Oatmeal swells when it absorbs water and is sticky enough to bind into a near-continuous mass. Water that is not actually absorbed into a grain is trapped by swollen soft, sticky goo. But it is still water, not chemically combined, and if you heat it to 100 C, it will volatilize with great expansion, stretching the goo as it does so, but the bubbles can't make it all the way out.

In a way, it's like a slow motion popcorn, or a speeded-up rising of yeast bread. You can do something similar with a marshmallow.

I found that if you exactly repeat a recipe, the microwave timer could be precisely set to swell the oatmeal to fill the bowl, and not overflow. Then you let it cool down and do it again.


Why do large cations stabilise large anions and small cations stabilise small anions?

For more detailed info, search "cation-anion radius ratio" or "Pauling's rules". Below is my review of this idea.

This ratio is ${R_{C}/R_{A}}$ (C = cation, A=anion).

Small cation (I mean, too small) attracts anions in such a way that they come too close to each other and repulsive force comes into play. This happens when ratio is <0,155. So for larger anion, larger cation suits better. Ratio can be 0.155-1 for stable compounds, and this ratio also determines coordination number and type of void! (Check the picture below, recall that since <0.155 ratio is for unstable compounds, you have empty "example" line there).


Solubility of ortho- and para-nitrophenol in benzene

It turns out that the opposite of what happens in polar solvents takes place when a non-polar solvent is used. At the same temperature, o-nitrophenol is more soluble in benzene than it's m and p isomers.
Sidgwick et al.1 did a study of this and obtained the following results. (Note that the solvent they used was toluene and not benzene, but they are similar enough that the same principles would apply.)

EDIT: The original version of this answer wrongly interpreted the graph and stated that o-nitrophenol would be less soluble in benzene than the other isomers. That was an error on my part. Thank you to Mathew Mahindaratne for pointing it out!
References:

Nevil Vincent Sidgwick, William James Spurrell, Thomas Ellis Davies, “CXXXII.—The Solubility of the Nitrophenols and Other Isomeric Disubstitution Products of Benzene,” J. Chem. Soc., Trans. 1915, 107, 1202–1213. doi:10.1039/CT9150701202.


Sigma  bonds  are  defined as  having their  electron  density  along the  bond  axis, while  pi  bonds  have  their  electron  density  above and  below  the  bond  axis. What this mean is that pi bonds cannot rotate the same way as sigma bonds since rotation would break the pi bond interaction. See the picture below for clarification.

If you're wondering why it rotates then the more  freely  that  a  bond  rotates, the  more  favorable the entropy.  In  general, the  more  kinds  of  motions  and  the  more  unconstrained  those motions are, the  more  favorable the  entropy.


How to differentiate the notation of a compounds charge and an anions oxidations state?

make sure to read up about that topic. The concept of charge and oxidation state are closely related but aren't the same.

Usually oxidation state is written with roman numbers (that's how I learned it) and are not explicitely written like in the example you showed.

Charges on the other hand are always written (if mentionned) on the upper right side of the anion.

I am far from being an expert, but thinks this might be an answer to your question.


This is just a metathesis (double replacement) reaction.

We know that nitrates are soluble, and alkali metals are soluble (at least in these general cases).

Let's ignore balancing for now, and do that last.

$\ce{Pb(NO3)2 + KI -> Pb^{+2} + NO3^- + K+ + I-}$

$\ce{Pb^{+2} + NO3- + K+ + I- -> PbI2 + KNO3}$

But the $\ce{KNO3}$ is also soluble, and would dissociate.

$\ce{Pb^{+2} + NO3- + K+ + I- -> PbI2 + K+ + NO3-}$

Evidently, we can cancel out the spectator ions $\ce{K+}$ and $\ce{NO3-}$ and are left with a net ionic equation (which is easily balanced).

$\ce{Pb^{+2} + 2I- -> PbI2}$


Does computational chemistry include neutrons in ground state calculations?

Depends on what you mean by 'computational chemistry'.
In the most common sense of the term, where the aim of a given calculation is determination of the electronic structure of a particular system, each entire nucleus is treated as a classical point charge/mass.  Thus, neutrons are not explicitly accounted for, aside from their effect on the various nuclear masses in, say, vibrational frequency analysis or in estimates of the kinetic isotope effect.
If you're referring to quantum models of the nucleus, though, then this statement of yours is incorrect:

... most particles in computational chemistry [are] treated as point charges.

In these nuclear computations, if the nucleons were being treated classically it would be correct to say that they are treated as point particles since, as you correctly note, neutrons carry no charge, and proper representation of both the neutrons and protons is necessary in order to obtain an accurate model.
However, as noted quite correctly by Mithoron in a comment, for quantum modeling of the nucleus, the protons and neutrons are not treated as particles but instead as contributors to the nuclear wavefunction.


In Proton Affinities of the Alkali Hydroxides J. Am. Chem. Soc., 1969, 91, pp 2810–2811, it is shown that gas phase affinities in $\pu{kcal/mol}$ are:

  $\ce{LiOH}$: 240.7
  $\ce{NaOH}$: 247.6
  $\ce{KOH}$: 262.6
  $\ce{CsOH}$: 269.2

So, yes, $\ce{LiOH}$ is weaker, at least in the gas phase.

In aqueous solution, ACIDITY FUNCTIONS FOR STRONGLY BASIC SOLUTIONS Chemical Reviews volume 66, pages 119-131:

  the order of basicity for the same molarity of aqueous solution, $\ce{LiOH < NaOH < KOH}$.

For example, for a $\pu{1M}$ solution, the acidity function (H) is:

  $\ce{LiOH}$: 13.48
  $\ce{NaOH}$: 14.01, 14.16 (two different measurement techniques)
  $\ce{KOH}$: 14.17

So in both the gas and aqueous phases, $\ce{LiOH}$ is a weaker base than $\ce{NaOH}$.


Why is alcohol boiled with HBr for preparing alkyl bromide?

Assuming i) a primary alcohol $\ce{R-OH}$ to be converted into the corresponding primary alkyl halogenide $\ce{R-X}$ ii) by $\ce{HX}$ acid only, I would expect a decrease of reactivity to be observed in this line: $\ce{HI} > \ce{HBr} > \ce{HCl} > \ce{HF}$, with $\ce{HF}$ of lowest reactivity.  This is based on the decrease of the nucleophilicity of the anions of these acids.
Of course, halogenations with these four acids are reactions of equilibria, and these may be driven forward by increasing the concentration of the reactants, for example.  And I do recall two-step protocols where the alcohol is activated by sulfuric acid before adding hydrobromic acid.  But in both variants, rather $\ce{H3O+}$ then a neutral $\ce{OH}$ would leave the molecule.
(However secondary and tertiary alcohols that are of higher reactivity than primary ones under these acidic conditions, are more prone to isomerization, too.  Simultaneously, as hinted by ssavec, instead of a mechanism of type $\mathrm{S_{N}1}$, the of  $\mathrm{S_{N}2}$ may become more favorable.)
For the preparation of alkyl chlorides, the addition of (dry) $\ce{ZnCl2}$ in presence of $\ce{HCl}$ is sometimes used.  More easy to handle are routes relying on $\ce{PCl3}$ or $\ce{SOCl2}$, for example.


As @Maurice pointed out, the result of a neutralization reaction is not impacted by the dissociation percentage.

Is $\pu{[}\ce{HA}{] = 10^{−3} M}$ or is $\pu{[}\ce{H+}{] = 10^{−3}}$?

$\pu{[}\ce{HA}{] = 10^{-3}}$.
Always remember, neutralization in aqueous medium depends upon the number of equivalents of acid and base:
$$N_1V_1=N_2V_2 $$  Since $V_1=V_2$, therefore, $\pu{[}\ce{HA}{]=[}\ce{NaOH}{]}$.

Why is it wrong to say that since $\ce{NaOH}$ completely dissociates, and solution is neutral, $[\ce{OH−}]$ $\pu{= 10^−3 M = [}\ce{H+}]$?

The secret lies in the name of the concept itself. Chemical equilibrium. To attain an equilibrium state, all favorable chemical reactions are first carried out. The remaining products/ions constitute the equilibrium state. Coming back to the question, the neutralisation reaction between $\ce{HA}$ and $\ce{NaOH}$ is carried out first:
$$\ce{HA + NaOH ->NaA +H2O}$$
The resultant ions constitute the equilibrium, which, are none unless you consider the hydrolysis of the resultant salt of a weak acid/strong base.


Calculating mass and volume of two different substances required for one solution

We have direct formula for calculating the weight of the substance as,
$$\text{Weight}=\frac{\text{Molarity} \times \text {molecular wt} \times \text{volume in mL}}{1000}$$
$$=\frac{0.02\times 250 \times 100}{1000}$$
$$=0.5 \text{ g}$$

So to prepare 0.02 $ \mathrm{mol~L}^{-1}~\mathrm{Cu}^{2+}$ ,0.5 g of copper(II) sulfate pentahydrate should be dissolved in 100 mL water.

To find the volume of ammonia,

Initial molarity = 2 M

Initial volume= ?

Final molarity = 0.2 M

Final volume = 100 mL

We have formula,
 $$ \text {Initial molarity} \times \text{Initial volume} = \text {Final molarity} \times \text{Final volume}$$
$$\text{Initial volume} = \frac{0.2 \times 100}{2}$$
$$= 10 mL$$

Hence, 10 mL of 2 $\mathrm{mol~L}^{-1} \mathrm{NH}_3$ should be added in 100 mL water to obtain 0.2 $\mathrm{mol~L}^{-1} \mathrm{NH}_3$.


Understanding FHF- molecular orbital diagram?

So we are treating the two fluorines as ligand groups on the right hand side and we introduce a Hydrogen on the left. We observe that each fluorine has a 2S, 2px, 2py and 2pz. If we take the 2pz as being along the internuclear axis then we assign it with sigma symmetry, while the x and y p orbitals have pi symmetry.

It is the 2pz that can interact with the Hydrogens 1S because it has the correct symmetry. It turns out that the symmetry of that orbital is ag. This creates as you say the bonding and anti bonding pair with the H's 1S. It is important to note that the bonding and anti bonding pair are of the same symmetry.

The hydrogen only has the 1S orbital (symmetry ag) and no others, so every other orbital of a symmetry that is not ag is non-bonding since it has no respective partner in the FHF environment.

With regard to the assignation of symmetry to an orbital you must have an understanding of SALCs (symmetry adapted linear combinations of molecular orbitals), if you perform a projection operator method then you will generate the symmetrys ... however this is a bit advanced and not something expected at undergraduate level (to my knowledge).

I hope that helps :)


Antiaromaticity doesn't exist in states that are observable. Its extended presence is a contradiction to its definition. Instead the double bond system exists as an independent system instead of being conjugated. This effect is known as the Jahn-Teller Distortion. A very famous example of said effect is cyclo-butadiene which exists in the form of a rectangular compound as explained in this answer
Proof that this is also the case in biphenylene is as follows:
Taken from the wikipedia page on biphenylene, (emphasis mine)

[...] Questions of bond alternation and ring currents have been investigated repeatedly. Both X-ray diffraction and electron diffraction studies show a considerable alternation of bond lengths, with the bridging bonds between the benzenoid rings having the unusually great length of $\pu{1.524 Å}$. The separation of the rings is also reflected by the absence of the transmission of NMR substituent effects through the central [4n] ring. However, more sensitive NMR evidence, and particularly the shifting of proton resonances to high field, does indicate the existence of electron delocalization in the central [4n] ring. [...]

The reference stated in the Wikipedia article is Tetrahedron 1964, 20 (5), 1179–1184, which states:

Taking into consideration the fact that the molecule is planar and that the two six-membered rings are known to be $\pu{1.52 A}$ apart and assuming that the charge density around the molecule is uniform, we calculated the chemical shifts between the ortho and meta hydrogens due only to ring currents in two cases, i.e., two ring currents (0.13 ppm) and three ring currents (0.34 ppm). The chemical shift for the two ring-current case is in good agreement with the experimentally determined value (0.105 ppm), but it should be borne in mind that ample physical chemical evidence exists to show that the six-membered rings are not entirely independent electronic systems.


Is it stupid to put used acid into a pool after cleaning the chlorinator?

The bicarbonate buffer capacity of the pool  water probably would deal with the remaining acid, but the pH would IMHO get lower.

But returning calcium to pool is not wise, as it will cause calcium accumulation and progressively shortening periods of calcium precipitation.


Is there s-character in the HOMO of dinitrogen?

There is a significant portion of s-p mixing in dinitrogen.
Please see this great answer by Wildcat on Molecular orbital (MO) diagram for $\ce{N2}$ and $\ce{N2^-}$.
For the impatient, I'll quote an image from myself from How to rationalise with MO theory that CO is a two-electron donor through carbon?

I have marked these interactions with the dotted lines. To show these valence MO in all their glory, I've performed a calculation on the RI-BP86/def2-SVP level of theory with Orca.

The analysis (main atomic orbitals) of these orbitals showed, above they are shown in reverse order:
MO  3 (occ.=2.00, E=-1.022230)
  N2 (S on atom N1) [0.416], N16 (S on atom N2) [0.416]

MO  4 (occ.=2.00, E=-0.485050)
  N2 (S on atom N1) [0.435], N16 (S on atom N2) [-0.435],
  N3 (S on atom N1) [0.427], N17 (S on atom N2) [-0.427],
  N15 (S on atom N2) [-0.201], N1 (S on atom N1) [0.201],
  N5 (Px on atom N1) [0.200], N19 (Px on atom N2) [0.200],
  N8 (Px on atom N1) [-0.066], N22 (Px on atom N2) [-0.066],
  N13 (D+2 on atom N1) [-0.015], N27 (D+2 on atom N2) [0.015],
  N10 (D 0 on atom N1) [0.009], N24 (D 0 on atom N2) [-0.009]

MO  5 (occ.=2.00, E=-0.421640)
  N4 (Pz on atom N1) [-0.422], N18 (Pz on atom N2) [-0.422],
  N7 (Pz on atom N1) [0.225], N21 (Pz on atom N2) [0.225],
  N6 (Py on atom N1) [-0.164], N20 (Py on atom N2) [-0.164],
  N9 (Py on atom N1) [0.087], N23 (Py on atom N2) [0.087],
  N11 (D+1 on atom N1) [0.033], N25 (D+1 on atom N2) [-0.033],
  N14 (D-2 on atom N1) [0.013], N28 (D-2 on atom N2) [-0.013]

MO  6 (occ.=2.00, E=-0.421640)
  N6 (Py on atom N1) [-0.422], N20 (Py on atom N2) [-0.422],
  N9 (Py on atom N1) [0.225], N23 (Py on atom N2) [0.225],
  N18 (Pz on atom N2) [0.164], N4 (Pz on atom N1) [0.164],
  N7 (Pz on atom N1) [-0.087], N21 (Pz on atom N2) [-0.087],
  N14 (D-2 on atom N1) [0.033], N28 (D-2 on atom N2) [-0.033],
  N11 (D+1 on atom N1) [-0.013], N25 (D+1 on atom N2) [0.013]

MO  7 (occ.=2.00, E=-0.371870)
  N5 (Px on atom N1) [-0.452], N19 (Px on atom N2) [0.452],
  N3 (S on atom N1) [-0.299], N17 (S on atom N2) [-0.299],
  N8 (Px on atom N1) [0.194], N22 (Px on atom N2) [-0.194],
  N16 (S on atom N2) [-0.134], N2 (S on atom N1) [-0.134],
  N15 (S on atom N2) [-0.074], N1 (S on atom N1) [-0.074],
  N13 (D+2 on atom N1) [0.016], N27 (D+2 on atom N2) [0.016],
  N10 (D 0 on atom N1) [-0.009], N24 (D 0 on atom N2) [-0.009]

MO  8 (occ.=0.00, E=-0.071630)
  N7 (Pz on atom N1) [-0.528], N21 (Pz on atom N2) [0.528],
  N4 (Pz on atom N1) [0.513], N18 (Pz on atom N2) [-0.513],
  N9 (Py on atom N1) [-0.032], N23 (Py on atom N2) [0.032],
  N20 (Py on atom N2) [-0.031], N6 (Py on atom N1) [0.031],
  N11 (D+1 on atom N1) [0.013], N25 (D+1 on atom N2) [0.013],
  N14 (D-2 on atom N1) [0.001], N28 (D-2 on atom N2) [0.001]

MO  9 (occ.=0.00, E=-0.071630)
  N9 (Py on atom N1) [0.528], N23 (Py on atom N2) [-0.528],
  N6 (Py on atom N1) [-0.513], N20 (Py on atom N2) [0.513],
  N7 (Pz on atom N1) [-0.032], N21 (Pz on atom N2) [0.032],
  N18 (Pz on atom N2) [-0.031], N4 (Pz on atom N1) [0.031],
  N14 (D-2 on atom N1) [-0.013], N28 (D-2 on atom N2) [-0.013],
  N11 (D+1 on atom N1) [0.001], N25 (D+1 on atom N2) [0.001]

MO 10 (occ.=0.00, E=0.456580)
  N3 (S on atom N1) [-3.853], N17 (S on atom N2) [3.853],
  N8 (Px on atom N1) [-2.584], N22 (Px on atom N2) [-2.584],
  N16 (S on atom N2) [0.543], N2 (S on atom N1) [-0.543],
  N5 (Px on atom N1) [0.266], N19 (Px on atom N2) [0.266],
  N15 (S on atom N2) [0.162], N1 (S on atom N1) [-0.162],
  N13 (D+2 on atom N1) [0.035], N27 (D+2 on atom N2) [-0.035],
  N10 (D 0 on atom N1) [-0.020], N24 (D 0 on atom N2) [0.020]

The main concern is MO 7, which shows significant s contribution.


Hydroxide formation: reaction of group 13 elements with water

Can anyone tell me why this is the case?

Though the elements reside in same group, they react with water and form different products. This is mainly due to inert pair effect.

Boron

$$\ce{2B + 3H2O ->[700-800 C] B2O3 + 3H2}$$

  Boron react with water to produce boron trioxide(III) and hydrogen.
  This reaction takes place at a temperature of 700-800°C.(source)

Aluminium

$$\ce{2Al + 6H2O → 2Al(OH)3 + 3H2}$$

Gallium

$$\ce{2Ga + 6H2O → 2Ga(OH)3 + 3H2 ~~~~~~(1)}$$

$$\ce{2Ga + 4H2O ->[350 C] 2GaO(OH) + 3H2O~~~~~(2)}$$

  Gallium react with water to produce gallium oxohydroxide and water.
  This reaction takes place at a temperature near 350°C. (source)

Thallium

$$\ce{2Tl + H2O ->[atm. O2 ~~~50-70 C] 2TlOH}$$

  Thallium react with oxygen and water to produce thallium(I) hydroxide.
  This reaction takes place at a temperature of 50-70°C. (source)


Thermodynamically Reversible Processes: How does an infinitesimal step make a process reversible and how is maximum work derived from the process?

I'll give a mathematical answer, and an analogy also for ease of thought. These are my own intuitions, so they may work for you.

My mathematical reasoning:

A lot of thermodynamic processes are phrased very carefully in terms of reversibility. Now my intuition of this is that reversibility is a consequence of the maths we must use.  Thermodynamics likes to look at continuous processes such that we can use integrals instead of summations in the calculations: this stands to reason since classical thermodynamics treats a collection of particles as moving, say, a piston continuously from one place to another, not in discrete steps.

Thus, one way to think about it is to think of reversibility as being necessary for an integral formulation. If we allow ourselves to try and think in infinitesimal steps, moving an infinitesimal step forward is just an infinitesimal away from going backwards and hence this is described as being reversible. In short, if a process really happens continuously then we must impose the condition of reversibility on it, otherwise the mathematics fails.

My analogy:

Why this gives maximum work can be seen by analogy with a real life classical system: the tug-of-war. If you're in a tug of war with some of your friends who are of equal size to you and of equal strength, you're going to put in maximum work in pulling the rope. If one side is of unequal size and strength it becomes much easier for the stronger side.

To be more serious, if your particles move your piston irreversibly (i.e. they never move backwards along the way) then no work has been expended in going backwards—only forwards. Reversible systems belie the work they've had to do since the physical outcome we actually see—the final outcome—shows the piston in the same place as the irreversible case, however what we haven't seen is the system battling backwards and forwards.

To conclude: if we imagine a set amount of work needs to be done to move a piston irreversibly from a point A to a point B, then the reversible case must provide maximum work since some of the time the system has battled going backwards and has still ended up in the same place as the irreversible case.

EDIT: How to Achieve Equilibrium and still have Movement

I think the source of your confusion lies in the statement 'it was at the end of the step of the equilibrium'. I know what you mean by this, but I don't think it's precise. Equilibrium doesn't mean to describe position as such, but the oscillation of position. Even if you've physically pushed your piston a distance ds, it can still be in equilibrium by the very fact that it can still move forwards and backwards in its next spot.

So in fact the periodicity of position is maintained at each step, which is what we mean by the equilibrium is maintained.

If we imagine a spring oscillating backwards and forwards, the movement of oscillation is what we mean here by a dynamic equilibrium. As long as this oscillation is maintained this equilibrium exists, however if we imagine our particles as springs, then the spring must be extended to reach its next spot. Hence in this sense we have equilibrium and movement can still occur.


Oxygen in water: How is it single if it's always double?

I think you are confusing something here: Molecular oxygen is most commonly found as the gas we know, $\ce{O2}$ and ozone, $\ce{O3}$. However, this doesn't limit it to only that. Oxygen can appear in a wide variety of (bio-) chemicals and there, it only needs two bonds to other atoms (or a double bond to one atom).

Good examples for this are:

The aforementioned water $\ce{H2O}$ where the oxygen is bonded to two hydrogen atoms.
One of my most loved molecules: Ethanol. Here, the oxygen is bonded to a hydrogen and a carbon atom.
In an ester, which usually are quite fragrant compounds, there actually are two oxygen atoms. But they are not connected to each other. Instead, one has a double bond and the other two bonds, each of them to carbon respectively.

I hope this has made it clear that oxygen doesn't have to bond to itself (but it can, see peroxides), and that oxygen doesn't have to come in pair of twos for the molecule to not fall apart.


Safety/health hazards: working with 2-aminobenzimidazole, iminoesters, and 1,3,5-triazine

Looking up the chemical you're working with is always a good idea, but you need to look closer and evaluate the specific danger those chemicals present to you. What you want to look at are the danger symbols and the hazard and precautionary statements. Those will tell you the specific dangers a chemical presents.

If you look up your first substance, 2-aminobenzimidazole, it looks pretty tame for organic chemistry standards. It has an exclamation mark as danger symbol, which doesn't tell you much, but the absence of a skull and crossbones and a health hazard symbol tell you that this probably won't kill you immediately. This doesn't mean the stuff is harmless, but it is also not extremely toxic. If you look at the hazard statements of this chemical you'll find that you shouldn't get it into your eyes, on your skin or breathe it. All of that is a good idea when working with any chemical, and this is exactly what you should learn when studying chemistry. This is a really tame chemical compared to a lot of stuff organic chemists work regularly with, but you still should avoid exposing yourself to it.

Your second item is an entire category of chemicals, I'll skip it for this reason. The third one is a bit more dangerous than the first one, it has a corrosive and a health hazard danger symbol. You want to avoid contact with it, but that's what all your protective equipment like safety goggles, your lab coat and your fume hood are for.

I'm not saying those chemicals are harmless, they're not. But they are less dangerous than many other common chemicals in an organic chemistry lab.


What are the organic reactions that are taking place in this reaction?

Here is what I think is going on:

$Step$ $1$: As the OP correctly identified is a Knoevenagel condensation to give A diethyl 4-cyclohexyl-benzalmalonate.
$Step$ $2$ is an example of little-used ester hydrolysis using cyanide ion to give initially the acyl cyanide which is unstable in aq EtOH and gives the diacid B
$Step$ $3$ is loss of one carboxy group by acid catalysed decarboxylation to give the cinnamic acid C
$Step$ $4$ formation of the mixed anhydride D
$Step$ $5$ cyclisation through formation of an aluminium-acyl complex literature example here to give the final product 6-cyclohexyl-Inden-1-one E


Why is p-methoxyphenol less acidic than phenol?

Deprotonation creates a negative charge on the phenolic oxygen. The better you can stabilise this charge, the more favored this process is going to be.

You are right in saying that the charge can be stabilised by delocalisation in the aromatic ring. However, the electrons of the methoxy oxygen can also be delocalised in the ring (to a much lesser extend, of course). This has the effect of making the ring richer in electrons. It will thus tend to less receive additional electrons, and thus stabilise less the formal negative charge. (However, keep in mind that the ring does still stabilise the negative charge, making p-methoxyphenol still much more acidic than alcohols with pKa values around 15)


What happens if you add NaCl during the neutralization reaction between HCl and NaOH?

You yourself said that there is "no equilibrium", but then forgot that Le Chatelier's principle only applies to any system at equilibrium!  Recall its definition:

  Le Chatelier's principle: a principle stating that if a constraint (such as a change in pressure, temperature, or concentration of a reactant) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint.

In your case, as also MaxW pointed out above, $\ce{NaOH}$, $\ce{HCl}$ and $\ce{NaCl}$ are all completely ionized in water (and their ions are hydrated/solvated by water molecules). There is no such thing as an "equilibrium" in this case, as the reaction, being an ionic reaction, completely proceeds in the forward direction.


Different Equivalent Weight for the same compound

Equivalent weights are considering a redox equation with one mole equivalent of electrons. But some redox reactions require more than one mole equivalent of electrons to be equilibrated.

Thus, if you have the second reaction that leads to a weight equivalent of $1/2$ of that for the first reaction, it means that the reaction 2 has consumed $1/2$ of  $B$ compared to reaction 1 for one mole equivalent of electrons.

You still cannot proceed to the molecular weights on these basis, but you can now try to consider that the most likely case is that the first one involved only one electron and the second one two electrons. Thus you can write that this is consistent with:

First equation:

$$\ce{A \rightarrow B + e-}$$

(equilibrated by $\ce{\frac 14 O_{2} + e- \rightarrow {1\over 2} O^2-}$).

And for the second equation:

$$\ce{B + 2 e- -> M}$$

(equilibrated with $\ce{H2 + O^2- -> H2O + 2 e-}$).

One this is said, one should consider the molar weight of $\ce M$ to be twice the equivalent weight of $\ce M$ calculated from the second equation, leading to a molar weight of 63.5 g/mol. Quite unsurprisingly, this is the molar weight of copper, so $M = Cu$. And going backwards, $\ce{B = CuO}$ with copper in its second oxidation state. We can check this, as the molecular weight of CuO is 76.5 g/mol, exactly the equivalent weight calculated for reaction 1. And thus $\ce{A = Cu2O}$ with copper in its first oxidation state.


Effect of temperature in the free radical polymerization of acrylic acid

Yes, this is expected.

With higher temperature, there is a larger concentration of radicals in the polymerization mixture, and initially the polymerization goes faster than at lower temperatures.  This leads to a faster increase in viscosity, which slows the ability of molecules to diffuse in solution.  This means it is more difficult for unreacted monomer to find growing radicals, and for unreacted initiator to find monomer.  Radicals are highly reactive so when they can't react the have a habit of engaging in termination reactions which are nonproductive regarding conversion.

This is normally called autoacceleration, and is fairly well described in Odian's 'Principles of Polymerization'.  Generally, you do not need to heat radical polymerizations beyond the bare minimum to decompose your thermal initiator.


What happens when C17H35COO-, a soap, reacts with an acidic solution?

Your equation is correct. Hydrochloric acid ($\ce{HCl}$) is a stronger acid than stearic acid ($\ce{C17H35COOH}$),i.e. it is dissociated into $\ce{H+}$ and $\ce{X-}$, the anionic rest, to a greater degree. Therefore, it protonates the stearate ion. One can also say that the conjugated base of stearic acid (the stearate ion) is a stronger base than chloride, the conjugated base of hydrochloric acid.


The CH3 and C2H5 groups are both smaller than the rest of the cycle. So these two groups can be forgotten and this molecule is Z.


Well, the answer to your question, without any complications, is pretty simple

Whatever can be forced into planarity in order to bear the $sp^2$ carbocation will undergo $E1$, this can be easily achieved in cyclohexane, as it can easily have a twisted conformation which is relatively less stable, but yet can bear the carbocation. Same is not the case with bridgehead carbons.


Energies of atomic orbitals on molecular orbital diagrams

Ben Norris's comment is correct.

An estimate of the atomic energies for orbitals may be gleaned from ionization energies, for example here:

http://periodictable.com/Properties/A/IonizationEnergies.html

Otherwise, the energies need to be calculated quantum mechanically.


Gaussian vs. Lorentz/Cauchy distributions for artificial broadening

I think that any of these functions, Gaussian, Lorenzian, Voight, will do, they do not differ by much but the Lorenzian has a slightly wider base and in overlapping spectra this may increase background too much. I assume that you will convolute your stick spectrum to get your 'real' spectrum, in this case you can zero pad the transform, it makes nicer looking spectra but adds no more information. Perhaps the most important point to check is  the spectral resolution of the instrument you are simulating spectra from, and adjust the width of you broadening function appropriately.


There are many, here are some examples

Ion interchange reactions of various types, such as precipitation

$\ce{NaCl + AgNO3 -> AgCl \downarrow + NaNO3 }$
Polar molecule insert/ejection. The most common case is hydrolysis and reverse reactions.

$\ce{PCl5 + 4 H2O ->  H3PO4 + 5 HCl}$
Reorganization of bonds between atoms of same type, such as catalytic benzene synthesis from acetylene

$\ce{3 C2H2 -> C6H6}$
Ligand exchange

$\ce{ [Fe(H2O)6]^{3+} + 6CN- -> [Fe(CN)6]^{3-} + 6 H2O}$
polar oligomerization

$\ce{ H2O + n CH2O -> HO-(-CH2-O-)_{n}-H}$


In Goodman and Gilman what does R,S-citalopram refer to?

It seems fairly clear that the authors intended to refer to the racemic mixture there. You're right that a comma would normally be used to separate the locants of distinct stereocenters when multiple ones are present, but then the individual chiral carbons would have to be numbered, which obviously isn't the case here.


In short its kinetically stabilised, attack at S is close to impossible due to steric hindrance and rigidity of FSF angles. Thermodynamically with water it is very unstable with respect to HF$_{aq}$ and H$_2$SO$_4$. Creepy ...


How does a halogen react in an electrophilic substitution reaction

A is correct because the question specifically says an electrophilic substitution reaction (D will also react, but it's an electrophilic addition reaction). However, you also need a Lewis acid catalyst like FeCl$_3$. This is because, while the delocalised system of electrons in benzene is indeed electron rich, the fact that it's aromatic (there is a system of pi electrons delocalised throughout the ring) makes it very stable.
The mechanism of the bromination of benzene is shown below (image from Wikipedia as I don't have ChemDraw on this computer). The mechanism is the same for Cl$_2$.


Can recent developments improve the total synthesis of B-12?

Almost certainly, yes. Applications of modern organic synthesis methods have drastically shortened the synthetic routes for such previously "daunting" targets such as strychnine see Vanderwal, tetracyclines Myers  and pyrroloindole alkaloids Movassaghi, to name but a few - and recent efforts by Baran and White on C-H activation of alkanes have already yielded more efficient syntheses of many complex polycyclic terpenes. There is no reason to think that significant improvement could not occur for chlorophyll or B12. In fact, the absence of chlorophyll or Vitamin B12 from modern synthetic efforts represents a bit of a blind spot; it could well be that new chemistry could be discovered in the process. Although that 1,16 hydride shift by Eschenmoser will be hard to beat :-)


How can I derive the preferred oxidation states of a transition metal from its electronic configuration?

Firstly you need to appreciate that the 4s orbital is at higher energy than the 3d orbitals (despite the filling order suggesting the oposite). This means that the two 4s electrons are lost first giving a valency of two (which is common to all the first row transition elements). The next point is that one of the 3d orbitals is fully ocupied so one of these two electrons will be preferentialy released due to the repulsion of its pair. This results in a valency of three. These are by far the most common valencies of iron. In fact compounds of 4, 5, and 6 oxidation states (valencies) are known but are very unstable, so of little importance.


I will try to summarize what Karl is saying about the ideal gases.
Suppose that you measure the temperature of a gas contained in a canister. Suppose that you measure the temperature $T_i$

Using your equation that you derived from $PV=nRT$, we should have $$T_i=\frac{P_iV_i}{nR}$$
Suppose that we then apply work to the canister. Since it is an ideal gas the only way that we can change its energy is through a change in volume ,pressure, or number of moles. So that is what we are going to do. Suppose that we compress the canister so that the pressure of the gas changes. Thus, our new canister will have a new pressure $P_f$ and a new volume $V_f$.
This volume $V_f$, however, is related to $V_i$ through the ideal gas equation $$P_iV_i=P_fV_f$$
Hence we have that $$V_f=\frac{P_iV_i}{P_f}$$
Lets substitute this equation to find the new temperature of our compressed canister:

$$T_{compressed}=\frac{P_f\frac{P_iV_i}{P_f}}{nR}$$
Which gives us the initial equation. Hence the temperature of the ideal doesn't change when it is compressed.

I will post on what happens when you increase the number of moles.


Dehydration of carboxylic acids using P2O5

$\ce{P2O5}$ is a solid. Hence, it is easy to separate from reaction products. Moreover, $\ce{P2O5}$ is a stronger dehydrating agent than $\ce{H2SO4}$


Is the hydrogen atom attached to the benzylic carbon in benzaldehyde acidic?

Yes and no.

What you discuss is only possible if the $\ce{C-H}$ bond is perpendicular to the ring plane. However, since the carbonyl group is planar and better conjugation can be achieved with the $\ce{C=O}$ bond participating in resonance with the benzene ring the predominant configuration of unsubstituted benzaldehyde should be planar, i.e. the $\ce{C-H}$ bond being in the ring plane.

If you introduce steric hindrance, e.g. 2,6-di-tert-butylbenzaldehyde, the chance of the carbonyl group turning out-of-plane is much greater resulting in a better overlap of the $\ce{C-H}$ bond with the ring system and thus a lower $\mathrm{p}K_\mathrm{a}$.


Stability of meta ethyl and meta methyl benzene carbocations respectively

We know that effect of neither hyperconjugation nor resonance is observed if the groups are attached in meta position. Hence, stability of structures I and IV will be decided on the basis of inductive effect only. Alkyl groups exert +I effect, and ethyl group exerts greater +I than methyl, hence stabilizing the electron deficient benzylic carbocation more. So, IV>I.
Now note that both of these structures must be less stable than II and III, because these are stabilized by both hyperconjugation and inductive effect, and hyperconjugation is a more stabilizing phenomenon than inductive effect. Since II has more $\alpha$-H than III(3 and 2, respectively), there would be more hyperconjugative structures for II, hence more stability. Hence, finally, the order is I<IV<III<II. This is the first option.


Do colligative properties depend on the number of solute molecules?

It's not good notation on the TAs part. IUPAC defines a molecule as having to be electrically neutral. I see what they are intending, but solute particles would be more correct. This is why, as you noted, compounds that dissolve into ions produce a greater effect than the number of original molecules would suggest and molecules that aggregate in solution have a smaller effect. Equations for colligative properties depend on the van't Hoff factor, which essentially converts from the original number of molecules to the number of distinct particles.


Anion-exchanged compound - smaller 1H NMR integrations

On face value, there is nothing here that necessarily suggests a change in relaxation times. Before thinking too hard about what chemistry is happening, you should assure us that your data is reliable and consistent. You should provide details of your experimental conditions; are you using a calibrated π/2 pulse for excitation or a shorter pulse approximation? What relaxation delay are you using? Have you measured T1 relaxations for these nuclei? You could consider repeating your experiment using a much greater relaxation rate between scans to see whether the integrations change.

These are more interesting phenomena in the spectra that need rationalising; loss of fine structure for downfield peaks for instance.

As for your question though, there are lots of examples of counterions changing relaxation times. This is a major tool used for probing outer sphere complexation, and can be used to determine how anions and cations interact around the ligand. It is used widely in protein structure (complexation with paramagnetic cations) and investigations with ionic liquids. If this is a counterion-induced relaxation change, you should be able to account for this by repeating your NMR experiment using appropriate parameters.


Comparing heat of hydrogenation per mole of butenes and 1,3-butadiene

a comparison without actual data should be possible. 1,3-butadiene is
  conjugated so it is the most stable compound and therefore its heat of
  hydrogenation should be the smallest

No, 1,3-butadiene is not the most stable of these compounds, either for the whole molecule or on a per double bond basis (see below).  The actual numbers (see table below) are too close, we need to look at some data in order to answer the question correctly.

  The question asks us to select the compound with the smallest heat of
  hydrogenation per mole out of the following:

  So would we consider each heat of hydrogenation "per mole of hydrogen"
  or "per mole of the given compound"?

Here is some data that will help answer the question.

(image source)

On a "per mole of given compound", trans-2-butene has the lowest heat of hydrogenation, 27.6 kcal/mole.

On a "per mole of hydrogen", 1,3-butadiene has a heat of hydrogenation between 28.2 (56.5/2) and 28.6 (57.1/2) kcal/mole, so trans-2-butene still has the lowest heat of hydrogenation at 27.6 kcal/mole.

So in this particular list of compounds, it doesn't matter whether the question is asking per mole of hydrogen or per mole of given compound, the answer is the same, trans-2-butene.

At first I thought "trick question", but then later thought, not really.  The questioner did his homework and wanted the student to work out the answer both ways and see that since the you get the same answer whether you answer on a per mole compound or per mole hydrogen, you don't need to know whether the question is being asked on a per mole compound or per mole hydrogen basis.


The crystal structure is trigonal, so has very low symmetry and high chance of twinning or other line defects forming during, say, Czochralski type growth methods.  My guess would be the good optical quality stuff sat and annealed in situ for a long time to clear most of the twins/dislocations out.

Support for this comes from A.J. Gratz et al., Geochimica et Cosmochimica Acta 57 491-495 (1993) where they observed, with AFM, calcite growth occuring primarily on screw dislocations.


How do we get g and E values for various levels to calculate electronic partition function

$E_n$ represents the energy of the $n$th electronic state relative to the ground state (i.e. the ground-state electronic energy $E_0 = 0$). Often, electronic states are very high in energy and the excited-state energies $E_1, E_2, \cdots$ are far greater than $k_\mathrm{B}T$, such that $\mathrm{e}^{-E_n/k_\mathrm{B}T} \approx 0$.
Under these conditions, the sum can be collapsed to just a ground-state term
$$q_\mathrm{elec} = \sum_n g_n \mathrm{e}^{-E_n/k_\mathrm{B}T} \approx g_0 \mathrm{e}^{-E_0/k_\mathrm{B}T} = g_0$$
where the latter equality is because $E_0 = 0$. $g_0$ is the degeneracy of the electronic ground state: to find this out you will need to draw some MO diagrams and/or construct the term symbols.
If the ground state is nondegenerate (e.g. $\ce{H2O}$) the whole thing is just equal to 1. Otherwise, for an atom with ground-state term symbol $^{2S+1}L_J$, the degeneracy is $2J + 1$. Similar considerations can be applied to molecules.
However, you will have to watch out for cases where there are low-lying energy states where $E_n$ is on the order of $k_\mathrm{B}T$. I am not aware of an easy way to predict this, though – you will have to know some typical examples, failing which you will just have to look it up, as porphyrin suggests. The NIST WebBook is a good starting place for this.


Decarboxylation of THCA during CO2 Extraction of Cannabis

If this problem has not already been solved by the author, then here: Supercritical extraction, by nature, breaks the carbon bond in the dissolution phase, which is why you are even seeing your biomass decarbed. It's not a heat thing, it's a function of SCF CO2.


You are correct suggesting that 1 μg/kg implies 1 ppb, however the reverse is not true. For instance, 1 ppb can also be 1 nmol/mol, and the reader will never have a chance to deduce which one is it unless you explicitly define the usage of the "parts per something" in the text.
This clutters the manuscript with redundant notes and causes overall confusion.

IUPAC also lists all similar symbols (ppm, ppt, ppb etc.) as deprecated; from IUPAC's “Green Book” [1, p. 98]:

  Although ppm, ppb, ppt and alike are widely used in various applications of
  analytical and environmental chemistry, it is suggested to abandon completely their use because of the ambiguities involved. These units are unnecessary and can be easily replaced by SI-compatible quantities such as pmol/mol (picomole per mole), which are unambiguous. The last column contains suggested replacements (similar replacements can be formulated as mg/g, μg/g, pg/g etc.).

  $$
\begin{array}{lllll}
\hline
\text{Name} & \text{Symbol} & \text{Value} & \text{Examples} & \text{Replacement} \\
\hline
\ldots & & & & \\
\text{part per billion} & \text{ppb} & 10^{-9} & \text{The air quality standard for ozone is a} & \pu{nmol/mol} \\
  & & & \text{volume fraction of}~\varphi = 120~\text{ppb} & \\
\ldots & & & & \\
\hline
\end{array}
$$

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. (PDF)


Which reactions occur during the oxidation and dissolution of bromide/bromine?

A possibilty is that elementary bromine formed -- causing the the yellow color -- then solubilized in water yielding bromine water.

(source)

It does not dissolve well (2.8% ref -- but do not forget the high density (about 3.10 g/mL) of liquid bromine here). It then may disproportionate into then colorless hypobromit and bromide

$\ce{Br2 + H2O -> Br- + BrO- + 2H+}$ (ref.)

especially under light, forming an acidic solution.  The hypobromit may act as an oxidizer.

To draw an analogy, chlorine gas does dissolve into water, too (ref, ref).

Proceed with caution since bromine is corrosive, though.


How to calculate the standard Gibbs energy at equilibrium?

I think the author of this problem forgot to add units to $K_c$ since it's not a dimensionless entity:

$$K_c = \frac{[\ce{NO}]^2}{[\ce{N2O}][\ce{O2}]^{0.5}}$$

and should be $K_c = \pu{1.7e-13 mol^{0.5} L^{-0.5}}$. In general

$$[K_c] = \mathrm{dim}(c)^{Δn}$$

where square brackets denote the dimensions of the quantity $K_c$, $c$ is concentration and $Δn$ is the difference in the amounts between gaseous products and reactants, e.g. here

$$Δn = 2 - (1 + 0.5) = 0.5$$

On the other hand, the equilibrium constant $K$ you use for determining the standard Gibbs energy, must be dimensionless. Since we are dealing with gases only, the easiest way is to use $K_p$, which is dimensionless as required (when normalized to the standard state of pressure $p^\circ = \pu{1 bar}$):

$$K_p = \frac{\left(\frac{p(\ce{NO})}{p^\circ}\right)^2}{\left(\frac{p(\ce{N2O})}{p^\circ}\right) \left(\frac{p(\ce{O2})}{p^\circ}\right)^{0.5}}$$

so that in general

$$[K_p] = \mathrm{dim}(p)^{Δn}\cdot \mathrm{dim}(p^\circ)^{-Δn}$$

$K_p$ and $K_c$ are related (via the ideal gas law):

$$K_p = K_c (RT)^{Δn}$$

So, the equilibrium constant is

$$
\begin{align}
K &= K_p(p^\circ)^{-Δn}\\
  &= K_c(RT)^{Δn}(p^\circ)^{-Δn} \\
  &= \pu{1.7e-13 mol^{0.5} L^{-0.5}}\cdot(\pu{8.314e-2 L bar K−1 mol−1}\cdot\pu{298 K})^{0.5}(\pu{1 bar})^{-0.5} \\
  &= \pu{8.5e-13}
\end{align}
$$

Note that here I used gas constant  expressed as $\pu{8.314e-2 L bar K−1 mol−1}$ since in this case all dimensions are cancelled out and $K$ is left dimensionless. Now we can finally find the standard Gibbs energy:

$$
\begin{align}
Δ G^\circ &= -RT\ln K\\
  &= -\pu{8.314 J mol-1 K-1}\cdot\pu{298 K}\cdot\ln\left(\pu{8.5e-13}\right)\\
  &= \pu{68.9 kJ mol-1}
\end{align}
$$

Here the product before the logarithm includes $R = \pu{8.314 J mol-1 K-1}$ to get answer in $\pu{kJ mol-1}$ straight away.


Ed has already explained the main ideas. Stokes shift is associated with the concept when you excite a substance with light (or electromagnetic radiation). In neon discharge tubes you are using electrical energy to excite the neon atoms. All the emission you see is from neon atoms. Also recall, that molecular emission has band like structure. Neon light spectrum is nowhere close to this band structure.

Buy a small pocket spectroscope or build your own and see how the spectra look like. The black and white picture has not done justice to the actual extremely beautiful spectrum of Swan bands (observed due to diatomic carbon species).

  a more LUMO state, resonating with high energy electricity due to the high stability of noble gases, and then contract back to a more HOMO state, releasing red light, which has a much lower frequency?

This has no scientific meaning, because molecules do not resonate with electricity or electrical energy. I do not know the exact mechanism how electrical energy or even heat is "transformed" into electronic energy, but electrical discharge is quite efficient at exciting atoms or molecules in the gas phase. I am sure it is quite complex, which involves ionization, collisions, recombination etc.


How to find the molecular weight of a given gas based on its density?

Indeed you can use Avogadro's law here, but I guess it is useful to give a little bit more insight in Avogadro's Law. What you can do is apply the ideal gas law:

$$p V = n R T $$

Here $n$ is the amount of substance of gas, which you can also write as:
$$ n=\frac{m}{M},$$
where $m$ is the total mass of the gas and $M$ is the molecular mass.
You can rewrite the ideal gas law now to:
$$p M = \frac{m}{V} R T,$$
where you should recognize that $\frac{m}{V}=\rho$.

The gas constant $R$ is (as the name shows) a constant so you can write
\begin{align}
  R &=\frac{p M}{\rho T} &
  \rightarrow            &&
  \frac{p_1 M_1}{\rho_1 T_1}&=\frac{p_2 M_2}{\rho_2 T_2}
\end{align}

And therefore, for equal pressure and temperature you will find that
$$\frac{M_1}{\rho_1}=\frac{M_2}{\rho_2},$$
which means that if the density of gas 2 (the unknown gas $\ce{X}$) is $10$ times as high, the molecular mass should be as well.

What is interesting to add is that this assumes the ideal gas law to be applicable. If, for example, your gases would be at high pressure (e.g. $\pu{200 bar}$) this is typically not a good assumption so the molecular mass that you would find for a $10$ times as dense gas will not be equal to $10$ times the molecular mass of hydrogen.


Acid dissociation constant and equilibrium constant

An acid dissociation constant is a particular example of an equilibrium constant.
For the reaction:
$$\ce{CH3COOH +H2O <=>CH3COO- +H3O+}$$
The equilibrium constant would be:
$$\newcommand{\ac}[1]{\text{activity of }\ce{#1}}{\bf K_{eq}}=\frac{\ac{CH3COO-}\cdot\ac{H3O+}}{\ac{CH3COOH}\cdot\ac{H2O}}$$
And the acid dissociation constant would be:
$${\bf K_a}=\ce{\frac{[CH3COO-][H3O+]}{[CH3COOH]}}$$
Now the relation between both would be:
$$\newcommand{\acc}[1]{\gamma_{\ce{#1}}}{\bf K_{eq}}={\bf K_a}\underbrace{\frac{\acc{CH3COO-}\acc{H3O+}}{\acc{CH3COOH}\acc{H2O}}}_{\Gamma_0}\frac1{\ce{[H2O]}}$$
Now:
$${\bf K_a}={\bf K_{eq}}[\ce{H2O}]\Gamma_0^{-1}$$
So you're missing some factors.They're not exactly the same.


Spectroscopy: What is an instrument response function?

Let me provide the intuitive idea so that you take the lead from here and understand the notion behind "instrument response" function. Any analytical/physical chemist, who uses an intrument, should have some basic idea of the instrument response. After that you should look up formal concepts of convolution on Youtube.
Whenever you collect experimental data using an instrument, you would like to have no distortions induced by the instrument in your data. For example, if your eyesight is weak, and you are trying to read a road sign, but it appears blurred to you, your eyes and brain have "convoluted" the true image of the road sign. Therefore, your eye/brain response (=instrument response) is to blurr the image and this is what you do not want. At least, one should be aware of this fact that the true image is not a blurred picture.
Hence people who wish to study lineshapes, linewidths in a spectrum they would like to know how their spectrometer set-up is distorting the true experimental result. In an ideal case there should be no distortion but a real instrument can introduce its own effects which we call as convolution.
Let us see a example to illustrate what Gaussian broadening does to a "true" signal shape, using an light absorbance detector (of a chromatograph). I did this just to understand the response function of that instrument.
Imagine that a light source is turning on and off periodically. It should look like a square wave to a light detector (signal is up when the light is on and it is down when the light of off). However, if you have increasing Gaussian broadening done by the instrument, you begin to loose the true shape of the square wave. If you have too much of Gaussian broadening, the square wave, appears to be completely distorted. Look at the top and the bottom figure. The top shows the relatively true signal shape, and the bottom figure shows the effect of Gaussian broadening on the same signal. All true information is lost.
Hope that highlights the importance of instrument response function in analytical chemistry/instrumental analysis.


Is there a potentiometric method for monitoring salt formation in organic solvent?

Add methanol with excess $\ce{H2SO4}$ to your solution. In acid solution to can use HPLC to determine the concentration of your mystery goo compound.
Add water to ppt whatever organic goo you're making.
Mechanically separate your organic goo (filter, centrifuge.). Instead of 2-3 maybe solvent extraction?
Back-titrate excess $\ce{H2SO4}$ in water/methanol solution with $\ce{NaOH}$ and phenolphthalein.

There has to be a wealth of information on the titration of "dicarboxylic acids" in methanol.


Calculating the concentration of product in a solution based on the peak area of HPLC

Your advisor is making a pretty dangerous assumption (although sometimes pharma people do it). The assumption is that the impurity is structurally similar to the main compound and its UV-Vis spectrum is almost similar i.e., their extinction coefficients are the same.
The idea in your formula is that the calibration curve for A will be same as B. Well that might be a first order approximation when the impurity is known to be structurally similar.
If B is a known compound, you should buy a separate standard, and make a calibration curve.


It's not that ionic bonds are weaker than hydrogen bonds, but that the energy cost of breaking the ionic bonds is well compensated by the gain in solvation energy when you put NaCl in solution.  Solvation energy drives the spontaneity of this dissolution process, because of entropic considerations.


You may ease the droplets of water to "fall further down" across the oil by adding some of dishwasher into the water.

As just tested with a glass filled with some of sunflower oil, the water-dishwater droplets reaching the surface air./.oil will deform and flatten when they fall from about 30 cm above on said surface. Eventually however, cohesion and surface tension of water contracts these and they pass across the oil in droplike shape to reach the bottom of the glass where multiple water-dishwater droplets may coalesce.

Limitation:  Since dishwasher is a surfactant, however, it renders the two phases miscible in each other.  Stirring the content of this glas will distroy the bounderies of the two condensed phases.  (Which might be not your concern, though, reading "I could not find any cheap heavy compound miscible with oil [fit for this purpose]".)  It won't create you a thing similar like a "sand clock" which may be used again and again.

Addition since the original post was edited:  James Guides indeed added a good thought to the table.  Perhaps you do not need a liquid A imiscible to an other liquid B to study the fall of droplets, and you may use spheres instead.  Depending on perspective, this then is a problem of fluid mechanics (mécanique des fluides) as in applied biochemistry (e.g., here) and rheology (e.g., sphere viscosimeter).  In a steady state, the settling velocity $V_s$ may be described as

\begin{equation}
V_s = \frac{2}{9} \frac{r^2 g (\varrho_p - \varrho_l)}{\mu}
\end{equation}

with
$r$ the Stokes radius of the particle
$g$ earth's gravitational accleration
$\varrho$ the densities of of the particle ($\varrho_p$), or of the liquid ($\varrho_l$), and
$\mu$, the dynamic fluid viscosity (which depends even more on temperature, than $\varrho$).

Other than the two fields mentioned above, this relation is of importance havesting small crystals for X-ray diffraction analysis.  From there I know about documenting examples as in the screen photo below

taken for a large, a medium, or a small sphere of Teflon falling in a column of glycerol.  It equally allows to understand why you can't harvest small objects in a viscous medium from a bottom of a container with a spoon and have to resort to a fork.  (And with 2 mm of diameter, Brownian motion wont inhibit yet their settlment.)


is it possible for covalent compounds to produce colour during flame test?

The colors produced in flame tests are the result of electrons becoming excited and returning to their ground states, releasing photons of characteristic colors. Different colors from flame tests result from different energy gaps between the electron shells in different atoms. Since flame tests determine properties of individual atoms, they cannot be used to determine if a compound is ionic or covalent. In fact, bonds are normally broken when a compound is placed into a flame, so it would be nearly impossible to use a flame test to find anything about the bonding nature of a compound. Glucose does not produce a color when it is burned because it just undergoes combustion - more or less the same reaction that is creating the flame in the first place.

On the other hand, there are other ways to determine if a compound is ionic or covalent. Ionic compounds normally have ridiculously high melting points, and if you put an ionic compound into distilled water, it will conduct electricity. There are lots of tools to determine if a compound is covalent or ionic; however, a flame test is not one of them.


Calculating [HI] given Kc, [H2], and [I2] for the reaction 2HI  H2+I2


The last time I calibrated an absorbance spectrometer, it was one I made from parts (for an undergraduate Instrumental Analysis teaching laboratory) and I used a 0.0509 M holmium chloride solution I prepared. The visible region spectrum I obtained is here:

Holmium solutions are well known as UV-Vis calibration standards and you can purchase from, e.g., www.starnacells.com. Below is a screenshot from the Starna Cells web site, indicating that their ready-made sealed holmium reference standard cuvette cells have NIST traceability. Evidently, they are good down to 241.0 nm. I would have purchased one, if I had had the money at the time, but the solution I prepared worked well and was not particularly expensive. Erbium and praseodymium solutions also work well.

Sharp peaks in an aqueous solution at room temperature are always nice. Maybe this will help you!


Experimetally :- one reactant is kept constant and concentration of other is changed  and accordingly rate of reaction is measured...then dependence with thta reactant is measured.....and then other one  is kept  constant and ist one is changed then  similarly rate of reaction and thus dependence is measured....... We cant determine the actual dependence of rate of reaction on the concentration of reactants therotically ......we have tk calculate it and then we can formulate the rate law or order of reaction


Conversion factor for conductivity of an aqueous solution

I think you have badly confused yourself by mixing the concept of a property of a solution with the units used to express this property.
Total dissolved solids (TDS) is a property of a solution. Theoretically, it can have any units kg/ kg of a solution, part per million, percentage, or whatever you like. However, people have standardized what they mean by TDS is usually mg solids/L of solution or even ppm.
In the same way, EC is electrical conductivity. It is a property of a solution. It can be expressed in units of microSiemens/cm or any multiple of it. EC itself is not a unit.


Hydrolysis of Sucrose over Sn1Ac or Sn2Ac

The mechanism supplied in the question statement is wrong according to this paper, where the evidence favored Carbon-Oxygen bond cleavage on the fructosyl-side instead of the glucosyl-side of the bridge Oxygen. As that reference states, "the initial products of the hydrolysis are a-D-glucose and a fructosyl carboxonium ion"

But don't feel bad, many textbooks have shown the same mechanism (detailed in the Discussion section of the paper). It's possible that the reaction mechanism changes with conditions, but I know of no evidence for that.

However, the same paper discusses that the mechanism was thought to depend on the type of sugar undergoing bond cleavage:

Results from a computational study show that SN1 mechanisms via a carbocation on fructose and glucose are both feasible, with the one on fructose slightly more likely.

Some other points:

1) The question statement refers to "primary" carbocations twice, and so do three of the comments. But in this reaction the 2 alternatives are a secondary or tertiary carboxoniun ion (for this terminology see Olah (1998) Onium Ions or ref 11 of the above paper).

2) It makes sense that an SN1-type mechanism would take place on the fructosyl-side of the bridge Oxygen since the resulting carboxoniun is tertiary rather than secondary.

Now, the real question is: How would you experimentally determine the mechanism? That is where the science comes in. The fact that so many textbook authors have presented this classic reaction with different mechanisms should be a big lesson to all students.  That expert authors would go to the extent of drawing detailed reaction mechanisms in a textbook that are not backed up by experimental reality, shows how mistrustful we should be of any non-experiment-based opinions. And this is for a classic, textbook example.  How much more unreliable "expert opinions" are for much more complicated things!


Honestly, I don't see how anyone could have predicted this outcome before seeing the actual $\ce{pK_a}$ data.  If you look up the acidities of methanal and ethanal, you find $\ce{pK_a}$ = 13.27 and 13.57 respectively.  A pretty small difference to begin with. As the picture below illustrates, in the case of methanal we are removing the proton directly bound to the carbonyl carbon to produce the anion pictured below.

Now if someone asked, "would pulling the same proton off from ethanal be easier or harder?", you could answer "harder" because that methyl group (being electron releasing) that is attached to the carbonyl carbon would act to slightly destabilize the anion with the negative charge on carbon.  But it turns out that when ethanal is treated with base a different proton is removed!  The proton that is on the methyl group (said to be alpha to the carbonyl carbon) is the proton that is removed.  Without knowing the actual $\ce{pK_{a}{'s}}$ beforehand, how could you know that the anion involved in the methanal case would be ever so slightly more stable than the resonance structures involved in the ethanal case?

Edit: Thanks to Greg E and Nicolau Saker Neto for pointing out an error in the earlier version of the figure and the corresponding text respectively.


What is the chemical equation for wine fermentation?

Generally there are two types of fermentation in biochemical processes:

homolactic fermentation
alcoholic fermentation

Both these processes occur under anaerobic conditions to replenish NAD+
by the reduction of pyruvate in an extension of the glycolytic pathway.

The overall reaction in muscles is:
$$\ce{Glucose + 2ADP + 2Pi -> 2 lactate + 2ATP + 2H2O + 2H^+}$$
For alcoholic fermentation:

Glucose is first converted to pyruvate by glycolysis, and the pyruvate is converted to ethanol and $\ce{CO2}$ in a two step process:

Note: For the second one we just started with pyruvate (to show mechanism and enzymes involved)Otherwise, for alcoholic fermentation, the net reaction is:

$$\ce{Glucose + 2ADP +2Pi + 2H+ -> 2CO2 + 2C2H5OH +2ATP +2H2O}$$

Both homolactic and alcoholic fermentation have the same function: the anaerobic regeneration of NAD+ for continued glycolysis. Their main difference is in their metabolic products

Having said that lets go back to experimental wine fermentation, it is essentially alcoholic fermentation (2).

Enzymes in yeast (zymase a mixture of enzymes) and  cozymase  a mixture of cofactors-coenzymes such as NAD+ , ATP, and ADP, as well as metal ions) help carry out the fermentation reactions.


This is known as the chelate effect. Complexes with bi- or polydentate ligands — chelating ligands to be exact — are observed to be more stable than their nonchelating counterparts.
To understand this, we need to compare apples with apples, so your example is not the best one. (However, based on the trend we can still give a prediction; more on that later.) Instead, I will base my argument on comparing $\ce{[Fe(NH3)6]^3+}$ and $\ce{[Fe(en)3]^3+}$, where $\ce{en}$ is ethylenediamine $\ce{H2N-CH2-CH2-NH2}$.
Consider the following equilibrium:
$$\ce{[Fe(NH3)6]^3+ + 3 en <=> [Fe(en)3]^3+ + 6 NH3}\tag{1}$$
To determine which side is favoured (‘more stable’), we need to take a look at the equilibrium constant $K$. This is defined as
$$K = \frac{[\ce{[Fe(en)3]^3+}][\ce{NH3}]^6}{[\ce{[Fe(NH3)6]^3+}][\ce{en}]^3}\tag{2}$$
But to get a hint on its numeric value without measuring it, we need to apply the Gibbs-Helmholtz equation:
$$\Delta_\mathrm{r}G^⦵ = - RT \ln K\tag{3}$$
Equation (3) tells us that $\Delta_\mathrm{r}G^⦵$ will be negative (exergonic) if $\ln K$ is positive ($R$ and $T$ are positive by definition) — so if $K >1$, i.e. the product side is more favoured than the reactant side. So far no new information. We need to apply another equation, the definition of the Gibbs free energy:
$$\Delta_\mathrm{r}G^⦵ = \Delta_\mathrm{r}H^⦵ - T\Delta_\mathrm{r}S^⦵ \tag{4}$$
For the enthalpic term $\Delta_\mathrm{r}H^⦵$, we need to ask ourselves which bonds are broken and which are reformed. We break six $\ce{Fe\bond{<-}N}$ bonds and form six $\ce{Fe\bond{<-}N}$ bonds over the course of the reaction. Thus we can assume:
$$\Delta_\mathrm{r}H^⦵ \approx 0\tag{5}$$
Therefore:
$$\Delta_\mathrm{r}G^⦵ = - T\Delta_\mathrm{r}S^⦵ \tag{4'}$$
The entropic term $\Delta_\mathrm{r}S^⦵$ is most influenced by the number of particles in solution. The reactant side has four free particles in solution while the product side has seven. Since $7>4$, we can assume the product side’s entropy to be larger. Thus giving:
$$\begin{align}\Delta_\mathrm{r}S^⦵ > 0\tag{6}\\
- T \Delta_\mathrm{r}S^⦵ < 0\tag{7}\\
\Delta_\mathrm{r}G^⦵ < 0\tag{4''}\end{align}$$
Therefore, the reaction as written in equation (1) is exergonic, the products are more stable then the reactants.
Applying this to the problem, we see that most of the ligands are similar enough to not warrant any other stronger effects. Thus, the complex with three bidentate, chelating ligands is most stable.


From paragraph of textbook.
At the electrode-electrolyte interface the metal ions in electrolyte leak out, leaving a negative charge on electrode side of interface. While metal ions migrate at the other side of the interface from the electrolyte solution.
This eventually comes to a stop when the equilibrium is reached. That is the rate of negative charge build-up is equal to the positive charge. Like you say at this point there is no more drive for the reaction to go one way or the other.
However, now because of the charge separation at the interface, there is a potential difference between both sides of this interface which leads to the reduction potential we read out which is not zero. This is where the Nernst equation comes in handy.


Photolysis of N-bromo compound in acidic conditions (radical cyclisation)

This is the Hofmann-Löffler-Freytag reaction, which was later examined by E. J. Corey (J. Am. Chem. Soc., 1960, 82, 1657-1668,
DOI).

When the reaction is performed in sulfuric acid, it is conceivable that the nitrogen atom of the N-Bromoalkane is protonated. Homolytic cleavage of the $\ce{N-Br}$ thus probably leads to a radical cation.

The latter then abstracts a $\delta$-hydrogen atom, resulting in a radical centre in the side chain. It has been suggested that this centre might recombine with the bromo radical previously released. With other words: $\ce{H}$ and $\ce{Br}$ have changed their position. Deprotonation of the ammonium cation and a subsequent nucleophilic substitution, releasing $\ce{Br-}$ would then result in the formation of the indolizidinium skeleton.


It depends on the context

"Substance" is not a tightly defined term so the argument is largely linguistic not about chemistry. Unless, that is, you are talking about observable chemical properties and use the word. Then, the context of use matters and you might need to be careful in how the word is used.

With gases it doesn't much matter how much stuff you have as the individual molecules don't interact much. But for solids and liquids when you talk about "substance" the apparent chemical properties may depend on how many molecules or atoms you have. So you might need to be careful how many of the components are present. The reason why I say this is that some collections of small numbers of components have different properties to larger collections.

Nanoparticles of gold, for example, have different properties to bulk metallic gold. Some show much stronger colours that metallic gold (eg small colloidal gold clusters are used to impart a red colour to some specialty glass products). So the small clusters have different properties to bulk gold and you don't see the bulk properties until the amount of gold is large enough.

So although the substance of both bulk gold and colloidal clusters of gold atoms might be said to be the same, it is probably better to be more specific and talk about colloidal clusters as being a different "substance" to bulk gold. But this isn't universally agreed as usage of substance is not precisely defined.

In short, if you are talking purely about what something is made from then the amount doesn't matter and you can say substance for any amount but if you are talkingg about the properties of the material it can be better to be specific as the properties do change with the amount of the components involved.


Sodium bisulphate (NaHSO4): an acid
Sodium bisulphiTe (NaHSO3): an antioxidant
Sodium metabisulphiTe (Na2S2O5): another antioxidant

In short, bisulphate and bisulphite are not interchangeable, but bisulphite and metabisulphite are.

Sodium metabisulphite is prepared from sodium bisulphite via dehydration, as in this equation:
$$\ce{2 NaHSO3 -> Na2S2O5 + H2O }$$
It's reversible in aqueous solution. The assumption that 30 mg of metabisulphite contain the same amount of sulphite as 30 mg of bisulphite is OK, the error is small, about 10 %.  The correct amount would be 27 mg of metabisulphite.

You'd ask yourself if pasteurization (75 deg. C, 30 seconds) would work to deactivate the polyphenoloxidase without affecting flavour too much.

I'd also suggest searching Pubmed (it's a decent alternative to commercial citation databases) for such keywords as "polyphenol oxidase" (that's the enzyme that causes browning), "inhibition" and "sulfite".

If you want hands-on answers I'd suggest going to the library and get (via interlibrary loan) a copy of this paper: Plant Foods Hum Nutr. 1995, 47, 245-56, "The control of polyphenol oxidase activity in fruits and vegetables. A study of the interactions between the chemical compounds used and heat treatment."

  From the abstract: ... Interactions between the use of ascorbic acid,
  citric acid, EDTA, sodium metabisulphite and heat treatment (70
  degrees C for 2 min) in the control of PPO activity were studied in
  avocado ..., banana ..., apple ..., pear ..., peach ..., potato ...,
  eggplant ..., mushroom ... and
  hearts-of-palm ... . The results demonstrated that
  PPO of avocado and eggplant was most resistant to inhibition by the
  methods used. ... The results indicated that,
  with the exception of PPO from avocado, the most adequate alternative
  method to substitute for the use of SO2 in the control of PPO was a
  combination of ascorbic acid, citric acid and heat treatment.

This shows that how to get on top of browning while still having a presentable product is a tough problem and an active area of research.  Also, avocado seems a specially hard case.


Products of o-aminobenzaldehyde thermal treatment with acetone and further with sulfuric acid

This is an example of the Friedländer synthesis of quinolines Wikipedia article  here.

Under catalysis by hydroxide acetone undergoes an aldol condensation with the aldehyde group to give 3, which is your compound A ($\ce{R2=Me, R3=H}$). Under the catalysis of sulfuric acid A loses water to give the quinoline 7 (for your B $\ce{R2 = Me, R3=H}$).

Several reviews of the Friedländer exist, this is a good one Chem. rev. here.

Image is from ref 1.


This just means that the same type of reaction may proceed by two different mechanisms depending on the exact chemical being used. It's not a particularly widely used term (it is not a "standard" phrase such as wave-particle duality) and there are many other ways of saying the same thing ("divergence", "pathway", etc.).

As an extremely simple example, alkyl bromides can react with nucleophiles via either an SN1 or SN2 pathway, depending on the degree of substitution of the carbon bonded to bromine. This is an example of how a different substrate may follow a different mechanism, although the end result is the same (replacement of Br with the nucleophile).

The papers you cited are just slightly more sophisticated examples of the same concept.


How to remove CH3 labels from a molecule's image? [OpenBabel in Python]

obConversion.AddOption("C",openbabel.OBConversion.OUTOPTIONS)

Referenced openbabel documentation. Setting it to "C" will remove terminal carbons.


Determining chirality after considering conformational interconversion in a cyclohexane

Your analysis is correct. The stipulation "taking into consideration conformational interconversions" means that conformational interconversion is fast. That's important because compounds that we consider achiral can have chiral conformations. I used the example of butane in a previous question.

In this problem, three compounds are given that have multiple stereocenters, compounds b, c, and d. A common way to decide whether a compound with multiple stereocenters is chiral is to look for planes or points of symmetry within the compound. Compound c has a plane of symmetry, so it is clearly achiral. Neither of compounds b or d has a plane of symmetry in the drawn conformations, which means that those conformations are chiral. However, as you suggest, chair flip of b gives the mirror image of the original drawing, which means the original and mirror image are equivalent, assuming that the two conformations can interconvert.

There are many cases where conformations cannot interconvert, and that leads to very interesting compounds that don't even require a stereocenter to be chiral (again see earlier question).


Kinetics of a simultaneous parallel radioactive decay

The question has already been solved by Yashwini and the answer given is correct.$^2$ A more intuitive and specific to question explanation would follow here.
Now, the two reactions given are:
\begin{array}{cc}
\require{cancel}
\ce{A -> P} &(t_{1/2} = 9\,\mathrm h) \\
\ce{A -> Q} &(t_{1/2} = 4.5\,\mathrm h) \\
\end{array}
Now using the rate law, we get,
\begin{align}
-\frac{\mathrm{d}[A]}{\mathrm{d}t}&=k_\mathrm P [A] \tag{1} \\
-\frac{\mathrm{d}[A]}{\mathrm{d}t}&=k_\mathrm Q [A] \tag{2} \\
\end{align}
The rate constant for a first order reaction having a half life of $t_{1/2}$ is defined as:
$$k=\frac{\ln 2}{t_{1/2}} \tag{3}$$
Now, substituting the given values of $t_{1/2}$ into the equations, we get $2k_\mathrm P = k_\mathrm Q$ (since $k\, \alpha \frac{1}{t_{1/2}})$
Now, intuitively since both reactions take place together, it would mean that for every one mole of P formed, two moles of Q forms. Therefore, for every mole of P formed, three moles of A react (since one mole is required for each mole of P and Q).
Now, we add the rate laws ($1$) and $(2)$, since the reactions take place simultaneously, to get:
$$-\frac{\mathrm{d}[A]}{\mathrm{d}t}=(k_\mathrm P +k_\mathrm Q) [A] \tag{4} $$
Now, since using the relation between $k_\mathrm{P}$ and $k_\mathrm{Q}$, we get $k_\mathrm{P} + k_\mathrm{Q} = 3k_\mathrm{P}$
Therefore using the integrated rate law for a first order reaction on equation $(4)$, we get:
$$A=A_0e^{-3k_\mathrm Pt} $$
Now, the amount of $A$ used here would be $A_0 -A$, and we get that value to be:
$$A_\text{used}=A_0\left(1-e^{-3k_\mathrm Pt}\right)$$
Now, as we have previously noted, for every three moles of A used, two moles  Q are formed. This means that the amount of Q now in the mixture would be two thirds of $A_\text{used}$. Therefore the amount of Q would be:
$$Q=\frac{2A_0\left(1-e^{-3k_\mathrm Pt}\right)}{3}$$
Now, we are given the condition, $Q = 2A$, substituting values of $Q$ and $A$ into the given relation we get:
$$\begin{align}
\frac{\cancel{2A_0}\left(1-e^{-3k_\mathrm Pt}\right)}{3} &= \cancel{2A_0}\left(e^{-3k_\mathrm Pt}\right) \\
\implies 1 -e^{-3k_\mathrm Pt} &= 3e^{-3k_\mathrm Pt} \\
\implies 4e^{-3k_\mathrm Pt} &= 1
\end{align}$$
Solving for $t$, we get:
\begin{align}
3k_\mathrm Pt&=2\ln 2 \\ \\
t&=\frac{2\ln 2}{3k_\mathrm P}\\
\end{align}
Now, using equation $(3)$, we get the rate constant $k_\mathrm P$ to be $\frac{\ln 2}{9}$. Substituting this value into the expression for time, we get:
$$t=\frac{2 \cancel{\ln 2}}{\cancel{3} \frac{\cancel{\ln 2}}{\cancelto{3}{9}}}$$
Therefore, time taken for this condition to happen is:
$$t=2\times 3 = 6\ \mathrm h$$


Why is the hydrolysis of ATP to AMP more energetic than the hydrolysis of ATP to ADP?

Dr. Larry Moran (Professor Emeritus in the Department of Biochemistry at the University of Toronto) has posted an nice article about the Free Energy of ATP Hydrolysis. Accordingly, some important concepts in biochemistry may be widely misunderstood and/or not well described in most textbooks. One of them is the free energy of ATP hydrolysis:

The traditional standard Gibbs free energy changes do not apply in biochemical system because there are $\ce{Mg^2+}$ ions, ionic strength, and $\mathrm{pH}$ involve in the transformation. The traditional standard Gibbs free energy changes are at $\pu{25 ^\circ C}$ $(\pu{298 K})$ and $\mathrm{pH} \ 7$. For reactions like ATP hydrolysis, we want a new "standard" that includes free $\ce{Mg^2+}$ ion concentration of $\pu{10^{-3} M}$ and an ionic strength $(I)$ of $\pu{0.25 M}$ (Ref.1).
The transformed Gibbs energy (denoted as $G'$) is defined by the following Legendre transform of the Gibbs energy $G$:
$$G' = G - n'(\ce{H+})\mu(\ce{H+}) - n'(\ce{Mg^2+})\mu(\ce{Mg^2+}) \tag1$$
where $\mu(\ce{H+})$ and $\mu(\ce{Mg^2+})$ are the specified chemical potential of $\ce{H+}$ and  $\ce{Mg^2+}$ at equilibrium, respectively, and $n'(\ce{H+})$ and $n'(\ce{Mg^2+})$ are the total amount of these ions (free and bound) in the system. Accordingly, following are the calculated new values for specific reactions:
$$
\begin{array}{l|c}
\text{Biochemical reaction} & \Delta G'_\text{hydrolysis} \ (\pu{kJ mol-1})  \\
\hline
\ce{ATP + H2O -> ADP + P_$i$ + H+} & -32  \\
\ce{ATP + H2O -> AMP + PP_$i$ + H+} & -45  \\
\ce{AMP + H2O -> Adinosine + P_$i$ + H+} & -13  \\
\ce{PP_$i$ + H2O -> 2P_$i$} & -29  \\
\hline
\end{array}
$$

References:

Robert A. Alberty, Robert N. Goldberg, “Standard thermodynamic formation properties for the adenosine 5'-triphosphate series,” Biochemistry 1992, 31(43), 10610–10615 (https://doi.org/10.1021/bi00158a025).


It's Silver(I) Hydroxide

The solution was very alkaline, as evident by it's pH value of 14. Although  $\ce{AgOH}$ is sparingly soluble, with a $K_{\text{sp}}$ of $\pu{6.8×10^{-9}}$, under alkaline conditions it precipitates as a brown solid.

This is a picture of precipitating silver hydroxide. Is it what you saw as well?


The 'expected' product comes from analogy of addition of strong acids like HBr or acid (often sulfuric acid) catalyzed hydration. However, The pKa of HOBr is 8.7 (ref), much lower than those conditions, so it's not reasonable to make that analogy. Put another way, the base in those reactions is an alkene, which is very weak. Only a very strong acid will protonate an alkene.

In some quick SciFinder searching, I haven't found any overwhelmingly convincing evidence for the regioselectivity observed in the product, but it does seem reasonable to me by a mechanism similar to what Aditya Dev suggests.

I look at HOBr as an oxygen with a leaving group (analogous to Br-Br). Attack by the alkene on the oxygen releases bromide and gives the tertiary carbocation. Kick back by the oxygen closes a protonated epoxide, which is analogous to the bromonium. In the bromonium case, these are considered resonance structures. I don't know if that's definitely the case here (oxygen is much smaller than bromine), but I've written it that way to continue the analogy. In any case, protonated epoxides are typically attacked at the site best able to stabilize positive charge, which is the tertiary carbon. The resulting bromohydrin is the 'observed' product.


What is the electron configuration for Og with a charge of 2-?

According to SCF Relativistic Hartree–Fock Calculations on the Superheavy Elements 118–131 J. Chem. Phys. 53, 2397 (1970), for neutral atoms:

119 $8s$

120 $8s^2$

121 $8p8s^2$

122 $7d8p8s^2$

123 $6f7d8p8s^2$

124 $6f^38p8s^2$

125 $5g6f^38p8s^2$

126 $5g^26f^28p^28s^2$

127 $5g^36f^28p^28s^2$

128 $5g^46f^28p^28s^2$

129 $5g^56f^28p^28s^2$

130 $5g^66f^28p^28s^2$

131 $5g^76f^28p^28s^2$

Later calculations reported in Electronic Configurations of Superheavy Elements J. Phys. Soc. Jpn. 65, pp. 3175-3179 (1996) are different starting with element 122:

122 $8p^28s^2$

123 $6f7d8p8s^2$

124 $6f^28p^28s^2$

125 $6f^48p8s^2$

126 $5g6f^48p8s^2$

127 $5g^26f^38p^28s^2$

128 $5g^36f^38p^28s^2$

129 $5g^46f^38p^28s^2$

130 $5g^56f^38p^28s^2$

131 $5g^66f^38p^28s^2$

However, Electronic structure of eka-thorium (element 122)
compared with thorium J. Phys. B: At. Mol. Opt. Phys. 35 (2002) 1693–1700 finds:

122 $7d8p8s^2$

specifically agreeing with the 1970 paper and disagreeing with the 1996 paper.

See A suggested periodic table up to Z[=]172, based on Dirac–Fock
calculations on atoms and ions Phys. Chem. Chem. Phys., 2011, 13, 161–168 for additional information.


I'm not too sure, but let me try. In the case of the F3- ion, the effective bond order is 1 (see attached images), not 2. In other words, the only effective bonding MO holding the entire thing together is a 3-centre 2-electron bond (think linear H3+ and FHF-).
I presume you know how the F2 molecule MO diagram looks like, so I'll just explain what F3- looks like. Using the D2h character table and forming our reducible representations, we get the following:

Note s-p mixing occurs in the Ag and B1u orbitals, as shown:

The MO diagram for F3- would look disastrously messy, so I'll simply show how the molecular orbitals look like and how they're ordered.

The first thing you'll probably notice is how the central fluorine isn't allowed to s-p mix anymore; although the symmetry remains D2h the now-presence of a central atom means the s and p orbitals of the central fluorine are no longer of the same symmetry, and no longer interact. That, in addition to the fact that the sole bond that is effectively holding the ion together is a 3-centre 2-electron bond, makes answering your question rather difficult.
I think a way to consider it would be in terms of the formation of a Lewis adduct; in other words, the pz electrons in the F- ion interact with the empty LUMO of the F2 molecule while all the lower-energy orbitals form bonding/antibonding sets with the lower-energy F2 MOs, forming a new HOMO and LUMO from the "HOMO" of F- and LUMO of F2.
Hope this helped.


Fluorine the element is very reactive, but that doesn't mean its compounds are

Fluorine the gas is a very reactive element because it loves to take electrons from other compounds. But this is irrelevant to whether compounds containing fluorine are reactive. Consider the more common chlorine. Chlorine gas is less reactive than fluorine, but is still very reactive. It is a dangerous and corrosive compound that has been (and probably is still) used as a military weapon. But you encounter very unreactive compounds of chlorine every day. Common salt is a chlorine-containing ionic compound that makes up a significant portion of the electrolytes in your body; PVC is a fairly inert plastic made from a polymer containing chlorine carbon bonds.

The very reactivity of fluorine makes some of its compounds very unreactive. F-C bonds, for example, are very strong and often very unreactive even compared to other carbon-halogen bonds. PTFE (a long carbon chain with all the hydrogens replaced by fluorine) is a very inert polymer that is safe enough to use at the high temperatures achieved in frying pans, for example.

Some fluorine compounds are, however, very biologically active but not because their fluorines are "reactive". Fluoroacetic acid and its salts, for example, are dangerous (and naturally occurring) poisons. But their mechanism of action relies on their stability not their reactivity: they are similar enough to acetic acid that they get involved in the standard citric acid cycle but end up blocking one of the key enzymes to further reactions because the product is too stable.

Whether or not fluorophenibut is "reactive" in the body in the sense you mean is unlikely. It is a neurodepressant similar to gabapentin and pregabalin, though little is known about its safety.

Complicated molecules can't be judged as dangerous, safe or reactive just because of the elements they contain. You need to know the detailed and specific ways they interact with bodily processes. Prozac, for example, is a widely used anti-depressant that contains fluorine. Just because it does doesn't make it reactive or dangerous.


Configuration change at chiral carbon in SN2 reaction

I am going to use ‘Pseudo-Fischer’ structures in this post, i.e. anything vertical should be considered behind the PC screen plane and anything horizontal in front of it.

Your original molecule and the hypothetic $\mathrm{S_N2}$-type inversion is here:

   Pr            Pr
   |             |
 H-C--Me ---> Me-C--Br
   |             |
   Et            Et

  (A)           (C)

If you determine both compounds’ absolute configuration, naturally we will have (R)-A but look closely at C and realise that it is also (R)-C. This is because of one of two fallacies students often have with CIP priorities (I’m presenting the other one here just for completeness):

  A chemical inversion of a specific stereocentre does not have to mean an inversion of the stereodescriptor.

  Just because a stereocentre does not participate in a reaction does not mean its stereodescriptor remains identical.

In our case, we can easily arrive at that conclusion by the two-step process you outlined. First, invert the original structure A to give B, then replace (under retention) the hydrogen in B to give C.

   Pr            Pr            Pr
   |             |             |
 H-C--Me ---> Me-C--H  ---> Me-C--Br
   |             |             |
   Et            Et            Et

  (A)           (B)           (C)

We already established that A is (R)-A. Going to be, this is a strict inversion and only in this case of a strict inversion will the fallacy of bullet point one actually hold true; thus, it is (S)-B. Finally, we replace the atom of lowest priority with one that will automatically become the highest priority. To explain what happens requires further elaboration.

First, assume without loss of generalisation that whichever group of B had second priority (in our case: ethyl) is pointing upwards in a Pseudo-Fischer structure while whichever group originally had third priority in B (here: methyl) is pointing downwards. This gives us depiction 1.

   Et
   |
Pr-C--H
   |
   Me

  (1)

As we have the highest priority on the left above and the lowest priority on the right above, it is easy to see that this compound has an (S) configuration. If the hydrogen and propyl groups had been switched, we would have had (R). Now, perform the replacement in a manner such that the former fourth priority becomes first. The original third priority is now fourth, second became third and first became second. This is shown in depiction 2.

   Et
   |
Pr-C--Br
   |
   Me

  (2)

Now, since our lowest priority is pointing downwards, to assign a stereodescriptor in 2 we must look from above. The first and second priorities are above the PC screen plane and on the left and right-hand sides. The third priority is behind the plane and centred. It is rather easy to see that in the case of 2 as drawn, with the right-hand group having a higher priority than the left-hand group, the stereodescriptor is (R). If they were reversed, it would be (S). We see that replacing the fourth priority with something of highest priority will invert the stereochemistry.

What happens if you replace the lowest priority with something of intermediate priority (i.e. something that will come in second or third)? The third case is easy; you are just switching two groups. Switching two groups will always give you a formal inversion. What if our replacement group takes second priority? I refer back to depiction 1 but this time replace hydrogen with $\ce{BEt} = \ce{(pin)B-C2H4}$ giving depiction 3.

   Et
   |
Pr-C--BEt
   |
   Me

  (3)

Again, we must look from above. The highest priority is on the left-hand sinde in 3, the second is on the right-hand side, the third is in the centre and behind. Our journey is counterclockwise or (S). Had the original configuration been inverted it would be identical. Thus, replacing a fourth priority with something of second priority retains the stereochemistry.

This is a one step analysis on a per-case basis. A.K. showed the usage of the switching method (two groups switch at a time) to its fullest; thereby, these independently established rules can also be derived as was shown.


Am I correctly showing a hydride shift on the alkyl halide?

When you solvolyze the starting bromide a secondary carbocation would be generated.  Since a secondary carbocation is not particularly stable, the molecule will explore other pathways such as hydrogen shift, alkyl shift, etc., in order to produce a more stable carbocation.  The figure below, explores these options for your starting bromide.  Realize that although I've drawn a discreet secondary carbocation that then undergoes rearrangement, in reality cation generation and alkyl or hydrogen shift may occur in a more or less concerted fashion.

In this case there are 3 options (actually there are 4 options, but the fourth one would produce a primary carbocation, so I left it out - can you find it?), a, b and c. Pathways a and c produce secondary carbocations; however pathway b, involving hydrogen shift, produces a more stable tertiary carbocation.  Capture of the tertiary carbocation from pathway c by solvent should produce the expected product.


What exchange timescales are needed to cause decoupling in NMR?

You need to differentiate what gets averaged, a CS difference or a scalar coupling.

Methyl protons have a different chemical shift depending on the current conformation, but they rotate so fast this gets averaged out perfectly, and they have a very sharp signal. They show no coupling between themselves because it´s an $AB_2$ system with CS difference zero, so $A_3$ actually.

$J$ coupling to a methyl group (e.g. the ethyl group in ethanol) gives sharp lines and clear splitting, because the coupling varies very rapidly (with the methyl rotation) around a stable and finite value, and it only depends on the occupation of the spin states in both groups.

The coupling methylene to OH is inverted every time the proton gets exchanged for one with different spin state, so it averages to zero. The exchange happens so often that the methylene group still gives a sharp quartett from the methyl.

The OH group coalesces with the water peak. Because the exchange happens randomly and differently for each molecule, the coherence in your spin ensemble deteriorates, hence the line broadening. Note: It´s not a splitting that get´s averaged out, but a change in chemical shift.

If the proton exchange was even faster, the line would sharpen again at some point I guess. Just like the methyl protons themselves. If the exchange was even slower (milliseconds), the line would broaden so much it became invisible. If you have a really dry sample, the exchange becomes so slow it does not happen on the NMR time scale, the line becomes sharp, and you even see the $^3J$ coupling to the methylene group.


The intensivity of a variable doesn't depend on the homogeneity of the system.

Consider system A, a metal rod of length 2 with a gradient of temperature :

This metal rod has an average temperature, which you could calculate.

Now consider system B, a metal rod of length 4 with the same gradient of temperature :

This metal rod also has an average temperature. Furthermore, it is the same temperature as system A. (if you are unsure about that try to pick a gradient (I arbitrarily chose T = exp(-x) and T = exp(-x/2) for the graphs) and do the calculations yourself)

System B is bigger (it would be twice as massive) but its temperature is the same. Ergo the temperature doesn't depend on the size of the system and is an intensive variable.


Why is density of chloroform greater than water?

The density of a substance depends on its molecular density and packing fraction (where you need to be consistent in definiting the molar volume in calculating both attributes).
However you are looking for a simpler explanation that gets the general trends right.   Just looking at molecular mass is too simple—larger molecules have greater mass, but take up more space.  So what would be a simple model that works reasonably well?
I suspect the main determinant of density is the average mass of a molecule's atoms, and that relative atomic sizes, as well as packing density, while important, are secondary contributors.
Given this, let's construct our very rough model for predicting relative densities by comparing just the average atomic masses of the molecule's atoms, and see if that, by itself, can explain most of the trends.
Of course, since we are ignoring these other factors, our simple model is imperfect and thus, as expected, will have many exceptions (one of which is included, and bolded, below). Nevertheless, in spite of the very simple nature of this model, it does surprisingly well in matching the general trends (at least for the substances shown here).
In each case I have calculated the average atomic mass of the atoms (u/atom) from $\frac{\text{molecular mass}}{\text{no. of atoms}}$:
Hexane ($\ce{C6H14}$): 86.18/20 = 4.3 u/atom; density = 0.661 g/$\text{cm}^3$
Octane ($\ce{C8H18}$): 114.23/26 = 4.4 u/atom; density = 0.703 g/$\text{cm}^3$
Water ($\ce{H2O}$):  18.015/3 = 6.0 u/atom; density = 0.997 g/$\text{cm}^3$ (at 25$^\circ$C)
Methyl chloride ($\ce{CH3Cl}$):  50.485/5 = 10.1 u/atom; density = 0.911 g/$\textbf{cm}^{\bf{3}}$   (note the exception to the trend; it works well in comparing like compounds—this one vs. the ones below—but fails when comparing very unlike compounds—this one vs. water—when the average atomic masses of the atoms are too close)
Methylene chloride ($\ce{CH2Cl2})$: 84.927/5 = 17.0 u/atom; density = 1.325 g/$\text{cm}^3$
Chloroform ($\ce{CHCl3})$: 119.369/5 = 23.9 u/atom; density = 1.492 g/$\text{cm}^3$
Carbon tetrachloride ($\ce{CCl4}$): 153.8118/5 = 30.8 u/atom; density = 1.594 g/$\text{cm}^3$
Carbon tetrabromide ($\ce{CBr4}$): 331.627/5 = 66.3 u/atom; density = 3.42 g/$\text{cm}^3$ (m.p. = 89$^\circ$C, so it's solid at room temperature)
Carbon tetraiodide ($\ce{CI4}$): 519.628/5 = 103.9 u/atom; density = 6.36 g/$\text{cm}^3$ (m.p. = 168$^\circ$C, so it's solid at room temperature)
Mercury ($\ce{Hg}$): 200.592/1 = 200.6 u/atom; density = 13.534 g/$\text{cm}^3$


Dispersion forces - what determines which molecule will have its electron cloud repelled?

You are correct that the electron clouds would repel each other equally.

The induction as shown in your second picture would require another outside entity, polarizing one of the He atoms, which would then affect the other.

At an arbitrarily show distance, we might describe a molecular orbital formation between the two helium nuclei, as shown here:

(source: mhhe.com)

In this case, without the external perubation, the orbital shape (and hence the electron clouds) is (are) quite symmetrically distorted.

However, it becomes nonesense to describe a system of two atoms.  In liquid helium there are manifold other charge fields distorting these nice shapes, and you may get some asymmetry as shown in your picture above.


Why is a comma used to separate hydrogen gas and platinum when using cell notation?

Since $\ce{Pt}$ is not a reactant and only a catalyst, one only wants to write $\ce{H_2}$ on the reactant side. When it is specified, since it is not involved in the reaction, the | is not warranted since Pt doesn't show up in the reaction as the comment by TAR86 pointed out.

Generally, the electrode material is not entered into the shorhand notation since the material of the electrode doesn't change the thermodynamic potential. In this case, the choice of $\ce{Pt}$ is critical and that is why it is specified. The potential of the electrode would be affected by the details of the measurement device with any other electrode material that is a worse catalyst.


Is it possible, in principle, for a second row homonuclear diatomic system to show a bond order of 4 in an excited state?

The normal excited state is one in which a single electron is promoted from a lower orbital into one of the anti-bonding ones. Thus the situation is initially $\cdots g^2$ for 2 electrons in a ground state orbital g (need not be the highest orbital but often is), and, say, after absorbing a photon of the correct energy the situation is $\cdots g^1a^1$ where a is an anti-bonding orbital. This combination of orbitals is called the excited state. (If the electron spin remains paired it is called a singlet excited state, if unpaired ('parallel') it is called a triplet excited state). As one electron is now in an anti-bonding orbital the molecule has overall weaker bonding. (However, the NO excited state may be an exception).

The excited state always decays back to the ground state (it in not a stationary state of the hamiltonian) with a lifetime that depends on the particular molecule or atom and between different molecules the lifetime
can have  a huge range, from less than a picosecond to several seconds.


Why is lutetium part of the lanthanide series?

The assignment of elements to groups or series in the periodic table is not based exclusively on the electron configurations concerned. Helium for example is an s-block element yet is placed above Ne in the p-block due its noble gas properties.

Lutetium is part of the lanthanide series, notwithstanding its electron configuration, because its properties and those of its compounds are similar to those of the other lanthanide elements.

Lawrencium is assigned to the actinides for basically the same reason. Experiments have so far failed to find any chemistry-based evidence for Lr having a p electron. While it may have a p electron in the gas phase it probably has a d electron in the condensed phase.

Whether either Lu and Lr, or La and Ac are assigned to group 3 will not make any difference to the composition of the lanthanides or actinides, since both of these are horizontal series rather than vertical groups.