and $\pu{100 cm-1}$. Two overlapping A stretches are clearly evident at $292$ and $\pu{286 cm-1}$, but no B stretch is apparent. The two E symmetry stretches expected for the anion of a $C_2$ site, but not for a $C_4$ site, are evident at $281$ and $\pu{271 cm-1}$. [...] With the aid of low-temperature data, the four E modes predicted for $\ce{InCl5^2-}$ at $C_2$ sites are found at $144$, $136$, $122$, and $\pu{103 cm-1}$ in the single-crystal data. With this reinterpretation of the spectrum it is no longer necessary to assume arbitrarily the presence of a lattice mode in this region.

  In summary, the vibrational data for the tetraethylammonium salts of $\ce{InCl5^2-}$ and $\ce{TlCl5^2-}$ indicate that these complexes reside on sites lacking full $C_4$ symmetry. The most straightforward conclusion from the vibrational data is that the $\ce{InCl5^2-}$ ions have local $C_2$ symmetry, which would be inconsistent with the previous structure determination [1]. [...]

  The structure of $\ce{[(C2H5)4N]2[InCl5]}$, as previously reported, consisted of centrosymmetrically related $\ce{InCl5^2-}$ ions situated about the
  Wyckoff c positions of $C_4$ symmetry in space group $C_\mathrm{4h}-P4/n$.
  Coaxial with the $\ce{InCl5^2-}$ ions are disordered $\ce{(C2H5)4N+}$ ions
  also situated about the Wyckoff c positions. The N atoms of the two remaining $\ce{(C2H5)4N+}$ ions in the unit cell occupy the centrosymmetrically related Wyckoff b sites of $S_4$ symmetry, the cation as a whole again being disordered. There are two disturbing features of this solution and its subsequent refinement: the disorder of the $\ce{(C2H5)4N+}$ ions and the significant residual electron density in the region of the basal Cl atoms of the tetragonal-pyramidal anion.

It is demonstrated further that for the ordered model the most satisfying crystallographic interpretation coexists with $C_2$ rather than $C_4$ symmetry. A correlation of allowed point symmetries implies switching from $P4/n$ to $P\bar{4}$ space group, thus reducing $R$-factor ($R_1$) from $9.5\%$ to $6.7\%$.

  During the refinement in $P\bar{4}$, small but significant changes
  occur in the basal plane of the $\ce{InCl5^2-}$ ion. The basal Cl atoms
  have moved by $\pu{-0.5 A}$ from their fourfold symmetric positions in $P4/n$, leading to variations in the bond distances and angles about the In atom (Figure 3) that are qualitatively sufficient to explain the vibrational results.

Geometry index $\tau_5$ for $\ce{[InCl5]}$ fragment can be used to determine formal coordination environment as follows:

$$\tau_5 = \frac{\beta - \alpha}{60^\circ},$$

where $\alpha$, $\beta$ - two greatest valence angles of the coordination center ($\angle \ce{Cl - In - Cl}$, $\alpha < \beta$).

\begin{align}
\begin{cases}
\tau_5 &= 0 \qquad &\text{square pyramidal geometry} \\
\tau_5 &= 1 \qquad &\text{trigonal bipyramidal geometry}
\end{cases}
\end{align}

For the original $\ce{[NEt4]2[InCl5]}$ structure refined in [3] I've taken angular values from their supplementary materials. It turned out that changing space group to $P\bar{4}$ still results in dominating character of square pyramidal geometry of $\ce{[InCl5]^2-}$, though one must remember that the rotation axis of highest order is $C_2$, not $C_4$:

$$\alpha = (150.6 \pm 5.0)^\circ, \beta = (154.1 \pm 6.0)^\circ, \bar{\tau_5} = \frac{154.1^\circ - 150.6^\circ}{60^\circ} = 0.06; \tau_5 \in [0.04; 0.24]$$

Paper [4] mentions that for isoelectronic $\ce{[SnCl5]-}$-contining structures with various cations only trigonal-bipyramidal geometries were reported, and suggests higher influence of crystal packing on $\ce{[InCl5]^2-}$ geometry. This can also be explained by higher flexibility of $\ce{MX5}$ fragment in $\ce{[PPh4]2[InCl5]}$ in comparison with $\ce{[PPh4][SnCl5]}$ [5] due to greater average bond length: $d(\ce{In - Cl}) = \pu{2.5 A}$, $d(\ce{Sn - Cl}) = \pu{2.3 A}$.

As for auxiliary experimental data, all crystal structures including $\ce{[InCl5]^2-}$ fragment determined in the past two decade completely support VSEPR theory that predicts trigonal bipyramidal $D_\mathrm{3h}$ symmetry (with one intermediate exception of $\tau_5 \approx 50\%$).

bis(Tetraphenylphosphonium) pentachloro-indate(III) [4], trigonal bipyramidal geometry:
$$\alpha = 124.24^\circ, \beta = 174.74^\circ, \tau_5 = \frac{174.74^\circ - 124.24^\circ}{60^\circ} = 0.84$$

$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF8000}{\Large\bullet}~\ce{P}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#A67573}{\Large\bullet}~\ce{In}$.

bis(Diphenyldichlorophosphonium) pentachloro-indate(III) [6], trigonal bipyramidal geometry:
$$\alpha = 120.93^\circ, \beta = 179.29^\circ, \tau_5 = \frac{179.29^\circ - 120.93^\circ}{60^\circ} = 0.97$$

$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF8000}{\Large\bullet}~\ce{P}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#A67573}{\Large\bullet}~\ce{In}$.

tetrakis($\mu_3$-Selenido)-tetrakis($\mu_2$-1,5-bis(diphenylphosphino)pentane)-deca-gold(I) pentachloro-indate(III) [7], intermediate geometry:
$$\alpha = 138.45^\circ, \beta = 166.50^\circ, \tau_5 = \frac{166.50^\circ - 138.45^\circ}{60^\circ} = 0.47$$

$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF8000}{\Large\bullet}~\ce{P}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#FFA100}{\Large\bullet}~\ce{Se}$;
$\color{#A67573}{\Large\bullet}~\ce{In}$;
$\color{#FFD123}{\Large\bullet}~\ce{Au}$.

tetrakis($\mu_2$-chloro)-chloro-tetrakis(triphenylphosphine-P)-di-copper(I)-indium(III) tetrahydrofuran solvate [8], trigonal bipyramidal geometry (though authors refer to it as to "quasi square-pyramidal" coordination based on a single slightly shorter $\ce{In - Cl}$ bond distance $(d(\ce{In-Cl_\mathrm{ap}}) = \pu{2.36 A})$, shortest $d(\ce{In-Cl_\mathrm{eq}}) = \pu{2.42 A})$:
$$\alpha = 123.90^\circ, \beta = 172.89^\circ, \tau_5 = \frac{172.89^\circ - 123.90^\circ}{60^\circ} = 0.82$$

$\color{#EEEEEE}{\Large\bullet}~\ce{H}$;
$\color{#909090}{\Large\bullet}~\ce{C}$;
$\color{#FF8000}{\Large\bullet}~\ce{P}$;
$\color{#1FF01F}{\Large\bullet}~\ce{Cl}$;
$\color{#C88033}{\Large\bullet}~\ce{Cu}$;
$\color{#A67573}{\Large\bullet}~\ce{In}$.

Bibliography

Brown, D. S.; Einstein, F. W. B.; Tuck, D. G. Inorganic Chemistry 1969, 8 (1), 14–18. DOI 10.1021/ic50071a004.
Adams, D. M.; Smardzewski, R. R. Journal of the Chemical Society A: Inorganic, Physical, Theoretical 1971, 714. DOI 10.1039/j19710000714.
Joy, G.; Gaughan, A. P.; Wharf, I.; Shriver, D. F.; Dougherty, J. P. Inorganic Chemistry 1975, 14 (8), 1795–1801. DOI 10.1021/ic50150a011.
Bubenheim, W.; Frenzen, G.; Müller, U. Acta Crystallographica Section C 1995, 51 (6), 1120–1124. DOI 10.1107/S0108270194011789.
Müller, U.; Siekmann, J. F. Acta Cryst C 1996, 52 (2), 330–333. DOI 10.1107/S0108270195011073.
Taraba, J.; Zak, Z. Inorganic Chemistry 2003, 42 (11), 3591–3594. DOI 10.1021/ic034091n.
Olkowska-Oetzel, J.; Sevillano, P.; Eichhöfer, A.; Fenske, D. European Journal of Inorganic Chemistry, 2004 (5), 1100–1106. DOI 10.1002/ejic.200300774.
Zhang, X.-Z.; Song, Y.-W.; HuiWu, F.; Zhang, Q.-F. Zeitschrift für Naturforschung B 2007, 62 (6). DOI 10.1515/znb-2007-0605.


Why is a racemic mixture formed in the Diels-Alder cycloaddition?

The concerted Diels-Alder reaction has in principle eight different possible transition states. Depending on the substrates some of them will lead to the same products.

Below you can see an overview of the (non-optimised) transition state arrangements for the reaction of isoprene and propene.

From these images you should be able to determine that the following behave like enantiomers:

(a) and (g)
(b) and (h)
(c) and (e)
(d) and (f)

When you think a little bit longer about it, you will also see that (a) and (h) will form the same product. This is also true for (b) and (g), (c) and (f), as well as (d) and (e).
In total, you will therefore obtain four products two different regio-isomers and their corresponding enantiomer each.


I will try to offer a detailed way to propagate errors here, but before beginning it's worth asking what your audience is likely to expect "error bars" to mean.  In your field are error bars usually determined simply by replicate analyses of the same sample?  Or replicate samples of the same experiment or field site?  If so, it's probably best just to stick with what your audience is likely to expect.  That said...

The theoretical best way to fit your calibration curve would be with weighted linear least-squares algorithms.  In that algorithm, in addition to the {x, y} data, which in your case is $\{\ce{Sr_{known}, Sr_{meas}}\}$, you also supply weights corresponding to the uncertainty in the y values.  The weight for each y data point would be $\frac{1}{\sigma^2}$, which you could calculate from the $2\sigma$ uncertainty given by your instrumentation for that data point.  In practice, if you did this, I doubt that the parameters of your calibration would change very much at all from what you did previously.  But formally the weighted variant of least-squares is required if the uncertainty in y depends on y.  See wikipedia for more.
After doing the (weighted) least squares calibration regression, you can find the uncertainty in the estimation using the calibration equation.  $\sigma_{est} = \sqrt{\frac{\sum{\left(y_{pred}-y_{meas} \right)^2}}{N}}$.  In that equation $y_{pred}$ is the calibration-predicted Sr concentration and $y_{meas}$ is the measured Sr concentration, and for you $N$, the number of data points, is 37.  This uncertainty is one part of the uncertainty you'll eventually report, and reflects the uncertainty in your calibration curve.  Based on the R-squared you reported, my guess is that this uncertainty will be low.
After all that (!), you need to estimate the other major source of uncertainty, the uncertainty in the measurement of your unknowns ($\sigma_{meas}$).  There are several OK methods for doing this, the best of which would be replicate analysis of several samples.  It might be OK to use the $1\sigma$ value calculated from the $2\sigma$ uncertainty reported by your instrument too.
Finally, combine the uncertainties with the standard equation for combining multiple sources of uncertainty $\sigma_{total} = \sqrt{(\sigma_{meas})^2 + (\sigma_{est})^2}$.


Why does the image of cyclo[18]carbon look like a nonagon?

The first thing to say is that I'm not sure where that image is taken from; it's neither in the original article nor in the supporting information to the article. Therefore, it appears to be more of an "artist's impression" rather than an actual atomic force microscopy (AFM) image, which is what was reported in the paper.

Nevertheless, the actual AFM images of $\ce{C18}$ are in Figs. 3Q and 3R. They are referred to as "AFM far" and "AFM close" respectively because of the height of the probe ($\Delta z$):

One can indeed see that there is 9-fold symmetry (technically, $D_\mathrm{9h}$). This implies that $\ce{C18}$ has a 'polyyne' structure in which there are two different types of bonds $\ce{-C#C-C#C-\phantom{}}$, rather than a 'cumulene' structure in which every bond is equivalent $\ce{=C=C=C=C=}$ (prior to this, computational studies had been equivocal as to which form was more stable).

The bright spots within the ring do not correspond to carbon atoms, but rather to carbon–carbon triple bonds. This is consistent with the AFM images obtained for other similar intermediates in the synthesis of cyclo[18]carbon. In the authors' own words:

  Assigning the bright features in the “AFM far” images to the location of triple bonds, we observed curved polyyne segments with the expected number of triple bonds: 5 in $\ce{C22O4}$ and 8 in $\ce{C20O2}$. At small tip height, we observed sharp bond-like features with corners at the assigned positions of triple bonds and straight lines in between. This contrast was explained by CO tip relaxation, in that maxima in the potential energy landscape, from which the tip apex was repelled, were located above the triple bonds because of their high electron density. In between these maxima, ridges in the potential landscape led to straight bond-like features.

(The two bright spots outside the ring are due to individual $\ce{CO}$ molecules.)


Why an asymmetric geometry with sp3d and sp3d3 hybridization?

As already stated in the comments, there is a considerable debate as to whether d-orbitals can be considered to take part in main group elements’ hybridisation at all.

d-Orbital hybridisation was invented by Linus Pauling to explain the bonds in sulfate and sulfur trioxide because he didn’t like the idea of separating charges. However, d-orbitals are far removed energetically compared to s- and p-orbitals — also the reason why 4s is filled before 3d. Technically, before a d-orbital hybridisation can be considered, the next shell s-orbital should be hybridised in first, following the path of lowest energy.

Conveniently, all the cases you presented in your question can be adequately explained using only s- and p-orbitals. See the links for $\ce{SF6}$, $\ce{PCl5}$ and $\ce{IF7}$ respectively. Many thanks to ron for providing them in his comments!


The stance is currently an even mix of hogwash conspiracy theories, pseudoscience and scams.

Even if the field had merit - which it most likely doesn't, the number of anti-scientific "papers" on the subject has ruined it for a long time.


Lithium compounds with good solubility in organic solvents?

Under inert conditions you may try dissolving pure metallic lithium. Lithium forms naphtalenide salt , I don't see why di-isopropyl-naphthalene should be any different.

Another option would be use of 12-crawn-4, which make a lot of salts lipo-soluble. Presence of TMEDA (tetramethylethilendiamine) should affect solubility of many lithium salts as well, though I would focus on salts like benzoate or perchlorate.


Why does copper precipitate out as a pink solid?

What you describe as pink is the actual colour of copper. The colour of a clean, solid surface of high-purity copper is typically salmon red. In your case, the appearance may be further affected by the particle size of the precipitate.

The familiar reddish-brown colour of copper is caused by copper(I) oxide, which is slowly formed on the metal surface.


How Many mL of 1.0M HCl Would be Required to Completely Decompose Sucrose into Glucose and Fructose?

The hydrolysis of sucrose is acid-catalysed, so HCl will not be in the chemical equation (it will be listed above the arrow). In theory, this means that you could have little HCl, as it will never be used up and will keep reacting to decompose all of your sucrose. So it isn't so important how much you use, but you should still use enough that it won't take ages to decompose all the molecules. If you're interested you could look up mechanisms for acid catalysis, but you wouldn't be expected to know this at high school level.

To find the chemical equation for the decomposition, you can add up the number of elements of fructose + glucose and compare to sucrose. You will find that two hydrogens and one oxygen is needed, so it is simply a hydrolysis reaction (adding a water molecule).