text:
Balancing a redox equation with only one product
Is it possible to balance this equation using the half reaction method?
$$\ce{Au + Cl2 + HCl⟶HAuCl4}$$
I know that the equation balanced would be:
$$\ce{2Au + 3Cl2 + 2HCl⟶2HAuCl4}$$
However, how would the half reactions end up?

You actually have different possibilities of setting up half-reactions. One way is to note that formally, gold goes from a $\pm 0$ to a $\mathrm{+III}$ oxidation state, thus it can be written as:
$$\ce{Au -> [Au^3+] + 3 e-}\tag{Ox}$$
$$\ce{Cl2 + 2 e- -> 2 Cl-}\tag{Red}$$
Which gives you:
$$\ce{2 Au + 3 Cl2 -> 2 [Au^3+] + 6 Cl- -> 2 AuCl3}\tag{Redox}$$
And, if additional $\ce{HCl}$ is present:
$$\ce{AuCl3 + HCl -> H[AuCl4]}\tag{Aurate}$$
Since $\ce{HCl}$ does not participate in the redox part of the reaction, it does not occur until after the redox part.
Naturally, $\ce{[Au^3+]}$ is not isolateable under these conditions, hence the square brackets.


text:
Is it the Na+ or the Cl- that gives table salt its taste?
What gives it that salty taste? I'm wondering if it's the $\ce{Na+}$ or the $\ce{Cl-}$. I know that $\ce{KCl}$ also tastes salty, and other salts have different flavours,  but where does the saltiness in $\ce{NaCl}$ come from? I mean, it makes sense that it's the $\ce{Cl-}$ ion, but it could also be that $\ce{K+}$ and $\ce{Na+}$ have similar flavours.

Mostly Cl-. Taste some baking soda, it is not quite salty.

The salt receptors are sodium ionic channels, i.e. peptides which are selectively permeable for Na ions [1].
Ionic current causes receptor activation and neural responce.
Anion plays minor role here, imposing upon salty taste. The result may appear acidic, bitter and even sweet.
For example, I believe that sulphates are overally bitter.
Carbonates or alkali probably just burn the receptors and may have random taste.

Sodium cyanide is reported to taste like common salt, I am not inclined to find out if the story is true or not,

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text:
How is the numbering in phenylpropenes?
Just wondering why the name of this compound is 3-phenylpropene? I understand the concept of phenyl groups but I'm confused where the 3 came from, because if it is referring to the double bond in propene, then shouldn't it be numbered 2?

3 is referring to phenyl. The double bond is at 1. In propene it can't be otherwise.

Obviously, there are three carbons in propene to which a phenyl group might be attached. Let's have a look and name the compounds:

1 with the phenyl group at C-2 of propene is 2-phenylpropene, also known as $\alpha$-methylstyrene.
2 with the phenyl group at C-1 of propene is 1-phenylpropene. To be exact, the compound drawn is (E)-1-phenylpropene. E is for entgegen, that is on opposite sides of the double bond.
3 is 3-phenylpropene, also known as allylbenzene.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the ring or the chain can be the senior parent structure. For the preferred IUPAC name, the ring is selected as the senior parent structure. In general nomenclature, however, the context may favour the ring or the chain.
For the name that is given in the question, the propene chain is taken as the senior parent structure. Accordingly, the benzene ring is considered to be a substituent (i.e. a phenyl group).
The propene chain is numbered from one end to the other, the direction being chosen so as to give the lower locants to the double bond. (Only if a choice remains, low locants are given to prefixes that indicate substituent groups.) Thus, the numbering of the propene chain leads to prop-1-ene:

Accordingly, in this case, the substituent is assigned the locant ‘3’:

Since this substituent is a phenyl group, the complete systematic name is 3-phenylprop-1-ene.

However, note that this is not the preferred IUPAC name. As mentioned above, the ring is selected as the senior parent structure for the preferred name. Therefore, the preferred name in this case is (prop-2-en-1-yl)benzene.

Note that the standalone name “prop-1-ene” for propene (PIN) does not follow P-14.3.4.2 (d).

When a simple substituent like, $\ce{-OH}$, $\ce{-NO2}$ is attached to the benzene ring then benzene is considered as the parent chain. But when a complex substituent like a long ($>6$) carbon chain, or a group with double bond/triple bond or another substituent is attached to the benzene ring, then benzene is considered the substituent, and named as phenyl group.
Here, the propene part is the main chain. Double bond gets more priority than phenyl substituent, so numbering is started from that end. Phenyl group gets number 3. So the more correct name should be 3-phenyl-1-propene. However, writing 3-phenylpropene does not cause any ambiguity (as the numbering is started from the double bond end)

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Why is the fluorescence spectrum of pyrene time-dependent?
I've got a solution of pyrene and I excite it with a laser. If I measure the emission spectrum 1 ns after absorption, a well‐structured emission with a small Stoke’s shift is observed.
However, if the emission is measured with a delay of 100 ns, a broad, featureless spectrum with a much larger Stoke’s shift is obtained.
Is this purely because pyrene has had some time to undergo solvent reorganization and thus the individual excited pyrene molecules have larger differences or is there something else at play?

Did you observe any dependence on the pyrene concentration?

I think that you have rediscovered the emission of pyrene exciplexes :)
As you have correctly assigned, the structured fluorescence spectrum shortly after the pulse is the emission of pyrene monomers.
Given sufficient time (100 ns), an excited molecule of pyrene diffuses towards a pyrene molecule in its ground state and forms an exciplex.
Fluorescence emission from these exciplexes is known to be broad and unstructured.

awesome. it's a 10^-3 M solution of pyrene. 100 ns should be enough for the excited molecule to meet another pyrine molecule.

At this (high) concentration the emission to the red part of the spectrum (relatively speaking) is due to pyrene excimer (one excited pyrene in a face to face dimer with a ground state pyrene). You should be able to see the rise-time of the emission as the excimer is formed if you choose the wavelength carefully. Excimers have a repulsive ground state but a potential well in the excited state. see J. Birks, Rep. Prog. Phys. 1975, 38, 903-974.  Excimers refers to the case when both molecules are of the same type, exciplexes when they are different.


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Why can't electrolysis utilize specific ions for reduction/oxidation
In the electrolysis of water, reduction half equation is $$\ce{2H2O + 2e- -> H2 + 2OH-}$$ with a standard $E$ value of $-0.83\ \mathrm V$.
If $\ce{H+}$ is present in water, why can $$\ce{2H+ + 2e- -> H2}$$ not occur, with potential $0\ \mathrm V$, which is more favourable?

Related: Electrolysis in aqueous solution – which equations to use to predict product at each electrode?

See, this is pretty much the same reaction. The potential is not always exactly what they say in the reference book. It depends on concentrations. The potential of $\ce{2H+ + 2e- -> H2}$ is $0 V$ only in standard conditions, that is, when the concentration of $\ce{H+}$ is 1 mol/L. When it is lower, the potential is different (see Nernst equation). Indeed, $\ce{H+}$ is present in pure water, but the concentration is quite low ($10^{-7}$ mol/L). If you plug that concentration into the Nernst equation, you'll get the same value that is listed for $\ce{2H2O + 2e- -> H2 + 2OH-}$.


text:
Functional difference of Benedict's solution and Fehling's solution
Chemically, Benedict's solution and Fehling's solution are very similar, with copper sulfate as a source of copper (II) ions, sodium carbonate, and sodium hydroxide respectively as alkalis, and sodium citrate and sodium potassium tartarate respectively as chelators. They are also both used to test aldehyde groups which can reduce the copper (II) ions into copper (I) ions, giving a precipitate of $\ce{Cu2O}$. Hence, they are often used in testing for the presence of aldehydes, and by extension, reducing sugars.
Are there any differences in function between the two reagents, perhaps in sensitivity, types of compounds that test positive, etc?

Fehling's solution consists of two parts, Fehling's A and Fehling's B. Fehling's A consists of aqueous copper(II) sulfate, while Fehling's B consists of aqueous sodium potassium tartrate ($\ce{NaK(C4H4O6)}$). Benedict's reagent consists of sodium carbonate, sodium citrate ($\ce{Na2(C6H6O7)}$), and copper(II) sulfate dissolved in water. It's pretty clear now, that the main difference between the two reagents is the complexing agent (citrate in Benedict's and tartrate in Fehling's).
Fehling's reagent was invented by Hermann von Fehling in 1849. You notice that this solution is given as two separate solutions and is only mixed together while performing a test. This is because the complex formed when mixed ($\ce{[Cu(C4O6)2]}$) is kinetically favored, and that $\ce{Cu(OH)2}$ will eventually precipitate out due to being favored thermodynamically. That's why the two solutions are stored, and only mixed during testing.

On the other hand, Benedict's reagent was discovered by Stanley Rossiter Benedict somewhere in between 1904 to 1936. Thus Benedict's reagent was developed after Fehling's was. The citrate complex in Benedict's reagent is much more stable that $\ce{Cu(OH)2}$, so Benedict's reagent doesn't precipitate on standing for long, unlike Fehling's. This is quite advantageous since you wouldn't have to mix two solutions together to perform the test. This is where Benedict's reagent is quite handy.

Both the reagents have a similar way of oxidizing aldehydes. The copper complex oxidized the aldehyde to a carboxylic acid, and while reducing itself from $\ce{Cu^2+}$ to $\ce{Cu+}$
($E^\circ = \pu{0.154 V}$). This reduction potential is sufficient to oxidize aldehydes to carboxylic acids. There is not much difference here though.
Neither of the reagents can oxidize aromatic aldehydes, due to the fact that they lack α-hydrogens, and thus oxidizing them requires stronger oxidizers (like Tollens’ reagent).
In short, Benedict's reagent is kind of like an "improved" version of Fehling's reagent, which is much more stable due to the difference in the complexing agent.

I have thought citrate complexes prefer chelation by 2 carboxyls and the hydroxyl, with the remaining carboxyl free, as bonds by all 3 carboxyls leads to unfavourable 7-member rings,


text:
What happens when prop-2-en-1-ol reacts with concentrated hydrogen bromide
What happens when prop-2-en-1-ol (allyl alcohol) reacts with concentrated hydrogen bromide?
My Idea: I thought that in the first step protonation of the hydroxyl group will take place followed by formation of a carbocation. Then the attack of bromide ion from hydrogen bromide takes place and that would give me 3-bromoprop-1-ene. Also, I thought since the double bond still remains Markovnikov addition at the double bond might also take place.
But the reaction results into something different
Here is a picture of the above reaction:

The answer has been given as option C. Could someone point out where I might be going wrong or how should the reaction actually proceed?

Please don't use MathJax in titles. In particular you don't need it to type the name of a chemical.

NOOO! Stop proposing primary (even allylic carboncations)! This is a normal hydrobromination of a alkene. It should follow anti-Markovnikov rules.

@Zhe Hydrobromination of alkenes follow Markovnikov rules. Furthermore, $\ce{HBr}$ is a strong acid and perfectly capable of protonating hydroxy groups. Whether the mechanism happens to me $\mathrm{S_N2}$, $\mathrm{S_N2'}$ or $\mathrm{S_N1}$ is of little to no importance.

Ugh. Undo. I learned the history wrong. Markovnikov addition it is... https://en.wikipedia.org/wiki/Morris_S._Kharasch#Proposal_for_anti-Markovnikov_addition:_The_peroxide_effect Thanks, @Jan.

Organic Syntheses, Coll. Vol. I, pg. 27, Gilman & Blatt, ed., describes the conversion of allyl alcohol to allyl bromide with 48% HBr (from Br2, SO2 and water) and H2SO4 on a 4 mole scale with a 92-96% yield.

Rather than trying to rationalise what the question wants to know, I ran a quick SciFinder search. One of the first results was the following reaction, reported for the perdeuterated allyl alcohol by Thiem, Mohn and Heesing in 1985:[1]

Scheme 1: Reaction of perdeuterated allyl alcohol with $\ce{HBr}$. All implicit hydrogens in this scheme are deuterium atoms.[1]
Thiem, Mohn and Heesing do not go into details regarding the mechanism, but as I commented it really doesn’t matter whether the reaction follows a $\mathrm{S_N1}$, $\mathrm{S_N2}$ or $\mathrm{S_N2'}$ mechanism, since the final product will always be the same. In any case, the first step will be protonation of the hydroxy group to turn it into a better leaving group.
Another one of the first results was the following, reported by Karki and Magolan in 2015:[2]

Scheme 2: Reaction of normal allylic alcohol with $48~\%\ \ce{HBr}$ in DMSO/chloroform.[2]
Their proposed mechanism includes the in situ creation of $\ce{Me2S^+-Br}$ which will then react with the double bond giving a bromonium ion; this will then be captured by a second bromide to give 2,3-dibromopropan-1-ol.
I was unable to find single-step reactions of allyl alcohol with hydrogen bromide leading to either 2-bromopropan-1-ol (which would imply a single Markovnikov addition of $\ce{HBr}$ onto the double bond), 1,2-dibromopropane (which would imply Markovnikov hydrobromination of the double bond of allyl bromide) or 1,3-dibromoprop-1-ene (where I have no clue how exactly that would be generated from allyl alcohol in the first place). Funnily enough, you state that this last product, C according to your scheme be the correct answer. I challenge the exam.

References:
[1]: J. Thiem, H. Mohn, A. Heesing, Synthesis 1985, 775. DOI: 10.1055/s-1985-31344.
[2]: M. Karki, J. Magolan, J. Org. Chem. 2015, 80, 3701. DOI: 10.1021/acs.joc.5b00211.

Can you form 1,2,3-tribromopropane via a combination of the two reactions and then eliminate one of the bromides? Not sure why it wouldn't eliminate anything else though...

@Zhe The elimination would require basic coditions to the best of my knowledge; they aren’t available in concentrated $\ce{HBr}$ …

Was thinking more $\mathrm{E}1$ type mechanism and that would probably require refluxing solvent like in example 1...

@Zhe $\mathrm{E1}$ still requires a base of some sort; something you just don’t have in refluxing $\ce{HBr}$ …

I found a patent dating back to the 1930s in which the dichloro homologue of C was produced in almost quantitative yield via Hydrochlorination. Long reaction time (100 hours) at 50-70C. When it was discovered, it was a surprise (they say) but I presume people didn't leave hot-plate stirrers going for days....
With this in mind, I decided to find out if Hydroiodination would yield 1,3-diiodoprope-1-ene (C analogue). Will hydroiodination live up to the theory?
It turns out that it's a rather difficult compound to come by with only 1 synthesis offered and that was hard to find:
To a solution of (Z)-1-iodo-3-methanesulfonyloxy-1-propene (3.40 g, 13.0 mmol) in acetone (18.6 ml) was added sodium iodide (1.94 g. 13.0 mmol) at 0 OC. The mixtute was stirred at RT for 2h. diluted with ether, and filtered The filtrate was washed with 10% aqueous Na2S203 and brine, dried (NazS04). and concentrated. The product was purified by silica gel chromatography (hexane) to give (3.19 g, 83%) as a dark red oil.
If I have missed a simple synthesis then I can only apologize but the stuff doesn't have a CAS. Searching for (E/Z ignored) '1,3-diiodoprop-1-ene' yields almost nothing so try '1,3-diidopropene' which if you think about it, describes it! IUPAC can be no help.


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Dynamic equilibrium of gases

$$\ce{2SO2(g) + O2(g) <=> 2SO3(g)}$$
  The equilibrium constant for the reaction above is $0.13$ at $\pu{830^\circ C}$.
  In one experiment $\pu{2.00 mol}$ of $\ce{SO2}$ and $\pu{2.00 mol}$ of $\ce{O2}$ were initially present in a flask until it reaches dynamic equilibrium. What must the total pressure of equilibrium be in order to have an 80.0% yield of $\ce{SO3}$?

How do I approach this type of problem?  Is it correct to assume that the total pressure of the equilibrium would be the pressure from the both sides of the equation (pressure of $\ce{SO2}$ + pressure of $\ce{O2}$ + pressure of $\ce{SO3}$)? And how do I find these different pressures?
I am thinking of using the method: $K_\mathrm{p}=0.3=\Large\frac{p^2_\ce{SO3}}{p^2_\ce{SO2}\times p_\ce{O2}}$
Hence, to find the respective pressure of the gasses, I attempted the RICE table, (Reaction, initial mol, change in mol, equilibrium point). How can I proceed further ?

The total pressure is equal to the sum of all three partial pressures. You should be able to write convert this information into a relationship that you can substitute into the equilibrium expression.

For an ideal gas mixture the total pressure is the sum of the partial pressures of all the gases in the mixture. To make intuitive sense of this, consider that since it is a gas mixture, all the gases essentially have access to the whole container, and so will homogenously fill it. Each gas wil the exert a certain pressure on the container walls. The sum the pressures from all the contributing gases will then equal the total pressure the container walls experience.
We also have the relationship $$p_\text{x}/p_\text{total}=n_\text{x}/n_\text{total}$$
The potential yield for the rxn is 2 mol ($\ce{SO2}$ limiting reactant). To get a 80% yield, we therefore have 1.6 mol $\ce{SO3}$ at equilibrium. Therefore there will be 0.4 mol $\ce{SO2}$ and 1.2 mol $\ce{O2}$ left at equilibrium. $n_\text{total}=3.2$
$n_\ce{SO2}/n_\text{total}=0.4/3.2=0.125$
$n_\ce{SO3}/n_\text{total}=1.6/3.2=0.5$
$n_\ce{O2}/n_\text{total}=1.2/3.2=0.375$
To find the partial pressures of the individual gases, we combine the ratios above with relationship 1
$p_\ce{SO2}=0.125\times p_\text{total}$
$p_\ce{SO3}=0.5\times p_\text{total}$
$p_\ce{O2}=0.375\times p_\text{total}$
As you know, the equilibrium can be expressed with partial pressures, and is constant for all changes in pressure. $K_\mathrm{p}=p_\ce{SO_3}^2/p_\ce{SO_2}^2p_\ce{O_2}=0.13$  at 830 Celsius
Plugging in the partial pressures and solving for p(total):
$(0.5x)^2/((0.125x)^2\times0.375x)=0.13$
$\implies x=328.205$ (The unit can be given by the units of the eq. constant.)


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Nuclear reactions involve naked nuclei only. True or false?
In nuclear reactions are all the species involved, naked nuclei that is, are they all atoms that have been completely stripped of their electrons? If not, then shouldn't an alpha decay result in the formation of an anion and shouldn't a beta decay result in a cation?

Alpha decay results in an energetic particle leaving the nucleus, and fast enough that it is unlikely to pick up the necessary electrons to be neutral. So, yes, the atom left behind has two spare electrons (plus/minus any ejected by interactions with the departing alpha), and will rapidly give at least one up. But, the question is: what are you really asking about?

If this process goes on and there is accumulation of charges taking place shouldn't it be evident in the form of some drastic property changes of the material for example, a solid specie decaying should break out of its solid state due to the building repulsive interaction in the atomic level.

Nuclear reactions involve energies far greater than anything in chemistry. Whether or not the nuclei are stripped of their electrons, does not matter at all (except for electron capture, of course; to capture an electron, you must have one around in the first place).
So yes, a beta decay of a neutral atom does indeed result in a cation. As for alpha decay, here we have a massive particle leaving, so the remaining nucleus would be shoved in the opposite direction with considerable force, no doubt enough to break it free of any chemical bonds and strip it of quite a few outer electrons. If not for this factor, the process would leave behind an anion.

Will the recoil of the parent nucleus always release the exact number of electrons ? If so, what governs this?

The recoil will release some random number of electrons, depending on the energy of the alpha particle, mass of the remaining nucleus, ionization energies of the element, and who knows what else. There is nothing "exact" about it, whatever that might mean.

Isn't  there a possibility that the post-decay product is not electrically neutral? Hence,If this process goes on and there is accumulation of charges taking place shouldn't it be evident in the form of some drastic property changes of the material for example, a solid specie decaying should break out of its solid state due to the building repulsive interaction in the atomic level.

The product is almost certainly not neutral, but as soon as the dust settles down, all runaway charges rush towards each other via the phenomenon known as electric current, and neutralize everything pretty soon (or not so soon, depending on the material). That being said, creeping change of nearly all properties in radioactive materials is a known effect, and a huge problem in certain technologies.


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Derivation of Ideal Gas Law with Boltzmann's Constant
I was reading about the kinetic theory of gases and I came to know about this particular equation which piqued my interest.
I know about the ideal gas law which gives us $pV = nRT$.
But this one contained Boltzmann's Constant and it goes like this $PV = Nk_\mathrm{B}T$, where  $k_\mathrm{B}$ is the Boltzmann constant and  $T$ the absolute temperature.
Could I get some insights to how this form of the ideal gas law wasderived?
And how was the value of $k_\mathrm{B}$ and $R$ determined?

$R$ is defined as $k_\mathrm{B}N_\mathrm{A}$ where $N_\mathrm{A}$ is Avogradro's number. Notice that these forms are equivalent once you realize that $N = N_\mathrm{A}n$.
I think $k_\mathrm{B}$ is measured experimentally. I learned the derivation via statistical mechanics.

Both sides of the equation specify the total kinetic energy of the closed system. The left side in terms of the aggregate macro measurables of pressure and volume. The right side in terms of the aggregate average of micro molecular mass and squared velocity. The Boltzmann constant links average molecular kinetic energy to the absolute temperature scale. The zero of the absolute temperature scale is chosen by extrapolation to zero kinetic energy and cant be reached in practice.

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text:
RCOH and RCHO Nomenclature
In terms of nomenclature, is there a difference between RCHO and RCOH when referring to an aldehyde?

$\ce{RCHO}$ is preferred. The problem with $\ce{RCOH}$ is that it gives the misleading impression that the hydrogen is attached to the oxygen. In fact, I don't believe I've ever seen a reference in literature to an aldehyde using the latter representation.

The Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), include recommendations for contracted labels (labels representing more than one non-hydrogen atom) in structure diagrams. In particular

GR-2.3.8 Ordering of multiple symbols attached to the same atom
  If a single atom is bound to both hydrogen atoms and non-hydrogen atoms, any hydrogen atoms should be presented adjacent to the first atom, followed by the others, reading outward from the bond.

Therefore, a contracted label representing an aldehyde group should be written as $\ce{-CHO}$, since the carbon atom is bound to a hydrogen atom and a non-hydrogen (oxygen) atom, and the hydrogen should be presented adjacent to the first (carbon) atom, followed by the other (oxygen) atom.


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Does the distance between the atoms and arrangement of a crystal affect malleability?
I'm designing an aircraft, and certain sections need to hold under a lot of stress. I know that it is the ways the metal crystallizes that effects the malleability, but does the distance between the atoms/ crystal arrangement also affect malleability?

This is a very broad question that veers rapidly away from chemistry and deeply into metallurgy and materials science. Crystal structure, grain size, alloying elements, multiple phases, annealing time, quenching and quench rate, etc., etc. all come in to play.

In general, malleability is determined by how easily metal crystals can be moved around. Anything that gets in the way, such as another substance blocking crystal movement, makes the metal less malleable.
For example, pure iron is malleable, but add some carbon to molten iron, and the carbon precipitates out on cooling, blocking crystal movement and creating much harder and more brittle carbon steel. Think of it as a two-phase material.


text:
What is the difference between activity and reactivity when discussing elements?
I screwed up on a true/false question that stated that lithium is the most active alkali metal. I answered false, because I remembered that the lower you go in this group, the more explosive the element is.
It seems, however, I was thinking of reactivity and not activity. I know about the activity series, and it's something I wouldn't like to memorize, but I would a deeper explanation on the difference between activity and reactivity. Why do we have these two terms?

see http://chemistry.stackexchange.com/questions/53907/can-someone-explain-the-unintuitive-ordering-in-the-metallic-activity-series

Activity is an ordered series, showing which element will replace another element. For example, in a solution of copper sulfate in water, a piece of iron quickly becomes coated with copper as the iron displaces copper from the solution because iron is more active. No explosion took place.
$\ce{Fe(solid iron) + CuSO4 (dissolved in water) -> Cu(solid copper) + FeSO4 (dissolved)}$
Reactivity is less clearly defined. In general, it means how easily some chemical change takes place. For example, chlorine, oxygen and sodium are all very reactive, easily attaching to other elements or replacing elements in compounds. Argon and helium are not reactive, and rarely, if ever, combine with other substances. Nitrogen triiodide, $\ce{NI3}$ and acetylene, $\ce{C2H2}$, are very reactive, but in a way opposite that of sodium or oxygen: rather than attaching to other substances, forming molecules, these compounds want to break up their molecules. In fact, even touching $\ce{NI3}$ will make it decompose violently... Now there's your explosion.

I actually learnt it as the "reactivity series", so this question kind of confused me. Not saying you're wrong, I'm sure it's just different terminology in different locales... But in any case, surely the heavier alkali metals should be higher in the series than lithium, or am I being stupid?

@orthocresol, I agree on the alkali metals being more reactive, from Li to Fr, i.e. electrons are lost more easily with increasing atomic radius. (The converse is true for halogens, so no elemental electron acceptor stronger than fluorine exists.) Re: use of English in chemistry, it is most imprecise. We can't even agree on aluminium vs. aluminum, or sulfur vs. sulphur, and that's just an elementary example.


text:
Change in internal energy of a gas upon expansion against a constant external pressure
I have a question about the following problem which I got wrong:
During expansion of its volume from $\mathrm{1.00\,L}$ to $\mathrm{10.00\,L}$ against a constant external pressure of $\mathrm{2.00\,atm}$, a gas absorbs $\mathrm{200\,J}$ of energy as heat. Calculate the change in internal energy of the gas.
First, I used the equation: change in $E = q + w$. I calculated work ($-P \Delta V$) and found it to be
$$-18 = -2 \cdot (10 - 1)$$. Then, I added $-18 + q$, the heat absorbed ($\mathrm{200\,J}$). This gave me a result of $\mathrm{-182\,J}$, which is wrong by a large factor. My textbook gives the answer of $\mathrm{-1623\,J}$, but I am not sure where I am going wrong...
Do you know where I made a mistake? Thank you!

My initial guess is units: atm probably need to be converted to Pa. 2 atm = 202.65 kPa and 1623/202.65 = 8, which makes me suspicious (you almost never get integers!). I don't have time to work it through right now but check your units.

You did everything correctly with the exception of converting the units of your result for work.
First, calculate the work in joules:
$$w = -p\Delta V = (-2\,\mathrm{atm}) \times 9\,\mathrm{L} =
\left(-18\,\mathrm{L\cdot atm}\right)
\left({101.325\,\mathrm{J}\over\mathrm{L\cdot atm}}\right) = -1823.85\,\mathrm{J}$$
Second, add the heat absorbed (already in joules):
$$\Delta E = 200\,\mathrm{J} - 1823.85\,\mathrm{J} = -1623.85\,\mathrm{J}$$
Third, always remember to check your units!


text:
Sulfinate R/S configuration
How to determine the configuration of the attached chiral compound? It can be represented with two Lewis structures, in the one to the left, the =O substituent has the biggest priority, while in the right one, the -OMe substituent has the bigger priority. In the former case the configuration would be R, in the second S.

Somewhat related. At least, in the way I attempted to ask the question xD

This reference draws the resonance structure without charge separation. I'm trying to think if that going to cover the general cases though...

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the trigonal pyramidal centre of the sulfinate group is converted to a tetrahedral centre using a phantom atom of low priority.

P-93.3.3.2 Trigonal pyramid
The configuration of molecules containing a trigonal pyramidal center (TPY-3) is described in a similar way to that of tetrahedral centers (T-4) described above in P-92 (see IR-9.3.4.3, ref. 12). The tetrahedral configuration is achieved by adding a phantom atom (0) to the central atom perpendicular to the base of the pyramid. (…)
Traditionally, sulfoxides have been considered as a tetrahedral system composed of a central atom, ligands, and a lone pair of electrons (or phantom atom). No polyhedral symbol is used.

P-93.3.4.1 The chirality symbols ‘R/S’.
The stereodescriptors ‘R’ and ‘S’ (as defined in P-92.2) are used to indicate the absolute configuration of a trigonal pyramidal system discussed in P-93.3.3.2 (see Rule IR-9.3.4.3, ref. 12). A phantom atom of low priority, and not a pair of electrons, is used to create the tetrahedral configuration permitting the use of ‘R/S’ stereodescriptors in the manner described for tetrahedral stereogenic centers. As no locants are present, the name following the stereodescriptor is placed in parentheses, or bracket, according to the required nesting order.

The examples given for this rule in the Blue Book include ethyl (R)-(4-nitrobenzene-1-sulfinate), which confirms that this rule is applicable to the similar compound methyl (S)-methanesulfinate, which is given in the question.
Notably, an equivalent rule also exists in the current version of the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book).

IR-9.3.4.3 The R/S convention for trigonal pyramidal centres
Molecules containing a trigonal pyramidal centre (TPY-3) may exist as a pair of stereoisomers. The configuration of this centre can be described in a similar way to that of a tetrahedral centre. This is achieved through notional placement of a ‘phantom atom’ of low priority in the coordination site that would create a tetrahedral centre from a trigonal pyramidal centre. The centre can then be identified as R or S by the methods described above.
The use of some bonding theories leads to the placement of a lone pair on a trigonal pyramidal centre. If this is done, the absolute configuration of the centre is also described by the R/S convention, in this case by placing the ‘phantom atom’ in the site that is occupied by the lone pair. Examples of this practice may be found in the description of absolute configurations for sulfoxides in which the alkyl substituents are different.

After placement of the phantom atom, the R/S convention for tetrahedral centres can be used. Rules for the tetrahedral configuration of elements other than carbon are given in Section P-93.2 of the Blue Book. In particular, the rule for compounds containing sulfur refers to the rule for similar phosphorus compounds.

P-93.2.5 Sulfates, sulfonates, and related compounds
Sulfates, sulfonates, and related anions are treated in the same way as phosphate anions (see P-93.2.4). (…) Sulfoxides are discussed in P-93.3.3.2.

The corresponding rule for compounds containing phosphorus reads as follows.

P-93.2.4 Phosphates, phosphonates, and related compounds
The ‘$\ce{P=O}$’ bond, as conventionally written in phosphates, phosphonates, and related compounds, is considered as a single bond, as there are already four atoms or groups in the tetrahedral configuration. Similarly, the formal arrangement of charges is not considered when determining the configuration of a chiral molecule. As the stereodescriptors ‘R’ and ‘S’ describe the entire structure, either a salt or an ester, the full name is placed in parentheses to denote the global configuration.

Therefore, the analogous ‘$\ce{S=O}$’ bond in the sulfinate group is considered only as a single bond since the tetrahedral centre of the sulfinate group with an attached phantom atom already has four groups. (Note that this is a deviation from the usual Sequence Rules for the Cahn-Ingold-Prelog (CIP) System. Using the Sequence Rules, double bonds are normally split into two bonds with duplicate representations of the atoms at the other end of the double bond.)
Thus, for the compound given in the question, the $\ce{-O-CH3}$ group ranks higher than the $\ce{=O}$ group. This leads to the configuration S and the systematic name methyl (S)-methanesulfinate.

So would it be allowed in the context of this question to add a phantom atom with atomic number $0$ to an enolate oxygen if I am not sure of a definite counterion?

These are two resonance forms.  Neither is the real structure, rather the real structure is a hybrid of the two.  Therefore they must give the same answer, since there aren't really two things here.
Each structure has the same four groups around the S: a methyl group, a lone pair, an oxygen (of the S=O) and a methoxy group. Priority is based on the atomic number of the first attached atom, and lone pairs are always lowest priority.  So the priority should be methoxy, oxygen, methyl, and lone pair, making the structure S.
One might wonder about the priority of the oxygen of S=O vs. the methoxy group.  The decision procedure is to work away from the proposed chiral atom, following the bonding wherever it leads in an every expanding tree-like network until a difference is found (or not).  Going from the S to the O in S=O we next encounter lone pairs, which have the lowest possible priority (S -> O -> :).  Going from the S to the O to the methyl (i.e. the methoxy group), we encounter a C at the same level (step) as we encounter a lone pair on the other oxygen (S -> O -> C).  C beats a lone pair, so methoxy wins over the O of S=O.

So, extended octets are forbidden when determining configuration. If you find any kind of proof (reference) I will accept your answer.

It's more that extended octets are not really relevant to the question of configuration. Chirality results from several structural phenomenon, in this case a chiral center.  These are by their very nature tetrahedral, so there are 4 objects surrounding the central atom.  The mode of bonding is not part of the definition.  Later I will see if I can find a reference that supports this.

I've meant that they are probably forbidden by convention.

Look at my reference in the comment to the OP's question. That's from IUPAC and there's a sulfoxide in there.

They are not really resonance forms. The doubly-bonded structure only contributes very minute amounts to the overall picture. It can be disregarded without any significant loss.


text:
Is diffuse reflectance measurement superior to transmittance for determining band gap energy?
Some papers report band gap estimation using Diffuse Reflectance Measurement (e.g. here).
Why would the results of diffuse reflectance be superior to regular transmittance measurement? And why diffuse reflection and not specular reflection?
(source)


text:
How does symmetry of half filled and fully filled orbitals lead to stability
I googled a lot about why symmetry of half filled and fully filled orbitals decreases their energy but every time it is repeated that symmetry leads to stability. So the question is why symmetry leads to stability? What is the actual logic behind it?

Symmetry does not lead to stability. In some cases, when your orbital is half-filled, certain kinds of symmetry lead to instability (see Jahn-Teller effect). In other cases, symmetry is mostly irrelevant to stability.

Actually in the books I have read and also on the internet, "symmetry" word is used widely for accounting the stability without any explanation "Why" . Thank You for reply.


text:
Five-pointed crown alkane C25H40: does it exist?
My 4 year old granddaughter, playing with molecular models, began making a cyclopentane ring, each of whose carbons bore a spiro-connected cyclopentane ring of its own , in the manner of spiro[4.4]nonane.  If she hadn't run out of carbon atoms first, the result would have been a pentagonal crown, with the five outer cyclopentane rings perpendicular to the central one.

I wonder if such an alkane has been synthesized. Its molecular formula would be $\ce{C25H40}$. I can't think of why the molecule would be unstable or sterically impossible, but I couldn't find it in a Google search on $\ce{C25H40}$.
In particular,  I wonder if the five outer rings would be exactly perpendicular to the center one (eclipsed), or somewhat staggered.

Looks somewhat strained, but possible. The outer rings themselves won't be exactly planar, let alone perpendicular.

These polyspiro compounds such as the one you gave are known as [m,n]rotanes, where m is the size of the central ring and n is the size of the decorating rings. So, your compound would be called [5,5]-rotane.
A Reaxys search unfortunately doesn't show anything for [5,5]-rotane, but in general, many such compounds have been made. A lot of the articles are in German or French (there are quite a few such articles). However, I did find one in English: J. Am. Chem. Soc. 1998, 120 (2), 317–328.
The authors synthesised and obtained X-ray structures of a number of rotanes, among them [5,4]rotane, which is the closest to the [5,5] that you are looking for. I quote:

In [5.4]rotane (7), the cyclopentane ring adopts a slightly distorted envelope conformation: C(1), C(5), C(9), and C(17) form an almost perfect plane (mean deviation 0.021 Å), while the flap atom C(13) is located 0.684 Å above this plane. The cyclobutane rings at C(13) and C(9) are nearly planar (φ = 9.7°) and planar (φ = 0.2°), respectively. The remaining three cyclobutane rings are moderately puckered (φ = 23.1, 26.5, and 28.2°).

The torsion angle φ of cyclobutane refers to the angle between the two planes that define the molecule. A completely planar four-membered ring would have φ = 0°, and an increasing value of φ indicates greater deviation from planarity:

As you can see in the X-ray structure, the five attached cyclobutane rings do eclipse each other to a certain extent: the obvious exception is the ring fused to C(13). This is not a very surprising conclusion at all. The envelope conformation of unsubstituted cyclopentane is known to suffer from torsional strain (i.e. unfavourable eclipsing interactions between substituents). Carey and Sundberg's Advanced Organic Chemistry, Part A writes (pp 162–3):

There is minimal angle strain in cyclopentane, but considerable torsional strain is present. Cyclopentane is nonplanar and the two minimum energy geometries are the envelope and the half-chair. [...] The planar portions of both conformations have torsional strain owing to C−H and C−C bond eclipsing.

There's a ton more information in the paper, and I'd recommend giving it a read if you are interested - I just tried to pick out the parts directly relevant to your question here, since you indicated that you were interested in the conformation of such a compound.

Thanks.  I'll have a look.  Another source of bizarre ring structures, not including [5.5]-rotane, is http://onlinelibrary.wiley.com/doi/10.1002/9780470682531.pat0337/full?pat0337-eo-0155

Fascinating molecule. I am not offering anything substantially, new or original here; merely here to confirm what has already been said. I couldn't find any evidence that suggests this particular molecule has been synthesised.

In particular, I wonder if the five outer rings would be exactly perpendicular to the center one (eclipsed), or somewhat staggered.

Now to specifically address this part of the question, (and also to complement @orthocresol's great answer), I drew out your molecule in Avogadro and performed a molecular mechanics geometry optimisation using the MMFF94s force field.
The visualisations of the results are given below ( I tried to take some screenshots from different perspectives)
(note: @Martin pointed out some funny business going on with the hydrogens in the old pictures, and I took a good, hard look at the file once again and saw the odd bridging hydrogen too. Anyway, I redid the calculation, and here's some new visualisations). I still stand by the conclsuions I drew earlier, though.

Anyway, the ring strain in general doesn't seem to terrible. The "ornamental" rings are not planar, and do look staggered to me. I also reckon this is chiral ($C_1$) symmetry.

Scifinder, the online tool of the Chemical Abstracts Service of the American Chemical Society shows no results. Thus, unfortunately, to the best of any chemist’s knowledge this beautiful molecule has not yet been synthesised or isolated.

The topologically closest analogue with a well-established crystal structure I found is 3a-Hydroxy-trispiro(octahydropentalene-1,1':2,1'':3,1'''-tris(cyclopentane))-2'''-one ($\ce{C20H30O2}$). Its structure contains 3 vertex-sharing and one edge-sharing cyclopentane units (1):

Colorless product was obtained during photochemical 2+2 cyclodimerization of 2-cyclopentylidenecyclopentanone with several tertiary amines (triethylamine, N-methyldiisopropylamine and N-ethyldiisopropylamine) in acetonitrile solution upon irradiation with medium-pressure Hg-lamp.
(1) Ho, T.-I.; Peng, S.-M.; Lee, G.-H.; Lee, H.-C.; Hwang, T.-C. Acta Cryst C 1986, 42 (12), 1842–1844. DOI: 10.1107/S0108270186090339


text:
How to prepare KF/Al2O3 i.e. potassium fluoride on alumina (/on aluminum oxide)
It's surprisingly hard to find an answer to this. One book I found referenced a method wherein one would simply mix (with a mortar I assume?) equal (another source said 40:60) portions of Al2O3 and KF and that's it.

In Bull. Chem. Soc. Jpn., 1983, 56, 1885-1886 (PDF), the authors (Takashi Ando and coworkers) give the following method:

Alumina (30 g, neutral alumina, type 90 for column chromatography, activity I, 70-230 mesh, Merck Jpn.) was mixed with $\ce{KF}$ (20 g) in 200 mL of water and the water was removed at 50-60 °C in a rotary evaporator. The impregnated alumina was then dried in a vacuum drying oven at 75 °C for several hours.

In another publication, J. Chem. Soc., Perkin Trans 2, 1986, 1133-1139, by the same main author, which focuses on the properties of the $\ce{KF}/\ce{Al2O3}$ system, a "dry preparation" by grinding $\ce{KF}$ and $\ce{Al2O3}$ under a nitrogen atmosphere is mentioned.


text:
The Rayleigh-Ritz variation method
I am currently reading Atkins and Friedman's "Molecular Quantum Mechanics" (4th ed), looking at the Rayleigh-Ritz variation method. Starting from the Schrödinger equation $\hat{H}\psi = E \psi$, we get the "Rayleigh ratio"
$$
E = \frac{\int \psi^*\hat{H}\psi d\tau}{\int \psi^*\psi d\tau}
$$
Setting a trial function to be the following linear combination (assuming real coefficients)
$$
\psi_{trial} = \sum_i c_i\psi_i
$$
we find that
$$
E = \frac{\sum_{i,j} c_ic_j H_{ij}}{\sum_{i,j} c_ic_j S_{ij}}
$$
where $H_{ij} = \int \psi_{trial}^*\hat{H}\psi_{trial}d\tau$ and $S_{ij} = \int \psi_{trial}^*\psi_{trial}d\tau$.
Now the goal is to minimize the expression for $E$. We should therefore "differentiate with respect to each coefficient in turn and set $\partial E / \partial c_k = 0$ in each case". Using the quotient rule, I get the following
$$
\frac{\partial E}{\partial c_k} = \frac{ \sum_{i,j} c_ic_jS_{ij} \cdot \frac{\partial}{\partial c_k} \left( \sum_{i,j}c_ic_jH_{ij} \right)  }{\left( \sum_{i,j}c_ic_jS_{ij} \right)^2} - \frac{   \sum_{i,j} c_ic_jH_{ij} \cdot \frac{\partial}{\partial c_k} \left( \sum_{i,j}c_ic_jS_{ij} \right)   }{\left( \sum_{i,j}c_ic_jS_{ij} \right)^2} = 0
$$
However, I am not sure how to simplify this to obtain the correct expression, which is
$$
\frac{\partial E}{\partial c_k}_{Correct} = \frac{\sum_j c_j (H_{kj} - ES_{kj})}{\sum_{i,j} c_ic_jS_{ij}} + \frac{\sum_i c_i (H_{ik} - ES_{ik})}{\sum_{i,j} c_ic_jS_{ij}} = 0
$$
Clearly I can cancel a factor in the first term of my expression, but I don't know how to handle the derivatives with respect to $c_k$.

You have a bunch of partials with respect to $c_{k}$. If neither $i$ or $j$ equals $k$, that term disappears...

Instead of doing the whole derivation, I will provide the following two hints usefull in the derivation:

Use the quotient rule (as is was already done by OP).
$$
  \left({\frac {f}{g}}\right)'={\frac {f'g-fg'}{g^{2}}} \, ,
$$
where in this particular case $f = \sum_{i,j} c_i c_j H_{ij}$ and $g = \sum_{i,j} c_i c_j S_{ij}$.
Then, when taking the derivatives of the above mentioned $f$ and $g$, take into account that for each and every $c_i c_j X_{ij}$ term (where $X$ is either $H$ or $S$) there are only three possibilities:

$i = k$ in which case $\partial (c_i c_j X_{ij})/\partial c_k = c_j X_{kj}$;
$j = k$ in which case $\partial (c_i c_j X_{ij})/\partial c_k = c_i X_{ik}$;
$i \neq k$ and $j \neq k$ in which case $\partial (c_i c_j X_{ij})/\partial c_k = 0$.


text:
Formaldehyde from Urea?
I was wondering if it is possible to make formaldehyde (1) from urea (2).

I noticed that they have a similar structure, but the two hydrogen atoms in formaldehyde are replaced with two amide groups in urea. Would it be possible to remove those groups and replace them with hydrogen?

Technically yes, but it would require much more than one step.

Do you have any suggestions?

This is more a proposition than an answer but after a lot of search on the net, in publications and also with the help of @orthocresol, @Jan and @jerepierre, I propose a reaction that will work. Also the question ask is really unusual, because it is obviously awfull to want to synthetize formaldehyde from urea.
You will find first in this "answer" the best way we found to make it. When I'll have time, I'll add all ideas I had and explain why it was not a good point.

A good question raised by this question is, "is it possible to "cut" the $\ce{C-N}$ bond to replace it by a $\ce{C-H}$ bond?" and if possible in one step. For the moment nobody knows.

The best way we found :
For the moment I can't explain you why I chose to make the N-alkylation but I know I got it in many of my organic chemistry lessons that this groupement will leave the molecule involving LAH so I wanted to do that. Also we are not at one more step from urea to get formaldehyde.
For the other part formaldehyde is much more reactive than the amide function because on the "diamide" both lone pair of nitrogen atoms are involve in mesomeric forms with the oxygen. Then the carbon is really not electrophilic and LAH will react then one the most reactive function in the reactor which will be the aldehyde.
And then the Formox process because it is use in the industry so I'm sure it will work.

After you will find ideas I had but that won't work too. I'll explain as much as as possible when it will be possible. Sometime it will be only because by experiments nobody have ever succeeded to do it.

This one won't work, one of us, I don't remember who didn't find anything in SciFinder about the possibility to make Eschweiler-Clarke reaction from an amide and especially urea.
This is really awfully awkward because you need formaldehyde itself to make the synthesis of formaldehyde. I was not thinking at that when I start but few minutes later it appears to me that I haven't realize before ^^

At the begining I thought it was great to use $\ce{HCl}$ in excess with $\ce{NaNO_2}$ and then get the carbonic acid $\ce{HO(CO)OH}$, because there is nice ways to reduce this acid and its derivatives but just one time. You can find informations in this publication. But when you have $\ce{H(CO)OH}$ I have really no idea of what to do, even $\ce{LiAlH_4}$ won't work. So at the moment I just have the first possible reaction.
Well also in terms of atoms economy using urea to synthetise formaldehyde is absolutely awkward too.

I hope you find some fun here, I keep thinking about it but it looks to be tricky. Have a good day.

Eschweiler-Clarke is typically for amines, not amide-type nitrogens. Is there reference for this? How is the formaldehyde not going to be reduced by LAH?

@jerepierre This is not an answer as one can understand "answer" this is a proposition. I'm only a chemical engineering student. But I have this reaction in my organic lesson and this is not write anything about "R" in "R-NH2" so I can't answer you. If you use the correct amount of LAH things must work. However I'm note sure if it is possible to reduce formaldehyde with LAH.

@jerepierre what do you think so if I do first a Wittig to "protect" the ketone function, I do my to steps like I did and at this HCl to remove all the LAH and after ozonolisis ?

@jerepierre I am guessing that he is banking on the tetrahedral intermediate being stable enough (a la Weinreb amide) such that aq workup protonates it to form a hemiaminal & destroys the excess LAH. I don't really buy it though. Reaxys has nothing on this transformation. I agree that Eschweiler-Clarke is odd here. Someone I know used to methylate carbamates with NaH/MeI.

@jerepierre http://www.sciencedirect.com/science/article/pii/S0040403900768948 if you think it must be a good way I can improve my answer with your remark about it and use a reaction of this publication.

@9-BBN The Eschweiler-Clarke reference is sufficient to make me believe that part might work. However, the reduction is extremely speculative. Stopping a reduction at the aldehyde oxidation state is not a generally successful process. Simply controlling the stoichiometry does not work for an ester->aldehyde (over reduced to alcohol), and I have no reason to believe it will work here. I understand this is a proposed synthesis. A good proposed synthesis requires careful thought, consideration of possible ways it may fail, and evidence to support its success.

The Formox process is all well and good for industrial production, but worth noting that in the lab you generally don't want to be doing reactions at $300\ ^\circ\mathrm{C}$.

@orthocresol who said we don't want to do it in industrial scale ? :P


text:
Can a table salt + vinegar mixture be explosive?
I remember having read in a book, Ask Me Anything (Dorling Kindersley), that a combination of copious amounts of table salt and vinegar is explosive.
Now this was a really long while back, and I can't even find my copy of the book anymore. But I'm quite sure there was a mention of salt+vinegar being explosive.
Simply knowing the composition of table salt and vinegar doesn't seem to help in any way. I really don't see any reason for some sodium, chlorine, acetate and hydrogen ions floating around in a bowl to simply react in some godforsaken way and explode. Heck, all salt+vinegar seems like is a good Pringles flavor. I did take into consideration possible impurities in table salt, like other sodium, potassium, calcium and magnesium salts, but even then, I don't see why the explosion would occur at all. I though of iodized salt too, but still, an explosion seems a little far-fetched.
Now the book did mention 'large' quantities of salt and vinegar, maybe there could be something to that?
Is the book correct? If so, what concentration of salt and vinegar have to be used?
I did conduct an internet search, but nothing really turned up.
[ I suspect, the book might've meant baking soda instead of table-salt. Baking soda combined with vinegar in a closed plastic bottle could pass for an 'explosion'. ]

If you put both into a closed aluminium flask, you might evolve some hydrogen, and you could ignite the mixture with desidual air in the bottle (it won't ignite by itself below  500 °C). But calling the mixture itself explosive is ridiculous nonsense. And also the baking soda won't explode. A closed vessel might rupture violently, but there can be no explosion.

If you reallllly wanted an explosion, molten salt would probably do the job. Although, I doubt whether you had any acetic acid would make a difference.

Well, I suppose if your "copious quantities" were large enough -- say, several times the mass of the Sun -- they would collapse under self-gravitation to form a supernova. But the same could be said of either salt or vinegar by itself. Or any other form of matter.

Of Course NaCl is the Salt of strong acid and strong base . It is very stable . Even if you take any amount of NaCl with Vinegar it will not do any explosion .
It may be the baking soda (NaHCO3 ) and Vinegar , As this reaction is a neutralization reaction (Exothermic) and accompanied by the formation of CO2 , If enough concentration is taken it can burst a plastic bottle on applying some pressure .
https://www.youtube.com/watch?v=UclxSfPoapc

why I got two down votes ? Is their any mistakes ?

Aditya, your answer seems completely correct to me. I can only guess that you got down votes because you discussed a potentially dangerous reaction without mentioning safety. This reaction is not extremely dangerous, but if the plastic bottle bursts whilst you are close to it, it can cause pretty bad cuts or eye injury. Another hazard may occur if the gas pressure is not quite enough to burst the bottle; if you don't have a safe way to render it safe, it may then burst later at a random time, when the temperature changes. (The safest method in this case is to shoot it with an air rifle.)


text:
In the Favorskii rearrangement, why doesn't the base grab the alpha hydrogen of the carbon bearing the halogen?
In the Favorskii rearrangement, the base (like hydroxide) always abstracts the alpha hydrogen of the carbon not bearing the halogen. But to me the hydrogens on the carbon directly connected to the halogen should be more acidic due to the electron pulling inductive effect of the halogen (which indeed does happen in the haloform reaction). This should result in an aldol condensation type reaction. Why is this not so?

The crux of the matter is that deprotonation opposite the halogen leads to a faster (intramolecular) reaction, so that's what you're going to observe. Never take the existence of a mechanism to mean that those steps are the only things that can happen in solution. Undoubtedly some deprotonation at the halogen position does occur. Whether it leads to a sufficiently fast reaction is another matter altogether.

I agree with orthocresol.
Both alpha positions will be roughly similar in acidity but the intramolecular reaction that occurs when the non-halogenated alpha position gets deprotonated will be just much faster than the intermolecular aldol condensation that could in principle happen on the other side.

Since it is much easier to discuss based on actual schemes, see scheme 1 for a general Favorskii rearrangement.

Scheme 1: General Favorskii rearrangement reaction conditions. Image taken from Wikipedia where a full list of authors is available.
And because it is also helpful, scheme 2 shows one of the most accepted mechanisms of the Favorskii rearrangement — the one used by both Wikipedia and Kürti-Czakó.[1] One should note, however, that a second mechanistic proposal[2] explains the reaction outcome well and is favoured by some calculations.[3] Because it does not require any deprotonation, it is outside of the scope of this answer.

Scheme 2: ‘Cyclopropanone’-mechanism of the Favorskii rearrangement. Image taken from Wikipedia where a full list of authors is available.
Your question asks why carbon 6 (from the usual numbering of 2-chlorocyclohexanone) is deprotonated to form the corresponding enolate and not carbon 2.
The most important notice here is the base used $\ce{NaOH}$ and the conditions employed (not mentioned in the image, but an example on organic-chemistry.org mentions room temperature). For comparison, table 1 shows a few $\mathrm{p}K_\mathrm{a}$ values from the Bordwell data — unfortunately, an α-chloroketone is not included in this data.
$$\textbf{Table 1: }\mathrm{p}K_\mathrm{a}\text{ values of different relevant acids in }\ce{H2O}\\
\begin{array}{ccc}\hline
\text{acid} & \mathrm{p}K_\mathrm{a}\left(\ce{H2O}\right) & \mathrm{p}K_\mathrm{a} \left(\ce{DMSO}\right)\\ \hline
\ce{Me2CO} & 19.3 & 26.5\\
\ce{PhCOMe} & 18.3 & 24.7\\
\ce{PhCO-CH2F} & & 21.7\\
\ce{Et2CO} & & 27.1\\
\text{3,3-dimethylcyclohexanone} & & 24.8\\
\ce{H2O} & 15.7^* & 31.4\\ \hline \end{array}\\
^*\text{ Other sources say 14.}$$
Obviously from this data, the choice of solvent is import when considering the acidity of a given substance — note the marked decrease in acidity of water in $\ce{DMSO}$. However generally, water is a weaker acid than carbonyl compounds; Professor Mayr taught us to estimate $\mathrm{p}K_\mathrm{a}$ values: alcohols $\approx 15$, ketones $\approx 20$. That means, the deprotonation is in no way complete; it is in equilibrium. This equilibrium will lead to a thermodynamic product distribution: carbon 2 will be deprotonated slightly more than carbon 6, but the most significant species will be the unenolised ketone.
Now as soon as carbon 6 has enolised, it is able to attack carbon 2 to form the bicyclo[3.1.0]hexan-6-one. This is an intramolecular attack and thus very fast. Once this has happened, the formation of the hydrate (6-hydroxybicyclo[3.1.0]hexanolate) is favoured due to the strained three-membered ring. And the subsequent intramolecular migration is also very fast. This leads to a stable product with little or no means for any back reaction; as soon as the initial attack has taken place, things are pretty much settled.
On the other hand, an aldol addition would not only require the enolate (doesn’t matter which one) to be around for some time; it is also inherently prone to retro-aldol dissociation. Usually in organic chemistry, aldol reactions are performed selectively by using kinetic deprotonation (e.g. $\ce{LDA}$; $\mathrm{p}K_\mathrm{a}(\ce{HN(iPr)2}) \approx 35$) ensuring that the forward reaction is strongly favoured. Hydroxide bases may be used in aldol reactions, but then generally the reaction is heated to reflux and the final product is the condensation rather than the addition.

Thus, to sum it up:

Both enolisations occur
The conditions do not favour aldol addition; they will revert readily
Once the Favorskii path has been entered, the rearrangement is basically irreversible.

Notes and References:
[1]: L. Kürti and B. Czako: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 164.
[2]: The alternative, semi-benzilic mechanism is initiated by the nucleophilic attack of a hydroxide ion on the carbonyl group. The thus formed 2-chloro-1-hydroxycyclohexanolate anion undergoes 1,2-migration with chloride displacement to form the product cyclopentyl carboxylic acid in a single step.
[3]: V. Moliner, R. Castillo, V. S. Safont, M. Oliva, S. Bohn, I. Tuñón, J. Andrés, J. Am. Chem. Soc. 1997, 119, 1941. DOI: 10.1021/ja962571q.

But why does this not happen in the haloform reaction? There the base deprotonates the alpha carbon having the halogen, and not the other alpha carbon(once the first halogen has been added)

Btw. both of your used images are in the public domain, there is no need for citing the source ;)

I have doubts about the proposed reaction mechanism and find them supported by J. Am. Chem. Soc., 1997, 119 (8), pp 1941–1947. Tl;dr: I would have assumed nucleophilic attack at the carbonyl carbon first, then rearrangement with dissociation of chloride.

@Martin-マーチン Hm, interesting. Kürti-Czakó show the mechanism that I wrote down here. But to be perfectly honest, the semi-benzilic rearrangement looks very plausible, too; especially in the light of other, similar rearrangements. I just skimmed across the paper right to the conclusions, but I’ll take a deeper look at it later™.

Before I left the office I found a paper with a title among the lines of 'a new mechanism for the Fav. Re.' I have not had the time to read it. I think it's safe to assume that no ultimate truth and it probably depends heavily on many different factors, like base, solvent, temperature, ring size, etc. When I did a few calculations (chex), the PA of the 2 or 6 position were not very much different, certainly within the accuracy of the method.

@Martin-マーチン After having fully read the paper you presented, I have to state doubts. 1) They used 2-chlorocyclobutanone as a starting material. The cyclopropanone mechanism thus requires the formation of bicyclo[1.1.0]butan-2-one which I consider a very strained intermediate. I wonder what the results would have been had they calculated starting from a larger ring size (e.g. cyclohexanone or even cyclopentanone). 2) I wouldn’t consider TS3B to be the lowest or best transition state on the way from I2B to P: Having a second hydroxide there seems much more plausible.

Of course, I am aware that the presence of a second hydroxide changes quite a lot with respect to the overall calculations: it needs to be accounted for all the way. On the other hand, the concerted 1,2-water addition doesn’t exactly look like the most favourable way. Nonetheless, they do make important points on the semi-benzilic mechanism in the introduction that I feel obliged to include it.


text:
Metal reagent in dehalogenation of vicinal dihalides
In the dehalogenation of vicinal dihalides why can't sodium be taken as a reagent?
I understand that zinc is taken because it is reactive and rate of reaction would be considerably high and it doesn't form complexes. Because of this many other metals can't be used including magnesium as it forms complexes. But I don't understand why sodium can't be taken.

I don't think you mean "catalyst". Also, please mention the complexes you are talking about (just one or two examples).

You mean something like the following right
$$
\mathrm{R-X ~+~ Na \longrightarrow R\cdot ~+ ~NaX}
$$
and we add a proton source like an alcohole to transfer an H atom to $\mathrm{R\cdot}$:
$$
\mathrm{R\cdot~ + ~R'-H} \longrightarrow \mathrm{R-H}
$$
Well, its in principle possible. But a) you don't want an explosion right?, so be careful as this will probably be very exothermic b) it will probably be really messy as side products like
$$
2~\mathrm{R\cdot} \longrightarrow \mathrm{R-R}
$$
can be formed. Also the whole approach is not really tolerant to pretty much any functional group and will just tear the substrate apart in many cases. The other alternatives (e.g. Zinc, $\mathrm{SmI_2}$) are just much milder, easier to use and much more versatile. But sure, try it out have fun (less than 0.5 g sodium should still be relatively harmless) ; )

Dehalogenation can be effected with many metals, e.g. zinc, magnesium, lithium and sodium in ammonia. Other reagents are sodium naphthalenide, phenyllithium, lithium aluminum hydride, chromium(II) chloride, sodium iodide, sodium sulfide and sodium selenide.
Usually good yields are obtained in these reactions; however, since the best route to the starting material is halogen addition to an alkene, the preparative value is limited. An advantage over dehydrohalogenation is the fixed position of the formed double bond.
The mechanisms of the dehalogenations are complex and depend on the particular conditions. The reactions are sometimes stereospecifically anti, but in other cases no stereospecificity has been observed.


text:
What would be the mechanism for this condensation reaction?
Below is the reaction in question. Will it go like an aldol consensation would, where the base would deprotonate the methyl group of the benzoxazole moiety, the negative charge would be transferred over to the nitrogen and this anion would then bond with the aldehyde?
In any way, a drawing of the mechanism would be really helpful.
A bonus question: is this a specific type of condensation? (i.e. like aldol)

Your 2-methyl-1,3-benzoxazole is a masked acetic acid.
Remember that one way to synthesize oxazoles actually starts from aldehydes (and uses TosMIC).
So, yes, the reaction is an aldol condensation where the $\ce{CH}$-active component is carboxylic acid in disguise.

I think that's correct. You should depronate on the methyl group and perform a normal aldol condensation. Interesting choice of base here. I guess the authors must've tuned this for yield.
The best name here would be aldol, since the nucleophile here is very much like an enolate...


text:
Polyatomics -How Do They Work?
In class, we recently learned about Ionic nomenclature. Specifically, polyatomics and monoatomics. Even though our teacher did a decent job (I guess) of explaining it to us, I'm still very confused.
I have multiply questions:

Question: How would you know what type a compound is when given only the formula?

For example: Our teacher started with $\text{AuCl}$ and wrote down Gold (I) Chloride as the nomenclature. How did he get from $\text{AuCl}$ to Gold (I)?
There are two possibilities: Gold (I) and Gold (III).
Another example: Another example would be $\text{Pb}_3\left(\text C_6\text H_5\text{O}_7\right)$. Which apparently corresponds with Lead (II) Citrate. What does the II inside the parenthesis mean?
And below, I have provided some more examples. Please leave some for me! :P

Examples: Write the formulas for the following compounds:

Copper (II) Chloride

Vandium (III) Selenide

Beryllium Oxide

Potassium Permanganate

Ammonium Nitrate

Tin (II) Sulfite

Chromium (IV) Cyanide

Aluminum Arsenide

I'm lost at this point. Any help and a few suggestions would help a lot!

Has your teacher yet talked about (preferred) oxidation states of the elements in compounds? He should, because this all makes no sense at all without knowing that oxygen is usually -II, halogens -I, alkali metals +I etc.

@Karl That doesn't seem too familiar. Maybe he has taught us, but I wasn't listening, maybe he didn't... But either way, I don't quite remember that.

Well, you will be screwed in chemistry without oxidation states, so check on it.

Our teacher started with $\ce{AuCl}$ and wrote down Gold (I) Chloride as the nomenclature. How did he get from $\ce{AuCl}$ to Gold (I)?
There are two possibilities: Gold (I) and Gold (III).

True, but there is only one possibility for chloride ;)
Karl's comment under your question gives the crucial hint:
Start with the part that you know and that is unambiguous!
Let's have a look at some examples:

Chloride is (always) $\ce{Cl-}$; it has a charge of (-1).
If your teacher would have written $\ce{AuCl3}$, you would know that there are 3 chloride ions with a total charge of $3\times(-1)=-3$. Consequently, you would know that you're having $\ce{Au(III)}$ here. The copper (II) cloride example belongs to the same category and if you like, you can try iron(II) bromide, iron(III) chloride and copper(I) iodide.

Karl also mentioned oxygen. You can safely assume one or more $\ce{O^{2-}}$ units in your metal oxides, depending on the oxidation state of the metal. Can you assign the oxidation state in $\ce{Fe2O3}$?

Nitrite, nitrate, sulfite, sulfate are some anions that you should memorize sooner or later. You might know that the names that start with nitr~ refer to nitrogen, the sulf~ refers to sulfur. All these anions contain various number of oxygen atoms.


text:
Why do people consider caffeine ineffective after its half-life?
Assuming you take $\pu{200mg}$ of caffeine, it's half life is roughly 6 hours. So after 6 hours there should still be $\pu{100mg}$ of caffeine in your body.
Then why is it generally said that after six hours caffeine's effect should wear off? Shouldn't it only be diminished since there is still $\pu{100mg}$ of caffeine in your body?

You should probably ask this on Biology.SE I think. Probably not many chemists know the exact mechanism by which caffeine stimulates the human body to know why the effects wear off over time (I certainly don't). My guess is that it has something to do with diminished signaling of some hormone (?) that doesn't signal linearly with concentration of caffeine.

Perhaps see the caffeine wikipedia page, and read the sections titled pharmacokinetics and pharmacodynamics. These contain the information needed probably, but understanding that information to be able to answer your question is beyond me.

Personal experience: Some years ago, I mowed our lawn on a hot summer day. It took about 2 hours, and I sweated up a storm. To compensate, every 20 minutes or so, I came into the house and swigged down some Pepsi Cola. After I was finished, I noticed that the 2 liter bottle of Pepsi was also finished, and I was tired, so I sat down.
With nothing else going on, I noted that my heart was skipping beats. 1,2,3,skip. 1,2,3,skip. Extra systole. Nothing to get excited about. But never this skippy. So I listened, and checked, and after about 15 minutes, I got 1,2,3,4,skip. 1,2,3,4,skip. Interesting. And about every 15 minutes, my heart would go another beat before the skip. It finally got out to 14 before skipping, then went all crazy: 12, 17, 20, then finally stopped skipping.
It took 2 hours to consume the Pepsi, and about 4 hours to overcome the observable effects of the caffeine. There is certainly a non-linearity involved with the excretion and with the effect of the caffeine on my heart, but yeah, the half life says something, but there is a diminution of effect, not a go/no go.
And I never again drank that much caffeinated beverage all at once.


text:
What is the source of discrepancy in the calculation of Standard State Gibbs Free Energy change of the reaction of HCl (l) dissolving in water?
I have calculated $\Delta_\text{r} G^o$ two different ways and got completely different results. Why? What assumptions have I made incorrectly, particularly in the first calculation shown below?
Okay, so everyone knows that liquid Hydrochloric acid is a strong acid so it completely dissociates in water according to the following chemical equation:
$\ce{HCl (l) + H2O (l) -> H3O+ (aq) + Cl- (aq)}$
If I used tabulated values of Standard State Gibbs Free Energy of formation $\Delta_\text{f} G^o$ to calculate $\Delta_\text{r} G^o$, standard State Gibbs Free Energy formation change of the reaction.
Standard State Gibbs Free Energies of formation:
$\begin{array}{lr}
\ce{HCl (l)} & \pu{-131.17 kJ mol^{-1}}\\
\ce{H2O (l)} & \pu{-237.19 kJ mol^{-1}}\\
\ce{H3O+(aq)} & \pu{-237.19 kJ mol^{-1}}\\
\ce{Cl-} & \pu{-131.17 kJ mol^{-1}}
\end{array}$
Source (I used this one because I couldn't find any that listed $\ce{HCl}$ as a liquid and not a gas):
http://www.vias.org/genchem/standard_ent...
Then, if you have one mole of each constituent according to the chemical equation, the standard state Gibbs Free Energy of the reaction = 0 kJ
by means of calculating ΔGoreaction = ΣΔGoformation of products - ΣΔGoformation of reactants
So ΔGoreaction should be less than zero if the reaction is favorable! But we know that HCl completely dissociates in water into Chloride ion and hydronium. What is going on here!?
If I calculate the standard state Gibbs Free Energy of the chemical reaction a different way.
That is that
$\Delta G^\circ_\text{reaction} = -R \cdot T \cdot \ln(\ce{Keq})$ at equilibrium, where
for the acid $\ce{HCl}$, $\ce{Keq = Ka} = \pu{1.3e6}$,
Then the Gibbs Free Energy change of the chemical reaction at standard state
$\Delta G^\circ_\text{reaction} = (\pu{-8.314 J K-1 mole-1}) \cdot (\pu{1 mole}) \cdot (\pu{298 K}) \cdot \ln(\pu{1.3e6})$
$= \pu{-34.88 kJ}$,
which is less than zero; therefore, the reaction is favorable. What is going on!? See the difference in the calculations!?
It seems that the Ka of strong acids like HCl is usually assumed to be too large for most websites to even bother displaying. But, I found a table of Ka's of strong acids here: http://depts.washington.edu/eooptic/links/acidstrength.html

Your second calculation is correct, to the extent the pKa value is correct - the extremes are difficult to measure.
The first calculation misses one crucial thing and that is the Gibbs free energy of mixing. In most cases this value is negative. You can get some basic intuition here. The values of $\Delta G_{mix}$ are a non-linear and often non-intuitive function of mole fraction of a mixture and can be challenging to predict.
I do not have experimental data at hand. However, in case of HCl and H$_2$O it is reasonable to assume that the enthalpic part will be negative (HCl is polar, favourable interaction with water). While this might be partially compensated for by entropy, the mixing itself is likely to be dominant, keeping the entropic part positive.
$$\Delta G_{mix} = \Delta H_{mix} - T\Delta S_{mix} $$
As a result, the Gibbs free energy will be negative and should account for the discrepancy between the two calculations.

Using the $\ce{\Delta{G^{\circ}}}$ value of liquid $\ce{HCl}$ ($-131.17$ kJ/mol) is wrong. Simply put, we are indeed interested in gaseous $\ce{HCl}$. The normal boiling point of $\ce{HCl}$ is $-85^\circ$ C. On the other hand, the standard state of Gibbs Free Energy is $25^\circ$ C for what you are calculating. (Having liquid $\ce{HCl}$ entails water being below that temperature as well, in which case it would freeze. Additionally, the equilibrium constant for the dissociation of $\ce{HCl}$ is also measured at $25^\circ$ C). Finally, though this is unrelated to your question, if $\ce{HCl}$ was liquid, it would not be included in the $K_a$ expression.
Obtaining the $\ce{\Delta{G^{\circ}}}$ value of $-95.30$ kJ/mol for $\ce{HCl}$ $(g)$, we find that $\ce{\Delta{G^{\circ}}}$ for the reaction is $-35.87$ kJ/mol. Applying the equation $K = e^{-\ce{\frac{\Delta{G^{\circ}}}{RT}}}$, we find that $K_a$ for $\ce{HCl}$ is $1.9 * 10^6$ (within magnitude of your given value).
What this looks like physically is that $\ce{HCl}$ gas is bubbled up into water, and it dissolves (dissociates) into the water as it mixes.


text:
Different names for Fidaxomicin (OPT-80)?
Is OPT in OPT-80 an abbreviation? If so, what is the full name?

Fidaxomicin is the International nonproprietary name (INN) of a pharmaceutical, which was/is distributed by different companies under different trade names.
To my knowledge, OPT-80 was a provisional name/signature used by Optimer Pharmaceuticals Inc.. The drug was later sold as Dificid.


text:
Reaction mechanism for ozonolysis
I have been trying to understand the reaction mechanism for ozonolysis but I am not able to understand what happens after the intermediate trioxolane is formed. I have searched online but didn't really get any intuitive understanding. Can you provide an explanation of how the reaction proceeds after trioxolane is formed?

You might want to run further online searches for ozonolysis + reductive workup (typically using zinc metal or dimethyl sulfide) and ozonolysis + oxidative workup (typically with hydrogen peroxide).

@KlausWarzecha I tried that but most sites give very brief information about them. Can you share any link which you feel is good enough to explain it?

Try searching for the Criegee mechanism (not Criegee rearrangement).  This is well established for ozonolysis of mono, di- and tri-substituted alkenes. Try here Angew. Chem. Int. Ed., 1975, 87, 745

https://sites.google.com/site/mechanismmordor/_/rsrc/1391158462221/first-year-mechanisms/ozonolysis/Criegee%20mechanism.jpg

http://chemistry.stackexchange.com/questions/7117/ozonolysis-conditions-for-benzene


text:
reduce size of zinc dust
I am working with zinc dust (particles of 10um). I want to make a stable dispersion of this dust with arabic gum. The problem is that the particles are still too big, the solution rapidly settles.... My idea to decrease particles size was to use HCl to corrode Zn into Zn2+. And then stopping the reaction by adding soda. I thought getting something light but instead I have a precipitate less soluble than my first zn solution. What happens ? What would be the best strategy to decrease zn particles size ?

I might suggest ball milling, rather than a chemical route.


text:
Why doesn't a C4H8 alkane exist?
This structure:
     H   H
     |   |
 H - C - C - H
     |   |
 H - C - C - H
     |   |
     H   H

Which would be $\ce{C4H8}$, like butene, but an alkane, because it's functional group is C - C.

Um....cyclobutane is a commercially available liquefied gas. Am I missing something here?

With the missing bond added, your structure is fine and part of a family known as cycloalkanes.

https://en.wikipedia.org/wiki/Cyclobutane

It does exist. The formula $\ce{C_{n}H_{2n+2}}$ is only applicable to straight chain alkanes. They can't have any degrees of unsaturation, which is defined as a multiple bond or a ring.


text:
Why do we have glycogen in the muscles and not glucose?
Why do we have glycogen in the muscles and not glucose?
I know this is in contrast to liver cells, which, on demand, readily do break down their stored glycogen into glucose and send it through the blood stream as fuel for other organs. Glycogen is also a suitable storage substance due to its insolubility in water, which means it does not affect the osmotic pressure of a cell.

I'm voting to close this question as off-topic because it's a biology-focused biochemistry question.

I think you already answered your own question.  If the glycogen, either in liver or muscle, were all converted to glucose at once it would have a disruptive osmotic effect. But muscles also readily break down glycogen, when they need glucose to contract.  They just don't ship the glucose to other organs the way liver does.


text:
Why aren't methanal and ethanal also ketones?
Why aren't the aldehydes methanal/formaldehyde and ethanal/acetaldehyde also ketones? Ketones are defined by replacing any H - C - H in an alkane with C=O, which is true for both of these: methanal replaces two hydrogen atoms in methane with C=O, while ethanal does it with ethane.

They are not ketones because the carbonyl group is at a terminal position. In a ketone, both neighbouring atoms of the carbon  in $\ce{C=O}$ need to be carbon atoms. In the aldehydes, one of the neighbours is a hydrogen atom.

That definition of ketones is wrong!

Aldehydes deserve a different name because they have different properties, for example being easily oxidized to acids, as in the Benedict test which distinguishes "reducing" sugars like glucose (an aldehyde) from non-reducing ones like fructose (a ketone).  The relation between aldehydes and ketones is like the relation between alcohols and ethers, both of which can be described as R-O-R, but the properties are significantly different when one of the R's is a hydrogen.

According to the Glossary of class names of organic compounds and reactivity intermediates based on structure (IUPAC Recommendations 1995), ketones are defined as follows:

ketones:
  Compounds in which a carbonyl group is bonded to two carbon atoms: $\ce{R2C=O}$ (neither $\ce{R}$ may be $\ce{H}$). NOC Rule C-311.1. (…)

The mentioned nomenclature of organic chemistry (NOC) Rule C-311.1 refers to the obsolete 1979 IUPAC recommendations:

311.1 – The generic name “ketone” is given to compounds containing an oxygen atom doubly bound to a single carbon atom with the carbonyl group $\ce{>C=O}$ joined to two carbon atoms. (…)

The corresponding rule in the also obsolete 1993 IUPAC recommendations reads:

R-5.6.2.1 Ketones. The generic term “ketone” refers to compounds containing a carbonyl group, $\ce{>C=O}$, joined to two carbon atoms. (…)

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) refers to the definition of the 1995 glossary of class names:

P-64.1.1 Ketones are defined classically as compounds in which a carbonyl group is bonded to two carbon atoms: $\ce{R2CO}$ (neither $\ce{R}$ may be $\ce{H}$). (…)

All definitions include the criterion that the carbonyl group is bonded to two carbon atoms. Since in methanal (formaldehyde) as well as in ethanal (acetaldehyde) the carbonyl group is not bonded to two carbon atoms, these compounds are not ketones.

Functional groups are identified based on their characteristic reactions. Aldehydes and ketones are structurally similar, and reactivity is certainly closely related for both groups. However, there are reactions where the outcome is very different for aldehydes than for ketones. Here is one of them:

Oxidation

Aldehydes react with a variety of oxidizing agents to form carboxylic acids (formaldehyde is often over-oxidized to carbon dioxide).
$$\ce{RC(O)H ->[\ce{CrO3}][\ce{H2SO4}] RCO2H}$$
Ketones, on the other hand, are hard to oxidize. Acetone (dimethyl ketone, 2-propanone) is sometimes used as a cosolvent in the chromic acid oxidation (above).
$$
\ce{RC(O)R ->[\ce{CrO3}][\ce{H2SO4}]} \text{no reaction}
$$
That is not to say that ketones cannot be oxidized, but different reagents are needed:
$$\ce{RC(O)R ->[\ce{RCO3H}] RCO2R}$$


text:
difference between glyceraldehyde-3-phosphate and 3-phosphoglycerate
In the photosynthetic dark reaction there are two types of reactions: light-dependent and light-independent.  In the light-independent reactions during reduction 3-phosphoglycerate is converted to glyceraldehyde-3-phosphate.  I don't find any difference between them because both have a 3 carbon chain attached to a phosphate group. Can you please clear it up for me because my book says that the latter one is a sugar.  Why isn't the first one sugar too?

The difference between the two molecules is highlighted in red.

The functional group in 3-phosphoglycerate is a carboxylic acid. That in glyceraldehyde-3-phosphate is an aldehyde.
Sugars have the general formula $\ce{C_$n$H_{$2n$}O_$n$}$. To accomplish this specific ratio of carbon to hydrogen to oxygen, the requirement is that one of the carbons must have a carbonyl group (i.e. forms an aldehyde or ketone), whereas the remainder of the carbons have one hydroxyl group each. For example, the open-chain structure of glucose is:

The carbonyl group is in red and the hydroxyl groups are in blue.
Therefore, glyceraldehyde-3-phosphate is simply a phosphorylated sugar: C-1 is the aldehydic carbon and C-2 and C-3 have hydroxyl groups, the latter of which is phosphorylated.
3-Phosphoglycerate is one oxidation state too high.

Which makes 3-phosphoglycerate an aldonic acid ;)


text:
Determining change in entropy and enthalpy during mixing of gases

Question taken from KVPY 2011. Answer key is A.
Logically it makes sense that $\Delta S_{\text{system}}>0$, because the process is spontaneous and as per second law, this means $\Delta S_{\text{universe}}>0)$. System is isolated ($\Delta S_{\text{surrounding}}=0$), hence system's entropy must be increasing.
However, it is also given that $$\Delta S_{\text{system}}=\frac{\Delta q}T$$ So does this mean that heat is transferred in some way? How is that possible for an isolated system?
Also, if there is no heat transfer, we can conclude that $\Delta H = 0$. But is that true?

Your equation for $\Delta S_{system}$ is incorrect.  The correct equation is $\Delta S_{system}=\int{\frac{dq_{rev}}{T}}$.  I hope you understand the difference, and the fact that the subscript "rev" refers exclusively to a reversible path between the initial and final states of the system.  To get that, you need to forget entirely about the actual irreversible path, and instead focus on the two end states.  You then have to devise a reversible path between these same two states so that you can determine the value of the integral.  For your problem, this recipe will yield a positive value for $\Delta S_{system}.$
Regarding the enthalpy change, you are aware that the enthalpy of mixing of ideal gases is zero, correct?

Gases mix due to the increase in entropy as each one expands into the new volume.
When the partition is removed the gases expand into the other half of the vessel. Because the gases are ideal they are treated as point particles and so   effectively each gas expands into a void as a free expansion which is an irreversible process. Thus we can treat each gas as if it were in the vessel on its own. As expansion is in effect into a vacuum the gas does no $pV$ work but the entropy increases. The entropy change is
$$\Delta S = nR\ln\left(\frac{V_2}{V_1}\right)  \tag {1}$$
for the expansion of $n$ moles from volume $V_1$ to $V_2$. It is twice this when both gases are considered.
As the system is isolated and consists of an ideal gas the temperature defines the internal energy $U$ which is constant, or $\Delta U=0 $. Since
$\Delta H=\Delta U+p\Delta V$
and as no work is done then
$\Delta H = 0$ .
To calculate the entropy change it is necessary to imagine a reversible process by which the gas now filling both vessels can be returned to its initial condition. This is possible because the entropy change only depends on the  difference in the values between the final and initial states and not on the path taken between them (state function). So if the entropy change on compression can be calculated it  will have same value (except for its sign) as that for expansion between the same two volumes $V_1, V_2$ and at the same constant temperature.
This can be done by imagining that the gas is reversibly compressed (it does not matter how this is done) from $V_2$ to $V_1$ and it is also kept in contact with a thermal reservoir so that the compression is isothermal. (We use a reservoir as the gas must be returned to exactly the same conditions, $T$ and $p$ as before expansion).  For a finite reversible change where $w$ is the work done on the gas and $q$ the heat absorbed by the reservoir then
$$q_\text{rev} = -w = \int_{V_2}^{V_1}p\,\mathrm dV$$
Using the gas law and substituting for $p$ produces
$$q_\text{rev} = -nRT\ln\left(\frac{V_2}{V_1}\right)$$
As the entropy is usually written as
$$\mathrm dS = \frac{\delta q_\text{rev}}{T} \text{ or equivalently } \Delta S = \int \frac{1}{T}\,\mathrm dq_\text{rev}$$
this leads to eqn $\text{(1)}$ but with the opposite sign as its the reverse process to expansion.
The table of the entropy production is as follows. The $\Delta S$ is that of eqn. $\text{(1)}$.
$$\begin{matrix}  & \Delta \text{ gas entropy}  & \Delta \text{ reservoir entropy} & \text{total } \Delta\text{ entropy produced}\\
\text{irrev expansion} &  +\Delta S & 0 & +\Delta S \\
\text{rev compression} & -\Delta S & +\Delta S & 0\\
\text {total both steps} & 0 & +\Delta S & +\Delta S\\
\end{matrix} \ $$
The total entropy change for the gas is zero as it has been returned to its original state. No net entropy was produced in the reversible compression since it was taken up by the reservoir, the initial entropy production was in the irreversible process of free expansion.