Atomic radius and nuclear effective charge in a group

Just so that we are on the same page before I attempt to answer this question, the definition of effective nuclear charge we are considering Zeff=Z−σ, where Z is the nuclear charge (number of protons) and σ is the shielding factor of the inner shell electrons. The overall effective nuclear charge experienced by the valence electron in question is Zeff.

You are right in thinking that the effective nuclear charge increases down as a group as the increasing nuclear charge has a greater effect on Zeff than the shielding effect of more inner electron shells. This, as you quite rightly mention, should suggest that because Zeff increases down a group, a greater force of attraction would be experienced between the outer shell valence electrons and the nucleus, thus decreasing the atomic radius. However, what we have failed to consider is the size of the electron orbitals in question. The principal quantum number, n, of electron orbitals affects their size. The quantum mechanical description shows that the 1s orbital of a hydrogen atom is a lot smaller in radius than the 2s orbital of a lithium atom for example. This comparison can be further extended where the 3s orbitals are larger in radius than 2s and these are larger than 1s. Thus we have a quantum mechanical picture whereby the principle quantum number of electron orbitals affects their radius.

Essentially we have two competing factors that affect the trend of atomic radii in question:

The effective nuclear charge, Zeff, increases down a group which draws electrons closer towards the nucleus, decreasing atomic radius.
The principal quantum number, n, of electron orbitals that increases down a group and due to the quantum mechanical nature of electrons, the radius of these electron orbitals increases with increasing n, thus increasing in size as you go down the group.

Simply put, the factor of increasing n down a group is greater than the effect caused by the increasing Zeff, thus causing the atomic radius of atoms to increase down a group in the periodic table.


Why is tetramethylammonium hydroxide a stronger base than tri- or dimethyl amine?

Between 1 and 2, 2 is more basic as the 3 methyl groups would produce a +I effect resulting in better availability of the lone pair than in option 2.

I agree that more methyl groups would result in a greater +I effect.

  But the book says it is 3. I don't understand why. I do make out that it is ionic, but how does this fact make it more basic than option 1?

The fully methylated ammonium part is not basic. It has no easily labile proton. As a result, tetramethylammonium hydroxide acts as a plain hydroxide salt, like sodium hydroxide. Hydroxide ion is much more basic than ammonia (at least in water).
So we've eliminated the amines with lone pairs as contenders for "most basic" because we've established hydroxide to be much more basic than an amine, regardless of the extent of the +I effect in an amine.

We're left with $\ce{NaOH}$ and tetramethylammonium hydroxide. The only difference between the two is the cation.

$\ce{Na+}$ is electron-withdrawing (electronegative); it has a positive charge. This lessens (stabilizes) the negative charge density on the hydroxide portion, reducing the basicity of the hydroxide ion slightly.

$\ce{(H3C)4N+}$ has a similar effect on the hydroxide ion part. However, tetramethylammonium ion has four inductively donating methyl groups to lessen its electron-withdrawing tendency. I suspect this is what your book is getting at.

Some speculation follows. In addition, $\ce{Na+}$ is likely smaller than $\ce{(H3C)4N+}$, which has a lot of bulk associated with its four methyl groups. This likely allows $\ce{Na+}$ ions to more effectively solvate and thereby stabilize hydroxide ions in solution. This again decreases basicity, but I'm unsure of how much of an effect this has, since the solution molecules will be present in a much greater quantity than the solute ions.


Why is (tri-)ammonium phosphate ((NH4)3PO4) so unstable?

Ammonium phosphates are solids between 0 and 75°C. Because ammonium
  hydroxide $(\ce{NH4OH})$ is a much weaker base than the common metal
  hydroxides, ammonium phosphates are comparatively unstable.  Both
  triammonium phosphate ($\ce{(NH4)3PO4}$) and the double salt
  ($\ce{(NH4)3PO4.2(NH4)2HPO4}$) are unstable at room temperature and
  evolve ammonia to form diammonium phosphate. Even the commercial mono-
  and diammonium phosphates exhibit an ammonium vapor pressure in the
  solid form and in solution. Although MAP is more stable than DAP, it
  decomposes at high temperature to give ammonia and polyphosphoric
  acid. Diammonium phosphate decomposes to give ammonia and monoammonium
  phosphate at around 70°C.

Source: (2005) Ammonium Phosphates and Ammonium Sulfate. In: Synthetic Nitrogen Products. Springer, Boston, MA. DOI: 10.1007/0-306-48639-3_12.


As the comments imply, having a center of symmetry makes a molecule achiral.  This is because, geometrically, inverting through the center equals a mirror reflection plus a 180° rotation.  If you start with an atom at point $(x,y,z)$ and you reflect it through the $xy$ plane, the atom is now at $(x,y,-z)$.  Now rotate 180° around the $z$ axis and you're at $(-x,-y,-z)$ just like inverting through the center.
Your molecule has no symmetry with respect to just a mirror reflection or just a 180° rotation.  (It might look like you could rotate 180° within the plane of the paper, but the stereochemistry is different.)  But it does have a symmetry with respect to the combination of the two, which is that center of inversion, so you can match the mirror image to the original by allowing the rotation.  That makes the molecule meso.


Can I determine purity of isopropanol with a hydrometer?

There is an image of a table provided by Chu and Thompson [1] listing isopropanol water solution densities:

The essential data part is (weight % vs density):

  77.87 0.8361
  89.84 0.8069
  99.91 0.7808

There is density gradient -0.0026 / 1 w%, resp. -0.00052 / 0.2 w%, determining purity by density via hydrometers is rather illusory.

Some success could be gained by pycnometers — using special glass flasks with a glass cap with a hollow. You would weight the flask, the flask with water and the flask with isopropanol and you would calculate the density.

In any case, as differences are minimal and as liquids have considerable thermal dilation, one must carefully watch the liquid temperature.

In case of the pycnometry measurements, the glass and both liquids must have the same temperature, the best the one for which density is tabelized.

References

Chu, K.-Y.; Thompson, A. R. Densities and Refractive Indices of Alcohol-Water Solutions of n-Propyl, Isopropyl, and Methyl Alcohols. J. Chem. Eng. Data 1962, 7 (3), 358–360. https://doi.org/10.1021/je60014a011.


Who came up with the box and arrow diagrams for electron configurations?

This image is from Atomic Spectra and Atomic Structure by Gerhard Herzberg, John William Tranter Spinks - 1944. It does not really have boxes (just boxes of a table), but spin +1/2 and spin -1/2 are shown as arrows. Of course, I am not claiming this is the first usage, just that in 1944 it existed.


Balancing a REDOX reaction using the oxidation number method, in acidic solution

Whenever you balance a redox reaction in acidic or basic solution, you have to account for interactions with the solution itself. The easiest way of doing this is by the half-reaction method.

This is a key detail that I find a lot of students miss - you cannot balance a reaction like this without considering the solution, because the solution itself is participating. Your clue that this is happening is the phrase "in acidic (or basic) solution" - any time you see that in a redox problem, you should be thinking "half-reaction method." Another clue is that H and O appear on only one side of the equation. If you are familiar with the algebraic method of balancing equations, another clue is that the system of equations would be unsolvable.

At any rate, once you know that you need to use the half-reaction method, the next step is to use it!

This tutorial has a good description of the method, your book probably does as well, and I'll give a brief description here for good measure.

These are the major steps:

Split the reaction into two half-reactions. Then, for each reaction
Balance everything except for H and O
Balance O by adding $\ce{H2O}$ (water is available in solution, so this is legal)
Balance H by adding $\ce{H^+}$ (it's an acidic solution - so this is also legal)
Balance charge by adding electrons ($\ce{e^-}$)
Multiply the half-reactions by the appropriate coefficients so that when they are added together, the charges (electrons) cancel - this is necessary because electrons are not created or destroyed in chemical reactions.
Add the equations together, and if the solution is basic, add $\ce{OH^-}$ to both sides to remove all protons.

I'll use an example from my copy of Brown, LeMay, Bursten and Murphy's "Chemistry: The Central Science" to demonstrate (all phases are aqueous):

Balance the following reaction in acidic solution:

$\ce{MnO4^- + C2O4^2- -> Mn^2+ + CO2}$

Break the reaction into two half-reactions:

$\ce{MnO4^- -> Mn^2+}$

$\ce{C2O4^2- ->CO2}$

It's not always easy to see how to do this. The quick way is to make sure that the elements that are not O and H are in their own equations. The proper way is to figure out what is being reduced and what is being oxidized, then split them so that the oxidation happens in one reaction and the reduction happens in the other. In this case, Mn is reduced and C is oxidized.

Balance everything except for H and O:

$\ce{MnO4^- -> Mn^2+}$

$\ce{C2O4^2- -> 2CO2}$

This step is usually pretty easy.

Balance O by adding $\ce{H2O}$:

$\ce{MnO4^- -> Mn^2+ + 4H2O }$

$\ce{C2O4^2- -> 2CO2}$

We start with O because we are using water to balance it. Each time we add a water, we also add two H, so it doesn't make sense to start with H.

Balance H by adding $\ce{H^+}$:

$\ce{8H^+ + MnO4^- -> Mn^2+ + 4H2O }$

$\ce{C2O4^2- -> 2CO2}$

Remember that we can do this because it's in acidic solution. You are probably wondering what we do for basic solution. The answer is - use protons! There is a trick we can do at the end to "fix" it for basic solutions, and this is much easier than iterating between $\ce{OH^-}$ and $\ce{H2O}$

Balance charge by adding electrons:

$\ce{5e^- + 8H^+ + MnO4^- -> Mn^2+ + 4H2O }$

$\ce{C2O4^2- -> 2CO2 + 2e^-}$

While this is easy to understand, I have found that in practice, a majority of mistakes happen here. Be careful when counting charges - remember to multiply by the coefficients, and remember that the charges don't have to add up to zero yet - they just have to have the same sum on both sides.

Multiply the half-reactions by the appropriate coefficients so that when they are added together, the charges (electrons) cancel:

$\ce{10e^- + 16H^+ + 2MnO4^- -> 2Mn^2+ + 8H2O }$

$\ce{5C2O4^2- -> 10CO2 + 10e^-}$

This part is also prone to error - be careful and remember to multiply every species by the coefficent.

Add the equations together

$\ce{16H^+ + 2MnO4^- + 5C2O4^2- -> 2Mn^2+ + 8H2O + 10CO2 }$

If the electrons don't cancel out here, then you made a mistake - go back and check each step.

If the solution is basic, at this point you would add $\ce{OH-}$ to both sides of the equation until the 16 protons were eliminated. We can add anything we want as long as it's to both sides (since this is, after all, just algebra). Each $\ce{OH^-}$ will react with a proton to form one water molecule, and so the waters on the right would cancel out some of the waters on the left. The result would be a similar equation, but with $\ce{OH^-}$ instead of $\ce{H^+}$.

Try out your problem using these steps, and if you get stuck, update your question.


Stability of organophosphorous vs organophosphates?

Setting the scene for your question

In Environmental Chemistry there are several pathways a chemical might travel through including:

being subjected to the biochemistry of an animal
photodegradation
chemical degradation (perhaps it slowly reacts with water, etc.)
uptake as a metabolite in plants
degradation by soil microorganisms

and this is ignoring aquatic and air pathways.

Say, within a mouse or a fly, a possible pathway for degradation for some insecticide is cleavage of the P-O bond. How they have determined this in the past is noteworthy; a synthetic chemist would isotope label an atom in the molecule, say $^{33}$P-labeled or $^{14}$C-labeled, and he purified it before finally determining its specific activity. Then a biochemist would dose a mouse with a solution of the compound dissolved in an appropriate solvent, such as olive oil. Urine and feces were collected after administration. An aliquot of the collected urine, for example, would have its radioactivity measured using scintillation counting. The organo-soluble species would be extracted from the rest of the urine and the distribution of radioactivity could then be determined between the aqueous and organic layers. Then one could investigate what were present by TLC if you're old school -- they would actually run, via cochromatography, all the possible metabolite products they could independently think of and synthesize to identify a TLC spot! (That is a lot of work.) These days we can cheat with HPLC-MS. So, you might ask, "are results from this generalizable?" I have to tell you that they generalize in a very rough way. Enzymes are responsible for all the nifty biochemistry and they tend to be rather selective towards a given substrate and, in many cases, chirality even matters! What they are picky about is a tough problem that is tailored to each individual enzyme. One can make all kinds of generalizations about the chemical nature of the bonds and some possible predictions from it, but those won't necessarily hold up in an animal model with a different molecule.

So nitty-gritty mechanisms might not be what we want to look at. Instead we'll step back and take a broader view of this landscape. We need to look at bigger items in terms of physical properties, namely:

1) Persistence: what is its half-life in soil, etc. This is a matter of degradation, whereas the next is all about movement of an intact molecule. A simple chemical way to test this is how they resist hydrolysis.

2) Mobility: over time, does the compound tend to sit or spread itself around?
For example, factors that influence this are:

sorption: how much does our chemical bind to our soil?
water solubility: how easy it is to move in water for our chemical?
vaporizability: how likely is the material to evaporate away from its source?

A famous organophosphate is malathion. The partition coefficient, according to the wiki page, is 2.36. This implies its solubility is greater in organics than water, but some of it will most certainly dissolve in water. I think, based on its size, it is reasonable to assume it has a negligible vapor pressure too. These factors suggest it will have a low mobility: it will tend to stick to organics, won't appreciably evaporate away and will take quite a lot of water to wash it away. Contrast this with a famous organochlorine, DDT, which has an even higher partition coefficient (on the order of 10, so even more sticky and less soluble in water) and an even higher mass (so probably a lower vapor pressure) and thus an even lower mobility. Indeed, organophosphates largely replaced organochlorines as pesticides, in part, because they persist for shorter periods of times in the environment.

Many of these organophosphates were tested by us and their history is relevant. World Wars I and II ushered in a new era of chemical warfare agents and pesticides, notably of the classes of organochlorines, organophosphates and carbamates. Some organophosphates hardly persist at all, such as tetraethyl pyrophosphate, but they are pretty toxic to mammals. Many other organophosphates were screened for low toxicity towards mammals and increased persistence. We actually sought longer-lasting half-lives for these compounds! It turns out that we see alkyl phosphates with shorter lives than the longer-lived aryl species.

We tend to see certain trends within the different organophosphates, such as the water solubility being greater for the less complex and more polar compounds. An ethyl- is less soluble than a methyl- derivative and an aryl-substituted species get partition coefficients approaching those of the organochlorines.

"All this is fine and interesting, but it doesn't have much to do with my direct question," you might say. And most of the above just sets the scene for my answer to your question. Your specific query, I believe, requires us to go back to a nitty-gritty detail which is really only relevant in the case of your comparison between a phosphate (P-O-C) and an organophosphorus compound (P-C).

My answer to your question

In the biochemistry of most life, phosphates are pretty ubiquitous. As a result, many enzymes have evolved to manipulate them in many different forms; one class of these enzymes is called phosphatases and they assist dephosphorylation reactions. They certainly cannot manipulate just any ol' phosphate (hence why more complex phosphates have longer residues). We could say organophosphates are more "natural" than your typical organophosphorus compound. There is, therefore, a greater chance of having a tool -- an enzyme -- that can pick your more "natural" looking compound apart than your harder to find P-C bond.

As you've indicated, these phosphates are actually pretty stable. It's why DNA is built using them! But when you compare the bond lengths between P-C and P-O and O-C, there's a decent chance of finding that a P-C bond length is longer than a P-O or O-C. This leads me to believe that there are many cases in which P-O and O-C are actually stronger than P-C! The same could easily be said in the other way too! This, I think, is why we have to step-back and look at the bigger picture: the exact physical properties vary considerably from one compound to another and paint a landscape that says the issue is complicated. Perhaps, instead of it being a mere thermodynamic issue dealing with a comparison of the general bonding energies, it has more to do with the kinetic advantage of having enzymes equipped to slice the P-O-C rather than P-C. Of course, your mileage may vary depending on your specific examples.


Mono chloro substituted products of naphthalene

Your chemistry teacher is partially correct by saying that naphthalene molecule is symmetrical about the axis passing through the $\ce{C-C}$ bond where the benzene rings are fused (going through the common carbons to both rings). Your chemistry teacher has forgot to mention that naphthalene molecule is also symmetrical about the axis perpendicular to the axis passing through the $\ce{C-C}$ bond where the benzene rings are fused (See the top structure of the diagram below). Thus, each quarter of the naphthalene molecule is equal and each quarter contains only two carbon atoms (disregarding common $\ce{C-C}$ bond). I named them as 1 and 2, because they are symmetrical. Therefore, only two different mono-substituted derivatives are possible from naphthalene. If substituent is chloro, then only 1-chloronaphthalene and 2-chloronaphthalene are possible, and so on.


Clarification in the mechanism for Molisch's test for glucose

YUSUF HASAN: 5-Hydroxymethyl furfural (5-HMF) 1 when protonated on the carbonyl oxygen becomes a reactive electrophile. Addition of phenol 2 at the reactive para-position affords 3 which deprotonates at the para-position, rearomatizing the ring to a phenol and liberating a proton. Secondary alcohol 4 is protonated on oxygen with loss of water to form stabilized cation 5. A second equivalent of phenol adds to the cation as previously described leading to the triarylmethane 6. The methane hydrogen is susceptible to oxidation by atomspheric oxygen, which is in a triplet state (unpaired electrons behaving as a free radical). Resonance stabilized radical 8 is formed along with the hydroperoxy radical 7, which can abstract the hydrogen from triarylmethane 6 forming more of radical 8 and hydrogen peroxide. Oxidation of radical 8 with oxygen gives the resonance stabilized carbocation 9 and superoxide anion, aka, superoxide radical anion 10. This species can be protonated by the the strong acid conditions of the Molisch test to form more of the hydroperoxy radical 7.


What determines how favorable (exothermic) the electron affinity of a certain element is?

The electron affinity (EA) of an atom (A) is defined as the minimal energy that is required to free an electron (e$^-$) from the associated anion (A$^-$)

$$
\text{EA: }\qquad\text{A}^-\text{(g)}\rightarrow \text{A(g)}+\text{e}^-,
$$
in other words, the electron affinity is the ionization energy of the associated anion.

A large positive EA means that the anion A$^-$ is stable, while a negative EA indicates that the anion ${\text{A}^-}$ is unstable (such as He$^-$). Halogens have the largest electronegativity as the associated anion obtains a completely filled shell.

In general, if an atom A obtains a full/half full shell/sub shell by addition of an additional electron, then A will have a large EA, while A will have a small EA if it already has a full/half full shell/sub shell.

In addition there are some competing effects along a group in the periodic table. Consider the EA of the halogens for example:

Atom:      F       Cl      Br      I
EA /eV     3.40    3.61    3.36    3.06

(i) The smaller the principle quantum number $n$ of the highest sub shell, the larger is the attraction between the nucleus and electrons and the larger is EA. This would predict the trend to be EA(F) > EA(Cl) > EA(Br) > EA(I).

(ii) The smaller the principle quantum number $n$ of the highest sub shell, the larger is the electron repulsion the smaller is EA. The expected trend is EA(F) < EA(Cl) < EA(Br) < EA(I).

As these effects are in competition with each other, there is a maximum in the observed trend for the halogens around chlorine. Where the maximum occurs, is unfortunately difficult to predict.


Difference between subshell and sub-subshell?

As you pointed out the the s, p, d, and f denoted subshells of atoms.

The term sub-subshells is very uncommon. I did find a couple of uses.

Materials Science
By G. K. Narula, K. S. Narula, V. K. Gupta

pdf file about Modern Atomic Mechanical Theory, page 24

What the authors are noting is that the 2p subshell will have three orbitals which can support two electrons each. Each of these three orbitals is being referred to as a sub-subshell.

As I said it seems a very uncommon term, and I'd suggest you avoid using it.


Wasn't Rutherford's assumption of gold foil to be made of single layer wrong?

You were confused about the result of the experiment. It did not measure the size of the atom, but the size of the nuclei. Nucleus is the area of concentrated positive charge in the core of the atom, and is only a very small fraction the size of the atom itself.

And the result, $10^{-15}\ \mathrm{m}$, was the size of the nuclei, while the size of gold atoms has the order of magnitude of ångströms ($1\ \mathrm{\mathring{A}}=10^{-10}\ \mathrm{m}$).

The measurement comes from the analysis deflection angle of alpha particles when hitting the nuclei. The one-layer-of-atom assumption leaded to Rutherford to use a mathematical model for one single collision to analyze the angular distribution. Because the nucleus is so small, even if there were a small number of layers, most alpha particles will still only have one collision and the result will be correct.

However, the alpha particle was passing through some hundreds of atoms and many particles actually were deflected multiple times. The calculation would be much more complicated. Even if he knew the number of layers of atoms and tried to do it the right way, he might not be able to arrive at the correct number. Of course people back then would not be able to know that, but nevertheless, the existence of nucleus is still evident and Rutherford did almost get the order of magnitude right.


Why is the pKa (COOH) of alanine and glycine similar but the pKa (NH3) different?

I found the following pKa values for glycine and alanine from the CRC Handbook of Chemistry and Physics via Sigma-Aldrich:

Glycine $\ce{COOH}$: 2.34
Alanine $\ce{COOH}$: 2.34

Glycine $\ce{NH3+}$: 9.60
Alanine $\ce{NH3+}$: 9.69

First, although the values for the carboxylic acid are identical as reported, you would see a difference if you found a value reported with enough precision. The carboxylic acid is not immune to changes in the structure of the rest of the molecule.

Second, the difference between 9.60 and 9.69 is pretty small. In most practical situations, the two amino acids would behave identically with regard to the $\ce{NH3+}$ pKa.

The fact that the magnitude of the difference is larger for the $\ce{NH3+}$ pKa values can likely be attributed to proximity. In this case, the relevant proton is 2 bonds away from the $\alpha$ carbon, whereas the proton in the $\ce{COOH}$ group is 3 bonds away.


The stoichiometric coefficients is something else than you suggest.
stoichiometric coefficients:
The number of molecules of a given component that participate in the reaction as written.
The indexes $x, y, z$ represent quantities of atoms in a compound. So they do not have any units.
@Q2 Do you know what are $c, M$ and $x$ in the equation? If you do, you can answer your question. If you do not, than you should find it out.


Sodium benzoate in soft drinks reacting to produce benzene?

The reaction taking place is described in article Benzene production from decarboxylation of benzoic acid in the presence of ascorbic acid and a transition-metal catalyst and its importance for food safety recently reviewed in Evaluation of Accelerated UV and Thermal Testing for Benzene Formation in Beverages Containing Benzoate and Ascorbic Acid.

To summarize, the reaction CAN occur, namely at elevated temperatures (>50 °C) and UV-light and in presence of transition metal. Even if the reaction takes place, the amount of benzene formed is either below detection limit, or below limit for drinking water.

For the end user it means, that in extremely unlucky situation, the benzene can form in quantity, which is not considered safe to drink in unlimited amount for the whole life (1 ppb). As such, I would consider the benzene formation as very weak argument against such beverages (mainly if compared with other problems, such as sugar content and associated obesity).


Effective nuclear charge and Ionization energy

Effective nuclear charge $Z_{\text{eff}}$ is given by Slater's Rules.

According to these rules, for $\ce{C}$, effective nuclear charge experienced by an outer shell electron is:
$$Z_{\text{eff}} = 6 - 0.85\times 2 - 0.35\times 3 = 3.25$$

For $\ce{C+}$, effective nuclear charge is
$$Z_{\text{eff}} = 6 - 0.85\times 2 - 0.35\times 2=3.6$$

Since the effective charge for $\ce{C+}$ is more than that on $\ce{C}$, the second ionisation energy is more than that of the first.


If you go by thermodynamic conventions,

$$\begin{align}q > 0 &\implies \text{ heat is gained by system}\\
q < 0 &\implies \text{ heat is lost by system} \end{align}$$

You initially considered water as your system, and it lost heat, therefore  $q$ was negative.
Now when you do calculation for ice, it becomes your system, and the heat is gained by ice, $\implies$ positive $q$


Is there a point at which Ethanol (E10) fuel becomes harmful to gas tanks or engines if not used?

Well first off, pure ethanol is hygroscopic; it attracts water, to the point that it will pull it out of the air. Ethanol and gasoline will mix, but ethanol, gasoline and water will not; the ethanol-water mixture will come out of solution and settle on the bottom of your tank. Add a little oxygen to the mix, and you get rust. However, the more common side effect of this is more immediate; turn the car on, and the fuel pump will draw the water from the bottom of the tank into the engine, where it will promptly kill it, and require a costly dry-out process.

Down here in the south, where it's hot and humid most of the year, you hear a news story or two about this every year, usually when a gas station didn't properly purge its storage tank on a regular basis. The regular addition of thousands of gallons of fresh ethanol fuel "recharges" this hygroscopic process, causing a rather large pool of water to form at the bottom of the tank which the station is supposed to drain regularly. If they don't, and you fill up when the store's tank is just a little too low, or just a little too soon after the tanker's refilled it, the pump will draw a water-gas emulsion, or even straight water, into your tank, and the water will settle to the bottom.

The usual fix for long-term storage of ethanol fuel is to add a stabilizer, which will contain a high concentration of isopropanol. The main thing that will do is form an azeotropic mixture with water, meaning that its components will evaporate at the same rate, so when the isopropanol evaporates, it takes the water with it. This same mixture, in sufficient concentration, will also burn, so while it's not the greatest thing for your engine it can at least be flushed out the normal way.

Second, gasoline, as you probably well know, isn't a pure substance; it was originally what was left over after the refinery had extracted the "useful" heavier compounds (kerosene, diesel fuel, lubricating oil, paraffin waxes) out of crude oil by fractional distillation. Nowadays it's much more carefully purpose-made by thermal decomposition of heavier alkanes, but it's still an amalgam of relatively lightweight carbon compounds (usually between 6-12 carbons, with the dividing line for "octane rating" drawn between heptane and octane), plus additives and pollutants. It contains three components of interest; sulfur (a naturally occurring pollutant and difficult to remove completely; "summer blend" gas in the U.S. is more expensive because they have to get rid of more of it to prevent smog and acid rain), oxygen (modern fuels are oxygenated to make them more clean-burning), and ethanol (added to make the fuel burn cleaner and be more sustainable to produce than neat gasoline).

In the presence of water (pulled in by the ethanol) and oxygen, sulfur naturally forms sulfuric acid by a variation of the "contact process"; it oxidizes easily to sulfur dioxide, then more slowly to sulfur trioxide, which when dissolved in water becomes sulfuric acid. This reaction, specifically the formation of sulfur trioxide, is sped up in industrial processes by heat and by contact with catalyst metals such as vanadium oxide (hence "contact process"), but it will also happen on its own without any special treatment (i.e. acid rain).

This sulfuric acid will do a lot of corrosion damage, reacting with the tank's sheet metal producing hydrates of iron sulfate. When burned, those sulfates decompose into iron oxides and sulfur dioxide again, which do more damage to your exhaust system on the way out (rust is a catalyst for more rust, and we've discussed sulfur dioxide's contribution to the party in detail). In addition, sulfuric acid is even more hygroscopic than ethanol; at concentrations in water as low as 10%, it'll attract more water. Lastly, acids and alcohols form esters, in our case ethyl sulfate, and esters and alcohols form ethers, here diethyl ether (with the sulfuric acid and water as a byproduct). These are reversible and ultimately circular reactions, which will form an equilibrium of roughly equal concentrations of the intermediates. A dehydration reaction to produce ethylene is also possible in very high acid-to-water concentrations of sulfuric acid (possible when filling up in the winter, when it's dry and the gas has more sulfur). Ethylene will form a number of intermediate products with just about anything in that gas tank, some better for your engine than others (ethylbenzene is an antiknock compound; methyl ethyl ketone is a solvent and a pollutant byproduct of combustion engines).


Why is methyl iodide more dense than ethyl iodide, even though the latter has more carbon atoms?

For the same reason why a bucket of sand will weigh more than a bucket of rocks.
Smaller molecules will result in a more dense, thus more heavy substance.


How much energy is needed to break a water molecule?

The answer is given by the standard enthalpy of formation, $\Delta H_\mathrm f^\circ$ for water:

$$\begin{alignat}{2}
\ce{H2 + 1/2O2 &-> H2O(l)}\quad&\Delta H_\mathrm f^\circ&=-285.8\ \mathrm{kJ/mol}\\[6pt]
\ce{H2 + 1/2O2 &-> H2O(g)}\quad&\Delta H_\mathrm f^\circ&=-241.8\ \mathrm{kJ/mol}
\end{alignat}$$

The energy required for the following reaction, your point of interest:
$\ce{H2O -> H2 + 1/2O2}$ is simply a positive value for either one of the above $\Delta H_\mathrm f^\circ$ values depending on the state of the reactant. It means that per mole water splitting requires at least $241.8\ \mathrm{kJ}$ of energy.

Any method(thermal,electrolysis etc) followed, this much of energy will be always required. When using electrolysis, it may be convenient to express it in $\mathrm{eV}$.

Regarding calculation, I have to mention that $\Delta H_\mathrm f^\circ$s are experimental values. To know how they are calculated, here is a nice example of $\ce{NaCl}$.


Equivalent weight in case of disproportionation reaction

The $n$-factor of a compound in a reaction gives the number of moles of electron lost or gained by it (during oxidation and reduction respectively) per mole of the given compound. In a disproportionation reaction, the compound undergoing both the oxidation and reduction is the same.

It is important to understand that the $n$-factor depends not just on the change in oxidation state of the element but also the number of moles of the element undergoing that change.
This is the reason why we cannot simply take the average of the two $n$-factors as the $n$-factor of the disproportionate. It is necessary to also consider how much of the reactant is getting oxidised and how much of it is getting reduced, something along the lines of a weighted mean.

It is quite simple to derive the expression you have found for the $n$-factor of a disproportionated compound. Consider the following general reaction:

$$\ce{nA -> xB + $(n - x)$C}$$

Where $x$ moles of $\ce{A}$ (for which $n$-factor is to be calculated) oxidises to $\ce{B}$ and the rest of it reduces to give $\ce{C}$. Let the $n$-factors for the oxidation and reduction respectively be $n_1$ and $n_2$.  Which means $n_1 x$ moles of electrons go into oxidation, and $(n - x)n_2$ moles into reduction.

We have 2 equations: one by conservation of electrons,
$$\begin{align*}
n_1 x &= n_2(n - x) \\[7pt]
\implies x &= \frac{n_2\cdot n}{n_1 + n_2}
\end{align*}$$
and the other from the definition of $n$-factor:
$$\begin{align*}
n_f &= \frac{\text{moles of electrons transferred}}{\text{moles of reactant}} \\
&= \frac{n_1 x}{n} \\
&= \frac{n_1 n_2}{n_1 + n_2}
\end{align*}$$

An interesting observation here is that upon simplifying the obtained equation, you get the formula for the equivalent resistance of resistors connected parallelly in an electric circuit. The correspondence between them is intuitive.
$$ \frac{1}{n_f} = \frac{1}{n_1} + \frac{1}{n_2} $$


What is a good way for a laymen to purify alcohol to near 200 proof?

Distillation of ethanol yields 95% ethanol and 5% water. That is the best one can do unless you don't want to drink it afterwards (it will have a little benzene, a known carcinogen). The Epsom salts may remove some water, but I don't think it will remove all the water, not to mention that some of the Epson salt will dissolve into solution.


Making oxygen from iron oxide (aka rust) with sodium chloride?

This reaction isn't feasible and won't likely occur.

Why not?

Rust and Salt are very stable, and happy with their current state. You can see this better by looking at each product.

Rust forms from Iron and Oxygen

  4Fe + 3O2 → 2Fe2O3

Salt is formed from Sodium and Chlorine

  Na + Cl → NaCl

The formation of both of these products is very favorable because both molecules want to bond together to complete their valence shells.

They do not want to break their bonds, and doing so is possible but requires a lot of energy making it unfavorable. This is why we don't see salt or rust breaking down into its elements without a high amount of energy. This all applies for the overall reaction. All the reactants are already happy and don't want to break down making this reaction, possible, but not feasible due to the high amount of energy that it takes.


Melting and boiling points of benzene and fluorobenzene

The answer really doesn't have much to do with fluorine.  Beacall's 1928 The melting-points of benzene derivatives explains on the first page that it has long been known "that the more symmetrical the constitution of a benzene derivative, the greater the resistance it offers to the passage from the solid to the liquid state".

In the more modern Effect of molecular symmetry on melting temperature and solubility:

  As a general rule, symmetrical molecules in crystalline form have higher melting temperatures and exhibit lower solubilities compared with molecules of similar structure but with lower symmetry. Symmetry in a molecule imparts a positive amount of residual entropy in the solid phase (i.e., more possible arrangements leading to the same structure). This means that the entropy of a crystal of symmetric molecules is greater than the entropy of crystal of a similar, but non-symmetric molecule.

Benzene and hexafluorbenzene have almost the same melting point, while monofluoro, monochloro, monobromo benzene all have much lower melting points.

Both benzene and hexafluoro-benezne have a very stable "herringbone" crystal structure, where the edge of one ring approaches the center of another ring in a perpendicular manner.

So realizing that the cyrstal structure of benzene is unusually stablized, it is understandable that the melting point of benzene is higher than monofluorobenzene, but the boiling point is not.


Why, during aspirin synthesis in acidic conditions, anydride carbonyl oxygen is protonated rather than salicylic acid carbonyl oxygen ?

Mechanisms are explanative, and not predictive.

The first step of the reaction mechanism is protonation of the anhydride because that is the first step of the reaction.  That just is.  If the anhydride is not protonated, step 2 simply won't happen.

This does not mean that the salicyclic acid is not protonated.  It just means that the protonation of salicyclic acid is not necessary to explain how acetylsalicyclate forms.  Proton exchanges are at equilibrium.  It is likely that the salycilic acid is protonated during this process.  However, that protonated salicyclic acid may then protonate acetic anhydride and proceed as indicated in your link.  For the purposes of describing the mechanism, it is not necessary to distinguish where the proton that protonates the acetic anhydride comes from.

So your question really is self-answered, where you say, roughly, 'Such a protonation does not lead to product directly'.


In electrolysis, does it make any difference if the electrolyte is an molten, aqueous or dilute aqueous solution?

In molten sodium chloride, the only available ions are $\ce{Na^+}$ and $\ce{Cl^-}$. With electrolysis, the products will be sodium metal and chlorine gas.

In an aqueous solution, the sodium chloride still moves the charges from one electrode to the other, but the potential for electrolysis of water is less than $\ce{Na^+}$ or $\ce{Cl^-}$. So the products of a solution of sodium chloride is hydrogen gas and oxygen gas.


Does calcium nitrate decahydrate dehydrate or decompose upon heating to 100 degrees Celsius?

Calcium nitrate exists under three forms : 1) $\ce{Ca(NO3)}_2 $ which melts at 361°C; 2) $\ce{Ca(NO3)}_2$$·\ 3 $$\ce{H_2O}$, which is not commercial and melts at 51°C ; 3) $\ce{Ca(NO3)}_2$$·\ 4 $$\ce{H_2O}$ which is commercially available and  melts at 43°C. It looses its water at 132°C and is deliquescent. The decahydrate that you mention does not exist.


A radical cation is simply a free radical with a positive charge. Similarly a radical anion is a free radical with a negative charge.

In other words, radical cations and radical anions are subsets of the free radicals.

That said, the term free radical itself is not entirely clear cut, as illustrated in this question.


Baking soda is sodium hydrogencarbonate, $\ce{NaHCO3}$, also known as sodium bicarbonate.

The only toxicological data in the GESTIS database on hazardous substances
 is a LD50 of 4220 mg/kg for the oral uptake by rats. There are no data for dermal uptake.

In water, $\ce{NaHCO3}$ is weakly alkaline. I do not know whether it might cause skin irritations, particularly in more sensitive regions, such as armpits.

In occupational safety, I'd recommend to avoid skin contact and use personal protection equipment when handling it.


Is the kinetic energy turning into activation energy when the reaction takes place?

You got it right: when the reaction takes place, the kinetic energy of the reacting particles turns smaller because some of it is (sort of) spent on activation energy. It is just that the "when" in question is incredibly short.

Imagine two tennis balls flying at great speed and hitting each other head on. Sure enough, they will bounce off each other and fly in the opposite directions with the same speed. Some time into the collision, there must be an instant when they are not flying anywhere; their speed is 0, and so is kinetic energy. All energy they possessed has converted into the potential energy of elastic deformation, only to return into kinetic energy again really soon.

Same thing with molecules, except they are way smaller and fly way faster. Also, there might or might not be a reaction, and if it happens, then the kinetic energy of the molecules after the collision will not be the same as before. It may be greater, or it may be smaller; that's how exothermic and endothermic reactions happen.

So it goes.


Can any compound be a Lewis acid but not be a Brønsted acid?

Certainly.  There are many compounds that can accept a pair of electrons from a Lewis base, yet don't have any Brønsted-Lowry acid (proton) moiety. The two are really rather different things, although there are of course many compounds that are both Lewis and Brønsted-Lowry acids ($\ce{Al(OH)3}$ for example).

Some examples of Lewis acids that are not also Brønsted-Lowry include:

$\ce{BF3}$
$\ce{(CH3)3B}$
$\ce{AlCl3}$
Metal cations like $\ce{Li+}$, $\ce{Na+}$, $\ce{Mg^2+}$, etc.


Can a meso compund be rotated in ANY sort of way in order to show its achirality?

I think what you may be asking is: could there could be a nominally "meso" compound that has restricted rotation about its central single bond, causing it to exist in the form of two non-interconvertible stereoisomers. Example- start with meso-tartaric acid, and make the di-tert butyl ether from the OH groups and di-tert-butyl ester from the COOH groups. Maybe these groups would be bulky enough to restrict rotation. You could mentally rotate around the central C-C bond until the molecule has a plane of symmetry, so it is still a nominally "meso" compound, but physically you could separate the two stereoisomers.


Mechanism for Synthesis of Tetranitromethane from Acetic Anhydride and Fuming Nitric Acid

The Organic Synthesis preparation you cited for the preparation of tetranitromethane has some clues that are useful in providing a viable mechanism. Upon slowly adding acetic anhydride (Ac2O) to anhydrous HNO3 with cooling an exotherm is produced. The reaction mixture is allowed to sit at room temperature for 7 days. Assuming that they investigators did not go on vacation for a week, I presume that the formation of tetranitromethane (TNM) is slow and that the exotherm is due to the formation of the mixed anhydride 2. [The nitronium cation is also a possible electrophile as well as the mixed anhydride 2.] The enol of acetic anhydride can undergo substitution to afford species 3 and then repetitively on to the trinitro mixed anhydride 4. Anhydride exchange with acetic acid (or nitric acid) affords trinitroacetic acid 5 which decarboxylates to enol 6 (aci-form). Tautomerization of 6 can form trinitromethane. Footnote 3 in the Organic Synthesis link mentions that trinitromethane 7 can be converted to tetranitromethane 8 with HNO3. The transformation of 6 to 8 does not preclude nitration via mixed anhydride 2.


Diluting methanol in water: is it exothermic?

When the methanol and water are separate, they both exhibit hydrogen bonding with themselves. When the water and methanol are mixed together, some of the existing hydrogen bonding (water-water or methanol-methanol) is disturbed and now there is hydrogen bonding between water and methanol. This new hydrogen bonding pair (water-methanol) releases energy when it forms as it is more favorable than maintaining only water-water and methanol-methanol hydrogen bonding; the temperature increase you observe is the result of the released energy warming the solution.


How do you calculate the concentrations of multiple reactants in a certain point in time?

$$\frac{dA}{dt}=-kAB$$
$$\frac{dB}{dt}=-kAB$$subject to initial conditions $A=A_0$ and $B=B_0$ at time t = 0.  Let x = number of moles of A that react = number of moles of B that react.  Then,$$\frac{d(A_0-x)}{dt}=-k(A_0-x)(B_0-x)$$and$$\frac{d(B_0-x)}{dt}=-k(A_0-x)(B_0-x)$$Therefore, from both equations  $$\frac{dx}{dt}=k(A_0-x)(B_0-x)$$ subject to the initial condition x = 0 at t = 0.  This differential equation can readily be integrated using separation of variables and partial fractions.


How to find pH by mixing two solutions of different concentrations?

You are right in assuming the two acids are strong acids and therefore completely dissociated, but you can easily look that also up, e.g. here.
Therefore you can assume, that the concentration of protons is equal to the concentration of the acid.
\begin{align}
c_{\ce{HI}}(\ce{H+}) &=  c(\ce{HI}) &
c_{\ce{HCl}}(\ce{H+}) &=  c(\ce{HCl}) &
\end{align}

Now you need to determine the amount of the protons in each solution with $$n=c\cdot V.$$

You have to add the amount of protons in each solution and determine the combined volume.
\begin{align}
n_\mathrm{tot}(\ce{H+}) &= n_{\ce{HI}}(\ce{H+}) + n_{\ce{HCl}}(\ce{H+})&
V_\mathrm{tot} &= V(\ce{HI}) + V(\ce{HCl})
\end{align}

To calculate the pH of the resulting concentration, just take the negative logarithm.
$$\mathrm{pH} = -\log c_\mathrm{tot}(\ce{H+}) = -\log\left[\frac{n_\mathrm{tot}(\ce{H+})}{V_\mathrm{tot}}\right]$$

   Your solution should be $\mathrm{pH}=1.06$, rounded to three significant figures.

The approach you chose can only work, if you have equal volumes of the two solutions.


Why does the standard enthalpy of formation diverge so far from the standard Gibbs free energy of formation for some substances?

Continuing from the comments above.

The difference is due to the entropy appearing in $\Delta G$ as the energy $-T\Delta S$. Classical thermodynamics cannot tell us what the entropy is in molecular terms. This is discovered using statistical mechanics via Boltzmann's entropy equation $S=k\ln(\Omega)$ where $\Omega$ is the number of arrangements of particles into their energy levels, sometimes called 'complexions'.

Different types of molecule have different natures, because they contain different number of atoms, which means that there are different numbers of molecular vibrations, or they contain atoms of different types, which means that vibrations are of different frequencies to one another. Hence molecules of different types have a unique set of energy levels (ignoring for simplicity degenerate vibrations).

As energy is added, the number of ways of filling up the energy levels is clearly different between different types of molecules because the stack of energy levels differs one to another. Single bonds between heavy atoms, for example, have lower frequencies, than those between lighter atoms, and so more levels are populated at the same temperature than in the heavier atom molecule. In this case the entropy is greater than that for a molecule with widely spaced levels.

The same type of argument can be made for rotational and translational energy levels, and can also be made for phase changes, such as solid-solid, or for melting or vaporisation. Which of these are to be counted depends upon whether the substance is gas, liquid, or solid at the required temperature.

Notes:

Technically the entropy is calculated as
$$S=R\ln(Z)+RT\left ( \frac{\partial \ln(Z)}{\partial T}  \right)_V$$

where Z is the partition function and $Z=\Sigma _i g_i \exp(-\epsilon _i/(kT))$ where $\epsilon_i$ is the energy of level i with degeneracy $g_i$ and k is the Boltzmann constant.


Sign conventions can make 'work' confusing, but the question here asks for the work done by the gas. When a gas expands, it's doing work (it's losing energy), so it's actually $W = + P \, \mathrm dV$, inverting all the inequality signs. So option B) is correct.


Why is Frost's Circle an extension of Huckel's 4n+2 rule? What physics leads to the emergent property that the circle can predict relative MOs?

The MO energies are obtained from Hückel theory: see Pi molecular orbitals of polyenes (and the Internet in general, there are many excellent expositions) for more details.

Regarding the Frost circles, they do not actually have any meaning: they are simply visual representations of the energies obtained from Hückel theory for cyclic polyenes. The last line of porphyrin's answer is relevant here:

  In the case of circular polyenes the solutions are $E_j=\alpha + 2\beta \cos(2\pi j/n)$.

where $n$ is the number of p-orbitals (e.g. for benzene $n = 6$) and $j$ is an integer satisfying $1 \leq j \leq n$. (The convention $0 \leq j \leq n-1$ is also sometimes seen, but it is mathematically equivalent.)

I am not a mathematician so some of the notation may be less than ideal, but the idea is as follows. Note that if you draw a circle with radius $r$, then the coordinates of any point on the circle $P$ may be described as $(r\sin\theta, -r\cos\theta)$ where $\theta$ is the (clockwise) angle that $\overrightarrow{OP}$ makes with the negative $y$-axis $\overrightarrow{OY}$. For example, here is a circle with $r = 2$. The angle $\theta$ here is equal to $\pi/4$, so the coordinates of the point $P$ are given by $(2\sin(\pi/4), -2\cos(\pi/4)) = (\sqrt{2}, -\sqrt{2})$.

Now we simply need to choose a circle with radius $r = -2\beta$ (note that $\beta < 0$, hence we need the negative sign) and select the six points on the circle where $\theta = 2\pi j/n$. The $y$-coordinates of these six points will then be the six MO energies. I have illustrated this for $n = 6$ and $\beta = 1$. We need to pick the points with

$$\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$$

and these are $A$, $B$, $C$, $D$, $E$, and $F$ in order. Note that $\angle YOA = \pi/3$, $\angle YOB = 2\pi/3$, and so on. And note that whatever the value of $n$, there will always be the case where $j = n$ and hence $\theta = 2\pi$: that corresponds to the vertex at the bottom of the circle.


What is the role of POCl₃ in the reaction to form (NPCl2)3 (hexachlorophosphazene)?

If your goal was to prepare hexachlorophosphazene $\ce{(NPCl2)_3}$,

then I'm not sure why $\ce{POCl3}$ was added.  The synthesis involving $\ce{POCl3}$ certainly works, but hexachlorophosphazene can be prepared more simply by heating phosphorous pentachloride and ammonium chloride in chlorobenzene.  Careful sublimation affords the desired product, pure and in high yield.  This route is more economical and less toxic since it avoids the use of $\ce{POCl3}$.

$\ce{POCl3}$ is typically added to the reaction when one is interested in preparing higher polymers of $\ce{(NPCl2)_{n}}$.  In this case, use of $\ce{POCl3}$ allows you to prepare and isolate $\ce{Cl3P=N-POCl2}~(1)$ in high yield and purity (ref. 1).  The purity is important in obtaining high molecular weight polymers of $\ce{(NPCl2)_{n}}$.  Condensation of $(1)$ to produce hexachlorophosphazene typically occurs when reaction temperatures are maintained well below $\pu{200^\circ C}$, while high molecular weight polymers are produced at temperatures over $\pu{200^\circ  C}$.

Reference:

Synthesis and Characterizations of Poly(organophosphazenes) By Roger De Jaeger, Mario Gleria. See page 26 here


Predicting isotope patterns in mass spectrometry

The first step is to write all the possible combinations of H atoms: $$\mathrm{H^1H^1~~ H^1H^2~~ H^1H^3 ~~H^2H^1 ~~H^2H^2~~ H^2H^3~~ H^3H^1~~ H^3H^2 ~~H^3H^3}$$ Now it is important to note that each of these parings have an equal probability of forming and exist in equal quantities in the mixture. Their molar mass are respectively: $$2, 3, 4, 3, 4, 5, 4, 5, 6$$ So there is $1$ way to make $2$, $2$ ways to make $3$, $3$ ways to make $4$, $2$ ways to make $5$ and only $1$ way to make $6$. Since each way has an equal probability of occurring, then the ratio of the molar masses should be in the ratio -  $~1:2:3:2:1$.

Hence for peaks of mass = $2, 3, 4, 5, 6$ ratio of heights = $1:2:3:2:1$

The reason why there is no peak for mass $1$, I am not too sure. However I think it is because if the hydrogen molecule was to break up, it would form a proton and a H radical. Both of these are extremely reactive and unstable and will probably just get lost in the machine and never actually get recorded. This could make sense as the mass spectra for $\ce{HCl}$ or $\ce{HBr}$ doesn't include a peak for mass $1$. Also in mass spectra for hydrocarbons, such as heptane, the smallest mass peak recorded is usually 29 which is the ethyl group. Therefore any thing lower than that (such as the methyl group) is probably too unstable.


Diagonal relationships in the Periodic Table

What I remember from studying C-P 'diagonal relationships' (for multiple bonds e.g. carbene vs phosphinidene, alkyne vs phosphaalkyne) is that similar electronegativity also played a role, also by affecting the valence orbitals.

However, perhaps it is better viewed from a different vantage point:
the main group elements behave rather similarly from 3rd row onwards, but there is a distinct difference between the 2nd and 3rd row elements and the explanation is pretty straightforward:
The 2s and 2p orbitals are roughly similar in 'size' since the 2p orbitals are the first p-orbitals and are pretty compact. Therefore 2s and 2p orbitals mix happily.
Going to the third row, the 3p orbitals need to be orthogonal to the 2p orbitals and as a consequence they are much more diffuse than the 3s orbitals, resulting in less efficient s-p hybridization. This also manifests itself in the so-called 'inert pair' effect whereby e.g. Si will resist hybridization in favor of a lone pair. There is some beautiful work by Knutzelnigg from the 80s (?) describing these effects.
This 2nd-3rd row discrepancy is somewhat counterbalanced if you step to right, increasing Z and thereby contracting the orbitals giving rise to this apparent diagonal relationship.

Similar arguments make that for the 3d TMs the hybridization is more efficient for 4s-3d than for the 4d/5d TMs. Consequently, 3d TMs tend to behave differently from their heavier sisters, which in turn are closer together (although for the 5d's the relativistic effects start to kick in more heavily...).


Firstly, $\ce{NaBH4}$ is not a Lewis acid because it cannot accept any more electrons. Aqueous $\ce{NaBH4}$ is a commonly used reducing agent often for reduction of carbonyls to their respective alcohols.

The mechanism of Lewis acid catalysis depends on the reaction but often Lewis acids are used to generate cationic electrophiles such as in the Friedel-Crafts reaction.


A glass of water with ice-cubes in it. Where's the water the coldest; at the top or bottom?

Interesting question!

A few things first:

  As the ice melts, it cools the water around it.

Technically, the ice cube melts because the water cools down. This may sound ridiculous at first, but you must consider the fact that the ice melts because it has drawn "heat" (energy) from its surroundings. The "surroundings" being the air and water that surround it (but the water's more important since it's a better conductor of thermal energy).

  Given that cold water is denser than hot water, I would presume that the cold water would sink to the bottom...but it would warm as it sinks, reducing the density.

You're right, cold water is denser than hot water. It is helpful to note that it shouldn't be too cold though. As the temperature of water drops to 4 °C, the density of water gradually increases. However, as the temperature drops below 4 °C the density of water actually begins to decrease and water in this range easily "floats" over water in the room temperature range.

  Meanwhile, the ice is still melting and giving off its cold to the surrounding water.

Ice isn't giving off its "cold", rather, it takes in the water's "heat" (thermal energy).

Back to your question.

As Max mentions in his answer, you have done a particularly good job of indicating what physical parameters we're dealing with; the really important ones being the temperature of ice, temperature of water (at the time you put the ice in) and the quantity of ice used (at least with respect to the water).

But assuming you're drinking water (originally at room temperature) out of a 250 ml styrofoam or plastic cup, and you used two (normal-sized) ice cubes and that you began drinking the water a minute after you plonk in the ice cubes, the water should be colder at the top than at the bottom.

Consider minute, imaginary layers/regions/packets of water in the cup (thinking about this in terms of water "packets" rather than water molecules is easier to comprehend). Also, think of the cup as having three (crudely demarcated) regions: Top, middle, and bottom.

Packets of water immediately adjacent to the ice cubes are in thermal equilibrium with the outermost regions of the ice. However, these packets soon gain some thermal energy from other water packets that are adjacent to them. So as these packets slowly rise in temperature, from zero degrees to past 4 °C, they sink and new packets occupy locations adjacent to the ice. The cycle repeats for as long as the ice is there.

Now, as those packets of ice sink, they gain more thermal energy from the packets of water they come in contact with on their way down. This, coupled with the viscous effects of water results in the mild "warming up" of the sinking packets.

Now since they warm up a bit, they tend to rise back up. Back at the top, they're get cooled and sink again. This process repeats for as long as the ice remains in the water.

Take a step back, and you'll see that the middle of the cup ought to be cold, the bottom of the cup ought to be colder, and the top of the cup is the coldest.

So even if the ice cubes aren't actually touching your lips, you'll find sipping the water at the top to be colder than sucking out water from the bottom through a straw.


Why does adding salt to boiling water cause it to flare up suddenly?

The salt does not instantly dissolve so the surface of the crystals suddenly provides a lot of nucleation sites for the water to form vapour - hence the surge as it boils from these surfaces. The white is due to turbulence from the boiling just like waterfalls look white.


Here are some rules i remember from organic chemistry in Collage.
When trying to distinguish between nucleophile and base,

Look for resonance
If atom is involved in resonsance, this makes ato/group more nucleophilic,
if electrons are more localized its more basic.
Eg. A carboxylate is a nucleophile, the -ve charge is delocalized (i.e - charge is  spread out across carbonyl and oxyanion. Compared to nitrogen on an amine,  electron are restricted.

Sterics,
more sterics less nucleophilic.
Smaller radius atoms, valence electrons are more localized in space, thus more basic. Big radius atoms, poor localization relative to small, thus more nucleophilic.
More electronegastive, less basic
Electronegativity is a measure of the nuclear positive charge effect on near -ve charges.
The more electronegative the less basic. Because the atom doesnt want to give up its electrons. Why? Because it has high electronegativity
Conduction effects
If there are electronegative atoms within 3 bonds (taking into account electronegativities, and proximity) more nucleophilic less basic.

Nitrogens are usually bases, oxygens nucleophiles. I swear it was always like that. Halogens are weak nucleophiles.

Will edit for legibility when i get to computer


Miller indices: How does one translates the orientation to the origin of the unit cell?

Let's say your structure contains a short piece of helical DNA, and it (the helix axis) starts at point P1 and goes to point P2. You want to know the Miller index associated with that direction. So you move the entire helix (without rotating) so that P1 is now at the origin. The new position of P2, lets call it P2(prime), will be at:

$$ P_2' = P_2 - P_1 $$

Just checking that is is a translation (I have to subtract the same vector from the entire structure, otherwise I'm rotating or warping it):

$$ P_1' = P_1 - P_1 = (0,0,0)$$

Again, the pair of two points still points in the same direction, but now $P_1$ is at the origin. Taking the fractional coordinates (i.e. along the unit cell axes) of $P_2'$, you can turn those into Miller indices $h$, $k$, and $l$ by multiplying with a large number and rounding to the next integer. Generally, the direction will be off because of the rounding. However, in special cases, e.g. if your DNA is on some crystallographic symmetry axis, the Miller indices will match your direction exactly.


A very typical situation in organic chemistry is the following:

$\ce{A -> undisired products}$
$\ce{A + B -> target product}$

Here it is easy to see that increasing the concentration of $\ce{B}$, we will get a higher yield of the target product.
The other reaction can be of many kinds.
$\ce{A}$ can decompose in the reaction medium, can react with itself, with the solvent, with byproducts form the desired reaction, etc.


You are correct in the first step - the decarboxylation gives the anion of the chloroalkene.

It is important to remember that such soda lime reactions are run at high temperature and CaO is a powerful drying agent, so there will be no water present to quench the anion. What happens next would be obvious if you were dealing with an alkane; there is a leaving group beta to an anion so there would be an elimination with loss of Cl-. This is what happens; Cl- leaves to give the triple bond (again remember that this is a high temperature reaction so plenty of energy for forming unstable species). This is clearly unstable within a [6] ring so it dimerises by a 2+2 process to give the product drawn.


Why are elimination reactions in organic chemistry known as beta eliminations?

In an elimination reaction we typically have a leaving group (denoted by "X" in the figure below),  The carbon that the leaving group is attached to is referred to as the "alpha" carbon.  If a hydrogen is removed from the "beta" carbon, then we have formed an olefin and the reaction is called a beta elimination.

[reference]

Sometimes an alpha hydrogen can be removed.  In this case both the hydrogen and leaving group were attached to the same carbon and a carbene is formed.


Regioselectivity of bromination of alkenes

With bromination, the first step is the formation of an adduct complex, which can be explained that the π-bond polarises the dibromide. The next step is the formation of the bromonium ion, which carries an overall positive charge. A symmetry-changing induction like substituents will always result in asymmetric carbon-bromine bond lengths in the bromonium ion. Since it is a cation, similar rules for the stability/instability as for carbocations apply. Therefore the higher substituted carbon will have the longer $\ce{C-Br}$ bond length, and the higher charge (more positive/less negative). This makes it a better target for the residual nucleophile $\ce{Br-}$ to attack there.

Here is the bromonium ion for the reaction
$$
\ce{\underset{alkene}{H3C-CH=CH2} + Br2 ->
[\underset{bromonium ion}{H3C-CH(Br)CH2}]+ + Br- ->
\underset{$trans-$alkane}{H3C-CHBr-CH2Br}}
$$
calculated at the DF-b97d3/def2-SVP level of theory, charges with NBO6, to illustrate the concept:

Additionally the HOMO has the higher coefficient at the lesser substituted carbon, while the LUMO has the higher coefficient at the higher substituted carbon.

From this, you can see that both quotes from the textbooks are correct.


What does the letter R represent in chemical structures?

Depending on the context, the letter R can stand for several things. From our friends teaching organic chemistry at UCLA:

  R group: An abbreviation for any group in which a carbon or hydrogen atom is attached to the rest of the molecule. Sometimes used more loosely, to include other elements such as halogens, oxygen, or nitrogen. R is an abbreviation for radical, when the term radical applied to a portion of a complete molecule (not necessarily a free radical), such as a methyl group. Should not be confused with R (the gas constant), R (the one-letter abbreviation for the amino acid arginine) or R (a designation of absolute configuration).

In the image from the Wikipedia article you have referenced, the letter R is as defined in the first section of the above quote - that is, it stands for "any group in which a carbon or hydrogen atom is attached to the rest of the molecule."


Cleavage of ethers using strong acids (classically $\ce{HI}$ but $\ce{HBr}$ should work as well) can proceed by either an $S_N1$ or $S_N2$ mechanism. In this case the cation formed would be primary or phenyl, neither of which are stable, and so the reaction proceeds via an $S_N2$ mechanism.

The oxygen is protonated in an equilibrium reaction. The bromide ion then attacks at the least hindered carbon and the $\ce{C-O}$ bond breaks in a concerted $S_N2$ mechanism.

(source)


Have a look at this paper

Summarizing:

put food coloring to 35% isopropyl alcohol in water.
add ammonium sulfate

What happens ? Once ammonium sulfate added and the solution mixed, the ammonium sulfate dissolves, pushing the isopropyl alcohol out of the aqueous solution (and the food coloring as well).

A separate layer thus forms, that is clearly visible thanks to the food coloring.

The weak point of the whole story is that isopropyl alcohol is flammable...

PS: I would have preferred to reply to @Michiel's answer and @Nicolau Saker Neto's comment: but I do not have enough reputation to do that...


Why does two nails of the same material not produce electricity when placed in saltwater?

First, you need to understand what is electricity. Electricity is simply the flow of electrons. Now, what I really suspect you are asking is why these electrons flow.

Electrons flow in an electric current due to a potential difference which is the difference in potential energy of the electron at two different places. Like everything else, electrons like to lower their potential energy (imagine water falling in a waterfall) hence in an electric current, electrons move because they are moving from a place of high potential energy to a place of lower potential energy.

Now when you place two of the same nails in a salt solution, there is no potential difference between the two nails as the potential energy of the electron moving from the two nails remain the same (since they are made of the same material).

However when you place two different metals into salt water, such as aluminium and copper, there is now a potential difference between the two metals. Hence this will cause the electrons to move creating electricity (an electric current).

For more detail about the voltage change and what the chemical reactions that are actually occurring and you google electrochemical cells and electrochemical series. Hyperphysics and Chemguide are both very good resources which provide in-depth explanation on what is actually going on.


Why is Cu2HgI4 filtered while it is hot?

Copper tetraiodomercurate, $\ce{Cu2HgI4}$, is typically prepared by heating combined suspensions of copper(I) iodide, $\ce{CuI}$ and mercury(II) iodide, $\ce{HgI2}$.

The iodides are prepared by (a) reacting copper(II) sulfate, $\ce{CuSO4*5 H2O}$ with potassium iodide, $\ce{KI}$, and (b) mercury(II) nitrate, $\ce{Hg(NO3)2}$ with with potassium iodide.

In order to avoid contamination of the final reaction product with one (or both) of these iodides, it is essential to filter off while the reaction mixture is still hot.


How can I check the sodium content of food?

Sodium is a difficult ion to estimate at home.

Since most sodium salts are water-soluble, colorimetric or spectroscopic methods are often employed, both of which I assume are not within arms' reach.

Some alternatives:

Speaking of sodium estimation in foodstuff, it is usually determined by Mohr's
method which involves the precipitation titration of $\ce{Cl-}$ ions. This is done under the assumption that sodium is mainly present as sodium chloride. Additionally, it will also register $\ce{Cl-}$ from other chlorides, and hence may give an elevated reading. You might be able to do this at home. Here is a page
to help you with the process.

It is done in the presence of chromate ion ($\ce{CrO4^2-}$), and $\ce{Ag+}$ ions are added gradually. This precipitation happens initially:

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

When the solution runs out of $\ce{Cl-}$ ions, precipitation of the red-brown $\ce{Ag2CrO4}$ begins:

$$\ce{Ag+ + CrO4^2- <=> Ag2CrO4(s)}$$

This signals the end point.
There are sodium ion meters, you can get them for ~300$.
There is a volumetric estimation[1] for sodium directly, but it is not expedient to do it. It is based on the precipitation of sodium zinc uranyl acetate.

Conclusions

IMO it is difficult to do this as a home project, as it will require stuff that is not usually found at home to do it precisely.

References

[1] J. Am. Chem. Soc. 1931,  53, 9, 3288-3291 DOI: 10.1021/ja01360a011


How does activated carbon trap impurities?

Never forget to interpret in terms of physical/chemical forces, by order of energy: van der Waals, hydrophobic (pseudo forces), hydrogen bonds, disulfide bridges, covalent, electrostatic (for the most frequent).

Indeed, in charcoal, there is a large number of $\pi$-electrons that can make bonds, in particular. And yes, the porosity is important.

But perhaps could you have a look at Boehm H. P. Some aspects of the surface of carbon blacks and other carbons, Carbon, 1994, 32(5), 759-769. DOI 10.1016/0008-6223(94)90031-0.


If LPG gas burners can reach temperatures above 1700 °C, then how do HCA and PAH not develop in extreme amounts during cooking?

It is the temperature of the pan that matters not the flame

The flame temperature is irrelevant if you are cooking in a vessel. the only temperature you need to worry about is the temperature of the surface of the pan (or–even more importantly–the temperature of the meat). The surface of the pan will rarely get above around 220 °C if you are monitoring it. An unattended pan can get hotter, but, if you let it get too hot, your food will rapidly burn and will not be edible. Conversely, if the food is palatable, you probably haven't heated it enough to get lots of nasty HCAs and PAHs.

But the research that caused your worry is also frequently overstated. Yes, large quantities of PAHs or HCAs may be nasty, but the amounts in food–even food cooked on an open flame barbecue–are very small and there have been no convincing studies showing a notable effect on health. There are some studies showing  a very small effect of meat on health (diseases such as bowel cancer have been linked to some meats but the studies are statistically weak and the effects are very small. Moreover these studies link to meat not cooking products but would probably have spotted any effects based on nasty cooking by-products). It is worth remembering that people have been cooking on open flames since we invented fire. If that were really dangerous, there would be strong evidence of harm and/or primitive man would have developed good defences to avoid the harm (as milk-drinkers evolved lactose tolerance in adults because their diets consisted of a lot of dairy products).

There have been many food scares based on observations of known nasties in cooked food. Acrolein and Nitites, for example. None of these have been shown to have any notable measurable effect on people in the concentrations present in the diet.


What is the exact definition of Salinity?

The term 'salinity' is measure of 'saltiness', particularly used in reference to sea water.

The exact scientific definition has evolved over time to incorporate our ever growing understanding of physics and chemistry.

In 1889 the International Council for the Exploration of the Sea (ICES) used a measure of salinity based on the total dissolved solids of salt (g/kg) found in seawater after evaporation and drying at 480$^{o}$C.

By taking samples of ocean water from around the world and at different depths, oceanographers have determined that whilst the concentration of salts varies, the relative ionic composition remains essentially constant. The average salinity of the world's ocean water is usually stated as 35 g/kg, with the relative ionic composition of this "average sea water" made up as follows:

The major cations present in 1 kg of sea water are Na$^{+}$ (10.7653 g), Mg$^{2+}$ (1.2942 g), Ca$^{2+}$ (0.4117 g), K$^{+}$ (0.3991 g) and Sr$^{2+}$ (0.0079 g).

The major anions present in 1kg of seawater are Cl$^{-}$ (19.3534 g), SO$_{4}$$^{2-}$ (2.7124 g), HCO$_{3}$$^{-}$ (0.1178 g), CO$_{3}$$^{2-}$ (0.0122 g), Br$^{-}$ (0.0674 g), F$^{-}$ (0.0013 g), B(OH)$_{4}$$^{-}$ (0.0066 g).

The process of determining salinity from evaporation & drying can be quite difficult, particularly on board ships.

In 1963, the ICES adopted a new definition based on "chlorinity" measurement (using titration). The "average sea water" with ionic composition given above was determined to have chlorinity of 19.374%. By measuring the relative chlorinity of a sample of sea water, the salinity could be calculated using the formula S = 1.80655 Cl.

In 1978, with the advent of modern electronics, the United Nations Educational, Scientific and Cultural Organization (UNESCO), International Council for the Exploration of the Sea (ICES), Scientific Committee on Oceanic Research (SCOR) and International Association for the Physical Sciences of the Oceans (IAPSO) adopted a method based on electrical conductivity measurements compensated for temperature. The Practical Salinity Scale of 1978 (PSS78) defines S (a dimensionless quantity) from conductivity and temperature measurements, in the range 2 $\leq$ S $\leq$ 42, by the following formula:

S = 0.0080 - 0.1692$K_{15}$$^{1/2}$ + 25.3851$K_{15}$ + 14.0941$K_{15}$$^{3/2}$ - 7.0261$K_{15}$$^{2}$ + 2.7081$K_{15}$$^{5/2}$

Where $K_{15}$ is the ratio of conductivity of the sea water sample to that of potassium chloride (KCl) solution of mass fraction 32.4356x10$^{-3}$, at 15$^{o}$C and atmospheric pressure.

A formula is also provided for compensation of the conductivity measurement at different temperatures (see http://unesdoc.unesco.org/images/0004/000479/047932eb.pdf)

In 2010, the Intergovernmental Oceanographic Commission released "the international thermodynamic equation of seawater (TEOS-10)" in which it recommended a return to the use of "Absolute Salinity" to take into account variations in the composition of sea water which can occur for example in regions where rivers run into the ocean or where some other regional variability results in "anomalous" sea water composition.

The rational for this approach is based on understanding the effect each chemical component of seawater has on its thermodynamic properties such as density, pressure, temperature and Gibbs function. Various measures of salinity are defined and their relationships with thermodynamic properties described.

The Added-Mass Salinity, which incorporates the Reference Composition Salinity of "standard sea water" (the Practical Salinity Scale) with an adjustment for the Absolute Salinity Anomaly (mass fraction of material which must be added to standard seawater to achieve the concentration levels in the sample) is sometimes difficult to determine in-situ.

Modern techniques for determining Absolute Salinity involve taking various other measurements such as conductivity, temperature, pressure and density by incorporating measurements from sensitive electronic and optical instruments with data-driven computer algorithms.

The Density Salinity, which is the salinity argument in the thermodynamic expression for density which gives the sample's actual measured density, can be used where accurate measurements of density are available, such as a vibrating-tube densimeter or optical refractometer. When combined with Practical Salinity, temperature and pressure measurements, this can provide an accurate and easily determined measure of the amount of dissolved material in seawater (within a certain range).

Composition anomalies can be estimated by computer algorithms which take into account longitude, latitude and depth data. The accuracy of such algorithms is dependent upon the collection of accurate data from field measurements and laboratories across the world and is therefore an evolving process.

(For more details, see: http://www.teos-10.org/).


Why does gallium nitride allow an AESA radar to be more powerful than a radar with elements made of gallium arsenide?

The short answer is that GaN transistors have a much higher breakdown voltage and saturation velocity than GaAs, allowing them to operate at higher voltages at high frequency, handling more power. The large band gap (~2x GaAs) means that GaN still behaves like a semiconductor, even at relatively high temperatures, allowing for higher operating powers.

GaN has really been a breakthrough in semiconductor technology. It's what allows us to have 405 nm violet laser diodes and could potentially replace magnetrons in microwave ovens. NXP has a little document here comparing GaN with a few older technologies that's a quick read.


The point is not how large the possibility is.
There is a possibility that the flask has been contaminated, so you should have taken a new flask.
I suppose you did not have a spare one with you.
Then, the best should have been to sterilize again that flask, but that is not easy in the field.
I mean, it is much easier to take more sterile flasks with you than to sterilize them in the field.

Now, if this sample shows that there are no bacteria, you can conclude that there are no bacteria in the water.
But if there is some concentration of bacteria, you can not know if they come from the water or from the soil or how much come from each part.


Using the electrical potential created by the electron transport chain to power electrical systems

There are fuel cells that run on alcohols such as ethanol or methanol and on glucose.

That said, the energy density in current Li-ion cells is quite high - a 18650 cell holds ~30,000 joules, about 9 kCal. By comparison, a  "biochemically" powered shotgun shell (well, the nitrate and carbon of gunpowder can come from bio sources) packs ~2,000 J.


Delocalised electrons and stability of benzene

In physics (or you can google it) you will learn about the particle in a box experiment.  At the atomic level, if you confine a particle to a box, it will have some energy.  It turns out, if you make the box longer, the energy of the particle will be lowered, if you make the box smaller, the particle will have a higher energy.  The same concept applies to benzene.  Basically you start with 3 separate acetylene molecules in which the pi electrons are confined to a 2 carbon segment.  Once you trimerize the acetylene and form benzene, the pi electrons can now roam over 6 carbon atoms, a much longer "box" if you will.  So the pi electrons in benzene will be stabilized (lower energy, longer box) compared to the pi electrons in the starting acetylene (higher energy, shorter box).  Molecular Orbital theory tells us the same thing in a slightly different manner.  In each acetylene (let's just focus on the 2 pi electrons in the acetylene that will be the pi electrons in benzene) we have 2 pi electrons in the HOMO and none in the LUMO.  As we bring these 3 acetylenes closer together to form benzene, the orbitals interact.  The 3 initially degenerate HOMOs in the 3 acetylene molecules split (they are no longer degenerate) as they move closer to becoming benzene and a net energetic stabilization results.


Salt Bridge for Microbial Fuel Cell: Full or Partly Full?

Does the salt bridge connecting the anode (benthic mud) and cathode
  (salt water) have to be full of the agar-salt solution, or can it be
  partly full

There should be no air gaps or cracks so that the circuit does not break anywhere. Take a narrow PVC pipe if your concern is using too much of agar-salt solution.


Are 3-center-2 electron bonds and and banana bonds same?

The phrase "banana bond" is usually reserved for situations where steric conditions prevent the orbitals of two atoms from aligning properly so that the bond appears "bent" and is weaker than an optimally aligned bond would be.  Cyclopropane is the poster child for such bonds, diborane is not because this description of banana involves only two atomic centers.

It's a different concept from three-center two-electron bonds, where typically the bonds are well aligned but the three atoms bonded together are not always collinear.  This is what diborane does.  The atoms in the diborane bonds are not collinear because we need two such bonds, not because either bond suffers from imperfect overlap.


What is the range of sizes for molecular systems?

Well apportioned, mixed, and cured epoxy glue is a single covalent molecule.  Single crystal white diamond is a single covalently-bonded molecule.  On the small end, a dihydrogen molecule, is 0.7414 Å (ortho or para, your choice to trim the decimal).

The largest human chromosome is ~220 million base pairs, or about 1.5x10^11 g/mol.


You alluded to the answer when you mention activation energy. Kinetically the equilibrium constant is $K_e = k_f/k_b$ where $k_f$ and $k_b$ are the forward are reverse reaction rate constants in the reaction $\ce{A <=> B}$. The reason that there is a finite and not zero back reaction rate constant, is that the activation barrier going B to A is not infinitely high. Quite the opposite in fact and  while larger than the barrier A to B, (if the reaction is exothermic), may nevertheless be of such a magnitude that the products can return to reactants and thus an equilibrium is established. In this case molecules are continually transforming between one another. The ratio of concentrations at equilibrium is the equilibrium constant.
The reason that the activation barrier can be surmounted is that there is a Boltzmann distribution of energies in all molecules and this means that the number of molecules in either A or B with energy $E_i$ is given by $n_i=n_0\exp(-E_i/(k_BT))$. This is the Boltzmann distribution for molecules with energy $E_i$ and $k_B$ is the Boltzmann constant.
The total number of molecules of type A is given by the sum of those in each level $E_i$ from level 0 to the maximum energy level, which is
$$\frac{n_A}{n_0}=\sum_i \exp(-E^A_i/(k_BT))$$
The equation for molecules of type B with energy $E^B_i$ is
$$\frac{n_B}{n_0}=\sum_i \exp(-(E^B_i+\Delta E_0)/(k_BT))=\sum_i \exp(-E^B_i/(k_BT))\exp(-\Delta E_0/(k_BT))$$
where $\Delta E_0$ is the difference in energy between the lowest levels of A and B.
At equilibrium the constant K is the ratio of the two concentrations which is therefore the same as the ratio of the two populations and is
$$K=\frac{Z_A}{Z_B}\exp(-\Delta E_0/(k_BT))$$
where  Z represents the summations and are called the partition functions. This equation represents the equilibrium constant in terms of the molecules energy levels.
The reason that there is a Boltzmann distribution, but a fixed energy at a give temperature, is that by far the the most likely way of populating a range of energy levels is according to the Boltzmann distribution. In fact the chance of having a Boltzmann distribution is greater than all other possible ways added together.
In classical thermodynamic terms, a plot of free energy G vs extent of reaction $\zeta$ (zeta) which ranges from 0 (entirely A) to 1 (entirely B), has a minimum at some intermediate value of $\zeta$ and this value naturally depends on the particular reaction.  The minimum value is where the reaction is at equilibrium and at the minimum the slope of G vs. $\zeta$ is zero which is also when $(\partial G/\partial \zeta)_{T,P} = 0$.
The minimum in G occurs because the entropy term $-TS$ has a minimum as the mole fraction of B increases (at constant T). Initially it is the entropy of A alone in solution, but at equilibrium is a mixture of A and B. If the reaction goes entirely to B then the entropy is that of B alone. The entropy of a mixture is naturally greater than that of one species thus $-TS$ has a minimum vs. $\zeta$.
If $dn_A$ and $dn_B$ of A and B molecules react the the infinitesimal change in zeta is $d\zeta = dn_A-dn_B$. The total change in free energy is $dG=\mu_Adn_A+\mu_Bdn_B$ where $\mu$ is the chemical potential or the free energy / mole. From these two equations the change in free energy of the reaction is
$$dG=(\mu_b-\mu_A)d\zeta$$
The reaction proceeds until $(\partial G/\partial \zeta)_{T,P} = 0$ and thus to proceed further expressions for the chemical potential are needed. For a perfect gas $\mu=\mu^0+RT\ln(p) $  thus
$$(\partial G/\partial \zeta)_{T,P} = \mu^0_B-\mu^0_A+RT\ln\frac{p_B}{p_A}$$
which can be recast as
$$(\partial G/\partial \zeta)_{T,P} = \Delta G^0+RT\ln\frac{p_B}{p_A}$$
and at equilibrium as the derivative is zero
$$\Delta G^0=-RT\ln\frac{p_B}{p_A}=-RT\ln K_p$$
which connect the equilibrium constant ($K_P$ for the gas phase reaction), with the free energy.

Edit:
Notes: The shape of the plot of G vs. extent of reaction can be shown by considering a general reaction of the form $\ce{A + B <=> 2C }$ . The Gibbs function is then
$$G = n_a\mu_a + n_b\mu_b + n_c\mu_c $$
for $n_i$ moles of species i with chemical potential $\mu_i$.  If we assume that the reaction is one of perfect gases then at total pressure P and with mole fraction x, the chemical potential of species i is
$$\mu _i = \mu_i^0+RT\ln(P)+RT\ln(x_i)  $$
where $\mu_i^0$ is a function of temperature only.  G can now be written as
$$G= n_a\mu_a^0 + n_b\mu_b^0 + n_c\mu_c^0+RT\ln(P)+RT(n_a\ln(x_a)     +n_b\ln(x_b)+n_c\ln(x_c) )$$
If we assume that the total pressure is 1 atm. then the $RT\ln(P)=0$ and can be removed from the equation. Using the relationship between number of moles $n_a=n_b$ and $n_c=2(1-n_a)$ gives
$$G= n_a(\mu_a^0 + \mu_b^0-2\mu_c^0) + 2\mu_c^0 + 2RT(n_a\ln(n_a/2)     +(1-n_a)\ln(1-n_a) )$$
The $\mu^0$ are the properties of the pure components so do not change during the course of the reaction which means that G is a function of $n_a$ only.
A plot of $G(n_a)-G(0)$ vs. $n_a$ is shown below.

Plotting this way gives a value of zero when $n=0$, which is the condition should all reactants be converted into product, point S. The equilibrium value is at point E. The energy T-R is that due to mixing reactants and R-E that extra energy of mixing when C is formed. The dashed diagonal line is $n_a(\mu_a^0 + \mu_b^0- 2\mu_c^0)$ which is how the reaction would behave if no entropy of mixing is included.
The important point is that there is a minimum in the free energy curve because the creation of a mixture is an irreversible process and this mixing produces a decrease in free energy.
The equilibrium constant is calculated using $-RT\ln(K_p)=2\mu_c^0-\mu_a^0-\mu_b^0$


In general, the problem is a mathematical one. What you are asking for is the 'equivalence classes' of the atoms under the symmetry group. As pointed out in the comments, this may or may not be chemically important.

One simple way to divide the atoms into equivalence classes is to use Morgan numbers. The algorithm is particularly simple:

Label the atoms with their degree (so terminal carbons are "1")
Update each atom's label by adding the labels of the neighbouring atoms
Repeat until the number of different labels is constant

So, for a linear hydrocarbon, the labels start out as [1, 2, 2, ..., 2, 1] and update to [3, 5, 6, 6, ..., 6, 5, 3]. After a couple more rounds, you get a pattern of labels like [a, b, c, d, d, c, b, a].

This works reasonably well, although there are more complex algorithms for determining non-equivalent atoms. As you have worked out, the atoms beyond the immediate neighbourhood are necessary.


Why does the carbonyl group in an acid anhydride have two stretching frequencies?

The two observed C=O frequencies are due to the symmetric and asymmetric stretching modes of the anhydride.

  Source: Introduction to Spectroscopy, Pavia and Lampman

You can see that the lower frequency symmetric stretch occurs where both C=O bonds are lengthening and shortening in tandem, whilst the higher frequency asymmetric stretch occurs when one C=O group is lengthening as the other is shortening.  You can see an example of this kind of spectral pattern in the example below.

  Source: Introduction to Spectroscopy, Pavia and Lampman

This explains why you get two frequencies for C=O bond in acid anhydrides, whilst ketones/aldehydes etc. only give a single IR frequency for the C=O bond.


In chemistry what do they mean by disorder when in the context of entropy?

What you have said about entropy is generally correct. For a given energy it is related to the number of accessible states the system can be in. If a system has 100 energetically accessible configurations then we would say it has more entropy than a system with only 10. You can call this disorder because you have no knowledge of which of the 100 (or 10) states the system is actually in.

That said, entropy has a very well define mathematical meaning. I'll start with the punchline and then breakdown what it means. Entropy ($S$) is equal to the natural log of the multiplicity function ($W$, or $g$ sometimes) times boltzmann's constant ($k_b$):
\begin{equation}
S = k_b \ln(W)
\end{equation}

The multiplicity function $W$ is the number of states in the system of a given energy. So what does this mean? Let's imagine we have a magnet with three unpaired electrons which can either be pointed up or down. (e.g. $\uparrow \uparrow \uparrow $,  $\uparrow \downarrow \uparrow$, etc. ) Now let's say that there's a magnetic field applied so that up and down have different energies. Each $\uparrow$ adds $\epsilon$ to the energy while each $\downarrow$ subtracts $\epsilon$ from  the energy. This system can now exist in states ranging from +3$\epsilon$ to -3$\epsilon$.

Let's look at the E = +3$\epsilon$ and -3$\epsilon$ states first. In both cases, W($\pm 3\epsilon$) =1 becasue there is only one configuration for each energy, $\uparrow \uparrow \uparrow $ and $\downarrow \downarrow \downarrow$  respectively. Thus, using the formula, S = $k_b \ln(1) = 0$. Thinking of entropy as disorder, this makes sense. If the energy is $\pm3\epsilon$ then we know exactly which state the system is in and there is no chance whatsoever.

Now let's look at the case of E = $\pm \epsilon$. Here, we have multiple options for each state. In the E = +$\epsilon$ there are three possible states: $\uparrow \uparrow \downarrow$, $\uparrow \downarrow \uparrow$, and $\downarrow \uparrow \uparrow$. Similarly, for the E = -$\epsilon$ case there are also three states: $\uparrow \downarrow \downarrow$, $\downarrow \uparrow \downarrow$, and $\downarrow \downarrow \uparrow$. Thus, $W(\pm \epsilon) = 3$, and the entropy is $S = k_b \ln(3)$

$k_b$ is called The Boltzmann Constant and is just a scaling constant. It doesn't contain any physics and exists only because we happen to use different units for temperature and energy. In SI it's equal to ~$1.38×10^{−23}$ J/K, but you can equivalently measure temperature in energy units and then $k_b =1$ (with no units).

The physical insight is that entropy entirely a consequence system being allowed in multiple states.


How does composition of air affect vapor saturation limit?

I agree with "repurposer"'s comment that the saturation limit would be independent of the composition of the air.

It would just be a function of the partial vapour pressure of water.

$K_{p}=p_{H_2O}$, where $K_p$ is the equilibrium constant in pressure terms.

Consider this case similar to that of aqueous tension - it depends only on the temperature of the system, and not the other (non-reacting) components.
Whenever we want to find the pressure of the dry gas, i.e. pressure of the mixture of gases $-$ the pressure due to the water vapour.

Hope this helps!


Why is it that every snowflake is unique?

It turns out that 2 identical snowflakes have been observed, but...

Two Identical Snowflakes

Although when we think "snowflake" we usually picture an object with 6-fold radial symmetry, snowflakes actually come in many different shapes (reference 1).  Many sites report that in 1988 an NCAR researcher found 2 identical snowflakes of the hollow columnar variety (reference 2 - see "Uniqueness" section, reference 3).  This pattern is much simpler than the 6-fold radial design, so the odds of finding 2 identical flakes are vastly improved.

But

If you wanted to confine the question to snowflakes of the 6-fold radial variety, then an argument can be made that 2 identical snowflakes will not be found.

Due to the vapor pressure difference between ice and water, vapor is pushed from water towards the ice (hence, the ice grows) in an ice-water droplet falling through the atmosphere.  As it falls through different temperature zones, the rate of ice crystal growth will change leading to different dendritic patterns along the radial arms.  Further, due to perturbations caused by temperature and wind gradients across a single crystal, each radial arm will have a slightly different pattern.

The mathematical argument against identical (6-fold radial) snowflakes is, as Jon Custer suggested in his comment, based on factorial probabilities.  The argument goes something like this.  If you have 10 different pictures to put in an album you could arrange them in 10! (10 factorial) or 3,628,800 different ways.  If you look at a snowflake under magnification, there are easily over a hundred different features that are identifiable.  In this case there are over 100! or greater than $10^{158}$ possible arrangements of these identifiable features.  Estimates are that there are around $10^{80}$ atoms in the universe.  Each visible feature in a snowflake contains a large number of water molecules and an even larger number of atoms.  Hence, the number of possible arrangements is far, far greater than the number of available atoms, making it extremely improbable that a duplicate of a complex snowflake would be created.

If you're still not convinced, we can add another level of probabilistic protection.  Remember that there are isotopes of hydrogen and oxygen.  The probability of finding two 6-fold radial snowflakes with the same dendritic pattern and the same isotopic substitution pattern is even closer to zero.


Conditions for choosing quantifying and qualifying ion in LC-MS/MS

The limit of detection (LOD) in these kinds of assays depends on both the noise and the signal.  Usually it is defined as lowest concentration analyzed such that the signal (peak height) is three standard deviations higher than the noise.  Exactly how to define "the noise" is a critical consideration -- many different sources provide different definitions and ideas.  The best measurement of "noise" is probably in a method blank at the exact RT window you are using for quantification.  Ideally the method blank would have a sample matrix identical to your real samples, but guaranteed to be missing the target compound.

It's a little tough to know from the information provided in your question how exactly you are calculating your LODs.  If for the moment we assume you are happy with how your LODs are computed, then I would pick the method with the lowest LOD.

It's also worth mentioning that a standard practice in targeted LC-MS/MS is to monitor two separate transitions per compound.  One of those is called the "quantifier" and it is this signal that is used to compute peak areas for quantification.  The other is called the "qualifier".  The ratio of the qualifier peak area to the quantifier peak area should be monitored for all analyses, and it should be approximately the same for all samples (that contain measuremable compound).  If this ratio deviates wildly, it's a sign that the quantifier ion may not be reliable for a particular sample.


How to make equal density pool dye without diffusion

Whole milk with food coloring solved my problem.  Whole milk has a specific gravity of 1.01, which allowed it to settle to the bottom around the main drains.  The diffusion rate was extremely slow over regular pool dye, which allowed the dyed milk to linger around the drains for an extended period of time (5+ minutes) to allow for the detection of the slow leak.


Two different containers full up with pure water and salt water in osmosis. What happens?

In your described scenario, there will be osmotic flow in an attempt to equalize the chemical potential of the solvent (water in this case, assuming the membrane is only permeable to water). The effect of the osmotic flow will be that pure water from container 1 will flow into container 2. Since this is a closed system, one will observe the liquid level rise in container 2 and fall in container 1, and the amount of displacement will be a function of how different the chemical potential of the water is between container 1 and container 2. There is no classical force that will push or pull the liquid from one container to another, osmotic flow is an instance of diffusion which is governed by the concentration gradient that is present.