What happens to the chlorine in epichlorohydrin when a two-part epoxy is air-cured?

There is little if any chlorine in a epichlorohydrin / bis phenol A expoxy resin, it is normally lost during production as sodium chloride into the aqueous phase while the resin forms an organic phase.

The normal way to make the resin is to react bisphenol A with sodium hydroxide and the epichlorohydrin. What happens is that a mixture of the three reagents and water is heated up together. The phenol groups on the bisphenol A are deprotonated and then they attack the epichlorohydrin to then form 2,2'-(((propane-2,2-diylbis(4,1-phenylene))bis(oxy))bis(methylene))bis(oxirane). The epoxide (oxirane) groups in this then react with phenolate anions from new bisphenol A molecules to form 4,4'-((((((propane-2,2-diylbis(4,1-phenylene))bis(oxy))bis(2-hydroxypropane-3,1-diyl))bis(oxy))bis(4,1-phenylene))bis(propane-2,2-diyl))diphenol. The unreacted phenol groups on this then are reacted with more epichlorohydrin to form new reactive groups.

This reaction continues on and on to form a polymer, it is normal to use a small excess of epichlorohydrin to give a epoxy terminiated polymer (resin). This is then reacted with a cross linking agent such as an amine to form the final epoxy. As a result the chlorine content of the epoxy should be very very low.


Why doesn't pH = pKa1 in the buffer zone for this titration?

Your acid dissociation constants are $10^{-0.6}$ and $10^{-1.74}$ whereas your thiosulfate species have only a total molar concentration of $10^{-2.00}$. When the dissociation constants of an acidic solute are greater than the solute concentration the solute is rendered like a strong acid by dilution.  If you have 0.01 M of an acid with a $pK_a$ of 1.74, and you use the equilibrium relation to determine how much of the acid is dissociated, you find that in fact most of the acid is dissociated.

The proposed use of the H-H equation then does not work because this equation assumes that aqueous hydrogen or hydroxide ions are only a small part of the total solute concentration.  When the solute becomes like a strong acid (or strong base) as above, it floods the solution with aqueous hydrogen or hydroxide ions and the assumption needed for the H-H equation fails.


First, take a look at classical physics. The angular momentum of a particle rotating in a plane is defined as
$$L = I \omega$$
and its kinetic energy is
$$E = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}.$$

So if you formulate your energy in terms of the angular momentum of your rotating particle, you arrive at the inverse relation.

In analogy to the classical picture, the eigenvalues of the rotational Schrödinger equation,
$$ E = hcBJ(J+1),$$
likewise depend quadratically on the angular momentum quantum number, and thus have a similar inverse dependence on the moment of inertia through $B$.


Why do complex ions not emit light from de-excitations of electrons?

The fate of electronically excited states is not immediately obvious.

First of all the rate at which the electron emits a photon and falls down a level is dependent upon the energy difference between the two levels, it is only an expectation value of a macroscopic observable, on an individual atomic scale it is impossible to predict when the electron in the upper level will emit a photon, however multiple instances of the same measurement will tend towards the expectation value. The direct result of this is that there is a path difference between the emitted photon and the 'rest' of the photons that were not absorbed by the complex and the result is a different interference between photons in the transmitted light compared to the incident light and hence a new colour is observed.

There is another fate of excited electrons, the Franck-Condon principle invokes the very plausible assumption that the nuclei do not respond immediately (at least on the time scale of electronic transitions) to the new equilibrium bond position that is determined by the new electronic state of the complex. This means the excited electronic state will most likely not be in the lowest vibrational state of the new equilibrium bond configuration, the splitting of vibrational energy levels is orders of magnitude lower than those of electronic transitions and hence the same law that governs electronic emission applies, the electron is much more likely to fall down these vibrational energy levels first, without emitting photons of visible light.

The electron will then be expected to, by the time it is occupying the lowest vibrational state of the excited electronic state, invoke a transition down to the ground state energy level. The energy of the excited state is described in a basic form by
$$H=E^*+v$$

where $H$ is the energy of the absorbed photon, $E^*$ is the energy of the excited state in the lowest vibrational mode and $v$ is the contribution from the initial vibrational state of the excited complex.


How to make acetic acid more volatile so that it can be removed by vigorous aeration of an acetic acid-water mixture?

Aeration does not seem to be the way to go.
If there are no other ions in the water, electrolysis (Kolbe reaction) would convert the $\ce{HOAc}$ to ethane and carbon dioxide:
$$\ce{2CH3CO2^- -> 2e^- + 2CO2 + H3C-CH3}$$
This process could probably be arranged to be done in a pipe with electrodes on the sides, so that the water was treated as it was used, rather than in a huge tank, where diffusion would slow down the completion of the reaction.


Would the reactions in the solution be affected?

Yes, they can be.  Here are some nice examples.  Electric fields also exert a significant effect on electron transfer reactions.  Basically you would expect charged molecules to be more affected than polar molecules, which should be more affected than non-polar, but polarizable molecules.  It's just a question of how big a field you can produce and how polarized and long-lived are the various entities (starting materials, intermediates, etc.) that exist along the reaction coordinate.


Is it correct to say that H2SO4 is an acid in this reaction?

Do you get the same reaction with sodium sulfate?

I do not think so, at least under "normal" conditions.

When the sulfuric acid acts as an oxidizing agent giving off $\ce{SO_2}$, the excess oxygen comes off as oxide ions which must be somehow put into more stable form.  The protons from the sulphuric acid do that by turning them to water.

Getting the reaction to go requires the sulfuric acid to act as both an acid and an oxidizing agent.


Why does 3-bromopropanoic acid have least pKb value when compared to 3-chloropropanoic acid, 3-fluropropanoic acid and fluoroacetic acid?

The correct order of PKb is
fluoroacetic acid> 3-fluoropropanoic acid>3-chloropropanoic acid>3-bromopropanoic acid.
To find the order of pKb, first find the order of pKa which can be found by checking the stability of negative charge on carboxylate anion once it donated proton.
Halogens are electron withdrawing groups. The halogen which will withdraw the most electron density will stabilise the anion the best. Fluorine, being the most electronegative withdraws the most electron density thus reducing the charge on oxygen and stabilising the anion, therefore having the least pKa and maximum pKb.


How to make permanent ink that appears after some time?

Depending on what you want to do (what do you want to do?) you might be looking for gall ink.  Briefly, the chemistry works like this: the recipe consists of soluble polyphenols (typically gallic acid) and a Fe(II) salt.  The preparation must be stored in an air-tight container and even then decomposes over time.  As soon as it is applied to paper and exposed to air, the Fe(II) is oxidized to Fe(III), which forms insoluble deeply coloured complexes with your polyphenol.  The permanent colour takes about a day to develop.  That's why often a non-permanent dye is added so people can see what they are writing.


Well, interesting question, I would re-word your question as "How can one determine an atomic mass with high accuracy using gravimetric analysis? Precision simply implies that you repeated a measurement and it agrees very well with itself even though it may be wrong. What we desire is high precision and high accuracy.

Almost all the naturally occurring elements had their atomic masses (used synonymously as atomic weight in older books) determined by gravimetry. For that you need to read Richard Theodore's Nobel Prize winning work who meticulously measured atomic weights for many many elements using gravimetry at Harvard in the early 1900s.

Do you know what was the experimental definition of atomic mass which made gravimetry possible? It is weight of an element which combined with 1 part of hydrogen or 16 parts of oxygen (we have to choose one of these but not both). Very sadly, nobody teaches this today so most chemistry students are oblivious to it and yet they can regurgitate the 1/12 of C-12 definition. This definition was invented for some other reasons rather than experimentally measuring atomic weights in the 1960s to resolve disputes among physicists and chemists.

So if I assign a weight of 16 units to oxygen and 65.4 parts of zinc combine with one oxygen in a 1:1 ratio (ZnO), we would assign an atomic weight of Zn as 65.4. Arbitrary as it may sound, this was the experimental way. What Richard did is that he used precipitation reactions using silver chloride and correlated that with oxygen 16.

I provide you with an exercise from Pauling's General Chemistry (forget about moles, Avogadro's, g/mol, molar mass etc.) to solve this problem. Let us go to 1940s, we have the above definition that atomic weight is weight of an element which combines with 16 parts of oxygen.

  In a determination of the atomic weight of iron, 7.59712 g of
  carefully purified ferric oxide, $\ce{Fe2O3}$ , was reduced by heating
  in a stream of hydrogen, and found to yield 5.31364 g of metallic
  iron. Given the atomic weight of oxygen as 15.9994, to what value of
  the atomic weight of iron does this result lead?

Hint: Try to find what weight of iron would combine with 15.9994 parts of oxygen and note that there are two iron atoms and three oxygen atoms.


Since probably only ppm levels of denatonium benzoate are in the compressed air, there are ppm levels or lower of denatonium in your flour grinder.  At that concentration, water should dissolve denatonium benzoate.  Is that a viable way to clean your grinder?

According to this solubility table, the best solvent for denatonium benzoate is methanol.  However I would not recommend using methanol on anything you are using for food.  Ethanol is only slightly less good.  Maybe you can wash your grinder in some cheap vodka a few times?

Here is a random thread from Google Groups where someone claims to have identified a vendor for non-bitterized air duster cans.


Knowing that CI is variational for any basis set, why is coupled-cluster variational in the full correlation limit?

The CI (configuration interaction) energy, given by Eqs. 3.9-3.10 of this, is an upper bound to the exact ground state energy (i.e. CI is variational). Coupled cluster energies are not generally variational, as explained in the above linked question, except in the case of CC($N$) for an $N$-electron system. The reason is because CC($N$) for an $N$-electron system is equivalent to FCI (full configuration interaction) which is an example of CI, and all CI energies are variational.

Since I was asked for a publication as a reference, a good one is this, in which it is shown that CC and CI become the exact same in the full-correlation limit.


Free software to draw Newman projections

You can use GChemPaint, which is available on the repositories of many popular distributions. Among its many components for drawing structures, it has a specific tool for Newman's projections.
You can add H, O and many other elements by simply right-clicking on the molecule, and it even allows you to search your molecule in databases like PubChem and ChemSpider.
A useful video tutorial can be found in spokentutorial.org

3.11 To Add a Newman Projection
To add a Newman projection, use the following tool:

This tool allows to add a Newman projection. The changes become permanent only when the mouse button is released.
To add a Newman projection, click on the background. Contrarily to other tools existing atoms are never used by this tool. It creates new atoms of the current element.
When the Newman projection tool is active, the toolbox looks like:

Options for the main bond

Bond length
The default length of the new projected bond. This has no influence ion the drawing, but might be important when exporting to a 3d model.

Bond order
The bond order.

Options for each atom

Bonds number
Number of other bonds for the atom.

First bond angle
The direction of the first bond. Angles are measured counter clockwise from horizontal right.

Angle between bonds
The angle between two consecutive bonds.

Warning
This version of GChemPaint does not correctly support Newman projections with eclipsed bonds.

Source: GChemPaint manual, pages 41-43.


Prevent crystal growth by adding a chemical to a solution?

Spray it as a fine mist into a drying chamber - the solvent will evaporate and you are left with a powder.
No chemical involved though,  but still process chemistry.


The comments pretty much sum up whats going on, but here is a more in-depth explanation to clear up any confusion for the OP.
There is a lot more behind this but here is the (very) basic principle:
Raman spectroscopy uses monochromatic light to excite molecules. These photons interact inelastically with the molecule (i.e. they gain or lose kinetic energy) and are then scattered. Some detector analyzes the scattered light and determines vibrational energy levels by calculating how much kinetic energy the photon has gained/lost which corresponds to the difference in energy of the vibrational modes.
When a black object is hit with a laser, the photons are absorbed. The absorbed energy is released as heat (This is why black cars are significantly hotter inside than white cars when left in the sun). Maybe some scattering does occur, but who knows if it is detectable. There are some articles on trying to determine the Raman spectra of black objects [Here is one].
Do some research; look at the Journal of Raman Spectroscopy if it peaks your curiosity. I'm not saying it is impossible, but it might require a lot of effort and understanding of the fundamental principles of this technique.


Why is the OO-H dissociation is unlikely for the Caro's Acid?

One possibility is the formation of an intra-"molecular" hydrogen bond between the peroxy-bonded hydrogen and a third oxygen atom.  The structure shown in Wikipedia for the acid suggests this interaction.  The involved hydrogen is curled up into the five-membered ring formed by such a hydrogen bond, while the other hydrogen atom is freely exposed and ready to launch.  Potassium peroxymonosulfate corroborates shows only the non-peroxo hydrogen dissociated.

Hydrogen peroxide does not form this sort of intramolecular hydrogen bond, and as we expect from induction considerations the proton dissociates from this molecule more readily than from water.


I'd really reccomend the free Marvin suite from ChemAxon.

MarvinSketch (the actual drawing package) hasn't disappointed me in the past and has a wide feature set.  In addition to just being able to draw compounds in Lewis format, the software includes plugins to name what you have drawn (systematic/traditional), predict properties, change atom and bond properties, generate stereoisomers and much more that I haven't got the experience to do anything more than experiment with!

You can view what you have drawn in Lewis form and also as a 3D structure should you wish to.  It supports many chemical drawing filetypes.

Here is the product page, including brochure and technical details.  It is Java based so works on Mac, Windows, Linux and is even available on the web (but is quite slow to generate properties etc. - I'd reccomend doing this locally).


Relationship between auto-ignition temperature with pressure and fuel

1) One could argue that yes. All other things being equal, larger molecules will clearly have lower vapor pressures at the same temperature so it is reasonable that there is a limit at which the fuel-air mixture is too lean to support autoignition.

2) After reading the paper I can only conclude that they are referring to the above idea that autoignition temperatures increase for larger compounds due to the need to increase the vapor pressure to reach an appropriate fuel concentration that supports autoignition.

3) Larger volumes have lower surface-area-to-volume ratios. The lower surface area means a the walls of a vessel play less of a role in the combustion chemistry leading to a lower AIT. These surface effects are detailed by Babrauskas with references to those who have observed this.

Babrauskas, V.
 of Fire Protection Engineers, S. (Ed.)
 Ignition Handbook
 Fire Science Publishers,
2003


At what temperature CO2 becomes solid at normal pressure?

The phase diagram you have shown is correct and shows the sublimation point at $1~\mathrm{atm}$ as $-78.5~\mathrm{^\circ C}$ . So if you have a pure $\ce{CO2}$ atmosphere, it will solidify at $-78.5~\mathrm{^\circ C}$.

Temperatues in Antarctica are known to get below $-78.5~\mathrm{^\circ C}$ so you might wonder why whe don't find dry ice in Antartica (like we do at the Martian poles [Note: they are a lot colder]). The reason is that the partial pressure of $\ce{CO2}$ in the atmosphere is so low that at $-78.5~\mathrm{^\circ C}$, dry ice is forming but it is subliming faster than it forms because the vapor pressure of $\ce{CO2}$ in the atmosphere above it is well below the saturation vapor pressure required for ice to accumulate.

  The vapor pressure must reach the saturation vapor pressure for dew or frost to form. This happens at the dew point or frost point temperature, which is dependent on atmospheric pressure and the absolute amount of vapor in the air. As atmospheric temperature increases, the dew/frost point temperature increases. As atmospheric pressure increases, the vapor pressure increases. At very low temperatures, the dew/frost point temperature is very low.

http://wattsupwiththat.com/2009/06/13/results-lab-experiment-regarding-co2-snow-in-antarctica-at-113%C2%B0f-80-5%C2%B0c-not-possible/

https://stevengoddard.wordpress.com/2014/06/03/antarctica-gets-cold-enough-to-freeze-co2/


Specifically how dangerous is a 10g chemical sample of pure arsenic and how do you properly dispose of it?

Treat this with very great care; don't let it come into contact with you or anyone else as it is readily absorbed through the skin, lungs and gastrointestinal tract.

Arsenic is the classic poison known also to the ancients. Ten grammes is a huge quantity as it is deadly in quantities of approximately  0.1 g in the blood stream. The oxide and compounds such as the  arsenate, are far more poisonous than the metal as they replace phosphate in the body (which is harder for the pure metal to do) so someone can be poisoned directly by taking a large dose or by gradual exposure over a period of time.

For a fascinating read on this and other poisons look up 'A is for Arsenic: The Poisons of Agatha Christie' by K. Harkup, publ. Bloomsbury. '


What are weak acids?

Examples of weak acids in water solution would be acetic acid (main component of vinegar) and ammonium ion (conjugate acid of ammonia).

  I understand that an acid is weak if it partial dissociates and breaks down into ionic compounds but they never stated why.

An acid is weak if not all of the acid molecules ionize into hydrogen protons and its conjugate base in a particular solvent system. Alternately, if we were to use the broader, Brønsted definition, an acid is weak if it does not completely or nearly completely donate its proton to some base.

Acetic acid is a weak acid in water. Put acetic acid into pure liquid ammonia and now acetic acid will fully dissociate. The proton will be lost from acetic acid to ammonia and this is a favorable reaction. Makes sense, because the only thing that's changed is the solvent that's grabbing the proton - and ammonia is only a few billion times stronger as a base than water.

  That and do the strong acids dissociate at all. Like if they were in a different solution or do they just stay themselves all the time.

Strong acids by definition ionize to an extent of 100% or nearly 100%. Strong acids will give up their protons completely to the strongest base in the system.

Note that in the above text I have used the word "ionize" to describe molecular acids because by definition when these acids react, they ionize. Or form ions. On the other hand salts of acids and bases such as sodium hydroxide are said to "dissociate," because ionic compounds are made of ions, and it wouldn't make much sense to say that ions ionize or become ions.


Wurtz reaction with two different alkyl halide

The alkyl groups mentioned could be primary, secondary or tertiary
The last one's will not be going by Wurtz reaction (these will rather undergo elimination to give alkenes); only primary & secondary can.
Even here, we can have a mixture of products:


Explain the nomenclature of 7-amino-4-methylcoumarin

This is one of the tricky parts in the nomenclature of fused cyclic systems. Excerpt from page 180 of the 1998 IUPAC Nomenclature of fused and bridged fused ring systems
(I have added bold to the piece of importance to answer your question):

  The numbering of the peripheral atoms in a particular orientation starts from the uppermost ring. If there is more than one uppermost ring the ring furthest to the right is chosen. Numbering starts from the non-fusion atom most counterclockwise in the ring selected, and numbering proceeds in a clockwise direction around the system, including fusion heteroatoms. Fusion carbon atoms are not included; each fusion carbon atom is given the same number as the immediately preceding position modified by a roman letter a, b, c, etc. If
  heteroatoms are identified by replacement nomenclature all atoms retain the numbers of the parent
  hydrocarbon.

If one considers coumarin, this is not the IUPAC nomenclature but the numbering rule remains the same. Below is an example of the numbering scheme for coumarin which summarizes why the substituents are numbered by this manner on coumarin.


Yes, you are correct: an -ate prefix is used when the complex has a net negative charge and the name should be triamminetrinitritocobaltate(I) ion

Reference:

  If the overall coordination complex is an anion, the ending "-ate" is attached to the metal center. Some metals also change to their Latin names in this situation. The rest of the metals simply have -ate added to the end (cobaltate, nickelate, zincate, osmate, cadmate, platinate, mercurate, etc. Note that the -ate tends to replace -um or -ium, if present).

Chemistry Libretexts - Nomenclature of Coordination compounds


Hydroxide is bad for this process. You will have additional problems if hydroxide anion is present.

Silver cation reacts with hydroxide anion to form silver hydroxide, which spontaneously decomposes into silver oxide:

$$\begin{aligned}\ce{Ag+ + OH-}&\ce{ -> AgOH}\\
\ce{2AgOH}&\ce{ -> H2O +Ag2O}\end{aligned}$$

In addition to shutting down the oxidation of silver, hydroxide precipitates any silver cation that does dissolve.

For this reason, the electroplating process is probably run at a slightly acidic pH, and the competing oxidation reaction is not of hydroxide, but of water (reduction potential 1.229 V):

$$\ce{2H2O -> O2 + 4H+ + 4e-}\ E^\circ = +1.229\mathrm{\ V}$$

So long as you don't apply too much of an overpotential to the electroplating operating, silver will oxidize before water.


Your idea seems mostly sound, but one thing you should consider is the solubility of calcium hydroxide (a.k.a. slaked lime).  It is fairly low.

That low solubility provides you with an opportunity.  If you have calcium chloride and sodium hydroxide, you can mix saturated solutions of both and a white precipitate should form.  The precipitate is calcium hydroxide.  You can then remove the precipitated calcium hydroxide, wash it with ice cold water, and then you'll have $\ce{Ca(OH)2}$!  A YouTube demo of this idea provides some more details.

$$\ce{CaCl2(aq) +  2NaOH(aq) -> 2NaCl(aq) + Ca(OH)2(s)}$$

I think this method is preferred because the low solubility of calcium hydroxide compared to sodium hydroxide naturally limits its alkalinity and corrosiveness.  It would be too easy to use undesirable (and possibly even dangerous) levels of NaOH if you made a mistake in how much you used for nixtamalification.  Additionally, simply mixing NaOH and calcium chloride and using the resulting milky liquid as a substitute for calcium hydroxide will have far too much salt (NaCl) present to be a good substitute.  Or conversely, if you lower the amount of NaOH and calcium chloride to where the saltiness is tolerable, you will have too little alkalinity.


When water undergoes electrolysis where in the beaker does the water molecule break apart?

The water molecules form long networks of hydrogens bonds; let's imagine one such H-bonded chain that stretches from the anode to the cathode. Reduction will occur at the cathode, so the nearest water molecule (on the end of our chain) will be deprotonated and that proton will be reduced to $H\cdot$ radical, which can combine with a nearby $H\cdot$ to form the reduction product, molecular hydrogen.

Now, the water molecule that was deprotonated can re-protonate itself with the hydrogen of its neighbor (the second $H_2O$ in our hypothetical chain), likely the very hydrogen that they were sharing as an H-bond. This leaves the second water molecule deprotonated, so it fully claims the proton the third molecule was offering in their H-bond.

The process continues, with the negative charge effectively traveling the length of the chain toward the anode as each water molecule in turn briefly becomes hydroxide. This "proton hopping" behavior is known as the Grotthuss Mechanism. This animation can give you a sense of how this phenomenon happens, but note that the animation shows an acidic mechanism (extra proton, positive charge traveling) rather than the basic mechanism I described (a missing proton, with the negative charge traveling). Both are likely at play during the electrolysis of water. (Actually, the two mechanisms probably cannot be distinguished due to the simultaneous nature of the redox process, which I'm about to describe).

At the anode, hydroxide will be oxidized to $O_2$ through a few steps that aren't necessarily relevant to your question. The big idea is that the hydroxide will be created from a water molecule at the anode end of the chain of H-bonded molecules, and it will probably happen nearly simultaneously as a water molecule at the cathode end is deprotonated to form the proton reactant. Most probably both ends of our H-bonded chain of water molecules will react at more or less the same time.

If you like metaphors, think of it as a line of people each holding a basketball. Everybody passes their basketball to the right, and received a basketball from the person on their left... Except the people standing on the two ends. One will throw their basketball into the hoop, and the other will be torn to shreds and evolve molecular oxygen.


Distinguishing stereoisomers using reagents

The only reagent listed that will distinguish between cis and trans diols is acetone. In the presence of catalytic acid it will form an acetal with the cis diol. This cannot occur with the trans diol as the two OH groups are too far apart. Try drawing the cis and trans diol in 3D.


The Nernst equation is as follows,
\begin{align*}
    E_\text{red}&=E_\text{red}^{0}-\frac{RT}{nF}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)}.
\end{align*}
where $c$ is the concentration of the species specified (usually the chemical activity, $a$, is used, but the equations follow the same principle). In the first case, $n=2$. So,

\begin{align*}
    E_\text{red}&=E_\text{red}^{0}-\frac{RT}{2F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)}.
\end{align*}

Suppose we double the coefficients in the equation. Now, $n=4$. However, the concentrations of all the species have also doubled. This requires the reaction quotient (the stuff inside the natural logarithm) to be squared. Therefore,
\begin{align*}
    E_\text{red}&=E_\text{red}^{0}-\frac{RT}{4F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)^2}.
\end{align*}

The laws of logarithms means that we can move the power of 2 to be a multiplication of the logarithm,

\begin{align*}
    E_\text{red}&=E_\text{red}^{0}-\frac{2RT}{4F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)},\\\\
    &=E_\text{red}^{0}-\frac{RT}{2F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)},
\end{align*}
which is the exact equation we had previously, thus, nothing has "changed".

So to answer your equation. Yes, $n=4$, but the equation does not change, as the reaction quotient alters accordingly, also.


Ultimately, explosions are just a type of spontaneous reaction, and even endothermic reactions can be spontaneous, so long as entropy increases enough over the course of the reaction to compensate it. Near room temperature, however, the entropic factor is usually small compared to the enthalpic factor. This is why there are comparatively few examples of reactions which produce strong cooling upon mixing reagents.

That said, none of this strictly precludes slightly-exothermic or endothermic explosions, to the point that the concept has been legitimately investigated. As I mentioned in the comments, there is even a specific term for these types of explosions; entropic explosions.

For a spontaneous reaction to be explosive, there is one property it must have above all others: the reaction must happen very quickly. Now we are dealing with kinetics, which is much harder to analyse theoretically. It is possible, however, to point out another factor that makes endothermic explosions difficult.

By far, most reactions speed up when temperature is increased. In a conventional, exothermic explosive, this establishes a positive feedback effect, where heat from the reaction causes it to react faster, which produces more heat, which causes even faster reaction, and so on, facilitating a runaway reaction and subsequent detonation. If the exotherm is weak, then there is a larger chance that the explosion will fizzle out before enough heat is produce to get things going, or the reaction might be too slow to explode and instead steadily combusts. For an endothermic explosive, very much the opposite happens, and a negative feedback is established; once started, the reaction would decrease the temperature of the explosive, hindering propagation.

So, while there are no arguments which completely forbid slightly-exothermic or endothermic explosions, it seems to be a rather unlikely scenario in practice. Wikipedia only lists one possible contender for an entropic explosion, which is triacetone triperoxide. The original article making the claim is mostly theoretical, and there seem to be some rebuttals to it, so even that is not certain. Of course, we may be in the middle of a giant selection bias effect, because almost all research into explosives is focused on getting more energetic materials, not less. Maybe there are examples waiting to be found, but no one has had the interest in doing so!

As a sidenote, explosions actually don't necessarily produce gas. There are some rare examples, such as cuprous acetylide ($\ce{Cu2C2}$) and silver acetylide ($\ce{Ag2C2}$) where all the products are solids. Also, there are examples of non-chemical explosions which are only slightly exothermic or even endothermic. As MaxW mentions, a pressure explosion from a liquefied gas will be endothermic, as energy is consumed to evaporate the liquid. Another interesting example of an explosion which is probably only slightly exothermic is a Prince Rupert's drop being set off.


Reactivity of different alcohols with NaNH2 and CH3I

I suspect, possibly just to a minor extent, that you have also managed to produce an expected photolysis product here (see below).
Interestingly, the photo sensitive methyl iodide, in the presence of a suitable catalyst ($\ce{CO2, Cu,}$..) with lab light (rich in UV) may have commenced in a radical reaction chain in the presence of ethanol:
$$\ce{CH3I ->[$h \nu$] ^.CH3 + I^.}$$
$$\ce{C2H5OH + I^. -> HI + C2H5O^.}$$
$$\ce{C2H5O^. + C2H5O^. -> (C2H5)2O2}$$
$$\ce{(C2H5)2O2 + ^.CH3 -> (C2H5)2O + CH3O^.}$$
....(the reaction chain continues)
where, I have stopped at the formation of Ether, albeit possibly created in very small amounts but detectable by its smell, nevertheless.
Repeat the experiment controlling for light/catalyst and see if you can still smell the Ether.
My advice, any time working with a photoreactive iodides, be mindful of light exposure.
[EDIT] The implication of my analysis, check your product yields, reagents and procedure, as a minor photo-induced side product may not (but still could) be connected to your results. In particular, I would start with the $\ce{NaNH2}$ as per Wikipedia:

$\ce{NaNH2 + H2O → NH3 + NaOH}$
$\ce{4 NaNH2 + 7 O2 → 2 Na2O + 4 NO2 + 4 H2O}$

which suggests that either exposure to moisture/water/air can degrade the compound.


Can a compound with indirectly bonded carbon and silicon atoms within the same aromatic system be called organosilicon compound?

Organosilicon compounds fall under the broader category of Organometallic compounds. It is a requirement that in organometallic compounds, atleast one $\ce{C-M}$ bond should exist. Since Si is only bonded to oxygen atoms in your compound, the given compound is not an organosilicon compound.

The following is also mentioned in the wikipedia page for Organosilicon compounds, under the image:

  A carbon–silicon bond present in all organosilicon compounds


How to finish silanization process correctly?

Sol-gel preparation techniques have many variations. The contact time -beads to the solution - determines thickness of the sol-gel layer. Here you can try to shorten the dipping (coating) time. Glass is usually dried at 60°C for 30 mins and then "cured" at 500 - 550°C form 5 s up to 2 hours. I would suggest to find out an article and follow its preparation method.


If a sample to be carbon-dated is wrapped in a piece of paper, will you still be able to effectively measure the radioactivity of the sample?

Hardly, with or without paper wrap.

Most of the radiation will be absorbed by the sample itself, you'll have a hard time modelling the absorption to derive the actual remaining activity of the carbon. Artifacts are usually not a piece of completely homogeneous material. Also you'll have a hard time finding anyone with a suitable Geiger counter, if you want to do more than guesswork.)

Modern radiocarbon methods are destructive, you take a number of tiny (<1 mg), representative samples out of your artifact, and directly do mass spectrometry on them, without waiting for radioactive decay.


What anions does the loss of a proton from HOCN and HNCO produce? Are they same or different?

When they are bonded to hydrogen that is $\ce{HOCN}$ $\ce{HNCO}$. The two compounds will be different:

From Wikipedia

  Organic cyanates are called isocyanates when there is a $\ce{C−NCO}$ bond and cyanates when there is a $\ce{C−OCN}$ bond.

Due to resonance after deprotonation the two anions will be the same.


What causes the lowering of vapour pressure in volatile/nonvolatile solvent mixtures?

I think that the second explanation is the correct one. A simple experiment can prove it: if you place a porous cover over the pure solvent, its vapour pressure will not change. In alternative, you can place a few corks and observe the same result (no variation in the vapour pressure). In both cases surface sites are blocked, but the vapor pressure will obviously remain unaltered. Therefore the reason cannot be that the solute molecules are hindering the solvent molecule escape by occupying the solution surface.

The force that is driving the solvent molecules to the vapour phase is the difference between the entropy of the liquid and the vapour phases: when the molecule escape the liquid there is an entropy gain. Adding a non-volatile solute will increase the entropy of the liquid phase without having an effect on the entropy of the vapour phrase (given that the solute is non-volatile). This will lower the difference in entropy between the two phases, which is the driving force that make vaporisation happen, therefore there will be a lower number of solvent particles in the vapour phase and a consequent lower vapour pressure.

Textbooks like Chemistry: The Molecular Nature of Matter and Change (Silberberg, Amateis) or Physical Chemistry for the Biosciences (Chang) provides the same explanation. In addition, there is a short paper/article on this topic[1] where the author highlights how only very few freshman chemistry texts give an adequate explanation of this phenomenon by making use of a entropy-based argument.

     1 Vapor Pressure Lowering by Nonvolatile Solutes, G. D. Peckham, J. Chem. Educ., 1998, 75 (6), 787


Electronegativity of an atom is due to the pull of electrons by the nucleus of atom.

Applying electric potential will not change the force the nucleus applies on the electron and hence it will not change the electronegativity.

By applying electric potential, the electron can be pulled out of the atom, not because it decreased the electronegativity but because it could apply enough force on the electron to overcome the electrostatic attraction by the nucleus and hence let it come out of the atom.

ps: If you had this doubt, then probably you are not clear with the actual concept of electronegativity. You might want to read up on that.


Why does Ethylene Glycol have higher boiling point than Propylene Glycol?

This is all about linearity of the molecule and steric hindrance. Consider the linear structure of ethylene glycol versus the branched structure of propylene glycol. EG can "stack" more easily and therefore engage in more intermolecular bonding than PG. More attraction means more energy required to vaporize. On the other hand, 1,3 propane diol, which is the linear isomer of propylene glycol, has a higher boiling boiling point than EG at around 213C. To further illustrate - 1,4 butanediol (linear) has a higher boiling point than 1,3 butanediol (branched).


Which pH indicator will give a dramatic color change with small changes in the amount of carbon dioxide dissolved in water?

Try neutral red. It's transition region is from 6.8 to 8.0 > red to yellow

Also check this wikipage for other options.


Why is 2-pyridone so much more acidic than a regular amide?

The main factor is aromaticity.

Let's look at the main mesomeric forms of DMAc and 2-pyridone.

For DMAc the true structure is somewhere in the middle. We know the C-N bond cannot rotate freely but the barrier is quite low. If we look at structure 2 we can imagine that this structure contributes a lot to the acidity of the amide, since we got a protonated imine like structure, while on the left it's more like an amine.

For 2-pyridone the right structure (4) is aromatic, so it is quite stable. In this case this mesomeric structure will contribute to the real structure much more, which means it is closer to the right structure than in case of DMAc. But in structure 4 we basically got a protonated pyridine. Which is very acidic. It also tautomerizes to 2-Hydroxypyridine, in that case we have a phenol like structure, which is quite acidic too.


Metallic osmium is harmless but it reacts with oxygen at room temperature, forming volatile osmium tetroxide. Any other osmium compounds are also converted to the tetroxide if oxygen is present. This makes osmium tetroxide the main source of contact with the environment and human body. Exposure to osmium tetroxide cause harm to the skin, eyes, and respiratory system.

So, why makes osmium tetroxide so harmful? (TL;Dr: Because to its oxidization ability)

There is a paper1 which studied the effect of osmium tetroxide in human body and how it can be a lethal bioweapon. The following is the text from the paper:

Osmium tetroxide is highly volatile and penetrates skin readily, and
is very toxic by inhalation, ingestion, and skin contact. Airborne low
concentrations of osmium tetroxide vapor can cause lung congestion and
skin or eye damage.
[...]
Osmium tetroxide is a rapid, indiscriminative oxidizer that does not distinguish between organic tissue and inorganic materials. An
inhalational toxicity study with rabbits proved futile, because of the
rapid reduction of $\ce{OsO4}$ by the skin, hair, mucous membranes,
etc., as well as by the chamber walls. Inhalation, ingestion,
contact with skin and with mucous membranes may all result in severe
consequences. Due to its high vapor pressure (7 mmHg at 20 °C), most
exposures are to vapor. These can cause severe chemical burns to the
eyes, skin and respiratory tract. Very short-term contact with the
vapor may cause lacrimation, accompanied by cough, headaches and
dizziness. $\ce{OsO4}$ may cause irreversible blindness by turning the
cornea black. Symptoms may not be noticeable until several hours after
the exposures. Another severe delayed effect following inhalational
exposure is acute lung injury, which may be followed by
non-cardiogenic pulmonary edema Direct contact with osmium tetroxide
solution will turn the skin black (severe chemical burns due to strong
oxidizing properties). Painful burns or contact dermatitis may
result, depending on the concentration.

The paper compares the LD50 value of osmium tetroxide with other chemical warfare agent and thus proves how lethal it can be as a bioweapon:

The paper also notes that osmium tetroxide can rapidly reduced to relatively inert compounds by e.g. ascorbic acid or polyunsaturated vegetable oils (such as corn oil which contains double/triple bonds. $\ce{OsO4}$ can attack those bonds and can convert to non-toxic vicinal diols).
Reference

Makarovsky, Igor & Markel, Gal & Hoffman, Azik & Schein, Ophir & Finkelstien, Arseny & Brosh-Nissimov, Tal & Tashma, Zeev & Dushnitsky, Tsvika & Eisenkraft, Arik. (2007). Osmium Tetroxide: A New Kind of Weapon. The Israel Medical Association journal : IMAJ. 9. 750-2.
McLaughlin AIG, Milton R, Perry KMA. (1946). Toxic manifestations of osmium tetroxide. Br J Ind Med 3:183-186.


discover of the neutron, what happen to 2+ charge

Corrected: "James Chadwik identified the neutron by exposing a Beryllium foil to  α-rays (He+2) and showing that the previously discovered emitted collision product were massive neutrally charged particles." The experimental set-up he used (see:http://rspa.royalsocietypublishing.org/content/royprsa/136/830/692.full.pdf) did not determine where any electrons came from, so while electrical charge can be assumed to balance, he couldn't claim any direct knowledge of the source or fate of any electrons in his experiment. For us, today, balancing a nuclear reaction's charge distribution on both sides of the arrow is simply "bookkeeping", it implies facts that were not as well supported in 1932 as they are today. He avoided the temptation to over-interpret his experimental data.


Anomalous Electronic Configuration of Thorium

For first four actinide elements $\ce{Th}$, $\ce{Pa}$, $\ce{U}$ and $\ce{Np}$, the difference in energy between $\mathrm{5f}$ and $\mathrm{6d}$ orbitals is small. Thus in elements electrons may occupy the $\mathrm{5f}$ or $\mathrm{6d}$ levels or sometimes both.

Later in actinide series the $\mathrm{5f}$ do become appreciable lower in energy, thus from $\ce{Pa}$ onward the $\mathrm{5f}$ shells fill in regular way.


Does the addition of acid change alkalinity?

Calcium carbonate has a solubility of $\pu{13 mg/L}$ at room temperature. Adding acid to this solution will dissolve the calcium carbonate, by destroying the carbonate via the equilibria
$$\ce{CO3^2- <=>[+ H+][- H+] HCO3- <=>[+ H+][- H+] H2CO3 -> H2O + CO2,}$$
with the carbon dioxide having a tendency to escape the solution.


The pentagonal pyramidal structure of XeOF5-

$\ce{XeF6}$ is not octahedral, but distorted (or capped octahedral), one of the possible arrangement for seven electron pairs.

Wikipedia provides the following, quoted from a paper by Konrad Seppelt in Accounts of Chemical Research[1]:

  ...the structure is best described in terms of a mobile electron pair that moves over the faces and edges of the octahedron and thus distorts it in a dynamic manner...

The molecular geometry of $\ce{XeF6}$ is thus not truly an octahedron, nor is it static. In fact, the referenced article by Seppelt indicates that $\ce{XeF6}$ is only monomeric in the gas phase. In solution and the solid phase it is tetrameric $\ce{Xe4F24}$ clusters. The solid phase appears to be more like clusters of $\ce{[XeF5+ F^{-}]}$ with the fluoride ions bridging two $\ce{XeF5+}$ centers. Thus, in the solid, every $\ce{Xe}$ atom has 7 bonds to fluorine, with two that are more ionic. The structure for this scenario looks like the pentagonal pyramid.

Reference:

  Seppelt, Konrad (June 1979). "Recent Developments in the Chemistry of Some Electronegative Elements". Accounts of Chemical Research 12 (6): 211–216. doi:10.1021/ar50138a004


Is solubility coefficient affected if ion data is given in Ksp?

Let the concentration of $\ce{Cr^{3+}}$ be $t$ then the concentration of $\ce{OH-}$ will be $3t$.

So, the site says that writing the concentration as $3t$ is not necessary.

What it is basically doing is considering $3t$ equal to some variable $s$, and solving the question. There is nothing wrong in this, is it?

Now, for example we had not done that and calculated the value of $t$ by using the equation

$K_{\mathrm{sp}}=[\ce{Cr^{3+}}] \cdot (3t)^3$

You will then multiply the derived value of $t$ with 3 to get $\ce{[OH-]}$.

From which you will find the $\mathrm{pH}$. So the site says:

"Why not calculate the value of $3t$ directly by using a variable $s$?"

Therefore: $s=3t$

Therefore: $\mathrm{pOH}=-\log{s}$

What you are thinking is also correct, and the answer with that will be:
$\mathrm{pOH}=-\log{3t}$

So you both are doing the exact same thing here.


Why can a face-centered cubic lattice not be redrawn as a body-centered tetragonal lattice?

OK, let's get it straight. A lattice is just a periodic set of points. For the sake of brevity, I'll talk about 2D lattices; trust me, with 3D it is pretty much the same, only more complicated.

Initially the lattices were just like that: periodic patterns of one infinitely repeating parallelogram, called the unit cell. Other than that, they were without form, and void; and darkness was upon the face of the deep. The very word "centered" was not invented yet.

Then people started to categorize lattices by symmetry.

First of all, there are those with $90^\circ$ rotational, or 4-fold symmetry (we may call them "tetragonal", or maybe "square"), and those with 6-fold symmetry (we may call them "hexagonal"). We put them aside and look deeper at the rest.

You see that the first one of these has no symmetry to speak about (other than the $180^\circ$ rotational symmetry, which all 2D lattices have by birthright). The link above calls such lattices "oblique"; I don't like the word, but whatever. The second one has mirror symmetry with respect to the lines running along the sides of the cell; this is an example of so-called "rectangular" lattice. And the third one has mirror lines along the diagonals of the cell. This just does not sit well with us; we'd rather have them along the sides. (We will meet these diagonal mirror lines later anyway, when we'll deal with the tetragonal and hexagonal lattices; but that's another story.) So we select another unit cell, twice bigger in area than the original, and call it centered.

See, the lattice (i.e., the arrangement of blue dots) is still the same; we couldn't and didn't make it more symmetric, or less symmetric, or in any slightest way different, for that matter. We just started to think of it in a different way.

And so the centered rectangular cell entered the list of 2D Bravais lattices. See, it did not get there because it can't be decomposed into some primitive cell (it surely can, as any centered cell can), or because other cells can't be centered (they surely can, only it would make no sense). It is there because we wanted our symmetry elements to be aligned in a certain way.

Like I said, in 3D we have the same story, only somewhat more complicated. Indeed, the face-centered cubic cell can be thought of as body-centered tetragonal (and also as some primitive non-centered cell, too); it is just that the cubic cell has 4-fold rotation axes running along all sides, and in a tetragonal cell some of them would run diagonally, which is not considered comme il faut.

My apologies to those who think they'd be better off with the diagonal mirrors and without centering. Anyway, the convention has been in use for well over a century now, so maybe it's not that bad after all.


Although Nicolau Saker Neto's answer provides the essential, I wish here to propose two emblematic examples for the situations cited in your question:

First case: two steps process, the first being the slowest.

Reaction between $\ce{H2O2}$ and $\ce{HBr}$:

\begin{align}
\ce{H2O2 + H+ +Br- &-> HOBr + H2O}  && \text{slow}\\
\ce{HOBr + H+ +Br- &-> Br2 + H2O}   && \text{fast}\\
\end{align}

The rate of this reaction, determined by the slow step, equals:

$$
\frac{d[\ce{Br2}]}{dt}=k_{1}\cdot[\ce{H2O2}]\cdot[\ce{H+}]\cdot[\ce{Br-}]
$$

This is therefore a third order reaction.  Stoichiometric coefficients of the slow elementary step yield the specific order in each of the three reacting species.

Second case: two steps process, the first being the fastest.

Reaction between $\ce{NO}$ and $\ce{O2}$:

The initial equilibrium is rapidly established:

\begin{align}
\ce{2NO &<=> N2O2}      && \text{fast}\\
\ce{N2O2 + O2 &-> 2NO2} && \text{slow}\\
\end{align}

In this case, the rate is:

$$
\frac{d[\ce{NO2}]}{dt}=k_{2}\cdot[\ce{N2O2}]\cdot[\ce{O2}]
$$

This is therefore a second order reaction, with specific orders coinciding, as usual, with the stoichiometric coefficients in the slow step. Using the equilibrium constant $K$ of the first step, one can finally express the rate of the global equation as a function of $[\ce{NO}]$ and $[\ce{O2}]$, say:

$$
\frac{d[\ce{NO2}]}{dt}=k_{2}\cdot K \cdot[\ce{NO}]^{2}\cdot[\ce{O2}]
$$

Summarizing, as you correctly say, the stoichiometric coefficient of a given reactant, in the rate determining elementary step ("slow" step), directly yields the order by which the concentration of that reactant is raised in the rate equation.


Which has stronger hydrogen bonds: water or ice?

The quick answer is that ice has stronger hydrogen bonds than liquid water on average.$\require{mediawiki-texvc}%for angstrom unit$

As has already been brought up, this question is not easy to answer comprehensively as it may very well depend on temperature and pressure in ways that are not immediately obvious. That being said, I will answer this question for the most obvious case which is looking at hydrogen bond strength when water is just barely a liquid (just above $\pu{0~^\circ C}$) and when water is just barely frozen (just below $\pu{0~^\circ C}$). Also, as you may or may not know, ice freezes into many different forms which all have different lattice arrangements. I cannot speak to unusual forms of ice such as the one with a cubic lattice, but what I say is immediately applicable to forms like proton-ordered hexagonal ice (I think this is ice IX). Regular ice is called ice Ih.

It is worth looking at the very comprehensive reference [1], which discusses details of water from the dimer all the way to the many forms of ice which can form. The important points for us are only the following structural values, however:

In ice Ih, the average $\ce{O-H}$ bond length is $\pu{1.01 \AA}$ and the $\ce{HOH}$ bond angle is $109.5^\circ$. The trend which has been observed experimentally is that in going from the gas-phase water monomer, through the clusters, to the liquid, and finally to ice, the bond angle slowly increases to this tetrahedral value. Additionally, the average bond length in the liquid is shorter than in the solid, but you will have a hard time finding reliable reported values of this number from experiment because it is very difficult to determine the position of hydrogen atoms from scattering experiments in non-solid phases (this is mostly because of an ambiguity as to which oxygen atoms a hydrogen should be associated with in the scattering data). Nonetheless, there are reliable average $\ce{O-O}$ distances from scattering experiments.

In the liquid, the average $\ce{O-O}$ distance is $\pu{2.85 \AA}$ while in ice Ih it shrinks to $\pu{2.74 \AA}$. Intuitively, it is very hard to understand how this would happen if it is not simply due to increased hydrogen bond strength.

We can look at an even simpler case though and see that increased hydrogen bond strength causes the $\ce{O-O}$ distance to decrease in ice. This example is a comparison of the water dimer and water trimer. In the water dimer, the $\ce{O-O}$ distance is $\pu{2.976 \AA}$. In the water trimer, the $\ce{O-O}$ distance is about $\pu{2.82 \AA}$. This decrease in distance and the corresponding red-shift in vibrational frequency are given physical meaning in the Ref. [2].

Ref. [2] gets at a very important point for us, one which Ref. [1] also uses to explain the shrinking in $\ce{O-O}$ distances in ice. This is that hydrogen-bonding in water highly cooperative. What do I mean by this? This is exactly the topic of Ref. [2]. The key idea is this: when you allow three water molecules to mutually hydrogen bond, as in the ring shape of the water trimer, the increase in attractive interaction strength is more than simply three times the interaction strength of a water dimer. By attractive interaction strength, I mean that if you subtract out the initial decrease in binding energy which comes from distorting the water monomers. Again, all of this is laid out in detail in Ref. [2], but we learn the general rule that many-body interactions in water are very important.

So, finally, I feel we have said enough to explain why the knowledge that a shrinking $\ce{O-O}$ distance and widening $\ce{HOH}$ angle tells us that ice has stronger hydrogen bonds than liquid water. In the first case, the decreasing $\ce{O-O}$ distance means that every pair of monomers experiences greater nuclear repulsion. This must be at least offset by a greater hydrogen bond strength in order for this geometry to be favorable. Part of the way this repulsion is offset is that as the system cools and forms a solid, the hydrogen-bonds are allowed to become more ideal. This is why the bond angle widens. The tetrahedral arrangement of atoms means that the $\ce{OH--H}$ hydrogen-bond angle is much closer to 180 degrees on average than in the liquid where it frequently deviates fairly significantly.

There is one last point to make. It can be estimated both from accurate simulations and from experiment what is the average dipole moment of a water monomer in bulk liquid and in ice. It is found that the dipole increases going from the liquid to ice. This additionally tells us that the average charge on each hydrogen atom must have increased from in the liquid. We are guaranteed this because we know the bond angle increased, which should generally decrease the dipole moment (a linear configuration has no dipole). Instead, the bond angle increases, so we get more ideal hydrogen bonds, and the charge increases, so the hydrogen bonds would be stronger even if they were no more ideal than in the liquid!

Hopefully this rather long-winded answer helps you understand how we can take the evidence at hand to arrive at the conclusion that ice has stronger hydrogen bonds than liquid water.

Now to the question of different forms of ice and different temperatures and pressures. I would think that pretty much across the board, every form of ice will follow the same trends we have seen here and hence have stronger hydrogen bonds. One strange case is cubic ice, which has $\ce{OHO}$ angles somewhere close to 90 degrees. This is very strange, but should lead to a large increase in the dipole of each molecule and probably increase in positive charge on each hydrogen atom, so even then the hydrogen bonds should be stronger than in the liquid.

There may be temperatures and pressures where some of these arguments break down, but the $\ce{O-O}$ distance should smoothly decrease to that in the liquid and the angle should close to that in the liquid as the pressure increases, so one would expect the hydrogen bond strength to decrease to that in the liquid and never overshoot it.

References:

Ludwig, R. Water: From Clusters to the Bulk. Angew. Chem. Int. Ed. 2001, 40 (10), 1808-1827. DOI: 10.1002/1521-3773(20010518)40:10<1808::AID-ANIE1808>3.0.CO;2-1.
Xantheas, S. S. Cooperativity and hydrogen bonding network in water clusters. Chemical Physics 2000, 258 (2-3), 225-231. DOI: 10.1016/S0301-0104(00)00189-0.


Why would I not see a reaction between aluminum and tin(IV) chloride?

You need to activate the $\ce{Al}$ to remove the tenacious oxide layer that protects it from reaction. This can be done by a quick wash with dil $\ce{HCl}$, rinse with $\ce{EtOH}$ and finally $\ce{Et2O}$ before adding to the tin chloride.


I'd say the diagram from Atkins "Inorganic Chemistry" is in error.

I went to what I consider the "definitive source" for such Walsh diagrams: Orbital Interactions in Chemistry by Albright, Burdett and Whangbo.

Here's their diagram for $\ce{H2S}$ (the linear on the left, the bent on the right):

They don't readily indicate the method they used for the calculation, but on this energy scale the difference between the $1\sigma_g$ and $1a_1$ is minimal. They show that the $1a_1$ is lower in energy since the interaction with the central p orbital will be stabilizing.

As to the $1b_2$ rising above the $2a_1$ in the bent structure -- the diagram from Albright, Burdett, and Whangbo agree that there's no "crossing."


As mentioned in the comments, $\ce{SbF5}$ in the gas phase has a rather high symmetry, the point group looks like $D_\mathrm{3h}$ to me, containing a principal three-fold rotational symmetry axis (along the axial bonds) and three two-fold axis (along each of the equatorial bonds), among other operations. As mentioned in this text, a dipole moment can only coincide with a rotational axis. Since there are four such axis in the molecule, it cannot have a dipole moment.

Note however, that each of the bonds is certainly polar. They just cancel out, as could be proven by vectorial addition of the bond dipole moment.


Is ion activity in a salt solution pH-dependent?

Considering the effect of $\text{pH}$ on ion activity, one should take into account that $\ce{K+}$ and $\ce{Cl-}$ are respectively conjugated acid of a strong base and conjugated base of a strong acid. They are therefore both weak, with no influence on the $\text{pH}$.

As a consequence, I would basically assume that also the other way round is true and that $\text{pH}$ has no effect on their activity.


What is a quadrupole moment of a molecule and how does it arise?

This explanation is geared at someone with one year of chemistry. It captures the gist but is not rigorous:

To test for monopoles, dipoles or quadrupoles in a molecule or ion, assign charges or partial charges to each atoms. For partial charges, δ+ and δ- is fine (if you want estimated values for partial charges of simple molecules, use a tool like MolCalc).

To test for monopoles (ions), add up all the charges. If they are non-zero, it's an ion.

If the net charge is zero, test for presence of a dipole: If you can draw a line through the molecule such that more negative partial charges are on one side of the line and more positive partial charges are on the other side of the line, it has a dipole moment. An example for water is below.

If it does not have a dipole, you can check for a quadrupole moment: If you can draw an "X" through the molecule such that more negative partial charges are in opposite quadrants and more positive partial charges are in the other two quadrants, it has a quadrupole moment.

The animated GIF shows the model of carbon dioxide rotated in the presence of such an "X" (or "+" in this orientation if you will). In one orientation (where it is diagonal), the negative partial charges of oxygen are in opposite quadrants, while the positive partial charge is in the center (neither quadrant), showing that it has a quadrupole moment.

For a definition of the quadrupole moment (i.e. how to calculate it from a given charge distribution), take a look at this section of the wikipedia article on multipole expansions, which I found when reading the answer to Can nonpolar molecules exhibit dipole-dipole forces?.

  How is it measured for a particular molecule?

The accepted answer for this question cites a paper describing one method.


Reaction of acyl chloride with excess Grignard reagent

The following paragraph from chem.libretext.org1 sums up the answer to your question:

Acid chlorides react with Grignard reagents to produce tertiary alcohols. Two equivalents of the Grignard reagent are needed because the first equivalent reacts to form a ketone which then reacts with the second equivalent. Because of the high reactivity of the Grigard reagent, the reaction can NOT be stopped at the ketone.

Coming to your specific case the reaction would proceed with the following mechanism:

Reference:
(1) Acid Halide Chemistry https://chem.libretexts.org/@go/page/183124 (accessed May 24, 2021).


Is there a relation between the ionization energy and the electron affinity?

Okay, here's what I did (and I highly encourage everybody to do this along with this tutorial-style answer, just so you get the feel for prepping and looking at data from the world wild web).

Data Acquisition

First, I went and copied the data from the tables at Wikipedia: This one for the ionization energies and this one for the electron affinities.

Next I had to clean up this data somewhat. For the electron affinity data I first had to get rid of all the unnecessary data columns. This was quickly done with awk:

   awk '{print $1, $2, $4}' raw-affinity.dat > affinity.dat

Next, I filtered out all the lines containing "-" and "?" using inverse search with grep:

    grep -v "?" affinity.dat | grep -v "-" > affinity2.dat

And finally I got rid of all these pesky statistical uncertainties with either one or two numbers between parentheses with sed:

    sed -e 's/(.)//g' -e 's/(..)//g' affinity2.dat > affinity3.dat

Now the data file looks somewhat like this:

1 H  0.754195
3 Li 0.618049
5 B  0.279723
6 C  1.262118

Now for the ionization data file. Since we only want to look at the first ionization energy we can scrap all the other data. Again, we'll use awk for this:

    awk '{print $1, $2}' raw-ionization.dat > ionization1.dat

Next we'll filter out all the data we don't want in there (again with grep and inverse searches):

    cat ionization1.dat | grep -v "use" | grep -v "WEL" | grep -v "talk" > ionization2.dat

In this file we still have some nasty linebreaks... So I loaded the file in vim and substituted the line break + "CRC" with nothing and saved it as ionization3.dat:

    :%s/\nCRC//g
    :saveas ionization3.dat

I also corrected the value for At manually in vim while I was at it, because it wasn't entered like all the other data in the Wikipedia table.

The ionization data file now looks like this:

1 H  13.59844
2 He 24.58741
3 Li 5.39172
4 Be 9.3227

All right, so we have the data nicely prepared and we're ready for the next step, which is

Data Analysis Shenanigans

I did everything in python using the numpy and matplotlib packages. The code used can be found here.

I didn't really notice some large trend of something you might expect. The plot gets pretty messy somehow. The darker the scatter points, the heavier the element.

Maybe someone else notices a pattern that I'm missing right now.


No ! There are no method for plating aluminum. Thousands of people have tried. In vain. It is impossible to prevent the formation of a thin layer of aluminum oxide which is waterproof at the surface of the metal. The only way of preventing it is to dip a piece of aluminum in a solution of mercuric chloride. This produces a reaction $$\ce{2 Al + 3 HgCl2 ->  3 Hg + 2 AlCl_3}$$ and the mercury makes an alloy (or amalgam) with not yet oxidized aluminum, and this prevents the formation of the continuous oxide layer. But this alloy does not prevent the oxidation of Al in contact with air. The oxidation of the amalgamated aluminum proceeds with such a speed that the growth of the aluminum oxide filaments can be seen by the naked eye. In a couple of minutes, white filaments of oxide grow to become one or two centimeters long. This extraordinary phenomena stops when the mercury has been removed from the surface by the growing oxide filaments. Of course, today, it is forbidden to carry out such experiments, because of the mercury toxicity. Fifty years ago, it was allowed. I do remember.


Major product of a cyclopropanol with silver ions

A common test for haloalkanes (save fluoroalkanes) is reaction with $\ce{AgNO3}$, which initially yields the corresponding silver halide and a carbocation intermediate. The formation of the insoluble salt is the driving force for the reaction:

On a primary haloalkane this carbocation would be quenched with water, but adjacent to a three-membered ring this carbocation is primed for ring expansion. In addition to reducing ring strain, the resultant carbocation is even further stabilized by the oxygen atom:

Deprotonation gives the final substituted cyclobutanone product.


Good question... would a plasma qualify? Metallic luster is due to a sea of electrons, free to move and to reflect (well, re-radiate) incident radiation. A plasma can be created in a gas, such as the interstellar medium, or in Earth's atmosphere. The Kennelly-Heaviside layer is such a "shiny", reflective layer, causing skip signal propagation to extend the range of radio communication. So if you saw in radio wavelenghts, rather than the visual spectrum,  plasma is shiny!

Another possibility would be a Bose-Einstein condensate... but the act of illuminating it tends to destroy it.


Under which criteria HNO3 is a concentrated or a diluted acid?

According to the European regulation on classification, labelling and packaging of substances and mixtures, the classification of nitric acid depends on concentration.

At a concentration of 20 % ≤ c, nitric acid is classified as skin corrosive category 1A. (For a pure substance on the basis of the results of animal testing, this category corresponds to a substance that produces destruction of skin tissue, namely, visible necrosis through the epidermis and into the dermis, in at least 1 of 3 tested animals where responses are noted
following up to 3 minutes exposure and up to 1 hour observation.)

At a concentration of 5 % ≤ c < 20 %, nitric acid is classified as skin corrosive category 1B. (This category corresponds to a substance that produces destruction of skin tissue in at least 1 of 3 tested animals where responses are described following exposure between 3 minutes and 1 hour and observations up to 14 days.)

Furthermore, at a concentration of 65 % ≤ c, nitric acid is classified as oxidising liquid category 3.

Therefore, nitric acid is labelled in accordance with the Globally Harmonized System of Classification and Labelling of Chemicals (GHS) as follows:

Concentration 65 % ≤ c

H272: May intensify fire; oxidiser
H314: Causes severe skin burns and eye damage

Concentration 20 % ≤ c < 65 %

H314: Causes severe skin burns and eye damage

Concentration 5 % ≤ c < 20 %

H314: Causes severe skin burns and eye damage


How do I decide the parent chain of an organic compound that has a ring as well as a side chain while naming the compound?

If you have a simple chain with carbon atoms and an aromatic ring, the priority has the ring, not matter how the chain is long. This rule is NOT valid when you have a functional group: if it is in the linear chain, it has the priority, and the numeration starts from the carbon more near at the functional group, for example
$\ce{CH3-CH2-CO-CH2-CH2-CH3}$. The numeration starts from left, so you write hexan-3-one.


Briefly, it's a matter of cationic radii values and crystal packing at given $p, T$. Without referring to the detailed review of the corresponding crystal structures, you can imagine $\ce{M^{II}2C}$ as the derivative of methane $\ce{CH4}$, whereas $\ce{M^{II}C2}$ can be seen as the derivative of acetylene $\ce{C2H2}$. Smaller cation is the best fit for the "richer" tetrahedral coordination environment of carbon, whereas larger cation of the same charge won't be able to do so (sterical hindrance). That's why $\ce{Be2C}$ is the primary carbide of beryllium, $\ce{Mg2C}$ is also known to exist at higher pressures, (magnesium, along with lithium, otherwise forms an allene-based $\ce{Mg2C3}$ or $\ce{Li4C3}$ 2), but bulkier cations $\ce{Ca^2+, Sr^2+, Ba^2+}$ only form 1/2-carbides. Aluminium, on the other hand, forms $\ce{Al4C3}$ that hydrolyses to methane. This is a good example of the diagonal relationship between $\ce{Be}$ and $\ce{Al}$, and also between $\ce{Li}$ and $\ce{Mg}$.


Phase description for a substance dissolved in a solvent other than water?

IUPAC “Green Book” recommends to use $\mathrm{sln}$ for denoting a solution in general [1, p. 54], referring to earlier Recommendations 1981. Appendix No. IV to Manual of Symbols and Terminology for Physicochemical Quantities and Units [2, pp. 1240–1242].
This has been extensively covered in the following posts:

What is the standard way to denote physical states in a chemical reaction? (Meta)
Phase abbreviations for non-aqueous solutions

On the contrary, abbreviation $\mathrm{sol}$ is used to denote a process (typically a sub- or superscript to a symbol for a thermodynamic quantity), not state of matter [1, pp. 59–60]:

2.11.1 Other symbols and conventions in chemical thermodynamics
(i) Symbols used as subscripts to denote a physical chemical process or reaction
These symbols should be printed in Roman (upright) type, without a full stop (period).
$$
\begin{array}{ll}
\ldots \\
\text{solution (of solute in solvent)} & \mathrm{sol} \\
\ldots
\end{array}
$$

Another reason not to use $\mathrm{sol}$ for denoting a state of aggregation is its ambiguity: $\mathrm{sol}$ is often used as an abbreviation for “solid”, as listed in Appendix 10-2 Abbreviations, Acronyms, and Symbols in ACS Style Guide [3, p. 197]

$$
\begin{array}{ll}
\ldots \\
\text{sol} & \text{solid} \\
\text{soln} & \text{solution} \\
\ldots
\end{array}
$$

Also, specialized literature on electrochemistry of non-aqueous solutions uses $\mathrm{sv}$ to denote solvation process separately, see e.g. [4, p. 27]:

$ΔH^\circ_\mathrm{sv},$ $ΔS^\circ_\mathrm{sv},$ $ΔG^\circ_\mathrm{sv}$: Enthalpy, entropy, and Gibbs energy of solvation of the electrolyte

With this obligatory standardization part, the simplest reaction can be written as
$$\ce{CuI(s) ->[CH3CN] Cu^+(sln) + I^-(sln)}$$
However, copper(I) is extremely stable in acetonitrile due to solvation [4, p. 33]:

Interactions by Back-Donation from $\mathrm{d^{10}}$-Cation to Solvent Molecules
Acetonitrile (AN) has relatively small DN and usually solvates rather weakly to
metal ions. However, it solvates very strongly to $\ce{Cu+},$ $\ce{Ag+}$ and $\ce{Au+},$ which are univalent $\mathrm{d^{10}}$-metal ions.
This is because these metal ions have an ability to back-donate their electrons into a π*-antibonding orbital of the $\ce{CN}$ group of AN, as shown by $\ce{CH3C#N\overleftarrow{\ce{:\bond{->}A}}g+}.$
As a result, $\ce{Cu+}$ and $\ce{Ag+}$ in AN are stable and not easily reduced to metal, while the weakly solvated $\ce{Cu^2+}$ is very easily reduced to $\ce{Cu+}$ … , making $\ce{Cu^2+}$ in AN a strong oxidizing agent.

and may theoretically be coordinated by up to five acetonitrile ligands [5] with C.N. 4 being the most common unless there are other ligands.
Copper(I) and iodide don't seem to form a complex in acetonitrile [6, p. 194]:

Also in tetraethyl ammonium iodide and thiocyanate solutions, the dissociation is seemingly complete, … while a slight association is indicated for bromide … and, especially, chloride.

To sum it up, upon dissolving of copper(I) iodide in acetnitrile the formation of tetrakis(acetonitrile-N)copper(I) is the most likely scenario so that the final reaction can be written as
$$\ce{CuI(s) + 4 CH3CN(l) ->[CH3CN] [Cu(CH3CN)4]^+(sln) + I^-(sln)}$$
It should be noted that in a presence of better coordinating ligands $\ce{L}$ composition of the complex may, of course, vary $(\ce{[Cu(CH3CN)_xL_y]^z+}).$
Also, copper(I) iodide may form various coordination polymers based on $\ce{(CuI)_n}$ cluster core [7, 8, 9].
References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
Cox, J. D. Notation for States and Processes, Significance of the Word Standard in Chemical Thermodynamics, and Remarks on Commonly Tabulated Forms of Thermodynamic Functions. Pure and Applied Chemistry 1982, 54 (6), 1239–1250. https://doi.org/10/d783th. (Free Access)
The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. ISBN 978-0-8412-3999-9.
Izutsu, K. Electrochemistry in Nonaqueous Solutions; John Wiley & Sons, 2005. ISBN 978-3-527-60065-6.
Vitale, G.; Valina, A. B.; Huang, H.; Amunugama, R.; Rodgers, M. T. Solvation of Copper Ions by Acetonitrile. Structures and Sequential Binding Energies of $\ce{Cu + (CH3CN)_x},$ $x = 1−5,$ from Collision-Induced Dissociation and Theoretical Studies. J. Phys. Chem. A 2001, 105 (50), 11351–11364. https://doi.org/10/ckvsxw.
Ahrland, S.; Nilsson, K.; Tagesson, B.; Haaland, A.; Schilling, B. E. R.; Seip, R.; Taugbøl, K. Thermodynamics of the Copper(I) Halide and Thiocyanate Complex Formation in Acetonitrile. Acta Chem. Scand. 1983, 37a, 193–201. https://doi.org/10/bqvn5f.
Knorr, M.; Guyon, F.; Khatyr, A.; Strohmann, C.; Allain, M.; Aly, S. M.; Lapprand, A.; Fortin, D.; Harvey, P. D. Construction of $\ce{(CuX)_{2n}}$  Cluster-Containing ($\ce{X} = \ce{Br}, \ce{I}$; $n = 1, 2$) Coordination Polymers Assembled by Dithioethers $\ce{ArS(CH2)_mSAr}$ ($\ce{Ar} = \ce{Ph},$ $\ce{\textit{p-}Tol}$; $m = 3, 5$): Effect of the Spacer Length, Aryl Group, and Metal-to-Ligand Ratio on the Dimensionality, Cluster Nuclearity, and the Luminescence Properties of the Metal–Organic Frameworks. Inorg. Chem. 2012, 51 (18), 9917–9934. https://doi.org/10/f37kdx.
Knorr, M.; Khatyr, A.; Dini Aleo, A.; El Yaagoubi, A.; Strohmann, C.; Kubicki, M. M.; Rousselin, Y.; Aly, S. M.; Fortin, D.; Lapprand, A.; et al. Copper(I) Halides $(\ce{X} = \ce{Br}, \ce{I})$ Coordinated to Bis(Arylthio)Methane Ligands: Aryl Substitution and Halide Effects on the Dimensionality, Cluster Size, and Luminescence Properties of the Coordination Polymers. Crystal Growth & Design 2014, 14 (11), 5373–5387. https://doi.org/10/f6pnff.
Knorr, M.; Bonnot, A.; Lapprand, A.; Khatyr, A.; Strohmann, C.; Kubicki, M. M.; Rousselin, Y.; Harvey, P. D. Reactivity of $\ce{CuI}$ and $\ce{CuBr}$ toward Dialkyl Sulfides RSR: From Discrete Molecular $\ce{Cu4I4S4}$ and $\ce{Cu8I8S6}$ Clusters to Luminescent Copper(I) Coordination Polymers. Inorg. Chem. 2015, 54 (8), 4076–4093. https://doi.org/10/f7cjk2.


Why does the rate of the reaction decrease with an increase in temperature?

It's strange, but there are mechanisms whereby this type of thing can happen.  Basically you have a rate determining step that involves an intermediate species, and that rate determining step must have the usual positive temperature dependence.  But the preceding equilibrium that produces the intermediate species could have a negative temperature dependence that dominates.

Let's translate this into concrete terms for the reaction given.  A quite plausible mechanism would look like this:

$\ce{NO + O2 <=> ONOO}$ Fast, reversible

$\ce{ONOO + NO -> 2NO2}$ Rate determining step

The equilibrium in the first step is given by

$K_\mathrm{p1} =\dfrac{p^\mathrm{eq}(\ce{ONOO})}{p^\mathrm{eq}(\ce{NO})p^\mathrm{eq}(\ce{O2})}$

Assuming $\ce{ONOO}$ is a minor species, it's formation hardly affects the equilibrium partial pressures of the other species at any time, so we may just use $p(\ce{NO})$ as the nominal nitric oxide partial pressure and similarly for the molecular oxygen.  Then we can solve for the partial pressure of the $\ce{ONOO}$ intermediate:

${p^\mathrm{eq}(\ce{ONOO})}=K_\mathrm{p1}(p(\ce{NO}))(p(\ce{O2})$

Now put this result into the rate determining step with rate

$r_2=k_2{p^\mathrm{eq}(\ce{ONOO})p^\mathrm{eq}(\ce{NO})}=k_2K_\mathrm{p1}(p(\ce{NO})^2)(p(\ce{O2})$

Now for the eye-opener:  put in the temperature dependence of the equilibrium and rate constants

$K_\mathrm{p1}=K_\mathrm{p1}^0\exp(-\Delta H_1/RT)$

$k_2=k_2^0\exp(-E_{a2}/RT)$

So then

$r_2=k_2^0K_\mathrm{p1}^0\exp(-\dfrac{\color{blue}{E_{a2}+\Delta H_1}}{RT})p(\ce{NO})^2p(\ce{O2})$

The activation energy then contains two terms, one from the inherent rate in Step 2 and the other from the equilibrium in Step 1.  The latter is negative if Step 1 is exothermic, as it might well be if a bond is being (temporarily) formed between the reactants.  Then if this enthalpy term is the absolutely larger part of the sum ... we see a reversed temperature dependence!


What kind of bonds can form between benzene and transition metals in π-complexes?

In theory benzene and a transition metal could form σ, π, δ, and even φ bonds as the symmetry allows this.

Here are the orbitals of benzene, calculated at the PBE0/def2-TZVPP level of theory in the D6h  point group with Gamess.

The (valence) orbitals of chromium on the same level of theory in the same point group are as follows:

The linear combination of these orbitals would yield orbitals that have the above mentioned bond symmetries. We will ignore for now that they will most likely not occupied in a complex like bis(benzene)chromium.

benzene a2u + Cr s → σ
benzene a2u + Cr dz² → σ
benzene e1g + Cr dxz → π
benzene e1g + Cr dyz → π
benzene e2u + Cr dxy → δ
benzene e2u + Cr dx²-y² → δ

If we look at f-orbitals (that is of course not very feasible for chromium), we could construct another bond type:

benzene a1g + Cr fx(x2−3y2) → φ

The necessary orbital looks like this:

Of course which of these bonds will form or not depends on the metal in question, the occupation of the orbitals and more. After all, there are also p orbitals which can overlap to form σ and π bonds.

Colour code: Occupied orbitals are shown in blue and orange, while virtual orbitals are shown in yellow and red. Phases have been assigned randomly. The total density of the molecule uses a violet contour.


Is water a better nucleophile than bromide?

That is not true. As you noted bromine has a negative charge (and also lower electronegativity than oxygen) and is therefore the better nucleophile (even under these aqueous conditions). Halohydrins are not even usual the result of bromination in water. Special conditions are required to prevent too much $\ce{Br-}$ from attacking the bromonium cation (e.g. certain pH, adding additional reagents as $\ce{AgOH}$ or using $\ce{HOBr}$, see the section Addition Reactions Initiated by Electrophilic Halogen in W. Reusch's online Organic Chemistry book). You would usually expect the dihalide product since $\ce{Br-}$ is already close to the bromonium cation as it forms.


Which intermolecular interaction is dependent on the inverse cube of distance?

2

The ion-ion or charge-charge interaction is given by

$$
V_{\text{cc}} = \frac{Z_{i}Z_{j}e^{2}}{4\pi\epsilon_{0}} \frac{1}{|\mathbf{R}_{i} - \mathbf{R}_{j}|}
$$

and is therefore proportional to $R^{-1}$.

3

The ion-dipole or charge-(permanent) dipole interaction is given by

$$
V_{\text{cd}} = \frac{Z_{i}e}{4\pi\epsilon_{0}} \frac{\mathbf{R}_{ij} \cdot \vec{\mu}_{j}}{R_{ij}^{3}}
$$

and is therefore proportional to $R^{-2}$.

4

"London force" is often meant to encapsulate any of the van der Waals forces (here, here), but in particular the dispersion force. All of the above have a $R^{-6}$ dependence.

Here is a general table that covers each of the interactions so far:

\begin{array}{lll}
\hline
\text{type of molecular unit} & \text{type of force} & n~\text{in}~R^{-n} \\
\hline
\text{ions}                        & \text{coulombic}               & 1 \\
\text{ion - polar molecule}        & \text{ion - dipole}            & 2 \\
\text{two polar molecules}         & \text{dipole - dipole}         & 3 \\
\text{ion - nonpolar molecule}     & \text{ion - induced dipole}    & 4 \\
\text{polar and nonpolar molecule} & \text{dipole - induced dipole} & 6 \\
\text{nonpolar molecules}          & \text{dispersion}              & 6 \\
\hline
\end{array}

1

From Wikipedia:

  The hydrogen bond is often described as an electrostatic dipole-dipole interaction. However, it also has some features of covalent bonding: it is directional and strong, produces interatomic distances shorter than the sum of the van der Waals radii, and usually involves a limited number of interaction partners, which can be interpreted as a type of valence. These covalent features are more substantial when acceptors bind hydrogens from more electronegative donors.

This implies that hydrogen bonding interactions may not behave exactly as $R^{-3}$. Consider the water dimer:

which is a prototypical example of a hydrogen-bonded complex. From Stone, using density functional-based symmetry-adapted perturbation theory (SAPT-DFT) in the aug-cc-pVQZ basis,
\begin{array}{lr}
\hline
\text{Type of interaction} & \text{Energy (kJ/mol)} \\
\hline
\text{Electrostatic}                 & -33.4 \\
\text{Exchange-repulsion}            &  31.5 \\
\text{Dispersion}                    & -12.2 \\
\text{Induction}                     &  -8.0 \\
\text{Charge transfer}               &  -5.9 \\
\text{Exchange-dispersion}           &   2.3 \\
\text{Exchange-induction}            &   7.8 \\
\delta_{\text{HF}}~\text{correction} &  -3.4 \\
\hline
\text{Total}                         & -21.5 \\
\hline
\end{array}
the electrostatic component is clearly the largest contributor, but after taking the absolute value of all interaction terms, is about 32% of the total interaction. Another approach for decomposing the interaction energy is based on absolutely localized molecular orbitals (ALMO-EDA), which at the $\omega$B97M-V/def2-QZVPP level gives
\begin{array}{lr}
\hline
\text{Type of interaction} & \text{Energy (kJ/mol)} \\
\hline
\text{Electrostatics}  & -65.7 \\
\text{Pauli repulsion} &  65.0 \\
\text{Dispersion}      &  -7.7 \\
\text{Polarization}    &  -4.6 \\
\text{Charge transfer} &  -7.9 \\
\hline
\text{Total}           & -20.9 \\
\hline
\end{array}
SAPT-DFT and ALMO-EDA are in good agreement, though charge transfer is more important in ALMO-EDA primarily due to the lackluster definition of charge transfer within SAPT. Electrostatics are now about 44% of the total interaction. For reference, here is how the additional terms decay (Stone, Table 1.2):
\begin{array}{lccl}
\hline
\text{Contribution} & \text{Additive?} & \text{Sign} & \text{Comment} \\
\hline
\textbf{Long-range}~(U \sim R^{-n}) & & & \\
\text{Electrostatic} & \text{Yes}     & \pm & \text{Strong orientation dependence} \\
\text{Induction}     & \text{No}      & -   &                                      \\
\text{Dispersion}    & \text{approx.} & -   & \text{Always present}                \\
\text{Resonance}     & \text{No}      & \pm & \text{Degenerate states only}        \\
\text{Magnetic}      & \text{Yes}     & \pm & \text{Very small}                    \\
\hline
\textbf{Short-range}~(U \sim e^{-\alpha R}) & & & \\
\text{Exchange-repulsion}  & \text{approx.} & + & \text{Dominates at very short range} \\
\text{Exchange-induction}  & \text{approx.} & - &                                      \\
\text{Exchange-dispersion} & \text{approx.} & - &                                      \\
\text{Charge transfer}     & \text{No}      & - & \text{Donor-acceptor interaction}    \\
\hline
\end{array}

Answer

Although components of hydrogen bonding interactions may have a $R^{-3}$ distance dependence, there are significant contributors to the hydrogen bond interaction other than electrostatics that do not decay as $R^{-3}$; in general, it is a complex interaction and its definition is still hotly debated. For that reason, the best answer is (3), the ion-dipole interaction.

References

Masahiro Yamamoto, Kyoto University: charge-charge, charge-dipole, dipole-charge, dipole-dipole interaction
Stephen Lower, Simon Fraser University: Chem1 Virtual Textbook, Interactions between Molecular Units
Anthony Stone: The Theory of Intermolecular Forces
ALMO-EDA calculations: geometry is B97-D/def2-TZVP, EDA input as comment


Cyclopentadiene/etc. hydrophobicity after nitrogen atom substitution at a double bond?

The cyclic unsaturated imine (Schiff base) will be more hydrophilic for the nitrogen lone pair hydrogen bonding and being a Lewis base toward water.  You do know that CpH [4 + 2} Diels-Alder reaction reversibly thermally dimerizes, right?  The endo-dimer is an exo-flip then hydrogenation away from JP-10 cruise missile fuel.

The cyclic imine, ${2H}$-pyrrole is an isomer of pyrrole.  One expects it will be labile toward proton migration in base.

http://www.chemspider.com/Chemical-Structure.7330.html
http://www.chemspider.com/Chemical-Structure.5256848.html
Click the Properties tab, etc.


Where does the characteristic peak for phtalic acid diethyl ester at m/z 149 originate?

First of all, you should indicate if you consider the fragmentation in electron ionization or in other ionization conditions. Considering it is the electron ionization conditions and that you refer to this spectrum, one can postulate for a reasonable mechanism, which can be derived from what is known on the fragmentation of benzoic acid and esters. I am not aware of any complete study of phtalate ester fragmentation, but I did not carry a complete litterature search.

In the figure below is a proposal for the mechanism leading to the formation of two possible structures for the $m/z$ 149 ion.

If one consider that is fragments as do benzoate esters, through radical induced $\alpha$-bond cleavage leading to the loss of $\ce{CH3CH2O^.}$ followed by standard decarbonylation of the acylium ion. For phtalate esters, the question at this point is whether the structure will remain as I or might undergo isomerisation to more stable structures.

I have provided a few alternative structures: simple addition of the neighbouring carbonyl would lead to structure II which although it has some constraints, allows a good stabilization of the positive charge both in the ester and by the aromatic ring. Other structures can be postulated, such as III after addition of the ethanoyl oxygen to the aromatic cycle. This four membered ring can reopen to lead to the acylium structure IV which can in turn undergo decarbonylation to form the ion $m/z$ 121, also present in the spectrum. So there is some indirect evidence that these structures might be involved in the fragmentation process.

Finally, in order to explain the formation of ions at $m/z$ 105 and 77, one can also propose a last isomer for ion $m/z$ 149, structure V, formed after 1,5 hydride transfer. This one can quite easily lead to loss of acetaldehyde, eventually followed by decarbonylation to lead to ion $m/z$ 77.

If phtalates present specific reactivity, the other option is that ion A at $m/z$ 177 can cyclize to structure B which is a very good candidate for ethene loss leading also to an ion at $m/z$ 149.

The question then becomes which of these two $m/z$ 149 is the actual ion. These two isobaric ions could be resolved by high resolution experiments, but after looking in publicly available databases, I did not find any high resolution spectra. Deuterium labelling of the ethyl groups could also prove handy, but once again I did not find any spectra of diethylphtalate labelled on the ethyl groups. (The usual diethylphtalate used in quantitation experiments seems to be the $d_{4}$ isotopomer, with deuterium on the aromatic ring.) Finally one can also have a look at isomers and related ions to see how they fragment. And indeed if one takes the dipropylphtalate ester, the ion at $m/z$ 149 is present (produced after propanoyl loss followed by propene loss) and there is no strong ion at $m/z 163$ which would tend to be a good indication of ethene loss for diethylphtalate. But, on the other hand, both the meta and para isomers of diethylphtalate have a much weaker ion at $m/z$ 149, which cannot be accounted so easily by an equivalent to intermediate B. So I would probably favor the concurrent occurence of both pathways, with a ratio that still needs to be measured.

In addition to the answer above, it should have been quite immediate that a McLafferty rearrangement was not an option: since there are no nitrogen, the nitrogen rule allows to conclude directly that the odd mass of the ion at m/z 149 indicates that it is an even electron cation, as opposed to radical cations produced by McLafferty rearrangements.


In chemistry, you don't solve things like that; you just recall the right class of reactions (Hofmann elimination in this case) and apply that knowledge. You can hardly expect to rediscover that from the first principles. Knowledge about other classes of organic compounds and their behavior in the presence of a base is not particularly helpful either. It is not until much later that those seemingly random pieces of data start to fit into a bigger picture.

As for the mechanism, it is all there in your second picture. Indeed, that $\ce{OH-}$ would readily snatch a proton from an ammonium cation $\ce{NH4+}$, only there is no ammonium, it is a substituted ammonium, and we don't have any H linked directly to N. So we go for the next best thing and grab a proton from some carbon in $\beta$ position, simultaneously with the formation of a double bond.

Note the reversed regiochemical preferences as compared to Zaitsev-type elimination, which has a different mechanism, usually involving the appearance of a carbocation. (Not that it matters much in your case, since we only have one option anyway.)


Magnitude of Lennard-Jones force evaluated with molecular dynamics simulation

The following plots the force for reasonable distances

Intuitively, if you are implementing the verlet algorithm then the step size will be $\delta x \propto a\delta t^2$ (ignoring velocities). If you want a step size of order ~$f \times \sigma$ (where f is a scaling factor) then you'd like to satisfy a relation such as $\delta t < \sqrt{\frac{f\sigma}{a}}$ (if ignoring velocities). The timestep cutoff $\delta t$ computed according to this equation with a force of $\pu{10^-12 N}$ and f=0.1 is 0.3 psec. But 0.3 psec is still a relatively long timestep. Clearly the problem is not with acceleration (provided the timestep is shorter than the cutoff just calculated), but maybe there is one with the velocities. You can proceed in the same way but solving the complete quadratic verlet equation for the timestep in terms of the step size, and inserting a value for the rms speed based on the desired T and step size, to arrive at a timestep cutoff. Using this reasoning I came up with a cutoff $\delta t < \pu{ 15 fsec}$
One problem with a large step size (and discretization of the trajectory in general) is that the energy is not conserved and can easily blow up (along with the temperature): if you take a step of order $\sigma$ from the potential well minimum toward the maximum in the potential then the atoms will suddenly find themselves in the strongly repulsive high energy region.
You may want to check Ref. 1 which discusses this to some extent, and provides this advice:

One wants to find the largest time step that will maintain the conservation of energy.

Reference

Sangrak Kim. Issues on the Choice of a Proper Time Step in Molecular Dynamics. Physics  Procedia 53 (2014) 60–62.
doi: 10.1016/j.phpro.2014.06.027


How do you explain the formula Fe3O4 using the ionic compound theory?

Iron forms both $\ce{Fe^2+}$ and $\ce{Fe^3+}$ ions. The first combine with oxygen ions to $\ce{FeO}$; the second $\ce{Fe2O3}$.

$\ce{Fe3O4}$ can be thought of as a mixture of both compounds in equal amounts. So $\ce{Fe3O4}$ contains both Fe(II) and Fe(III) ions; twice as much Fe(III) as Fe(II).

In the picture below, showing the crystal structure, green atoms are Fe(II) and brown atoms Fe(III). All Fe(III) atoms are linked with six O atoms, sharing 'half an electron' with each oxygen; all Fe(II) atoms are linked with four O atoms, and all O atoms are linked with four Fe atoms.

(source: Manoj B. Gawande, Paula S. Branco, and Rajender S. Varma, Chem. Soc. Rev. 2013, 42, 3371-3393. DOI: 10.1039/C3CS35480F See also: ResearchGate
)


Silver nitrate mirror on 3d printed sla resin

My experience with glass reduced-silver mirrors is that bonding is poor, and that unprotected Ag rapidly degrades in air (a matter of days, in some cases). The bond to organic surfaces, I believe, might be worse.

If you have access to a fairly good vacuum pump (less than 10-4 Torr), it's not hard to aluminize. Watch a demo online.

You might also request a quote from an aluminization service, such as those in this list. In 2010, a 200 mm diameter mirror was said to cost about US$70 to aluminize. This British service has an online price list, starting at £20.


Simpler explanations, can by their nature be less accurate than a detailed description, but are helpful in beginning to understand things.  Getting to the heart of your question:
“How is the aromaticity in graphene different from the aromaticity in benzene? . . .  but is it possible to find a simpler if less accurate way to explain the difference,”
Perhaps a productive way to approach your question [portion given above] is to consider the molecular structure of coronene and circumcoronene while contrasting them to the interpretation of aromaticity of graphene given in your cited article.
 circumcoronene
 CORONENE
Graphene from article cited in question.
According to the computational method employed “adaptive natural density partitioning method” [we can choose to agree or not agree with their interpretation as indicated by a comment to your post by @A.K.], the aromaticity of graphene is centralized within the set of hexagonal rings.  To quote the authors “Aromaticity in graphene is local with two π-electrons delocalized over every hexagon ring.”  The salient point here being the “delocalization” is within and isolated to or restricted to [my words] a six-membered ring.
As discrete molecules, the aromaticity in coronene and circumcoronene, as depicted the PubChem structures, is delocalized throughout the entire discrete molecule and not restricted to a discrete set of hexagonally bonded carbon atoms.
In other words, it comes down to electron density distribution.  In graphene, according to the authors of the article cited, the areas of highest electrons density are within the center of the six-membered ring while in coronene and circumcoronene the electron density is more uniformly distributed over the entire molecule.
Ref:
https://pubchem.ncbi.nlm.nih.gov/compound/coronene
https://pubchem.ncbi.nlm.nih.gov/compound/Circumcoronene


Do molecules containing only carbon exist?

They do form.  Besides fullerenes, there are carbon molecules with varying numbers of atoms around some stars. Such molecular forms are, of course, less stable than the macromolecular structures we more commonly see under ambient conditions.


NMR population difference from a pulse of known flip angle

immediately after the pulse:

$M_z = M_0cos(\beta )$

$M_x = M_0sin(\beta )$

$M_y = 0$

so for example after a 90 degree pulse, all the magnetization has been rotated into the x-y plane, and after a 180 pulse, the magnetization is in the -z direction.


How many dichlorinated products including stereoisomers can be isolated when (S)-2-chlorobutane reacts with Cl2 /hv?

Elaborating on what Waylander said, The C-2 carbon chlorine should be in S configuration so both (III) and (IV) structures can't be correct and only one with S configuration(IV) will be considered and rotating that will result in the other structure.


The radiocarbon method is in principle fairly simple. $^{14}\ce{C}$ is an isotope of Carbon that has a relative abundance of around $10^{-10} \%$ on earth.

Addition by @MaxW:

  $^{14}\ce{C}$ is formed in the upper atmosphere from $^{14}\ce{N}$ so
  that the relative amount of $^{14}\ce{C}$ in the atmospheric
  $\ce{CO_2}$ is reasonably constant. Since plants get most of their
  carbon from atmospheric $\ce{CO_2}$, the relative amount of
  $^{14}\ce{C}$ to $\ce{^{12}C}$ is thus constant.

It is also not stable and decays $^{14}\ce{C ->}^{14}\ce{N} + e + v$.

Now a living organism based on carbon (like us) always has a stable amount of $^{14}\ce{C}$ in his body, because he has to eat and so there is a stable balance between $^{12}\ce{C}$, $^{13}\ce{C}$ and $^{14}\ce{C}$.

Now if that organism dies, it does not eat anymore, so there is no new income of $^{14}\ce{C}$. Since that decays and the other $\ce{C}$-isotopes don't, over the years, the percentage of $^{14}\ce{C}$ is getting lower and lower at a constant rate (you can calculate that rate with the half-time of $^{14}\ce{C}$).

Now when you measure the relative amount of $^{14}\ce{C}$ in a skeleton, you know since when it hasn't eaten anymore, so you know how old it is. The method can be used around 600-50k years to the past.

There are different methods of measuring isotope amounts. You can measure how fast a radioactive decay still is and calculate from that the amount of $^{14}\ce{C}$, which is the method from Libby.

You can also use Mass-Spectroscopy, because $^{14}\ce{C}$ is heavier than  $^{12}\ce{C}$ and $^{13}\ce{C}$ you can "easily" measure the amount of $^{14}\ce{C}$.

There are other methods, you can look this up in Wikipedia.


In the absence of other chemicals you can utilize the fact that nitrates(V) are much stronger oxidizing agents than phosphates(V).
You can soak a piece of newspaper in a saturated solution of the salt, let it dry well in a ventilated place and try to ignite it.

If the salt was $\ce{KNO3}$, it will burn vigorously and fast (that's why kids are often using paper soaked in sodium nitrate as primitive fuse for home-made firecrackers and rockets).

$\ce{KH2PO4}$ doesn't promote combustion of paper, so it will burn normally or even slower.