Why do we use [H⁺] and [OH⁻] in balancing redox reactions instead of using [H₃O⁺] and [OH⁻]?

It is by far not universally agreed upon. My year 10 chemistry teacher taught me to use $\ce{H3O+}$ in balancing. The result is a larger number of water molecules on the other side of the equation but everything else is the same.

Many chemists will prefer $\ce{H+}$ because it is shorter and simpler, that’s all.


Numbering starts from the side that has more locants or from the side that has a more preferable group?

Concerning the numbering of locants for substituents, the most important simplified criteria are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The $\ce{-OH}$ group is the principal characteristic group in this case. It is expressed as suffix ‘ol’.
Therefore, low locants are assigned first to the $\ce{-OH}$ groups according to Rule (c), which yields ethane-1,1-diol.
After that, a low locant is assigned to the chloro groups according to Rule (f); however, the only possible numbering left is 2,2,2-trichloroethane-1,1-diol.


Number of isomers in a 3-sugar oligosaccharide

This question is interesting and neglected. Bob made a sensible attempt at solving the problem. I took a different approach. [The number of isomers 12,288 was found in Stryer’s Biochemistry (Google: stryer 12288 isomers)]. Of course, oligo- means a few. I presume this means three in this case. But can D-glucose, D-galactose and D-mannose be used as many times as you want? That would mean 27 (33 = 27) permutations which would eventually make the total much greater than 12,288. However, if each hexose is to be used once, then six (6! = 6 factorial) permutations are possible for the location of the hexoses in the oligosaccharide. Division of 12,288 by 6 removes this issue from the question and leaves 2,048 as the number to justify. Now 2048 = 211. The problem reduces to finding 11 stereocenters, if possible!
Let us assume as a point of departure that diagramatic structure ABC (top of the illustration; The connectivity C1-C4 is arbitrary.) shows where bonds are formed between the C1 anomeric carbons and the hydroxyls at C2, C3, C4 and C6 of another hexose. Another pair of assumptions is that all rings are 6-membered (pyranose) and that all hexoses, as defined, are in the D-series. Finally, assume for the time being that there are no C1-O-C1 connections as typified  in figs. D and E.
Rather than define C1 of hexose A as having two stereocenters (α and β anomers), this assignment will be considered in a different way. Hexose A is given a value of zero because no attachments may be made at its hydroxyl groups at C2, C3, C4 and C6. Hexose B has four hydroxyl groups (blue dots) that can form two anomeric, glycosidic bonds at C1 with hexose A. The AB connection creates 24 (16) stereoisomers. In a similar fashion the hydroxyls at C2, C3, C4 and C6 of hexose C (red dots) form another 16 stereoisomers with C1 of hexose B. The total count thus far is 3! x 28 = 6 x 256 = 1536. This result is only one-eightth of the goal of 12,288 or short by a factor of 23. To this point the eight hydroxyl groups considered can only form two configurations each.
Now the anomeric hydroxyl of hexose C when bonded to the anomeric carbon of hexose B will give not two stereoisomers but rather four (αα, αβ, βα, ββ)(see fig. E). Similarly, the same relationship exists between hexoses B and A (see fig. D). These configurations add a factor of 16 (24) to the total. This reasoning leads to a value of (3!) x (212) = 6 x 4096 = 24,576.

If the original analysis considered a structure such as ABC having a mutarotating C1-OH, then the total might be 12,288. However, ABC is a connectivity map, not a chemical structure.

What may be concluded from all of this? The premises set down here may not be those applied by the proposers of 12,288. There may be a flaw in this, or their, analysis. The answer may be 12,288, 24,576 or something else.


My background is macromolecular crystallography; while experimental data was kept locked in the research laboratories in the 1980s, they are now widely available through the international protein database. For example, you can view the electron density of crambin at http://www.rcsb.org/3d-view/1EJG.

The database for smaller structures is the Cambridge Structural Database. They have implemented depositing experimental data (structure factors) in the 2000s, but I did not find an efficient way to search for those structures that come with structure factors.

Charge density maps are often visualized as 2D-sections of the experimental density minus the density expected from isolated atoms (deformation density). One software program to calculate these is MoPro. Another program is CrystalExplorer, which shows deformation densities in 3D. In all cases, you would have to have access to the structure factors to make your own figures.

  Is there any collection of charge density maps for small molecules? Any good data at all?

I don't think so.


Reaction of solid silver sulfide on warming with dilute nitric acid

Nitric acid is a strong oxidising agent and it will oxidise both the reactants and products. Thus, a reaction involving nitric acid is not a simple double-displacement reaction but a redox reaction where nitric acid itself get reduced to a couple of nitrogen oxides depending on concentration of the acid and the other reactants gets oxidised. According to [1]:

  Nitric acid, when of concentration above 5%, dissolves
  precipitated silver sulphide rapidly. Very strong acid yields silver
  sulphate alone, while acid of lower concentration forms some nitrate
  in proportion to its dilution. 15-20%, acid yields the
  maximum, about 95 per cent, nitrate, greater dilution again resulting
  in a smaller percentage. Heating the acid, or prolonging its time of
  action is unfavorable to the formation of nitrate.

Another paper[2] suggests formation of normal silver(I) sulfate through a series of exotic sulfates which is confirmed by the change in color: -

  [...]the acid additions used correspond to 75%, 150% and 200% of the
  amount required stoichiometrically based on the following reaction:

  $$\ce{3Ag2S + 8HNO3 <=>> Ag2SO4 + 8NO + 4H2O}$$

  [...]

  The distinct colors of the various phases involved in this system
  (black silver sulfide ($\ce{Ag2S}$), brown silver sulphide sulphate
  ($\ce{Ag6S3O4}$), white to light yellow silver sulfate ($\ce{Ag2SO4}$)
  and bright yellow sulfur($\ce{S}$)) are a further indicator of this
  behaviour; the leach went from brown ($\ce{Ag6S3O4}$) to white
  ($\ce{Ag2SO4}$) with increased pressure and temperature and some
  residue samples showed evidence of the presence of elemental sulphur.
  This transition from $\ce{Ag2S}$ to $\ce{Ag6S3O4}$ to $\ce{Ag2SO4}$
  shows the progressive oxidation of sulphur in the leach solids as the
  leaching reaction goes to completion.

In conclusion, reaction of silver sulfide and nitric acid will give different products depending on stoichiometry of the reactants, concentration of acid, effect of temperature and pressure etc. As you noted correctly, if the concentration of nitric acid is high enough, it may even produce $\ce{NO2}$ gas.

$$\ce{Ag2S + 10HNO3 → 2AgNO3 + H2SO4 + 8NO2 + 4H2O}$$

References:

Gruener, H. Silver Nitrate Formed by the Action of Nitric Acid on Silver Sulphide. J. Am. Chem. Soc. 1910, 32 (9), 1030–1032. DOI: 10.1021/ja01927a002.
Holloway, P.; Merriam, K.; Etsell, T. Nitric acid leaching of silver sulphide precipitates. Hydrometallurgy 2004, 74 (3-4), 213–220. DOI: 10.1016/j.hydromet.2004.05.003.


How does the addition of acetone cause a shift from hexaaquacobalt(II) to tetrachloridocobaltate(II)?

The great problem is that the ability of a chemical substance to act in some way, is controlled by the activity of the substance and not by the concentration of it. Oftein people consider the concnetration in a first approximation where the activity coefficent of everything is assumed to be one. But for a good prediction or good undertsanding we need the activity coefficents.

for a species X

activity (a) = concentration x activity coefficent (gamma)

We need to understand that as the composition of the solvent changes or the temperture changes the activity coefficents will change.

I think that the answer relates to the relative dielectric constant of the medium, as you replace water with ethanol (or acetone) or increase the temperature then the relative dielectric constant decreases.

By lowering the relative dielectric constant you will make the coordination of the anions to the cations more favourable. I think that this is the reason why a solution of cobalt chloride in ethanol/water is thermochromic. It is pink when cold and blue when hot.

The replacement of the water with a less coordinating solvent will also help to shift the equilibrium in favour of the blue tetrachloro anionic complex.

If you look at the Debye-Huckel equation for the activity coefficient and make sure that you look at what is inside the A term, you will find that when the relative dielectric constant changes that the activity coefficients of ions will change. I think that the change of colour that you see is due to a change in activity coefficient.

The extended Debye-Huckel equation in its full glory is

$$\log (\gamma i) = \left(\frac{-Z_i^2 e^3}{4 \pi \left({\epsilon_0 \epsilon_r k_\mathrm B T}^{1.5}\right)}\right)\frac 1{1 + r_i(8 \pi n e^2/\epsilon_0 \epsilon_r k_\mathrm B T)^{0.5}}(n/2)^{0.5}$$

$Z_i$ is the charge on the ion of interest
$e$ is the electronic charge
$\epsilon_0$ is the vacuum permittivity
$\epsilon_r$ is the relative dielectric constant
$k_\mathrm B$ is Boltzmann’s constant
$T$ is the temperature in kelvin
$n$ is the number of ions in one cubic meter.
$r_i$ is the sum of the radii of the cation and anion which are in contact.

I have done the calculations for the activity coefficient of chloride in mixtures of water and acetone for a 500 μM solution of cobalt(II) chloride at 298 K in mixtures of water and acetone. As the acetone replaces the water the activity coefficient of the chloride goes down, I think that this will be due to the electrostatic interactions of the chloride anions with the cobalt cations (ion pairing):

I repeated the calculation for the activity coefficients of the cobalt ion and the $\ce{CoCl4^2-}$ ion. Bear in mind that the sum of the radii for $\ce{Co^2+}$ and $\ce{CoCl4^2-}$ will be greater than the sum of the radii for $\ce{Co^2+}$ and $\ce{Cl-}$.

Now if we consider the equation

$$K_\mathrm{eq} = a\frac{\ce{CoCl4^2-}}{a(\ce{Co^2+}){a\ce{(Cl-)^4}}}$$

Now if $K_\mathrm{eq}$ is a constant regardless of the solvent system, then as

$$a(\ce{Cl-}) = [\ce{Cl-}] \times \text{activity coefficient of } \ce{Cl-}$$

$$a(\ce{CoCl4^2-}) = [\ce{CoCl4^2-}] \times \text{activity coefficient of } \ce{CoCl4^2-}$$

$$a(\ce{Co^2+}) = [\ce{Co^2+}] \times \text{activity coefficient of } \ce{Co^2+}$$

Then if we have a new equation which is

$$\text{Random greek symbol} = \text{activity coefficient } \ce{CoCl4^2-} / {\text{activity coefficient } \ce{Co^2+} \times (\text{activity coefficient } \ce{Cl-})^4}$$

If we multiply "random greek symbol" with $K_\mathrm{eq}$' we will get a new equilibrium constant for the system.

($K_\mathrm{eq}$' is the $K$ value determined in water where it is an apparent constant - one calculated making the assumption that all activity coefficients are one)

We can get a new equilibrium constant $K$ (solvent composition) which we can plot against the fraction of the solvent mixture which is acetone. Here we have the graph.

Here K' is the true equilibrium constant which would be for a system in water with a zero concentration of cobalt chloride.

We can see as the acetone content increases that the cobalt cations are more likely to react with the chloride anions to form $\ce{CoCl4}$ dianions than they are in the aqueous system.

If you read the papers by people like Bromley[3]The Journal of Chemical Thermodynamics 1972, 4 (5), 669–673. who clearly expresses the ideas of Guggenheim you will find a specific ion interaction term in the activity coefficient equation (Bromley equation). I am sure that the value of the specific ion interaction constant for cobalt and chloride will increase as the water is replaced with acetone.

Sadly I very much doubt if the constants for water/acetone mixtures exist, most of the research effort has gone into aqueous only systems. You could make a lot of the measurements by doing UV/vis spectroscopy on solutions of chloride salts in acetone/water mixtures. But it will be a complex problem, I know that while sodium chloride is a strong electrolyte in water it is a weak one in acetone. I know a classic reaction in organic chemistry for the formation of alkyl iodides is to treat an alkyl bromide or alkyl chloride with sodium iodide in acetone. The larger anion size makes sodium iodide more soluble (and a stronger electroyte) in acetone than sodium chloride/sodium bromide.

If you look at the supporting information for [1], you will see at the end an example of thermochromism. Here mixtures of water, cobalt chloride and the lactic acid / choline chloride eutectic are shown both hot and cold.

This effect has also been reported with nickel chloride in deep eutectic solvents made from choline chloride by some chemists in China. Similar effects were seen for nickel by [2]. They have observed thermochromism which I think is due to a change in the coordination chemistry which in turn is induced by the temperature induced change in relative dielectric constant.

References:

Foreman, M. R. S. J.; Holgersson, S.; Mcphee, C.; Tyumentsev, M. S. Activity coefficients in deep eutectic solvents: implications for the solvent extraction of metals. New J. Chem. 2018, 42 (3), 2006–2012. DOI: 10.1039/c7nj03736h.
Gu, C.; Tu, J. Thermochromic behavior of chloro-nickel(II) in deep eutectic solvents and their application in thermochromic composite films. RSC Adv. 2011, 1 (7), 1220. DOI: 10.1039/C1RA00345C.
Bromley, L. A. Approximate individual ion values of β (or B) in extended Debye-Hückel theory for uni-univalent aqueous solutions at 298.15 K. The Journal of Chemical Thermodynamics 1972, 4 (5), 669–673. DOI: 10.1016/0021-9614(72)90038-9.


Why is there no electron swapping after excitation?

On the first part: it will, or won't, in a subsequent, separate process.

If you excite the system and promote an electron from the lowest orbital to the next orbital, the immediate result is that you have a lowest orbital populated by one electron and a higher orbital populated by two electrons. It is an excited state, and will tend to relax back to the fundamental state if there is an available mechanism to do so - but that will be a second distinct event after the excitation. In fact, the excited state can, depending on the circumstances, last long.

The second part makes no sense. The nature of the interaction between nucleus and electron is not gravitational (it's electromagnetic), and the lowest energy level isn't "below" in any spatial sense. The gravitational analogy is misleading.


Which ion of iron is produced in a reaction between iron and copper(II) sulfate?

When "doing" science, usually you try to (in-)validate a model of the workings of the world. In the case of chemists, this world is mostly confined to the scale of molecules.

That said, let's proceed with

A Theoretical Model

Listed below are some standard reduction potentials I've gathered from the Handbook of Chemistry and Physics, 95th ed. (probably paywalled for you):
$$ \begin{align}
&&E^0_\text{red} / \mathrm{V}\\
\hline
\ce{Fe^{2+} + 2e-  & <=> Fe} & -0.447 \\
\ce{Fe^{3+} + 3e-  & <=> Fe} & -0.037 \\
\ce{Cu^{2+} + 2e-  & <=> Cu} & 0.342 \\
\end{align} $$

Now we can calculate the potential (and the resulting electromotive force EMF) for each of the reactions:
$$ \begin{align}
E^0_\text{(Fe2)} = E^{0}_\text{red}(\text{red}) - E^0_\text{red}(\text{ox}) &= 0.342~\mathrm{V} - (-0.447~\mathrm{V}) = 0.789~\mathrm{V} \\
E^0_\text{(Fe3)} &= 0.342~\mathrm{V} - (-0.037~\mathrm{V}) = 0.379~\mathrm{V}
\end{align} $$

It is clear that the reaction (Fe2) releases more energy and thus should be preferred over the other reaction. This means that we expect a yield of 100% Cu and not 150% Cu.

Now that we have gathered some theoretical understanding of the matter, let's proceed to the interesting part, which is

Confirmation With Experimental Results

You have correctly calculated the theoretical yields for copper to be 0.89 g and 1.33 g, respectively. The corresponding amounts of copper are 0.014 mol and 0.021 mol.

You have measured the amount of copper produced, and it clocks in at 0.78 g. This is already a strong indication that we should side with (Fe2) on this one. But just to really be sure, let's look at the error that we would have if it was (Fe3):
$$ \varepsilon = \frac{\Delta x}{x} = 41.4\% $$
This is a very high error, especially when compared to the error for (Fe2), which is only 12.1%. Assuming that you did not perform the reaction with pure Fe but a mixture of iron with some iron oxides as contaminants, did not use an analytical scale and performed the experiment in fairly uncontrolled conditions (are you sure the reaction was complete when you aborted?) this error is not as bad as it seems.

With the experimental results in our pocket, we can now move on to the final part:

The Conclusion

We have shown, using theoretical predictions and experimental confirmation, that the reaction observed was the oxidation of iron to iron(II) under reduction of copper(II) to copper.

By the way, most of the calculations I did with an iPython Notebook, which you can view here. I might have messed something up, you never know.


Short answer, no. There is not a "twist-chair."

To convince yourself, it probably helps to make a physical model with a chemistry model kit. If you try to twist the chair, you can't do it without significantly moving at least one atom, in which case, the conformation is basically a "skew" or "twist boat." Try it.

The boat and chair conformations are indeed symmetric and achiral. The twist-boat or skew conformation has point group $D_{2}$ and is indeed chiral.

A normal half-chair will have an axis of symmetry and is chiral.

An envelope conformation will have a plane of symmetry and thus be achiral.

I've found that some of the best depictions of six-member-ring conformations come from pyranose sugars. Because of the hydroxyl groups and oxygen atoms, there are 38 distinct conformations (2 chairs, 6 boats, 6 skew-boats, 12 half-chairs, and 12 envelopes).

From Wikipedia:

Obviously, in less-substituted cyclohexane, there are fewer unique shapes, but it helps to see the different ring-puckering or pseudo-rotation shapes.


The first question to always ask when you are trying to build an MO diagram is what point group does my molecule fall into?  In this case, the $\ce{CH2}$ molecule has C2V symmetry (like the water example shown above).  Each of the atomic orbitals (AO's) on carbon (left) and the combination of $1s$ orbitals on the hydrogen atoms (right) has a symmetry label that corresponds to one of the irreducible representations of the point group.

You can determine by symmetry label for the AO's by operating on the orbital with each of the symmetry operators in the C2V group (E, C2, σ, σ').  For example, the 2py orbital is in the plane of the nuclei and so it is symmetric with respect to E and σ' but anti-symmetric with respect to C2 and σ and has b2 symmetry from the C2V character table.

For the hydrogens, you have to consider the combination of both 1s orbitals at the same time.  The a1 combination is when the phase of the wavefunction is the same and the b2 combination is when the phase is opposite (i.e. H1 is positive and H2 is negative or vice versa).

AO's that have the same symmetry can combine to form MO's with that symmetry label.  In the above example, the 2s and 2pz orbitals on oxygen have a1 symmetry, and they combine with the a1 combination of hydrogen 1s orbitals to form three MO's with a1 symmetry.  The 2py (b2 symmetry) orbital combines with the b2 combination of the hydrogen atoms to form two MO's. Finally, the 2px orbital remains unchanged because there are no other SLAC's with b1 symmetry.

The MO's are qualitatively ordered based on the number of nodes that are found when you combine the SLAC's.  The numbers in front of the MO's differentiate the orbitals with the same symmetry.


What is Polymer Environmental Resin (PER)?

PER is essentially PVC which has been plasticized and stabilized with acetyl tributyl citrate, instead of the host of phthalate-based plasticizers such as bis-diethylhexyl phthalate (DEHP) whose endocrine-disrupting properties give PVC its well-deserved bad reputation.

Citrate salts tend to be readily water soluble, and therefore this material is touted to be more biodegradeable than phthalate-plasticized PVC. I certainly cannot find any information to refute that claim. ATBC will further degrade into other acetates and butyrates, most of which are relatively common and harmless in the environment (though butyl compounds are infamously stinky; butyric acid is a favorite less-lethal offensive-odor weapon of law enforcement and groups like the Sea Shepherds). All PVC will eventually degrade as the plasticizer leaches out over time, unlike other types of plastics that are so ridiculously inert they'll likely be the fossils of our modern era, but again, the phthalates just aren't a great idea to have in the environment.

Of course, neither is the vinyl chloride monomer. What many proponents of PER fail to mention is that the monomer is worse for your health than the plasticizers. It's carcinogenic in relatively small doses, for one thing. It's also toxic to the liver in even smaller doses, measured in the low parts per million of body mass for an acute exposure. Lastly, the monomer is extremely flammable, though this is mainly an industrial hazard; the polymer is actually among the more flame-retardant commercial plastics, due to its chlorine content (in other words it's already highly oxidized, as opposed to something like polyethylene). None of this changes when the plasticizer does.

In short, while PER is a step forward, there are still plenty of inherent hazards in the production, long-term use and disposal of PER products. PER is also a relative specialty product, whose primary use is a foam product marketed for children's toys and yoga mats. It's not a silver bullet to replace PVC wholesale.


This depends on the medium in which you do the electrolysis. In the case of a basic medium your equation would be more correct, but in the case of an acidic medium your answer would be wrong, since the concentration of $\ce{OH-}$ ions would be far too small.

Generally: In acidic medium you balance your redox equations using $\ce{H+}$ and $\ce{H2O}$, in basic medium you balance with $\ce{OH-}$ and $\ce{H2O}$.

Note: Since you generate $\ce{H+}$ ions it is possible to argue that starting with a solution at $\text{pH}=7$ gives you an acidic medium, resulting in the equation in the book.


Calcium chloride hexahydrate has separated in jar - is it beyond saving?

Calcium chloride is incredibly hygroscopic (it sucks in water), this property makes it incredibly useful (for instance when used in a desiccator or in a drying tube to prevent water from getting into a reaction).

The wet $\ce{CaCl2}$ can be turned back into a solid  by heating (ideally under vacuum), however you can end up with a mixture of different compounds (for instance heating very wet $\ce{CaCl2}$ has been known to give $\ce{Ca(OH)2}$).  It can also be recrystallised from ethanol to give the di-hydrate, which in turn may be dehydrated by heating, however fresh $\ce{CaCl2}$ is very inexpensive, and so I would suggest buying a new bottle, and ensuring it is sealed well after use to prevent it taking up moisture from the atmosphere.


Do polycarbonate and/or polyurethane have latex?

We have a general rule against "medical questions," but I think yours is different.

First off, latex is a form of poly-isoprene. The "isoprene" part here is the key chemical identity - the letter (so to speak) that repeats in the polymer.

So latex is chemically quite distinct from polyurethane and polycarbonate. Similarly, polyethylene and polypropylene are also distinct.

In short, from a chemical perspective, all of these materials are all different.

Now, that's not to say that:

You may have allergies to multiple polymers / plastics.
Manufacturers may blend multiple chemical components into one object.
Some latex residue may end up on various objects.

For example, many people in healthcare (as you know) use latex gloves, often with latex powder. Small amounts of the latex powder can remain on a wheelchair - causing problems for sensitive individuals like yourself.

If you're looking for a smartphone, you should be able to find one with a metal / glass case (e.g., iPhone, but also others). In that case, the amount of plastic (polymer) should be minimal and you (or a friend) can thoroughly wipe down the device before you use it (e.g, the wrapping plastic might cause sensitivity).


Unexpected result (sodium hypochlorite & steel wool)

After the edits by the original post, the OP clarified that they used a hypochlorite (not an oxygen bleach as stated earlier) and left a piece of steel wool in it. Due to corrosive nature of chloride/hypochlorite, the steel wool crumbled to iron (III) oxides. When you gently heat rust in highly alkaline medium in the presence of hypochlorite ion, it forms a purplish sodium ferrate. This has iron in the (+6) oxidation state.  Over time the purple color disappeared because ferrate slowly decomposes back to lower iron oxidation states in water solution, releasing its excess oxygen.  Ferrates are more stable in alkaline solution than in neutral or acidic ones, so absorption of carbon dioxide over time, making the solution less alkaline, could have promoted this decomposition.

This decomposition of ferrates by acids, carbonic acids, and acidic salts was known as early as 1890s as noted in a review by Moser in Journal für Praktische Chemie 1897, 56, 425 (see relevant comments). Ferrates were known in 1700s.

Also note that hydrogen peroxide and ferrate decompose each other immediately, just like permanganate ion in aqueous solution.

The quartz like crystals are nothing but sodium chloride with perhaps some sodium carbonate.

Image from: https://www.youtube.com/watch?v=5u8LOfTrO6w


Is there a convention as to what order to write reactants and products in a reaction?

There is no convention.

There might be some cases where one choice simply makes more sense to people. For example I have yet to see anybody write

$$\ce{NaCl(s) <=>[H2O] Cl-(aq) + Na+(aq)},$$

but there is no convention dictating that you have to write $\ce{Na+}$ before $\ce{Cl-}$.


The formula only works when the orbital is directed along the bond line between two atoms. Since no orbitals can give you a $60^o$ bond angle, the orbitals are directed outside the triangle, forming a “banana” bond and the formula is being used inappropriately. So the negative sign in your Question #1 is not acceptable. We must do something different.
Let’s look at the H-C-H bond angle. The H-C-H bond angle in cyclopropane is $115^o$ Ref 1. Using that angle, and assuming that the orbitals are directed exactly between the atoms, s = 29.7% for the C-H bond, which is broken to form the carbanion. The acidity of cyclopropane is also indicated by its pKa: 46.
Cyclopropane is interesting because you might think, starting with atomic orbitals, that such a structure would not form, since the smallest angle you could imagine for a carbon atom would be $90^o$; how could you possibly get $60^o$?
Well, that's where your imagination comes into play. Don't go to negative s-bonds (that's too imaginative). Consider how the simple (i.e., positive) addition of individual atomic orbitals could lead to something like the bonding that you do observe:

With simple $sp^3$ orbitals, you can point one pair from each carbon just outside the ring (there will still be considerable overlap: Figure b) and each carbon will be attached to 2 hydrogens above and below the ring with the other two $sp^3$ orbitals. These ring bonds are called banana bonds - not directly in line with the atoms, but bonding nevertheless. Figure a) shows a more complicated bond scheme where the carbon is hybridized $sp^2p$ with the $p$ obital in the plane of the carbons: you get two bonds and one antibond (the whites overlap, the reds overlap, and the red-white antibonds). But then you get three $sp^2$ orbitals poking right into the ring to form a 3-atom bond - again, the bonding is not in line with the atoms. (The hydrogens are bound with $sp^2$ bonds, above and below the ring. You might use the equation to determine the s and p character of these C-H bonds if you know the H-C-H angle (Ref 1).
A very interesting (11 slides) presentation about the bonding is made at Ref 2.
Ref 1: https://cyclopropanelasalle.weebly.com/structure.html
Ref 2: https://www.slideshare.net/DanielMorton26/using-molecular-orbital-theory-to-explain-bonding-in-cyclopropane
But the upshot is that all (most?) equations have limited ranges of usefulness. Whenever some experimenter makes something that goes beyond, a new equation is needed.
Cyclohexane, at the other end of your question, has a pKa of 52 (very non-acidic!), and a probable H-C-H bond angle of $109.5^o$, and s-character in the C-H bond of 25% (note that the molecule has a chair structure and is not planar).
Other pKa’s of interest are:
acetylene, pKa =15, C-H bond is $sp$, 50% s
ethylene, pKa = 44, C-H bond is $sp^2$, 33% s
methane, pKa = 48, C-H bond is $sp^3$, 25% s
(interesting that cyclohexane is similarly hybridized but even less acidic)
Perhaps the best answer to the stability question is that the smaller rings are more acidic, i.e., have more stable anions because the acidic proton has more s-character in the more strained ring.
Question #2 discussed sign in your link. “Sign” (+/-) in the equation for s-character is a measure of amount: how much s-character, in %, is in the bond. But when describing orbitals, we use + and - as indicators of the polarity of the wavefunction, not the amount of s or p character. Plusses combine in such a way that electrons flow from one orbital to another without hindrance, in a bond. A - polarity next to a + requires a node: no electrons there at the node, therefore no bonding. So s-orbitals are designated with one sign (either + or -, depending on whether it is bonding with an adjacent atom; higher levels have internal nodes, but that is no concern for us here). p-Orbitals have a node at the nucleus which is indicated by calling one lobe + and the other lobe -. Bonding molecular orbitals have the same sign of the wavefunction adjacent; in the picture, red-red and blue-blue (in (c)) are bonding; red-blue (in (a) and (b)) is antibonding because the wavefunction splits between the atoms.


Which is the suitable process for ball milling?

Wet ball milling is much more effective due to Rehbinder effect producing a disjoining pressure.  Adsorption layers from surfactants that arise in microcracks significantly reduce the energy expenditure necessary to crush solid materials as a result of a decrease in their free surface energy upon contact with a medium containing substances capable of adsorption. Effect is observed in solids and porous materials, e.g. metals, ionic, covalent and molecular mono- and polycrystalline substances, glasses, polymers, both partially crystallized and amorphous.

This reduces milling time and wear of the working surface, at the same time improving homogeneity of the grinding fraction. The same effect is also used at the lab scale when, for example, metal oxides are dispersed in a mortar much more effectively when acetone is added. Sometimes after such treatment the reaction even takes place in a solid phase.


Have a look at some non-linear reaction schemes such as gas phase radical chain reactions that may lead to explosion. The propagation steps in the hydrogen/bromine gas phase reaction do what you suggest:

$$\begin{align}
\ce{H^. + Br2 &-> HBr + Br^.} \\
\ce{Br^. +H2 &-> HBr + H^.}
\end{align}$$


Asides from a general search on Google Scholar, where, unfortunately, many resources are behind paywalls, here are some book references that may be of use for your broad-ranging request:

Methods in Agricultural Chemical Analysis: A Practical Handbook (Google eBook) (Faithfull, 2002) - some of this is available online through Google Books

Fertilizer Manual (UNIDO, 1998) - this, according to its contents has some information about the chemistry of frtilizers.

Chemistry For Agriculture And Ecology (Mido and Satake, 1994), A lot of this book goes back to basics, but chapter 18 is specifically about the chemistry of fertilizers.

Agricultural Chemicals and the Environment (Hester and Harrison, 1996) - once again, has some information about the chemical processes of fertilizers, with a focus on environmental concerns.

I hope these are useful.


Why does entropy increase when the difference in temperatures is decreased?

Consider two tanks of water. One hot, one cold.

You open a valve between the two. The temperature between the two bleeds, and the hot tank loses entropy, and the cold tank gains entropy. However, entropy is also gained just by this process happening, as it would be impossible to force the 'cold' tank to retransfer the heat back to the 'hot' tank.

Also, you must consider something else. The two conjoined systems now have more molecules than either system alone, increasing the 'randomness' as a whole. Considering dice, it's like having two sets of two, then one set of four. The set of four has more possibilities than just the set of two.


Silver "polish" with aluminum foil left a hard yellow coat

The source of the "hard yellowish coating" may be Silver carbonate (Ag2CO3), as per Wikipedia:

Silver carbonate is yellow but typical samples are grayish due to the presence of elemental silver.

I suspect the teapot in question is "electroplated Britannia metal" per Wikipedia with a nickel alloy as the base metal, as to quote:

After the development of electroplating with silver in 1846, Britannia metal was widely used as the base metal for silver-plated household goods and cutlery.[5] The abbreviation EPBM on such items denotes "electroplated Britannia metal". Britannia metal was generally used as a cheaper alternative to electroplated nickel silver (EPNS) which is more durable.

The Baking Soda will interact with any soluble Ag ions to form the unstable Silver bicarbonate which breaks down to Ag2CO3 on warming.
The question is what is the source of Ag+?
My speculation is that the Al foil was completely coated given the surface area of the teapot, and the nickel-based alloy is notoriously cathodic (hence, the use of Nickel alloys in coins). So now, the Silver itself can be subject to an anodic attack in the presence of H+ (from amphoteric NaHCO3) and O2 (from air exposure), resulting in Silver ions.
This was a somewhat unique situation, hence the successful use of the Al/NaHCO3 process on the Silver-plated flatware.


It depends how you define the surface of an atom. Atoms maintain no surface in the normal sense; only regions of space where you have a better chance of finding electrons. So in fact it is not correct to say they have a true shape at all.
Shapes of Atomic Orbitals
However if you plot the region of higher probability of finding electrons in an atom you can obtain something like this:

These are the shapes of the first five atomic orbitals: $1\mathrm{s}$, $2\mathrm{s}$, $2\mathrm{p}_x$, $2\mathrm{p}_y$, and $2\mathrm{p}_z$ from Wikipedia. $1\mathrm{s}$ orbital is sphere-shaped but other orbitals have more complex shapes so atoms with many electrons have orbitals very different from a sphere.
Shapes of Molecular Orbitals
Molecules have more electrons and so even more orbitals. They can have very strange "shapes". I've calculated for you with GAMESS and Avogadro the LUMO 2$b_2$ water's orbital that is one of the molecular orbitals of water. This is the result:

Shapes of Atomic Constituents
For answering your question in the comment: In fact even protons, neutrons and electrons don't have a real shape due to wave-particle duality. However we can assume in many cases that neutrons are particles (so we suppose a spherical symmetry) but the  de Broglie hypothesis states that they have also a wavelength!


Is there an otherwise liquid material that has angle of repose?

Such materials are called Bingham plastics.  An angle of repose means that the material must withstand a non-zero shear stress (from its own weight) while still exhibiting zero shear rate.

Examples include toothpaste, ketchup, mayonnaise, latex paint, and concentrated suspensions of coal particles in a otherwise newtonian liquid.  I'm not sure if any of these materials meets your requirement that it consist "of individual molecules".  I don't think a single-component, single-phase, non-polymeric substance will ever exhibit Bingham plastic behavior, although (I might be wrong and) I can't cite any theoretical reason why this is so.


Is this really a comparison of the absolute energies in the reaction energy profile for ester formation?

These authors are comparing relative energies. Look at what they set as their reference... the products. They could have easily made "starting materials" as their 0.0 reference point and stated that the products of the acyl chloride reaction is lower in energy relative to the anhydride reaction mechanism.


I don’t consider your working to be arbitrary at all. Let’s first consider the β decay including the associated $\ce{OH}$ part of $\ce{HTO}$:

$$\ce{_1^3THO -> _2^3He^2+ + OH- + e-^ + \bar{\nu_e}}\tag{1}$$

As you suggested, the electron created in equation (1) won’t stick around, it is emitted into the environment, leaving us with a very reactive $\ce{He^2+}$ ion. Obviously, that is a strong oxidant as it wants to get reduced itself. The closest electron sources are $\ce{OH-}$ and $\ce{HTO}$, so we can consider a number of other reactions. However, it makes sense to organise it in a simple redox reaction. The reduced component is obviously going to be $\ce{He^2+}$ which only leaves $\ce{HTO}$ being the oxidised component — and since both hydrogen and tritium are in the $\mathrm{+I}$ oxidation state in $\ce{HTO}$, it must be oxygen which is oxidised to either peroxide or elemental oxygen. Indeed, peroxides are not stable and typically disproportionate into oxygen and water anyway, so we might as well just consider the formation of oxygen. Here goes:

$$\textbf{Reduction:}\\
\begin{gather}\ce{He^2+ + 2 e- -> He^{\pm 0}}\tag{2}\end{gather}$$
$$\textbf{Oxidation:}\\
\begin{align}\ce{2HTO^{-II} &-> O2^{\pm0} + 4 e-}\tag{3.1}\\
\ce{2 HTO & -> O2 + 4 e- + 2 H+ + 2 T+}\tag{3.2}\end{align}$$
$$\textbf{Redox:}\\
\begin{gather}\ce{2 He^2+ + 2 HTO -> 2 He + O2 + 2 H+ + 2 T+}\tag{4}\end{gather}$$

Equation (3) and (4) would be similar if I had chosen to oxidise only to $\ce{H2O2}$ and then would have been followed by an equation (5) to show the disproportionation of hydrogen peroxide:

$$\ce{2 H2O2 -> O2 + 2 H2O}\tag{5}$$

In the end, you will arrive more or less at the same substances.

Note that to end up with hydrogen gas, you would have needed a reductive environment capable of reducing $\ce{H^{+I}}$ to $\ce{H2^{\pm0}}$ — not the case if you have a severly electron-desiring ion like $\ce{He^2+}$ around.


Why is the reaction between potassium permanganate and hydrogen peroxide spontaneous?

For both half-reactions, the actual potentials depend on $\mathrm{pH}$; however, the values given in the question apply to different $\mathrm{pH}$.

The given reaction of hydrogen peroxide and the corresponding potential apply to $\mathrm{pH=0}$ or $\left[\ce{H+}\right]=1$:

$$\ce{O2 + 2 H+ + 2e- <=> H2O2}\qquad E^\circ=0.695\ \mathrm V$$

Whereas the given reaction of permanganate and the corresponding potential apply to $\mathrm{pH=14}$ or $\left[\ce{OH-}\right]=1$:

$$\ce{MnO4- + 2 H2O + 3e- <=> MnO2 + 4 OH-}\qquad E^\circ=0.595\ \mathrm V$$

The corresponding reaction and potential for $\mathrm{pH=0}$ or $\left[\ce{H+}\right]=1$ are:

$$\ce{MnO4- + 4 H+ + 3e- <=> MnO2 + 2H2O}\qquad E^\circ=1.695\ \mathrm V$$


How to buy 1g of enzyme that is sold in units/mg?

Sigma Aldrich defines their units as:

  One pyrogallol unit will form 1.0 mg purpurogallin from pyrogallol in 20 sec at pH 6.0 at 20 °C.

Assuming you're doing a biological imaging experiment, these activity values tell you much more about what you're getting than mass.

From a casual internet search it appears that HRP is pretty expensive. I would consider whether you actually need so much of it for your research.


Harnessing phase separation to easily make Li film

You are thinking in the wrong direction. Meniscus is not about pressure at all.

You seem to be imagining external hydrostatic pressure as a huge invisible flat lid that we put on top of your molten lithium and kinda force it to stay flat. This is very much not so. The pressure is the same all around, thanks to the Pascal's law. It does not affect the shape. It presses on the curved sides of the meniscus with the same force as it does on top. The very notion of "flat" is simply not there.

If anything, it is gravity force that reduces the meniscus. In effect, adding a liquid on top will partially negate the gravity and thus make things worse. Keep in mind that such liquid, if it is found (I still can't think of any, and I'm not a novice in chemistry, mind you) can't be much less dense than lithium, which almost brings us to the conditions of Plateau's oil drop experiment, with lithium playing the role of oil. Not quite the outcome you wanted, is it?

It is the surface tension that matters, and adding liquid might change that in an unpredictable way. Then again, you can achieve the same result without any extra liquid. Just use different material for the container. I guess using metal instead of glass would make the meniscus bend in the opposite direction. With some tweaking, it might be possible to find the material with wetting angle of $90^\circ$ that would make lithium surface perfectly flat, but I don't think it is worth the effort.

So it goes.


is the adsorption of resveratrol within wine or grapes linked with the time it was kept in your mouth?

This article on the bioavailability or resveratrol seems to partly support your assumption: Keeping it in the mouth for a bit longer rather than swallowing directly might help to facilitate the uptake.

However, the solubility of resveratrol in water is apparently as low as 30 mg/L.


Is zero-point vibrational energy an intensive or extensive property?

Mathematically, an extensive property $f$ is one for which $f(\lambda x,\lambda y,z) = \lambda f(x,y,z)$ for extensive variables $x$, $y$, and intensive variable $z$. For example, the Helmholtz free energy $A(T,\lambda V,\lambda N) = \lambda A(T,V,N)$ is extensive.

Following this definition, the zero-point energy $$E_\text{ZPVE}(\{\omega_c\},\lambda n) = \lambda E_\text{ZPVE}(\{\omega_c\},n),$$ where $n$ is the number of normal modes, is extensive in the thermodynamic limit $N \to \infty$.

The polarizability should be able to be treated similarly.

It bears noting that a lot of macroscopic extensive properties are not extensive in the microscopic regime. For example, if we consider two identical systems, each with an energy $U$, then the combined system has energy $2U$ plus an additional energetic term $U_\text{sys-sys}$, accounting for superficial energetic interactions between the two subsystems. The surface, and hence the superficial energy term, scales as one dimension removed of the volume, and likewise vanishes in the thermodynamic limit.

Furthermore, it's perfectly valid to have properties that are neither extensive nor intensive, like the number of microstates that a system possesses.


What is the unit of the Rydberg Constant?

To convert between joules (units of energy) and inverse centimetres (conventional units of wavenumber) simply requires multiplication by a factor of $hc$. This is because

$$E = h\nu = \frac{hc}{\lambda} = hc\bar\nu$$

Of course, they are not literally equal, but they are physically equivalent.


Order of positive mesomeric effect of halides

$\ce{2p-2p}$ orbitals overlaps better than $\ce{2p-3p}$, $\ce{2p-4p}$, $\ce{2p-5p}$. Although $\ce{F}$ is more electronegative, but it is also smaller in size. Both $\ce{F}$ and $\ce{C}$ (next to it) have $\ce{2p}$ orbitals to overlap.


Which of the following is a secondary alkyl halide

Well, no. Here is the quote from Wikipedia:

  From the structural perspective, haloalkanes can be classified
  according to the connectivity of the carbon atom to which the halogen
  is attached. In primary (1°) haloalkanes, the carbon that carries the
  halogen atom is only attached to one other alkyl group. An example is
  chloroethane ($\ce{CH3CH2Cl}$). In secondary (2°) haloalkanes, the carbon
  that carries the halogen atom has two $\ce{C–C}$ bonds. In tertiary (3°)
  haloalkanes, the carbon that carries the halogen atom has three $\ce{C–C}$
  bonds.

So, the presence of $\ce{CH2Cl}$, $\ce{CHCl}$, and $\ce{CCl}$ are the indicators of primary, secondary, and tertiary haloalkane respectively.

Here is b with the $\ce{C}$ atom to which $\ce{Cl}$ is attached as well as its two $\ce{C–C}$ bonds highlighted in red.


To qualify as «crystalline», there must be a minimal unique pattern (unit cell) to construct the solid only by applying regular, periodic translations. You may predict the outcome of the translation both in proximity, as well as in far distance reliably.  In an amorphous state, there is no such predictability.

For two, anisotropic means that a vectorial property depends on the orientation within the sample.  In the case of an amorphous sample (as contrasting to crystalline) like glass there is no orientational difference.  As an example, the speed of light in glass is independent on the direction you look at the sample; thus, glass is an isotropic material.

In a crystal, however, the speed of light depends on the relative orientation of the sample, thus crystals generally are anisotropic.  Depending on the symmetry of the underlying Bravais lattice, there may be two, or three special directions of observation along those vectorial properties are equal to each other (e.g., uni- and biaxial crystals in birefringence).  Then, these directions of observation are isotropic to each other; allong all other directions of observation, however, the crystal's vectorial properties are anisotropic.


Why is the cathode of a lithium-ion battery metallic, which is the opposite of most other batteries?

This is not entirely true.  Wikipedia lists several possible cathode materials and they are oxides.  They may be "metallic" is the sense of being metallic conductors (some oxides are so), but they are ceramic compounds.
In secondary batteries the anode is typically Li-intercalated carbon (which is also a good conductor).  Lithium intercalates readily into carbon, allowing a safer and less side-reaction prone choice than pure lithium, whereas most other metals do not.


Does broader peak mean stronger absorber in UV/VIS?

The classical concept of  oscillator strength $f$ is useful here, but it should be used only in a qualitative way. This is defined as $f=a\int\epsilon_\nu d\nu$ where $a$ is set of units with value $4.3 10^{-9}$ if the extinction coefficient is in units  dm$^3$/mol/cm and $\nu$ is in wavenumbers. The maximum value of $f$ is unity and is close to this for an allowed transition in a dye molecule, such as rhodamine, but can be as small as $10^{-9}$ for a forbidden transition.

In your case if two molecules have similar $\epsilon$ at some wavelength (or frequency) but one has a narrow transition compared to the other, then the one with the narrow transition will have a smaller integral and hence the smaller oscillator strength. In this sense only is the transition 'weaker', i.e. it is not the probability of  absorbing at a given wavelength that is being considered but that over all wavelengths, so it really depends on how you want to define a 'strong' vs a 'weak' transition.  In QM the transition moment integral is used instead of oscillator strength and gives a proper description of absorption.


Why is the compound 3,4-dihydroxy-3-cyclobutene-1,2-dione more acidic than acetic acid?

This molecule is commonly known as Squaric acid. The Wikipedia entry gives the possible resonance structures of the conjugate base (added here for convenience):

You can see that there are four possible resonance structures for the conjugate base, resulting from the proximity of two ketone groups and two hydroxyl groups around the ring. This proximity allows movement of electrons around the square ring and between the ring and each oxygen atom.

On the other hand, acetic acid (or rather, the conjugate acetate anion) has only two resonance structures, as seen here:

Since the principle is that a molecule with more resonance structures is more stable than a molecule with less resonance structures, squaric acid anion is more stable than acetate anion. Since a more stable conjugate base makes an acid stronger, squaric acid is considered to be stronger than acetic acid.

This conclusion is also evident by the Pka values. This table provides Pka values for many acids, including squaric, acetic and formic acid. You can see that squaric acid has two values for two acidic protons - 1.5 for the first proton and 3.5 for the second proton. acetic acid has a Pka value of 4.75 and formic acid has a Pka value of 3.75.
This means that the strongest acid of the three is squaric acid, followed by formic acid, and the least acidic of them is acetic acid.

Squaric acid > Formic acid > Acetic acid


Converting a CIF file to a .com or .gjf file

In general, OpenBabel may convert some of the information of a .cif into a Gaussian .gjf.  If your are fine with using the CLI, the command would follow the syntax pattern of
obabel -icif input.cif -ogjf -O output.gjf
The parameters -icif and -ogjf hereby instruct OpenBabel about the expected input / output format of the files, while -O actually triggers the program to write the output into a permanent record.  Further information may be obtained here.
Depending on the computation intended, the crosses Mercury displays may be very valuable information.  It is Mercury's way to tell the user that the structure is disordered, i.e. there is more than one orientation of a molecule in the unit cell.  For example, that a conformational flexible $\ce{-CH3}$ or $\ce{-CF3}$ group has been refined into more than one position (think staggered / eclipsed conformation of butane, there are more than one energetic minima, and one is the absolute one.)  But this orientational disorder may extend over larger portions and complete molecules, too (e.g., small solvent molecules).  It would be better if you could display more of the structure, and tell a bit in addition about the intended computation.
Do you know to read a .cif file about site occupancy factors?  E.g., for molecules with an up / down or ($\uparrow$ / $\downarrow$) orientational disorder and only two realizations, these often are refined to add up to $p(\uparrow{}) + p(\downarrow) = 1.00$.


Is blending 15% methanol with petrol a good idea?

I think it's important to first understand why methanol is added to petrol to begin with, outside of what I will discuss later on as some of the benefits to atmospheric emissions.

For internal combustion engines, petrol should be of a high "octane rating". This is a measure of the compression ratio of the fuel, and it needs to be high in order for the engine to work efficiently. If the petrol has too low of an octane rating, then something called knocking can occur, which over time can cause very bad engine damage.

In order to improve the octane rating of petrol then, a variety of compounds can be added that boost the overall octane rating (appropriately named octane boosters). In the past this has included tetraethyl lead, which when burned during combustion lead to the accumulation of lead in the atmosphere, which was enormously damaging to the environment and to human health. Methanol, and other alcohols like ethanol, are one such example of an octane booster.

Before I discuss the effects on engine performance, I will say that the use of using methanol blended fuel does have positive environmental impacts over the use of diesel or just straight petrol. Combustion efficiency is improved, thus leading to lower emissions of $\ce{NO}$ and $\ce{NO2}$ (collectively called $\ce{NOx}$ gases), which has positive implications for the atmosphere overall (lower $\ce{NOx}$ generally leads to lower ground level ozone, but this is dependent on VOC emissions of course!), as well as to lower particulate emissions. They also do not contain sulfur, and thus blended mixes have a lower sulfur load - that means less sulfur dioxide emissions.

In terms of redesigning engines on the basic, fundamental level - no change is required. Any internal combustion engine can use blended fuels, it just has implications of the lifetime of the engine. However, methanol can act as a corrosive on certain components of the engine. As blended fuels are increasingly seen as a good way for governments to reduce their $\ce{NOx}$ emissions - particularly after the problem with diesel emissions over the last decade - it seems likely that manufacturers will respond to this by redesigning engines to resist the corrosive effect of methanol within the blend. For now, though, having a small ratio of methanol to petrol - and it seems in this case, India has designed for 15% methanol blend, which should have little impact on engine damage through corrosion - is sufficient.

One last thing to note, however, that may not have come up. Some people might wonder why we don't just a 100% methanol fuel once we design engines that don't corrode due to it. The simple answer is - it's not as fuel efficient as petrol!


Compare the work done during dissolution of carbides

The relevant equations are:

$\ce{Mg2C3(s) + 4 H2O (aq) -> C3H4 (g) + 2Mg(OH)2 (aq)}$

$\ce{Al4C3 (s) + 12H2O(aq) -> 3CH4(g) + 4Al(OH)3(aq)  }$

$\ce{CaC2(s) + 2H2O(aq) -> C2H2(g) + Ca(OH)2(aq)}$

As you can see, most number of gaseous moles are released in the 2nd reaction while equal number of gaseous moles are released in the first and third reaction, therefore maximum work is done in the first reaction under same conditions of temperature and pressure.

Note that:

In open vessel reaction, gases expand against a constant external atmospheric pressure, thereby doing work on the surroundings , thus the work done $= P_{\text{atm}}\Delta V$ is independent of the gas's pressure and only dependent on $\Delta V$

Also, recall that Avogadro's law states:

  For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.

Hence, here,  $\Delta V$ only depends on the moles of the gas.


Why do we still need to know about the Rankine temperature scale?

Rankine is commonly used in the aerospace industry in the United States.

Rankine is to Fahrenheit what Kelvin is for Celsius. So when people in the United States were creating programs and using equations that needed an absolute temperature, they used Rankine before Celsius became dominate for scientific calculations. The reason people still sometimes use it in the aerospace industry is that there are a lot of programs that were developed using Rankine, so in order to be compatible with those old programs, it's often simpler to just use Rankine in the new programs too.

Thus, as an intern at NASA, I was writing simulations that used Rankine in 2009.


From this source broadenings occur due to

  Doppler effect
  This effect arises because atoms will have different components of velocity along the line of observation.

  Lorentz effect
  This effect occurs as a result of the concentration of foreign atoms present in the environment of the emitting or absorbing atoms. The magnitude of the broadening varies with the pressure of the foreign gases and their physical properties.
  Quenching effect
  In a low-pressure spectral source, quenching collision can occur in flames as the result of the presence of foreign gas molecules with vibration levels very close to the excited state of the resonance line.
  Self absorption or self-reversal effect
  The atoms of the same kind as that emitting radiation will absorb maximum radiation at the center of the line than at the wings, resulting in the change of shape of the line as well as its intensity. This effect becomes serious if the vapor, which is absorbing radiation is considerably cooler than that which is emitting radiation.

As Greg pointed out in the comment, You can find more detailed explanation on the topic here

Pressure broadening should be dependent on the concentration of atoms. The other type of broadening shouldn't be dependent on the concentration


Extraction of glucose and fructose from golden syrup

Extraction in terms of selectively extracting either glucose or fructose from the syrup is going to be difficult to impossible, at least as far as achieving any reasonable degree of purity is concerned.  The glucose-fructose separation method preferred by Dow Chemical is called ligand exchange chromatography.  According to this publication:

  Unlike the macroporous structure of DOWEX™ resins used in deashing,
  DOWEX MONOSPHERE™ 99 separation resins are gel beads which have a
  smooth, uniform surface.  While macroporous resins are opaque, gel
  resins are translucent.  Chromatographic separation resins are
  functionalized and the beads contain a significant amount of water.
  While going through the column sections, the sugars to be separated
  dissolve in the water contained within the beads.  Inside the  bead,
  the dissolved sugars interact with the calcium ions held by the resin.
  Fructose, glucose and water form weak ligand complexes with the
  calcium ion.  A stronger interaction in the fructose/calcium ion
  complex than in the glucose/calcium ion complex is the basis of the
  mechanism of separation of fructose from glucose.

As this is a readily available commercial product designed for use on massive scales, constructing a simple separation column based on these separation resins as described in the publication above should be a very doable, cost-effective approach to your problem.


Carbon/silicon pentacovalent intermediate during substitution

Bonds involving carbon are typically shorter than bonds involving silicon.  Therefore a pentacoordinated carbon species would be sterically more crowded around the central carbon atom than its pentacoordinated silicon analogue would be. This crowding would raise the energy of pentacoordinate carbon compared to pentacoordinate silicon; which is consistent with the observation that the SN2 reaction with carbon produces a higher energy pentacoordinate transition state, while in the case of silicon a more stable pentacoordinate intermediate can be observed.
\begin{array}{lcc}
\hline
\ce{Bond} & d(\ce{X=C})/\pu{pm} & d(\ce{X=Si})/\pu{pm} \\
\hline
\ce{X-O} & 143 & 163 \\
\ce{X-Cl} & 177 & 202 \\
\ce{X-Br} & 194 & 215 \\
\ce{X-I} & 214 & 243 \\
\hline
\end{array}
Recasting this into MO terms, we would say that both the carbon and silicon pentacoordinate species involve a trigonal bipyramid structure where the two ligands (the "attacking" and "leaving" groups in the carbon case) occupy axial positions in the trigonal bipyramid arrangement.

(source)
The figure above makes it easy to see that we have a hypercoordinated species involving a 3-center 4-electron bond. Since a silicon p-orbital is larger than a carbon p-orbital, the 4 electrons are spread out over a larger volume in the silicon case, reducing electron-electron repulsion.  This makes the pentacoordinate silicon case more stable than pentacoordinate carbon.


How does temperature affect the concentration of flavonoids in orange juice?

Orange juice is often subjected to temperature variation during processing, pasteurization, storage and refrigeration at 4$^{\circ}$C. According to some research, there appears to be a dramatic decrease in the concentration of certain flavonoids at storage temperatures especially above 20$^{\circ}$C.

A variety of flavonoids found in orange juice includes didymin, eriocitrin, hespertin, narirutin, poncirin, apigenin and neoeriocitrin.

Effective analytical methods used to identify and quantify individual flavonoids include gas chromatography mass spectroscopy (GCMS) and high-performance liquid chromatography (HPLC).

Whilst chromatography is the most effective method of identifying specific flavonoids, you may be interested to know that ultra-violet spectrophotometry has been used to measure total flavonoid content before & after treatment.


How to calculate molar mass of (6)UO2+2D and (6)UO2OH+D?

These notations signify someone performed calculations in Visual MINTEQ accounting for dissolved organic matter (DOM) using a NICA-Donnan modeling method.
From the manual nicadonnan.pdf bundled with the installer:

In Visual MINTEQ, a “D” suffix is used to identify a diffuse-layer species such as a Donnan species.
[…]
The numbers 6, 7, 8 and 9 have no conceptual significance; they are only used by Visual MINTEQ to distinguish different humic components, and they reflect the order in which the components appear on the NICA-Donnan menu (with 6 as the starting number).

Considering the default numbering layout

and that FA1 and FA2 refer to fulvic acids containing carboxylic (FA1) and phenolic groups (FA2), OP's species can be interpreted as such:
(6)UO2+2D(aq): Weakly (electrostatically) bound uranyl $\ce{UO2^2+}$ to fulvic acid in the aqueous phase.
(6)UO2OH+D(aq): Weakly (electrostatically) bound uranyl monohydroxide $\ce{UO2(OH)+}$ to fulvic acid in the aqueous phase.
FA1-UO2(6)(aq): Organically complexed uranyl $\ce{UO2^2+}$ to carboxylic fulvate in the aqueous phase.
Several examples from the manual:

FA1-Zn(aq), FA2-Zn(aq): Organically complexed Zn to dissolved fulvic acid. Sites 1 and 2  refer to carboxylic and phenolic functional groups, respectively.
FA1-Zn(s), FA2-Zn(s): Organically complexed Zn to fulvic acid in the solid phase.
HA1-Zn(s), HA2-Zn(s): Organically complexed Zn to humic acid in the solid phase.
(8)Zn+2D: Weakly (electrostatically) bound Zn to dissolved fulvic acid.
Zn+2D(s)(6): Weakly (electrostatically) bound Zn to fulvic acid in the solid phase.
Zn+2D(s)(7): Weakly (electrostatically) bound Zn to humic acid in the solid phase.


I think that the main purpose (/ historical advantage) of steam distillation was to avoid the exposure of sensitive compounds to sustained heat by either lowering the required distillation temperature, or the distillation duration (at equal temperature), or both, by using steam as a carrier.

The solution that contains the compound to be distillated doesn't necessarily initially contain water itself, though even if it does, there is a significant advantage in that the duration of exposure to high temperatures is as short as the transport time until it is condensed (since the water is boiled separately, and not together with the compound). In traditional distillation the compound has to be exposed to heat for the time necessary to reach the solvant's  evaporation point AND the transport time, which is significantly longer than the transport time alone.


Is it possible to calculate chromacity coordinates using absorbance values?

This is a partial answer to your question.  I have used chromaticity coordinates for various scientific endeavours (air quality mainly).  I found that it is a very powerful means of determining colour differences.

A bit of searching and there is somewhat of a precedent for using chromaticity for blood plasma.  An example is the article Features of Color Characteristics of Blood Plasma of Patients with Liver Cirrhosis in the Colorimetric System XYZ as Compared with Healthy People (Dolomatov et al. 2012) - even though it may be a different context to your study, but does show that it can and has been done before.  Specifically, their method was

  by colorimetric method, based on absorption electromagnetic radiation in visible area.

There are several more examples of how it can be used.


SMILES vs. Graph representation in deep learning

I agree that there seems to be a shift towards using graph representations over SMILES strings. I personally think this is a good thing, and I'll try to explain why, but even if there's nothing inherently better about graph representations of molecules, there is a very practical reason people are moving towards graph representations.

So, first of all, as you point out, both representations suffer from an issue in that they are necessarily a reduction from the complete amount of information needed to specify a molecular structure. For instance, neither a graph nor a SMILES string could distinguish between two isomers which differ only in the direction in which a free $\ce{O-H}$ group points. In fact, a bare graph cannot tell the difference between the boat, chair, or planar hexane molecules as they all have identical connectivity. Nonetheless, a graph is a much more flexible representation than a SMILES string because it is quite common to add weights to edges, which could be bond distances, or add parameters to nodes, which could describe angles to other nodes. So, you can include some information which is in the Cartesian coordinates, without having the problems associated with deriving your molecular representation from Cartesian coordinates.

This hints at the very practical reason why people are using graph representations. Basically, graphs are used in machine learning models in many fields other than chemistry. So, it's very nice to not have to reinvent the wheel, especially when people have so much success with things like transfer learning, where you just take a pre-trained model and re-train it for your own purposes.

Also, very often the first step in training a neural network is making some transformation to your data. For instance, graph convolutional neural networks have been successful in many tasks, so why not just use a convolutional filter on your graph representation of a molecule? You could do this with a SMILES string, but you would probably first just transform the string into something resembling a graph.

As to your specific points about chirality and aromaticity, etc. All of this information can be attached to a graph via parameters belonging to each node, although I would personally avoid giving information that isn't strictly necessary. That is, there is nothing special about a bond which is in an aromatic ring. You need to provide enough data that the model can learn about this on its own. If you tell it that bonds in aromatic rings are special enough to get another parameter, this is probably going to bias the model in some unforeseen way. Chirality is easily handled by a simple parameter attached to each node.

Ultimately, though, the representation depends a lot on the problem you are trying to solve. For instance, if you are trying to learn a representation of the potential energy surface, then graphs can work quite well. What is more common perhaps, is to use so-called atom-centered symmetry functions. In this case, the actual features are abstract vectors which are guaranteed to have the relevant symmetries and smoothness needed in the potential energy surface.

If you're doing something more like a classification problem, then using a representation like a SMILES string might be perfectly suitable.

TL;DR

Graphs are a more flexible representation which are commonly used in fields outside of chemistry. Hence, being able to draw on the knowledge of other disciplines is a huge plus, especially when you're collaborating with computer scientists who know a lot about machine learning and nothing about chemistry.


How do I figure out the hybridization of a particular atom in a molecule?

If you can assign the total electron geometry (geometry of all electron domains, not just bonding domains) on the central atom using VSEPR, then you can always automatically assign hybridization. Hybridization was invented to make quantum mechanical bonding theories work better with known empirical geometries. If you know one, then you always know the other.

Linear - $\ce{sp}$ - the hybridization of one $\ce{s}$ and one $\ce{p}$ orbital produce two hybrid orbitals oriented $180^\circ$ apart.
Trigonal planar - $\ce{sp^2}$ - the hybridization of one $\ce{s}$ and two $\ce{p}$ orbitals produce three hybrid orbitals oriented $120^\circ$ from each other all in the same plane.
Tetrahedral - $\ce{sp^3}$ - the hybridization of one $\ce{s}$ and three $\ce{p}$ orbitals produce four hybrid orbitals oriented toward the points of a regular tetrahedron, $109.5^\circ$ apart.
Trigonal bipyramidal - $\ce{dsp^3}$ or $\ce{sp^3d}$ - the hybridization of one $\ce{s}$, three $\ce{p}$, and one $\ce{d}$ orbitals produce five hybrid orbitals oriented in this weird shape: three equatorial hybrid orbitals oriented $120^\circ$ from each other all in the same plane and two axial orbitals oriented $180^\circ$ apart, orthogonal to the equatorial orbitals.
Octahedral - $\ce{d^2sp^3}$ or $\ce{sp^3d^2}$ - the hybridization of one $\ce{s}$, three $\ce{p}$, and two $\ce{d}$ orbitals produce six hybrid orbitals oriented toward the points of a regular octahedron $90^\circ$ apart.

I assume you haven't learned any of the geometries above steric number 6 (since they are rare), but they each correspond to a specific hybridization also.

  $\ce{NH3}$

For $\ce{NH3}$, which category does it fit in above? Remember to count the lone pair as an electron domain for determining total electron geometry. Since the sample question says $\ce{NH3}$ is $\ce{sp^3}$, then $\ce{NH3}$ must be tetrahedral. Make sure you can figure out how $\ce{NH3}$ has tetrahedral electron geometry.

  For $\ce{H2CO}$

Start by drawing the Lewis structure. The least electronegative atom that is not a hydrogen goes in the center (unless you have been given structural arrangement).
Determine the number of electron domains on the central atom.
Determine the electron geometry using VSEPR. Correlate the geometry with the hybridization.
Practice until you can do this quickly.


That reaction will definitely not yield a phenol.
Phenols are not usually prepared directly from benzene, but from benzene derivatives, such as toluene or chlorobenzene:

Oxidation of toluene, as developed by Dow Chemical:

$\ce{C6H5CH3 + 2 O2 → C6H5OH + CO2 + H2O}$

Hydrolysis of chlorobenzene, using base or steam (Raschig–Hooker process):

$\ce{C6H5Cl + H2O → C6H5OH + HCl}$

Actually, there are processes that allow us to obtain it directly from benzene, but these are rarely used:

Direct oxidation of benzene with nitrous oxide, a potentially "green" process:

$\ce{C6H6 + N2O → C6H5OH + N2}$

The most widely used process for industrial production (95%) involves the partial oxidation of cumene (isopropylbenzene) via the Hock rearrangement:

$\ce{C6H5CH(CH3)2 + O2 → C6H5OH + (CH3)2CO}$

Source of the reactions: Wikipedia English page for phenol


Can ionic compounds dissolved in organic solvent conduct electricity?

Sure they can.  Even if most of the ions are in ion pairs or other complexes, they can still move by being transferred from one complex to another.  All you need is to get the ionic compound to dissolve, which in most cases requires a polar solvent with which the ions will not react.
A rather unexpected case involves magnesocene, whose bonding is a mixture of ionic and covalent.  When dissolved in ether or THF, magnesocene forms enough ions not only to impart significant conductivity in solution, but also to be proposed for use in batteries:

Unlike ferrocene, magnesocene displays slight dissociation and subsequent ion association in polar, electron-donating solvents (such as ether and THF).[1]
$\displaystyle {\ce {MgCp2 <=> MgCp+ + Cp-}}$
$\displaystyle {\ce {MgCp2 + MgCp+ <=> Mg2Cp3+}}$
$\displaystyle {\ce {MgCp2 + Cp- <=> MgCp3-}}$

Note the single charge on all ionic species.  This is to be expected in organic solvents with their weaker polarity and ion solvation power relative to water.
Cited reference:
1.
Schwarz, Rainer; Pejic, Marijana; Fischer, Philipp; Marinaro, Mario; Jörissen, Ludwig; Wachtler, Mario, "Magnesocene-Based Electrolytes: A New Class of Electrolytes for Magnesium Batteries," Angewandte Chemie International Edition 2016, 55(48), 14958–14962.  https://onlinelibrary.wiley.com/doi/abs/10.1002/anie.201606448


Why is nitric acid such a strong oxidizing agent?

Because, unlike in other metal dissolution reactions, the $\ce{H+}$ of $\ce{HNO3}$ isn't reduced—the $\ce{NO3-}$ is. The following data and balanced reactions are taken from Wikipedia:

\begin{align}
  \ce{NO3- + 4 H+ + 3 e- &-> NO  + 2 H2O} &
    E^\circ_\mathrm{red} &= \pu{0.96 V} \\
  \ce{NO3- + 2 H+ +   e- &-> NO2 +   H2O} &
    E^\circ_\mathrm{red} &= \pu{0.79 V} \\
  \ce{Ag+(aq) + e-       &-> Ag(s)} &
    E^\circ_\mathrm{red} &= \pu{0.799 V}
\end{align}

Since the standard reduction potential (SRP) of the $\ce{NO2}$ reaction looks smaller than that of $\ce{Ag+}$ (I may be wrong, but by significant digits it can't be greater than $0.799$), one can conclude that it's the $\ce{NO}$ reaction that's occurring here. And the $\ce{NO}$ reaction has a large enough SRP to oxidise $\ce{Ag+}$.

Usually nitrogen compounds are pretty versatile when it comes to redox reactions, since nitrogen shows many oxidation states. So the simple reason for why $\ce{HNO3}$ is so strong an oxidising agent (with respect to other acids) is that it has a different, better path available to it to get reduced.

Note that the exact reduction path (i.e. final reduction products/oxidation state) depends upon the concentration of nitric acid—so much that copper can be oxidised in three different ways.


Many inexpensive spectrophotometers cannot measure absorbances over 1.5 or so reliably. It's possible that you're out of the measurement range of your instrument. Try diluting your samples with water first. If you dilute 1:1, then the absorbance value of your sample is double what you measure (assuming the Beer-Lambert law is valid in this situation)

Also, I imagine molasses is very absorbent at that wavelength so it might be obscuring any change from the yeast and it may not be possible to measure the yeast growth by absorbance, with such a matrix.


Are ionic bonds stronger than covalent bonds?

In a fully covalent bond, you only have the mixing of the orbitals, as you said. But it's not only that, you also have coulomb interactions between the nuclei, that slightly raise the energy of the molecular orbitals, which results in asymmetric energies.

In a fully ionic bond you also have a coulomb interaction, but this time with the different sign, because you have a positive and a negative charge. These want to be as close together as possible, resulting in a stronger bond.

Generally speaking: Coulomb interactions have a greater impact than molecular orbitals on the energy, and thus the strength, of a bond.

It's almost like saying that electromagnetic forces are stronger than gravitational forces.


How to control and vanish termiticide/pesticide fumes from room

As stated by bon in the comments, this is probably something you should contact someone who is actually supposed to be able to answer this question about. The fumes might just be an annoyance, or they could be highly and immediately dangerous.

One place to start is with the manufacturer. Indeed, the manufacturer is probably the best place to go in general with chemical safety concerns. This chemical seems to be produced by a number of firms, so I was not able to locate a specific one, but a google search shows a number of possible contacts. This MSDS provides contact information for a Dow subsidiary. I suspect they will be willing to help, but if they refuse to because they were not the manufacturers of your specific product, you will need to ask the carpenter for the documentation. In a number of locations, the law specifies he must be able to provide an MSDS.

Alternatively, you might wish to contact your local fire department (non-emergency line). They can probably point you to someone local whose job it is to handle potential chemical contamination.

I will note that the MSDS I cited does not suggest extreme danger. However, it is still best to contact a producer or specialist of some sort.


Do bulky bases participate in substitution reactions?

It is very tough, but cases of t-butoxide acting as a nucleophile are known, e.g. J. Org. Chem. 2011, 76 (24), 10129–10139:

  n-Butyl-tert-butylether. To a stirred solution of 10 g of potassium tert-butoxide (89 mmol) in 100 mL of anhydrous dimethylsulfoxide was added 12.1 g of n-butylbromide (89 mmol) dropwise at 10 °C. The reaction mixture was allowed to warm and stand under stirring for 8 h at room temperature. The mixture was distilled under vacuum at −20 °C. The crude was redistilled at normal pressure to collect 6 g (52%) of the fraction distilled at 75−78 °C as a colorless liquid.

Mechanistically, I don't see any other real possibility by which this occurs apart from an SN2 reaction. Of course, there will be significant amounts of competing elimination, and that is probably why the yield is not amazing.


SN1 vs SN2 in low concentration of nucleophile

Let us assume that the competing reaction is a Sn2 based pathway.

Sn2=k[substrate][nucleophile]

Sn1=k'[substrate]

Relative rate of products formed=(ie.. Sn1 to Sn2 product)=k'/(k[nucleophile]

Sn1 Reaction doesnot depend on concentration of the nucleophile , so more the concentration of nucleophile, the relative rate of Sn1 product diminishes.

Hence ,when the competing reaction is a Sn2 pathway, lesser concentration of nucleophile would favour the Sn1 pathway.

Hope this helps.


Is it possible to measure pH without physically coming into contact with a solution?

Interesting analytical chemistry problem. Currently, it is not possible to accurately determine pH in a remote fashion. Spectroscopy is the technique of choice for remote sensing, but the H$^+$ is not that spectroscopy friendly i.e., it is a quiet ion. It will respond to radiofrequency in a powerful magnetic field but the sensitivity is quite poor in an NMR experiment by current standards. You can Google "measurement of pH by NMR". In general what appears to be pure to an organic chemist by NMR experiment, analytical chemists call it a dirty sample because there a lot of minor impurities.

A quick literature search shows that that satellite data can be used to make an only an educated guess about the ocean pH. It not accurate but good enough for a ball-park estimate.

Remote sensing of ocean pH by R. Sabia, D. Fernández-Prieto, J. Shutler, C. Donlon, P. Land and N. Reul, "Remote sensing of surface ocean pH exploiting sea surface salinity satellite observations," 2015 IEEE International Geoscience and Remote Sensing Symposium (IGARSS), Milan, 2015, pp. 106-109.

  Within this context, the purpose of the ESA “Pathfinders-OA” project is to quantitatively and routinely estimate surface ocean pH by means of satellite observations in several ocean regions. Satellite Ocean Colour, Sea Surface Temperature and Sea Surface Salinity data (with an emphasis on the latter) will be exploited. A proper merging of these different datasets will allow to compute at least two independent proxies among the seawater carbonate system parameters and therefore obtain the best educated guess of the surface ocean pH.


Determining geometry of trisubstituted alkene

As @Zhe points out, its not possibly to definitively answer your questions without knowing the structure of the olefin, as its important what is around the olefin, as well as just how many protons are attacked to the olefin.

If you read many of the original Grubbs' papers (and indeed any papers using a Grubbs' metathesis to make tri-substituted alkenes), the common method to determine the geometry is to use the nuclear Overhauser effect, or nOe for short.

The nOe identifies interactions through space (as opposed to standard NMR where we're looking at the interactions through bonds. By recording the nOe, one can essentially 'measure' how close two protons are, which in this case would allow you to distinguish E and Z.

Practically speaking, there are several different NMR experiments that one can run to observe the nOe.

The 1D nOe (1D-NOESY) difference experiment involves irradiation of a particular signal (for instance the one alkene proton), to allow measurement of what other protons are proximal in space. This experiment has the advantage of being quick to run, but in order to gain useful information, you have to run several of them (irradiate the alkene proton, then run a different 1D-NOESY in order to make sure that the proton that the alkene proton is 'seeing' also 'sees' the alkene proton
The 2D nOe (2D-NOESY) experiment shows all nOe enhancements in a molecule, this is incredibly useful and often used in structure elucidation, but can be slow to run, especially if the amount of sample available is limited.


The crystal structure of lithium tetraethylaluminate has been published (DOI)
and as far as I can tell from the abstract, there is no indication for a dimeric anion here. It might be save to assume that the balanced equation therefore can be written as:

$\ce{2 LiCH2CH3 + Al2(CH2CH3)6 -> 2 Al(CH2CH3)4^{\delta-}\,Li^{\delta+}}$


η1 or sigma complex involving cyclopentadienyl anion

Even though depictions of $\eta^1$-coordination might suggest a $\mathrm{sp^3}$-carbon donating, there is little evidence, that it actually is. Taking the example you have presented in your answer from Casey et. al.[1] I have run a geometry optimisation to obtain molecular orbitals at the DF-BP86/def2-SVP level of theory. The following geometry parameters around the coordinating carbon already suggest only little pyramidalisation:
\begin{array}{lrr}\hline
\text{Parameter} & \text{Experimental [1]} & \text{DF-BP86/def2-SVP}\\\hline
\mathbf{d}(\ce{Re-C^1}) & 236.0 & 240.2\\
\mathbf{d}(\ce{Re-H})   & 288.7 & 282.6\\
\mathbf{d}(\ce{Re-C^2}) & 317.8 & 321.3\\
\mathbf{d}(\ce{Re-C^5}) & 315.6 & 315.6\\
\angle(\ce{Re-C^1-H})   & 114.3 & 100.7\\
\angle(\ce{Re-C^1-C^2}) & 110.8 & 109.5\\
\angle(\ce{Re-C^1-C^5}) & 108.5 & 106.4\\
\angle(\triangle[\ce{C^1C^2C^5}]\ce{-H}) &
                        122.8   & 139.1\\
\text{RMSD} & 0.0 & 0.1\\\hline
\end{array}

We know that crystal structures are not really that accurate with protons, hence I'd put a little more trust into the calculation at this point. The paper does not give a value for any protons.
Here is a picture of the calculated structure.

When we look at the canonical MO we will see that all (occupied) π-obitals of the Cp-ligand are still there. Obviously they are distorted, but clearly recognisable.

The second orbital is the HOMO-1, which is the main contribution for the coordination, and the last is the HOMO (click for large). The only s-contribution to the HOMO-1 is from the carbon of the $trans\text{-}\ce{CO}$-ligand (and some hydrogen).
In output numbers, missing from 1 are minor contributions (<0.05):

 Alpha occ 81 OE=-0.276 is C39-p=0.19 C42-p=0.12 C41-p=0.12 C43-p=0.11 C40-p=0.11
 Alpha occ 87 OE=-0.185 is C39-p=0.26 C42-p=0.23 C41-p=0.22 Re24-d=0.08 C27-s=0.05
 Alpha occ 88 OE=-0.153 is C43-p=0.32 C40-p=0.32 C41-p=0.13 C42-p=0.12

When we localise the orbitals with the Natural Bond Orbital (NBO) analysis, we find a lone-pair at $\ce{c^1}$ (blue/orange) which donates into the anti-bonding $\ce{Re-^{$trans$}CO}$ orbital (yellow/red):

The composition of the lone pair orbital is

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 36. (1.27157) LP ( 1) C 39            s(  8.77%)p10.40( 91.23%)d 0.00(  0.00%)

The low occupancy is normal for donor-acceptor-complexes with some covalent character.

TL;DR Most $\eta^1$ complexes are in first approximation π-coordination complexes like the definition suggests[2]. This may change significantly when moieties are altered.

Further thoughts:
Wikipedia makes one important statement (it rarely does, but here it is true)3:

  Molecules with polyhapto ligands are often fluxional, also known as stereochemically non-rigid.

You can expect that in $\eta^1$-ligands the metal can easily migrate from one carbon to another, as the MO show. I would expect some kind of pseudo-rotation.

Ref.

Charles P. Casey, Joseph M. O'Connor, William D. Jones, Kenneth J. Haller. Organometallics 1983, 2 (4), 535–538. (link via doi, pdf via researchgate.net) Crystallographic data via CCDC number: 1118829.
η (eta or hapto) in inorganic nomenclature
Wikipedia: Hapticity and fluxionality


What home products are reactive with glass?

Some rust and stain removers (like Whink) contain hydrofluoric acid* at low concentrations (1-2.3%) Here is the MSDS for Whink: https://www.whink.com/cmssites/ws0811www.whink.com/uploads/Documents/Rust%20Stain%20Remover%20SDS.pdf

Now, at those concentrations it would probably take a while to dissolve silica, but it would be arguably quicker than a $\ce{NaOH}$ cleaning product.

*I do not recommend playing around with $\ce{HF}$ and its products unless you are experienced with them.


The reaction proceeds by SN2, which is highly accelerated at the benzylic position, due to stabilisation of the transition state by conjugation with the aromatic ring.

So you'll have to replace the benzylic bromine.

References: Benzyl chloride: chemtube3d


As you note, there are some general rules about solubility. For example, acetone is a highly polar organic molecule, so it's not surprising it has high solubility/mixibility in water. The "like dissolves like" rule is generally based on polarity and size/shape considerations, and it's a fairly good rule.

But solubility is complex because you have two big sources of driving force:

enthalpy: The molecular and intermolecular interactions between the solvent and solute are favorable. Consider methanol dissolving in water: it forms hydrogen bonds with the water, so the enthalpy is favorable.
entropy: This is often the larger driving force. You create a far more disordered result with a mixture than with two pure components.

While most chemists develop reasonably accurate intuitions of enthalpy, the entropic effects are harder to predict.

Many efforts have been made to predict solubility, both qualitatively and quantitatively. The comments above give some good links. Consider that this problem is important to chemical synthesis broadly, but also to drug delivery (i.e., can a drug be given orally and dissolve in the stomach, can a drug go through the blood-brain barrier?)

Perhaps the most successful is the Abraham general solubility model which attempts to predict solubility of organic compounds in 84 different solvents. Here’s the article.

Basically, it breaks down solubility into multiple components (grabbed from Andrew Lang and Jean-Claude Bradley):

E is the solute excess molar refractivity in units of (cubic cm per mol)/10. It represents the solute's polarizability and gives a measure of the ability of a solute to interact with a solvent through n- and $\pi$- electron pairs.
S is the solute dipolarity/polarizability. It gives a measure of the solute's ability to stabilize a charge or dipole
A is the overall (summation) hydrogen bond acidity. The hydrogen bond acidity descriptor measures the extent of hydrogen bonding by the solute in a basic solvent.
B is the overall (summation) hydrogen bond basicity. The hydrogen bond basicity descriptor measures of the extent of hydrogen bonding by the solute in an acidic solvent.
V is the McGowan characteristic volume in units of (cubic cm per mol)/100.

The nice thing about this model is that it makes sense -- you want a solute that matches lone pairs, hydrogen bond donating or accepting ability, polarizability, and volume for the solvent.

The catch is that surprisingly little data is available. The late Jean-Claude Bradley encouraged his students to make accurate solubility measurements, but also to publish them on the web. In this way, they gathered a large spreadsheet of "compound X dissolves Y amount in Z solvent."


Examples of evolving networks in chemistry

Synthetic Pathways

... is the only example I can think of right now.

As far back as 1971¹ people have been positing the possibility of an AI to search for novel synthesis paths, but only a couple of years ago a decent one was developed² which could interpret literature and create novel sets of steps for particular reactions -- for example, given a particular reactant and product, produce a set of steps which would produce one from the other. The particular focus of the paper for the 2012 one was to produce single-pot (multi-step) procedures to replace awkward separated multi-step syntheses, but in principle the procedure is generalisable to any path.

Sridharan, N.S., An Application of Artificial Intelligence to Organic Chemical Synthesis, Doctoral Thesis, State University of New York at Stony Brook, New York, July 1971.
C M Gothard et al, Angew. Chem., Int. Ed., 2012, 51, 7922 (DOI: 10.1002/anie.201202155) but there's a good write-up by Philip Ball in Chemistry World, August 2012 issue.


How are the LCAO ansatz and Slater determinant ansatz for the wavefunction connected?

I found the question somewhat confusing. I think it is because mixing up some concepts.

  I understand that for the basis set expansion in an orthonormal basis,
  one can obtain the coefficients by the inner product  $$
 \left<\phi_i|\Psi \right>=c_i $$

Please, think about what you mean by the coefficients and where the functions live, before continue reading.

As you thought this expression as an inner product (which is right), both functions must live in the same space so they have the same number of coordinates. What you are doing with that inner product is projecting $\Psi$ over the $\phi_i$. If $\{\phi_i\}$ is an orthogonal set, then you can make the expansion you wrote. But, if the are not, then such expansion is meaning less. You still can make expansions that recover $\Psi$  (in fact, an infinite number of them). So, I get confused about which coefficients are you referring to. As for example, $\phi_1$, appears in both: many Slater's det. and in many terms in one (or more) Slater's det., to ask  which is its coefficient in the Slater det. makes not sense, except if you want to expand the Slater det. into a LCAO which is not possible (but you still can project it).

See below.

  But how does this work if the wavefunction is multidimensional like the slater determinant wavefunction ?

First, note that even for 1-electron problems the problem is not mono-dimensional. So, the case of Slater's determinants is exactly the same. You only need to integrate over every coordinate.

  To my knowledge only basis functions of one coordinate are used in
  calculations, like 3D Gaussians.

3D-Gaussians have three coordinates. It is obvious using Cartesian coordinates. Under spherical coordinates they only depends on $r$, but they still are functions of $\theta$ and $\phi$. Although you can take only one coordinate ($r$) for its evaluation, it is not true for computing an arbitrary inner product.

  Assuming that i have a wavefunction given in slater determinant form,
  how would i find the coefficients of the basis functions ?

At first, take in mind that those coefficients are for expanding the Slater det. into the subspace spanned by your $\{\phi_i\}$. That is, it they are not an complete set, you only will get an approximation. You just have to perform those integrations. In your example, the $c_i$ will not be a function of $x_2$, because you have to integrate over every coordinate.

I am not sure if I understand correctly your doubts. Please, do not hesitate in ask for clarification. Maybe this answer could be useful.


Why do Zr, Sn, and Ti enolates form syn aldols selectively?

For most aldol reactions, the relative 1,2-diastereoselectivity is a consequence of the enolate geometry.*

Both tin and titanium enolates are easily formed, and have been shown to exist predominantly in the (Z) configuration.

When reacting a cyclic (Zimmerman Traxler) TS, its commonly observed that (Z) enolates afford 1,2-syn aldol adducts, due to the orientation of the enolate   (the (Z) enolate always has an axial R-group).

The following examples are instructive. In both cases (tin or titanium), we first form a (Z) enolate using the conditions shown. When the aldehyde (R-CHO) is introduced, chelation using the metal organises the Zimmerman Traxler TS. The lowest energy TS possible is the one in which the aldehyde is arranged with its bulky R group equatorial, however the enolate cannot put the methyl group (in this case) equatorial, as the enolate geometry enforces its axial positioning. Following both transition states through, we get the 1,2-syn aldol adduct in both cases.#

  Image taken from Andy Myers' CHEM115 Handout: Stereoselective, directed aldol reaction

The description given in Carey is fundamentally incorrect for the reactions shown (even if it does predict the correct product). Looking at the references he gives, they're all fairly old, before more recent experimental and computation studies have validated the cyclic TS model for these kinds of metal enolates.

*: Mukaiyama aldol reactions do not conform to this, since they occur via a fundamentally different transition state (an open TS) to the Zimmerman Traxler model commonly found for other aldol reactions.

#: In this case, we also get enantioselectivity via the use of a chiral auxiliary, and depending on the metal used the enantioselectivity can be switched around due to chelation or lack thereof between the auxiliary and the metal


Why is sand/glass polar? How does lattice structure change polarity?

I'm not sure where you get the idea that glass is linear. What makes a glass is the lack of long range ordering, though in fused silica, the approximate tetrahedral structure of $\ce{SiO4}$ is maintained (this is a 2D section, the fourth oxygens are not shown):

 Source

In normal glasses, there is no significant net dipole moment across the entire structure, but the reason we observe strong adhesive forces between water and glass is a surface phenomenon—in the bulk of the glass, every silicon atom is connected to four oxygen atoms and each oxygen atom to two silicon atoms, but what happens at a boundary/surface? There are oxygen atoms that only have one bond. When these are exposed to an aqueous solution, some will bear a negative charge, others will pick up a proton from water. The theory behind it is pretty complicated, but you can think of it as a simple acid-base reaction—at low pH more will be protonated:

$$\ce{Si-OH <=> Si-O- + H+}$$

In any case, protonated or not, silanol groups are very happy to associate with water, accounting for the concave meniscus.


How to demonstrate the safety of wearing leather shoes in the chemistry lab?

The simplest thing to do is to sacrifice two pairs of shoes, one with leather uppers and the other with canvas/cloth/whatever. Put something inside the shoes to represent "feet". Do something nasty to each pair, like pour concentrated acid over them (using secondary containment to avoid spills and splashes!). Then, extract your indicator and compare (assuming the different in damage to the shoes was not enough). You can also try lighting them on fire (though canvas shoes probably have fire-retardant compounds added), or seeing how long it takes water to soak through (leather will win here).

And, whatever you do, record it to video so that next time you only need to play the video and not buy more shoes. If this is for your students, I would make sure you appear in the video to increase "authenticity".


Well, if you don't make it in situ, you would have to add the nitrous acid you have already kept in your laboratory shelf, and surprise! It's not there! Because, it has already decomposed.

Nitrous acid ($\ce{HNO2}$) has an intermediate oxidation state of +3, and hence, it tends to disproportionate into more stable oxidation states:

$$\ce{2HNO2 -> NO2 + NO + H2O}$$

$\ce{NO2}$ itself dissolves in water to give $\ce{HNO3}$ and $\ce{HNO2}$. This gives an overall reaction in aqueous solution as:

$$\ce{3HNO2 -> HNO3 + 2NO + H2O}$$

This should give you a good idea on why we make nitrous acid within the reaction mixture. You can't store it beforehand. It must be made when it's time to use it.


At the equivalence point, you have an aqueous solution of weak base $\ce{Na}A$ at the concentration of $c$. The pH of the solution is given by the equation: $$\mathrm{pH}=\frac{1}{2}(\mathrm{p}K_a +\mathrm{p}K_w -\mathrm{p}c)$$
In your case: $\mathrm{pH}=\frac{1}{2}(4.0 +14 -2.0)=8.0$
Now, if you are not familiar with the above equation, you can write the reaction that could happen in the solution; the reaction of  $\ce{Na}A$, completely dissociated, with water:
$$A^- +\ce{ H2O <=> OH- +H}A$$
The constant of this equilibrium is: $$K=\frac{K_w}{K_a}= 10^{-10}$$

On the other hand,  you have:

$$K=\frac{x^2}{c-x}=\frac{x^2}{1.0\times10^{-2}-x}\approx\frac{x^2}{1.0\times10^{-2}} =1.0\times10^{-10}$$
Where $x$ is the concentration of ion hydroxide and the weak acid $\ce{H}A$.

We have, $x=1.0\times10^{-6}\mathrm{M}$.

$$[\ce{H+_\mathrm{(aq)}}]=\frac{10^{-14}}{1.0\times10^{-6}}\mathrm{M}$$
 $$\mathrm{pH}=8.0$$