Is there an agent that changes colour when exposed to air?

Many compounds change colour when exposed to 'air', however, this is usually due to the presence of either water or oxygen, rather than 'air'.

To give a few examples:

*when distilling a solvent such as THF over sodium/potassium, benzophenone is used as an indicator. In the absence of air, the indicator leaves the solution a deep purple colour, however if an excess of air is present, the indicator turns a clear/brown colour.

*drierite is an indicator often used in inert gas drying tubes. In its dry form, it is either deep blue or orange, however when wet it turns white/pale orange.

Several other indicators exist, however its worth noting that air is merely a mixture of multiple things, and as such any indicator will indicate for a specific component of air rather than air as a whole- with water and oxygen being the most common things to indicate for.


If the reagent is alcoholic KOH, then the reaction mechanism followed is E2 elimination (if steric/other conditions permit)

However, if the reagent is aqueous KOH, it is simply a substitution (SN) type reaction in which the Br atoms are replaced by -OH groups (not to mention, if conditions permit, i.e. SN is actually possible in the given molecule*)

*example: SN type reactions are usually forbidden in compounds like PhBr (bromobenzene) as the π cloud of the benzene ring repels the incoming nucleophile (SN² not possible) and formation of phenyl carbocation is not thermodynamically feasible due to its instability (SN¹ forbidden)

As far as your main question is concerned, i.e. difference between Saytzeff's rule and Elimination -

They are two completely different things, one is a rule to correctly predict major products of elimination reactions and the other is a reaction mechanism, which defines the pathway for the reaction to proceed through.

In fact, Saytzeff's rule states:
'In an elimination reaction, the most substituted product will be the most stable, and therefore the most favored.'

This is not always true, however. Major products are formed when the reaction intermediate is stabilized (product stability is usually secondary)
In case of eliminations, especially if the reaction is to proceed through an E1 pathway which involves formation of a carbocation, stability of the carbocation becomes a very important deciding factor in the product percentage. If the carbocation so formed (assuming reaction proceeds via predicted pathway. Practically, it is not fruitful to talk about such intermediates as they have a very small lifespan and reaction may proceed through other more favourable pathways) is highly destabilised even though it is more substituted, for example, carbon attached to three nitro groups (or any strong electron withdrawing groups for that matter) then Saytzeff's rule is violated and corresponding Hoffman product is found to be major. Of course, there are several other cases in which Hoffman product is found to be major and corresponding Saytzeff product is minor. I believe that discussion deserves a separate post altogether, so I won't elaborate on it here.

Hope it helped!


Is crushing an avocado a chemical or physical change?

I don't think so.  A chemical change implies changing molecular-level structure usually accompanied by transferring electrons.  When you squish an avocado you're only rearranging the fats, carbohydrates, and fiber in it - I don't think a chemical reaction occurs.

If you left it out, and it turned brown (oxidized), that would be a chemical change!


Why do many transparent plastics turn opaque when deformed?

The process of forming an opaque area when you apply stress to a polymer film is known as crazing.

Polymers are long chains and, when coated, often form tangled, amorphous, randomly aligned polymer regions.  When stress is applied to the film, the stress can be relieved by the polymer chains aligning.  In other words, energy introduced into the film by stressing it can be used to align some polymer chains.

Increasing the alignment in the stressed region means that the density of polymer in this region of the film will be different from the density in non-stressed regions.  This difference in density will cause a difference in light reflection and scattering accounting for the different optical properties between these two regions.

Further, within the stressed area of the film, mechanical stresses are produced between areas that are more and areas that are less aligned.  These stresses can be relieved through the formation of micro-cracks between the regions.  These small cracks also result in different light reflecting and scattering properties of the film.

Finally, as the polymer chains align, they are more likely to crystallize.  The formation of small crystallites within the stressed (aligned) region will also alter the light reflecting and scattering properties of the film.  Often, upon heating or standing at room temperature (for a longer time) the optical changes may reverse as the thermal energy supplied will:

cause the polymer structure to randomize back to its amorphous structure,
allow the polymer to "flow" and fill the micro-cracks, and\or
the crystallites can redissolve.

References: The following links provide more information about crazing and associated phenomena at different levels of complexity:

reference 1

reference 2

reference 3


Reduction of oximes with lithium aluminium hydride

Comprehensive Organic Synthesis, 2nd ed. writes (emphasis mine)1

  The reduction of oximes or oxime ethers and esters to primary amines is a useful synthetic transformation that is used as a substitute for direct reductive amination to convert ketones or aldehydes to primary amines. The reaction involves reduction of C=N bond and reductive cleavage of N–O bond. Thus, the transformation requires the use of strong hydride reagents or reagent
  combinations.

I would guess that after the addition of hydride to the $\ce{C=N}$ bond (incidentally, you should not draw it as free hydride) the next step is a substitution at nitrogen. $\ce{X-Y}$ $\sigma^*$ orbitals ($\ce{X},\ce{Y}$ being heteroatoms) tend to be low in energy and rather electrophilic, so this is probably no different.

If it helps you to feel better about the negative charges, you should probably draw the nitrogen as being coordinated to a lithium and the oxygen as being coordinated to an $\ce{AlH3}$ fragment, or some permutation thereof. The close association of the heteroatoms with metal fragments also encourages "hydride" to attack. Again it's not really hydride attacking but rather the $\ce{Al-H}$ $\sigma$ bond.

1 Abdel-Magid, A. F. Reduction of C=N to CH–NH by Metal Hydrides. In Comprehensive Organic Synthesis, 2nd ed.; Knochel, P., Molander, G. A., Eds.; DOI: 10.1016/B978-0-08-097742-3.00802-8.


A couple of points for starters:

to say anything about impurities you need to show the entire spectrum, typically from $0$ to $\pu{10ppm}$ for $\ce{^1H}$. Merely describing the existence of peaks is not enough.
you also need to make sure your baseline is a proper baseline as you need to use the integrals to calculate purity if you even can. Here again, $n_\mathrm{t} = 8$ is absolutely too little.

However, it still needs to be said that determining the purity in percent just from NMR spectra is a very non-trivial task. It can only be done, if you know exactly which impurities you have, if they all contain the corresponding NMR-active nucleus, if the NMR-active nucleus is sufficiently predominant for the read to be quantitative and if there is no mechanism by which a non-NMR-active nucleus may be introduced into your impurity. Especially the predominance of NMR active nuclei along with another few requirements for ‘nice’ NMR spectra means that purity via NMR can in practice only be measure for $\ce{^1H, ^19F, ^31P}$ and a few much less common elements.

The main deal breaker is typically the requirement to contain the NMR-active nucleus. You can have a beautifully clean $\ce{^1H}$ NMR spectrum but the actual substance can be as little as $1~\%$ pure: most notably if the impurities are inorganic salts and do not contain any hydrogen. In most non-carbohydrate organic chemistry, this is not so much of a problem, though.

For any substances that contain the NMR-active nucleus and are impurities, the first step is to find out exactly what the substance is. To be able to calculate anything from an NMR spectrum, you need to know the impurity’s molar mass as NMR readings can only give you relative amounts.

Finally, you also need to make sure that your impurity does not lose its NMR-active nucleus. For example, if you are measuring in $\ce{CD3OD}$ and the impurity happens to be $\ce{C3F7OH}$ (1,1,1,2,3,3,3-heptafluoropropan-2-ol), the reading of the final remaining signal (if it even remains) will not be accurate due to proton/deuterium exchange:

$$\ce{C3F7OH + CD3OD <=>> C3F7OD + CD3OH}\tag{1}$$

(The equilibrium is shifted towards the products due to the abundance of methanol as the solvent.)

This is why NMR analysis is typically only used to determine relative purity, e.g. if two diastereomers form during a reaction. These will have the same molar masses allowing for an easy comparison.

To actually perform this, you need to:

Identify the peaks belonging to an impurity, the impurity’s structure and the signals’ integrals.
Normalise the impurity’s integrals to a single proton equivalent (i.e. if your impurity is cyclohexane, divide the integral by $12$ since you have $12$ magnetically equivalent protons.)
Normalise your desired compound’s integrals so that a single proton has the integral 1.
Determine a ratio of amounts in moles. (e.g. $\pu{8mol}:\pu{2mol}$)
Calculate the corresponding masses of desired product and impurity.
Add up the masses you calculated to a total mass.
Divide your desired product’s mass by the total mass to get your purity. Multiply by $100~\%$ to get a percentage.

From the information you give and the spectrum you show it is absolutely impossible to tell you anything about the purity of your compound. Signal integrals themselves are also not helpful; if your impurity is huge in molar mass but only has few protons, or if it has multiple magnetically identical protons a simple integral reading will be way off.

The practical approach for an organic chemist to determine purity is to use chromatography (either flash or HPLC) until no signals can be detected that do not belong to the compound in question and then label it ‘pure’.


Grignard reaction on electron-poor benzyl bromides

This question has aged a bit with no other answers, so here I go.

The same nitro group that can oxidize Grignard reagents could be oxidizing the magnesium too, thus keeping it passivated.  What you need is not so much a milder EWG but one with less oxidizing potential.  Try your reaction with nonoxidizing EWGs like acetyl (watch for side reactions), a $\ce{CF3}$ group, or a chlorine atom attached to the aryl group.


There are a couple ways that people define bonding and antibonding orbitals. One of the simplest is to just look at the energy of the MOs formed compared to the AOs they were formed from: if the MO is lower in energy, it is bonding and if the MO is higher in energy, it is antibonding.

Like I said, this way of doing thing is relatively simplistic, but you will see it pretty commonly in textbooks on the matter. The method of course isn't perfect, in particular because it essentially assumes that all local bonds are affected equally by a particular orbital.

We can see this is not the case by looking at the $\pi_2$ orbital you show. While the orbital is listed as bonding, the interactions of the p-orbitals where a positive lobe is next to a negative love will locally decrease the bonding between those two atoms (ie the MO is antibonding with respect to those local bonds).

So, for a simple first approximation, you can use the energy of the MO compared to the AO, but for a more complete sense, you should look at the symmetry of the MO and see what effect it has on a given local bond.

In terms of the ordering of the MOs themselves, assuming you don't have experimental way to determine the ordering like ionization energies, you can heuristically use an approach like you suggest where you compare the number of nodes and less nodes will mean lower in energy, more nodes will be higher.

In terms of comparing to the original AOs, it can be tricky without experimental data, however there are again general qualitative ways of ordering the MOs with respect to their original AOs. For example, if you have 2 MOs formed from two overlapping AOs, one will have to be higher in energy and the other will have to be lower.

When you get more orbitals overlapping, the arrangements can vary, but the sum of the energies has to remain the same. So, for example, if you have 3 overlapping AOs, you could form 2 bonding and 1 antibonding MOs or vis versa so long as their combined energy is the same as before.

Beyond this, you really just have to read up on MO theory to get a sense of how qualitative energy ordering is done. There is a lot more to it than what I've written (or could write) and it's probably better to build your understanding of the theory from the ground up rather than getting it piecemeal.


$\ce{2Zn(NO3)2}$ can be interpreted differently, depending on the context.
The '2' outside of the parenthesis could either refer to two molecules or two moles of substance. In either case, the chemical formula tells you the same thing: that there are two units of $\ce{NO3-}$ for every unit $\ce{Zn^2+}$.

I would write it as zinc (II) nitrate, (read as 'zinc-two-nitrate) The roman numeral (II) refers to the oxidation state of the zinc ion, which is $+2$.


How does resonance fail in approximating chemical structures?

Resonance structures are one model that explains bonding, but not a very good one

Resonance structures are widely misunderstood. Individual resonance structures are not real compounds that exist, but an approximation that assumes all structures are made from bonds consisting of two electrons. The true structure of a compound is thought to be an average mixture of the various possible resonance structures consisting of simple bonds.

The archetypal example is benzene. If we model benzene as a structure consisting of "normal" 2-electron bonds, we get a ring consisting of alternating single and double bonds. This contradicts the observed structure where all the bonds are the same (and are somewhere between the character of single and double carbon-carbon bonds). The resonance model of benzene consists of two structures with the normal double and single bonds swapped, giving an average overall structure where all the bonds are an average of single and double bonds.

For structures with more complexity than benzene, the resonance structures are more complex and frequently don't make equal contributions, nor do they do a good job of predicting the real structures. Molecular orbital based approaches (which recognise the difference between pi-bonds and sigma bonds and do a much better job of recognising the fractional character of many real bonds) are far better.

That is why resonance structures "fail". They make drawing simple structures easier, but they don't do a good job of predicting real structures.


I did some brief research and I saw that sulfuric acid can dissolve iron:

Fe(s) + H$_2$SO$_4$(aq) → Fe$^{2+}$(aq) + SO$_4$$^{2-}$(aq) + H$_2$(g)

Of course, this would leave sulfate ions and perhaps some precautions would need to be taken because of hydrogen gas production. If you absolutely cant have any sulfate ions, you could precipitate them out by adding barium or silver ions to the solution, but that would require extra work and introduce new ions to the solution. Also note, the iron ions should be aqueous in +2 state but I read that if there is oxygen in the system (which there naturally would be), some iron(II) can oxidize to iron(III).


Active methylene group nucleophilic substitution

You have given correct suggestions for the first two part of the question. Now, the question is where would be the electrophilic attack of acid chloride taken place under Friedel-Crafts acylation conditions. As Ron correctly pointed out in his comment that there are only 2 possibilities to cyclization: That would be to the 2 or 8 positions of naphthalene nucleus, and it would certainly be the 2 position, which produces a 6-membered ring. The method is a named reaction, which is called Haworth Phenanthrene Synthesis (Ref.1 & 2). According to the abstract of Ref.2:

  Haworth synthesis is a multistep preparation of phenanthrenes from naphthalenes by means of the Friedel–Crafts acylation with succinic anhydride, followed by a Clemmensen reduction or Wolff–Kishner reduction, cyclization, reduction, and dehydrogenation. This synthesis is also referred to as the Haworth phenanthrene synthesis. Additionally, the same procedure used for the preparation of naphthalenes from benzenes is called the Haworth reaction. This reaction has been modified to react with 4‐pentenoic acid ester to form alkyl substituted phenanthrenes. This reaction has a wide application in preparation of fused aromatics, including naphthalenes and phenanthrenes.

The Haworth synthesis is depicted in following Scheme under the same title in top box:

You see after the reduction by $\ce{Zn/HCl}$, you get the same compound (4-(1-naphthyl)butanoic acid) you got after acid hydrolysis and heat in your scheme, as indicated in top yellow box. In Haworth reaction, $\ce{H2SO4}$ has been used instead of $\ce{AlCl3}$ for the convenience, the reaction under goes same pathway to give tetrahydrophenanthrene keto derivative by acylating 2-position of the naphthalene nucleus (see second yellow box). There was no evidence for the formation of 7-membered ring (by undergoing cyclization through 8-position).

However, if you had 4-(2-naphthyl)butanoic acid instead of 4-(1-naphthyl)butanoic acid, you certainly would get anthracene derivative (e.g., compound 4) as the major product with phenanthrene derivative (e.g., compound 5) as a trace product (See second box in the Scheme)(Ref.3).

References:

R. D. Haworth, “145. Syntheses of alkylphenanthrenes. Part I. 1-, 2-, 3-, and 4-Methylphenanthrenes,” J. Chem. Soc. 1932, 1125-1133 (DOI: 10.1039/JR9320001125).
Zerong Wang (Ed.), In Comprehensive Organic Name Reactions and Reagents – PART TWO; John Wiley & Sons, Inc.: New York, NY, 2010, “Chapter 299: Haworth Synthesis (Haworth Phenanthrene Synthesis),” (https://doi.org/10.1002/9780470638859.conrr299).
Z. Zhang, R. Sangaiah, A. Gold, L. M. Ball, “Synthesis of uniformly $\ce{^{13}C}$-labeled polycyclic aromatic hydrocarbons,” Org. Biomol. Chem. 2011, 9, 5431-5435 (DOI: 10.1039/C0OB01107J).


Why are Alcohol Dehydration Reactions Favored by High Temperatures?

Let's take: $\,\,\,\,\ce{ethene(g) + H2O(g) <=> ethanol(g)} $

The standard enthalpy for hydration of ethene is $\Delta H^\circ_r=-45\frac{KJ}{mol}$. So we have $\Delta H^\circ_r=45\frac{KJ}{mol}$ for dehydration, which means that it is endothermic.

The expression:

\begin{align}
\frac{K_2}{K_1} &= \exp{\left(\frac{\Delta T \Delta H}{RT_1T_2}\right)} \tag{1}\\
\end{align}

(This equation is valid when Ts are similar, because $\Delta H$ varies with $T$ and you are considering it constant.)

with $\Delta T= T_2-T_1$ and $\Delta H_r^\circ>0$ (dehydration) equation (1) gives $K_2>K_1$.

As $$K = \frac{[\ce{H2O}]\,[\ce{CH2CH2}]}{[\ce{CH3CH2OH}]}$$

and $K_2>K_1$. In consequence dehydration enthalpy is favoured by temperature-increase.


Negative retention for chloride in hydrolysate filtration

A combination of osmosis and Gibbs-Donnan effect is the correct explanation.

With a semi-permeable membrane such as the one you are using, or a cell membrane, or in dialysis, water can move freely as can chloride and other small ions, however large molecules such as proteins cannot.

A simplified explanation is this: in osmosis, the tendency is for the osmotic pressure and charge to equalise on either side of the membrane. Proteins usually have a negative charge and exert a large osmotic pressure. Thus, to balance this out, small anions will diffuse to the opposite side to where the proteins are. Water will diffuse until the osmolality is equal on either side of the membrane unless there is a hydrostatic pressure forcing it to the other side. In the absence of hydrostatic pressure, you will end up with a higher concentration chloride solution on one side and a higher concentration protein solution on the other side.

Here is a diagram:

If you assume that the small particles you can see are chloride and the big particles are your proteins, you can see that the chloride concentration in the permeate is higher than the retentate and higher even than the feed.


How to find enthalpy of solvation for gas phase cation using Hess's law?

One approach to compute the enthalpy change for the reaction
$$\ce{H2(g)->2H+(aq) + 2e-}$$
is to combine the bond dissociation enthalpy and hydrogen ionization energy (both refer to gas phase species) and the hydration enthalpy of the hydrogen ion:
$$\ce{H2(g)->2H(g)}~~~~~~~~{\Delta H = \pu{439 kJ/mol}  }\\
\ce{2H(g)->2H+(g) +2e-(g)}~~~~~~~~{\Delta H = \pu{262 kJ/mol}  }\\
\ce{2H+(g)->2H+(aq)}~~~~~~~~{\Delta H = \pu{-2260 kJ/mol}  }$$
If you add up the enthalpies you obtain as the solvation enthalpy of $\ce{Ca^{2+}}$ the value $\Delta_{\text{solv}} H = -1546 \pu{kJ/mol}$ which is quite reasonable (see eg a table here)
The result is clearly very different from my earlier answer which I am still pondering:
The catch is that $\Delta H^\circ=0$ for this reaction. The reason is because $\Delta G^\circ=0$ for the reaction (by definition, see standard reduction potential) at all temperatures, and then, since
$$\Delta H = -T^2\left( \frac{\partial (\Delta G/T)}{\partial T}\right)_p$$
it follows that  $\Delta H^\circ=0$. I suspect the problem with this approach is due to a missing term involving enthalpy of solvation of the electrons (or a similar term involving the electrons) but I haven't put my finger on it.


What is the "mechanism" of the decomposition of hydrogen peroxide?

Below is one way to write it that shows what the enzyme does and avoids the uncertainty of the details:

$$\ce{E + H2O2 -> E-O + H2O}\tag{1}$$
$$\ce{E-O + H2O2 -> E-O2 + H2O}\tag{2}$$
$$\ce{E-O2  -> E + O2}\tag{3}$$

In words, the enzyme binds to the first molecule of hydrogen peroxide, release one molecule of water and holds on to the other oxygen atom (via the iron bound to heme bound to the enzyme). Then, the second molecule of hydrogen peroxide binds to the enzyme, which again release water and holds on to the second oxygen atom. Finally, the enzyme releases the two oxygen atoms in the form of the dioxygen molecule.

Alternatively, you could write the following three steps, giving even less information about what is happening in the enzyme:

$$\ce{E + 2H2O2 -> [E\cdot H2O2\cdot H2O2]}\tag{A}$$
$$\ce{[E\cdot H2O2\cdot H2O2]-> [E\cdot H2O\cdot H2O\cdot O2]}\tag{B}$$
$$\ce{[E\cdot H2O\cdot H2O\cdot O2] -> E + 2H2O + 2O2}\tag{C}$$

In words, the enzyme first binds to the substrates, then catalyzes the reaction, and finally releases the products. While this in general might be a good way of describing the action of an enzyme, it is not quite correct for catalase, which releases one product before the second substrate binds. However, this might be the answer you are looking for at the IB level. I would ask the teacher for further clarification.


You are right and whoever wrote this quote "Alkaline metals and their salts impart characteristic colour to an oxidising flame. This is because the heat from the flame excites the outermost orbital electron to higher energy level. When excited electron comes back to the ground state, there is emission of radiation in the visible region of the spectrum. Beryllium and magnesium do not impart colour to flame because electrons in beryllium and magnesium metal are too strongly bound to get excited by flame." is mostly wrong about Mg and Be.

The first process in flame spectroscopy is to decompose the compounds in the flame into atoms. /This requires a lot of energy. In ordinary lab flames, like that of the Bunsen burner (methane+air) the temperature is so low that alkaline earth (Ca, Ba, Sr) emit light because of their compounds in the flame not by their atoms. Recall magnesium and beryllium forms refractory oxides, so the correct explanation is that the flame temperature of ordinary oxidizing Bunsen burner is so low that it is not able to break magnesium compounds and excite beryllium in the flame.

Now if you use a proper flame, say that of acetylene-air for Mg or better acetylene-nitrous oxide flame for Be, you should see emission mostly in the UV but also very low intensity visible lines. The visible lines could be detected by a sensitive detector rather than the eyes. Nitrous oxide flame is highly colored itself (sunset pink color) and one cannot see visible emission easily without using a monochromator.


How to calculate the oxidation state and the number of equivalents oxidized with YBa₂Cu₃O₇

Your fundamental assumption is that all of the oxygen is $\ce{O^2-}$. In oxide ceramics, however, it is possible for oxygen atoms to dissolve into the matrix. zirconia,  titania, and bismuth oxide can all easily accumulate non-stoichiometric oxygen. In fact this ability to "dissolve" oxygen into a solid matrix is how fuel cells are able to operate.

You are correct that the yttrium is $\ce{Y^{3+}}$, barium is $\ce{Ba^{2+}}$, and copper is $\ce{Cu^{2+}}$, but in YBCO the oxygen can form $\ce{O^{(2x-2)-}_x}$ sites as part of $\ce{Cu2O_x}$ ($1\leq x \leq 4$) clusters. Note: for $x = 2\ \ce{O^{2-}2}$ is not inherently a peroxide ion, you need cooper pairs of electrons to have superconductivity (see image below). Thus another way to present the structure of YBCO may be as $\ce{YBa2Cu3O5(O2)}$ or rather $\ce{YBa2Cu3O_{5+x}(O2)_{1-x}}$ or even $\ce{YBa2Cu3O_{3+3x}\nu^0_x(O4)_{1-x}}$ $(0 \leq x \leq 1)$ as the ceramic can be non-stoichiometric and produce oxygen vacancies which affect the superconductivity.


What is the formula for finding the average atomic mass of an element based on the abundance of its isotopes?

The average relative atomic mass of an element comprised of $n$ isotopes with relative atomic masses $A_i$ and relative fractional abundances $p_i$ is given by:

$$
A = p_1 A_1 + p_2 A_2 + \dots + p_n A_n = \sum\limits_{i=1}^n p_i A_i
$$

For example carbon:

\begin{array}{lrrr}
\text{Isotope} &
\text{Isotopic Mass $A$} &
\text{Abundance $p$} &
A\times p\\\hline
\ce{^{12}C}:   & \pu{12.000000 u} & 0.98892 &= \pu{11.867 u}\\
\ce{^{13}C}:   & \pu{13.003354 u} & 0.01108 &= \pu{00.144 u}
\end{array}

Since we have $p_1A_1$ and $p_2A_2$, we add those together to find $A$, therefore
 the chemical relative atomic mass of carbon is
$$A = \pu{00.144 u} + \pu{11.867 u} = \pu{12.0011 u}.$$


I think there are multiple reasons contributing.

First of all, the atomic orbitals of the f-block and their properties and interactions are complicated. You have a lot of electrons, and they are far away from the nucleus, on many differently oriented orbital lobes. (well, that is the visualisation at least, in reality atomic orbitals are described by quantum mechanical wavefunctions, and those get more complicated as well for the f orbitals)
Since you have a lot of electrons repelling each other, and on top of that there are relativistic effects coming into play, it is rather difficult to do quantum chemical calculations with these elements to determine their energy levels, etc.
A lot of "rules" (and approximations) valid for low atomic number elements stop working.
Overall, heavy elements have special properties, that cannot be explained by the simple rules and approximations used for light elements. As you have said there is a lot of difficult math in the background. The rules and visualisations are just there to be an  easy to grasp abstraction of the underlying physics. The side effect of this is that you need more and more exceptions as the system gets more complicated.

Second, they are usually rare and expensive elements.(actinides are also radioactive) They have no known biological role, and usually only have few significant uses. So usually one does not encounter them unless one is working in some niche field.


To start, it's important to differentiate kinetics (which describe the rate at which reactions happen) and thermodynamics (which describe what net reaction can happen, even if it doesn't happen on the time scale of the universe's existence).

Your equations are all thermodynamic equations, so let's start there. I'm going to assume that you meant standard enthalpy and entropy. Then if we multiply through by $-T$, we end up with the definition of standard Gibbs free energy:

$\Delta G^\circ = -T\Delta S_{tot}=\Delta H^\circ_{sys} - T\Delta S^\circ_{sys}$.

If $\Delta G^\circ<0$, we say the reaction is spontaneous, but only in the thermodynamic sense that it can happen given infinite time. An important point is that this statement refers to the net reaction involved. The reverse reaction (which necessarily has a positive $\Delta G^\circ$) can happen, but always at a rate less than that of the forward reaction, such that the net reaction observed is the forward reaction. Because we're talking about thermodynamics and not kinetics, we can't make any conclusion about what those rates will be in absolute terms, only that the forward will be relatively faster than the reverse.

A different way of saying that is that if we set up a situation in which all reactants and products are at standard states (generally defined as 1 M solutes, 1 bar gases, pure solids, pure liquids), the concentration changes over time will be increases in products and decreases in reactants until the reaction reaches equilibrium. For a reaction that reaches equilibrium quickly, both the forward and reverse reaction happen quickly, but the forward slightly more quickly. At equilibrium, both reactions are still occurring, but now at exactly the same rate.

For a concrete example, consider the dissolution of a soluble compound like NaCl. If we have a solution of 1 M NaCl in water and we add excess solid NaCl, more NaCl will dissolve, but even when the solution becomes saturated (equilibrium), ions are constantly leaving the solid and going into solution while other ions are depositing from solution to the solid at the same rate.

In your hypothetical case where a change in T results in $\Delta G^\circ$ becoming positive, the reactions again are still happening. The difference is just that the reverse is faster than the forward when things are at standard states. It turns out that dissolution of NaCl is one of those cases you described, where $\Delta H^\circ >0$ and $\Delta S^\circ>0$. That means that if you made a 1 M solution and then cooled it, at a certain point solid would precipitate out, meaning that you've gotten to the point where at the standard state, the reverse reaction is faster than the forward reaction. At this temp, $\Delta G^\circ$ for dissolution is positive. If you stay at that temp, you can wait infinite time, and the solid will never fully redissolve.

All of that said, it sounds like your question might really be about kinetics, that is, how much can you slow down a reaction by cooling it? That question has little to do with entropy and instead relates to activation energies.


You're getting 0 ox. state for carbon because of treating both carbons equally - 0 is mean average value from oxidation numbers of both carbons. The one in carboxylic group  has three bonds with more electronegative oxygen therefore has +3. The one in methyl group has -3 because it's connected with three hydrogens which have lower electronegativity.

See https://en.wikipedia.org/wiki/Oxidation_state#The_Lewis_structure - there's example with acetic acid!

In organic chemistry trying to get average value of ox. state usually doesn't make much sense as it's rare to have all carbons equivalent. @Shafter gave you good advice - without knowing the structure you were making mistake.


Fractional distillation of a product differs from the components listed in the SDS. What am I doing wrong?

Best Answer is this:
Your observations are consistent with the SDS for the reasons below.

A couple of issues see below.

Fractional distillation is much more complicated than you suppose. You have no way to control reflux ratio, (amount of distillate vs reflux) which right away will limit your separation. The column is not insulated or temperature controlled.
Vigreux columns are good for distilling one component from non-volatiles, but are not made for fractional distillations. You will need a much more efficient column, the cheapest choice being a vacuum jacketed Hempel column with some sort of packing, ie Raschig rings.
With the equipment you are using, a range of 10 C between ideal boiling points would be considered normal, you will not see the sharp cut offs you are expecting.
The first component is likely methanol, and the boiling point range you are seeing is due to the above reasons.
Ethylene glycol does not boil to almost 200 C, when you end the distillation, is there any liquid left in the distilling flask?
As far as I could quickly check, methanol and ethylene glycol do not form an azeotrope, but the tables I checked were on Wikipedia and not thorough, you might want to check that. The formation of an azeotrope definitely explains the observations especially if it boils in the range of 98C +/- 20 C
Not sure on the exact conditions or if the MSDS is wrong, but it is conceivable dimethoxyethane (dimethyl glycol) formed or was the original constituent, this has a boiling point of 85 C, it could have even formed in the distilling flask depending on conditions.
Note the MSDS or SDSs are they are now called are not complete, and there are usually other minor constituents that might appear and could possible catalyze the formation of dimethoxyethane, assuming the SDS was correct in the first place.
Distillation is not the appropriate method here, GC-MS would be the way to go.

Hope that helps! I suggest you purchase Handbook of Laboratory Distillation, it is the fundamental text for distillation in theory and practice, the price is well worth it.

In conclusion your observations could be explained by: azeotrope formation, incorrect SDS, formation of dimethoxyethane, and incorrect set up.


How does the composition of ice differ from that of water in a partly frozen solution?

It sounds like you're talking about freezing-point depression.  If the solid component of a system is pure (oft assumed so; crystals are good at excluding impurities), the equilibrium is shifted because the chemical potential of the solution drops.

For a system with two pure species, the chemical change from say, water ($\star$ denoting pure) to ice is in equilibrium when their chemical potentials (denoted $\mu$) are equal:

$$\mu_{\ce{H_2O}(\text{s})} = \mu^\star_{\ce{H_2O}(\text{l})}$$

Note: Chemical potentials are always dependent, to varying degrees, on both temperature and pressure, so I'm not going to bother writing $\mu(T, P)$ everywhere

On the following graph that's at $T_0$; the usual freezing point of water where the "pure water" and "ice" curves intersect.  If water exists at a temperature lower than $T_0$, its higher chemical potential translates to a higher free energy, so it's thermodynamically driven towards freezing.

(image source: a website that is currently down; retrieved via the Web Archive)

If something non-volatile (e.g. salt) is dissolved in the water, its chemical potential changes to $\mu_i^\star+RT\ln \chi_i$, where $\mu_i^\star$ is the chemical potential of the pure species $i$, $R$ is the gas constant, and $\chi_i$ is the molar fraction of species $i$.

Originally, the fraction was $1.0$ (it was pure), and because $\ln1 = 0$, that term was ignored.  But now, because some of the solution species is something else, $\chi_i$ drops, thus $\ln\chi_i$ becomes negative, so the term lowers the overall chemical potential.  Because the chemical potential is lower, its free energy is lower, so it is thermodynamically more stable at lower temperatures.  At some $T$, we will find that:

$$\mu_{\ce{H_2O}(\text{s})} = \mu_{\text{solution}} = \mu^\star_{\ce{H_2O}(\text{l})} + RT\ln\chi_i$$

And there, the two species will be in thermodynamic equilibrium.

For more, check this site, or Wikipedia articles on chemical potential and freezing point depression (a basic Physical Chemistry text book, or chapter in a Gen. Chem. one might be better; the WP articles are not that great to get started with).


Intuituve reason for the constant of proportionality 2/3 in pv=nRT=(2/3)U

The 2/3 factor is from the math of the kinetic theory of gases. Think of a cube with six faces (i.e. three orthogonal directions), and the particles have equal probability of hitting a face in the x, y or z direction.  The pressure is exerted equally amongst the three orthogonal directions, hence a fact of 1/3.

Wikipdedia also shows the steps in deriving the kinetic theory.


What is the order of the reaction with respect to NO?

The rate of that second reaction will indeed be first order with respect to $\ce{NO}$:

$$
R = R_2 = k_2[\ce{NO}][\ce{NOBr2}].
$$

The problem is, the overall order of the reaction is determined from the rate expression involving only the reactants.  The above rate expression is written in terms of $\ce{NO}$ (reactant) and $\ce{NOBr2}$ (intermediate), not in terms only of reactants.

To convert the reaction rate expression to one involving only reactants, exploit the fast equilibrium of the first reaction,

$$
K_1 = \frac{[\ce{NOBr2}]}{[\ce{NO}][\ce{Br2}]},
$$

to find an expression for $[\ce{NOBr2}]$ in terms only of $K_1$ and the reactants:

$$
[\ce{NOBr2}] = K_1[\ce{NO}][\ce{Br2}].
$$

The expression for $R = R_2$ can thus be rewritten as

$$
R=R_2 = k_2[\ce{NO}]\left(K_1[\ce{NO}][\ce{Br2}]\right) = k_2K_1[\ce{NO}]^2[\ce{Br2}],
$$

which is indeed 2$^{\mathrm{nd}}$ order in $\ce{NO}$.


In solid-phase peptide synthesis, why are the formation of aspartimides a problem but not "glutamides"?

Since you understand how aspartimides (succinimides) form and why are they a problem in solid-phase peptide synthesis (SPPS), I'm not going to elaborate it further. Your main question is why don't glutamic acid residues face the same problem? Then, my question is who said they (glutamic acid residues) won't? Of course, they do. One example is discussed in Ref.1. You didn't recognized this cycle formation because the cyclic form is called Glutarimide Formation (contrast to aspartimides, which is actually a succinimide). The abstract states that:

[...] We found that the sequence of Glu(Gly)-OAll leads to glutarimide formation, which disrupts the formation of cyclic RGD peptides. However, our results show that glutarimide formation is sequence dependent and can be inhibited by incorporating an amino acid like Lys(Boc) with steric hindrance from the protecting group. To prevent glutarimide formation, Ser(tBu) was used to replace the glycine in the GGG spacer adjacent to the residue of Glu, and a tailed cyclic RGD peptide, c[RGDfE(SGGKK-NH2)] was successfully obtained.

Initially, $\ce{c[RGDfE(GGGKK-NH2)]}$ was designed with $\ce{c(RGDfE)}$ linked through $\ce{Glu}$ to a tail consisting of a spacer of three $\ce{Gly}$ residues $\ce{(GGG)}$ and a linker of two $\ce{Lys}$ residues $\ce{(KK)}$. We found that the sequence of $\ce{Glu(Gly)-OAll}$ leads to glutarimide formation, which disrupts the formation of cyclic RGD peptides (Note: $\ce{Glu(Gly)-OAll}$ is glutamic acid(glusine) $\alpha$-allyl ester):

The succinimide and glutarimide formations are not just common in SPPS. They can be formed under physiological conditions and can disrupt the biological processes (e.g., Ref.2,3).
References:

Junmin Zhu and Roger E. Marchant, "Solid-Phase Synthesis of Tailed Cyclic RGD Peptides Using Glutamic Acid: Unexpected Glutarimide Formation," J. Peptide Sci. 2008, 14(6), 690–696 (DOI: https://doi.org/10.1002/psc.975).
Josipa GrzeticJosipa Grzetic and Jos Oomens, “Spectroscopic Identification of Cyclic Imide b2-Ions from Peptides Containing Gln and Asn Residues,” J. Am. Soc. Mass Spectrom. 2013, 24(8), 1228–1241 (DOI: https://doi.org/10.1007/s13361-013-0661-6).
Josipa GrzeticJosipa Grzetic and Jos Oomens, “Effect of the Asn side chain on the dissociation of deprotonated peptides elucidated by IRMPD spectroscopy,” International Journal of Mass Spectrometry 2013, 354-355, 70-77 (DOI: https://doi.org/10.1016/j.ijms.2013.05.005).


How to understand this form of writing the solution: (some salt • n H₂O)?

The salt's crystal lattice's repeating unit is constituted of n molecules of salt and m molecules of water. Such salts are called hydrates. Wikipedia even has a nice picture of a hydrated vs non hydrated salt.

You can have anhydrous ferrous chloride, as well as ferrous chloride tetrahydrate. The number indicates how many water molecules are present in the crystal lattice's unit cell.

EDIT

This has just crossed my mind: be careful when preparing solutions of metal complexes to make sure you know what compound you're working with. One gram of anhydrous salt contains more equivalents than a gram of hydrated salt. This information should be obvious on the bottle. If not, ask a lab tech if they can identify the hydrated/anhydrous salt by memory.


Can a clay brick waste in a soil be a source of minerals for plants?

In general clay is not a good source of nutrients. Clay is a class of hydrated alumina-silicates containing small amounts of metals. The alumina and silica are not plant nutrients, but some of the metals can be, in particular: potassium, magnesium, calcium, and iron.

Of these, only potassium is needed in large amounts by plants, and conversely, is fairly uncommon in clays. Calcium and iron are much more common, but are needed by plants in very small quantities.

For any mineral nutrient in any case I can think of, other sources would be better than clay.

Therefore, in the specific case where there was a mineral deficiency in the soil, and you didn't have another option available, clay could provide that nutrient.

However, clay can help plants in another manner. This article describes the ability of clay soil to attract and hold positive ions (metals that can be absorbed by plants are positive ions) due to the high surface area and negative surface charge of the clay particles. Since leaching can be a problem in nutrient-poor soils, having a soil component that "holds on" to those nutrients can be helpful.

Whether or not brick dust would be helpful to plant growth would depend on the specific case, in particular:

The type of plant(s)
The composition of the rest of the soil
The type of clay used to make the brick
The climate conditions (amount of rainfall, etc.)


Why does gum get harder to chew after a while?

I would be surprised if there was reticulation of the gum's polymers upon contact with your saliva. According to this manufacturer, plasticizers are used, and the plasticizers appear to be fats. This would mean that, upon chewing, the plasticizers would be removed from the gum, rendering it unpleasantly harder.

Plasticizers are used to render a polymer more pliable, more elastic, than it naturally is. This could get its own question, imo.


Why isn't the orbital angular momentum also considered while calculating the magnetic moments 3d transition elements?

Simplistically speaking orbital angular momentum is present when some conditions are satisfied:

A set of orbitals are degenerate;
These orbitals can be "interconverted" by rotation about a certain axis;
The set of orbitals is not empty, half-filled, or fully filled.

Physically speaking, the "rotation" of an electron from one orbital to the next generates orbital angular momentum. The rotation must not have an energy cost, which necessitates degeneracy. It also means that the set of orbitals must have at least one electron, as well as one empty space (a so-called hole) for the electron to "rotate" into. The hole must have the same spin as the electron, which means that a half-filled subshell doesn't allow for orbital angular momentum (since all the electrons have parallel spins).

This is most easily illustrated with the three 2p orbitals in an atom. Rotation of a $\mathrm{p}_x$ orbital by 90 degrees around the $y$-axis, for example, turns it into a $\mathrm{p}_z$ orbital.

Any atom with a $\mathrm{p}^1$, $\mathrm{p}^2$, $\mathrm{p}^4$, or $\mathrm{p}^5$ configuration therefore possesses nonzero angular momentum. This is also reflected in their term symbols where $L = 1$. This is said to be "one unit of angular momentum". An atom with a half-filled shell (for example, $\ce{N}$)  or a closed-shell $\mathrm{p}^6$ configuration (for example, $\ce{Ne}$) does not have any orbital angular momentum.

The five 3d orbitals in an atom also satisfy these requirements. It is less intuitive to see this as compared to the 2p orbitals, because we have taken specific linear combinations of the complex 3d orbitals and converted them into real orbitals. Nevertheless one can use quantum mechanics to prove that the 3d orbitals possess angular momentum. The maths can be found in Figgis and Hitchman's Ligand Field Theory and Its Applications (it is very early in the book, but I do not have it with me now and cannot remember offhand where it is). As such, for free 3d metal atoms or ions, the orbital angular momentum has to be incorporated into the calculation of the magnetic moment. Exceptions are, of course, $\mathrm{d^0}$, $\mathrm{d^5}$, and $\mathrm{d^{10}}$ ions (e.g. $\ce{Sc^3+}$, $\ce{Mn^2+}$, and $\ce{Zn^2+}$).

That is applicable for free ions. However, 3d metals do not commonly exist as free ions, but exist in coordination complexes with a ligand field. For simplicity we'll consider a perfect octahedral ligand field, which causes d orbitals to split into a $\mathrm{t_{2g}}$ ($\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$) and $\mathrm{e_{g}}$ ($\mathrm{d}_{z^2}$, $\mathrm{d}_{x^2-y^2}$) set. This means that the five d orbitals are no longer all degenerate, breaking the first condition stipulated above.

At a simplistic level, many textbooks will claim that this leads to a "quenching" of orbital angular momentum, which just means that there is no longer any orbital angular momentum, and as such, the spin-only formula applies. [Note that there are some second-order effects (quantum mechanical mixing of excited states into the ground state) that can lead to deviations from the spin-only formula.]

This is true to a certain extent. However, the five 3d orbitals are not split into individually nondegenerate orbitals; the three $\mathrm{t_{2g}}$ orbitals, for example, are still degenerate. On top of that, one can verify pictorially that it is possible to rotate the $\mathrm{t_{2g}}$ orbitals into one another. In this aspect the $\mathrm{t_{2g}}$ orbitals behave quite similarly to the three $\mathrm{p}$ orbitals in a free atom; this is called the "$\mathrm{t_2}$-$\mathrm{p}$ isomorphism". As such, the $\mathrm{t_{2g}}$ orbitals still have orbital angular momentum associated with them.

Any complex with unsymmetrical occupancy of the $\mathrm{t_{2g}}$ orbitals (i.e. $\mathrm{t_{2g}}^1$, $\mathrm{t_{2g}}^2$, $\mathrm{t_{2g}}^4$, $\mathrm{t_{2g}}^5$) therefore possesses orbital angular momentum and exhibit great deviations from the spin-only formula. These complexes all have a $\mathrm{T_{1g}}$ or $\mathrm{T_{2g}}$ ground state term symbol, so it is commonly said that "$\mathrm{T}$ terms possess one unit of orbital angular momentum". Spin-orbit coupling has to be invoked to obtain a better agreement between the calculated and measured magnetic moments.

As an example of how large the deviations can be, consider the compound $\ce{(NH4)2Co(SO4)2.2H2O}$. This features octahedral Co(II) which adopts a high-spin configuration $(\mathrm{t_{2g}})^5(\mathrm{e_g})^2$. From the spin-only formula, we would expect $\mu_\mathrm{so} = \sqrt{3(3+2} = 3.87$. However, the unsymmetrical population of the $\mathrm{t_{2g}}$ orbitals leads to the generation of orbital angular momentum, and at $300~\mathrm{K}$ the measured effective magnetic moment is actually $5.1$. (Data from here.)

A more detailed discussion can be found in any of the standard inorganic chemistry textbooks. Figgis and Hitchman's book (mentioned earlier) is more advanced and has a very comprehensive treatment of the subject.


Chemical formula of a compound crystallized in a cubic lattice

I would like to say your answer is right, but the question is wrong.

Remember that a lattice is "infinitely" repeated units in 3 dimensional space so you should be able to expand the unit cell in x, y and z axes, indefinitely.

Now imagine your cubic lattice with an $\ce{Y}$ atom in just 1 (out of 8) corners.

Expanding the unit cell in the x-axis will necessarily duplicate the $\ce{Y}$ atom in the x-axis because all expanded unit cells must be identical in composition. Similarly, expanding the unit cell in the y-axis will duplicate the $\ce{Y}$ atom in the y-axis, and expanding the unit cell in the z-axis will duplicate the $\ce{Y}$ atom in the z-axis.

In the end you will find the having one corner as $\ce{Y}$ atom will necessarily have ALL corners as $\ce{Y}$ atoms. In other words, $\ce{X}$ IS $\ce{Y}$, and the formula is necessarily $\ce{XZ_3}$ or $\ce{YZ_3}$

Unfortunately, I don'k think the designer of the question saw his fatal contradiction because this very question is seen in the (mock) Joint Entrance Examination (JEE) in India.

And $\ce{X_2YZ_24}$ is not in the options even in the wrong question.


How are diazonium salts prepared without self coupling

During diazonium ion preparation, first, sodium nitrite is mixed with hydrochloric acid to produce nitrous acid. The nitrous acid can be protonated under acidic conditions, which undergoes the loss of water and produce the nitrosonium ion ($\ce{O=N+}$; this reaction is usually performed at or around $\pu{0 ^\circ C}$):

An aromatic amine can attack the electrophilic nitrosonium ion to form a nitrosoamine $(\ce{Ar-NH-N=O})$.
The nitrosoamine can undergo a proton transfer (tautomerism) to give corresponding imminol $(\ce{Ar-N=N-OH})$.
Under acidic conditions, this $\ce{N=N-OH}$ group can be protonated and left as a water molecule to give the diazonium salt, which is resonance stabilized $(\ce{(Ar-N#N)+Cl-})$.

The key word here is all of these happen under acidic conditions. Therefore, the aromatic nucleus of aniline derivative is not active because of protonation. Also, during the synthesis, the starting aniline derivative is used as the limiting reagent. In addition, many of diazonium salts are unstable and hence, they are always prepared as needed under acidic conditions with good stirring, kept at or near $\pu{0 ^\circ C}$ to minimize the side reactions such as the reaction with water to produce a corresponding phenol until used in the coupling reaction (usually immediately).
The diazonium salt reacts as an electrophile with an electron-rich coupling component (a phenol or an aniline derivative), through an electrophilic aromatic substitution mechanism under basic conditions (specially when coupling reagent is a phenol derivative). To complete the synthesis of some azo dyes, it is also required heating.


Estimating magnitude of percentage and absolute error

When you take a small amount of a liquid in mL, you you usually use a pipette. On every pipette, at the top of it, the absolute error is written. As for the relative error, it is the ratio of the absolute error to the measured value. The percent error is the relative error multiplied by 100.


How can I make a fairly saturated copper (II) sulfate solution by heating?

I suggest to use a combination of heating, dissolving, shock cooling and filtrating off the precipitate.

As was already stated, copper(II) sulphate dissolves slowly and is better soluble in the heat. The exact temperature doesn’t matter, but why not just go up to $80$ or $100~\mathrm{^\circ C}$? This will also make the dissolution faster. You can look up the solubility of copper(II) sulphate to calculate how much you would need and use some more to be on the safe side.

Then, cool the solution down rather quickly (but not too quickly or the glass will crack). This will initially create a supersaturated solution but the supersaturated bit will crystallise out rather quickly. You can filter off this crystalline precipitate to get your saturated copper(II) sulphate solution. Do not filtrate too quickly, or the solution will still be too warm and slightly supersaturated.

Then finally remember that if you add too much ammonia, rather than a precipitate of $\ce{Cu(OH)2}$ you might get the redissolved $\ce{[Cu(NH3)4(H2O)2]^2+}$ tetraammindiaquacopper(II) complex. You can tell this by the intensive dark blue colour.


Something is not correct with your MO diagram.  Here is a correct MO diagram for NO

Note that in this diagram the oxygen atomic orbitals are lower in energy than the nitrogen AOs.  This makes sense since the AOs of the more electronegative element should always be at lower energy than the AOs of the more electropositive element (in your diagram the relative energies of the oxygen and nitrogen AOs are somehow reversed).  The energetic ordering of the starting AOs is important as we'll see in a minute.

The closer in energy a MO is to an AO from which it was formed, the more character of that AO it shows.  So let's examine the nature of the single electron in the $\ce{\pi_{2p}^{\ast}}$ orbital since this electron will be involved in forming the new bond when $\ce{NO}$ dimerizes.  This MO is formed from both the nitrogen and oxygen p atomic orbitals, but it is closer in energy to the nitrogen p AO.  Therefore, we would expect this electron to have more nitrogen and less oxygen character (in other words it will reside more around nitrogen than oxygen).  Because this electron has greater density on nitrogen, dimerization to form an N-N bond is more likely than dimerization to form an O-O bond.


Will adding up protons and electrons (without neutrons) create a new element?

Yes and no.

Elements are defined by the number of protons only. It does not matter if (say) a carbon nucleus has six or seven (or eight) neutrons, they will all react the same.*

With that, to create new elements, you would need to get up to some 115 or so protons fused together. But there is a reason for neutrons: all the positively charged protons in the nucleus repel each other electrostatically so neutrons are there to stabilise the nucleus — you can think of it as cushioning protons apart. For each element there is a set number of neutrons that will create a stable nucleus. E.g. for hydrogen (one proton) it is zero or one, for oxygen it is eight, for carbon six or seven, for tin it can be $62, 64, 65, 66, 67, 68, 69, 70, 72$ or $74$.

Nuclei with other neutron numbers decay radioactively on half-life scales from femtoseconds to millions of years. I would expect no-neutron nuclei (non-hydrogen) to be on the very short edge of that spectrum (although I’m no radiochemist).

*: There are reactions in certain systems that depend a lot on hydrogen being $\ce{^1H}$ rather than $\ce{^2H}$ or $\ce{D}$ because they rely on the tunneling effect whose probability decreases with mass (I think by the factor $m^2$). That’s also why one shouldn’t drink $\ce{D2O}$


How to separate sodium carbonate from sodium hydroxide?

There is a forum which discusses the similar problem of separating sodium hydroxide and sodium carbonate from drain cleaner. The solution is taken from there(slightly edited):

  This is relatively easy because sodium carbonate is practically
  insoluble in 50% sodium hydroxide solution so a glass frit type filter
  is needed to remover the insoluble sodium carbonate. Most commercial
  sodium hydroxide contains only a little sodium carbonate from
  absorption of atmospheric $\ce{CO2}$. So for carbonate free sodium
  hydroxide solution, make up one litre at a time of say 60% solution.
  To do this, add about 300g of sodium hydroxide pellets of flakes to
  500ml of ice-cold water in a large beaker and stirr until practically
  dissolved, cover and leave to cool . Allow it to cool to about 20-25 C
  and the add another 300g of sodium hydroxide and stirr until most has
  dissolved (much less heat is generated on the second addition), dilute
  to one litre, pour into a 1 L thick-walled polythene bottle and leave
  it to stand for about 1 month in a cool place and then decant the
  clear sodium hydroxide solution.

  If the material is say 50% sodium carbonate do the following:

  Add 500g of mixed sodium salts to 500ml of ice-cold water, stirr the
  slurry periodically for a couple of hours and then filter through a
  glass frit funnel (a vacuum filtration set-up will be needed) and wash
  the cake with a little water, say 50-100ml, to remove most of the
  remaining caustic soda; some sodium carbonate will dissolve but this
  will precipitate on the next addition. The cake will be almost pure
  sodium carbonate but you can always treat it with saturated sodium
  bicarbonate solution to ensure that it is.

  The clear filtrate should be placed in a calibrated vessel and another
  500g of material added and again stirred periodically for a couple of
  hours then filter and wash as before. Hopefully there should be about
  700-800ml of very strong sodium hydroxide solution. Place it in a
  polythene bottle and let it stand of a month so the sodium carbonate
  can crystallize out, decant the clear solution and dilute as required.

  NOTE:

  60% caustic soda solution is rather viscous and will not
  filter without suction even on a fairly coarse frit, 50% is still
  pretty viscous and difficult to filter.
  The sodium carbonate that
  crystallises out sinks very slowly hence the long standing time.
  Sodium carbonate dissolves rather slowly in water form some reason so
  with vacuum filtation the residence time of the wash water will
  actually dissolve very little sodium carbonate, too much wash water
  will dilute the solution below 50% and make the carbonate more
  soluble.
  Hot 50% caustic soda solution is extremely aggresive
  towards skin! When handling strong sodium hydroxide solutions, keep a
  bowl full of dilute citric around in case of emergency.

  There are other method such as treating a more dilute solution with
  slaked lime, filtering (difficult because the precipitate is fine
  grained) and evaporating down.


How to Predict the Formation of Racemic Mixtures or Single Products from a Diels Alder Reaction?

It is very simple: if you do not have any enantiopure compound present at the beginning of the reaction, the end product will either be inherently symmetric (contain a plane of centre of symmetry) or racemic. This will be because there are two symmetry-equivalent paths of attacks that will lead to the two enantiomeric products (if enantiomers are even generated) and neither of the two paths is inherently more probable. It is a great exercise to try getting both enantiomers for a stereouncontrolled reaction!

If you do have any enantiomeric excess in your reactants, your products will also have some enantiomeric excess. If the chiral centre is far from the reactive site, the ee will be low; if it is very close or part of the reactive site the ee will typically be much higher. This is because the transition states will now be diastereomeric rather than enantiomeric, meaning they differ in energy and one will be more likely.

While it is still a great exercise to draw the two different transition states leading to different products it can be much harder to distinguish the more likely major product; especially if it is a chiral catalyst giving enantioselectivity.


Non-bonding molecular orbital in triatomic hydrogen

In order to draw the MO diagram for the H3+ system, it helps to first thing about the MO diagram for a simple diatomic case, and then mix in the additional s-orbital.

The diagram below hopefully illustrates this technique (and can be found in many introductory textbooks).

First consider two s-orbitals (the right hand side of the diagram). They can either mix in a symmetric (in phase) or anti-symmetric (out of phase) combination.

The two atomic orbitals we initially combine can give rise to two molecular orbitals, as shown on the right hand side.

We can give them symmetry labels to help us identify what kinds of things these molecular orbitals can then combine with to form the new molecular orbitals for the triatomic system.

The labels g and u are related to whether the molecular orbital is symmetric or anti-symmetric with respect to an inversion centre.

Once our two molecular orbitals for the diatomic system are formed, we can mix these orbitals with another hydrogen s-orbital in order to generate the final MO diagram as per your question.

Only the g molecular orbital from the diatomic system is of the correct symmetry to interact. This mixing creates two molecular orbitals (the top and bottom ones) for the triatomic system. The middle, non-bonding molecular orbital is simply the orbital of u symmetry from the initial diatomic system, which is unable to interact since there is nothing of the correct symmetry for it to do so.


Bromination of ethyl substituted benzaldehydes

Your first query is genuine and the 2nd product for the first case is wrong. It should be, as you noted as well, 5-bromo-2-ethyl-benzaldehyde.

Regarding your second query, you're wrong. Recall the general rule that:

when more than one activators are present on the same benzene ring, the substitution is governed by the best activator alone
when an activator and a deactivator is present on the same benzene ring, the substitution is governed by the activator alone
when more than one deactivators are present on the same benzene ring, the substitution is governed by the weakest deactivator alone

So, here $\ce{Br}$ will go to the o/p position of ethyl group.

Note that in the high school context, I've assumed we're only talking about the major product, because otherwise, all types of crazy products are generally formed, but in smaller ratios


How does conductivity work for non-redoxed ions?

This can arise due to ion-ion interactions brought on by Coulombic fields.

A proper explanation is rooted in transport phenomena. Unfortunately chemists are largely not taught this, but chemical engineers do get the opportunity.

Under standard conditions, ions in a solution will take a random walk during diffusion. This results in no net movement of our ions over some span of time. Such a system perturbed by an applied electric field will cause ions to preferentially move in one direction, we say the ions will drift. We say they have a drift velocity. These velocities are affected by a number of parameters, as there is still continual collision, but we can say that the mobilities will be unequal for ions of unequal sizes.

Different mobilities mean concentration gradients form and charge separation too. Typically we use an electroneutrality field as a reasonable approximation to allow analytic solutions for various equations. This field depends on all the ionic fluxes present and as such will affect the total measurable current.

The result is simple: the introduction of any ion will cause a perturbation in the field that all the ions encounter in solution.


Why are imines more susceptible to reduction than carbonyl compounds?

Generally speaking, I think it is the reverse - carbonyls are reduced faster than imines with various sources of $\ce{H^-}$.  The reason usually cited (see pages 2-4 of this reference) for this behavior is that the imine double bond is less polarized than the carbonyl double bond.  In the figure below resonance structure II contributes less when X=N as compared to the case X=O, due to the decreased electronegativity of nitrogen compared to oxygen.

However, if the reaction is run under mildly acidic conditions, since the imine nitrogen is more basic than the carbonyl oxygen, more of the imine nitrogen will be protonated than the carbonyl oxygen.  This will make the protonated imine carbon much more electrophilic (positively polarized).  Consequently, under acidic conditions, an imine may be more reactive than a carbonyl to nucleophilic reduction (in addition to the above reference, also see this reference).


Van't Hoff factor for citric and phosphoric acids

The Van't Hoff factor isn't an intrinsic property of any species - it depends on the dissociation degree of an electrolyte. Both of those acids are weak and triprotic, meaning they can lose up to three protons in solution (depending on the pH). Therefore, the Van't Hoff factor will likewise change depending on the pH.

1) If the acid is the only species in solution, and since it's a weak acid, you need to use the same formula as for any weak electrolyte:

$i=1+\alpha$

You can figure out the dissociation factor, $\alpha$, from the dissociation constant:

$K_1=\cfrac{[H_2A^-][H^+]}{[H_3A]}=\cfrac{(\alpha C_{H_3A})^2}{(1-\alpha)C_{H_3A}}=\cfrac{\alpha^2}{1-\alpha}C_{H_3A}$

where $C_{H_3A}$ is the formal concentration of the acid.

2) If you have more species in solution, it will depend on the pH of the solution. In general, you'll have this expression:

$i=\cfrac{[H_3A]+[H_2A^-]+[HA^{2-}]+[A^{3-}]+[H^+]}{C_{H_3A}}$

After some tedious substitutions you'll get

$i=1+\cfrac{K_1}{[H^+]+K_1}+\cfrac{K_2}{[H^+]+K_2}+\cfrac{K_3}{[H^+]+K_3}$

Some relevant values:

$pH=0 \rightarrow i \approx 1$

$pH=pK_1 \rightarrow i = 1.5$

$pH=pK_2 \rightarrow i = 2.5$

$pH=pK_3 \rightarrow i = 3.5$

$pH=14 \rightarrow i \approx 4$

and it changes logarithmically in between.


Heisenberg's Uncertainty Principle's inequality

As andynitrox said in the comments :

  The HUP just gives a lower bound in the uncertainty of Δx and Δpx, it is of course possible to measure both quantities with a higher uncertainty than given by the HUP, but not less.

The inequality exists to show that, the uncertainty is atleast this much and in real-life cases, its much larger than this . (source)(provided by Mithoron)


Does hydrogen really have seven isotopes?

Yes, there are seven known isotopes of hydrogen, though only two ($\ce{^1H}$ and $\ce{^2H}$) are stable with respect to nuclear decay, and only three ($\ce{^1H}$, $\ce{^2H}$ and $\ce{^3H}$) exist/can be made in enough quantities to be relevant outside of nuclear physics. All other hydrogen isotopes have extremely small half-lives. The next most stable hydrogen isotope after $\ce{^3H}$ ($t_{1/2}=12.3\ yr$) seems to be $\ce{^5H}$ ($t_{1/2} \sim 10^{-21}\ s$). Given that even the fastest chemical reactions happen in timespans of the order of $10^{-15}\ s$, we can safely discard any influence of hydrogen isotopes heavier than $\ce{^3H}$ in chemistry.

Out of the currently 115 elements which have been confirmed to exist, there are approximately 254 isotopes which are considered to be indefinitely stable, or have half-lives so enormous (over trillions of times greater than the age of the Universe, 13.8 billion years) that no value could yet be determined. Theoretically, only 90 of those 254 isotopes are expected to be truly indefinitely stable (proton decay aside), but there are great difficulties in setting up experiments capable of measuring such rare decay events.

On top of the previous 254, another 34 isotopes have half-lives which have actually been measured in laboratory, but are still large enough (>50 million years) so that a small fraction of the isotope could conceivably have survived since the creation of the Earth 4.6 billion years ago, and may be of interest in geological timescales. There are also isotopes with smaller half-lives, but which are continuously replenished by the decay of heavier atoms or by the action of cosmic rays, so they can still be found on Earth today. In all, about 339 different isotopes can be found naturally on our planet.

However, as many as 3100-3300 different isotopes of the first 118 elements are claimed to have been detected in laboratories, most of them with very small half-lives (seconds or less). Nobody knows for sure how many elements and how many isotopes can exist, though the number is finite.


Why is tert-Butyl (tBu) often not listed as protecting group for alcohol?

Protecting group selection is a balancing act of ease of preparation and deprotection. In common practice, use of a protecting group adds two steps to a synthetic route. Installation and removing a tert-butyl group requires stronger (acidic) conditions. Of course, it's a rock while in place and inert to nearly all conditions short of strong acid.

Peptide synthesis is particularly amenable to use of tert-butyl protection of hydroxyl groups because (1) all of the suitably protected residues are commercially available and (2) the protecting groups have been designed to be globally deprotected with trifluoroacetic acid. These factors minimize the inherent inefficiency to protecting groups.

The workhorse protecting groups for alcohols are silyl ethers (e.g.trimethylsilyl, triethylsilyl, tertbutyldimethylsilyl, triisopropylsilyl, tertbutyldiphenylsilyl). The silyl ethers are easier to form and cleave than standard ethers and have excellent compatibility with reactions that require alcohol protection.

Specialized (carbon) ethers can be useful as well, including unsubstituted or substituted benzyl ether (removed by oxidative or reductive conditions) and acetals (removed by mild acid treatment).

For a more complete description of protecting groups (including tert-butyl), see the books by Greene and Wuts or Kocienski.


"Cathode rays" are simply beams of electrons. The name exists because cathode ray devices predate the discovery of the electron. The sharp shadows produced by Crookes tubes meant that something was travelling in a straight line from the cathode down the tube, so they were named cathode rays.

The vacuum doesn't have much to do with the production of electrons, per se, but without high vacuum, the electrons collide with gas particles before they travel very far. If there's enough gas, you get a plasma rather than a beam of electrons. The electron beam is produced by thermionic emission in hot cathodes—electrons are liberated from a hot filament when the thermal energy is enough for electrons to exceed the filament's work function. In cold cathodes, electrons are released by field emission, where a large applied electric field allows electrons to tunnel away from a metal electrode.

The glow of the glass envelope is luminescence—either fluorescence or phosphorescence (or both). Basically, the electrons striking the glass excite components of the glass to an elevated electronic state which releases light upon returning to the ground state. Normal glass isn't particularly luminescent so tubes designed to produce light (like a CRT) are coated with phosphors.


What will be the colour of HCl with methyl orange?

Methyl orange (MO), as you know, is an acid-base indicator.

The problem with those is that they do not distinguish between acidic and basic medium, but rather between different pHs. MO has a transition interval of 3.1-4.4, which means it's red below 3.1 and yellow over 4.4. Within this interval it's all about transition colors (red through orange to yellow).

Answering your question: a few drops of HCl would affect water differently. Let's say, you had 1 L of water and you added 3 drops (0.15 mL) of concentrated saturated HCl (~40% w/w). That leads us to  ~0.002 mol of HCl added and to pH lower than 3, so the MO would turn red. If you had diluted solutions, like 0.01M or so, the calculations would tell you the pH would still be higher than 4.4, so the MO would still be yellow.


We could. Mathematically there is nothing wrong with that - the overlap matrix is positive definite and therefore must have an inverse, and so your manipulation is perfectly correct mathematically. The problem is that $\mathbf{S}^{-1} \mathbf{F}$ is not a symmetric (or in the more general case Hermitian) matrix. Computationally it is MUCH more difficult to find accurate eigenvalues and eigenvectors of a general non-symmetric matrix than a symmetric one. Thus, when solving the problem on a computer, if one can cast one's algorithm in terms of diagonalisation of a symmetric rather than a non-symmetric matrix you are likely to get more correct answers, quicker. And this is what the standard manipulations achieve.


How can two orbitals constructively and destructively interfere simultaneously?

They do not simultaneously interfere constructively and destructively.

Let me explain. Since the time-independent Schrödinger equation is not separable for molecules due to inter-electronic interaction terms, we can seek solutions to this problem with the variational method.

Let's examine the case of the $\mathrm{H_2}$ molecule. The most natural/physically meaningful/simple choice is to use a trial function that consists of a linear combination of hydrogen 1s atomic orbitals:
$$E\left(c_{1},c_{2}\right)=\dfrac{\left\langle c_{1}\psi_{1}+c_{2}\psi_{2}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|c_{1}\psi_{1}+c_{2}\psi_{2}\right\rangle }{\left\langle c_{1}\psi_{1}+c_{2}\psi_{2}|c_{1}\psi_{1}+c_{2}\psi_{2}\right\rangle }=\dfrac{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}\left\langle \psi_{i}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\psi_{j}\right\rangle }{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}\left\langle \psi_{i}|\psi_{j}\right\rangle }$$

$$H_{ij}\equiv\left\langle \psi_{i}\left|\hat{\mathcal{H}}_{\mathrm{e}}\right|\psi_{j}\right\rangle$$

$$S_{ij}\equiv\left\langle \psi_{i}|\psi_{j}\right\rangle$$

$$E\left(c_{1},c_{2}\right)=\dfrac{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}H_{ij}}{{\displaystyle \sum_{i=1}^{2}}{\displaystyle \sum_{j=1}^{2}}c_{i}^{\star}c_{j}S_{ij}}$$

After minimizing the energy with respect to $c_1$ and $c_2$, you will end up with two energy levels, each one associated with a wave function:
$$E_{+}=\dfrac{H_{11}-H_{21}}{\left(1-S_{12}\right)}\qquad\psi_{+}=\dfrac{1}{\sqrt{2}}\left(\psi_{1}+\psi_{2}\right)$$

$$E_{-}=\dfrac{H_{11}+H_{21}}{\left(1+S_{12}\right)}\qquad\psi_{-}=\dfrac{1}{\sqrt{2}}\left(\psi_{1}-\psi_{2}\right)$$

In other words, by choosing to model the $\mathrm{H_2}$ molecule with two hydrogen $1s$ atomic orbitals, you find two discrete energy levels, each one associated with its wave function (one anti-bonding and one bonding). So this answers your questions:

The two atomic orbitals do not simultaneously interfere: The MOs are two different wave functions.
Of course the anti-bonding orbital exists when not occupied.

Note that the previous procedure can be generalized for all kind of atomic orbitals.

I want to clarify something about the meaning of the wave function since you're asking: As a result, is it correct to think that it's the orbitals themselves that are the waves, not the electrons?

Let's assume that your system consists of a single particle in a one-dimensional space. The wave function, $\Psi\left(x,t\right)$,
is a complex function that contains all the information about the system. But what is really physical meaningful is its modulus squared, $\left|\Psi\left(x,t\right)\right|^{2}$, which represents the position probability density. What is the position probability density? It's something that multiplied by a certain length (or volume if you're working in three dimensions), let's say dx, gives you the probability for finding the particle between $x$ and $x+dx$
$$P\left(x,x+dx\right)=\left|\Psi\left(x,t\right)\right|^{2}dx$$.
Of course, you can integrate the previous equation to find the probability over a certain region of space, $\Omega$
$$P_{\Omega}={\displaystyle \int_{\Omega}}\left|\Psi\left(x,t\right)\right|^{2}dx$$.


Difference between 'homogeneous mixture of several chemical compounds' & 'non-homogeneous system'

You are right that normally a non-homogeneous system would be one where there is boundary between two internally homogeneous phases (like milk which is basically oil and water mixed at a very fine scale).

But Fermi as also allowing for the possibility that there may be only one phase where the concentration of its components varies continuously in space (if there is more than one phase the variation is discontinuous by definition).

It certainly isn't a common situation if the phase in in equilibrium where we would expect things like diffusion to create a uniform concentration of things over time, at least in liquids and gases). But it is very common in the real world where equilibrium is rarely reached (imagine the state of a poorly stirred cup of tea where the concentrations of tea-stuff will be larger near the tea bag than far away from the tea bag).

It is also very common in places where it is hard to get to equilibrium which is often true in solids (I don't know whether this is what Fermi was thinking of). Pyrex glass, for example, is apparently a single phase but the structure is deliberately made different near the surface than far away from the surface to create tension in the surface layers that protect the glass from thermal shock. But the properties vary continuously.

The examples I can think of may not be relevant to what Fermi was saying as they mostly are metastable and won't be at equilibrium which is often what thermodynamics restricts itself to worrying about.


How does Polytetrafluoroethylene (Teflon) differ from Polyvinylidene fluoride (PVDF) as a protein binding membrane material?

Essentially you are asking why about the different surface properties of PTFE vs. PVDF.  Generally PTFE is more 'non stick' than PVDF, which is more non-stick than polyethene (PE).  PTFE is much more chemically resistant to corrosion than PVDF or PE.  You are correct to look at the chemical structure.  The more Fluorine atoms on the backbone, the more proteins will not be able to chemically latch onto the surface.  Measuring things like contact angle will confirm this.  I don't know how much the pore structure of the membranes affect things, but from a chemical standpoint, there are clear differences.


A few thoughts:

For starters, $\ce{Zn^{2+}}$ is also part of the reaction quotient, so the log term in the Nernst equation will not necessarily be zero if the hexacyanoferrate concentrations are equal, depending on the $\ce{Zn^{2+}}$ concentration:

$$E = E°+\frac{RT}{zF}\ln\frac{[\ce{Zn^2+}][\ce{Fe^2+}]^2}{[\ce{Fe^3+}]^2}$$
Depending on whether the reaction quotient is greater than or less than 1, the potential can have either a positive or negative temperature dependance.

Second, the Nernst equation uses activities not concentrations so depending on the ionic strength and other factors, the activities of the ions in solution may not be what you think. This deviation is greater for multiply charged ions and may be significant for concentrations over 1 mM.

Finally, since the hexacyanoferrate species are both soluble, I'm assuming you're using some kind of reference electrode for the second electrode. This is also temperature dependant in the same way as the rest of the cell.


Internal (Z-Matrix) to Cartesian coordinates conversion

Sure, it's just trigonometry. There are a variety of open source implementations.

For example:

InternalToCartesian from Open Babel
ChemKit
QCL Python

There are dozens in a search on GitHub if you don't like C++ or Python.

There's also a web app from Shodor for z-matrix conversion.


Is Changing between Allotropic forms a Physical or Chemical Process?

Well, it's a matter of choice and tradition. We traditionally classify phase transitions (such as melting of ice or evaporation of water) as physical changes. The conversion of ferrite into austenite is a phase transition, so it should formally be classified as a physical change. But yes, it's hair splitting indeed, especially when you consider the phase change from diamond (giant covalent lattice, sigma-bonds only) to fullerene (simple molecular lattice, sigma plus delocalised pi-bonds), which clearly shows some characteristics of a chemical transformation (the change in the bonding nature). Still, even such changes are formally "physical". The traditions do not go easily!


There are quite some molecules that are persistent (= live long enough in solution to allow for a spectroscopical characterization, but cannot be isolated) or even stable (= can be isolated, bottled and/or used as a starting material) that have an odd number n of electrons, however $n \ne 13$.

Among these radicals are the Gomberg radical (1), the Koelsch radical (2), and the TEMPO radical (3).


What is the effect of conjugation of a carbonyl group in IR spectroscopy?

It would shift the IR peak to lower wavenumbers. You can rationalise this by understanding the bonding in the carbonyl group; in essence, by conjugating the carbonyl, the $\ce{C=O}$ bond is weakened, which shifts the IR peak down. You can think of it in terms of, if there is little or no conjugation, the electrons are firmly based in the $\ce{C=O}$ bond. As soon as we introduce some resonance structure, or conjugation, now those electrons can delocalise over 3-4 atoms or even more. This weakens the $\ce{C=O}$ bond, moving the IR peak down to lower wavenumbers.


Can't nobelium form compounds like other lanthanides/actinides do?

Wikipedia reports that nobelium does form compounds, but unlike other later actinides it could adopt a +2 oxidation state as well as +3, which complicates identifying specific compounds (e.g. nobelium reacts with chlorine but the specific product is not known).  The +2 state is stabilized by the full shell configuration $[\ce{Rn}]5f^{14}$.  Experiments in aqueous solution seem to show a preference for the +2 oxidation state, and $\ce{No}^{2+}$ has been coprecipitated into solids.  But we have yet to definitively identify a neat nobelium compound in either specific oxidation state.


Why do we not include water in equilibrium expression calculations?

Usually when you have water in a reaction it is taken in excess, i.e in a much larger quantity compared to the other reactants. Ultimately the change in concentration of water is negligible. As an example you can consider the hydrolysis of ethyl acetate to ethanol and ethanoic acid. For better understanding consider the initial amount of ethyl acetate to be 0.01mol and that of water as 10 mol.
This however is valid only when water is taken in excess. If the amount of water and the other reactants is comparable, then water has to be included in the equilibrium expression.


How would I go about making an ethyl ester from vegetable oil?

The biodiesel process should work well, you will need just slightly more aggressive conditions (i.e. slightly more catalyst, slightly higher EtOH excess, and/or slightly longer time) to make ethyl esters instead of methyl esters, as ethanol is slightly less reactive than methanol.

For skin care, you will need higher purity than biodiesel processes usually provides.  I would imagine that the cosmetics industry uses vacuum distillation to purify their esters.  This would separate the ethyl esters from unreacted triacylglycerols, any base (or acid) catalyst left over from the reaction and soluble in the organic ester phase, organic-soluble glycerol, and organic-soluble impurities.


What does this three-number isotope notation mean?

The superscript '$3+$' denotes a normalised electric charge. Charges are quantised, that is they take on only certain discrete values.

Neutrons do not have a net charge ($q = 0 $).
Electrons and protons have the same absolute charge[footnote]. They have opposite, arbitrary signs. By historical convention, protons are assigned positive charges.

Hence,

$$\text{electric charge} = \text{no. of protons} - \text{no. of electrons.}$$

For shorthand purposes, we usually normalise the electric charge by dividing the total electric charge of our agreed system with the elementary charge $e$.[footnote] This provides us an opportunity to operate with integer values in chemistry; there are fractional charges for quarks which were discovered later but these particles cannot be observed separately by current understanding.

Note also that we write '$3+$' instead of '$+3$' since we are, again by convention, counting charges (that is $+ + + $), not denoting the mathematical normalised charge itself.

As of 2011, this is verified up to essentially $1$ part in $10^{21}$. The absolute charge of an electron (or charge of a proton) is called the elementary charge $e$, and

$$e=1.6021766208(98) \cdot 10^{-19}\ \mathrm{C}$$

where the parentheses designate a standard uncertainty (assuming a normal distribution, $\approx 0.68$ probability, coverage factor $k =1$).

ref: Mohr, P. J.; Newell, D. B.; Taylor, B. N. 'CODATA Recommended Values of the Fundamental Physical Constants: 2014'. Reviews of Modern Physics, 2016, 88 (3). DOI: 10.1103/RevModPhys.88.035009, freely available on arXiv here, also archived link.

  Disclaimer: originally, I claimed that proton–electron charge symmetry is verified up to $1$ part in $10^{26}$. This was wrong, and based on a naive approach. Again, see How accurately is it known that protons have the same charge as electrons? for a correction.


Your premise Wikipedia's statement about transition metals seems to be flawed.
(1) It isn't possible to have a transition metal that has the electronic structure $\mathrm{(n-1)d^{10} ns^2np^6}$. Such an electronic structure would imply a noble gas.
(2) Gold $\mathrm{[Xe]4f^{14} 5d^{10} 6s^1}$ and platinum $\mathrm{[Xe]4f^{14} 5d^{9} 6s^1}$ are the most noble metals and neither even has a $\mathrm{(n-1)d^{10} ns^2}$ configuration.

Granted these configurations are for the free atom, not the bulk metal...


What is the mechanism of alkane combustion?

The following, very general radical reactions apply to any common combustion reaction of organic material. Details vary largely.

Step 1

An energy spark of whatever kind turns triplet oxygen into its high energy singlet state$^*$, which typically strips a hydrogen atom off of some organic molecule
$\ce{R-CH3 + O2^* -> R-CH2. + HOO.}$
Now you have two radicals.

Step 2

The perhydroxy radical strips another hydrogen off somewhere,
$\ce{R-CH3 + .OOH -> R-CH2. + HOOH}$
while the alkyl radicals react with more oxygen (self-catalysed)
$\ce{R-CH2. + O2 -> R-CH2-OO.}$
$\ce{R-CH3 + .OO-CH2-R -> R-CH2. + HOO-CH2-R}$
, this incorporates $\ce{-OOH}$ hydroperoxide motives into more and more molecules. This is in itself a linear process, with a constant number of radicals. Very similar to free-radical polymerisation or organic radical-substitution reactions, and very much the same as formation of the infamous ether hydroperoxides.

Step 3

These peroxides are not stable at elevated temperature, and their O-O bond tends to split homolytically
$\ce{R-OO-H -> R-O. + .OH}$
Now the number of radicals grows rapidly, the combustion is gaining speed.

Step 4

As temperatures rise (see step 5!), esp. larger organic molecules start to break up also by themselves
$\ce{R-CH2-CH2-R -> R-CH2. + .H2C-R}$
(That's also a possible initiation step.) You get more radicals, which can react with more oxygen to form peroxides. Now the chain reaction snowballs!

Step 5

$\ce{HOOH -> HO. + .OH}$ gives more hydroxy radicals (also from step 3), which turn into water
$\ce{HO. + H3C-R -> H2O + .H2C-R}$
leaving another radical behind. Producing water is highly exothermic, this fuels the reactions creating radicals (in steps 3+4+5). Making radicals costs a lot of energy.
Steps 2-5 (it's really not a sequence) run in parallel and repeat as they produce more radicals and peroxides all the time. The reaction is usually limited by the inflow of fresh oxygen.

Step 6

Termination: Two carbon radicals can recombine, but at high temperature this does not really happen, because the energy from the recombination has to go somewhere. When temperatures are too low, this makes coke and smoke.
This picture is by no means complete. $\ce{CO2}$ would likely occur via decarboxylisation of $\ce{R-COO.}$ radicals, etc.  Organic radicals at high temperatures have more internal reactions where they split off smaller molecules, e.g. $\ce{R-CH2-CH2. -> R. + H2C=CH2}$. (The latter can be seen as depolymerisation of polyethylene.) A radical in the middle of a chain splits it in half ($\ce{R-CH2-CH2.-R -> R. + H2C=CHR}$, increasing the volatility.
And remember that this picture is terribly simplified. I'm by no means an expert on this subject, and the experts also have not agreed on all details yet, especially since it varies a lot.


In which state does the bifluoride anion exist?

Wikipedia page for for bifluoride anion. The question's answer appears to be true, with explanation and references.

The fluorohydrogenate group $\ce{-HF-}$ in anions such as bifluoride can assimilate a proton by recombination:

$$\ce{HF2- + H+ -> H2F2 -> 2 HF}$$

Because of this capture of a proton ($\ce{H+}$), bifluoride has basic character. Its conjugate acid is the reactive intermediate, μ-fluoro-fluorodihydrogen ($\ce{H2F2}$), which subsequently dissociates to become hydrogen fluoride. In solution, most bifluoride ions are dissociated.

$\ce{HF2- + 3H2O}$ is in a favored equilibrium with $\ce{2H3OF + HO-}$

This makes for an interesting oxygen coordination sphere re the other lone pair of electrons.  One presumes it is hydronium fluoride in which the fluoride is bonded to a proton with a delocalized charge.

I've worked with aqueous ammonium hydrogen bifluoride as a silicate etchant. The solution has tremendous surface tension.  A drop of Fluorad surfactant was necessary to have things work.  $\ce{HF}$ will reach out and eat you.  $\ce{NH4HF2}$ is fairly benign.  No skin contact!  Don't work in Pyrex.


How can we measure the atomic radius of an atom?

As you told it is not possible to measure the inter nuclear distance of single atom. But here internuclear distance does not mean diameter of a single atom but it means distance between nucleus of two atom of same element.

This internuclear distance can be determined by two methods:

X-rays method
Spectroscopy method

Note: Atomic radius is not a set value for a given atom. Eg. $\ce{Fe^{+2}}$ ion have different atomic radius than  $\ce{Fe+}$. Moreover,  $\ce{Fe^{+2}}$<  $\ce{Fe+}$


Can anions be reduced at the cathode in an electrochemical cell?

Theoretically, it is possible for anions to get reduced more. According to wikipedia's reduction potential table page, this reaction occurs at +0.08V vs SHE:

$\ce{S4O6^2− + 2 e− ⇌ 2 S2O3^2−}$

In your chloride example, though, you're trying to reduce $\ce{Cl^-}$ to (at least) $\ce{Cl^2-}$ and the potential for that reaction would be extremely negative. In most real-world scenarios, you would start reducing other things way before you hit that potential. Most likely you'd start reducing water at -0.8V.

$\ce{2H2O + 2 e− ⇌ H2(g) + 2 OH−}$

As an aside, be careful with your terminology. Reduction always happens at the cathode, that's what "cathode" means. Sometimes we get sloppy and don't switch the terms when we go from galvanic to electrolytic mode, but that doesn't change the chemistry. If electrons are flowing into an electrode, it's the cathode. I find the (+)/(-) labels on terminals more confusing than anything else.

Also, you said that

  for a galvanic cell, as electrons flow from the anode to the cathode, the cathode (+) becomes negatively charged due to the accumulation of electrons

This is backwards. The cathode does not become negatively charged because electrons have flowed into it. Rather, reduction occurs spontaneously at the cathode, leaving it electron-deficient. Electrons then flow from the anode to the cathode to compensate this charge imbalance and make both sides neutral again.


Application of the Woodward-Hoffman rules to a [14+2] cycloaddition (follow up)

For reactions in the excited state, a modified statement of the Woodward Hoffman rule applies:

  A pericyclic change in the first excited state is symmetry allowed when the total number of (4q+2)s and (4r)a components is even.

The only difference is that we're now looking for the total to be even rather than odd.


Nitrogen is in the second row with no $d$ orbital in the valence shell.  It obeys the octet rule and cannot have more than 8 electron.

There are exceptions to the octet rule.  Having less than 8 electron is less preferable but still possible, and is commonly seen in free radicals and cations. On the other hand, having more than 8 electron is extremely unfavorable for second period atoms. Such electronic structures can be found in extremely unstable species or excited states, such as the CH5 radical.

As a comparison, first shell atoms obey the duet rule while atoms in the 3rd shell and beyond may obey 18-electron rule, 12-electron rule or 8-electron rule.   However, 18-electron rule and 12-electron rules are much less strict than 8-electron rule and violations are common place.

According to valence bond theory, the electronic structure of a molecule is a combination of all possible resonance structure that you can write down, including structures with all possible formal charges and strange electron counts.  However, their contributions are not even.  some of them are more favorable than others.

For HNO3, in order to satisfy the octet rule, the nitrogen atom would form 1 double bond and 2 single bonds.  Based on octet rule alone, there are 3 possible resonance structures that are favorable.

However, the first two resonance structures are significantly more favorable than the third, because they have smaller amount of formal charges.  As a result, we usually only write the two dominant structures. The bond between OH and N is close to a normal single bond. The other two N-O bonds have bond order near 1.5.

It is also common to write a mixture of resonance structures as the hybrid form

Note that this representation is not a single Lewis structure, but a convenient way to represent many resonance structures in the same figure. It gives no information about the exact bond order or formal charges on individual atoms.  The dotted lines indicates that in some structures where that bond is a single bond and in others a double bond, and the order is somewhere between 1 and 2.


Low-volatile liquids that 'ball-up' on siliconoxide or perfluoroalkanes

Here it comes... mercury.  75-wt% Ga 25 wti%, mp. 15.5 C and

http://en.wikipedia.org/wiki/Galinstan

in general wet glass.  Maybe ionic liquids,

J. Am. Chem. Soc. 129(13), 3804 (2007), DOI: 10.1021/ja070169d
Aust. J. Chem. 57(2) 113-155 (2004)

http://www.researchgate.net/publication/233906460_Wetting_of_surfaces_with_ionic_liquids
http://dx.doi.org/10.1016/j.jcis.2008.10.019
http://ura.unisa.edu.au/R/?func=dbin-jump-full&object_id=63927

http://www.ppmp.pwr.wroc.pl/sorpa/file/eee7c6b05fb9497a9a59507ca0346479/download
[DMIM][Cl], 1,3-Dimethylimidazolium chloride, is reported 105.9 contact angle on glass.  That might be a typo - it's a huge number for the class of compound.

http://english.cas.cn/ST/BR/br_progress/201106/t20110622_71442.shtml


What is the importance of using a pKa value instead of a pKb value when describing drug chemistry?

For easier comparability:

Comparing two pKa values is much easier than comparing a pKa with a pKb value.


What happens to an acid when it completely dries out?

When an aqueous solution of an acid is dried, water evaporates and the $\ce{H+}$ ions crystallize with the anionic part of the acid and form a crystal.

If the acid used to make the solution was not of crystalline form then after evaporation of water the anion and cation would have combined to form a compound of some other type (may be a solid formed by co-ordinate bonds or a liquid or a gas like $\ce{CO2}$)

Initially, because of presence of $\ce{H2O}$ molecules, the $\ce{H+}$ ions were bonded with them to form $\ce{H3O+}$ molecules and the anionic part of the acid were solvated by these $\ce{H3O+}$ ions. This solvation happened because it liberated more energy than consumed in dissociation of the acid molecule (i.e. was thermodynamicaly favorable)

But, now since the water molecules are absent, the $\ce{H+}$ and $\ce{A-}$ (anions) will combine to form $\ce{AH (s)}$ as its formation will be favorable because there is no more water to compensate the energy required in there dissociation.

If there had been a solution of acid and base then the $\ce{H+}$ ions from the acid and $\ce{OH-}$ ions from the base would have already reacted to form $\ce{H2O}$ and hence the scenario will be the same as above instead we will also notice crystallization of the cation $\ce{(B+)}$ from the base $\ce{(BOH)}$ and the anion ($\ce{A-}$) from the acid ($\ce{AH}$) to form $\ce{BA (s)}$.

If it would had been an aqueous solution of only a base then also the water would have evaporated and crystals of the corresponding base would have formed.

To talk a bit more (though not necessary in the answer): Acidity is a measure of concentration of $\ce{H+}$ ions in water and now since there is no solution present but only the crystals, the crystals won't behave like an acidic solution until and unless solvent is added to it.


What impact does heating or cooling have on the oxidation reduction potential of water?

I see two questions here.  The one that is stated (temperature effects on ORP) and a hidden one about water ionizers.

First, the temperature effect on ORP.  The thermodynamics of oxidation and reduction reactions is governed by the Nernst equation:
$$E(ORP) = E_o - \frac{2.3RT}{nF}\times log\frac{[reductant] }{[oxidant]}$$

In this equation, R and F are the gas law and Faraday's constants, respectively, $E_o$ is the standard reduction potential of the system of interest, T is temperature, n is the number of electrons transferred in the reaction and the concentrations of the reductant and oxidant are found in the log term.  The coefficient $\frac{2.3RT}{nF}$ in front of the log term can be simplified to $\frac{0.00198\times T}{n}$ and when we perform dimensional analysis on the term we find that it will provide units in Volts.  If we are looking for the largest effect that temperature would have on ORP then we want to use the smallest value of n, which would be 1 electron, further simplifying the coefficient to approximately $0.002T$.  T has to be in Kelvin so if we take our two extreme values as the freezing (273 K) and boiling (373 K) points of water, we find that the difference in ORP between these two extremes (all else being the same) would be about 200 millivolts.

Now to the hidden question about water ionizers.  There's a lot of hype about these systems with little scientific evidence to back up the claims.  There is at least one website useful for exploring the science behind water ionizers that is (I think) accessible to the general public.  A number of claims made by the manufacturers of these products, such as ion streams and water clusters simply don't stand up to scientific experimentation, and consumers need to be wary of the health benefit claims.