First things first: $\ce{BF3}$ is not tetrahedral. It cannot be, as a tetrahedral structure requires a total of five atoms (one in the centre and four in the corners).[1] Instead, it is trigonal planar. However, that is of minor importance.

Secondly, fluorine is most certainly not sp3 hybridised. In general, hybridisation is determined by geometry and not vice-versa, thus we only need to establish the trigonal planar geometry to arrive at boron’s sp2 hybridisation. However, this doesn’t help us for terminal atoms as there is no ‘geometry’—only a single binding partner. In some cases, constraints may exist due to multiple bonds which require π-type bonding orbitals. Again, this is not the case for single bonds which rely solely on their σ orbital.

Instead, the basic assumption should be to assume the hybridisation as low as possible. In that way, fluorine’s s type orbital can be as low as possible in energy—which means a higher energy gain. It is easy and possible to assume fluorine bonds solely with a correctly aligned p orbital and so it does.

Note:

[1]: This sentence depends on the system of classification you were taught. I was taught only to label those molecules as tetrahedral which have five atoms that correspond to the four corners plus the centre. If there are four atoms as in the case of ammonia ($\ce{NH3}$), it is possible for the overall structure to be a tetrahedron with one corner removed. We were taught not to call this tetrahedral but a trigonal pyramid. Your teacher’s mileage may vary.


The strength of a Lewis base corresponds to the strength of its conjugate acid. As usual, a weak acid will give a stronger base, therefore parent acids with higher pKa's deprotonate to stronger Lewis bases. ammonium, $\ce{NH4+}$, has pKa=9.25, while phosphonium, $\ce{PH4+}$, is a much stronger acid with pKa=~-12. Basicity decreases down the group because proton affinity decreases as electronegativity decreases. The N-H bond is stronger than the P-H bond (hence $\ce{NH4+}$ is a weaker acid than $\ce{PH4+}$).

Lewis bases donate electrons from their HOMO, and for ammonia's MO, the HOMO, which corresponds to the lone pair (LP), is very similar in energy to the nitrogen AO, therefore the LP electron density can be considered to be more localized to the nitrogen. Presumably in a MO for an analogous species with a group 15 central atom lower on the periodic table (with lower electronegativity), the AO of the central atom would be higher in energy, so the HOMO electron density would be less concentrated on that atom.

Also, ammonia is more hybridized than phosphine. The s-p energy gap is greater for phosphorous, so the energy penalty of hybridization is higher, therefore the LP will have more pure s character than sp3 character, making it less directed.

Since the ammonia LP electron density is more concentrated on the central atom, and more directed, it makes it more nucleophilic.


What compound does this chemical formula represent?

The structure represents adrenaline or 'epinephrine' (en.wikipedia.org/wiki/Epinephrine). As in:

  Go out and have an adrenaline rush on your birthday!!

The image on wikipedia has a slightly rotated ring, so doesn't look identical.


Why don't periodic acids convert alkenes to 1,2 diols?

Sodium periodate can be used indirectly to form diols from alkene, see:
Emmanuvel, T. M. A. Shaikh, A. Sudalai, Org. Lett., 2005, 7, 5071-5074.
The mechanism for periodate cleavage of diols also strongly suggests the periodate is an electrophile.  This may explain why in general, alkenes are not nucleophilic enough to react under normal conditions without a co-reagent. The osmium tetroxide reaction is favoured as a [3+2] cycloaddition mechanism.


Why is wavenumber used in IR spectroscopy rather than wavelength?

Not only in IR spectroscopy. Wavenumber is unit of energy and therefore you can directly deduce the difference of energy between states.

In addition, humans like to think in acceptably small numbers (0.01 - 10,000). Wavenumber allows this for IR and conveniently supplements the eV unit in small energy separations range. Admittedly, the conversion factor of 8,065.73 won't win beauty contest.


How does propylene glycol/glycerin form formaldehyde/acetaldehyde when vaporized?

I think the reaction is essentially an incomplete combustion reaction.

In a vaporizer, the idea is to get the propylene glycol, glycerine, and and "flavor" molecules into the vapor phase (or into aerosol droplets) without chemically degrading them.

However, the way this is accomplished is with very high-temperature resistor coils - the idea is that if you have a small amount of heat in a confined space, and the liquid is distributed throughout an absorbent (like cotton, for example), then the heat transfer will happen very quickly and a large amount of liquid will be vaporized.

The problem is that when you have a very high temperature heat source, and there is lots of oxygen in comparison to the liquid, there is a good chance that at least some of the liquid will reach the combustion temperature instead of just vaporizing.

What happens is that the hydrocarbons in the propylene glycol and glycerin molecules are partially oxidized - oxygen reacts with them to "steal" electrons. If the combustion was complete, you would wind up with just carbon dioxide and water. When combustion is incomplete, you can get any number of compounds, two of which happen to be formaldehyde and acetaldehyde. Carbon monoxide is another one I would be concerned about.

To see how this works, take a look at proplyene glycol (all pictures are from wikipedia):

and glycerine:

Notice how they both have three carbon atoms in the "backbone" and either one or none $\ce{O-H}$ groups (called hydroxyls or alcohols) attached to each one.

Oxygen can react with any one of those carbons and will try to form $\ce{CO2}$ and water, the most thermodynamically stable products that are possible:

If there is plenty of oxygen, it will. However, if there is not enough, you will get some intermediate species - formaldehyde or acetaldehyde, for example.

Let's look at formaldehyde:

In this case, oxygen has reacted with the the other two carbons on the backbone, leaving one with a hydroxyl. The hydrogen on the oxygen goes somewhere (either water or one of the other pieces of the original molecule) and the leftover electrons end up in the oxygen-carbon bond, forming a double bond. This molecule is more thermodynamically stable than proplyene glycol or glycerol, but less stable than carbon dioxide.

Now let's look at acetaldehyde:

Notice the similarity - it's almost the same, there is just a $\ce{CH3}$ (methyl) group in place of one of the hydrogens. You could get this by reacting the first carbon on propylene glycol with oxygen, and then removing the hydrogen from the hydroxyl on the second carbon. This is again more stable than propylene glycol, but less stable than carbon dioxide - it is also an intermediate.

  Why does propylene glycol form more carbonylated products than glycerol?

This is a guess - but I think it likely has to do with the reaction kinetics. The reason I think this is true is that all of the intermediate products are less thermodynamically stable than carbon dioxide and water. Therefore, if a propylene glycol (PG) gives more of a given intermediate product than glycerine does, that product is "easier" to make for PG. This would happen if the activation energy for forming it is lower, or if the activation energy to go from the intermediate to carbon dioxide and water is higher compared to the intermediates formed by glycerine.

Note: Many people in other answers are pointing out that there are other mechanisms that could lead to these products, and that combustion is not likely at these temperatures. I think that given the mechanics of heat transfer we can't just rule it out, but I don't have any evidence for that. My answer is meant to show you how the reactant molecules contain the same structures as the products - I wouldn't put much faith in my mechanism (or anybody else's) without some real evidence.


How does chloroform react with cyclopentadiene in base?

The worry I had is that the normal way to make dichlorocarbene from chloroform is to treat it with a strong base, this will convert the cyclopentadiene into the cyclopentadienyl anion. This anion is an aromatic species which is unlikely to be able to react with the carbene.

But in one paper (Grignon-Dubois, M.; Dunogues, J.; Ahra, M. - Recueil des Travaux Chimiques des Pays-Bas, 1988, vol. 107, # 3, p. 216 - 225) it has been reported that chloroform, base and cyclopentadiene will react together to form the first intermidate.

I think that the reaction step showing the formation of the cyclopropane ring is perfectly reasonable. We need to consider how we would form the carbene. I know that it is possible to form carbene by a photochemical reaction on ketene which forms the carbene and carbon monoxide. I know from personal experience that chloroketene reacts in a [2+2] reaction with cyclopentadiene to form a cyclobutanone. So that route might not be a good idea.


Why do some reactions require specific pressures to happen?

ChemGuide has a good introductory article here.

The effects of increasing pressure and temperature are, to an extent, equivalent. Increased pressure leads to increased collisions and increased collision strength between molecules, allowing the (usually high) activation energy barrier to be overcome at a noticeable rate; at standard temperatures and pressures, for example, collisions between $\ce{CO}$ and $\ce{H2}$ are far too infrequent and involve far too little energy for the formation of $\ce{CH3OH}$.

Remark. How do we achieve high pressures and temperatures? We could pressurize a steel cylinder and use it as a reaction vessel, for one, but a novel approach involves sonochemistry. Ultrasound waves in solution lead to cavitation, the formation and rapid collapse of bubbles, which are highly pressurized and at high temperatures. The use of ultrasound can even open up new reaction pathways not usually available; for example, radical species are often transient, but the forcing conditions in these bubbles allow them to survive long enough to react.


Extract all structures of Gaussian 09 molecular dynamics calculation using babel?

If you have the latest copy of cclib installed, it can extract both the energies and geometries and write the geometries to an XYZ trajectory file that you can open with Avogadro or VMD.

To print the HF or DFT energies to the screen,

$ ccget scfenergies admp.out

but be warned that the energies are in eV, not hartree.

To dump geometries to XYZ files, here's a simplified version of a script I use all the time:

#!/usr/bin/env python

"""cclib_extract_geom.py: Extract geometries from a quantum chemical
output file using cclib.
"""

from __future__ import print_function

import os.path

from cclib.io import ccopen, ccwrite

def getargs():
    """Get command-line arguments."""

    import argparse

    parser = argparse.ArgumentParser()

    parser.add_argument('outputfilename', nargs='+')

    parser.add_argument('--trajectory',
                        action='store_true',
                        help="""Extract all possible geometries into a trajectory-like file?""")
    parser.add_argument('--suffix')

    args = parser.parse_args()

    return args

if __name__ == '__main__':

    args = getargs()

    for outputfilename in args.outputfilename:

        job = ccopen(outputfilename)
        data = job.parse()

        stub = os.path.splitext(outputfilename)[0]
        if args.suffix:
            xyzfilename = ''.join([stub, '.', args.suffix, '.xyz'])
        else:
            xyzfilename = ''.join([stub, '.xyz'])

        ccwrite(data, outputdest=xyzfilename, allgeom=args.trajectory)

To write the energies in to a text file in hartree, you might add something like

        from cclib.parser.utils import convertor # put me at the top of the file
        with open(stub + '.energies', 'w') as fh:
            for energy in data.scfenergies:
                fh.write(str(convertor(energy, 'eV', 'hartree')) + '\n')


The third Grignard reagent molecule could extract a proton from the methylene group which is in an electron withdrawing environment.  This type of enhanced acididity on carbon is referred to by the term "carbon acid" .

While this reference deals with carbonyl groups, conjugation with a phenyl ring can do the same thing to the methylene group.

We do not see any change of the methylene group in the final product because it's just deprotonated, and the acid workup brings the proton back.  We would, however, see the third Grignard reagent being protonated to its parent hydrocarbon, in this case methane which bubbles off.


Acidity data:

The true order of acidity is $\ce{Br = I > Cl >> F}$. Relative to unsubstituted benzoic acid, all of the ortho-halogen substituted benzoic acid derivatives have a greater acidity, clearly having some stabilising effect on the resulting benzoate anion.

  Data retrieved from Journal of Research of the National Bureau of Standards, 1958, 60, p569 - research paper 271. $\mathrm{p}K_\mathrm{a}$ values measured in water.

Rationalisation:

Polar effects All of the halogens exert a negative inductive (-I) effect, removing electron density from the ring. This makes the benzoic acid derivatives more acidic in all 4 cases since the resulting conjugate base is stabilised to a certain extent by the presence of the electron withdrawing halogen.
Steric effects As the halogen gets bigger, strain may force the carboxylate out of planarity with the aromatic ring. Whilst fluorine is roughly the same size as the hydrogen atom, bromine and iodine are both sterically more demanding.
Internal H-bonding This is suggested by another answer below, whilst this has been studied and shown to operate for protic groups such as alcohols and amines, in halogens it has little impact on stabilising the negative charge.

Overall, steric and polar effects operate in tandem to give an overall acidity for the benzoic acid. It's perhaps not intuitive which one will dominate (e.g. fluorine is far more electronegative, but also far smaller than iodine, so which is more acidic).

The table below, taken from the cited paper, attempts to separate out steric and polar effects through calculations on both the acid and its conjugate base. Sadly, they only look into $\ce{F}$ and $\ce{Cl}$ (no $\ce{Br}$ or $\ce{I}$), but it can be clearly seen that:

Hydrogen bonding is making no significant contribution to the acidity of the substituted benzoic acid
Polar and steric effects are both in operation, with sterics being far more important in the ground state (the carboxylic acid) and polar effects dominating for the conjugate base (the carboxylate).

  Table copied from New. J. Chem., 2004, 28, p67


The electron cloud distribution describes the probability of finding an electron in some region of space around the nucleus at some instant of time.

Imagine you have a camera that can take close-up shots of a hydrogen atom showing the electron at an instant in time. Also suppose that camera can keep snapping pictures, say, a thousand times a second, each time on the same image so that you get a multiple exposure. As the camera runs, the multiple images of the electron will begin to form a cloud-like picture which is dense in some places and thin in others. For the single electron of hydrogen, the densest part of this cloud will form a sphere with the nucleus at the center. As you travel inward toward the nucleus, or outward away from the nucleus, the cloud will get progressively thinner.

The thickest regions of this cloud are thick because the electron was found there more often than in the thin regions. So, if we were to take just one snapshot, this is where the electron would be most likely be found. That is, the probability of finding the electron at a given time is greatest where the cloud is densest.

Other elecrton clouds in other element's atoms have different shapes. Do understand that this is a model that explains some chemical behavior but it is not the current "final" model of how electrons behave in atoms.


To what state would copper wiring revert if abandoned for centuries?

Let me talk first about how iron corrodes (as in rusts) and then compare that to copper. The simple answer is that when iron corrodes, the iron oxide (iron which has chemically bonded with oxygen, which it loves to do) that forms on the surface of the pure iron underneath, does not seal the surface against further intrusion of more oxygen. When copper corrodes, the copper oxide does seal the surface protecting underlying copper against further corrosion. I don't know if iron oxide is simply more porous to oxygen than copper oxide is but I do know the following. Iron can chemically combine with oxygen in multiple ratios. When you examine a highly corroded iron surface you'll notice that the first layer of rust is very red. If you brush or maybe lightly sand that red layer of rust away you'll notice a black layer underneath it. That's still rust. If you sand your way through that you'll get to the pure iron underneath which has more of a dark silver color. The red rust has more oxygen in it than the black rust but can pass some of that oxygen to the black rust turning it red. Similarly, the black rust can pass some of its oxygen on to the pure iron underneath it. That process continues until oxygen can leach all the way through the iron. Chemically, iron has more than one oxidation state. Oxidized copper atoms are much more reluctant to release their oxygen to adjacent pure copper and act as a barrier to the intrusion of more oxygen.


Why do colligative properties depend only on number of solute particles?

It's a lie. Colligative properties do depend on the chemical nature of the solute and solvent - their interaction. The trick is: an ideal solution, in which the solute-solute intermolecular forces and solute-solvent interactions are more or less of the same character, has the property that colligative properties are dictated by composition given only in terms of amount of moles. However, any sufficiently diluted solution will behave somehow ideally, that is where the statement goes: for diluted solutions, colligative properties depend only on the number of moles of solute.


Result of glucose with acetic anhydride and pyridine?

Most probably it forms glucose pentaacetate. The acetic anhydride esterifies with all the alcohol groups on the glucose ring.


What functional groups are subsets of the Alkoxy group?

There is another Wkipedia article entitled "Alkoxy group".  In it they show the following molecules as containing alkoxy groups.

They include methylenedioxy, but not peroxide in their list.


Can inductive effect be the sole driving force for carbocation rearrangement?

The answer to this question is quite difficult to deduce logically. In the first place, the carbocation that is formed is secondary. Rearranging will only change the position of the carbocation, but it will be still secondary. But then there is also a stabilising +I effect.

Instead, if we take into account that that these carbocations are intermediates to form alkenes, a conclusion can be drawn. So, the best way is to consider the thermodynamic favour of the rearrangements.

Prior to rearrangement, the carbocation is secondary. If you consider that this carbocation is going to form an alkene, the major product from it (Saytzeff product) will be  4-methyl pent-2-ene, which is of the form $\ce{RCH=CHR}$. After rearrangement, this will be still a secondary carbocation, so there is a significant energy required for this hydride shift. This energy is not really compensated by the +I effect of the isopropyl group. But if we consider that this carbocation is an intermediate for the alkene formation, the major alkene (Saytzeff product) will be 2-methyl pent-2-ene, which is of the form $\ce{R2C=CHR}$, which is thermodynamically more favourable.

So it is seen that in the rearrangement, the energy of the intermediate is increased, but the overall energy of the product is decreased. So, I think that, at higher temperatures, this rearrangement can be possible, because the ultimate product (if we think this as an alkene) is thermodynamically favourable, while at normal temperature this rearrangement is energetically restricted a bit.


Dynamic equilibrium - effect of adding inert gas

At constant temperature, only changes to the concentration (technically activity) of a reactant or product, shift the equilibrium.

Adding inert gas does not change the concentration of any reactant or product, so it does not affect equilibrium.


The number "25" indeed comes from the fact that it was the 25th compound of the lysergic acid amide series or lysergamides. I found an article* by Frank J. Ayd, Jr. & Barry Blackwell who accounted Alexander Hoffman's investigation in his LSD synthesis:

A number of pharmacological experiments were carried out by Professor
Ernst Rothlin with this new compound, which was given the laboratory
code name LSD-25 because it was the twenty-fifth compound of the
lysergic acid amide series.

You can check the other compounds of the series in the Wikipedia article linked above. Most of them have their own specific names and not LSD-1, LSD-2 etc.

I also did find something in an e-book(Ref) which says that the number has something to do with date:

Dr. Hoffmann described the results of accidental ingestion of
LSD. It was in Switzerland in 1943, five years after Dr. Hoffmann and
an associate has synthesized lysergic acid diethylamide tartarate
(from ergot). The substance was labeled "25" because of the date: "May
2, 1938" (the date when LSD was synthesized(?)).

Ref.: LSD-25: a Factual Account: Layman's Guide to the Pharmacology, Physiology, Psychology, and Sociology of LSD, Louise G. Richards, U.S. Bureau of Narcotics and Dangerous Drugs, 1969
ಠ_ಠ

*I was researching about something else i.e "serendipity in Chemistry". While I was surfing the Internet, I stumbled upon this article and from that, I came to know about LSD, searched for the sources of their information and from there I got access to the above article and then proceeded to compile this answer without any prior intentions. What a beautiful irony!


If NO3- were to get oxidised to some compound in the way SO4 2- gets oxidised to S2O8 2-, what would be that compound?

First, let’s take a look at both sulfate and nitrate. They both show similar characteristics when you compare the oxidation states: remember that nitrogen is in group 15 so it has five valence electrons while sulfur is in group 16 so it has six — therefore, the highest reasonable oxidation states for these elements should be $\mathrm{+V}$ and $\mathrm{+VI}$, respectively.

$$\begin{align}&\ce{\overset{$\mathrm{+V}$}{N}\overset{$\mathrm{-II}$}{O3}^-}&&\ce{\overset{$\mathrm{+VI}$}{S}\overset{$\mathrm{-II}$}{O4}^2-}\end{align}$$

When oxidising sulfate $\ce{SO4^2-}$ to peroxidisulfate $\ce{S2O8^2-}$, sulfur cannot be oxidised any further; the only atom that can still be oxidised is oxygen. Hence, in an initial step, we generate $\ce{SO4^{.-}}$ with three oxygens of $\mathrm{-II}$ and a fourth of $\mathrm{-I}$. Two of these then dimerise forming a peroxo-bond.

$$\ce{^-\overset{$\mathrm{-II}$}{O3}\overset{$\mathrm{+VI}$}{S}-\overset{$\mathrm{-I}$}{O}-\overset{$\mathrm{-I}$}{O}-\overset{$\mathrm{+VI}$}{S}\overset{$\mathrm{-II}$}{O3}^-}$$

Looking at nitrate, the only obvious theoretical would be to do the same, i.e. remove one oxygen electron to intermediately generate $\ce{NO3^.}$ and then have this dimerise to $\ce{O2N-O-O-NO2}$ or $\ce{N2O6}$. Interestingly, this would be a neutral compound unlike $\ce{S2O8^2-}$ which retains its charge. I don’t know whether it has been observed or not, but I would predict its formation from what I know.


Reactivity of thioesters with respect to nucleophilic attack

According to these lecture notes from Gonzaga University:

  The question arises over why thioesters occupy a prominent place in metabolism while oxygen esters play a relatively minor role. One answer is thermodynamic. The sulfur in the thioester linkage is less able to participate in electron delocalization through the acyl group (the usual explanation given for this is poor overlap between the 2p orbital of carbon and the 3p orbital of sulfur), and this makes the thioester bond less stable than the ester bond. Hence, the thioester bond has a more negative standard free energy of hydrolysis (−7.5 kcal/mole vs. about −5 kcal/mole for most oxygen esters). In many cases, it appears that thioesters are more reactive than oxygen esters, undergoing more facile nucleophilic displacement reactions at the acyl group. The reactivity of a thioester at the alpha carbon compares favorably with that of ketones.

...

  Because resonance structure 2c [see web page - CF] makes a more significant contribution for a thioester than an analogous structure for an ester, the acyl carbon is more positive, hence more susceptible to nucleophilic attack. An attacking nucleophile would be more readily acylated by a thioester than it would be by an ester. Furthermore, the C-S bond is weaker than the C-O bond, and the thiolate (or thiol, if protonated) is a better leaving group than alkoxide (or alcohol). These factors all make a thioester in general a better acylating agent than an ester.

So the answer doesn't seem to involve d orbitals explicitly, but instead the difference between 2p and 3p orbitals (as well as other factors).


What happened to this 10 year old unopened plastic bottle?

Since the drink is not carbonated, no liquid has been escaping the bottle, and the components don't seem to be able to participate in any kind of chemical reactions involving gaseous products, I assume the gas that's been leaked is air.

I also cannot imagine any kind of chemical reaction involving consumption of this amount of air leading to such a severe bottle deformation. My theory is that air simply escaped over time through the bottleneck. There are 2 parts made of 2 materials with different thermal expansion coefficients:

the bottle itself (polyethylene #1 (PET), or polyethylene terephthalate (PETE));
the screw cap (polypropylene (PP)).

Due to the construction of the joint, both parts basically act as a pressure equalization valve. I suspect that the coefficients of thermal expansion are chosen in such a way that when heated, excess gas would slowly escape in order to minimize the risk of explosion when the storing conditions for the finished product is not met. So, when the bottle is heated, the air is slowly leaking out. Once the temperature drops, the "valve" closes and the process stops. Over the course of 10 years, there surely were many temperature fluctuations making air leave the bottle little-by-little.


Does the mechanism of AChE inhibition by Isoparathion depend on chirality?

It would appear that the mechanism of inhibition of AChE by both stereoisomers of isoparathion (Or, at least, its methyl analog) proceeds through the expulsion of the 4-nitrophenoxy leaving group. Several papers have studied the inhibition kinetics of isomalathion and compared the results to those of isoparathion. All of them seem to agree that, at least for isoparathion methyl, AChE inhibition proceeds through the expulsion of the same leaving group, but the O,S-dimethyl phosphorylated AChE would be chiral, as is expected. Furthermore, the inhibition, reactivation, and aging reaction rate constants would also differ depending on the stereoisomer of isoparathion.
References:
Berkman, C. E., Quinn, D. A., & Thompson, C. M. (1993). Interaction of acetylcholinesterase with the enantiomers of malaoxon and isomalathion. Chemical Research in Toxicology, 6(5), 724–730. doi:10.1021/tx00035a019
https://pubs.acs.org/doi/pdf/10.1021/tx00023a004
Jianmongkol, S., Marable, B. R., Berkman, C. E., Talley, T. T., Thompson, C. M., & Richardson, R. J. (1999). Kinetic Evidence for Different Mechanisms of Acetylcholinesterase Inhibition by (1R)- and (1S)-Stereoisomers of Isomalathion. Toxicology and Applied Pharmacology, 155(1), 43–53. doi:10.1006/taap.1998.8608


You are thinking too far ahead for a simple reaction. First of all, $\ce{I-}$ is not a very strong base to abstract a protone from alcohol. $\ce{I-}$ is the conjugate base of very strong acid, $\ce{HI}$. Yet, $\ce{I-}$ is a very good nucleophile. Therefore, the condition given ($\ce{NaI}$ in anhydrous acetone; $\ce{NaI}$ dissolves in anhydrous acetone and therefore exists as $\ce{I-}$ in solution) is suitable only for nucleophilic substitution reaction and also favored $\mathrm{S_N2}$ mechanism. Since the byproduct form ($\ce{NaCl}$) is not soluble in acetone, the reaction favors only the forward reaction:
$$\ce{R-Cl + Na+ + I- -> R-I + NaCl_{(s)}}$$
Since it is a $\mathrm{S_N2}$ reaction, inversion is the result ($(1R,2S)$ to $(1R,2R)$ with closed to $100\% e.e.$).
However, note that there is a neighboring group participation (NGP) possible in 1,2-halohydrins in certain reactions (Ref.1). As OP suggested, if NGP is contributed here (without strong base I doubt it though), the product from NGP is going to be the retention product with $(1R,2S)$ configuration. Thus, total $\%e.e.$ of final product is $\lt 100\% e.e.$.
References:

T. A. Geissman, Richard I. Akawie, "Rearrangements of Halomagnesium Derivatives of Halohydrins," J. Am. Chem. Soc. 1951, 73(5), 1993–1998 (https://doi.org/10.1021/ja01149a029).


White growth on unopened bottle of chlorotrimethylsilane

Chlorotrimethylsilane, like other organosilicon compounds, is easily attacked by nucleophiles and readily hydrolyzes in air yielding hydrogen chloride:
$$\ce{2 Me3SiCl(l) + H2O(l) -> Me3Si-O-SiMe3(l) + 2 HCl(g)}$$
Hydrogen chloride reacting with traces of ammonia $\ce{NH3}$ in the air builds up white solid ammonium chloride $\ce{NH4Cl}$ depositing on the nearby surfaces.
Also, the final product of hydrolysis for the majority of organosilicon compounds is silicon dioxide $\ce{SiO2}$ (sol-gel process similar to olation). Fine powder of silicon dioxide is white as well.


Can heat capacity at constant volume for a gas decrease as temperature increases?

As temperature increases, it activates the rotational and vibrational degrees of freedom, so $\ce{Cv}$ increases with increase in temperature. The graph $\ce{CO2}$ would be:


Will trenbolone acetate in the presence of methanol yield trenbolone acetate if distilled water is subsequently added thereafter?

The method will not convert trenbolone acetate to trenbolone, the ester is not cleaved under these conditions.

It is however possible to separate the steroid from a mixture with other, water-soluble compounds this way.

Unless you are a veterinarian and use trenbolone or its acetate as a growth promoter for cattle, you might want to consider the legal situation in your country. As far as Germany is concerned, the possession of as much as 2 g might lead to real trouble. Here, the non-punishable, non-significant amount is in the 150 mg range (for its use in sports).


Why doesn't an sp3s hybridisation exist?

The counter question that might help you rule this out is: "Which 3s orbital?" Note that atomic Nitrogen doesn't have a 3s orbital as N itself belongs to the 2nd period(n=2).

Also note that in hybridization, Atomic orbitals of same or almost similar energies intermix to form same number of new Hybrid orbitals of same shap and equivalent energy.
That being said, you cant expect 2s or 2p orbitals to intermix with a 3s orbital
(if it was a case. Since this case isn't possible with N)


Why does chlorine undergo disproportionation in alkaline media?

At constant pressure and temperature, the spontaneity of all chemical reactions is controlled by the difference in Gibbs free energy ($\Delta G$), as described by the following equation:

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

Where $\Delta H$ is the enthalpy change of reaction, $\Delta S$ is the entropy change, and $T$ is the absolute temperature. For a reaction to take place spontaneously, the free energy change must be negative (i.e., $\Delta G < 0$). In the specific reaction you're asking about, it's very probable that the change in entropy is negative, since a gas is being consumed and an equal total number of moles products are being formed as moles of reactants, so that the end result is an increase in order with respect to the reaction mixture. However, the formation of products results in a net release of heat (i.e., $\Delta H$ is negative), which increases the entropy of the surroundings sufficiently, such that, overall, entropy is increased in a global sense.

Ultimately, what drives this reactions forward is its exothermic nature (under standard conditions). In this reaction, the formation of stable chemical bonds releases energy as heat. If one were to attempt this reaction at very high temperatures, it would eventually fail, as the increase of the $T\Delta S$ term would at some point dwarf the enthalpy change.

There are other factors at work as well, including chemical equilibrium. By Le Chatelier's principle, the process can be made favorable by, e.g., maintaining a continual excess of reactants, or removing products formed from the reaction mixture as it proceeds.

Edit: Here's one source which indicates 348cal/g energy released, to put a specific number on the process. You could also calculate the enthalpy change using tables of standard thermodynamic data.


There are a lot of organic chemicals[citation needed].

Due to the incomprehensibly vast number of chemical species and variation between them, almost every classification system is domain-specific.

In terms of pure chemistry, the best you can do is usually to organize them according to what functional groups are in play; this will usually reveal something about the fundamental chemical interactions available to the species in question, but this is not necessarily useful or enlightening:

In many applications, such as as antifreeze, fuel or a solvent, methanol and ethanol are (largely) interchangeable: They both have short carbon chains attached to a single alcohol group (-OH), their boiling points, dipole moments, chemical energy and other properties are similar, and they mostly react in the same way on a chemical level.

In other applications, however, they are as  night and day: In biology, methanol is a potent poison, and ethanol is a mostly harmless intoxicant.

Organic chemistry is, in short, notoriously resilient to generally useful classification systems.


Let me add a bit more to the answers already given. As has been said, $\mathrm{pH}$ is nothing but a measure of the activity of protons ($\ce{H+}$) in a solvent - $\displaystyle\mathrm{pH} = -\log_{10} \ce{\ a_{H_{\text{solvated}}^{+}}}$. In dilute solutions, a solute's activity is approximately equal to its concentration, and so you can get away with saying $\mathrm{pH} = -\log_{10} \ce{[H_{solvated}^{+}]}$.

The "normal" aqueous $\mathrm{pH}$ scale that goes from 0-14 delimits the region where both $\ce{[H^{+}]}$ and $\ce{[OH^{-}]}$ are lower than $\pu{1 mol L^{-1}}$, which is about the upper limit where concentrations and activities of solutes in solution are approximately the same and they can be used interchangeably without introducing too much error.

(The reason this is true is because a solute in a very dilute solution behaves as if it surrounded by an infinite amount of solvent; there are so few solute species that they interact very little with each other on average. This view breaks down at high concentrations because now the solute species aren't far enough apart on average, and so the intermolecular interactions aren't quite the same.)

Not only is it rarer to talk about pH for very highly concentrated solutions, because activity has a much more subtle definition than concentration, but also because you have to produce quite concentrated solutions to get unusual values of pH. For example, very approximately, to get a litre of an aqueous solution of $\mathrm{pH} = -1$ using hydrogen perchlorate ($\ce{HClO4}$, a very strong monoprotic acid) you would need to dissolve around $\pu{1 kg}$ of $\ce{HClO4}$ in a few hundred $\pu{mL}$ of water. A litre of $\mathrm{pH} = 15$ aqueous solution using potassium hydroxide ($\ce{KOH}$) would also require several hundred grams of the hydroxide in about as much water. It's so much acid/base in a small amount of solvent that many acids/bases can't even dissolve well enough to reach the required concentrations.

There is also another way to come across unusual values of pH. We almost always talk about acids in bases in water, but they also exist in other media. Notice that in the definition of pH, no direct reference is made to water. It just happens that, for water, we have:

$$\ce{H2O (l) <=> H^+ (aq) + OH^{-} (aq)} \\ K^{\pu{25^\circ C}}_{\text{autodissociation}}=k^{\pu{25^\circ C}}_\mathrm{w}=a_{\ce{H^+(aq)}} \times a_{\ce{OH^- (aq)} } \simeq \ce{[H^{+}]} \times \ce{[OH^{-}]}
 = 1.01\times 10^{-14}$$
$$\mathrm{pH}+\mathrm{pOH}=-\log\ k_\mathrm{w} \simeq 14 \ (25^\circ C)$$

However, for liquid ammonia, one has:

$$\ce{NH3{(l)} <=> H^+{(am)} + NH2^{-}{(am)}}
\\
K^{\pu{-50^\circ C}}_{\text{autodissociation}}=k^{\pu{-50^\circ C}}_\mathrm{am}=a_{\ce{H^+(am)}} \times a_{\ce{NH2^- (am)} } \simeq \ce{[H^{+}]} \times \ce{[NH2^{-}]} = 10^{-33}$$
$$\mathrm{pH}+\mathrm{p}\ce{NH2}=-\log\ k_{\mathrm{am}} \simeq 33 \ (-\pu{50^\circ C})$$

($\ce{(am)}$ stands for a substance solvated by ammonia). Hence, in liquid ammonia at $-50^\circ C$, the pH scale can easily go all the way from 0 to 33 (of course, it can go a little lower and higher still, but again now activities become important), and that neutral pH is actually 16.5.

For pure liquid hydrogen sulfate (some difficulties arise as $\ce{H2SO4}$ is a diprotic acid and tends to decompose itself when pure, but putting that aside and looking only at the first dissociation):

$$\ce{H2SO4{(l)} <=> H^+{(hs)} + HSO4^{-}{(hs)}}\\
K^{\pu{25^\circ C}}_{\text{autodissociation}}=k^{\pu{25^\circ C}}_\mathrm{hs}=a_{\ce{H^+(hs)}} \times a_{\ce{HSO4^- (hs)} } \simeq \ce{[H^{+}]} \times \ce{[HSO4-]} \simeq 10^{-3}$$
$$\mathrm{pH}+\mathrm{p}\ce{HSO4}=-\log\ k_{\mathrm{hs}} \simeq 3 \ (\pu{25^\circ C})$$

($\ce{(hs)}$ stands for a substance solvated by hydrogen sulfate). Thus, the pH in liquid hydrogen sulfate has a much smaller range than in water and ammonia, not going much further than interval from 0 to 3.

Notice also that not only does the range of $\mathrm{pH}$s change in each solvent, but there is no direct relationship between the values of pH between different solvents; formic acid in ammonia would behave as a strong acid and as such a $\pu{1 M}$ solution in ammonia would have a pH close to 0, but formic acid is actually a base in liquid hydrogen sulfate, and as such a $\pu{1 M}$ solution would have a pH above 1.5. An approximate comparison between the pH ranges and their relative positions in different solvents can be found in the figure below, from A Unified pH Scale for All Phases.


What is happening when KCl burns in a bunsen flame?

I wanted to elaborate a little bit more so that the OP does not develop misconceptions. It is the potassium atoms not potassium ions that color the flame purple. While I was writing the answer, another poster emphasized the same point.
Caveat: In ordinary Bunsen burner type flames, emission from calcium, Sr, and Ba is from molecular compounds not atoms. The story of atomic emission is only true for alkali metals. The flame temperature is quite low in Bunsen burners. You need special flames to "see" the atomic emission of Ca, Ba, Sr.
Assume the processes when you introduce KCl in the flame:
a) The salt has to melt KCl (l)
b) It has to vaporize KCl (gas)
c) It has to decompose into constituents. Flames have free electrons, so potassium ions are converted to potassium atoms.
d) From the thermal energy of the flame, potassium atoms are excited. Once they de-excite, you see the violet/lilac color. It is an ultrafast process!
e) The flame temperature is so "low" on the scale of universal temperatures (stars, Sun) is that only 1-2% atoms get excited.
Now you may ask, what about the chlorine atoms, well, they emit in the deep ultraviolet and we cannot see them. The flame temperature is so low that it cannot excite chlorine atoms.
Keep in mind it is not a single wavelength. If you were lucky enough, you could use a pocket spectroscope and see red and violet lines (plus a persistent yellow line of sodium contamination).
Actually the strongest lines for potassium are 764 and 769 nm which are deep red. Many humans cannot see this deep red. So your flame would rather be deep red. Potassium atoms also emit less strongly at the other extreme end of the spectrum, corresponding to deep purple lines around 404 nm. The resulting "mix" of wavelengths appears to our eyes as lilac color!


Are there any gases more dense than liquids?

It depends on the conditions. Let's start decomposing your question in two related questions:

Denser gasses at SATP?

There are indeed some gasses that are quite dense. Sulfur hexafluoride has a density of 6.17 g/L  while tungsten hexafluoride of 12.4 g/L. But usually, they are not so dense compared to the density of liquids.

Lightest liquid at SATP?

The density of liquids largely depends on the atomic mass of the compounds. Hydrogen that is the compound with the lowest atomic mass has a density of 70.85 g/L, which is probably the lowest density you can find.

So at room temperature is not possible but if you increase the pressure the density of the gasses will increase while the density of the liquid won't increase much because liquids are not appreciably compressible.  It is hence theoretically possible to achieve a gas with a greater density compared to that one of a liquid, and also the coexistence of the two phases should be possible in certain conditions but the pressure required won't allow you to observe it.

Bubbles sinking down?

That won't happen also if there was a denser gas compared to a liquid. Bubble formation is quite a different process in this case inverting the gravity would be better! ;-)


Very interesting question - and thorough. But I would like to get proof that the spot is due to moisture. #1 Does the spot grow and shrink dependent on the relative humidity at that position? #2 If the spot is misted with a water sprayer, is it sharply defined? #3 If the spot is gently warmed with a space heater, does it disappear? #4 What color is the concrete? Very light (low iron), medium gray (medium iron), or dark, or brown?
If multiple cat pees have been located there, it's possible that attempts at cleaning or deodorizing have also been applied to the carpet, and have soaked thru and into the concrete. At this point, so much could have happened to the cement compounds that you and your parents should consider yourself fortunate that the deviation from perfection is only visual and not oderiferous! I have a spot on an outside patio that is dribbled upon from an aluminum gutter, only occasionally, when it rains (the end cap on the gutter has a tiny leak). The spot is about 4 inches in diameter, and is only visible at certain times (usually high humidity, like when rain has almost completely evaporated). Without a microscopic examination, I can't see anything unusual, but I suspect that some of the $Ca(OH)_2$ in the concrete has been dissolved, or perhaps carbonated more, or both, and the water in the pores at that spot is slower to evaporate.
If the discoloration is obnoxious, you might try painting (or dusting) that area with a non-paint paste of calcium carbonate in water. It would be harmless to the concrete, but provide some lightening. But then the area might be too white...


Bonding metal pieces using electroplating - what's the process called?

There's a 1974 Sandia labs paper on "Joining by Electroplating", abstract and citation available here.  They recommend tapering the joint and plating a triangular void closed, which brings up the important point that you need to let fresh electrolyte in to reach the joint as it builds up material.  For example, trying to electroplate two flat disks face-to-face would likely result in most of the deposit forming a ring around the disk edges, sealing a layer of corrosive electrolyte between the disks rather than securely welding them together.

The terminology for this sort of technology has rather permeable boundaries, but includes:

Electroplating, normally aiming to deposit a mostly uniform surface coating over an existing substrate.  Common in a variety of consumer goods, often for corrosion protection.
Electroforming, depositing material over a removable mandrel.  Common in micromanufacturing.
Electrowinning, extracting or purifying material.  Common in mining.


What are we seeing in this AC salt-water copper-wire experiment?

Is there really "a reaction of the copper with the salt in water" as the author of the video claims? Or is it a pure physical process where the copper "boils off" the wire?

Yes it reacts, no it doesn't boil. Essentially the wire corrodes at an accelerated rate where during the part of the cycle where the wire is the positive terminal the copper reacts as follows:

$$\ce{Cu^0 + 2OH- -> Cu^2+ + 2e- + 2OH- -> Cu(OH)2 + 2e- -> CuO + 2e- + H2O }$$
(expanded for clarity, not a reaction mechanism)

There is obviously oxygen generated too but I am only focusing on the copper part of the reaction. During the part of the cycle where the wire is the positive terminal the reverse of this reaction does not occur on the other half of the cycle and over time more and more copper will be consumed eventually consuming the entire wire.


How does Gaussian 09 normalize the excitation coefficients?


Role of NaI in Formation of Silyl Enol Ethers using Chloro(trimethyl)silane

The use of $\ce{NaI}$ in $\ce{MeCN}$ (a solvent in which $\ce{NaI}$ is quite soluble) is forming $\ce{Me3SiI}$ from the $\ce{Me3SiCl}$. The reason for doing this is $\ce{Me3SiI}$ is significantly more reactive with the triethylammonium enolate and will increase the yield of the TMS-enolate.

You are correct in thinking that the white fume you see when adding the $\ce{Me3SiCl}$ is $\ce{HCl}$. The product of aqueous hydrolysis of $\ce{Me3SiCl}$ is $\ce{HCl}$ plus hexamethyldisiloxane (Wikipedia):

$$\ce{2Me3SiCl + 2H2O -> 2HCl + Me3SiOSiMe3}$$


Can ascorbic acid be neutralized and keep it's elastic properties in dough?

There is high probability the desired properties of ascorbic acid are directly related to acid being acidic and in the free acid form, i.e. not in the form of the ascorbate.

It is hard to advise, not being familiar with this part of the applied food chemistry, it the effect of ascorbic acid happens in the cold dough conditioning stage and/or in the stage of the later thermal processing, in case of noodles.

If the sodium carbonate is added later to the dough, it is unclear, how well would be the dough mixed to neutralized the acid, and what the sideeffect would be.

Bakerpedia.com  explicitly advices against using ascorbic salts.

I highly advice to experiment, as theory will probably not provide clear advice.
I am not sure, if the acid is applicable to the noodle dough as much as for bread dough.


Coordinate bonds are covalent bonds where both bond electrons stem from only one of the two bond partners. They are formed when a Lewis base donates two electrons into accepting orbitals of the Lewis acid. That's not what you're looking at in this case, where both $\ce{P}$ and $\ce{O}$ contribute one electron to the common $\sigma$ bond. The interesting part is about everything that is to be described beyond the bond order of $1$. For example, everything that contributes to PO bonding beyond that single $\sigma$ bond.
I will further regard to phosphane oxides $\ce{P(=O)R3}$, where the question was originally asking about $\ce{P=O}$ bonds in $\ce{P4O10}$. The problem might be reduced to PO bonding in phosphoric acid, as $\ce{P4O10}$ is nothing more than its anhydride. The general observations regarding the $\ce{PO}$ bonding situation are unaffected. Kutzelnigg noticed in $1977$ already:

[The PO] bond is somewhere between single and triple and [...] only by chance it may happen to be a double bond.

What we know about the '$\ce{P=O}$ bond':

Its length is remarkably shorter than the P-O bond ($\pu{1.43 Å}$ compared to $\pu{1.60 Å}$ for $\ce{P4O10}$ according to Wikipedia).

It is highly polar.

Its bond strength of about $\pu{540 kJ/mol}$ is higher than for a corresponding single bond.

A critical review regarding the bonding situation in phosphane oxides was given by Gilheany. If it comes to Lewis valence formula, one might find three reasonable structures:

None of those is suitable to represent all theoretical and experimental findings regarding the characteristics of the $\ce{PO}$ bond!
Some scientists prefer to use the zwitterionic representation A, where the predominant $\sigma$ character of the $\ce{PO}$ bond is highlighted, while its dipolar character is also taken into account. If one chooses to write a Lewis structure as shown in B, one implies (besides one  $\sigma$ bond) the contribution of a $\pi$ bond between oxygen p orbitals and phosphorus d orbitals. However, the $3$d orbitals of phosphorus are no suitable $\pi$ accepting orbitals for $2$p orbitals of oxygen, as they are quite high in energy. Besides compatible symmetry, atomic orbitals (AOs) of the atoms in question require to be somewhat similar in energy to allow for a combination to form molecular orbitals (MOs).
If one allows $\pi$ back bonding from oxygen into $\sigma ^*$($\ce{P-R}$) (anti-bonding orbitals between P and its substituents R), one might come up with Lewis valence formula C. However, it should be noted that the $\pi$ bond order for $\ce{Me3PO}$ is only $0.7$ - and not $2$, what is implied by structure C!


Consider the simplest possible NOE experiment, with two dipole-coupled protons, the
target I and nearby proton S.

Before the experiment there are four populated spin states: $\alpha_{I}\alpha_{S}$, $\beta_{I}\alpha_{S}$, $\alpha_{I}\beta_{S}$ and $\beta_{I}\beta_{S}$, where $\alpha$ indicates $m_{i} = \frac{1}{2}$ and $\beta$ indicates $m_{i} = -\frac{1}{2}$. The relative populations of the states at thermal equilibrium (from Boltzmann distribution) are:

$\alpha_{I}\alpha_{S}: 1 + 2\Delta $
$\beta_{I}\alpha_{S}$,  $\alpha_{I}\beta_{S}: 1$
$\beta_{I}\beta_{S}: 1 - 2\Delta $

S is irradiated so that the population difference between the $m_{S} = \pm \frac{1}{2}$ states goes to zero. This results immediately in the following populations:

$\alpha_{I}\alpha_{S}, \beta_{I}\alpha_{S}: 1 + \Delta $
$\beta_{I}\beta_{S}, \alpha_{I}\beta_{S}: 1 - \Delta $

What happens next typically for small molecules is fast "flip-flip" cross-relaxation, restoring the $\alpha_{I}\alpha_{S}, \beta_{I}\beta_{S} $ states to their equilibrium populations. So we have the populations:

$\alpha_{I}\alpha_{S}: 1 + 2\Delta $
$\beta_{I}\alpha_{S}: 1 + \Delta$
$\alpha_{I}\beta_{S}: 1 - \Delta $
$\beta_{I}\beta_{S}: 1 - 2\Delta $

The population difference between the pairs $\alpha_{I}\beta_{S}, \alpha_{I}\alpha_{S}$ and  $\beta_{I}\beta_{S}, \beta_{I}\alpha_{S}$ has now increased by 50%, from $2\Delta$ to $3\Delta$. This corresponds to a 50% increase in intensity of the peak. Transition between these pairs obviously corresponds to spin-lattice relaxation.


Do some particles get ionised to 2+ or higher in a mass spectrometer or do they only get ionised to 1+?

In electron impact, you would see mostly +1 charged ions. Electron impact is a "harsh" way of ionization. However, +1 charge is not a universal rule in MS. The state of ionization depends on the ionization source. Later in your courses you might encounter electrospray ionization. There you might have +14 charges, etc.


Why is this β hydrogen in trans position to the leaving group?

Your assessment of which groups are trans to each other is correct but you have not taken into account that the molecule can adopt other conformers.
The classic confomation for E2 elimination is when the hydrogen and the leaving group are trans. This is because it maximises the orbital overlap between the $\ce{C-X \sigma ^*}$ and the $\ce{C-H \sigma}$ orbitals which are key to the reaction.

As shown in the question, the molecule will not undergo elimination without rotating into the appropriate conformation where the hydrogen and the leaving group are trans. This leads to the product shown.
Other points to note:

However, in the case of ammonium and sulphonate salts, the least substituted olefin is formed as a major product

This seems to be a complete red herring because there is no option for variably substituted products in this case (or in any other case of E2 elimination that I can think of?).
E2 elimination is also possible when the hydrogen and leaving group are syn, although it is usually less favourable. However, in this case the positive charge on the leaving group and the negative charge on the base might promote this pathway, leading to a different product with the two phenyl groups cis.


In Advanced Drug Delivery Reviews, 2001, 46, 3-26, (DOI) Lipinski and coworkers write:

  Too many hydrogen bond acceptor groups also hinder permeability across a membrane bi-layer. The sum of $\ce{N}$s and $\ce{O}$s is a rough measure of H bond accepting ability.

So, for the empirical rule of five it's simply the number, nothing more.

In the cited article, the situation for amides is addressed on page 19 in the context of log P calculations and correction factors that increase lipophilicity. Here, the authors write:

  PRX, a proximity correction factor for nitrogen
  and oxygen atoms that are close to one another
  topologically. [...] In addition, for each
  carboxamide group, we add an extra 1.0 and for
  each sulfonamide group, we add 2.0


Why do all radioactive decay series terminate at lead?

There are four main decay chains for actinides and superheavy elements. This is a simple consequence of the fact that one of the main processes to increase a heavy nucleus' stability is the emission of alpha particles, which have a mass number of 4 ($\ce{^4_2\alpha}$); notice that if you take the isotope's mass number and divide it by 4, the remainder of this division (0, 1, 2 or 3, corresponding to the $4n$, $4n+1$, $4n+2$ and $4n+3$ mass number decay chains respectively) stays constant under alpha, beta or gamma decay. There are decays which change the remainder (neutron emission, proton emission, spontaneous fission, etc) and therefore allow hopping between the main decay chains, but for simplicity these can be approximately ignored for the isotopes which are not too unstable or too heavy.

Only three out of these four main decay chains "stop" at lead, so already the statement in the question is incorrect. The $4n$, $4n+2$ and $4n+3$ chains "stop" at $\ce{^208_82Pb}$, $\ce{^206_82Pb}$ and $\ce{^207_82Pb}$, respectively. The $4n+1$ decay chain reaches $\ce{^209_82Pb}$, but this lead isotope is quite short-lived ($t_{1/2}=3.25\ \mathrm{h})$ and decays further to $\ce{^209_83Bi}$ where it "stops" (or at least seemed to until 2003).

Note that I said the chains "stopped" at these isotopes. That's because in reality they don't. It just happens that $\ce{^208_82Pb}$, $\ce{^209_83Bi}$, $\ce{^206_82Pb}$ and $\ce{^207_82Pb}$ are all very long-lived isotopes, with half-lives billions of times greater than the current age of the Universe. This means that any continued decay is severely bottlenecked at these points, so for almost all practical purposes, the decay chains stop there. However, rigorously speaking, if you were really patient there would be further decays in the sequence. In fact, theoretically the heaviest isotope which is not susceptible to any known mode of spontaneous radioactive decay (except proton decay) is an isotope of zirconium, $\ce{^92_40Zr}$, so the decay chains actually go on well past lead or bismuth. It just takes an immense amount of time for the decays to happen.


When the intermolecular forces in a liquid increase, does the equilibrium vapor pressure of the liquid at a given temperature increase or decrease?

Yeah your hunch is right. Higher the intermolecular forces between the liquid particles, harder it is for it to escape into the vapor phase, ie., you need more energy to convert it from liquid to the vapor
phase, in other words, higher its boiling point.

If it is harder for a liquid to escape into vapor, because it is held back into the liquid by the neighboring particles attraction, you have less vapor and hence low vapor pressure.

So in nutshell, if the intermolecular forces are increased, the equilibrium vapor pressure of the liquid will decrease.


What's the chemistry behind only charging a Lithium-ion battery to 80% capacity at most, to increase its lifespan?

Most commercially-available lithium-ion batteries use graphite as the anode (negative electrode, where the lithium ions go when the battery is charged) and $Li Co O_2$ as the cathode (positive electrode, where the lithium ions go when the battery is discharged). Both graphite and lithium cobalt oxide have a structure composed of layers stacked on top of one another; the lithium ions fit in between these layers in a process called intercalation.

The lithium sitting between the layers props the layers apart. If too much of the lithium is removed from either the cathode (over-charging) or the anode (over-discharging) then the layers can collapse back together and it becomes hard to put the lithium back in. Normally batteries have control circuitry to prevent the cells from over-charging or over-discharging but this circuitry is not always perfect, particularly for smaller, simpler battery packs.

That said, AFAIK the circuitry is already set to limit charging and discharging to well below the theoretical maximum capacity, so I would not recommend limiting the charge any further. Perhaps unscrupulous vendors will cut the amount of active material inside their batteries while keeping the rated capacity the same and therefore run the cells closer to their theoretical maximum, in which case keeping charge and discharge within their rated limits would be a good idea, but it should not be necessary for high-quality batteries.


Bond order has to do nothing with ionization energy, it is the bond dissociation energy which has to be dealt with bond order. For minimum ionization energy, the electron has to be lost from the highest energetic orbital (outer).

In $\ce{C2^-}$, the electron is lost from high energetic $\sigma(\ce{2p_x})$ (if you take the bonding along x-axis) orbital, whereas, in the other two molecules electrons are lost from low energetic  $\pi(\ce{2p_y})$ or $\pi (\ce{2p_z})$ orbitals. So $E_{\infty} - E_\text{initial}$ is higher for the latter case, which makes $\ce {C2^-}$ having lower ionization energy.


What's the difference between radial probability and probability density?

Probability density is probability per volume.  Probability density at a given point means probability per volume in the limit that the volume is infinitesimally small.

Radial probability distribution at a given radius is the probability per distance that the event occurs in a infinitesimally thin spherical shell at that radius. (The distance being the thickness of the shell).


Why Birch reduction of alkyne is E-selective?

It's not rotation but inversion (remember amine lone pair inversion) from a linear $sp$ form to trigonal planar $sp^2$ that occurs. Lone pair and radical electron are held close to atom and repulse each other a lot so stable to keep them in anti dihedral positions. If the lone pair next to a radical electron was less repulsive, then it would have a lower energy transition state and be more readily formed according to Hammond's Postulate. But experimental data indicates that that is not the case.

Picture Citation: Kurti (2019)


Yes, this compound is chiral.

The polycyclic backbone is called adamantane. It has $T_\mathrm{d}$ symmetry, meaning that as far as chirality goes, it behaves like a perfect tetrahedron, somewhat like methane does.

It also has the interesting property that if you extend the C-F, C-Cl, ... bonds inwards, they will all meet at the same point. Those red dotted lines form a perfect tetrahedron.

Now, since the adamantane backbone behaves exactly like a methane molecule - why don't we just replace it with one? [n.b. this is not 100% rigorous but works here.]

Hopefully you'll agree that this is chiral. It basically behaves like it has a chiral centre right in the middle of the adamantane backbone.

If you really wanted to be sure, you could build a mirror image and verify that they are non-superimposable. (Hint: they are.)

Each of the four substituted carbons is connected to four different groups, and theoretically one would expect $2^4 = 16$ stereoisomers. However, because three of the four groups (the polycyclic backbone) are held in a rigid, fixed geometry, this number is reduced to 2.

There is more discussion of this class of compounds here.


Energy cost of C−N rotation in hydroxamate group

Reference 1 reports the isomerization barrier for N‐methylacetohydroxamic acid. From the abstract:

  We have now determined the activation barrier to C–N rotation from major to minor conformation of a small secondary HA in $\ce{D2O}$ to be 67.3 kJ/mol. The HA rotational barrier scales with solvent polarity, as is observed in amides, although the HA barrier is less than that of a comparable tertiary amide in aqueous solution.

N‐methylacetohydroxamic acid (image from PubChem):

References

Stefanie P. Sippl, Paul B. White, Charles G. Fry, Sarah E. Volk, Lingxiao Ye, Heather L. Schenck. Kinetic and thermodynamic characterization of C–N bond rotation by N‐methylacetohydroxamic acid in aqueous media. Mag Res Chem 54 (2016), 46-50. DOI: 10.1002/mrc.4318.


Molybdenum/Tungsten alloy armor piercing round

Probably not. Molybdenum and Tungsten are both hardening alloys to steel, they perform the same function, though differently. I suspect a Mo - W alloy would have the fragility of both molybdenum and tungsten while not hardening each other especially much nor increasing density.

The answer is: Sure, it would be hard, heavy and likely armor piercing, but very inefficient use of both Mo and W. Just use W alone, or alloy it with elements that are lighter, but melt within reasonable ranges. A quick glance at wikipedia confirms that Fe, Ni and Co are usual alloying elements. Tungsten carbide is also used.

Creating this Mo-W alloy would also be very difficult, usually you solvate the high mp material in the liquid low mp material but these both do not melt within reasonable limits so you would have to use some seriously high temperature equipment, we are talking beyond bright white heat - or rely on very, very slow processes (diffusion - and these are large atoms...).


Why don't effervescent vitamin C tablets react with the vitamin C?

Acid-base reactions are fast and reversible (barring any additional reaction that occurs). That is, once you move the conjugate base of ascorbic acid to a low-pH environment (e.g. your stomach), it rapidly reverts to the protonated form. (Ascorbic acid does not undergo any pH-dependent further reactions, as opposed to carbonate, which when protonated to carbonic acid can decompose to carbon dioxide and water.)

Actually, as the pKa (roughly the "halfway pH" of acid/base reactions) of ascorbic acid is 4.2, the moment the protonated ascorbic acid enters your bloodstream (roughly pH 7.3), it immediately loses that proton again, reverting to the deprotonated ascorbate ion.

So in a practical sense, it really doesn't matter which form you take your Vitamin C in - either (protonated) ascorbic acid or (deprotonated) ascorbate acts in you body the same way, because they can rapidly interconvert based on the pH of the solution they happen to be in at the time.

(Because they typically work in buffered aqueous environments, biochemists typically play fast-and-loose with the difference between the protonation states of molecules. Unless you're explicitly dealing with pH-dependance, they're typically lumped all together, with the implicit understanding that at a given pH they'll adjust to the appropriate relative concentration. This can get confusing if you're a stickler for nomenclature.)


What is the correct number of stereo isomers in case of 3,6-dimethylpiperazine-2,5-dione?

Three is the correct answer.

As Ivan pointed out, there are no planes of symmetry, for example:

Ivan is also correct - the "RS" is the same as the "SR". You can prove this to yourself most easily using molecular models, but here is how to see it on the page. You just need to rotate the molecule by 180° in the plane of the page:

So, despite having chiral centres, this compound is achiral because it is superposable with its mirror image. "But wait, I was told that a molecule needs to have an internal plane of symmetry to be a meso compound" I hear you say. Unfortunately, this is an incomplete truth that many sources repeat. To be a meso compound, it can alternatively have a centre of symmetry, also known as an inversion centre. This is simply one point in space through which any atom is reflected through - ie. each atom is diametrically opposite an equivalent atom at an equal distance from this point.

The other two stereoisomers are the (R,R) and the (S,S). These are non-superposable mirror images, and hence enantiomers. Note these isomers have no plane of symmetry or centre of symmetry.

Once again, this is easier to see with physical models.


Both quantities are of the real gas.

Note that for the ideal gas behaviour, the compressibility factor $$Z=\frac{V_\text{real}}{V_\text{ideal}}=\frac{pV_\text{real}}{nRT}=1$$
 for any combination of $p, V, T$.

But real gases, in contrary to the ideal gas, have nonzero volume of molecules and there are intermolecular interactions.

The volume of molecules increases $Z$, as it makes the pressure higher, because the collision frequency is higher.

The cohesive forces between molecules decreases  $Z$. As molecule attraction, like hydrogen bonds or dipole interactions, decreases the effective number of molecules and therefore pressure.

This is reflected in the van der Waals equation - probably the simplest state equation of real gases, for not too high pressure:
$$\begin{align}
\left(p + \frac{an^2}{V^2}\right)\left(V - nb\right)&=nRT \\
\left(p + \frac{a}{V_\mathrm m^2}\right)\left(V_\mathrm m - b\right)&=RT
\end{align}$$

$$\begin{align}
Z &= \frac{p\cdot V_\text{real}}{nRT} \\
Z&=  \frac{p\cdot V_\text{real}}{\left(p + \frac{an^2}{V_\text{real}^2}\right)(V_\text{real} - nb)} \\
Z&=  \frac{1}{\left(1 + \frac{an^2}{p \cdot V_\text{real}^2}\right)\left(1 - \frac{nb}{V_\text{real}}\right)} \\
Z&=  \frac{1}{\left(1 + \frac{a}{pV_{\mathrm m,\text{real}}^2}\right)\left(1 - \frac{b}{V_{\mathrm m,\text{real}}}\right)}
\end{align}$$

For $|x|\ll1, 1/(1+x)=1-x$

For not too far from ideal behaviour, we can apply the above approximation.

$$\begin{align}
Z&=\left(1 - \frac{an^2}{pV_{\mathrm{real}}^2}\right)\left(1 + \frac{bn}{V_{\mathrm{real}}}\right) \\
Z&=\left(1 - \frac{a}{pV_{\mathrm m,\text{real}}^2}\right)\left(1 + \frac{b}{V_{\mathrm m,\text{real}}}\right) \\
\end{align}$$

We can therefore also afford to neglect minor terms.

$$\begin{align}
Z&=1 - \frac{an^2}{pV_{\mathrm{real}}^2}+ \frac{bn}{V_{\mathrm{real}}}\\
Z&=1 - \frac{a}{pV_{\mathrm m,\text{real}}^2}+ \frac{b}{V_{\mathrm m,\text{real}}}\\
\end{align}$$

To address clarified scenario, as a the same amounts of the ideal and real gas
are under the same temperature and pressure:

$$V_\text{ideal gas}=\frac{nRT}{p}$$
$$V_\text{real gas}=Z \frac{nRT}{p}$$

There is need to perform iteration for probably the easiest way to get the result.
With the $V_{\rm{ideal gas}}\ $ as the first approximation. The other option is to solve the root of the cubic equation what we probably do not want to.

$$\begin{align}
V_\text{real}&=Z(p, V_\text{real})\frac{nRT}{p}\\
V_\text{real}&=\frac{1}{\left(1 + \frac{an^2}{p \cdot V_\text{real}^2}\right)\left(1 - \frac{nb}{V_\text{real}}\right)} \frac{nRT}{p} \\
V_\text{real}&=\left( 1 - \frac{an^2}{pV_{\rm{real}}^2}+ \frac{bn}{V_{\rm{real}}}\right) \frac{nRT}{p} \\
\end{align}$$

The van der Waals equation can also be expressed in terms of reduced
properties

$$\left( P_r + \frac{3}{V_\rm{r}^2}\right) \left( V_{\rm r}-\frac{1}{3}\right) = \frac{8}{3}T_\mathrm r$$

This yields a critical compressibility factor $3/8$.

The values of pressure, temperature and volume are divided by the
respective critical values of the given gas.


How deep does a counter electrode need to penetrate the solution in electrochemical experiments?

TL;DR As long as you are not getting overloads on your instrument, it is submerged enough.

Longer answer:
Your potentiostat will increase the potential on your counter electrode until it achieves the desired potential across the working electrode(WE) reference electrode(RE) circuit. The exact potential applied on the counter will depend on a lot of factors like solution resistance, amount of current flowing on the working at the potential desired, material your counter is made of and finally, the area of the counter electrode.

If you end up with a situation where your potentiostat hits its electronic limits before the potential difference WE-RE is the desired value, you will start seeing overloads.

If you are curious, you can always put a voltmeter between your WE and your counter electrode to see how much of a potential your instrument applies.


Can two atoms with the same amount of electrons but differing electronic configurations be considered isoelectronic?

$(\ce{Y+})$ with configuration: $\ce{[Kr] 5s^1 4d^1}$ and strontium $(\ce{Sr^0})$ with the configuration $\ce{[Kr] 5s^2}$ are not isoelectronic. In order to be isoelectronic the two atoms would have to have the same electronic structure as well as the same number of valence electrons.

IUPAC Goldbook

  isoelectronic - Two or more molecular entities are described as isoelectronic if they have the same number of valence electrons and the same structure, i.e. number and connectivity of atoms, but differ in some of the elements involved.


Cheap metal to replace platinum in hydrogen fuel cell

We have tinkered with this a little. Currently, a handful of others have been experimenting with certain blends of metals, using lower costs to create equal reactions.

I have heard a lot of different ratios of Ni/Cu and Pd blends. But a good experimental blend (if your making a reasonable size batch), would comprise of a ZrO2–Pd(x)D core, with (x) varying as a slew of other metals.
 This article is regarding the direct deposition of hydrogen on palladium, probably a little more refined towards what you where searching.

While im not sure how intricate your going to get with your experimental cell, here is some good food for thought., If you can create a vessel with proper pressure safety, academic groups have been experimenting with super critical fluids, trying to test metallic blends as catalysts under states past standard classical critical conditions.


Are there ethanol and methanol analogues with more than one hydroxyl group?

Generally speaking, they're called diols when bearing two hydroxyl groups. If they're on the same C, this is called a geminal diol, if they are on neighboring Cs, they're called vicinal diols. The C on geminal diols has a higher oxidation state than that of an alcohol, i.e. which is why they could be seen as relatives of an aldehydes (or ketones).
In a diluted aqueous solution, formaldehyde $\ce{H2C=O}$ actually predominantly exists in its geminal diol form $\ce{H2C(OH)2}$. This is no longer true for 1,1-ethanediol (the hydrated form of acetaldehyde).

The simplest vicinal diol 1,2-ethanediol is better known under its trivial name ethylene glycol (a common ingredient of anti-freeze formulations).

Triols with the hydroxyl groups on the same C (with the oxidation state of carboxylic acids) are generally unstable.


Let us take the example of your equation:
$$\ce{2SO3<=>2SO2 + O2}$$

The stoichiometric coefficients represent two facts:

If 2 mole $\ce{SO3}$ reacts, it will give 2 mole $\ce{SO2}$ and 1 mole $\ce{O2}$ but one needs to remember here that all of the 2 moles will not react. But the number of moles that do dissociate, give products in the above ratio i.e. if initially have 2 moles of $\ce{SO3}$ and degree of dissociation is 50%, then we will have 1 mole of $\ce{SO3}$ reacting to give 1 mole of $\ce{SO2}$ and half mole of $\ce{O2}$
The stoichiometric coefficients also provide the powers to which the concentrations of each compound will be raised to in the expression for equilibrium constant $$K_\mathrm c = \frac{[\ce{SO2}]^2[\ce{O2}]}{[\ce{SO3}]^2}$$


Why rutile structure have primitive unit cell instead of body centered?

You said you read that $\ce{MgH2}$ has a structure of rutile. Actually it is $\alpha$-$\ce{MgH2}$, which forms at room temperature and atmospheric pressure and have rutile type structure. The crystal structure of $\alpha$-$\ce{MgD2}$ was studied as early as 1963 (Ref.1). The abstract states that:

  A magnesium deuteride preparation of composition $\ce{Mg(D_{0.9}H_{0.1})2}$ is found to be tetragonal with $a = 4.5025, c = \pu{3.0123 \mathring A }$. The atomic positions are: $\ce{2 Mg}$ in $(000)(\frac12 \frac12 \frac12)$, $\ce{4 D + H}$ in $\pm (xx0)(\frac12 +x, \frac12 –x, \frac12)$ with $x = 0.306 \pm 0.003$. The bond lengths are $\ce{Mg-6D} = \pu{1.95 \pm 0.02 \mathring A }$. The structure is of rutile type.

The authors state that the choice of $\ce{MgD2}$ in place of $\ce{MgH2}$ because hydrogen atoms had no measurable effects on diffraction intensities (note that they were doing neutron diffraction studies). These studies were a followup to confirm their previous studied on $\ce{MgH2}$ (Ellinger, et al. in 1955; Ref.2). In that studies on $\ce{MgH2}$ with X-ray diffraction, it has been shown that $\ce{MgH2}$ crystallizes in the tetragonal system with $a_\circ = \pu{4.5168 \mathring A}$ and $c_\circ = \pu{3.0205 \mathring A}$. The measured density of the crystals was $\pu{1.45 \pm 0.03 g/cm-3}$, which has shown that there are two molecules in the unit cell (Ref.2). The calculated X-ray density is $\pu{1.419 g/cm-3}$.

The last statement answers your question. The rutile type structure provide two $\ce{MgH2}$ molecules in the unit cell. There are one $\ce{Mg}$ atom in the center (shares with one unit cell) and eight other $\ce{Mg}$ atoms at corners, each of which shares it with eight unit cells. Thus, number of $\ce{Mg}$ atoms in a unit cell is: $1 \times 1 + 8 \times \frac {1}{8}=2$. There are six $\ce{H}$ atoms in the unit cell. If you look closely, you'd realize two of them in inside the unit cell (shares with one unit cell) and rest are on two opposite surfaces, two in each surface (two $\ce{H}$ atoms in each surface share with two unit cells). Thus, number of $\ce{H}$ atoms in a unit cell is: $2 \times 1 + 4 \times \frac {1}{2}=4$. Hence, number of $\ce{MgH2}$ molecules in a unit cell is $\ce{Mg2H4 # 2 MgH2}$.

  The X-ray powder pattern shows that two $\ce{Mg}$ atoms in the unit cell are in a body-centered configuration, i.e. $\ce{2 Mg}$ in $(000)(\frac12 \frac12 \frac12)$. A consideration of all tetragonal space groups shows that there is one and only one possible choice of positions for the four $\ce{H}$ atoms leading to a plausible structure. This is,

  $$\text{space group }  P4/mnm \ \left(D^{14}_{4h} \right)\\
\ce{2Mg} \text { in } (000)(\frac12 \frac12 \frac12)\\
\ce{4H} \text { in } \pm (XX0)\left(X + \frac12, \frac12-X, \frac12\right) \ \text{with } X=0.306$$

  In this structure (rutile type), each magnesium is coordinated to six hydrogens at a distance of $\pu{1.95 \mathring A}$ and each hydrogen is coordinated to three magnesiums. One $\ce{H-H}$ distance is $\pu{2.49 \mathring A}$ and the others are $\pu{2.76 \mathring A}$. The distance of $\pu{2.76 \mathring A}$ compares favorably with the diameter of $\ce{H-}$ ion ($\pu{2.72 \mathring A}$) as found in $\ce{LiH}$. One short $\ce{H-H}$ distance is characteristic of one anion-anion distance of the rutile type structure.

By these statements, it can be concluded that the structure of $\alpha$-$\ce{MgH2}$ is indeed rutile type (not body-centered cube).

In another point of view, one would also suggest that the structure of $\alpha$-$\ce{MgH2}$ is face-centered cube, based on equal distances of $\ce{6 Mg-H}$ (see $\color{blue}{\text{blue box}}$  in following diagram; opposite face $\ce{H}$ atoms are highlighted in blue, red, and gray for visualization). However, again, calculations lead to $\ce{MgH3}$ in the unit cell, which would rule out the possibility.

The above diagram shows the crystal structures of other modifications of $\ce{MgH2}$ along with $\alpha$-$\ce{MgH2}$ (Ref.3 & 4). These other modifications are formed in high pressure situations. For example, when low-pressure phase $\alpha$-$\ce{MgH2}$ was heated at $\pu{1070 K}$ and in $\pu{2 GPa}$ pressure, the high-pressure phase $\gamma$-$\ce{MgH2}$ was formed in the orthorhombic (space group: $Pbcn$) structure (Ref.3):

  The high-pressure phase $\gamma$-$\ce{MgH2}$ was formed by heating the low-pressure phase $\alpha$-$\ce{MgH2}$ in a multianvil press at $\pu{2 GPa}$ pressure to $\pu{1070 K}$ for $\pu{120 min}$ and successive rapid quenching. Investigation by X-ray and neutron powder diffraction on the deuteride at ambient conditions revealed that it crystallises with the orthorhombic $\alpha$-$\ce{PbO2}$ type structure (space group $Pbcn, Z=4, a=4.5213(3), b=5.4382(3), c= \pu{4.9337(3) \mathring A }$ (hydride); $a=4.5056(3), b=5.4212(3), c= \pu{4.9183(3) \mathring A}$ (deuteride) at $T=\pu{295 K}$). The deuterium atoms surround magnesium in a distorted octahedral configuration with bond distances $\ce{Mg-D} = 1.915(3), 1.943(3)$ and $\pu{2.004(3)  \mathring A }$. The rutile structure of $\alpha$-$\ce{MgH2}$ was re-evaluated.

The bonding nature of $\ce{MgH2}$ is analyzed in recent studies with the help of charge-density, charge-transfer, electron-localization-function, and Mulliken-population analyses, which clearly have shown that all polymorphs of $\ce{MgH2}$  are to be classified as ionic materials with $\ce{Mg}$ and $\ce{H}$ in nearly 2+ and 1−  states, respectively (Ref.4).

References:

W. H. Zachariasen, C. E. Holley, Jr., J. F. Stamper, Jr., “Neutron diffraction study of magnesium deuteride ,” Acta Chrystallographica 1963, 16(5), 352-353 (https://doi.org/10.1107/S0365110X63000967).
F. H. Ellinger, C. E. Holley, Jr., B. B. McInteer, D. Pavone, R. M. Potter, E. Staritzky, W. H. Zachariasen, “The Preparation and Some Properties of Magnesium Hydride,” J. Am. Chem. Soc. 1955, 77(9), 2647–2648 (https://doi.org/10.1021/ja01614a094).
M. Bortz, B. Bertheville, G. Böttger, K. Yvon, “Structure of the high pressure phase $\gamma$-$\ce{MgH2}$ by neutron powder diffraction,” Journal of Alloys and Compounds 1999, 287(1-2), L4-L6 (https://doi.org/10.1016/S0925-8388(99)00028-6).
P. Vajeeston, P. Ravindran, B. C. Hauback, H. Fjellvåg, A. Kjekshus, S. Furuseth, M. Hanfland, “Structural stability and pressure-induced phase transitions in $\ce{MgH2}$,” Phys. Rev. 2006, B73, 224102 (https://doi.org/10.1103/PhysRevB.73.224102).


5-methylhexa-1,3-diene reaction with HBr

In this reaction if peroxide is not present then reaction follows markovnikov addition in which carbonation is formed as intermediate. By stability due to resonance and 2° carbonation
5-methylhexa-1-ene cation with C+ at 3 is formed. Hence 3-bromo-5-methylhexa-1-ene is formed as product.


If you read the question carefully, they wish to precipitate only one ion, either copper or lead. The facts that must be in your mind are:

(i) Lead sulfate is insoluble in water but copper sulfate is.
(ii) Both lead sulfide and copper sulfide are water insoluble.


Does a neutral dimercury molecule exist?

There has been recent research on mercury dimers in its ground state. Bond energies, transition energies, band spectrum and other spectroscopic parameters has been calculated. Here are the abstracts of two papers on mercury dimer:

  The potential energy curve of the ground electronic state of the $\ce{Hg}$ dimer has been calculated using the CCSD(T) procedure and relativistic effective core potentials. The calculated binding energy ($\pu{0.047 eV}$) and equilibrium separation ($\pu{3.72 Å}$) are in excellent agreement with experiment. A variety of properties, including the second virial coefficient, rotational and vibrational spectroscopic constants, and vibrational energy levels, have been calculated using this interatomic potential and agreement with experiment is good overall.(Source)

  The mercury dimer is among the most weakly bound metal dimers and has been extensively studied. The ground state $\ce{O^+_g}$ dissociation energy has been considered to lie between $\pu{0.55 eV}$ ($\pu{440 cm−1}$) and $\pu{0.091 eV}$ ($\pu{730 cm−1}$). We report here a spectroscopic study of $\ce{Hg2}$ in a supersonic jet. The first optical transition, $\ce{1_u←O^+_g}$, was characterized by its fluorescence excitation spectrum and the binding energy of the ground state has been measured precisely through the threshold of collision induced dissociation of $\ce{Hg2 1_u}$ to $\ce{Hg(^1S_0)+Hg(^3P_0)}$.(Source)


Could blue flames arise within the hull of a burning ship?

In a Bunsen burner, the gas (methane or propane) may burn under two practical regimes : reductive or oxidizing conditions. In the reductive conditions, the flame is yellow and unstable, because the amount of oxygen is low with respect to the stoichiometric ratio. In the oxidative conditions, the flame is noisy, strong and stable, and it is due to an excess of oxygen in the mixed gases before the combustion. The oxidative flame is nearly colorless, but it has a characteristic blue cone at its base. This blue cone has a temperature of about $900°$C, and its spectrum has been analyzed by Hertzfeld in $1935$ : it shows a line spectrum typical of the diatomic "molecule" $\ce{C_2}$, with plenty of vibrational and even rotational sublevels. Maybe that was the origin of the blue flame described by Joseph Conrad. Why not ?


How to dilute a stock solution when the resulting solution has a smaller volume?

Maybe the question was posed this way to make you think practically.

Approach this problem as if you were in a lab and you actually had a bottle of $2.0~\mathrm{L}$ of a $0.85~\mathrm{mol/L}$ magnesium hydroxide stock solution, and you wanted to make a new solution of $200.0~\mathrm{mL}$ of $0.30~\mathrm{mol/L}$ magnesium hydroxide.

You do not need to dilute the entire stock, but just a portion of the stock solution.  Describe what volume of stock solution you would take out, and how you would dilute it, using the $$c_1 V_1 = c_2 V_2$$ formula, to reach $200.0~\mathrm{mL}$ of $0.30~\mathrm{mol/L}$ of magnesium hydroxide.


Concentration and Half Life of reactants A and B

The whole reaction isn't first order and that can be arrived at using one simple concept. The unit of rate constant $k$ is $\pu{M^{-1} s^{-1}}$ and not $\mathrm s^{-1}$.

Since reaction is first order with respect to both reactant A and B, I can write,
$$ k=\frac{1}{t} \ln{\frac{a}{a-x}} $$

This would have been correct if and only if the whole reaction was a first order reaction. Not a first order reaction with respect to each reactant individually. For example:
$$\ce{A->B}\quad \text{Rate}=k[A]$$
That is not the case here. Here the reaction would be better phrased as:
$$\ce{A + B->C}\quad \text{Rate} = k[A][B]$$
Which means that this is a second order reaction and not a first order reaction.
Now, the other issue is that this second order reaction isn't of the form $\ce{2A-> B}$ where the rate would have been relatively easy to find using a generic law and hence we move to a more basic proof for a reaction $\ce{A + B -> C}$. As given above, the rate law for this expression is "Rate $= k[A][B]$" Now we draw a RICE table for this generic reaction to get:
\begin{array}{|l|c|} \hline
\mathrm{R} & \mathrm A & \mathrm B &\mathrm C \\ \hline
\mathrm I & a & b & 0 \\
\mathrm C & -x & -x & x \\
\mathrm E & a-x & b-x & x \\ \hline
\end{array}
Hence, we can see that for a time t, the rate of reaction can be defined as:
$$ \text{Rate} = -\frac{\mathrm d[A]}{\mathrm dt}=k(a-x)(b-x)$$
Differentiating $[A]$ with respect to $t$, we get:
$$-\frac{\mathrm d[A]}{\mathrm dt} = \frac{\mathrm dx}{\mathrm dt}$$
Substituting this in the previous equation and integrating with respect to x, we get:
$$kt = \frac{1}{b-a}\ln \frac{a(b-x)}{b(a-x)}$$
Now, in the given question, we are asked to find the value of $[A]$ and $[B]$ after $\pu{100 s}$. Therefore, $t=\pu{100 s}$, $a = \pu{0.1 M}\,,b=\pu{6.93 M}$ and $k = \pu{2 \times 10^-3 M^{-1} s^{-1}}$.
Solve for x in this equation and find the value of $a-x$ and $b-x$.
This isn't a complete solution, but the rest is just manipulating the expression using mathematics and has no actual relation to kinetics.
A hint to make calculations simpler would arise from noticing that $a<<b$ and so $b-a$, $b-x$ can both be approximated to $b$.


Why aren't atomic radii calculated using Schrödinger's equation?

Erwin with his psi can do
Calculations quite a few.
But one thing has not been seen:
Just what does psi really mean?

(Erich Hückel, english translation by Felix Bloch)

It is not easy to perform the calculation that you called simple. The Hamilton operator $\hat{H}$ for a multi-body problem of $K$ cores and $N$ electrons

\[
\hat{H} = - \frac{1}{2}\sum_a^K\frac{1}{M_a}\Delta_a - \frac{1}{2}\sum_i^N\frac{1}{M_i}\Delta_i - \sum_i^N\sum_a^K \frac{Z_a}{r_{ai}} + \sum_{i < j}^N\frac{1}{r_{ij}}+ \sum_{a < b}^K\frac{Z_aZ_b}{R_{ab}}
\]

is still a beast for $K=1$ with all the core-electron and electron-electron interactions.

So, while the question is legitimate, the calculation simply cannot be done without a lot of approximations, such as the Hartree-Fock method.

By the way, for the determination of the atomic radii of solid metals, there's x-ray diffraction, which I wouldn't call crude.


Edit: step count is 15, longest linear sequence of 9 steps

Just to (hopefully) get the ball rolling, here's something I scribbled down. I am sure people here will have better answers. And I have not done a lot of literature research to see whether the steps are plausible, so there might be fatal flaws... but hopefully not; as far as I know/can tell, everything here is relatively sensible.

Retrosynthesis first. I chose a Horner–Wadsworth–Emmons reaction to make the (E)-double bond, and something similar to the Reissert synthesis for the indole. It's not exactly the same because I didn't acylate the benzylic carbon, but the initial disconnection is similar. This leads to the key building blocks:

I chose the C(sp2)-C(sp3) bonds in 2 to disconnect. One can be made by SNAr with the amazing SNAr substrate 1-fluoro-2-nitrobenzene (5), and the other by Pd cross-coupling (I'm guessing cheaper ways probably exist). I've never actually seen SNAr reactions with enolates as nucleophiles, but I guess that if it doesn't work, SRN1-type conditions should still make that step possible:

As for 3, I figured it could be made from a symmetrical intermediate:

Forward synthesis

Synthesis of 3, starting from cyclopentadiene:

  (i) $\ce{BH3.THF}$, then $\ce{H2O2, NaOH}$; (ii) TBSCl, imidazole; (iii) $\ce{O3}$, then $\ce{H2O2}$; (iv) $\ce{TMSCHN2}$; (v) enzymatic desymmetrisation (I did not look up specifically which, but I think it should be possible); (vi) 7, NaH

Next, the synthesis of 2, starting from methylglyoxal:

  (vii) ethylene glycol, TsOH; (viii) Pd cat., ligand, $\ce{NaO^{t}Bu}$, 6; (ix) NaH, 5

Forming the indole:

  (x) Zn, AcOH; (xi) $\ce{NaH, ^{i}PrI}$; (xii) aq. HCl

and finally the HWE, the TBS deprotection, and the last redox step, a diastereoselective boron reduction. I wasn't entirely sure if it would work on an α,β-unsaturated ketone, but Comp Org Synth II, section 8.01 has several examples of it being used on an α,β-unsaturated ketones.

  (xiii) LiHMDS, 3; (xiv) $\ce{TBAF, H2O}$; (xv) $\ce{Et2B(OMe), NaBH4}$


Is Kirchoffs law valid at all pressures?

It would say that you are making three separate mistakes:

The standard state pressure is any arbitrary pressure, not necessarily 1 bar, so there is no implication from the equation specifying standard state that it is only true at one particular pressure.
The important specification of being in the standard state is not the value of arbitrary pressure chosen, but that the substances are not mixed.  They are pure or in solvent as the single solute.  Also, for gases and solutions the states are fictitious.  In your derivation, you are not accounting for changes due to mixing and any changes due to being in real, rather than fictitious, states.
I don't see why you say to integrate Cv rather than Cp.


Is this a plausible reaction mechanism? (H2SO4 cyclization after Friedel-Crafts)

I would propose the mechanism illustrated below, which would seem plausible in concentrated sulfuric acid:


OK, new route. Retrosynthesis:

Denopamine 1 can be made by a reductive amination between amine 2 and aldehyde 3 (either of the two reductive amination disconnections are fairly obvious choices).

The requisite amine 2 contains 10 carbons and so it becomes necessary to make it using some other means - in this case, reduction of a nitroolefin 4, which can itself be made via a nitroaldol reaction from 3,4-dimethoxybenzaldehyde 5 and nitromethane 6.

The chiral α-hydroxyketone 3 can probably be made in numerous ways. Earlier I proposed organoocatalytic oxygenation with PhNO, but another possibility is to have an addition of "formaldehyde anion" to 4-hydroxybenzaldehyde 7. Nitromethane 6 was chosen as the acyl anion equivalent, as the asymmetric Henry reaction is well-established methodology, e.g. by the Evans group (J. Am. Chem. Soc. 2003, 125 (42), 12692–12693).

The forward synthesis is fairly straightforward:

5 steps from cheap starting materials, with longest linear sequence being 3 steps.


The molecule seen in the tattoo is a protein-based mammalian hormone called oxytocin, commonly called "the love hormone" because it is involved in several aspects of sexual reproduction.

Image Source


Why does the nitrile carbon experience more shielding than a carbonyl carbon in CNMR?

TL;DR: It's complicated and likely has to do with paramagnetic shielding effects rather than the usual arguments based on electron density, ring currents, and so on.
Firstly, because of low-lying π* orbitals, π-bonded carbons typically have lower excitation energies than purely σ-bonded carbons and are thus more deshielded. However, due to the presence of a π-bond order term, alkynes experience less deshielding due to paramagnetic effects. So the general order of chemical shifts is: alkenes > alkynes > alkanes.
Carbonyls are more deshielded than alkenes, and nitriles more deshielded than alkynes (both due to the paramagnetic contribution), but this doesn't overturn the existing bias of alkenes > alkynes.

It is instructive to start by comparing alkenes (~120 ppm) vs alkynes (~70 ppm). There is already a large difference between those two. Carbonyls are essentially deshielded alkenes, and nitriles essentially deshielded alkynes. And so it is no longer too surprising that carbonyls come in at a higher chemical shift than nitriles.
The only question left is to ask why sp2 carbons are generally more deshielded than sp carbons. Although it's tempting to just look purely at the hybridisation and s-percentage, this doesn't explain why the pattern is generally sp2 > sp > sp3. I should first point out that the physical factors controlling 13C shifts are (usually) markedly different from those controlling 1H shifts, so any text that tries to "carry over" those ideas is likely incomplete or incorrect. The reason for that is described in further detail here, and you can see Hans Reich's excellent website for more information about the various terms that contribute to the chemical shift.
With that out of the way, I will briefly quote from Gunther's NMR Spectroscopy, 3rd ed. (p 410):

For the paramagnetic contribution, early theoretical considerations [...] led to the expression:
$$\sigma_\mathrm{p}^i \approx -\frac{1}{\Delta E} \left(\frac{1}{r_i^3}\right)_\mathrm{2p_z} \left(\color{red}{Q_{ii}} + \sum_{j \neq i} Q_{ij}\right)$$
where $\Delta E$ is a mean electronic excitation energy [...]

I have coloured $Q_{ii}$ in red because it is not in Gunther's text (I suspect this is an error). I have added it to bring it in line with an article by Pople (Mol. Phys. 1964, 7 (4), 301–306).
Note first that the paramagnetic shielding is negative, so a larger magnitude corresponds to a less shielded nucleus, or a higher chemical shift. Now, the "excitation energy" for an alkane is probably a $\sigma \to \sigma^*$ excitation, which is quite large. On the other hand, for unsaturated compounds (alkenes / alkynes), there are low-lying $\pi^*$ orbitals, so $\Delta E$ becomes smaller and $\sigma_\mathrm p$ more negative.* This accounts for why unsaturated groups have higher chemical shifts than alkanes.
It also explains why carbonyls/nitriles are more deshielded than alkenes/alkynes respectively: because a heteroatom (O/N) has a high-energy lone pair, the primary excitation becomes a $n \to \pi^*$ transition which has a smaller $\Delta E$.

[...] $Q_{ij}$ is a bond order term that originates from the presence of $\pi$-bonds. [...] The bond order term superimposes additional changes that are most pronounced for [...] the $\ce{^{13}C}$ resonances of alkynes. For [alkynes], theory yields $\sum_{j\neq i} Q_{ij} = 0$ as in alkanes, whereas for [the central carbon of] allene $\sum_{j\neq i} Q_{ij} = 0.8$.

Gunther doesn't say exactly what this term would be for a plain old alkene (not allene), but Pople writes that it is 0.44. According to Pople, the $Q_{ii}$ term (in red) is equal to 2, so the overall term in parentheses is equal to 2 for alkynes and 2.44 for alkenes. This means that an alkyne carbon experiences a smaller (less negative) $\sigma_\mathrm p$, and consequently is less deshielded than an alkene carbon.

* Note also that $\pi \to \pi^*$ excitations are symmetry-forbidden in the context of paramagnetic shielding, so the primary excitation is probably a $\sigma \to \pi^*$ or $\pi \to \sigma^*$.


Start by considering that electron affinities and ionization energies are related. Thus $\mathrm{EA}$ of $\ce{S-}$ is equal to $-\mathrm{IE}$ of $\ce{S^2-}$. (Note the convention used here is to report the value of the signed energy change associated with the process (not the absolute value), see the Wikipedia for an explanation if necessary).

A good place to continue is reading chem.libretexts.org:

  the first electron affinity can be greater than or equal to zero or negative, depending on the electron configuration of the atom. In contrast, the second electron affinity is always positive because the increased electron–electron repulsions in a dianion are far greater than the attraction of the nucleus for the extra electrons.[...] Similarly, the formation of all common dianions (such as $\ce{S^2-}$) or trianions (such as $\ce{P^3−}$) is energetically unfavorable in the gas phase.

This means that the electron affinity of $\ce{S-}$ is positive. Since $\mathrm{IE}(\ce{S^2-}) = -\mathrm{EA}(\ce{S-})$, this means that  $\mathrm{IE}(\ce{S^2-}) < 0$.

Note in comparison that the first ionization energy of sulfur is $\pu{999.6 kJ/mol}$.

Therefore $\mathrm{IE}(\ce{S^2-}) < \mathrm{IE}(\ce{S})$, with $\mathrm{IE}(\ce{S^2-})$ exothermic.


Highly symmetric molecules tightly fit into crystal lattices, give high melting points for their composition.  Their small surface areas minimize London dispersion forces and van der Walls forces, giving high volatility.  They tend to sublime.  Compare mp and bp data for cyclohexane and benzene versus their perfluorinated derivatives. Less exotic: n-octane, iso-octane, 2,2,4-trimethylpentane, 2,2,3,3-tetramethylbutane, and cubane.

http://en.wikipedia.org/wiki/Neopentane
n-Pentane and isopentane versus neopentane.

Chromium, molybdenum, and tungsetn metal versus their respective hexacarbonyls.  $\ce{-SiMe3}$ and $\ce{-CF3}$ plus symmetry confer remarkable volatility.  $\ce{I(CF3)7}$ melts and boils around 0 C.  That is molecular weight 609.95 - and it has no static molecular structure (Bartell mechanism).

http://www.patentgenius.com/patent/3992424.html
US Patent 3992424


Can aluminium hydroxide be split into Al 3+ and OH-?

Aluminum hydroxide has poor solubility in water (solubility product $3 \times 10^{-34}$).

However, $\ce{Al(OH)3}$ is amphoteric and soluble in acids and bases. An excess of $\ce{OH-}$ will form tetrahydroxidoaluminate:

$\ce{Al(OH)3 + OH- <=> [Al(OH)4]-}$


How to determine the efficiency of intermolecular charge transfer?

Indeed, if both complex A-C and complex B-C experience CT emission, the efficiency of CT character present in the complexes may be responsible for the difference in PLQY. This however, will be the case if there is no reverse in CT polarity (for example if in both complexes C is the acceptor (and emitter), this may also the reason for porphyrin's comment). Hence if the criteria above are met and You should generally observe the same (or highly similar) emission from both of the complexes. That would also indicate that one species is responsible for the observed emission.

Calculating efficiency of CT. Although I have not seen extensive literature about CT efficiency, there are few papers which try to quantify the CT. The best example I could find is available here: https://onlinelibrary.wiley.com/doi/full/10.1002/jcc.24822. In that paper the authors are trying to quantify CT within a single molecule. Technically the same methodology should also apply for the intermolecular CT in A-C and B-C. Complaring the values obtained should provide some insight in the CT character between both species. Importantly, to do this calculation You need to know exactly how the complex looks. One way to obtain their geometry is to obtain X-Ray single crystal diffraction representations of A-C and B-C in the solid state (this however works only if A-C and B-C do not dissociate upon crystallization).

Godspeed.


Fermentation in the presence of UV light

Depends on the wavelength. Fermentation is basically metabolic activity by yeast and bacteria. UV-C (<280 nm) is pretty good at killing a lot of micro-organisms. Some species are more resistant than others, but it's safe to say that exposure to this kind of light will probably interfere with fermentation. Depending on the medium and how the light is applied (thickness of material, power, etc.), it may have more or less of an effect depending on how much light reaches the yeast or bacteria.