Calculating the value of equilibrium constant in terms of pressure

The problem lies in that our information of the system is in terms of concentration, so you actually calculated $K_c$.
$$\ce{2H2 +S2<=>2H2S}$$
$$\begin{array}{|l|l|l|}
\ce{H2 & S2 & H2S} \\ \hline
0.992\ \mathrm{M}& 0\ \mathrm{M}& 0.062\ \mathrm{M} \\
2(1.6\times 10^{-5}) & (1.6\times 10^{-5}) & -2(1.6\times 10^{-5}) \\
0.992+2(1.6\times 10^{-5})  &  1.6\times10^{-5}\ \mathrm{M}   &  0.062-2(1.6\times 10^{-5})\end{array}$$
$$K_c=\frac{(0.062\ \mathrm{M}-3.2\times 10^{-5}\ \mathrm{M})^2}{(1.6\times10^{-5}\ \mathrm{M})\cdot(0.992\ \mathrm{M}+3.2\times 10^{-5}\ \mathrm{M})^2}=245\ \mathrm{M^{-1}}$$
which is numerically the same as the value you obtained, except that I used a more precise molecular mass for $\ce{H2}$.
Using the relation $$K_p=K_c(RT)^{\Delta n},$$
where $$\Delta n=(\sum \text{coefficient of gaseous products})-(\sum\text{coefficient of gaseous reactants}),$$ we can calculate $K_p$:
$$K_p=(245\ \mathrm{M^{-1}})\cdot(8.3144598\ \mathrm{J\ K^{-1}\ mol^{-1}}\cdot 1670\ \mathrm{K})^{-1}=0.0176\ \mathrm{kPa^{-1}}$$


Why did the salt on the skin of my turkey react with the aluminum foil it was touching?

If you don't mind, I'll start with a more colloquial and hopfully not to lengthy explanation:

Precious metals, such gold, do not corrode under "household conditions", others, such as iron or aluminium do. Rusting of iron is a typical example. Here, the corroded surface often is very porous and the corrosion carries on until the whole piece of metal is turned to rust (oxidized).

In the case of aluminium, the initial corrosive layer is very tight and serves as a protection for the rest of the aluminium foil. Such a very thin and transparent protection layer often is artificially applied. The item are sold as Eloxal (= electrolytically oxidized aluminium).

Now comes the table salt, which is sodium chloride ($\ce{NaCl}$). Sodium chloride is hygroscopic, it "attracts" the water and that's why you sprinkled it on the turkey. It is also very soluble in water. When it dissolves, sodium $\ce{Na+}$ and chloride $\ce{Cl-}$ ions are formed.

The chloride ions attach and move into the protective oxide layer of the aluminium foil and finally induce the corrosion of the aluminium metal. Aluminium ions are released to the water (= the wet surface of the turkey).

For a more scientific, in-depth explanation of this corrosive processes known as pitting, you might want to have a look at this article or this review.

  Was my turkey poisoned with aluminum at this point?

There seem to be some crazy myths concerning the "toxicity" of aluminium (ions), often spread by opponents of immunization and other quacks.

Fact is that

a healthy human body is capable to excrete aluminium ions though the kidneys
the British Alzheimer Society does not see any convincing relation between the uptake of aluminium and Alzheimer's Disease
the cancer information service of the dkfz (Deutsches Krebsforschungszentrum, German Cancer Research Centre) refutes the aluminium induced breast cancer myth
many antacids do contain aluminium hydroxide as one ingredient. The effect of "poisoning" is not known.

In summary:

Yes, aluminium ions were released to the wet surface of the turkey.
There's no reason to assume that the turkey is "poisoned".


Why does CO2 diffuse through a butyl rubber membrane more readily than air?

The permeability of a gas through rubber depends mainly on its diffusivity and solubility in rubber.

$\ce{CO2}$ has a significantly higher solubility in rubber than $\ce{O2}$ and $\ce{N2}$, whereas the diffusion coefficients differ not that much.

The result is that carbon dioxide passes ordinary rubber about 5 times faster than oxygen and about 15 times faster than nitrogen[1, 2].


Which of these would most likely be a strong electrolyte?

Yes. You are right. Ionic compound are strong electrolytes if they are soluble in water. $\ce{CaF2, BaSO4}$ are ionic compounds, but they are not soluble in water. I would not consider them as strong electrolytes. On the other hand, ionic compounds are a part of the bigger family of the polar compounds. But polar compounds are not always strong electrolytes. $\ce{CO, H2S, HF, H2O}$ are polar, but they are not strong electrolytes.


Meta-directing nature of anilinium cations

This does not answer your question exactly, but I hope this can shed some light on the anomalous behaviour of anilinium cations.
First of all, it is not true that the para-product is always favoured with the $\ce{NH3+}$ ion, it depends on the reaction conditions. There are two effects to consider here—1) kinetic/thermodynamic control, and 2) rate determining step of the reaction. If the reaction is under thermodynamic constrol, the most stable isomer will form regardless of the directing property of the substituent. If it is under kinetic control, then you have to look at the rate determining step of the reaction. The electrophilic aromatic substitution is a two step process in general, and if the first step is rate determining, then the nucleophilicity of each postion of the ring will be the main factor. If the second step is rate determining then the energy of the intermediate will also matter. Obviously the ortho and para intermediates will be unstable as the positive charge is delocalized next to the electron withdrawing group. So, it's not always true that inductive effect  by itself cannot be meta-directing.
Anyway, back to your original question. Let's consider the nitration of aniline in presence of conc. $\ce{H2SO4}$.

Brickman and Ridd[1] has noted the relative ratios of products for different sulfuric acid concentrations.
$$\begin{array} {|r|r|}\hline \text{conc. of }\ce{H2SO4}(\%) & 85 & 92.4 & 96.4 & 100 \\ \hline ortho(\%) & 6 & - & - & - \\ \hline meta(\%) & 34 & 53 & 58 & 64 \\ \hline para(\%) & 59 & 47 & 42 & 36 \\ \hline  \end{array} $$
As you can see, the meta product becomes more favoured as the concentration of $\ce{H2SO4}$ increases.
What about the N-methylated anilines? In another paper, Brickman, Utley and Ridd[2] have mentioned the percentages of meta and para products (ortho does not form at high acid concentrations, probably due to the steric effect).
$$\begin{array} {|r|r|}\hline \text{conc. of }\ce{H2SO4}(\%) & 90.9 & 96.2 & 99.8 \\ \hline ortho(\%) & - & - & -  \\ \hline meta(\%) & 61 & 67.5 & 70.3 \\ \hline para(\%) & 39 & 70.3 & 29.7 \\ \hline  \end{array}$$
$$\text{For N-methylanilinium ion }\ce{Ph-NH2Me+}$$
Here, we again see the same trend that the increasing sulfuric acid concentration increases the meta-product percentage. However, the meta-product is now the major product.
$\ce{Ph-NHMe2+}$ also shows similar results. The N,N,N-triemthylanilinium ion $\ce{Ph-NMe3+}$ shows $89\%$ m-product and $11\%$ p-product at $98\%$ conc. $\ce{H2SO4}$.
In the two papers, the authors mention that the analysis of acid concentration vs. rate plot seems to indicate that the substitution predominantly happens on the anilinium cation i.e. the protonated state, so there is no possibility of the para-product arising from the reaction of the deprotonated compound. Sulfuric acid is a pretty strong acid, so it's unlikely that there is much deprotonated aniline left at such high concentration. The p/m ratio of aniline nitration changes by a factor of less than $2$, when the acid concentration is changed from $90$ to $98\%$, which reduces the proportion of deprotonated form by a factor of $31$. So, it's not possible that the para-product is coming from aniline, because then the rate profile would have been much more sensitive to the concentration of acid.
They have also ruled out significant hyperconjugative effects, by comparing the rates of the benzylammonium variants of all of these anilines. So, the effect must be mostly from inductive effects.
After a lengthy discussion, it is noted that the $\ce{-NH3+}$ group is much more solvated than the $\ce{-NHMe2+}$, $\ce{-NMe3+}$ etc. groups. So, it is possible that the solvent forms hydrogen bonds to the $\ce{-NH3+}$ group and allows the positive charge on it to spread away. This means that the inductive effect from $\ce{-NH3+}$ is actually weaker than the methylated groups. So, the methylated anilinium ions are simply more strongly averse to reaction on ortho/para positions than anilinium.
However, the authors note after that[2]:

The above explanation  accounts for the major differences in the substituent effects recorded in Table 3 but it is not completely satisfactory, for the extent of  para-substitution in the anilinium ion is  more than would  be  expected if the effect of replacing $\ce{-N+Me}$, by $\ce{-N+H}$, were simply to "scale down" the effect of the positive pole;  such an effect would  never  cause the para-position to be more reactive than one meta-position. This anomaly is also illustrated by the comparison of  the anilinium ion with the benzyltrimethyl-ammonium ion: the latter ion is more reactive by a factor of ca. 25 but gives less para- substitution. It appears that the protonated poles give significantly higher p/m ratios than expected  from their reactivity relative to the methylated poles.
The reason for this is not entirely clear ...

So, it's not completely known why aniline gives more para-products than its N-methylated derivatives.
This is only considering the nitration reaction. If you consider other reactions, there are varied directing nature. For example the sulfonation reaction (done with $\ce{H2SO4}$ and $\ce{SO3}$) produces almost entirely the para-product[3] (sulfanilic acid), and this has been attributed to the fact that the sulfonation is reversible (thermodynamic control). I haven't been able to find any data on halogenation or other substitutions on the anilinium ion.

M. Brickman, J.H. Ridd, J. Chem. Soc., 1965, 6845-6851
M. Brickman, J.H.P. Utley, J.H. Ridd, J. Chem. Soc., 1965, 6851-6857
J.J. Jacobs Jr., D.F. Othmer, A. Hokanson, Ind. Eng. Chem., 1943, 35, 3, 321–323


Your phenethylamine and bromobutane/methyl iodide are not going to be soluble in water but will dissolve in each other and react. If you use aq base and stir, possibly with gentle heating for bromobutane, then you will get your product. No need for sieves or drying agents. This is essentially a phase-transfer reaction but without a second solvent. As with any attempted amine alkylation, the danger is over reaction to give the quat. salt. If you find over reaction to be a significant problem then consider reductive amination.


Is there a difference between oxidation state and valency?

I think it will help you understand if you know the historical context. The concept of valency existed well before chemical bond, electrons, protons, etc. were known. It is a primitive concept from late 1860s. It referred to the capacity of an element to bind with hydrogen or any other element.
For example, if we say carbon has a valency of four, all it means that it can combine with 4 hydrogen atoms. If we have say oxygen has a valency of 2, it means it will combine with two hydrogen, the famous H$_2$O! Does this make it clear? One would determine valency by pure elemental analysis. If I have FeCl$_3$, iron has a valency (=combining power) of 3, because it can combine with three chlorine atoms. I am avoiding the word ion, because ions did not exist then! The valency of oxygen in peroxide is still 2 because each oxygen atom is combining with two atoms via single bonds. In this case, the other atom is another oxygen itself.
Now read the Wikipedia article Valence. Note that I did not assign any sign to valency
Oxidation number or state is a newer notion which came about when electrons were very well recognized to play a role in bonding and the atomic structure become relatively well understood. I quote it's definition from the OED

a number representing the degree of oxidation of an element in a
particular ion, molecule, or substance, being a notional electric
charge equal to the number of electrons lost (or gained, if the number
is negative) by an atom of the element in forming the species or
substance, all bonding being assumed to be completely ionic

Oxidation state/ number has a sign associated with it. Sometimes, valency and oxidation number can match in terms of their absolute values.
NH$_3$
Valency of nitrogen= 3
Oxidation state of nitrogen = -3


When there is an increase in temperature, how is the pH of the ocean affected? (Question is related to ocean acidification)

You have to compare the size of these counteracting effects.

The temperature increase slightly decreases the acidification caused by increased carbon dioxide partial pressure.

The shift of reaction constant is minimal.

From the Wikipedia carbon dioxide data page, Water solubility section, we can see for temperature  15 resp 17 $^\circ$C V/V solubility 1.019 resp. 0.956. That is 6.2% lower solubility.

Increased partial $\ce{CO2}$ pressure from 0.03 to 0.04% gives 33%, what is significantly more than by 6.2% decreased solubility.

Aside of that, higher $\ce{CO2}$ concentration and lower sea water $\mathrm{pH}$ are already observed.

This is directly affecting carbonate related equilibriums, particularly in context of forming of mollusk shells.

Some mollusks are already observed  having troubles in forming aragonite - the more soluble form of calcium carbonate.


Minimizing Catalyst Residue in Vegetable Oil Hydrogenation

Nickel is usually a good catalyst for this sort of hydrogenation as it's cheap and effective. As to residues, I'm assuming that alludes to residual catalyst; there are a few ways to mitigate against this. Initially, ensure you have minimal fines in the catalyst. The easiest way is to ask your supplier to ensure this. If it's not that sort of process and you just use off the shelf material, then settle your catalyst in water, give it a mix and remove supernatant after a few seconds. Fines will suspend for longer and can be removed in this way. Large particle size material will sink faster. By using the larger particles, you make it easier to remove the catalyst by filtration later. The definitive check is residue on ignition of the final samples. The downside to nickel in batch (as opposed to continuous or flow)  is that it's very dense and needs the right agitator type and speed to get it suspended effectively. However,  this is not normally an issue unless you have inappropriately sized the agitator and type e.g. anchor, to the vessel size and geometry. It may not be worth using precious metal (PM) catalysts for this sort of transformation, but they are easier to suspend when supported on carbon. The advantage of PM catalysts is lower loading.
In all references to catalysts above, they are of the heterogeneous type.


Strength identification in conjugate acid-base pairing

$\ce{H2O}$ is a weak acid.  Abstract a proton to give $\ce{HO^{-}}$, a strong base.  Abstract a proton from that to give $\ce{O^{2-}}$, a very strong base.  See a species as a whole, up and down for a property.

Try it with ammonia.  Feel free to wonder why $\ce{Li3N}$ is red.  Look what a very strong base can do,

$\ce{Li3N + 2 H2 → LiNH2 + 2 LiH}$ (reversible at 270°C)


What causes the smell in rotten meat-grease and how can I get rid of it thoroughly?

If the stench is caused by putrescine or cadaverine, then you are in luck! Both putrescine and cadaverine are amines ($\ce{RNH2}$), which react with acids to form water-soluble ammonium salts ($\ce{RNH3+}$). Most soaps, surfactants, and other cleaning products are basic to help remove fats and oils (by hydrolysis). Try white vinegar, which contains acetic acid. If that fails, you can buy muriatic acid (a dilute solution of hydrochloric acid), which is used to clean pools. Find it wherever pool supplies are found.

Caution: While both of these solutions are fine for the granite, muriatic acid may not be fine for your floor - test in a small hidden area first. Vinegar should be okay for the floor as long as you don't let it soak.

Putrescine:

Cadaverine:


For any instrument that quantifies via integration of a peak (GC, NMR, etc.), sample concentration (peak height) is important especially with low signal/noise ratios. Obviously if the peak isn't in the sweet spot - so to speak - (where its area is significantly above the baseline, but not too big that it saturates the detector) the quantification may be unreliable. Also peak shape is a consideration in integration. If the peak is too broad, much of the area under it can be obscured by the baseline. On GC/MS or GC/FID in particular I used to have problems with peaks 'tailing', that is a broad lingering asymmetry after the peak max which would often overlap with species with greater retention times. No es bueno for quantification. Baking out the column, finding a better heating profile/method usually did the trick.

Optimal methods are going to vary widely based the instrument you use, the solvent, what components you have in your mixture, there relative amounts and concentration etc.. A lot of trial and error. As a general rule though typically when I had 2 components that were close in rt, but wouldn't resolve, I'd move to a method with a much slower ramp for the column temp, upping the carrier gas if possible on your equipment might help, too.


$$\frac{r_A}{r_B} =\frac{1}{4}\implies\sqrt{\frac{M_B}{M_A}}=\frac{1}{4}$$
$$\therefore \frac{M_B}{M_A} = \frac{1}{16}$$
On dividing the given mass by molar mass, we get the respective number of moles.
Let the number of moles be denoted by $n$
$$\therefore {n_A} = \frac{m_{A}}{M_{A}}$$
$$\therefore {n_B} = \frac{m_{B}}{M_{B}}$$
$$\therefore \frac{n_{A}}{n_B} = \frac{m_{A}\cdot M_{B}}{M_{A}\cdot m_{B}} = \frac{m_{A}}{m_{B}} \cdot \frac{M_{B}}{M_{A}} = \frac{1}{24}$$

I do not see what's wrong.
Do you want the mole fractions of A and B?


The rule says that for many liquids the entropy of vaporization is almost the same, around ${\rm 85\,J K^{-1}} $.

The (partial) success of the rule is due to the fact that the entropy of a gas is so much larger than that of any liquid. Therefore, the entropy of the initial state (e.g. the liquid) usually makes no big difference.  However, if the liquid presents hydrogen bonding or any other kind of high ordered structure, its entropy will be particularly low and the entropy gain during vaporization will be larger than usual, too.


Can the hydroxyl group of 4-hydroxybenzoic acid react with phosphorus pentachloride?

a phenol will not react (or react very slowly) with $\ce{PCl5}$

That's not true. Phenol and $\ce{PCl5}$ do react, producing mixture of compounds of formula $\ce{PCl_n(C6H5O)_{5-n}}$ (with $n > 1$ ). It is even possible to produce $\ce{(C6H5O)5P}$![1]

However, unlike in alcohols, in phenols:

the carbon–oxygen bond is much stronger, having a bond order of more than one and being formed with $\mathrm{sp^2}$ hybridized orbitals
the formation of the corresponding transition state for nucleophilic substitution would destroy the aromaticity of the benzene ring.

Thus, the formation of chlorobenzene does not proceed at a significant rate.

It is still possible to replace the phenol group by a chloride atom in some cases. For example, it is possible to prepare picrylchloride (1,3,5-trinitrochlorobenzene) by reaction of pyridine picrate and $\ce{PCl5}$, since three nitro groups greatly increase the reactivity of the phenyl fragment in nucleophilic aromatic substitution reactions.[2] However, as a general rule, benzene derivatives are not active in nucleophilic substitution except if there are a lot of strong π-electron withdrawing groups in the ring.

As for 4-hydroxybenzoic acid, my guess is that it should react as both phenol (as above) and benzoic acid (which forms benzoyl chloride).

Addition

While I was unable to find precise information on para-hydroxybenzoic acid, there is some info on ortho-hydroxybenzoic (salicylic acid). $\ce{PCl5}$ replaces the $\ce{OH}$ fragment of the carboxylic acid group with chlorine. The phosphorus oxochloride then reacts with the phenolic group, producing an $\ce{O-POCl2}$ fragment. The information, however, does not look absolutely trustworthy. Still, assuming it is correct and extrapolating it onto para-hydroxybenzoic acid, I'd say that the product should be $\ce{Cl-C(O)-C6H4-O-P(O)Cl2}$.

References

[1]: Ramirez, F.; Bigler, A. J.; Smith, C. P. Pentaphenoxyphosphorane. J. Am. Chem. Soc. 1968, 90 (13), 3507–3511. DOI: 10.1021/ja01015a038.

[2]: Boyer, R.; Spencer, E. Y.; Wright, G. F. Can. J. Chem. 1946, 24b (5), 200–203. DOI: 10.1139/cjr46b-025.


Some compounds decompose before they reach the melting point. In the literature, you'll see something like "m.p. $40\ ^{\circ}\mathrm{C}$ (dec.)"

Ammonium chloride is an example.


How do I reduce the smell of store-bought liquid soap?

If the scent reaches out and touches you that much, it is likely volatile. Pour it in a pot, add a volume of water, heat on a hot plate outside your dwelling till the smell goes away. Evaporate to volume and check ingredients. You may want to add back a small amount of a desirable ingredient that boiled off. Raise the temperature slowly. A vigorous boil will make a real mess.


Good choice for heat-conductive coil submerged in sea water

My advice is to use copper and protect it with a sacrificial anode.  Copper would be the choice in a non corrosive environment and if you can protect it go for it.

However, if the coils will contact parts of the boat that are important than the coils should be aluminum.  Copper would cause the other metal parts to corrode faster.

Why not use rubber or plastic, something like the pipes used in drip irrigation?


Since your solvent is itself water, it makes no difference whether you use $\ce{H+}$ or $\ce{H3O+}$.
$\ce{H3O+}$ is basically the hydrated form of $\ce{H+}$. If you know, the oxygen atom in water contains two lone pairs. When it donates one of the lone pairs to the hydrogen atom which doesn't have any electrons, you get $\ce{H3O+}$.
So,
$\ce{H3O+}$ is not $\ce{H+}$
$\ce{H3O+}$ is $\ce{H+(aq)}$
This means that the aqueous form of $\ce{H+}$ is represented as $\ce{H3O+}$


If you're smoldering, you're generally not producing enough heat to vaporize anything except in a small local environment. Only the vaporized fuel will burn.


It's true that a precipitation reaction will always yield at least one solid product, but not every reaction that yields a solid is, strictly speaking, properly categorized as a precipitation reaction. Precipitation reactions are ones in which at least one of the reactants is in the aqueous phase (i.e., dissolved in water), and a solid forms on the product side which was not present on the reactant side. If all the reactants are solid, then a solid product forming should not be called a precipitate. To be called a precipitate, an insoluble product must form from within solution, either from a solid and solute interacting, or from an interaction strictly between solutes.

Examples of precipitation reactions:

$\ce{AgNO3_{(aq)} + NaCl_{(aq)} -> AgCl_{(s)} + NaNO3_{(aq)}}$

$\ce{CuCl2_{(aq)} + Zn_{(s)} -> ZnCl2_{(aq)} + Cu_{(s)}}$

Notice that at least one of the reactants was in the aqueous phase, and a new solid chemical species is produced on the product side. $\ce{AgCl_{(s)}}$ and $\ce{Cu_{(s)}}$, respectively, are the precipitates. Incidentally, the second reaction is also an example of a redox reaction, since it involved a change in the oxidation states of the reactants (here, zinc is oxidized while copper is reduced).

Identifying acid-base and redox reactions is a lengthier topic, so I suggest you consult a textbook or any of the numerous online resources.


The bag contains a supersaturated solution of some salt with a highly negative enthalpy of crystallisation, which could be something like sodium acetate ($\ce{Na^{+}CH3CO2^{-}}$). The dissolved salt would really like to come out of solution but lacks a nucleation site. Bending the metal disk produces a transient site where crystallisation can begin, and from there are chain reaction results, causing all of the solute to crystallise and release the heat associated with this crystallisation.

Your guess regarding the half-dissolved crystals is correct - the crystals that have not dissolved act as seed crystals for further crystal growth. You need to dissolve all of the crystals to prevent the process from spontaneously reversing.

Despite the popularity of this kind of reaction for modern self-heating food/drink packaging in Japan, I suspect the discovery of this kind of reaction is lost to antiquity, as for instance sodium acetate can be made by simply mixing salt and vinegar. I recall reading that the highly exothermic reaction of calcium oxide and water was used in self-heating cans of food back in World War I.


We do not decide either which. Hybridization is an observational theory - it is a theory crafted to fit observation, it is not made from the bottom up. That doesn't make it any less valuable (well... maybe it does...) but your question becomes rather easy to answer. We decide which orbitals become hybridized based on how we measure the molecule to be. We see the symmetry.

Hybridization is just a model. But then again so are most things we learn in chemistry. I would say you are not missing out on anything if you just skip learning it and go directly to molecular orbital theory. Your teacher might argue though, do what (s)he says.


How to test for halides without ammoniacal silver nitrate solution?

ChemGuide gives another method:

  If you add concentrated sulphuric acid to a solid sample of one of the halides you get these results:
  $$
\begin{array}{c|c}
\text{Ion present}&\text{Observation}\\\hline
\ce{F-}&\text{steamy acidic fumes}\\
\ce{Cl-}&\text{steamy acidic fumes}\\
\ce{Br-}&\text{steamy acidic fumes contaminated with brown bromine vapour}\\
\ce{I-}&\text{some steamy fumes, and lots of purple iodine vapour}\\
\end{array}
$$

Note that those "fumes" are of the corresponding hydrogen halides. Also, in the case of the iodide ion, "various red colours in the tube" are also observed.

This reaction occurs because concentrated sulphuric acid acts as an oxidizing agent. It isn't able to oxidize fluoride or chloride ions, but definitely oxidizes $\ce{Br-}$ and $\ce{I-}$ to $\ce{Br2}$ and $\ce{I2}$ respectively (details here)

Do note that, to distinguish between the fumes of $\ce{HF}$ and $\ce{HCl}$ above, you'll need to dissolve them in water and perform the silver nitrate test!


Explanation for increase in the number of oxidation states for transition elements from Sc to Mn of the 4th period

The oxidation state of a transition metal is an accounting device.  It's not the number of "lost electrons".  In potassium permanganate, which has Mn (VII), the manganese atom shares its seven 3d electrons with the ligand (oxygen) in what are essentially covalent bonds, and the "positive charge" associated with the metal is not very different than what it is for ionic calcium compounds.

Regarding "paired electrons are less likely to be involved in bonding than unpaired electrons" -- both paired and unpaired electrons in a transition metal can participate in bonding, provided they are in the outermost shell.  The paired electrons in inner shells are too deep in energy to take part in bonding.


Cis and Trans Isomerism in Resonance-Stabilized Compounds

Thanks for the picture, now I understand your question. All of the atoms in $\ce{N2O5}$ don't need to lie in a plane as is shown in the picture in your link.  At room temperature there would be rapid rotation about the 2 bonds connecting the central oxygen to the two nitrogen atoms, so there is no "fixed" geometry from which you could use the cis and trans references.  For example, imagine the conformation with one $\ce{NO2}$ group in the plane of the screen and the other perpendicular to the screen, even if each $\ce{NO2}$ group had one $\ce{N=O}$ double bond and one $\ce{N-O}$ single bond, the terms cis and trans would still have no meaning in this conformation. Further, as you noticed, all 4 bonds between nitrogen and the exterior oxygens are equivalent due to resonance, so again terms like cis and trans would have no meaning.  Finally, I can't imagine a situation where an $\ce{N=O}$ double bond (e.g. where the oxygen is unsubstituted) could have cis-trans isomers because the 2 lone pairs on the oxygen are equivalent.


As mentioned in the comments, this topic can become exceedingly complex if all the aspects of current-gen photochromic lenses are brought to the table (and I'm sure a lot of it is locked away in trade secrets). My goal here is to provide some insight into the chemical/physical mechanisms at work.

Photochromic sunglasses are a smart new application of very old knowledge. Silver salts have been isolated and manipulated for centuries, so it's surely been known they have a tendency to darken in light for a long time. This simple fact (plus a lot of auxiliary processes) was the basis of the main forms of photography all throughout the 19th and 20th century.

Because of this, the classic examples of photosensitive compounds are the silver halides ($\ce{AgX}$), particularly $\ce{AgCl}$, $\ce{AgBr}$ and $\ce{AgI}$. These compounds are colourless to slightly yellow when pure. Upon exposure to photons of appropriate frequency, the idealised sequence of events is:

$$1) \quad \ce{2 X- \xrightarrow{h\nu} X_2 + 2 e-}$$
$$2)\quad \ce{2 Ag+ + 2e^- -> 2 Ag^0}$$

So essentially, illumination of the silver halides triggers the production of elemental (i.e. metallic) silver. This is a photoredox process; we say the halide anions undergo photooxidation, and the silver cations undergo photoreduction. An interesting way to think of it is that the halides normally are not strong enough reductants to trigger the reduction of $\ce{Ag+}$ to $\ce{Ag^0}$. However, this is only true in the electronic ground state of these species - the halide ion $\ce{X-}$ in an electronic excited state has different chemical properties during its short life, and in particular it becomes a much stronger reductant, and now can reduce silver directly. This leads to another description, which can be more accurate in other contexts

$$1) \quad \ce{2 X- \xrightarrow{h\nu} 2X^{-*}}$$
$$2)\quad \ce{2 Ag+ + 2X^{-*} -> 2 Ag^0 + X_2}$$

Back to the silver. Instead of forming a shiny reflective film, the silver is formed as a multitude of tiny disconnected particles. As the photons bounce around chaotically in the maze of silver clusters, a large fraction of the incident light ends up getting absorbed (and turned into heat). On the macroscopic scale, this makes the material appear dark.

It's interesting to point out here that the element silver alone is not responsible for the darkening effect. The process starts with the absorption of light to generate a strongly reducing species, from which the silver ions then steal an electron to become silver metal. In principle, therefore, it is possible to tune this effect depending on what else is mixed with the silver ions. Indeed, as the halides get heavier, ionisation becomes easier (or alternatively, it becomes easier to reach the strongly reducing photoexcited state), and the silver halides become more photosensitive. However, it's possible to mix in something else entirely, such as electron-rich organic compounds, which could then surrender their electron to the silver ions when exposed to light. It's also not fundamental to use silver at all - all metals are dark when dispersed into very tiny clusters.

So we now put this knowledge together into one of the simplest examples of photochromic lenses. This technology may have been used in the past, but is now likely supplanted.

Like before, we start with a silver halide. For example, during the fabrication of glass, a small amount of $\ce{AgCl}$ can be added. Though it would look like otherwise ordinary glass, when struck with light, it can potentially trigger the formation of $\ce{Ag^0}$ and darken the glass.

The problem now is to make this process reversible. In the ideal reactions above, the formation of $\ce{Ag^0}$ would also require the formation of $\ce{Cl_2}$. These two compounds can in principle react to regenerate $\ce{AgCl}$ (after all, $\ce{AgCl}$ was the thermodynamic minimum, it had to be hit with a photon to turn into something else!). However, $\ce{Cl_2}$ could potentially trigger a number of other reaction, or move around in the glass (it's a gas in standard conditions), which makes the process unlikely to be reproducibly reversible.

To solve this problem, it's possible to add another pair of electrochemically active species. In particular, it turns out $\ce{CuCl}$ is a nice choice. Why? $\ce{AgCl}$ and $\ce{CuCl}$ can be mixed into glass without reacting, once again generating a colourless material. However, upon exposure to light, the following (idealised) reactions happen:

$$1) \quad \ce{ Cl- \xrightarrow{h\nu} Cl^{0} + e^-}$$
$$2) \quad \ce{Ag+ + e^- -> Ag^0 }$$
$$3) \quad \ce{Cu+ + Cl^0 -> Cu^2+ + Cl-}$$

Now the chloride is essentially a spectator species. The process could be rewritten as:

$$\text{Net ionic equation}: \quad \ce{Cu+ + Ag+ \xrightarrow{h\nu} Cu^2+ + Ag^0}$$

Over time, the tendency is for the $\ce{Cu^2+}$ to reoxidise the silver metal back completely to $\ce{Ag+}$, thus making the glass transparent again. This process is comparatively slow, however, and therefore while there is light shining on the glass, there is a buildup of $\ce{Ag^0}$, making it dark. All in all, very convenient, indeed.

Lastly, I'll briefly address why this process happens in sunlight, but not indoors. Unsurprisingly, this is an entirely intentional property. Good eye-wear glasses block invisible and damaging UV photons present in sunlight, but shouldn't block visible photons. Meanwhile, a good visible light source should create visible photons, and avoid wasting energy on harmful UV photons. The design of the materials used in both photochromic glasses and fluorescent lamps has therefore naturally lead to the situation where photochromic lenses should generally not darken with indoor light sources.

For a more hands-on approach to this whole topic, I highly recommend this excellent video.


There are at least two different radiationless pathways to wreak havoc on the DNA:

intercalaction: compounds like benzo[a]yrene or ethidium molecules slide between DNA strands and cause transcription errors
alkylation: compounds like N-Nitroso-N-methylurea may transfer alkyl groups to nucleobases

But the list is by no means complete.


Difference between vapor-phase transport and vapor-phase deposition?

The paper of interest is C.M. Wolfe et al., "Growth and Dislocation Structure of Single-Crystal Ga(As$_{1-x}$P$_{x}$)", Journal of Applied Physics 36(12) 3790 (1965).

The key paragraph is:

  The GaAsP of this investigation has been prepared
  by a closed-tube method utilizing a quartz ampoule
  which is cleaned and baked-out under vacuum near
  the softening point of quartz. The halide is introduced
  by subliming a metallic halide, such as PbCI2, PbI2,
  NH4CI, NH4I, etc.,ro into the reaction vessel. Typically,
  2.2 g of source GaAs and GaP, 4-8 mg of excess P,
  40-160 mg of halide, and a dopant are placed in an
  8-cm$^{3}$ ampoule as in Fig. 1. This vessel is then evacuated
  to 10$^{-6}$ Torr, sealed, and placed in a single-zone furnace
  at 950°C with a gradient of 5°-15° along the tube.
  Commercially available GaAs is used with GaP grown
  from high-purity elements by a method previously
  described.

So, yes, it is a vapor deposition technique, but uses a thermal gradient to evaporate from one place and redeposit in another place, controlled by the temperature.


Stacking two square (2D) layers to make a 3D close packed structure

In the picture, the cubic face-centered packing is shown, with layers of identical atoms colored in red and blue for better contrast. The unit cell shown with thin lines is the conventional face-centered setting. It has red spheres on the corner, and on the center of the bottom face (the faces not normal to the plain of the paper have blue spheres in their centers).

The unit cell shown with thick dashed lines is what you were asking about. It has square close packing of the red spheres, with square close packing of the blue spheres shifted below. (The third unit cell vector is not shown, and is the same for both, from red to red layer).

So to answer your question, just tilt your head by 45 degrees, and you have your "close square" packing rather than face-centered cubic. As you can see, the three unit cell vectors will not be of equal length anymore, so the unit cell is no longer cubic, and it is body-centered instead of face-centered. The volume is smaller by a factor 2, and the number of atoms per unit cell is smaller by a factor of 2 (2 atom per cell instead of 4 atoms per cell).

In exploring this question, I made use of this site: https://scilearn.sydney.edu.au/fychemistry/iChem/solids.cfm?page=ccp&type=cubic


Cooling Curves differences between Solution and Pure Solvent

During the phase change, the composition of the solution phase is changing from that of the original solution as the solidifying component is removed.  This doesn't happen for a pure liquid.  In a simple system, the slant continues until both components begin to solidify simultaneously - the eutectic point.  The temperature then flattens out because the composition of the liquid phase doesn't change.  It stays flat until no liquid remains.  This can be complicated if there are other solid phases or formation of new chemical components.


How can I synthesize phenylpiracetam from styrene or benzaldehyde?

This is a reasonably sound route. I have some observations:
The prep of nitrostyrene from benzaldehyde is well known, it is best to use freshly distilled benzaldehyde. Here are the conditions from org syn prep.
The sidechain can be added by treatment of the phenylpyrrolidine with $\ce{NaOMe/MeOH}$/Methyl chloroacetate then heating with $\ce{NH3/MeOH}$
(patent).
You are better off reducing the nitro group at the diester stage as it will spontaneously cyclise to the pyrrolidine. On hydrolysis it will then decarboxylate. In my experience, hydrogenation of alkyl nitro compounds does not give good yields. $\ce{Al/Hg}$ amalgam is better. Sodium hypophosphite /phosphinic acid with cat. 5% Pd/C in water/THF and ultrasound was recently reported here as an efficient method for reduction of alkyl nitro, and the paper contains an example of spontaneous cyclisation of the product amine onto a methyl ester.
A synthesis of phenylpiracetam is described here in this patent.


Reaction of SeCl4 with alkene and possible consecutive reaction?

We humans are still at best mediocre at predicting the outcome of chemical reactions. Even computers are at best a little bit better. If there is no literature on this compound (and Orthocresol commented that he found none), no experiment has likely been performed and we cannot be certain.
You could find a computational chemist of your choice with free time and the appropriate software and let them simulate this reaction. This would allow a slightly better prediction as it would answer the question ‘is there a reasonable reaction pathway that this compound could follow?’ However, even when simulating ‘obvious’ reactions we chemists might overlook problems with a simple-looking pathway that make this pathway non-accessible in the first place. Thus, even if the simulation looks promising, the only true answer is to go into the lab and try it out.
On paper, your scheme seems promising. You may require additional base (maybe a weak base like carbonate will do the trick, maybe you need selective oxygen deprotonation e.g. by sodium hydride) or it may happen during the chlorination reaction.
There may be other side reactions. Without knowing whether it is applicable in your case, I would attempt to reseach potential non-classical cations and how they might interfer, in case this reaction would follow a more $\mathrm{S_N1}$-style pathway.
In addition, I wonder if an elimination reaction to a chloro-norbonene might be a competing side-reaction.
I don’t think that a nucleophilic substitution on the other chloride would pose much of a problem as five-membered rings are typically kinetically preferred over six-membered rings.


Is thin layer chromatography suitable to characterize distillate samples?

Your statement is unnecessarily complicated and not easy to follow. You can use TLC as a quick way to assess the purity of a sample. It doesn't matter if your sample is a distillate or not. It only has to be non volatile as you said and soluble in some solvent or a mixture of solvents that can (the solvent/s) make it move on the TLC plate.

I assume you want to use distillation to separate a mixture and TLC in order to assess the purity of your fractions. Obviously TLC becomes an unsuitable method for low b.p. distillates


Like @orthocresol said in his comment, biomolecules are often numbered by convention. The 'traditional' numbering for 2-deoxystreptamine is shown below on the left, and IUPAC numbering (not used) is shown on the right:

The most likely rationale for the traditional numbering is that the compound is a diamine so the numbering starts from one of the amino substituents and goes towards the other one.

EDIT:
OK, now I understand exactly where lies your confusion. The molecule of 2-deoxystreptamine (2-DOS) has a plane of symmetry. If we are to number the molecule alone, it doesn't matter from which amine group we will start. But as you noticed, the kanamycin is asymmetrical and there are indeed two ways to number the ring of 2-DOS. To solve this problem, we need to look on the structures of other aminoglycosides and their biosynthetic pathway. Many of the antibiotics from this class are biosynthesized through an intermediate compound known as paromamine (I prefer Haworth projection):

The numbering is logical - if we would start from the other amine then the glucosamine substituent would be at atom 6 (against the usual rules of numbering where we prefer lower numbers for substituents). Most aminoglycosides can be divided into two groups: 4,5-disubstituted (neomycins, paromomycins) and 4,6-disubstituted (kanamycins among others). Therefore the red numbering in the original question is the right one, if we assume the synthetic pathway described in this article:

Park J.W. et al, "Discovery of parallel pathways of kanamycin biosynthesis allows antibiotic manipulation" Nat. Chem. Biol. 2011,  7(11):843-852 (DOI: 10.1038/nchembio.671).


Could someone please explain the difference between m/z and m/e in mass spectroscopy

$m$ is the atomic mass of the ion (in units ($u$); $1u=1.665402\times10^{-27} \mathrm{kg}$), $z$ is the number of its positive/negative charges. $z$ should not be used in interchange with $e$ because it is dimensionless, while $e$ has the unit $\mathrm{C}$ (Coulomb). For $\ce{[C7H7]^2+}$, $m/z$ is given by (the unit of $m$ is omitted):

$$m(\ce{C7H7})/2=(7 \cdot 12+7 \cdot 1)/2=91/2=45.5$$


Is it right to say that fluorescent lamps are actually fluorescent?

Fluorescent lamps do primarily work by fluorescence. According to this Wikipedia article:

  "The inner surface of the lamp is coated with a fluorescent (and often slightly phosphorescent) coating..."

First off, an electric current excites the outer electrons of gas atoms (typically mercury) within the lamp to a higher energy state. When the atoms relax back to their stable state, UV light is emitted. This UV light is then absorbed by a fluorescent material coating the inside of the lamp, which in turn fluoresces in the visible region of the EM spectrum. Some confusion could arise here as this coating is called a phosphor, even though the type of luminescence is fluorescence, not phosphorescence.

The rate at which the luminescence occurs is not important because there is a continuous supply of electric current. If it were necessary to provide light for any significant period of time after the power source was removed, then some type of phosphorescence would be necessary due to it's much longer lifetime as compared to fluorescence, as you pointed out. However, this is clearly not a requirement for a standard fluorescent lamp.


How to calculate the concentration of all relevant species in a buffer of a given pH?

The value given for the pH of the solution seems to be incorrect*.

As the reaction is $\ce{HA <=> H+ + A-}$, your initial assumption that $[\ce{H+}]=[\ce{A-}]$ is correct.

From your chart,

\begin{array}{l|ccc}
               & \ce{HA} &  \ce{H+}  &  \ce{A-} \\\hline
\text{Initial} & 0.25    &  0        &  0       \\
\text{Change}  &  -x     &  +x       &  +x      \\
\text{End}     & 0.25 -x &  +x       & +x       \\\hline
\end{array}

it follows that
$$K_a=\frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]} = \frac{x^2}{0.25 -x} = 1.74\cdot10^{-5}.$$

We can solve this for $x$:
\begin{align}
x^2 &= 1.74\cdot 10^{-5}(0.25-x)\\
x^2 + 1.74\cdot10^{-5} x - 4.35\cdot10^{-6} &= 0\\
x &= 0.00207698~\mathrm{M}\\
\end{align}

Thus, $[\ce{HA}] = 0.25~\mathrm{M} - 0.00207698~\mathrm{M} = 0.24792302~\mathrm{M}$ and $[\ce{H+}]=[\ce{A-}] = 0.00207698~\mathrm{M}$.

Also note that in your second attempt, the value $[\ce{A-}] = 4.35~\mathrm{M}$ makes no sense, as the amount of $\ce{A-}$ present should be in a 1:1 ratio to the amount of $\ce{HA}$ consumed in the reaction, and $[\ce{HA}] < 4.35~\mathrm{M}$.

The given $\ce{pH}$ is likely incorrect because the concentration of $[\ce{H+}]$ ions for the given molarity and concentration does not match with the given $\ce{pH}$, and because the given $K_a$ is close to values of $K_a$ provided online.


Br- and I- ions give layer test, but Cl- ion doesn't. Why?

Organic Layer test is a test involving redox reactions among halides. This test is used to detect in the presence of bromide ions and iodide ions in a solution (e.g., Lassaigne’s test solution).  Layer test is carried out by adding organic solvent such as carbon disulphide or chloroform in the presence of nitric acid and chlorine water to the given solution. For example, Lassaigne’s extract is acidified with dilute nitric acid and chlorine water is added to it. When organic solvent is added to this aqueous solution and shaken the mixture vigorously but carefully, the liberated bromine or iodine liberated by redox reaction of bromide or iodide ions with chlorine) would be traveled to organic layer to give distinguished color. In case bromine is present, it will give an orange layer while it will give a violet layer in case of presence of Iodine. Relevant redox reactions are given below:

$$\ce{Br- + Cl2 -> Br2 + Cl-}$$
$$\ce{I- + Cl2 -> I2 + Cl-}$$

The key is bromine and iodine are much soluble in organic solvents than in water. Since we are using chlorine as a oxidant, chloride ions are negated here.


Symmetry and stability of a molecule (Jahn-Teller Distortion)

The symmetry and stability are not directly related per se; is benzene more stable that naphthalene for example? Benzene is more 'symmetric' than the naphthalene but both are stable molecules; $D_{6h}$ vs $D_{2h}$ point groups.

Jahn  and Teller discovered that in a non-linear molecule in an orbitally degenerate electronic state, say $e_g$ or $t_{2g}$ in an octahedral complex, that there is an interaction between the electronic orbital motion and nuclear vibrational motion, which causes the minimum energy position  to differ from the symmetric structure. This motion lifts the orbital degeneracy and lowers the energy. Thus the lower symmetry state has lower energy.

An example might be the lengthening of a pair of axial bonds in a octahedral complex, $\ce{[CoF6]}^{3-}$ for example, however, the J_T theorem does not give any indication of how big the distortion might be, i.e. it does not predict the geometrical nature of the distortion. There is also a restriction which is that if the degenerate structure has a centre of symmetry so does the distorted structure.

note: The electronic spin of the molecule (singlet, triplet etc.) is not important it is the orbital degeneracy only that matters. The Jahn-Teller effect is a breakdown of the Born-Oppenheimer approximation which is based on separating nuclear and electronic motion.


A pH indicator is generally a weak acid and often work in a fashion similar to buffers in that most of the time both the acid and its conjugate base exist in significant concentration. From the Ka of the indicator you should therefore be able to find the "halfway" point of the indicator or the middle of the range of pH values over which the indicator operates.

Think about how indicators work. Indicators work simply by mixing colors. If I have an indicator which is black at pH = 0.0 and white at pH = 2.0, then I know that the protonated form of the indicator must be black-colored, and that the deprotonated form of the indicator must be white colored. This should make sense. As pH increase, hydroxide ion molarity increases, and hydroxide ion deprotonates other molecules.

Thus, at pH = 1.0, we should have a 50/50 mix of black and white molecules. In other words, the indicator shows a gray color at pH = 1.0. At pH = 1.0 we also have a buffer solution - we have the acid and conjugate base - and specifically in a 1:1 ratio.

By application of the Henderson-Hasselbach equation we will realize:

$\ce{pH = pK_{a} + log\frac{base}{conjugate~acid}}$ simplifies to ...

$\ce{pH = pK_{a}}$

I'll let you take the problem from here. Don't forget to apply common-sense considerations.


What is the criterion for filling up of electrons in molecular orbitals of increasing energy in simple diatomic species?

What is known for sure (and is mentioned in all the books I've seen so far) is that for homonuclear diatomics the "$\pi \rightarrow \sigma$ switch" happens when going from $\ce{N2}$ to $\ce{O2}$: for 14-electron case $\pi_{2p_{x,y}}$ orbitals are lower in energy then $\sigma_{2p_{z}}$, for 16-electon one it is other way around. The usual "explanation" is that moving from $\ce{Li2}$ to $\ce{F2}$ the successive decrease of the $s-p$ interaction between the $2s$ orbital of one atom and the $2p_z$ orbital of another atom decreases the energy of the $\sigma_{2p_{z}}$. But you have to understand that this is merely a manifestation of a known fact: we know how orbitals of these molecules are arranged from the molecular orbital theory calculations and use the notion of $s-p$ interaction to rationalize this knowledge.

The problem is that no information about 15-electron case can be inferred from the above mentioned knowledge about molecular orbitals of homonuclear diatomics from $\ce{Li2}$ to $\ce{F2}$. A priori we could not tell will $\sigma_{2p_{z}}$ orbital be lower in energy than $\pi_{2p_{x,y}}$ orbitals or higher then them for a 15-electron case. Besides, things coul be different for different 15-electron cases, such as $\ce{N2-}$, $\ce{O2+}$, $\ce{NO}$, etc.: for some $\sigma_{2p_{z}}$ orbital might be lower than $\pi_{2p_{x,y}}$ orbitals, for others - higher. There is no simple rule to use, one have to resort to molecular orbital theory calculations to tell the truth.


Pyrylium is aromatic. Following Hückel's rule, pyrylium is uninterrupted (no $\mathrm{sp^3}$ hybridization), all atoms have p-orbitals and there is 6 π electrons. the 6 π electrons indicate that Hückel's rule $(4n+2)$ is agreeable for pyrylium, $n = 1$.


Lithography: Is isopropanol a developer and/or a stopper?

Methyl isobutyl ketone is the solvent for PMMA - the alcohol is just the carrier. MIBK has low solubility in water, so if you stopped with water, you would probably leave more MIBK residue in your PMMA.


Are there plastic materials that turn into ceramics?

The set of compounds you are looking for are known as Thermosetting Polymers. A few examples of these are:

Bakelite: This is used for making switches and electrical enclosures
Melamine: This is commonly used for making food grade and microwaveable utensils and is an alternative to ceramic (at least in the food packaging industry)

You might also want to take a look at Thermoplastic Polymers, as they behave a bit like wet clay (they become soft on heating and harden on cooling and this can be done as many number of times as required)


Placing a drop of oleic acid on the surface of water contained in a pan is a relatively straightforward experiment that has been widely used to determine Avogadro's number.  Here are two links that provide good background on a) how to run the experiment, b) how to analyze the data, and 3) the science behind the experiment.  The two links are just variations on the same experiment, but if you read them both it will really enhance your understanding of the experiment.

link 1

link 2


Does "higher symmetry" mean any symmetry group of higher order?

"Higher symmetry" per se does not mean anything, unless you define it to mean something.

Depending on the context, it usually means one of the following things:

Any symmetry that is a supergroup of the group in question. In this case $C_{4v}$ is higher than $C_4$, while $C_6$ is incomparable to both. Bear with it. Partially ordered sets are quite a thing in math.
(rarely used) Any symmetry group having more elements than the group in question, so that $C_{4v}$ is higher than $C_6$, which in turn is higher than $C_4$.
(apparently, the meaning used by Cotton) Any symmetry group which is not a cyclic rotation group. In this case $C_{4v}$ is higher, while $C_4$ and $C_6$ are lower.

See, there is an intuition, after all. Worse yet, there are quite a few different kinds of intuition.

So it goes.


Why is bipolar resonating structure less stable than monopolar structure?

In the diagram below, the 'bipolar resonance structure' refers to the resonance structure with both positive and negative partial charges (left), while the monopolar resonance structure refers to the anionic form in the right. I don't think this terminology is appropriate, and I have never encountered it before. I will refer to the structures as 'protonated' (left) and 'deprotonated' (right).

The diagram above illustrates the scenario. While it is true that the resonance stabilization is greater for the deprotonated form, the protonated form is still overall more stable. The book is inaccurate in that regard.

The "resonance stabilization" refers to electrons being donated into empty orbitals within the molecule. In this case, there is a lone pair in the hydroxide oxygen atom being donated into the antibonding pi orbital of the C = O. A way to rationalize the effect is that, when the proton is attached to the oxygen, you have both the H and the O pulling on the electrons. When the H is removed the electrons will more readily flow into the C = O pi*. Since both O atoms are equivalent, you end up forming an orbital that looks like an extended pi bond.

The stabilization due to the resonance of the deprotonated form is of greater magnitude than the stabilization due to resonance of the protonated form. However, the proton-oxygen bond is more stabilizing than the resonance, so the protonated form is overall more stable.

Diagram soure: http://chemwiki.ucdavis.edu/Organic_Chemistry/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Physical_Properties_of_Carboxylic_Acids


What is the significance of the kink in the [A] vs T graph for a 2nd order reaction?

There's no significance to the shape, that's simply how any graph of the form $y = 1/(c+x)$ looks like ($c$ being any constant you like).
Even if the supposed "kink" has some significance, there isn't any significance to the number, especially since the number depends on your system anyway (it's dependent on the value of your rate constant as well as the initial concentration).


Will a solid object lose or gain atoms on standing over time (without being acted upon by external influences)?

Short answer: Yes they do.

Firstly, you need to arrive at a proper definition for an "object". Why you ask? Because it would otherwise present philosophical problems which become apparent when we begin to work with those objects.

  ... what is an object? Philosophers are always saying, “Well, just take a chair for example.” The moment they say that, you know that they do not know what they are talking about any more. What is a chair? Well, a chair is a certain thing over there … certain?, how certain? The atoms are evaporating from it from time to time—not many atoms, but a few—dirt falls on it and gets dissolved in the paint; so to define a chair precisely, to say exactly which atoms are chair, and which atoms are air, or which atoms are dirt, or which atoms are paint that belongs to the chair is impossible. So the mass of a chair can be defined only approximately. In the same way, to define the mass of a single object is impossible, because there are not any single, left-alone objects in the world—every object is a mixture of a lot of things, so we can deal with it only as a series of approximations and idealizations.

If you can't define an object (more specifically: If you can't define its location in space) exactly...you cannot know its mass exactly. If you do not have its exact mass, you cannot make an exact/complete description (be it physical or chemical) of the body. [In fact, our ability to provide accurate descriptions of bodies is because we make good approximations]

I get what you are trying to ask in your question, and it is per se, a reasonable question; but the example (your room) is (unfortunately) difficult to deal with...the single glaring issue being the inability to define an object in an absolute sense (I'm not blaming you, in fact, no one can define any real-life object exactly :P)

Just picking out some other holes in your "room" example: Is you room sealed (can it be considered a closed system)? Is your room evacuated (I mean a "Vacuum")? Are the different "objects" present in the room in contact with each other? Are there foreign particles present on the walls, ceiling and floor? You can see where this is going...

You go on to mention using a "pure glass/gold sphere" in order to circumvent some of the difficulties of defining an "object". Unfortunately, 100% purity is inconceivable (the main issue being the dust/oxides that are present at the surface). To neglect the presence of these layers would be to forgo reality...and the question (using the "pure" spheres) would be reduced to a purely theoretical proposition.

Trying to address you question, I'll do this in two ways: first I will deal with an idealized case and secondly, I shall deal with the abomination you call your "room" (introducing a few constraints of course, to keep things simpler) :D

First: The (highly) idealized version of your example:

You have a really large ("infinitely" large, if you will, since this is a thought-experiment) sealed region of space, and this "chamber" (as we shall now call it) shall be totally isolated from its surroundings: No energy transfer, no mass transfer, no (external) gravitational field, no (external) electric or magnetic fields...no nothing. The walls of the chamber have no substance and are infinitesimally thin. The inside of the chamber is maintained at an arbitrary temperature of 25°C.

Suppose your chamber has a small amount of a mono-atomic gas (take neon for example) in it. Now you introduce an "object" in the chamber instantaneously. The "object" is a 1000 atom x 1000 atom x 1000 atom cube of absolutely pure, solid gold (and we shall assume there's no oxide layer on the surface).

Now what happens?

Simply put; the neon atoms will impinge on the gold cube and some may end up "sticking" to it. The gold atoms will slowly (like reeaaally slowly) "diffuse" out of the original block and co-exist with the neon atoms as a gas (you can think of this as the "evaporation" of the solid gold cube). You'll observe that even though neon diffuses into the gold cube, and the gold atoms diffuse into the neon gas...over a period of time there is a net diffusion of gold atoms out of the block. Keep watching, and you'll find that you have a gaseous mixture of neon and gold, with no solid gold left to speak of. Handwaving the kinetics involved here, this is an example of the Second law of Thermodynamics at play. The entropy (or the "randomness") of the system goes on increasing over time, and will continue to do so until an equilibrium condition has been achieved. {You might also want to read about the Heat Death of the Universe  <--- Interesting thought}.

Even if you didn't have the neon in the chamber (I put it in there to "simulate" the air in your room), the cube would still "evaporate" into a gas, albeit, at a slightly faster rate.

Second: Your room (with a few restrictions):

I shall proceed to define your room (albeit with a few constraints). You room is, also sealed so that no mass can be exchanged between the inside of your room and the surroundings. Everything inside your room is still subject to external gravitational, electric and magnetic fields. The temperature inside your room is 25°C. You room is not dust-free either. Your room is now set.

Start the clock...and what happens?

The results are similar (albeit, a scaled down version) to the idealized case. Every "object" in your room loses atoms/molecules and gains some too. Like before, there is a net diffusion of particles out of the objects in your room...entropy increases yet again. Leave it there for a few hundred million years, and you'll find that the objects in your room will have changed significantly (remember, the time frame was in hundreds of millions  of years).

As @Ruslan pointed out in the comments: Unlike the ideal case where the gold atoms can evaporate from the cube and continue to spread ad infinitum...your furniture and computer don't have that kind of luxury (their ability to spread out in space is limited by the walls of your cramped room) and hence won't evaporate completely (but there is evaporation nonetheless).

To understand that last bit ("incomplete" evaporation in your room, but "complete" evaporation in the idealized chamber), it will help to consider an analogy with the evaporation of water:

Here's an excellent piece (The Feynman Lectures on Physics, Volume 1, Chapter 1) you must read through before we continue:

  What happens at the surface of the water? We shall now make the picture more complicated—and more realistic—by imagining that the surface is in air. The figure shows the surface of water in air.

  We see the water molecules as before, forming a body of liquid water, but now we also see the surface of the water. Above the surface we find a number of things: First of all there are water molecules, as in steam. This is water vapor, which is always found above liquid water. (There is an equilibrium between the steam vapor and the water which will be described later.) In addition we find some other molecules—here two oxygen atoms stuck together by themselves, forming an oxygen molecule, there two nitrogen atoms also stuck together to make a nitrogen molecule. Air consists almost entirely of nitrogen, oxygen, some water vapor, and lesser amounts of carbon dioxide, argon, and other things. So above the water surface is the air, a gas, containing some water vapor.

  Now what is happening in this picture? The molecules in the water are always jiggling around. From time to time, one on the surface happens to be hit a little harder than usual, and gets knocked away. It is hard to see that happening in the picture because it is a still picture. But we can imagine that one molecule near the surface has just been hit and is flying out, or perhaps another one has been hit and is flying out. Thus, molecule by molecule, the water disappears—it evaporates. But if we close the vessel above, after a while we shall find a large number of molecules of water amongst the air molecules. From time to time, one of these vapor molecules comes flying down to the water and gets stuck again.

  So we see that what looks like a dead, uninteresting thing—a glass of water with a cover, that has been sitting there for perhaps twenty years—really contains a dynamic and interesting phenomenon which is going on all the time. To our eyes, our crude eyes, nothing is changing, but if we could see it a billion times magnified, we would see that from its own point of view it is always changing: molecules are leaving the surface, molecules are coming back.
  Why do we see no change? Because just as many molecules are leaving as are coming back! In the long run “nothing happens.” If we then take the top of the vessel off and blow the moist air away, replacing it with dry air, then the number of molecules leaving is just the same as it was before, because this depends on the jiggling of the water, but the number coming back is greatly reduced because there are so many fewer water molecules above the water. Therefore there are more going out than coming in, and the water evaporates.

Now back to where I left off; why do we see partial evaporation in your room, but total evaporation in the "infinite" chamber?

Based on the excerpt, I'll try to develop an analogy for what's going on in 1) Your room and 2) The "infinite" chamber I mentioned first.

1) Analogy for what's going on in your room:

You fill a perfectly dry bottle halfway up with water and screw on the cap. The water molecules constantly evaporate from the surface and molecules of oxygen, nitrogen, etc from the air inside the bottle constantly impinge on the water surface and may get dissolved into the water as a result. However, their ability of the water molecules to spread out in space (increasing entropy) is limited by the bottle's boundaries. Though we have an initial net evaporation of water...we're soon faced with an equilibrium state (described in the excerpt) where there is no net evaporation of water. You're still left with water that is mostly in its original (liquid) state.

It's pretty much for reasons similar to this, that the object in your room won't completely evaporate. There simply isn't enough space for the particles to "comfortably" move out in to.

2) Analogy for what's going on in the idealized case:

You pour a glass of water on the side-walk. Unlike the case of the bottle, water molecules can evaporate and expand till infinity (well, it's "infinite"  relative to the the space you had in the bottle).

It's pretty much for reasons similar to this that the gold cube in the idealized chamber will completely evaporate.There is enough space for the particles to "comfortably" occupy outside of the solid that they were initially constrained to.

So yes... over time, all the "objects" in your room lose and gain particles over time. Ultimately however, the "object" you once knew will no longer be the same >:)


however numbering the bond 1 means there's ambiguity between it being between 1 and 2 and between 1 and 6

It might appear to you all of a sudden that there's an ambiguity, but actually there's none. Look at these two molecules carefully:

Note that the numbering for the molecule on the left will increase in the anticlockwise direction, while the numbering for the molecule on the right will increase in the clockwise direction. The pi bond in each case is named as a -1-en (and it should, because both the molecules are identical)

You seem to be assuming that the numbering for the molecule on the left also applies on the molecule on the right. But, that is not the case, as shown above.

The disubstituted cyclohexene ring complicates the scenario, but only slightly. Consider the case when functional groups of equal priority are present:

Now, recognize that - even before assigning a name - that both the molecules are different. Thus, their names will also be different. The molecule on the left begins numbering from the top $-\ce{COOH}$ and increases anticlockwise. The molecule on the right begins numbering from the lower $-\ce{COOH}$ and increases clockwise. The first pi bond is assigned a name "-1-en" while the other is assigned "-3-en".

If a lower priority functional group is present instead, like so:

the numbering will increase along the direction of the pi bond, beginning from the top $-\ce{COOH}$. Hence, their names will be 3-hydroxycyclohex-1-ene-1-carboxylic acid and 5-hydroxycyclohex-1-ene-1-carboxylic acid respectively.

A possible way to achieve a "-6-en" is to take equal priority suffix groups on positions 1, 2 and 4, and then place a pi bond between positions 1 and 6, like so:

It is named "cyclohex-6-ene-1,2,4-tricarboxylic acid" giving the suffix functional groups a lowest locant set of "1,2,4", consequently naming the alkene as a "-6-en". Actually, though, this is not 100% correct. A slightly technical detail is missing, which can be found in Loong's answer.


Is it correct to look at plane of symmetry in substituted cyclohexane for determining a meso compound?

Operatively:
A plane of symmetry in a 2-D projection of a cyclohexane is a sufficient condition.
Alternatively, and lengthy, you conduct an analysis involving conformers but not limited to the flipping of the ring, that means you must let the chloromethyl rotate, in your example.

In more details:
The answer to your title question is YES providing that the molecule isn't blocked in one conformer.
The latter situation requires low T, or the presence of more than one bulky substituent. At room temperature and within the context of common practice and exercises we can be sure that the condition is met.
I suspect that cases in which the situation can rapidly escalates towards complexity exist, as for heavy substitutions alters the conformational distribution, too.
The second question is basically
"How is that/why a plane of symmetry in a 2-D projection of conformers which themselves have no symmetry but are enantiomers does effectively work as a criterion?".
Well, a laconic but solid answer is that there is no other possible outcome.
There must be a formal mathematical way to express this fact; I am thinking of symmetry and projection operations.
Yet, I can't do better than visualise the situation, but this is again following the reasoning and depicting the sketch within the answer you have linked, although mentally.
It is the flipping between enantiomeric conformers that makes the two simultaneously equalise each other, and this necessarily requires the two stereocentres having opposite absolute configuration and being at opposite sides of a symmetry plane of the molecule's projection, otherwise they would forcedly be different isomers and there won't even be this discussion.
You can also think of the structure as having the axial and equatorial substituents in an averaged  intermediate position, as for the two limits can't be resolved.
Indeed, a 2-D projection is in a way a real representation of the molecule - not brutal alteration is operated. It just says what it's supposed to tell, i.e. how one would see the molecule as watched from above the projection plane.
During the flipping, one would see a rapid on-plane oscillation of the ax and eq substituents in an alternate fashion, and this require a plane of symmetry in order for the two to compensate their effects.
To make this visualisation clearer, I've sketched a rudimental figure showing the motion of ax / eq substituents and their average position as ×

Considering that the motion of the substituent can be taken as a handle for the entire conformational flipping, one can see the relevance of a symmetry plane of the projection. In such a picture there is no reason to expect optical activity, as if the molecule would be flat.
However, while the picture just above is a nice visual, the optical inactivity originates from the simultaneous presence of both enantiomers, and this should be right the reason for deprecating the old term internal raceme in favour of meso compound.


Why does LiAlH4 reduce an amide to an amine, but only reduce a ketone/aldehyde to an alcohol?

You've got it correct in your comment with the $\ce{N}$ lone pair assisting. The reason why the $\ce{N}$ lone pair kicks oxygen out, and not the other way round, is because oxygen-based groups are better leaving groups than nitrogen-based ones under the reaction conditions. (Just as a rough guide, you could compare $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 15.7$ and $\mathrm{p}K_\mathrm{a}(\ce{NH3}) = 38$.)

(Don't worry about how I've drawn the metals coordinating to oxygen - the exact details aren't important and sometimes aren't even fully understood.)

In the aldehyde reduction, there is no lone pair to assist in the expulsion of oxygen. So the intermediate alkoxide just sits there, until you add water or acid:

The aqueous workup serves two purposes: firstly to reprotonate the product (for example, with the aldehyde reduction, your final product is an alkoxide ion and adding water gets you the alcohol), and secondly to destroy any unreacted $\ce{LiAlH4}$. In fact, the aqueous workup is also carried out in the reduction of the ketone or aldehyde - we often just omit it because it's so common and it's tiresome to have to write it on every reaction we do. So even though it's not there, it's still implied.

If you wanted to reduce the carbonyl group of a ketone/aldehyde to a methylene group, the three most well-known methods available are the Wolff-Kishner reaction, the Clemmensen reduction, and the Mozingo reduction.


Carbohydrates: What does "typical nature" mean here?

Looks like your book's come up with a "localized" version of the D-L notation.

$\ce{D-L}$ notation in brief: A $\ce{D}$-sugar has the $\ce{-OH}$ group (on the stereogenic carbon atom on the parent chain, furthest away from the aldehyde group) facing right in the Fischer representation.
Likewise an $\ce{L}$-sugar has the $\ce{-OH}$ group facing left.

What your book's attempting to do is extend the $\ce{D-L}$ classification to individual carbon atoms in the parent chain.

The "3rd($\ce{L}$)" bit is supposed to mean, "3rd carbon atom has an $\ce{L}$-type configuration".

The "2,3($\ce{L}$)" bit is supposed to mean, "2nd and 3rd carbon atoms have an $\ce{L}$-type configuration".

You can figure out the rest ;-)

This appears to be some sort of memory aid to help you remember the configuration of the various kinds of 6-carbon sugars. The "typical nature" term was, in all likelihood, concocted by the authors (perhaps they thought it sounds better than "mnemonic"?)


Since most thermal greases are ceramic and/or metal suspended in a silicone or organic oil, the temperature the grease remains stable at will likely primarily be determined by the properties of that oil -- either its ease of evaporation, or decomposition temperature (or, I suppose, its flashpoint, but that seems unlikely).

As Ben Norris says in the comment above, your best bet is to contact the manufacturer or check their data sheets: for common CPU uses, most manufacturers will give you stable and extreme temperature ranges which are likely to be fairly conservative (for example, here), and they may be able to help with more esoteric data if you're using it for other purposes.


Does rubber use one or two sulfur atoms to crosslink?

According to the Polymer Database,
the sulfur bridges formed in the vulcanization process of natural latex are usually comprised of 4-6 sulfur atoms, although this polyssulfide chain can be shorter if the temperature is increased.

  http://polymerdatabase.com/polymer%20chemistry/Vulcanization.html


How do you name a bicyclo compound with a fused benzene ring?

Beside systematic nomenclature about organic structures, there are other systems in use.  One of them is the by Hantzsch and Widman which departs from a number of stem structures.  Some of these archetypical motifs are named systematically, others (still) retain their traditional name which does say little  to you about the structure (e.g., pyridine) until you learn them.  One (incomplete) compilation of these names, for example, is this one.
This being said, you either look up, or over time (i.e., training) recognize that the parent structure of your molecule is indene:

except that the double bond in the five-membered ring of the structure you showed is completely saturated.
You remember reactions on benzene, and recall about the importance to denote the sites in organic structures.  The approach to use numbers equally exists for larger compounds, too; a few positions are labeled.  While there are cases where you may shorten 1H-indene to indene, this no longer is the case to describe unambiguously the compound shown by you, which is 2,3-dihydro-1H-indene.


How does aminoguanidine interfere with the formation of advanced glycation endproducts?

I would expect the N-amino group of aminoguanidine to react with 3-deoxyglucosone at either of the adjacent, reactive carbonyl groups to form a stable guanidinylimine derivative (analogous to a hydrazone). This derivative, unlike 3-deoxyglucosone itself, would not be able to react with free amino groups on proteins and continue on the pathway to AGEs, but rather would be excreted as is.


SCF procedure performed for some quantum-chemical methods, but not others. Why?

Most computation methods write the wavefunction($\Psi$) in terms of Slater determinants ($\Phi$)

$\Psi = \sum_i^L c_i\Phi_i$

which in turn are made from MOs ($\phi$), which are linear combinations of basis functions ($\chi$)

$\phi_i = \sum_\mu^K C_{\mu i}\chi_\mu$

The molecular energy depends on the MO coefficients ($C_{\mu i}$) in a non-linear fashion, so if they are variationally optimized then you need to iterate (i.e. SCF). This is the case for HF (where $L$ = 1 and $c_1$ = 1) and MCSCF (where both $\{c_i \}$ and $\{C_{\mu i} \}$ are optimized).

For methods like CI, perturbation theory, and coupled cluster, the MO coefficients are not changed, so you don't need an SCF.


Can I use my extech pH 60 meter to measure soil pH if so how?

Take a soil sample and mix it with distilled water. The most common ratio used for soil-water pH is 1:1 soil:water.
Stir the soil and solution and wait 15 minutes to 1 hour to set the slurry.
Measure the pH (check the temperature first).


Why do SN2 reactions of alkyl halides proceed differently with KCN and AgCN?

The cyanide anion is an example of an ambident nucleophile, these are nucleophiles which, usually due to delocalisation, are able to attack an electrophile from two or more atoms. In the case of the cyanide anion reaction from carbon leads to nitriles, whilst reaction from the nitrogen leads to isocyanides.

To give the full IUPAC definition from the Gold Book:

  A description applied to a chemical species whose molecular entities each possess two alternative and strongly interacting distinguishable reactive centres, to either of which a bond may be made in a reaction: the centres must be connected in such a way that reaction at either site stops or greatly retards subsequent attack at the second site. The term is most commonly applied to conjugated nucleophiles. [...]

To answer your actual question about why $\ce{KCN}$ forms nitriles, whilst $\ce{AgCN}$ forms isocyanides, March's Organic Chemistry (7th ed., p 449) gives a pretty concise answer in which we consider the extent to which the cyanide anion is associated with the counter-cation in solution.

  All negatively charged nucleophiles must of course have a positive counterion. If this ion is $\ce{Ag+}$ (or some other ion that specifically helps in removing the leaving group), rather than the more usual $\ce{Na+}$ or $\ce{K+}$, then the transition state is more $\mathrm{S_N1}$ like. Therefore the use of $\ce{Ag+}$ promotes attack at the more electronegative atom. For example, alkyl halides treated with $\ce{NaCN}$ generally give mostly $\ce{RCN}$, but the use of $\ce{AgCN}$ increases the yield of isocyanides ($\ce{RNC}$).

                           [See footnote]

This, of course, is not a complete picture, as solvent effects greatly change the reactivity. In protic solvents (MeOH, to give a possible example) the most electronegative atoms (nitrogen in this case) will be solvated to a greater extent via H-bondng. In aprotic solvents (THF/DMF, to give possible examples) the anion isn't solvated quite so much (on either atom of the nucleophile), but the cation is somewhat more solvated, leaving the nucleophile more able to attack (via the most electronegative atom as this is the most nucleophilic in the absence of other factors).

Footnote: As an aside at this point, it's important to realise that these are not fully covalent bonds (in $\ce{AgCN}$ and $\ce{KCN}$), despite having some level of covalent character. We're somewhat bastardising the Winstein model of nucleophilic substitution in which we define $\mathrm{S_N1}$/$\mathrm{S_N2}$ based upon the separation between the cation and the anion at the point when substitution takes place, with the two mechanisms being extremes of a continuum:


How would one make a non-toxic pen out of osmium?

Osmium alloys with many metals, so it could be covered by an alloy. Osmiridium, an alloy with iridium, is even harder than pure osmium.

It would seem possible to dip the osmium pen blank in an iridium or osmiridium bath (at rather high temperature) to produce a long-lasting casing, rather like dipped chocolate cordials.

If you accept this answer, I'd be glad to receive an award of one of these pens. ;-)


Name of toxic liquid product that can be cleared with flocking agent

If your impurity is tritium, it can not be purified, whatever the method used. If your impurity is another substance, it can be purified by adding aluminum sulfate. Aluminum sulfate is slowly hydrolyzed in dilute solutions. And this phenomena produces a light and insoluble substance called aluminum hydroxide $\ce{Al(OH)3}$. The particles of this substance look a bit like snow flakes and have the property of fixing impurities on its surface, between its branches. So after a couple of hours, this aluminum hydroxide will be deposited on the ground of the river or the lake, with the impurities fixed inside its particles.


How does aniline react with HNO3 in the absence of sulfuric acid?

In the absence of sulfuric acid, aniline does react with nitric acid to form anilinium nitrate. It is observed as an intermediate product during the production of diazobenzene nitrate (benzenediazonium nitrate). But the species is too unstable for mass production and cannot be stored for long hours. Here is the description:

  20 g of aniline are placed in a beaker, well cooled, and “boiled-out”
  nitric acid, previously diluted with half its volume of water,
  carefully added till the mixture sets to a thick crystalline paste –
  aniline nitrate.[...]

Caution: The synthesis is not to be carried out due to high explosive nature of diazobenzene nitrate. Explode in dry state.


Which of the following is the most basic compound?

I agree with Doctor Insult's argument that "the stronger the base, weaker is the conjugate acid." In that sense, higher the $\mathrm{p}K_\mathrm{a}$ of acid, weaker is the acid, and hence, stronger is the conjugate base. The $\mathrm{p}K_\mathrm{a1}$ values of $\ce{HBr}$, $\ce{H2S}$, $\ce{H3P}$, and $\ce{H4Si}$ are -8.7, 6.89, 27, and ~35, respectively (Ref.1). Therefore, the order of the strength of their conjugate bases should be:
$$\ce{H3Si- \gt H2P- \gt HS- \gt Br-}$$
Hence, strongest base is $\ce{H3Si-}$.

Note: Ironically, this is the same decreasing order of electronegativity of center atoms $\ce{Si, P, S, Br}$: $1.90 \lt 2.19 \lt 2.58 \lt 2.96$, respectively.

References:

William L. Jolly, In Modern Inorganic Chemistry; 1st Edn.; McGraw-Hill, Inc.: New York, NY, 1984, p 177.


Striking examples where Kohn-Sham orbitals clearly have no physical meaning

When people say that Kohn-Sham orbitals bear no physical meaning, they mean it in the sense that nobody has proved mathematically that they mean anything. However, it has been empirically observed that many times, Kohn-Sham orbitals often do look very much like Hartree-Fock orbitals, which do have accepted physical interpretations in molecular orbital theory. In fact, the reference in the OP lends evidence to precisely this latter viewpoint.

To say that orbitals are "good" or "bad" is not really that meaningful in the first place. A basic fact that can be found in any electronic structure textbook is that in theories that use determinantal wavefunctions such as Hartree-Fock theory or Kohn-Sham DFT, the occupied orbitals form an invariant subspace in that any (unitary) rotation can be applied to the collection of occupied orbitals while leaving the overall density matrix unchanged. Since any observable you would care to construct is a functional of the density matrix in SCF theories, this means that individuals orbitals themselves aren't physical observables, and therefore interpretations of any orbitals should always be undertaken with caution.

Even the premise of this question is not quite true. The energies of Kohn-Sham orbitals are known to correspond to ionization energies and electron affinities of the true electronic system due to Janak's theorem, which is the DFT analogue of Koopmans' theorem. It would be exceedingly strange if the eigenvalues were meaningful while their corresponding eigenvectors were completely meaningless.


Aspartame is (a methyl ester of) a dipeptide. Therefore, you could use the nomenclature of amino acids to find a descriptive name for aspartame.

However, since you are asking about the correct order of the individual groups, apparently, you want to generate the fully systematic name according to nomenclature for organic chemistry.

Since there are various groups, you have to identify the principal characteristic group in accordance with the seniority of classes and suffixes first. In this case, the characteristic group with the highest seniority is the carboxylic acid group.

The principal characteristic group is chosen for citation at the end of a name by means of a suffix or a class name (or implied by a trivial name). Since there is only one carboxylic acid group in aspartame, the principal characteristic group also decides the choice of a principal chain. Thus, the parent structure of aspartame is butanoic acid.

The remaining substituents are expressed as prefixes in the name in alphabetical order. In this case, the remaining substituents are one amino group, one oxo group, and the complex substituent group [(2S)-1-methoxy-1-oxo-3-phenylpropan-2-yl]amino.

Therefore, the complete systematic name of aspartame is (3S)-3-amino-4-{[(2S)-1-methoxy-1-oxo-3-phenylpropan-2-yl]amino}-4-oxobutanoic acid.


Thermal stability of charged and uncharged alpha helices (the concept of alpha-helix capping)

This might relate to the concept of 'capping' alpha-helices, where the placement of beneficial charges at either end of a helix is shown to increase the overall stability of the protein.  See the article1 published in Nature.
In general, by introducing preferential charges at the ends of alpha-helices you stabilize the inherent dipole moment created by the alignment of amide groups in the protein backbone.

Here is a different citation (found here):

In 1988, Richardson and Richardson observed that certain amino acid
residues are stastically favored as the amino-terminal caps for the
alpha helices in 45 globular proteins. The introduction of a
negatively charged side chain at the N-cap, which can neutralize the
partial dipole created by the unpaired amide protons, has been shown
to increase stability in T4 lysozyme.

I also believe the figure in the question is a bit misleading - as the illustrated charges may be intended on side-chain groups. I therefore assume that this helix is a part of a protein, and that the illustration simply wants to highlight the fact that you place charged side-chain groups on the ends of the helix (i.e. capping) within a protein structure.
Reference

Serrano, L., Fersht, A. Capping and α-helix stability. Nature 342, 296–299 (1989). DOI: 10.1038/342296a0


What is the condensation point of water?

You're right - water vapor can exist at temperatures such as $\pu{50^\circ C}$. This is a phenomenon known as evaporative cooling, where molecules of water with higher kinetic energies tend to "release" themselves from the system, and as a result, less and less water molecules are held within that system.

Temperature is a sort of proxy for kinetic energy, and vice versa. Reference, also, the Maxwell-Boltzmann distribution, which reveals that at $\pu{100^\circ C}$, for example, not all molecules in the system possess the kinetic energy that they would at $\pu{100^\circ C}$, but rather the average exists at that temperature.

To answer your question, never in a real scenario.


Removing unreacted Mg metal in a Grignard reaction

Pure water would certainly be a bad idea, but a 1% solution of 1 M hydrochloric acid in ether would probably work without a vigorous reaction.


The chlorofluorocarbon reaction in the lab?

The catalytic stratospheric ozone destruction by CFC's and other compounds is almost certainly too complex to fully replicate in a laboratory with current technology.  The only laboratory of proper scale and conditions is the earth's atmosphere as a whole (no pun here, seriously ;).

Even the 1995 Nobel prize winners Crutzen, Rowland and Molina had made serious omissions from their original work in the 70's that are now known to be critical to the mechanisms of stratospheric ozone depletion.  While Crutzen is pretty much a pure theoretician, Molina and Rowland did much of their work in the laboratory and combined that work with atmospheric measurements.  The bottom line is that only pieces of the whole process can be controlled in a single or set of experiments.  One important factor, for example, that was omitted from their original work was the role played by condensed-phase surface process in mediating catalysis by active chlorine.  Specifically, without the different types of polar stratospheric clouds (i.e. water ice, nitric acid trihydrate,  liquid supercooled solutions of sulfuric acid, nitric acid and water, and other complex matrices) catalytic ozone destruction does not occur. So, although the net of their predictions in the 70's was correct, it was the real-world observations decades later, not laboratory confirmations, that sent them to Stockholm for their prize.

In 2010, an article was published called "Key compound of ozone destruction detected; Scientists disprove doubts in ozone hole chemistry.", Karlsruhe Institute of Technology (KIT).  ScienceDaily. ScienceDaily, 22 July 2010, excerpt available here.  The summary states:

  For the first time, scientists in Germany have successfully measured
  in the ozone layer the chlorine compound ClOOCl, which plays an
  important role in stratospheric ozone depletion. Doubts in the
  established models of polar ozone chemistry expressed by American
  researchers based on laboratory measurements are disproved by these
  new atmospheric observations.

This represents an even more recent case in which rigorous laboratory attempts to simulate stratospheric ozone depletion processes failed in the face of experiments and measurements made in-situ in the "real world" laboratory.

The article further points out:

  The ozone hole above the Antarctic and the destructive role of
  chlorofluorocarbons (CFC) and their decomposition products have become
  a synonym of both global environmental problems and their solution by
  concerted agreements worldwide. Scientific fundamental research into
  ozone chemistry of the atmosphere was the basis of international
  agreements, such as the Montreal Protocol of 1987, which has put
  limits on CFC production. The success of the political implementation
  of these scientific findings is reflected by the fact that the
  chlorine content of the atmosphere and, hence, the ozone destruction
  potential recently started to decrease slowly.

So, the upshot here is that it was primarily in-situ experiments and measurements that led to predictions of catalytic chlorine mediated ozone destruction being later confirmed by measurements. The same was true for formulation of a solution to the problem, which again in-situ measurements confirmed.

In short, laboratory experiments have repeatedly been outperformed by in-situ experimentation and observation in the case of stratospheric ozone depletion studies.  The same is true for many complex, global chemical and physical process; it simply is not always feasible to re-create a system of large scale and complexity in a laboratory.


A reversible exothermic reaction between a molecule in an excited triplet state and a reducing agent

We're looking at a case of PET (photoinduced electron transfer) with electron back transfer between an (electron) acceptor A and a donor D here.

For organic dyes, the ground state typically is a singlet state ($S_0)$, excitation of A yields an excited singlet state (typically $S_1$).

$\ce{A ->[h\nu]\  ^1A^{\ast}}$

Quinone based dyes (or aza-BODIPY derivatives) often have a high triplet quantum yield, the following ISC (intersystem crossing) to a triplet state is efficient then.

$\ce{^1A^{\ast} ->[ISC]\  ^3A^{\ast}}$

In the excited (triplet) state, electron transfer from a (ground state) donor may occur by filling the lower SOMO of the excited species:

$\ce{^3A^{\ast} + D -> A^{\bullet-} + D^{\bullet+}}$

Is the acceptor radical anion still in a triplet state? No, since there no longer an electron pair with opposite spins.

Q: So, what is this good for?

A: By electronic excitation, we managed to facilitate a redox reaction that would not have been possible in the ground state and generated a radical ion pair.

Radical ions are reactive intermediates.

For a spectroscopist, it can be favourable if the only reaction of the radical ion pair is  electron back transfer:

$\ce{A^{\bullet-} + D^{\bullet+} -> A + D}$

This regenerates acceptor and donor in their former spin states, which - as mentioned before - are typically singlet states in the case of organic molecules.

Practically, this means that no permanent bleaching occurs. You can excite the sample with laser pulses again and again to measure the transients in time-resolved uv-vis spectrosopy.
+
From a preparative perspective, this is the worst case: there is no product being formed.

We have examined more productive cases of PET in the context of PET-initiated radical cyclizations of isoprenoid polyalkenes a while ago and published about these, e.g. in J. Phys.Chem A, in J.Chem. Soc., Faraday Trans. and in Pure Appl. Chem.. (The last two should be freely accessible).

In the course of these reactions, an alkene is oxidized to the corresponding radical cation by the excited state of an electron acceptor. The radical cation
traps a nucleophile (water, alkanol), immediately deprotonates and thus form a radical, which undergoes 5-exo-trig cyclization to a stabilized dicyanomethyl radical. The latter is reduced to its corresponding anion by $A^{\bullet-}$ and eventually protonated to yield the product (as a pair of separatable diastereomers).


Is it correct to consider dissolving HCl gas in water as oxidation –reduction process?

Given the dissolution of $\ce{HCl}$ gas in water:

$$\ce{HCl(g) + H2O(l) <-->H3O+(aq) +Cl-(aq)}$$

The oxidation state of all hydrogen atoms is +1 and that of all chlorine atoms is -1. Thus, nothing has been oxidized or reduced, and no redox reaction has taken place.


Why do some chemical reactions require many steps?

The number of steps required to model a reaction really depends on what time scales you are interested in looking at. Let's take the process of burning natural gas with air. If you only care about the final products of the combustion after a really long time, like let's say hours, then a single step reaction would be a perfectly good model. In fact, in situations where you only care about the global heat release and maybe the speed of the reaction at one particular condition, single-step or few-step reactions are just fine.

But let's say you are really interested in what's happening. When we look at a global reaction, let's say:

$$ O_2 + 2H_2 \rightarrow 2H_2 O $$

We'll assume this is in air, but the nitrogen doesn't really react. Nice and simple, right? Well, for the reactions to occur, you need some $O_2$ molecules to collide with $H_2$ molecules. But it's not enough that they collide. They need to collide with enough energy to break the bonds between them. And it's possible that one of the molecules breaks at lower energy than the the other. When they break, they break into 2 O's or 2 H'2. So our simple reaction really starts out as:

$$ O_2 + M \rightarrow 2 O + M $$
and
$$ H_2 + M \rightarrow 2 H + M $$

where $M$ is any other molecule. Okay, so now we've got a mixture of $H$, $O$, $H_2$, $O_2$, and $N_2$ floating around and bumping into each other. If they bump together with enough energy, we can start forming other things. Things like $HO_2$, $OH$, $H_2O_2$ and of course, $H_2O$.

Because these reactions rely on collisions, you need to form some things before you can form other things. So before an $O$ and an $H$ can collide to form $OH$, the $O_2$ and $H_2$ have to collide with other things and break apart. So there's finite time scales with all of the reactions, and they tend to cascade through a series of reactions before reaching the final product. If you are looking at time scales similar to those of the reactions, then these details really matter.

Hydrogen reactions are super simple. Imagine how complex organic molecules are. Even things like methane or propane have hundreds or thousands of steps to go from the complex molecule through all of the radicals to the final species. And the time scales the reactions occur over are all based on how frequently the collisions between them occur, and how much energy those collisions have.


What are the benefits of using zero capillary voltage for discarding fractions in LC–ESI–MS/MS?

I think there is no perfect answer to this situation. In general when the flow is stopped to any detector in an HPLC experiment, there is a change in baseline, and when the flow is re-started, ghost peaks appear in chromatogram. I am talking about optical detectors such as UV, fluorescence, refractive index etc. Same thing happens in MS and perhaps LC ESI-MS is not an exception, in the sense that diverting the flow can lead to the appearance of so-called "ghost" peaks or raised baseline. I have seen that in GC-MS experiments where a change in flow causes weird baseline shifts. This is an interesting phenomenon worth investigating.
So the authors suggest that let us continue the flow, but let us stop the ionization process so that the junk does not go inside the quadrupoles and stay there forever. It is always easy the clean the external ESI system but the internal system needs professional service.
Someone talks about these problems in this forum Diverter valve issues