text:
How do I solve questions pertaining to rate of disappearance and appearance?

In a reversible reaction $\ce{2NO2 <=>[$k_1$][$k_2$] N2O4}$, the rate of disappearance of $\ce{NO2}$ is equal to:

$2\frac{k_1}{k_2}\ce{[NO2]}^2$
$2k_1\ce{[NO2]}^2-2k_2\ce{[N2O4]}$
$2k_1\ce{[NO2]}^2-k_2\ce{[N2O4]}$
$(2k_1-k_2)\ce{[NO2]}$

The answer, they say, is (2). I just don't understand how they got it. I find it difficult to solve these questions. Is the rate of disappearance the derivative of the concentration of the reactant divided by its coefficient in the reaction, or is it simply the derivative?
If someone could help me with the solution, it would be great. And please, don't assume I'm just picking up a random question from a book and asking it for fun without actually trying to do it. I have worked at it and I don't understand what to do.

In general, if you have a system of elementary reactions, the rate of appearance of a species $\ce{A}$ will be
$$\cfrac{\mathrm{d}\ce{[A]}}{\mathrm{d}t} = \sum\limits_i \nu_{\ce{A},i} r_i$$
where

$i$ is each reaction in the system
$\nu_{\ce{A},i}$ is the stoichiometric coefficient of species $\ce{A}$ in reaction $i$ (positive for products, negative for reagents)
$r_i$ is the rate for reaction $i$, which in turn will be calculated as a product of concentrations for all reagents $j$ times the kinetic coefficient $k_i$:

$$r_i = k_i \prod\limits_{j} [j]^{\nu_{j,i}}$$
In other words, there's a positive contribution to the rate of appearance for each reaction in which $\ce{A}$ is produced, and a negative contribution to the rate of appearance for each reaction in which $\ce{A}$ is consumed, and these contributions are equal to the rate of that reaction times the stoichiometric coefficient.
The rate of disappearance will simply be minus the rate of appearance, so the signs of the contributions will be the opposite.
In your example, we have two elementary reactions:
$$\ce{2NO ->[$k_1$] N2O4} \tag{1}$$
$$\ce{N2O4 ->[$k_2$] 2NO} \tag{2}$$
So, the rate of appearance of $\ce{N2O4}$ would be
$$\cfrac{\mathrm{d}\ce{[N2O4]}}{\mathrm{d}t} = r_1 - r_2 $$
Similarly, the rate of appearance of $\ce{NO}$ would be
$$\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = - 2 r_1 + 2 r_2$$
and the rate of disappearance of $\ce{NO}$ would be minus its rate of appearance:
$$-\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = 2 r_1 - 2 r_2$$
Since the rates for both reactions would be
$$r_1 = k_1 \ce{[NO]}^2$$
$$r_2 = k_2 \ce{[N2O4]}$$
the rate of disappearance for $\ce{NO}$ will be
$$-\cfrac{\mathrm{d}\ce{[NO]}}{\mathrm{d}t} = 2 k_1 \ce{[NO]}^2 - 2 k_2 \ce{[N2O4]}$$


text:
SN1 reactivity order for chloromethyl methyl ether and trityl chloride

Compare the reactivity order towards SN1 reaction for:

(2-chloropropan-2-yl)cyclopropane
$\ce{CH3OCH2Cl}$
$\ce{Ph3CCl}$

Owing to the exceptional stability of cyclopropyl ethyl carbocation (see), obviously it should be on the top.
We did this question in class and my teacher told me that option 2 is more reactive than option 3, because the carbocation formed in the case of 2 is more stable. However, I couldn't quite understand the reason for it.
Anyway, I searched more for this and found the following table in March's 7th edition, page 425:

According to this table $\ce{Ar3CX > CH3OCH2X}$ in terms of reactivity towards SN1.
Today, I went to my teacher with this data and asked him about it. He told me that things in March can be wrong sometimes, I should rather check some other sources and prove it to him.
So I'd like to know what is the actual order, is 2 more reactive than 3, or vice-versa? (please mention your source) And why?
I have also checked Peter Skyes' and Morrison Boyd's, Clayden's book but it's not given there.

This issue has been discussed  previously on the site but no concrete conclusion has been reached.
Jan's answer suggests that $\ce{CH3OCH2Cl}$ should be more reactive than trityl chloride.

Unfortunately, a table like this is not super helpful. What are the reaction conditions? Solvent? What is your nucleophile? All of these factors can influence the rate of reactivity. Also, it is incorrect to say that a stabilized cation increases the rate of an SN1 reaction. That is a shortcut obtained by applying the Hammond Postulate. Here, you are assuming that dissociation is rate limiting and endothermic, and therefore, the transition state is late and cation-like. Therefore, factors that stabilize cations will lower the energy of the transition state and increase rate.

The reaction the rate with 2 is higher than you thing is because the intermediate transition state is stabilized by donation of electron density from the adjacent oxygen. By the Hammond Postulate, the transition state is late and electron density will lower the barrier to reactivity.

In the table under Sn2, shouldn't the first three groups be in the opposite order because of steric hindrance?

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text:
Why only F, O and N form Hydrogen Bonds?
Why only fluorine, oxygen and nitrogen can form hydrogen-bonds with the hydrogen of another molecule?

That's a common misconception, that is thoroughly incorrect.

No, that's not necessary. Amongst the strongest hydrogen bonds are formed by $\ce{ N, O, F}$ because of their high electronegativity.
There's evidence that the ions $\ce{Cl-, I-, Br- }$ form hydrogen bonds that are much stronger than those of the covalently bonded atoms
$\ce{Cl}$ can form weak hydrogen bonds, but $\ce{Br}$ and $\ce{I}$ form very weak bonds if at all.
A sulfur atom can also function as weak acceptor., but the $\ce{SH-}$ ion forms much stronger bonds.
Hydrogen bonding has been directly observed between a negatively charged carbon and an $\ce{-OH}$ group in the same molecule
Isocyanides' ($\ce{R- N+ #C- }$) carbon atom can also act as an acceptor (forming a rather strong Hydrogen bond).

Reference: March's Advanced Organic Chemistry, Chapter 3: Bonding weaker than Covalent

Just to avoid future confusion: the NOF hydrogen bonds are not necessarily stronger than others either.

I think they are, why not? @LinearChristmas . Other atoms form hydrogen bonding but the strongest is formed by N O F atoms...

@santimirandarp While many strong and common hydrogen bonds are formed with NOF, it is not a sufficient criterion of assigning superiority when comparing strengths of two hydrogen bonds. In other words: while many strong H-bonds are indeed of the NOF variety, there are still cases where others are stronger. One must still carry out meticulous experiments, or perform accurate calculations to make such assignments. See my earlier comment on another question for a concrete example.

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text:
What is the chemical reason talc is used in water-based paint and not in solvent based paint?
As I understand, paints & coatings use talc predominately in water-based products and not in solvent-based products. Why is this? Is it because the solvent-based products don't need fillers? Or do they already have the characteristics that talc would provide?

Most water based paints are emulsions of plastics particles dispersed in water, it is common to use vinyl acetate to make polyvinyl acetate. To do this an emulsion of vinyl acetate in water is formed, then by adding a free radical initiator.
One purpose which fine clay can furfill in the paint polymer synthesis is to increase the stability of the emulsion of droplets of the liquid monomer. These clay particles will be incorperated into the final paint.
It is also common to add other inorganic substances such as barium sulfate and titanium dioxide. These are used to make the paint more opaque.
One classic organic based paint class is the alkyd paint, these are made by reacting acid anhydrides such as phthalic and maleic anhydride with a polyalcohol such as glycerol or 2,2-dimethylpropane-1,3-diol under acidic conditions. It is normal to distill water out of the reaction, often an unsaturated fatty acid is added in the form of a glycerol triester (vegtable oil).
The idea is that with glycerol the fatty acid ester will react in a transestification reaction to add pedent groups onto the main polymer. Then when the paint is exposed to oxygen a radical cross linking reaction occurs. The other main option I see is to add some maleic acid to the synthesis of the paint binder. The to add some styrene to the paint. This again will cross link with a radical reaction.
For the alkyd paint synthesis there is no need to add talc, these polymerisations are done under homogenous conditions by heating a close to solvent free mixture of monomers. Rather than being a radical polymerisation (example of an addition polymerisation) this reaction is a condensation polymersation where many small chains are converted into a few long chains.

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text:
Why is SN1 mechanism followed in diazo aromatic compounds' nucleophilic substitution?
There is considerable evidence that SN1 mechanism is followed in diazo aromatic compounds' nucleophilic substitution. I would like to know the reason for this and why is it preferred to SNAr mechanism even when the $\ce{N2+}$ could stabilize the carbanion-type intermediate through -I effect.
Does it have to do with the excellent leaving tendency of $\ce{N2+}$ but at the cost of formation of unstable phenyl cation?

Generally, you describe the leaving group as it would be it had left, in this case, dinitrogen ($\ce{N2}$). So, then you're looking at a heterolytic bond cleavage that is highly favorable against a bimolecular $S_{\mathrm{N}}Ar$ mechanism. Under concentrated conditions, the bimolecular mechanism might be competitive.


text:
For a given period, why is the boiling point of the halogen greater than that of the noble gas?
The boiling point of bromine, a halogen, is $\pu{58.8^\circ C}$, while the boiling point of krypton, the noble gas in the same period as bromine, is $\pu{-153.4 ^\circ C}$.
I thought that the larger the atomic radius of an element, the more loosely the electrons would be held by the nucleus. So it would be easier for the atom to develop an instantaneous dipole, strengthening the London forces between the atoms of the element, and increasing the boiling point of the substance.
Krypton has a larger atomic radius than bromine. Using my reasoning from above, I thought that the boiling point of krypton would be higher than that of bromine. However, bromine actually has a higher boiling point than krypton.
Why is this? And where is my reasoning incorrect?

Complete data for reference: notice that the boiling point of a halogen is always higher than that of the corresponding noble gas, and the difference increases down the group.
$$
\begin{array}{|c|c|c|}\hline
\text{Period}&\text{Halogen boiling point} (\pu{^\circ C})&\text{Noble gas boiling point} (\pu{^\circ C})\\\hline
2&−188.11&−246.046 \\\hline
3&−34.04&−185.848 \\\hline
4&58.8&−153.415 \\\hline
5&184.3&−108.099 \\\hline
\end{array}
$$

Bromine is found in nature as Br2 while krypton is just Kr.

FYI: Your listed value $(\pu{-7.2 ^\circ C})$ for the boiling point of bromine $(\ce{Br2})$ is incorrect. It is actually even higher value: $\pu{58.8 ^\circ C}$ look here.

The both values you listed are not relevant boiling points. They are melting points.

Just a thought: if the interactions in $Br_2$ molecules is governed by dip-ind interactions, bromine can form bigger induced dipole moments -cause the charge will get polarized in the molecule- which will be > than the one between krypton atoms.


text:
With which group does the middle oxygen atom of an anhydride remain, on hydrolysis?
With which group does the middle oxygen atom of an anhydride remain, on hydrolysis? As an example, let us say we have the following modification of ethanoic methanoic anhydride, with the middle oxygen atom being the $\ce{^18O}$ isotope:

On acid catalysed hydrolysis with normal water, is this isotope found in formic acid, or acetic acid, or in both (i.e., hydrolysis is random and hence the product contains both the acids with both types of oxygen atoms)?

I think it will be largely random, maybe with slight prevalence of one side.

This is an interesting question. Moreover, what if the given example was of some beta branched anhydride? Would then, due to steric repulsions, the $\ce{O^{18}}$ be dominantly present in one of the acid product? It would be even more interesting to see if some research has taken place on this topic.

@GaurangTandon The first step, i.e. the protonation of the middle oxygen is obviously clear, and I don't believe $H^+$ suffers any steric repulsion (I may be wrong, but I have never seen any example of that).  Then the answer would probably be something like "the better leaving group leaves". But then I have never seen this mentioned in any book so it might well be dependent on the conditions.

I'm not sure this is a meaningful question. Is there isotopic scrambling as a result of being in an acidic aqueous medium??

What I think is, the marked oxygen atom will stay with acetic acid because the carbon on the right side of the oxygen isotope has a slightly more δ+ charge than the carbon on the left due to the +I effect of methyl group attached to it. So, the marked oxygen should go with acetic acid instead of formic acid in the major product.

For the reaction the mechanism is usually tetrahedral. Only under acid catalysis does the SN1 mechanism occur and seldom even then.

The tetrahedral reaction mechanism of hydrolysis of anhydrides is similar to that for the hydrolysis of carboxylic esters which usually also proceeds through the tetrahedral mechanism.

Of the eight mechanisms given by Ingold for ester hydrolysis, seven have actually been observed in hydrolysis of carboxylic esters. The one that has not been observed is the BAC1 mechanism. The most common mechanisms are the BAC2 for basic catalysis and the AAC2 for acid catalysis, that is, the two tetrahedral mechanisms. Both involve acyl-oxygen cleavage.

So considering the mechanisms for ester hydrolysis and anhydride hydrolysis to be similar, it can be proved that the extent of δ+ charge on the carbon being attacked does play a major role in deciding the rate of reaction.
For instance, the rate of hydrolysis for the following esters is in the order shown:

So I think this should be enough to prove what I stated in the very beginning of my answer.

Source: Jerry March Organic Chemistry 6th edition, Organic chemistry by Paula Yurkanis Bruice.

Two users (Zhe and user55119) have already said the isotopic oxygen will be present in both the acids as a product. I appreciate your responses, but I would really recommend you back up your idea with some evidences/papers. Thank you.

@GaurangTandon that's the reason I especially mentioned "in the major pdt" in the last line of my answer.

Thanks for the additional details. The new addition seems to speak about the rate of hydrolysis, rather than the migration of any isotopic atom towards any group. So I fail to see the connection here, and I am not sure how they both relate to each other. Can you please tell how?

@GaurangTandon I talked about rate of hydrolysis to show that the attack will be more frequent on the carbon to the right of the isotopic oxygen rather than the left one and so the isotopic oxygen will majorly be with the acetic acid molecule.

The mixed anhydride, acetic formic anhydride (AFA), was first prepared by Béhal in 1899 who formed it by dissolving formic acid in acetic anhydride (AFM) in 1:1 mole ratio (Béhal, Compt. rend., 1899, 128, 1460)[1]. It was described as a liquid with an odor similar to that of acetic anhydride, boiling at about $\pu{29 ^\circ C}$ at a pressure of $\pu{18 torr}$. Similar to formic anhydride, which decomposes to formic acid and carbon monoxide ($\ce{CO}$) at room temperature within hours, AFA also decomposes slowly to acetic acid and $\ce{CO}$ by long standing at room temperature. For example, $\pu{50 mL}$ of AFM in a 100 mL volumetric flask with a rubber stopper had suddenly exploded after standing undisturbed for about 2 weeks [2]. Only $\ce{CO}$ (by flame test) and acetic acid in the flask were detected. The reaction was described as: $\ce{HC(=O)OC(=O)CH3 -> CO + CH3COOH}$.
The decomposition of AFA is accelerated by increasing temperature and types of catalyst used, and also depends upon the solvent used[1]: for instance, it is very fast in aprotic solvents such as toluene, carbon tetrachloride, and nitrobenzene, but slow in protic solvents such as ethanol, I-pentanol, and 2-methylpropanol (note that AFA undergoes formylation reaction of alcohols, instead of decomposition to $\ce{CO}$ and acetic acid; [3]). The catalyst for the decomposition could be either an acid (e.g., $\ce{H2SO4}$, $\ce{HNO3}$, $\ce{HF}$, $\ce{HCl}$, etc.) or tertiary amine bases (e.g., N,N-dimethylbenzeneamine, brucine, strychnine, nicotine, etc.), either of which produces $\ce{CO}$ and acetic acid. Note also that, a mixture of anhydrous $\ce{HF}$ and AFA at $\pu{0 ^\circ C}$ and atmospheric pressure produced formyl fluoride ($61\%$) with small amount of acetyl fluoride [4]. It’s noteworthy to mention that formyl fluoride is the only known stable halide of formic acid. The above reaction, however, depends on even relatively minor variations of conditions. For example, the reaction performed at room temperature instead of $\pu{0 ^\circ C}$ has yielded almost equimolar amounts of the formyl and acetyl fluorides [4]. Based on all of these evidences, it is safe to suggest that the most susceptible carbon center of AFA to a nucleophilic attack is the carbonyl carbon of its formyl group, but very sensitive to the reaction conditions.
Olah and others have done excellent work on AFA protonation [5]. According to $\ce{^1H}$ NMR evidence, they suggested rapid proton exchange between carbonyl oxygen atoms (observable at $\pu{-85 ^\circ C}$) of AFA and its eventual decomposition by excess super acids to protonated acetic and formic acids, acetyl cation, and formyl cation intermediate, which is not detected even at $\pu{-130 ^\circ C}$, but rapidly decomposed to $\ce{CO}$. Suggested mechanism and $\ce{^1H}$ NMR evidence are as follows:

Since, AFA gives formylated alcohols under similar condition in alcohols, the following mechanism can be suggested:

Yet, the major and minor product ratio is extremely depends on reaction conditions due to high sensitivity of AFA to its atmosphere.
Following example shows it can be used to selectively acylate (formylate) vide variety of heteroatoms if conditions are used accordingly [6].

References:

Strazzolini, P.; G. Giumanini, A.; Cauci, S. Acetic formic anhydride a review. Tetrahedron 1990, 46 (4), 1081-1118. DOI: 10.1016/S0040-4020(01)86676-X.
Schierz, E. R. The Catalytic Decomposition of Formic Acid in Acetic Anhudride: J. Am. Chem. Soc., 1923, 45(2), 455-468. DOI: 10.1021/ja01655a022.
Stevens, W.; van Es, A. Mixed Carboxylic Acid Anhydrides: II. Esterification of alcohols with the formic acid/acetic anhydride reaction mixture: Rec. Trav. Chim., 1964, 83(12), 1287-1293. DOI: 10.1002/recl.19640831212.
Olah, G. A.; Kuhn, S. J. Formylation with Formyl Fluoride: A New Aldehyde Synthesis and Formylation Method1. J. Am. Chem. Soc. 1960, 82 (9), 2380-2382. DOI: 10.1021/ja01494a065.
Olah, G. A.; Dunne, K.; Mo, Y. K.; Szilagyi, P. Stable carbocations. CXXVIII. Protonated acyclic carboxylic acid anhydrides and their cleavage to oxocarbenium ions. Question of the formyl cation in superacid media. J. Am. Chem. Soc. 1972, 94 (12), 4200-4205. DOI: 10.1021/ja00767a027.
Practical and Efficient Synthesis of N-Formylbenzotriazole: Org. Synth., 2013, 90, 358-366. DOI: 10.15227/orgsyn.090.0358.

As always, a very comprehensive answer! :D Just one thing, I don't understand how the first two paragraphs of your answer relate to the question being asked, as there is no mention of isotopic oxygen atoms in them. So, what purpose do they serve?

@Gaurang Tandon: We don't use isotope experiments in everyday lab. They are used to solve some problems in reaction mechanisms (e.g., deuterium isotope effect). In my paragraph, I generally talked about stability and reactivity of the compound without isotope label to give the expecting results to a reader.

This mixed anhydride is used to make formate esters of alcohols. Productive protonation will occur on the formate carbonyl oxygen followed by nucleophilic addition of water and loss of labeled acetic acid (either oxygen site; resonance) and unlabeled formic acid as the kinetically formed products. The $\ce{O^{18}}$ label in acetic acid will eventually exchange with water to form $\ce{O^{18}}$ water in an equilibrium mixture. Ultimately, the $\ce{O^{18}}$ label will also work its way into formic acid by acid-catalyzed exchange with $\ce{O^{18}}$ water.
References:
For the reactivity of acetic formic anhydride with nucleophiles, click here.
For the kinetics of O18 exchange between carboxylic acids and water, click here. [R. L. Redington, J. Phys. Chem., 1976, 80, 229.]

Sorry, but what is "productive protonation"?

There are three oxygens that can be protonated reversibly. Only protonation of the formyl oxygen leads to further reactions.

Really, the oxygens on carboxylic acids exchange with water?! Do you have a reference for that? I've never heard of that before; I'd like to read more.

@hBy2Py I assumed this is what Zhe meant by "isotopic scrambling" in his comment to the question, but now I'm not sure if that's actually what user55119 actually meant as well :/

@hBy2Py; Checkout the mechanism of Fischer esterification. If O-18 alcohol can incorporate and form labeled ester, why not O-18 water to form labeled carboxylic acid?

The oxygen would remain more often with the aceto group, but I'm sure you would get a mixture of products. The aldehyde moiety is more electronegative than the ketone moiety of the anhydride so would be preferentially attacked by the water molecule during hydrolysis. Also, the aldehyde portion would be more sterically accessible to the attack.

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text:
IUPAC Nomenclature of organic compounds
Can the prefixes like iso,neo be used in the IUPAC nomenclature of organic compounds and can all the common names of aromatic compounds be used in their IUPAC name?

Related: Can we use prefixes like iso, neo, etc in IUPAC nomenclature of organic compounds?

Prefixes -  I think the prefixes neo, iso etc. are used only in common names. There's a different system in IUPAC for naming things having a branched chain.
Aromatic Compounds - The only aromatic compound that I know of having same common and IUPAC name is benzene.

I haven't downvoted but it might be related to I think... The question is clear and the answer straightforward, possibly from the red book of IUPAC...; But your answer  fits pretty well as a comment!


text:
How does the presence of various NMR active nuclei in a compound affect coupling and signal multiplicity?
I am basically wondering how the presence of various NMR active nuclei present in a organic molecules affect the shift values and coupling states normally seen in a molecule lacking these nuclei.
Can NMR active metallic moieties in organometallic compounds such as magnesium 25 in a Grignard reagents couple with the spin of protons in H-NMR or carbon 13 atoms in C-NMR? Do the peaks show up in the same range and how are the shift values affected?
In a structure containing many active NMR nuclei (such as a structure containing appropriate isotopes of chlorine, phosphorus, and nitrogen) is special decoupling software or supplemental instrumentation needed to interpret these spectra or are there other techniques to readily discern structure?

I'd like to expand on the other answer a bit (at least on the question of coupling).  An NMR active nuclei X can couple with (and detectably split peaks of) another NMR active nuclei Y if X is both not too rare (as discussed in the other answer) AND does not have an extremely short relaxation time.
I'll focus on the common elements that an organic chemist might work with:  $\ce{H}$, $\ce{C}$, $\ce{O}$, $\ce{N}$, $\ce{S}$, $\ce{F}$, $\ce{Cl}$, $\ce{Br}$, $\ce{I}$, $\ce{P}$, and toss in $\ce{Mg}$ (as mentioned in the question), $\ce{Li}$ and $\ce{Si}$.
"The good"
NMR active isotopes with high natural abundance and not extremely short relaxation times: in our list $\ce{^1H}$, $\ce{^19F}$, $\ce{^31P}$, and, marginally, $\ce{^29Si}$ and $\ce{^6Li}$.
These all are decent NMR nuclei and abundant, so they definitely split peaks of nearby NMR active nuclei.

$\ce{^1H}$:  99.98% natural abundance, spin 1/2 (so 0 quadrupolar moment), T1
around 10 seconds for the standard.
$\ce{^19F}$:  100% natural abundance, spin 1/2, sharp peaks (thus
relaxation times not too short).
$\ce{^29Si}$:  A bit marginal.  Around 5% natural abundance (so the 'side
peaks' caused by splitting come from only 5% of the molecules and
thus will be small, called satellites), spin 1/2.
$\ce{^6Li}$:
Around 8% natural abundance, spin 1, but is an honorary spin 1/2 as
it has a very small quadrupolar moment.  Relaxation times are long.

These nuclei will split nearby NMR active nuclei (not many bonds away).  This is either a bug or a feature.  The splitting pattern can tell you a lot about the molecule along with other advantages you can get from internuclear coupling (I have a colleague who uses this all the time on fluorocarbons).  If you don't want the effect, though, you will have to use a decoupling pulse on these nuclei.
For instance, in the text "Organic Structures from 2D NMR Spectra" (Field, Li, Magill) the examples include many uncoupled experiments (Note:  Y{X} means Y is observed while uncoupling X): $\ce{^13C}\{\ce{^1H}\}$ (often), $\ce{^19F}\{\ce{^1H}\}$, $\ce{^13C}\{\ce{^1H, ^19F}\}$, $\ce{^1H}\{\ce{^31P}\}$, $\ce{^31P}\{\ce{^1H}\}$.
"The bad"
There are nuclei that, if enriched, would split nearby peaks (having not-too-short relaxation times), but have a low natural abundance.  These can be analyzed quite well using NMR, and may have peaks split by other active nuclei, but won't usually cause visible splitting in other nuclei.  Out of our list:  $\ce{^13C}$, $\ce{^15N}$.

$\ce{^13C}$:  Only 1% natural abundance, spin 1/2.  While it is often
examined either alone or in conjunction with other active nuclei, it
won't cause anything but the tiniest satellites in other nuclei.
$\ce{^15N}$:  Even worse.  Spin 1/2 but only 0.4% natural abundance.

"The ugly"
These might or might not have high natural abundance, but they have short relaxation times.  When relaxation times are very short, the nuclei is, in effect, self-decoupling.  Nuclei with very short relaxation times have, though the Heisenberg Uncertainty Principal, broad lines and thus often are pretty hard to work with in any NMR context.  They certainly don't show visible splitting of the signal of nearby nuclei.  From our list, we have:  $\ce{^14N}$, $\ce{^33S}$, $\ce{^35Cl}$, $\ce{^37Cl}$, $\ce{^79Br}$, $\ce{^81Br}$, $\ce{^127I}$, $\ce{^17O}$ and $\ce{^25Mg}$.  At least you don't have to use decoupling pulses on these nuclei!

$\ce{^14N}$:  Too bad.  Even though it's 99.6% abundant (the rest is $\ce{^15N}$), it
has spin 1 and a reasonably large quadrupolar moment, thus a short
relaxation time (often around 20 ms).  It might broaden the peaks of
nearby nuclei, but won't show coupling in any but the most symmetric
environments.
$\ce{^33S}$:  Ouch.  0.8% natural abundance, spin 3/2, high quadrupolar
moment, only NMR active sulfur nuclei.
$\ce{^35Cl}$, $\ce{^37Cl}$:  Between these, 100% abundant.  Both are spin 3/2 with
VERY large quadrupolar moments and short relaxation times.
$\ce{^79Br}$, $\ce{^81Br}$:  Same as the sad fate of chlorine.
$\ce{^25Mg}$:  Only active NMR nuclei of Mg, 6% abundance, spin 5/2, VERY
large quadrupolar moment.
$\ce{^17O}$:  The only NMR active oxygen nuclei.  Very rare (0.04% natural
abundance), spin 5/2 with a fairly large quadrupolar moment.
$\ce{^127I}$:  100% natural abundance, but VERY large quadrupolar moment.
Maybe also prone to CSA relaxation (I'm not sure).

COMMENTS on short relaxation in nuclei
Nuclei with very short relaxation times often get these via two relaxation processes.  One is quadrupolar relaxation (only for spin > 1/2) and chemical shift anisotropy (CSA) relaxation.
For high spin (>1/2) nuclei, relaxation can occur because charge on the nuclei is not symmetric (thus a quadrupolar moment) so that energy can be exchanged with an environment that is not highly symmetric.  The effectiveness of this type of relaxation rises as the square of the quadrupolar moments times the square of the applied magnetic field.  Spin 1/2 nuclei have quadrpolar moment zero.  Quadrupolar moments range all the way from values such as 0.0028 ($\ce{^2H}$), -0.0008 ($\ce{^6Li}$) to values such as 0.31 ($\ce{^79Br}$).
CSA:  Also caused by tumbling in an asymmetric environment.  Larger for nuclei with large shift ranges, proportional to gyromagnetic ratio squared times magnetic field strength squared.  Not really predominant in protons or any carbon but Quaternary carbon.
These relaxation mechanisms don't work if the environment of the nuclei is very (especially tetrahedral or octahedral) symmetric, in which case relaxation times can be long.
The perverse thing about these relaxation mechanisms is that they get worse (more decoupling) with higher field strength.  Usually anyone working with NMR dreams night and day of getting a higher field spectrometer.  However, if you are interested in a nuclei prone to CSA relaxation (say, $\ce{^199Hg}$ or $\ce{^195Pt}$, which are spin 1/2 and not prone to quadrupolar relaxation), and you want to see splitting of peaks (coupling) in nearby nuclei, you might have to move to a LOWER field NMR.
References
A good site with lots of information on NMR active nuclei is the NMR periodic table.

Indeed those NMR active nuclei will effect the coupling of intended signal, which is generally depend on the natural abundance of "foreign" NMR active nuclei. For example, $\ce{^{13}C}$-nuclei would not affect $\ce{^1H}$-NMR signals because of the low natural abundance of $\ce{^{13}C}$ ($\mathrm {\approx 1.12\%}$) compared to that of $\ce{^1H}$. Also keep in mind that the average width of a $\ce{^1H}$-NMR (if not working on specific compounds with higher range) is $\pu{10 ppm}$. You may able to see $\ce{^{13}C}$ satellite peaks only when you are working on concentrated samples. However, when you are taking  $\ce{^{13}C}$-NMR, the weight of the effect is reversed and you'd see $\ce{^{13}C-^1H}$ couplings on $\ce{^{13}C}$-NMR spectrum, if it was taken without applying decoupling techniques. The modern NMR techniques can resolve this problem for you, and even worked it for your advantage. See following three $\ce{^{13}C}$-NMR spectra of ethylbenzene (last spectrum shows only aliphatic region). First spectrum shows only singlet because it was subjected to regular decoupling. The last one shows gated-decoupled spetrum so that it coupled only with protons, which are directly attached to the relevant carbon. If you know the $n+1$ rule, you'd know carbon resonance at $\pu{15.5 ppm}$ is the methyl carbon of ethyl group and the resonance at $\pu{28.8 ppm}$ represents the methylene carbon of the same.


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How to change the direction of evaporation and thus location of condensation?
For a specific region of air, enclosed by a metal box in which a constituent is obviously gaseous $\ce{H2O}$, is there any way in which we can control the location of condensation?
Assume the box is heated, typically we will have condensed $\ce{H2O}$ formed at the ceiling, does anyone have any strategies of condensing water on the inner sides of the box, or maybe at the base?
I feel like I have a limited scope, with the outer-most idea being to stick anhydrous calcium chloride (dessicant) on the inner sides and have a ledge for the water to sit on.

Heat the ceiling, and water will condense at the base. It is as simple as that.

Forgot to mention.. Without the use of a heat/electrical source...

Make the side of the walls a lot cooler than the ceiling? Maybe strap some ice packs onto the walls

Why would you want to do this by the way?

The simple answer is that water evaporate where the concentration in the air is lower than the vapor pressure some liquid water at a given temperature. Then it will condense at a location where the temperature is lower that the saturation temperature of water at that partial pressure.
Industrially, though, one widely used technique to avoid condensation when the surface temperature is low enough for it to happen is to use hydrophobic coatings to prevent the water vapor from nucleating on certain surfaces. A quick google search will reveal lots of these for different applications.

is there any way in which we can control the location of condensation?

Condensation requires a nucleus to start. Droplets will form more readily on surface than in the gas phase, unless it contains impurities such as smoke.
In physics, cloud chambers were used to detect and trace the path of particles. The picture below, from Wikipedia's article on cloud chambers, shows the path of a positron:

The same idea is used at a bigger scale to make it rain by seeding the clouds, see for example this report.

does anyone have any strategies of condensing water on the inner sides of the box, or maybe at the base?

If you use different materials for the different sides of the box, they probably would be more or less efficient in acting as nuclei for condensation. That would be a way to control where condensation occurs even when the temperature is constant throughout.


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Reaction of anisole and succinic anhydride in presence of aluminium chloride

Predict the structures of the intermediates/products in the following reaction sequence:

In the first step, I know that aluminium chloride  is playing some role,  as otherwise hydronium ion would have reacted with succinic anhydride  to give succinic acid.
I thought of the following two possibilities:

Friedel Crafts: But till date, I have only seen Friedel Crafts type reactions and have not seen aluminium  chloride complex with either acid or anhydride. I could also not find it on Google.

Chloride formation: But I have also never seen aluminium chloride forming a chloride in reaction with anhydride. I am not sure if it does that, does it?

So, what is the reaction mechanism for the formation of (A)?
Source: Joint Entrance Exam (JEE) 1996 India

Aluminium chloride ($\ce{AlCl_3}$) is mainly used for this Friedel-Crafts alkylation or acylation, where this compound acts as catalyst, and also controls the production of the desired compound.
For example, in this case, the reactants are anisole ($\ce{Ph -OMe}$) and succinic anhydride. So, the main reaction which drives these reactants towards product is the nucleophilic attack by the para carbon (w.r.t the $\ce{OMe}$ group) of the phenyl ring at the carbonyl carbon of the succinic anhydride, and an addition-elimination of the attached oxygen of the ester part in succinic anhydride.
For this, we need the state of the reactants in a proper way. We have to make sure that, the oxygen of $\ce{OMe}$ does not attack prior to the phenyl ring, and also, the electrophilicity of the carbonyl carbon needs to be increased to much extent, which is generally not so much due to present resonance in the compound.
$\ce{AlCl_3}$ plays a special role over controlling these two phenomena. Being a Lewis acid, it forms a Lewis acid-base interaction with the oxygen of $\ce{OMe}$ and also with the oxygen atom of $\ce{C=O}$ group in succinic anyhdride. Due to this interaction, a δ+ charge is introduced on both oxygen atoms described before. This increased formal charge helps to increase the electrophilicity of the carbon due to increased -I effect of the attached oxygen and also reduces the chances of attacking by $\ce{OMe}$ group. This catalyses to form our desired product with the proper reaction between the reactants.

You claim that the reaction is initiated by a nucleophilic attack. In a medium where AlCl3 is available, what nucleophile would attack an acid anhydrite? The better approach is to initiate a reaction between the Anhydrite and the aluminum chloride, forming a carbocation like species, which undergoes EAS easily in the given conditions.

@ShekharUpadhyay  can I not see EAS be seen as nucleophilic attack of benzene derivatives on the electrophile??

@ShekharUpadhyay I saw what you are saying in Peter Sykes.

@KavitaJuneja no. There is a reason it's called Electrophilic Aromatic Substitution. The initiator is an Electrophile. Benzene, in itself, is not a system that would want to show nucleophilic activity. When it's electrons are disturbed, it loses its aromaticity. It won't spontaneously do this on it's own. An extremely strong Electrophile is necessary for the initiation of this reaction.

@ShekharUpadhyay Any source of your claim?

@KavitaJuneja This discussion is increasingly becoming a debate similar to whether the Sun moves around the Earth or the Earth moves around the sun. The answer is simple: motion is relative. It is only easier to analyse in certain frames. In this case, much easier to analyse it as an electrophillic attack. And as far as sources are concerned.... Jerry March should have mentioned this.

@ShekharUpadhyay yes.. you're right.

@KavitaJuneja I assume you're studying for JEE. Try Clayden or Bruice. Sanyal is also good.


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From where does sea water get salt?
I have learned that in primary classes  that sea water is salty and I had also wondered that, what makes it salty? and my question is, where does sea water gets its salt?

It doesn't. The salt is already there.

This probably should migrate to https://earthscience.stackexchange.com/ . However, see here https://oceanservice.noaa.gov/facts/whysalty.html

The important idea is that salt builds up in sea water over time as small amounts are carried there in rivers but evaporation takes water back out of the sea, and deposits fresh water as rain forming the flow in rivers, but leaves the salt behind.

Your comment has 7 up votes, could you please write a slightly more elaborate answer?

The salt in the oceans came originally from rocks. Rain can dissolve some of the minerals found in rocks, especially rain that is slightly acidic from dissolved $\ce{CO2}$. The chemicals in the minerals are broken down into ions in solution. Some of the ions will tend to precipitate out of solution, or be taken up by living things (for example calcium is removed from water by the shells of molluscs and diatoms). Other ions will tend to build up in the water. Two of the most soluble and common ions are sodium and chloride. Over time these built up in the ocean. The sodium chloride in sea water is what makes the seas salty.


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What red and blue area represent in a orbital software?
I have a software in my android device named hydrogen atom. It show orbital of hydrogen we have to set the quantum numbers and probability of finding particle. But what the blue and red colour area in below pictures represent.

To answer this question, we need to start from the beginning.

Movement of an electron is described by wave function. The best layman description for it is that wave function is a quantum equivalent of trajectory. It isn't exactly trajectory, but it serves as equivalent for quantum mechanics.
A trajectory repeating itself and wrapped around an object is called an orbit. A quantum mechanical equivalent is a wave function, remaining constant in time. Such a function for hydrogen atom is called orbital.
Wave function is a function. It means, that it can be positive and negative. (Well, in case of a stationary function, see below). It usually doesn't matter in simple chemistry, but once we start talking about orbital interactions it becomes important. In short, orbitals interact with formation of orbitals with reduced energy as long as lobes with same sign overlap. Thus, it is important to know sign of the wave function in lobes of the orbitals.
So, whenever you see an orbital depicted using two colors, it means two different signs of the wave function. Note, since math, it is not important which color depicts what sign. If one inverses sign of all lobes simultaneously, the result will depict the same system, actually.
Side note 1: the surfaces depicted in the picture likely are isosurfaces (i.e. the function in consideration has same value in all surface points) of square of wave function that enclose volume containing some fixed amount of electron density (usually somewhere around 90-95%). There is no good way to depict wave function/orbital in 3 dimensions really.
Side note 2: wave function has well defined sign only in case of stationary system (i.e. unchanged with time). Whenever we talk about system evolving in time, we need to consider arcane math of complex numbers.

and to add a little bit extra to your clear answer; sometimes the plus and minus are called 'parity' or 'phase'. This is important when bonding: in phase is bonding and out of phase is anti-bonding, most easily imagined with $\pi$ orbitals.


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Why does DMSO help facilitate halohydrin reactions?
It would make sense, due to bond energy, that water molecules would have a hard time competing with $\ce{Br}$ in a bromonium ion intermediate.
However, I was reading that by doing the reaction in DMSO, it encourages the protonation of the water molecules. How? What is the rational behind this?

Alkenes predominantly gives relevant 1,2-dihalo product when reacted with a halogen (e.g., $\ce{Br2}$) in a organic solvent such as chloroform or dichliromethane. If the reaction is performed in water, the product is the corresponding 1,2-halohydrin due to restriction applied by polar water molecules caging $\ce{Br-}$ ion. Organic molecules are sparingly soluble in water as solvent. Thus, the halohydrin reaction is often done in a mixture of organic solvent and water. For example, bromohydrin can be achieved by using N-bromosuccinimide as the electrophilic bromine source in $\ce{DMSO/H2O}$.
Use of $\ce{DMSO/H2O}$ as a solvent has few advantages:

The bromine source, N-bromosuccinimide (NBS) is readily soluble in this solvent system [Ref. 1]).
$\ce{DMSO}$ is also participated in reaction mechanism. Reference 1 has detailed labeling procedure using $\ce{DMSO^{18}}$ to show that isotopic $\ce{^{18}OH}$ was incorporated in $\ce{OH}$ group of bromohydrin [Ref. 1].
Evedence also suggest the participation of water in the medium, probably due to increasd basic nature of water in $\ce{DMSO}$ ($\mathrm{p}K_\mathrm{a}$ of water in $\ce{DMSO}$ is $31.4$ at $\pu{25 ^\circ C}$ [Ref. 2]).

Suggested mechanism for involvement of $\ce{DMSO}$ as follows:

References:

Bromohydrin Formation in Dimethyl Sulfoxide: D. R. Dalton, V. P. Dutta, D. C. Jones , J. Am. Chem. Soc., 1968, 90(20), 5498–5501 (https://pubs.acs.org/doi/abs/10.1021/ja01022a030).
Acidities of water and simple alcohols in dimethyl sulfoxide solution: W. N. Olmstead, Z. Margolin, F. G. Bordwell, J. Org. Chem., 1980, 45(16), 3295–3299 (https://pubs.acs.org/doi/abs/10.1021/jo01304a032).


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How to calculate molarity of a solution by solving the percentage?
Molarity is number of moles per litre of solution and we calculate it by simply dividing moles by volume. But, if moles are not given and the w/w or w/v percentage is given then, how can we solve for the molarity?
For example, if we have 2 % w/v solution of $\ce{NaOH}$, how would we find its molarity? Do we solve by dividing 2 by 40 (molar mass of $\ce{NaOH}$)?

Both of them are pretty straightforward. Don't go by the formulae, but instead, internalize the steps required here.
$2\%$ w/v solution has $\pu{2g}$ of solute in $\pu{100 ml}$ solution. That implies $\pu{20g}$ of solute in $\pu{1L}$ solution. $\pu{20g}$ solute is 0.5 moles of $\ce{NaOH}$ (why?). So, molarity is simply moles divided by volume of solution.
The $2\%$ w/w solution case is trickier, and you'll also be given the density of solution in this case.  $2\%$ w/w solution has $\pu{2g}$ solute in $\pu{100g}$ solution, or $\pu{20g}$ solute in $\pu{1000g}$ solution. If the density is $\pu{2gm L^-1}$ (assumed), then the volume of solution would be $\ce{500 mL}$. Now, you can proceed by the same logic as above.
In fact, the conversion of $\pu{100ml}$ volume to $\pu{1L}$ in case 1 wasn't even necessary, but I only did that to make the step-by-step calculations more approachable for you.

Do you mean I have to divide 0.5 by 100ml?If so ,then answer comes 0.005M, and my book says answer is 0.5M?Can you please explain it in a bit more simple words?

@RabikJohn As I said earlier, you either take the combination of (20g and 1 litre) or (2gram and 100ml). I took the former, and computed moles = 20g/40g=0.5 That gave me molarity = 0.5moles/1litre=0.5M. You can take the latter combination and will get the same answer.

I'd like to add one more situation to @Gaurang Tandon answer:
If you are dealing with liquid solute and a solvent (also a liquid), then there would be another way of giving percent concentrations: %v/v. This is rare because sometimes, volumes reduce when a solute and a solvent mixed, specifically when both are liquids. Still, it's worth considering:
For example, if the solution is said to be $10\%$ v/v, then it has $\pu{10 mL}$ of solute in $\pu{100 mL}$ of solution. That implies $\pu{100 mL}$ of solute in $\pu{1000 mL}$ or $\pu{1 L}$ of solution. If the density of solute is $x~ \mathrm{g\cdot mL^{-1}}$, then the mass of $\pu{100 mL}$ of solute is $(100~\mathrm{mL})\cdot (x~ \mathrm{g\cdot mL^{-1}})=100x~\mathrm{g}$. If molar mass of the solute is given (it should be), now you can find (how?) the number of moles of solute in $\pu{1 L}$ of solution, which is the molarity of your solution (moles per $\pu{1 L}$ of solution).


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Number of stereoisomers of tetramethylspirononane

The above compound – 2,3,7,8-tetramethylspiro[4.4]nonane – clearly exhibits both optical and geometrical isomerism. The question is, how many stereoisomers does it have in all?
As the number of possibilities are many, I couldn't really visualise the molecule's all possible stereoisomers properly. Could someone please help me count the total number of stereoisomers?
I figured that the carbon common to both rings is sp3 hybridised (tetrahedral geometry), and the two rings don't lie in the same plane.

Related: How to find the number of optically active and optically inactive isomers for a spiropentane derivative?

The compound 2,3,7,8-tetramethylspiro[4.4]nonane has five potential stereogenic or pseudoasymmetric centres: the four C atoms with the methyl substituents and the spiro atom. In total there are $2^5=32$ possible configurations. However, many of these combinations are actually identical compounds.
A simple and obviously exhaustive (but also exhausting) approach is to draw all 32 configurations and name them. (This approach may also be used to check the results if someone finds a more elegant solution.) The resulting names for the 32 combinations are:

(2R,3R,7R,8R)-2,3,7,8-tetramethylspiro[4.4]nonane (2 times)

(2R,3R,7R,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (8 times)

(2R,3R,7S,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (4 times)

(2R,3S,5r,7R,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (4 times)

(2R,3S,5s,7R,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (4 times)

(2R,3S,7S,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (8 times)

(2S,3S,7S,8S)-2,3,7,8-tetramethylspiro[4.4]nonane (2 times)

Therefore, there are seven different combinations.

It might be more meaningful to draw them using Newman projections, much as you would for an allene, just a thought.

For the middle spiro carbon, why did you draw the wedge and dash in cases 4 and 5, but not in any of the other cases? Wouldn't in all cases the carbon atom be sp3 hybrid, and hence, need a wedge and dash bond also?

@GaurangTandon In the cases 1, 2, 3, 6, and 7, the spiro atom is nonstereogenic; i.e. flipping of the spiro atom portion of the structure changes nothing otherwise. Single bonds attached to nonstereogenic atoms should normally be drawn as plain bonds (neither as solid wedged bonds nor as hashed wedged bonds). The use of stereobonds at nonstereogenic atoms should generally be avoided.

Here is another approach to @loong's solution to the seven stereoisomers of this tetramethyl spiro[4.4.0]nonane. [This question is akin to one recently asked about tetramethylspiropentanes].

Using the arbitrary numbering scheme shown above and assigning C1 as having the R-configuration, the eight permutations for C1-4 are listed on the left of the diagram. Only RSRS and RSSR have stereochemical designations at the spiro carbon (C5). Inside the black box are the eight structures. The R/S designations are in the order 1234(5). The three blue structures, RRRS, RRSR and RSRR, are identical and its enantiomer is RSSS, aka SSSR. The black RRRR isomer’s enantiomer is the SSSS isomer shown on the left. The green pair are enantiomers. The red isomer is a meso structure.
Summing up, there are three pairs of enantiomers and one meso compound. There are seven stereoisomers.
ADDENDUM: Structure A (the red meso isomer RRSS) is identical to its mirror image B by reflection  and rotation. There is no plane of symmetry.

The red structure is only a meso compound if the rings were flat, or if you'd allow the molecule to vibrate/relax into the same structure again.


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Metal phosphate buffers

A calibration solution contains $\pu{0.02M}$ $\ce{Cr^3+}$ and $\pu{0.02M}$ $\ce{Cu^2+}$. In order to control the solution $\mathrm{pH} = 12$, a phosphate buffer is used with the weak acid being $\ce{HPO4^2-}$ and the weak base being $\ce{PO4^3-}$. The maximum permissible change in $\mathrm{pH}$ is 0.2 units, how many moles of strong acid can be buffered per mole of weak base present?
  $\ce{H3PO4}$: $K_\mathrm{a3}$ = $4.2 \times 10^{-13}$

I have tried to formulate an equation for the reaction but I'm not sure which metal ion is/if both are taking part and therefore finding it hard to produce a $K_\mathrm{a}$ expression.

Neither metal ion is significant in repect of the buffer calculation

Shouldn't the formula $$\ce{pH = pK_{a3} +log \frac{[PO_4^3-]}{[HPO_4^2-]}}$$ be applied here !!

I think phosphate buffer is the solution here. Reason:
$K_\mathrm{sp}$ of $\ce{CrPO4}$ at $\pu{25 ^\circ C}$ is $2.4 \times 10^{-23}$ while that of $\ce{Cu3(PO4)2}$ at $\pu{25 ^\circ C}$ is $1.4 \times 10^{-37}$, thus both ions will precipitate since concentration of each ion is $\pu{0.02 M}$.


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What happens to promoters or poisons during a catalyzed reaction?

Catalysts are substances that influence reaction rates without undergoing changes in their mass and chemical composition, but may undergo physical changes.

Promoters, poison etc. affect the activity of the catalyst. What are some examples of promoters or poison undergoing a change (physical or chemical) during a chemical reaction?


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Attack of electrophile on toluene
Does hyperconjugation in toluene influence the attack of an electrophile in any way?
If not, where would the electrophile prefer to attack on toluene?
Looking at related answers, it seems like hyperconjugation leads to delocalisation of electrons over the ring, like in resonance. But hyperconjugation is the interaction between $\sigma$ and $π$ orbitals, how does it lead to delocalisation over the ring? I cannot understand it in terms of the molecular orbital theory.
Lastly, would the electrophile prefer to attack at Ortho position due to greater $+I$ effect and (if possible) hyperconjugation effect? Please correct me if I'm missing something.

Related: Inductive vs hyperconjuative effects in electrophilic aromatic substitution

sorry that was my mistake. the correct mechanism is this


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What changes occur when adding differing amounts of acids & bases?

In an experiment, $\pu{5 cm^3}$ of $\pu{1 mol/dm^3}$ sodium
  hydroxide are gradually added to $\pu{10 cm^3}$ of $\pu{1 mol/dm^3}$ hydrochloric acid containing methyl orange. What changes occur in the mixture?

The concentration of $\ce{H+}$ ions increase (decreases)
The methyl orange changes colour
More water molecules are formed (water is formed, but use of the word 'more'??)
A precipitate is formed (no ppt formed, $\ce{NaCl}$ is soluble)

I thought the answer would be (2). However when thinking about the question the methyl orange would be red in the acid and given the proportions of the reactants and the 1:1 mole ratio, the $\ce{HCl}$ would be in excess, so would it still change colour?

It is perhaps difficult to comment on the intended semantics of the option 3, but I think that is the most appropriate answer.
Option 1 and 4 are wrong because of the reasons you cite.
To comment upon option 2, we need to know the pH range over which this indicator changes colour. According to Wikipedia, an unequivocal shift from red to yellow occurs from a pH of $3.1$ to a pH of $4.4$. For our case, we would end up with a $\pu{\frac 13 M}$ solution of $\ce{HCl}$, which, assuming complete dissociation and negligible contribution of $\text{H}^+$ from water, we end up with a pH of around $0.48$ which is well in the no-colour-change region of the pH values.
Regarding option 3, the 'more' just refers the water molecules formed as a result of neutralisation reaction, which are additional to the ones already present in either of the mixing solutions. Along with the formation of 'new' $\ce{NaCl}$ molecules (as in they did not exist bonded to one another before), water molecules are also formed.
$\ce{NaOH + HCl -> NaCl + H2O}$, i.e. table salt and water

Undoubtedly,  use of the words more water molecules should be to indicate water molecules formed from the reaction in addition to the water  molecule already in these two aqueous solutions.


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Conversion of a permeability value from barrer to SI unit
I have done a permeability experiment on LDPE in the presence of air. At $\pu{30^\circ C}$, the permeability was found to be 1.846 barrer.
Now I am trying to apply this permeability value to a calculation whereby I am trying to find what the change in pressure of a dunnage bag is (it is a bag that has the shape of a pillow) after 5 minutes. I have the following info:
$$
\begin{align}
\text{Volume of bag}&= \pu{0.2064 m^3}\\
\text{Approximate area} &= \pu{1.35 m^2} \text{ (this is just a quick approximation)}\\
\text{Thickness of LDPE} &=\pu{ 0.042 *10^-3 m}\\
\text{Time }&= 5 \times \pu{60s} = \pu{300s}\\
\end{align}$$
Is it correct if I convert barrer to SI units in the following manner?
$$\pu{1 barrer} = 7.5005\times10^{-18}\ \mathrm{m^2\ s^{-1}\ Pa^{-1}}$$
By using the following formula, I was hoping to calculate the change in pressure. However, my values don't make any sense at all!
$\displaystyle\text{Permeability} = \frac{\text{Volume}\times\text{thickness}}{\text{Area}\times\text{time}\times\text{Change in Pressure}}$


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Why do we not generally have "enneane"?
Most numerical prefixes are based on Greek roots, but for "nine" we use the Latin root instead which seems inconsistent.  Thus, for instance, "nonane" instead of "enneane", or in mathematics "nonagon" instead of "enneagon" (although here the Greek form seems to have more traction) .  Why?  Can we use the Greek root as an alternative?

Why wouldn't "any-ane" be inspecific as to number?

Only if "non-ane" means it's not an alkane.

Personal opinion: I think 20th century nihilist would prefer spelling it nonane to enneane. These undertones may play more in the subtleties involved in word preference than one may think, also especially as in Rome\Greek myths there was often 9 main gods, "by the nine gods he swore", so 9 was a sacred number.

Perhaps not a complete answer, but this might put you on the right track. It looks as though a lot of the system we have today grew out of recommendations made by August Wilhelm von Hofmann in a paper titled On the Action of Trichloride of Phosphorus on the Salts of the Aromatic Monamines
He actually suggested using the Latin numerals as a prefix for the alkanes with n>3. So pentane was quintane, hexane was sextane, etc. Over time, the Latin prefixes fell out of favor and were replaced with the Greek one's except in the case of nonane (and quadrane which became butane rather than tetrane). A Yale website, which led me to this article and helped me answer a previous nomenclature question about the alkanes (Etymology of saturation degrees (-ane, -ene, -yne) in aliphatic compounds), posits that it might just be that enneane just doesn't sound as good. Plus the trend from the polygons probably reinforced keeping it around.
So not a full answer, but at least it shows that nonane wasn't chosen to break the trend, but remained in spite of new naming convention.
The author of the Yale page does mention that their source was The Name Game by Alex Nickon and Ernest F. Silversmith, so perhaps there is more information there.

I believe Tyberius gave an extremely reasonable answer for the question and it deserves to be the best answer. I just want to provide some supporting facts for his argument. For example, there are a lot of 9-membered-enes, -en-ynes, and -en-oic acid derivatives among marine natural products. For instance, siamol is a  biologically active natural product with given IUPAC name, 1,5-dimethyl-7-(2-methylprop-1-enyl)cyclonon-4-en-1-ol. If name with the Greek origin had been used, it should have been: 1,5-dimethyl-7-(2-methylprop-1-enyl)cycloenne-4-en-1-ol.
The compound $\mathrm{5D}$ (See Process A in scheme below) is given the IUPAC name, (E)-5-phenyl-3-(2,2,2-trifluoroethyl)cyclonon-4-enone. The Greek origin name would have been(E)-5-phenyl-3-(2,2,2-trifluoroethyl)cycloenne-4-enone. I think you got my point.

Note: Even though we use pent prefix for 5, we still use 'quintet' to describe splitting pattern of 5.
About nonagon in mathematics: Wikipedia states that

In geometry, a nonagon /ˈnɒnəɡɒn/ (or enneagon /ˈɛniːəɡɒn/) is a nine-sided polygon or 9-gon. The name "nonagon" is a prefix hybrid formation, from Latin (nonus, "ninth" + gonon), used equivalently, attested already in the 16th century in French nonogone and in English from the 17th century. The name "enneagon" comes from Greek enneagonon (εννεα, "nine" + γωνον (from γωνία = "corner")), and is arguably more correct, though somewhat less common than "nonagon".

Another reason why nonagon is more common than enneagon is that the latter can be confused with the most common way to pronounce $n$-agon (a generic word for a polygon with $n$ sides), a word which (arguably) a mathematician uses more often.

@FedericoPoloni I thought it was generally n-gon and that that the "a" came from the numeric prefix.

@MathewMahindaratne I think this is definitely the key to why "non" stuck around, since the ane, ene, and yne endings were already established and names like enneaene would have proved confusing.

@Federico usually the generic term is "$n$-gon", spelled and pronounced without the "a".  In fact "enneagon" has two vowel differences from "$n$-gon".


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Mechanism of lithium-halogen exchange of primary alkyl iodide
On Clayden's page 190 this reaction is given for preparation of organometallics:

What is the mechanism of this reaction? I expected the "almost carboanion" part of the $\ce{t-BuLi}$ to attack as a nucleophile and form a longer 7 membered chain? But instead, that almost anionic $\ce {t-Bu^-} $ has attached to iodine.
I realised that $\ce{t-Bu-}$ is too unstable because of electron releasing alkyls, but what is the exact mechanism of this reaction?

With primary alkyl iodides the (simplified) mechanism is likely similar to a $\mathrm{S_N2}$ process, but with attack of $\ce{t-Bu}$ at iodine instead of carbon.

Note that this is not necessarily the case for other substrates, e.g. aryl iodides, or aryl/alkyl bromides. In general lithium-halogen exchange can occur either via a polar mechanism (such as the one shown above) or via a radical mechanism (by two successive electron transfers from RLi to RX). There is also a solvent and temperature dependence. Why does the solvent matter? We know that ethereal solvents such as Et2O or THF are capable of acting as Lewis bases - or ligands for lithium - and possibly assist in deaggregation of the alkyllithium. This sort of thing can't be depicted easily with curly arrows, which is why I add the disclaimer that the mechanism above is (heavily) simplified.
In any case, this particular substrate (Tetrahedron Lett. 1986, 27 (17), 1861–1864) is fairly interesting because if the Li/I exchange were to occur via a radical pathway, the radical formed would cyclise onto the alkene via a 5-exo-trig pathway (even at −78 °C):

However, if you quench the reaction at −78 °C with methanol, no methylcyclopentane is observed, which rules out the intermediacy of a radical. [If you warm up the reaction before quenching then the organolithium can cyclise by nucleophilic attack onto the alkene.]
This particular case (and general mechanisms for Li/Hal exchange) are described quite thoroughly in Clayden's Organolithiums: Selectivity for Synthesis, Chapter 3.
There is also an excellent set of slides on Li/Hal exchange reactions, courtesy of the MacMillan group at Princeton; and also another set, courtesy of the Denmark group at the University of Illinois.


text:
Why HF can't reduce H2SO2?
I read that $\ce{H_2SO_2}$ is an extremely unstable acid, so shouldn't even a weak reducing agent like $\ce{HF}$ be able to reduce $\ce{H_2SO_2}$ to $\ce{S}$ or something along similar lines?

H2SO2 is not a thing at all, HF is not a reducing agent at all, and being unstable does not mean being able to react with everything around.

@IvanNeretin https://en.wikipedia.org/wiki/Sulfoxylic_acid

No offence to  Ivan Neretin, but I afraid, $\ce{H2SO2}$ (Sulfoxylic Acid) exists (Ref. 1), which states that:

Sulfoxylic acid ($\ce{HOSOH}$), a chemical intermediate roughly midway along the path between highly reduced ($\ce{H2S}$) and highly oxidized sulfur ($\ce{H2SO4}$), has been detected using Fourier transform microwave spectroscopy and double resonance techniques, guided by new high-level CCSD(T) quantum-chemical calculations of its molecular structure. Rotational spectra of the two most stable isomers of $\ce{HOSOH}$, the putative ground state with $C_2$ symmetry and the low-lying $C_s$ rotamer, have been measured to high precision up to $\pu{71 GHz}$, allowing accurate spectroscopic parameters to be derived for both isomers. $\ce{HOSOH}$ may play a role in atmospheric and interstellar chemistry, and the present work provides the essential data to enable remote sensing and/or radioastronomical searches for these species. Spectroscopic characterization of $\ce{HOSOH}$ suggests that other transient intermediates in the oxidation of $\ce{SO2}$ to $\ce{H2SO4}$ may be amenable to laboratory detection as well.

Most recently, using chemical trapping techniques, the additional three tautomers of sulfoxylic acid were identified and characterized (see scheme below; ref.2). These authors suggest that $\ce{HOSOH}$ may exhibit both nucleophilic and electrophilic reactivity, of which they attributed to interconversion of $\ce{HOSOH}$ between its tautomers (they used that phenomenon to trapped them). Theoretical calculations suggest that alcohol $16$ is the most stable, and peroxide $19$ is the least stable tautomer (see Scheme).

References:

Microwave Detection of Sulfoxylic Acid ($\ce{HOSOH}$): K. N. Crabtree, O. Martinez, L. Barreau, S. Thorwirth, M. C. McCarthy, J. Phys. Chem. A, 2013, 117(17), 3608–3613.
Chemical trapping and characterization of small oxoacids of sulfur (SOS) generated in aqueous oxidations of $\ce{H2S}$: M. R. Kumar, P. J. Farmer, Redox Biology, 2018, 14, 485–491.

Your answer says $\ce{H2SO2}$ exist. Fine. But OP wants to find out if $\ce{H2SO2}$ can be reduced or not.

No body knows it yet. See see the article published in 2018.

Well, that's what I call "doesn't exist".

I'm going with literature evidence. The authors did not study its properties very well but somebody would do in future.


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How to qualitatively compare bond strength?
Without any thermodynamic data, how would one compare the bond strength of one molecule to another?
I understand a major factor is bond order: triple bond strength > double bond strength > single bond strength. But what if the molecules have the same bond order.
Also, generally a higher electronegativity difference gives a stronger bond (Chemistry LibreTexts article). But what if the difference is insignificant?
For example, in an exam, I was asked to compare $\ce{O-O}$ vs $\ce{S-S}$, and $\ce{O=O}$ vs $\ce{S=S}$ bond strengths. How would I compare them?

Hmm, the second part seems to be a duplicate of Why is an S-S bond stronger than an O-O bond? Is it?

@GaurangTandon thanks for the link, it answers the example but I'm looking for a more general answer.

While I understand that you've received an answer to the query of O−O  vs S−S, and O=O vs S=S, I am not sure if there's a single (or even a proper exhaustive list of several) of catch-all rules for comparing the bond strength of any two given molecules. Hence, I'm voting to close as "too broad". Perhaps, if you could detail your question to focus on a specific subset of molecules?

You can compare the bond strengths of $\ce{O-O}$,  $\ce{S-S}$,  $\ce{S=S}$ and   $\ce{O=O}$ by considering all possible effects that can alter the bond strengths. The effects which needs to be considered here are extent of π-overlap between the orbitals and inter electronic repulsions between the bonding atoms.
Firstly, it is obvious that double bonds will be relatively stronger than the single bonds. So, it is for sure that  $\ce{O=O} $ and  $\ce{S=S}$ will have higher bond strengths than  $\ce{O-O}$ and  $\ce{S-S}$. Now, we have to compare between two double bonds and two single bonds.
In  $\ce{O-O}$, two similar $\ce{2p}$ orbitals of two oxygen atoms form head-on overlap (σ-bonding) with each other, whereas, in  $\ce{S-S}$, two similar $\ce{3p}$ orbitals overlap. As it is a head on overlap, the extent of overlapping doesn't differ significantly, but for $\ce{2p}$ orbitals being smaller in size, two highly electronically dense $\ce{O}$ atoms comes much close to each other which hugely increase the inter electronic repulsions between the atoms. But in $\ce{S}$, the two bonding atoms remains much far from each other, and the electronic charge density is not also too much on each $\ce{S}$ as the $\ce{3p}$ orbitals are large in size, where the charge can easily diffuse. For this reason, $\ce{S-S}$ bonds become much stronger than  $\ce{O-O}$ bonds. (The reason is similar for the bond dissociation energy order  to decrease as in $\ce{Cl-Cl} >  \ce{Br-Br} > \ce{F-F}$)
But if we compare $\ce{O=O}$ and $\ce{S=S}$, the charge density on oxygen is reduced as lone pairs on oxygen decreases, therefore that effect becomes negligible. In this case, the extent of π-bonding starts to dominate. Being small sized,  $\ce{2p\pi-2p\pi}$ overlap in oxygen becomes much more strong and effective rather than  $\ce{3p\pi-3p\pi}$ overlap (where extent of overlap is not so much) in $\ce{S=S}$. That's why the introduction of double bond becomes much more thermodynamically favourable for oxygen and therefore, $\ce{O=O}$ bond is stronger than  $\ce{S=S}$ bond.
Thus the overall order of bond strength becomes, $$\ce{O=O > S=S > S-S > O-O}$$


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Does dihydroimidazole show geometrical isomerism?

Does dihydroimidazole (above) exhibit geometrical isomerism? Does the fact that the lone pair is involved in conjugation with a double bond affect anything?
As far as I know, it does not exhibit geometrical isomerism. However, one of my colleagues pointed out that the two nitrogens' lone pairs and hydrogens can adopt cis and trans configuration with respect to each other.
Please provide a detailed explanation. If possible, please also cite your sources.

Of course it does show isomerism, like any cyclic compound with different substitutes. (Double bond is irrelevant.) It is just that the isomers can't be isolated because they switch into each other way too easily.

@santimirandarp Hydrogen atoms connect with nitrogens can be cis/trans to each other. This isn't particularly good example, because conjugation with double bond almost flattens amine "pyramids".

related https://chemistry.stackexchange.com/questions/30943/is-nitrogen-of-aniline-sp%c2%b2-or-sp%c2%b3-hybridized https://chemistry.stackexchange.com/questions/4390/how-to-rationalise-the-resonance-structures-and-hybridisation-of-the-nitrogen-in

@schrodinger_16 although Mithoron is quite rude, he is right. Nitrogens are somewhere between $sp_2$ and $sp_3$ hybridization. If you look carefully at $sp_3$ nitrogen, you will see that nitrogen's hydrogen can be in cis-trans configuration...(hope to be clear even if I am wrong)


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Is the hydrogenation of carbon dioxide to methane a solution or a problem?

The hydrogenation of $\ce{CO2}$ to $\ce{CH4}$ is thermodynamically a favored reaction at ambient to moderately high temperatures.
$$\ce{CO2 + 4H2 <=> CH4 + 2H2O (g)}\qquad G^\circ_{\pu{298K}}= \pu{-27 kcal/mol}$$
$\ce{CH4}$ (natural gas) is an excellent fuel for a variety of applications. It is also a good C1 feedstock for manufacturing other chemicals. This reaction can be used to recycle $\ce{CO2}$ and reduce its emission to the atmosphere.
So, why not use $\ce{H2}$ from photoelectrolysis of water to produce $\ce{CH4}$ and achieve multiple goals of storing solar energy, creating a useful feedstock and recycle atmospheric $\ce{CO2}$?

I said, $\ce{CH4}$ is 30 times worse than $\ce{CO2}$ to the atmosphere and also the photoelectrolysis of $\ce{H2}$ is still an expensive method that requires more time to achieve maturity.
Is it right? the other student on the panel said my answer was too vague.

The problem is that the carbon dioxide which is released from burning is often mixed with other things which include air. If we were to want to convert carbon dioxide into methane then the oxygen of the air would make it harder.
While one option might be chemical looping combustion to obtain a purer carbon dioxide stream we still have the problem of where would we get hydrogen from. The normal method in industry to get the hydrogen gas needed for things like hydrogen peroxide production is to use steam reforming of methane.
$\ce{H2O + CH4 -> CO + 3H2}$
Using the water gas shift reaction the carbon monoxide is converted into carbon dioxide and hydrogen (using more steam). My worry about an industrial process for the conversion of unwanted carbon dioxide into methane is that unless we use hydropower, solar power or nuclear power to obtain electricity for the production of hydrogen from water then we will never be able to make a net reduction in the amount of carbon dioxide as the process will form as much carbon dioxide as it consumes.
There is some electrochemistry in which carbon dioxide can be converted into other things by reduction at the cathode but I think that this is currently only at a very small scale.


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Number of total stereoisomers of a piperazine derivative

How do I find out the total number of stereoisomers of 3,6-dimethylpiperazine-2,5-dione?
As far as I can see, the compound has two chiral carbons. Would the nitrogens make a difference? They seem chiral as well (three different groups and a lone pair).
My book says there are 3 total stereoisomers, but I expected more. It has considered only two chiral carbons, and nothing has been said about the nitrogens. Why?

Nitrogen's lone pair is conjugated with C=O and effectively doesn't exist.

Oh, right! Good observation!

Is that the only reason? How can you be so sure? Do you have a source?

Yes I think that's the only reason. I can be so sure because I've learned some chemistry in my life. Source? What source?

The meso compound in this example is quite interesting looking...

I believe this is a sophomore organic chemistry text book question, so I won't elaborate possible stereoisomers involving lone pairs on nitrogen atoms. Nonetheless, 3,6-dimethylpiperazine-2,5-dione has 2 chiral carbon centers so that it should theoretically have $2^2 = 4$ stereoisomers. If you look at the molecule, it has center of symmetry (i):

Therefore, two of the isomers are identical and we call it meso-isomer. Thus, total isomers of this molecule are three: ($3R,6R$)-, ($3S,6S$)-, and  ($3R,6S$)- (($3S,6R$)- is essentially ($3R,6S$)-).


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Half life of Nitrofurantoin
Both wikipedia and Toxnet list the half-life of Nitrofurantoin as 20-60 minutes in the blood plasma. Can anyone explain how this is so short, especially for an antibiotic, and how it still has time to be effective.

This question may be better off on Biology.SE

Nah. We're just talking pharmacokinetics.

Answer to your question is within the same Wikipedia page, which states following:

Mechanism of action: Nitrofurantoin is concentrated in the urine, leading to higher and more effective levels in the urinary tract than in other tissues or compartments. With a $\pu{100 mg}$ oral dose, plasma levels are typically less than $\pu{1 \mu g/ml}$ while in the urine it reaches $\pu{200 \mu g/ml}$. The mechanism of action is unique and complex. The drug works by damaging bacterial DNA, since its reduced form is highly reactive. This is made possible by the rapid reduction of nitrofurantoin inside the bacterial cell by flavoproteins (nitrofuran reductase) to multiple reactive intermediates that attack ribosomal proteins, DNA,[38] respiration, pyruvate metabolism and other macromolecules within the cell. Nitrofurantoin exerts greater effects on bacterial cells than mammalian cells because bacterial cells activate the drug more rapidly. It is not known which of the actions of nitrofurantoin is primarily responsible for its bactericidal activity. The broad mechanism of action for this drug likely is responsible for the low development of resistance to its effects, as the drug affects many different processes important to the bacterial cell.

High availability of nitrofurantoin in the urine ($\pu{200 \mu g/ml}$) compared to that in plasma ($\pu{1 \mu g/ml}$) is due to its high solubility in the urine. At this concentration, it exceeds the requirement for minimum inhibitory concentration for bacteria ($\pu{32 \mu g/ml}$), and as a result, nitrofurantoin becomes bactericidal in urine at therapeutical doses ($\pu{100 mg/capsule}$). Keep in mind that reduced nitrofurantoin (its metabolite by nitrofuran reductase) is much more reactive than nitrofurantoin, thus shorter half-life make it highly lethal to bacteria.

Why is it effective with such a low half-life in the blood?

First, we need to see what it is used for. This drug is primarily used to treat urinary tract infections (UTI). Therefore, it needs to be delivered to the urinary tract. The short half-life is a benefit because the drug gets concentrated in the patient’s urine, which kills bacteria in the urinary tract.

How is it metabolized?
Bioavailability is 80%, with roughly 25% of nitrofurantoin excreted in
the urine unchanged. Except in patients with severe renal failure,
serum concentrations are almost undetectable, with peak levels of
1 µg/mL. This is because of destruction of nitrofurantoin in the
tissues and, in particular, a very rapid renal elimination by
glomerular filtration (20%) and tubular secretion, resulting in a
serum half-life of only 20 minutes in patients with normal renal
function.

Source: Wikipedia page you referenced and ScienceDirect

Note that the OP is asking  how  not  why. You may want to expand your answer.

Ok, will edit my answer @AvnishKabaj

"75% of the dose is rapidly metabolised by the liver, but 25% of the dose is excreted in the urine unchanged"

Thanks @Zhe I was without internet for the last 30 mins or so.

I don't feel the answer is acceptable - it just spouts out what Wikipedia says, and doesn't address my question. I want to know how it is effective with such a short half life - your answer also appears to be wrong as per the comment from @Zhe

@NoahP what I am trying to get at is that the antibiotic isn’t designed to treat blood infections: it treats UTI, which means it doesn’t need to be in the blood

Presumably it also has a very short half life in the urinary tract too though, so how is it effective there?

The main reason it has a short half life in the blood stream is that the liver metabolizes it quickly, whereas in the urine there is nothing to metabolize it. (It’s stable on its own)


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T shape molecule with three hydrogens
Is there any sort of AX3E2 molecule that consists of hydrogen atoms bonded to the main atom who is left over with two lone pairs of electrons? (Similar to ClF3 just with hydrogen instead of fluorine?)

Probably doesn't exist.

I've not thought about it thoroughly but I also suspect it does not exist, for reasons similar to why SH6 doesn't exist.

We do have an example, with hydrogen bonded to ... ruthenium. The ternary hydride $\ce{Mg3RuH3}$ has been characterized via X-ray and neutron powder diffraction (with hydrogen in the form of deuterium) by Bonhomme et al. 1. The T shape is slightly distorted in the Y-like direction, but clearly closer to an ideal T than to a trigonally symmetric structure and having no evidence of pyramidalization. The shape is depicted in the figure taken from the reference.

The electronic structure is a nontrivial problem. The authors in Ref. 1 suggest an ionic formulation $\ce{[Mg^{2+}]3[RuH3^{6-}]}$, with the anion dimerized via a ruthenium-ruthenium bond to meet the 18-electron rule. However, an analysis of electron distribution by Firman and Landis 2 suggests that the magnesium is actually magnesium(I) ($\ce{Mg2^{2+}}$) and the ruthenium complex $\ce{RuH3^{3-}}$ consists of a covalently bound $\ce{RuH^-}$ center bonded predominantly ionically with the two flanking hydride ligands. If the $\ce{RuH^-}$ center is considered a unit isoelectronic with a silver ion (they both have $46$ electrons), then the complete complex would be valence-isoelectronic with a species such as $\ce{AgCl2^-}$.
References

F. Bonhomme, K. Yvon, P. Fischer (1992). "Tetragonal trimagnesium ruthenium trideuteride, Mg3RuD3, containing dinuclear [Ru2D6]^{12-} complex anions". Journal of Alloys and Compounds, 186, 2, 309-314.
ISSN 0925-8388, https://doi.org/10.1016/0925-8388(92)90017-4.

Timothy K. Firman and Clark R. Landis 1998). "Structure and Electron Counting in Ternary Transition Metal Hydrides". J. Am. Chem. Soc. 120, 48, 12650–12656. https://doi.org/10.1021/ja982746r.


text:
What result should I have obtained in these tests? Also, what went wrong?
As part of a Chemistry practical, I had to five different tests for things, and note down the result. Unfortunately, at least several results were incorrect, and despite hours of research, I cannot fully determine what I was supposed to observe and why.

Part 1 – A test for an alcohol.

To approximately 1 cm3 of ethanol in a dry test tube, add a small piece of metallic sodium.
Record your observations.
Make sure that you dispose safely of any excess sodium using the beaker of ethanol provided.

The liquid turned cloudy white. I can't find any discernible reason why this would be, unless sodium ethoxide is milky white, precipitate or dissolved. Should it just have released bubbles?

Part 2 – A test for an aldehyde using Fehling’s solution.

In a clean test tube mix together equal volumes of Fehling's solution A and Fehling's solution B. The resultant Fehling's test reagent should be a clear dark blue solution.
Add 5 drops of this test reagent to 1 cm3 of sodium carbonate solution in a test tube containing a few anti-bumping granules and then add 1 cm3 of ethanal (or propanal) to this same test tube.
Warm the test tube gently for approximately two minutes in a beaker half-filled with hot water and then gradually bring the beaker of water to boiling and maintain this temperature for a few minutes.
Using the test tube holder, carefully lift the test tube out of the boiling water and allow its contents to stand for several minutes.
Record your observations.

The liquid turned murky green. My understanding is that it's supposed to be red? I'd love to have a chemical equation for this as well.

Part 3 – A test for an alkene (a test for unsaturation)

To approximately 1 cm3 of cyclohexene in a test tube, add an equal volume of bromine water and shake the contents of the tube vigorously from side to side.
Record your observations.

The liquid went colourless. This one went as expected, I believe. My understanding is that it went colourless because all the bromine reacted?

Part 4 – A test for a carboxylic acid

Place one small spatula measure of solid sodium hydrogencarbonate in a test tube and add to it approximately 2 cm3 of dilute ethanoic acid.
Record your observations.

The liquid produced bubbles, went cloudy, then went colourless. No explanation here.

Part 5 – A test for a haloalkane.

Using a teat pipette, add approximately 5 drops of 1-bromobutane to 1 cm3 of sodium hydroxide solution in a test tube. Warm the contents of the test tube for a few minutes, by placing it into a beaker filled with hot water at approximately 60 C.
Acidify the contents of the test tube by adding 2 cm3 of dilute nitric acid and then add 1 cm3 of silver nitrate solution.
Record your observations.

Produced a blue precipitate after the bromobutane was added, and went colourless after we added the silver nitrate solution. No explanation here either.
Not enough time has elapsed as of now, but I will put reputation on this, and award it to a sufficient answer as soon as I am able.

@GaurangTandon I'm at my question limit for Chemistry, and I really need all of these answered.

One additional reason your question may have attracted negative attention is that it was hard to read. In the future, try to avoid lots of bold text, and use more lists. I turned the questions into quotes using >.


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What does ΔG° refer to exactly?

My interpretation of this image is that $\Delta G^\circ$ is equal to the difference between $G^\circ$ of the products and $G^\circ$ of the reactants.
But, I've also read $\Delta G^\circ$ as being given for calculating the difference in free energy between a point when $Q=1$ and the equilibrium position. That is, $\Delta G^\circ=-RT\ln(K)$.
Which one of these is the accurate interpretation?

$\Delta_\mathrm{r} G^\circ = -RT\ln K$ is undoubtedly correct, but "the difference in free energy between a point when Q=1 and the equilibrium position" sounds suspect.

hmm, won't ΔG = ΔG° when Q= 1? is it more precise to say that ΔG° is equal to the ΔG btw q=1 and k? furthermore, im trying to visualize these equations against the curve above ( i wish the y-axis had some sort of values, even if just throw away for the point of illustration). anyway, is it incorrect to thing of q=1 as being a point being XkJ higher on the curve relative to the equilibrium point if ΔG° = -XkJ ?

$\Delta_\mathrm{r}G$ and $\Delta_\mathrm{r}G^\circ$ represent slopes of the curve (specifically, $\Delta_\mathrm{r}G$ refers to the slope at any one specified point and $\Delta_\mathrm{r}G^\circ$ refers to the slope when $Q=1$) but the Gibbs energy $G$ refers to the actual value of the curve. The thing is, there is no reason why $\Delta_\mathrm{r}G^\circ$, which is a slope, should relate to the difference in the value of $G$ at two points - unless the graph is perfectly linear, which it isn't. Note also that the units of $\Delta_\mathrm{r}G^\circ$ are kJ/mol, not kJ.

The main problem in this diagram, I suppose, is the conflation of two very different concepts: changes in the Gibbs free energy of the system (i.e. the difference between two values of the curve, which is correctly denoted $\Delta G$) with the reaction Gibbs free energy (which is a slope and is more properly denoted with $\Delta_\mathrm{r}G$). This might help: https://chemistry.stackexchange.com/q/50569/16683

if $\xi$ is the extent of reaction the slope of the graph is $\left(\partial G/\partial \xi\right){T,p}\equiv\Delta_rG=\Delta G_r^\mathrm{O}+RT\ln(Q)$, at equilibrium (point 3 on the plot)  $\Delta_r G =\left(\partial G/\partial \xi\right){T,p}=0$

... and I should have mentioned that at equilibrium $Q=K_e$ the equilibrium constant.

I reread your post in light of your comments above and have a much better understanding of the difference btw these terms. Thanks!


text:
Importance of higher order exchange terms?
The MP2 ground state energy of a molecule or solid can be written as
$$E^{(2)} = \frac{1}{2}\sum^{\text{occ.}}_{ij}\sum^{\text{virt.}}_{ab} \frac{\langle ij | ab \rangle [\langle ab|ij \rangle - \langle ab|ji\rangle]}{\varepsilon_i + \varepsilon_j - \varepsilon_a - \varepsilon_b}$$
where the first part of the numerator is commonly called "direct MP2" and the second part "exchange MP2", since i and j are exchanged.
However, for energy differences, it is often sufficient to calculate only the direct MP2 energy, since the contribution of the exchange part is mostly negligible. For the same reason the random phase approximation, which consists of these "direct terms" up to infinite order, works so well for many materials.
But are there good examples (materials or molecules) where the exchange term is important? I guess magnetic systems or situations where Pauli repulsion is important... but does someone know particular examples?

Complete guess, with a rather indirect supporting argument: $\ce{F2}$.

@hBy2Py It would be more conclusive to look at an energy difference, say its IP. Do you what kind of bonding behavior the cation displays?

IPs are a very good example, see e.g. here https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.112.096401. But this I already knew ;-)

I actually don't think this paper presents enough information to be conclusive. Would you say the differences between the GW0, TC-TC, and gamma methods are significant for carbon, MgO, ZnO, and CdO? It's a difference of 0.5 eV at most.

Exchange is important and is present in any system. To be more specific, in MP2 you already have exact exchange for the ground state from the zeroth-term(HF) while in the second-term (MP2) you have a correction to correlation plus some more exchange from the excited state.
MP2 allows the electrons to be partly in the unoccupied MO however, it is limited to double excitations. It turns out that you end up having a small correction to exchange and a big correction to correlation which is completely missing in HF. So the exchange correction will be small in every system due to the type of excitation treated in MP2.

Thank you for the answer. I know all that, but I believe that there are systems where the MP2 exchange is not just tiny and negligible. You wrote "exchange is important in any system" but you also wrote "the exchange correction will be small". Yes, mostly, for energy differences, you can disregard it. And I am wondering if there are systems where you can not disregard it, without introducing a significant error. Please note, that I only talk about the MP2 exchange (not exact exchange from Hartree-Fock).

Look at the S66 collection  (https://cdn-pubs.acs.org/doi/10.1021/ct2002946) that analyzes weak, non-covalent interactions.  Exchange is extremely important in many of these.


text:
Different machines testing for same thing
For a chemistry report, we were given a sample of water and a variety of machines to test it with - giving their respective results about whatever aspect they were testing (eg pH, metal ion concentration etc). However the issue is that two of the machines tested for the same thing (concentration of metals in the water). What should I do in this case, now that I have 2 sets of data about the same thing? Should I discard one? Should I keep both and use it for comparison (although the results do differ somewhat between them)? I feel that doing something like averaging would be very invalid, and it has me very conflicted.
If it helps, the machines in question are AAS and ICP.

If possible repeat your analysis to make sure that an error has not crept in that you were unaware of. This would invalidate averaging. Also read up about the two techniques, one may be inherently more accurate/sensitive than the other or one may be subject to interference  from some types of metal ions or other species.

There are often standard solutions you can run to ensure the instrument's giving you the values you'd expect. NIST has a lot of standard solutions, as do commercial providers. Porphyrin's suggestion is a good one from a theoretical perspective, and if you have experimental data you should be able to decide how best to handle your data.


text:
Do atoms exist in hybrid form in nature, or do they form hybrid orbitals during molecule formation?
Do atoms exist in hybrid form in nature, or do they form hybrid orbitals during molecule formation?
Take methane for example. Does carbon atom exist with $\ce{sp^3}$ hybrid orbitals in nature and then hydrogen forms bond with only $\ce{sp^3}$ carbon atoms?
Or does carbon form $\ce{sp^3}$ hybrid orbitals during the formation of $\ce{CH4}$?


text:
Strong activation of phenoxide ion compared to others
I was told that in phenoxide ion, benzene ring is very strongly activated. The order my teacher told me was phenol < amine < $2^\circ$-amine < $3^\circ$-amine < phenoxide (for activation of benzene).
Example: Reimer–Tiemann reaction happens only in sodium phenoxide
So, what is so special in phenoxide that makes it so much activating? Also, why does oxygen contribute more in resonance than nitrogen, when it is more electronegative than the latter?

Strongly activated in respect of which property?

for example rtr happens only in sodium phenoxide

The question is probably about reactions of type electrophilic aromatic substitution. But there are possible others, e.g. redox reactions.

I assume by your notes that your teacher was mentioning electrophilic aromatic substitution reactions. These are two-step reactions.  The product-determining step (whether it is ortho-, meta- or para-) in the substitution mechanism is the first step, which is also the slowest (high activation energy), and hence the rate determining step (see [energy Diagram] below).  The second step is relatively faster, because positively charged intermediate ion loose a proton to get back the resonance stabilized aromatic ring again. Accordingly, the rate of the reaction is roughly depend on the influence of each of the given substituents to stabilize the benzenonium intermediate (positively charged adduct). You can best see it by looking at the capability of each substituent to delocalize the positive charge within the benzenonium intermediate.

Each given substituent ($\ce{-OH, -NH2, -NHR, -NR2}$) except for the phenolate ion has a lone pair. Thus, each can share it with the benzenonium nucleus to reduce the positive charge, reducing the energy of intermediate. However, phenolate ion ($\ce{ph-O^-}$) has full negative charge, which can neutralize benzenonium intermediate completely, to give least energy among all substituents considered here.


text:
How to count free rotatable bonds
I am having difficulties to count the number of free rotating bonds for the Lipinski and Veber rules. What are the rules in doing so?

In Vemurafenib (shown above), which bonds are considered rotatable? Do the sulfonamide N-S bond, or the single bond linking two aryl groups, rotate?
Many thanks

The idea of a 'rotatable bond' is somewhat arbitrary, given that under the right conditions basically any bond can be rotated. When thinking about concepts in medicinal chemistry like rotatable bonds, its therefore important to make sure you're working with the definition used when defining the rule.
The original Veber papers (J. Med. Chem. 2002, 45, 2615) define what they consider to be a rotatable bond:

Rotatable bonds were defined as any single bond, not in a ring, bound to a nonterminal heavy (i.e., non-hydrogen) atom. Excluded from the count were amide C-N bonds because of their high rotational energy barrier.

Based on this rule, the Ar-Ar single bond would count as a rotatable bond, as would the Ar-N bond. The sulfonamide is generally excluded for the same reason as amide C-N bonds (although I can't find a quote by Veber that actually discusses this).
In reality, these properties are usually calculated automatically, for instance using ChEMBL.

To add to this resonance affects if a bond can rotate. An amide bond tautomerises into an imide readily so is planar and a amine substituent on an arene (e.g. aniline) is not fully sp3 (cf. https://en.wikipedia.org/wiki/Aniline#Pyramidalization)


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Aromaticity of two rings connected by double bond
I want to know how to decide aromaticity of two rings connected by a double bond.
For example consider the ring in which when double bond is broken generates tropylium cation and cyclopentadienyl anion. I know that is aromatic because the ions generated are themselves aromatic.
How will the aromaticity be affected when we get an antiaromatic one?
For example consider the ring in which when we break the double bond cyclopropenyl anion is generated.
What will be the overall aromaticity of the molecule in this case.
And in general what is the rule to decide the aromaticity of the rings connected by a double bond?

Actually, the breakage of double bond doesn't happen in the way you are thinking it does. Let me illustrate the phenomena what happens in such rings connected by double bonds.
If you have two rings connected by double bonds where the breakage of that will produce anti aromatic rings, the molecule itself becomes very inert towards that breakage. The molecule always tries to gain stability through any perturbations to it. That's why production of anti-aromatic rings becomes extremely difficult in those cases.
In your given case, the molecule becomes highly polar due to unsymmetrical distribution of electron cloud (i.e. more negative charge on cyclopentadienyl ring and more positive charge on the other ring). But in a case, where you are assuming that breakage of double bond gives anti-aromatic cyclopropenyl anion or cyclopentadienyl cation, that breakage becomes extremely impossible. Even if you would subject it to extreme conditions, the molecule will never go into that transition state as that will have immensely high energy. Instead it will always shift towards that transition state by which it can gain some stability, by forming non aromatic or aromatic rings.
If it can't stabilize by that also, it will then become almost inert towards any electrophile or any external reagent which tries to make it anti-aromatic. Even if you perform ozonolysis, it will remain as a perfect keto-group, no unsymmetrical distribution will be observed for these classes of rings.
Considering the number of π-electrons in continuous conjugation of the whole molecule, we can judge if the molecule is aromatic or anti-aromatic. But if the molecules breakage of one double bond gives it anti-aromaticity, it will try to stay in its own form, whatever it may be.


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Anomalous trend in formation enthalpy of alkali metal halides
I read that in alkali metal halides, the formation enthalpy for fluorides become less negative as we go down the group, whereas the reverse is true for chlorides, bromides and iodides.
Why is it so? In my opinion, the formation enthalpy should increase with the stability which according to me, will increase as the cation's size increases and anion's decreases.
Therefore the trend is understandable in chlorides and others, but what about fluorides? May be it is due to its extremely small size but I can't find a suitable reason.