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Is there a real detection limit in NMR spectroscopy?
From what I've read, my impression is that with a sufficient number of scans etc., one can get spectra from even very dilute samples (S:N ratio ~ $\sqrt{\text{NS}}$). Obviously, this is not always possible in practice and it is not necessarily efficient use of instrument time.
However, assuming that (1) we have infinite instrument time and (2) the sample is indefinitely stable, is there still a true detection limit? If there is, where does it arise from?
[Unrelated: this question originally came about because I was reading a paper by Gschwind on CEST (J. Am. Chem. Soc. 2018, 140, 1855), but I realise that that technique is dealing with short-lived intermediates – hence my insertion of assumption (2).]

I'm unsure of the consequences of (2), but practically speaking if a sample isn't done after a day or two of running, no amount of additional scans is going to salvage it - peaks begin to broaden and the extra S:N is offset by this (at least on the kind of 500/600 machine used for routine organic characterisation - I'm sure there are far more stable magnets that don't experience this).

With any instrument its usually the noise that limits detection not the signal, and by scanning repeatedly and averaging then S/N can be improved as the noise is (ideally) random and the signal is not. In NMR if the magenetisation of the sample could be increased then so would the sensitivity other things being constant.

I've seen natural product NMR spectra collected from ~5 μg samples, including 2D spectra, using very narrow tubes. I can't recall how long scanning took, though.

see Lee et al.  J Am Chem Soc. 2011 Jun 1;133(21):8062-5. doi: 10.1021/ja111613c. for a sensitivity increasing method. Also optically pumping Xenon with the sample  can lead to huge sensitivity increase, see M.M. Spence et al., Proc. Natl. Acad. Sci. USA 98(19) 10654 (Sept. 11, 2001).

This is an interesting problem which also appears in gamma spectroscopy. Both 1D NMR and normal gamma spectroscopy are related. A sample emits photons, the photons are then captured by a detector.
In the case of a modern FT NMR machine the nuclei are excited by a radiofrequency pulse, the nuclei in the sample then re-emit radiowaves and the sample relaxs back to the way it was before the pulse in the free induction decay. After capture of the signal a analouge-digitial converter (ADC) supplies it to a computer which then uses the fourier transform to convert it into a histrogram of intensity against frequency.
Gamma counting uses something similar a energy dispersive detector records a series of events, these events are then fed into a ADC whose output is recorded by the computer. Normally again as a histrogram of events at different energies.
In both NMR and gamma spectroscopy there are two sources of noise, I define noise as unwanted signals in a system or spectra.
If we use gamma spectrscopy as an example for a moment, we have random noise due to cosmic rays (things like muons) and effects like Compton scattering. This will cause the random appearance of events in the detector with random energies. This will have a similar effect on the appearance of the spectrum to the random noise which exists in the electronics of a NMR machine. If we were to lower the noise level of the amplifers in the signal path before the ADC in a FTNMR machine we would be able to record spectra faster.
The more times we run the experiment the smaller the effect of the random noise will be on the final spectrum. I once ran an experiment for three weeks to get a spectrum from a lump of trinitite I bought off ebay. It turned out to be genuine.
The question is about the measurement of a stable sample. When I was a PhD student I worked on some samples which were unstable. I suspect to this day that I managed to dissolve for a short time somethings in my NMR samples which then later formed a superinsoluble solid. Once I saw a peak appear in the FT-NMR spectrum at the expected chemical shift which then disappeared again. We can consider this side question also.
Now one problem I see in both systems is if some non random unwanted signal enters the spectrometer before the ADC.
Consider for a moment if we were to do NMR spectroscopy in a 1:1 mixture of CHCl3 and CDCl3. We would have a very strong chloroform peak, this would limit the sensitivity of the amplifiers which can be used before the ADC, by forcing the system to reduce the senstivity of the amplifiers before the ADC what will happen is that the effect of random noise in the ADC will become greater. This would make it harder to make the measurement.
Also with a NMR spectrometer we have a limit as to the number of bits in the computer memory which handles the output from the ADC. If we had eight bit memory. then when one of the points in the free induction decay was to reach 11111111 then the system should not attempt to pulse the sample to record any transient.
If we were to pulse the sample again and then try to increase the count in this point (in time) in the FID, then we would have a nonsensical situation which would cause mayhem with the fourier transform. I think it could then fail to work as intended
Also in gamma spectroscopy we would have the problem if we were to overflow the memory storing the output of the ADC we would start to get a distortion of the spectrum. I suspect that in gamma spectra the effect would be less. We would merely get the highest peak to become zero again. But I suspect that the rest of the spectrum would still be OK.


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Why the energies of orbitals in the same subshell decrease with increase of atomic number?
Why the energies of orbitals in the same subshell decrease with increase of atomic number? For eg., energy of 2s orbital of hydrogen atom is greater than energy of 2s orbital of lithium atom. What is the reason?

Hydrogen doesn't have a '2s orbital' or a 2s subshell. It only has a single s orbital in the 1s subshell.

@KianStevens an excited state of hydrogen can have the electron promoted to the 2s orbital.

@Tyberius Maybe so. I wasn't on about excited states however.

@KianStevens I guess my point is just that, whether experimentally or through the Schrodinger equation, you can measure the energy of other orbitals of hydrogen. In some sense, hydrogen is the only atom that has a 2s orbital.

I haven't ever studied the Schrödinger equation etc. so my knowledge is limited in that regard.

I think the point is that the atom "has" a 2s orbital, but in the ground state there's just no electron in that state.

Let example of H and He, He has two protons and attracts it's subshells or shells with more energy and it's shells acquire more stability than in H with only 1 proton

Short answer: more protons in the nucleus means higher nuclear charge which means stronger attraction between the electron and the nucleus.
Long answer: The energy of the orbital is driven by the Hamiltonian operator $\hat{H} = \hat{T} + \hat{V}$ where $\hat{V}$ is the potential energy. For a simple 1-electron atom/ion, the potential energy is the electrostatic attraction between the positively charged nucleus $Q$ and the negative electron $q$: $V = \frac{1}{4\pi\epsilon_0} \frac{Qq}{r}$. Lithium has 3 protons in the nucleus versus the 1 proton for hydrogen, so $Q$ is three times higher and the potential energy of the 1s orbital in $\ce{Li^{2+}}$ is roughly three times more negative than the 1s orbital in $\ce{H}$ (remember $q$ is negative).
If you have more than one electron, the potential energy becomes more complicated because you have to include the electrostatic repulsion from all the other electrons as well, but the general trend still holds, all else being equal.

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Does hydrogen peroxide - urea become more active when illuminated?
Dentists' tooth whitening systems and consumer kits use gels containing carbamide peroxide, also known as hydrogen peroxide - urea.
Dentists use a light activator: (image from Wikipedia).

Consumer kits typically include 35% carbamide peroxide gel. Some of them use an LED light too, while others don't.
Wikipedia seems to be in two minds about it, stating

Recent research has shown that the use of a light activator does not improve bleaching, has no measurable effect and most likely to increase the temperature of the associated tissues, resulting in damage

while also talking about which light is "best":

The ideal source of energy should be high energy to excite the peroxide molecules without overheating the pulp of the tooth

Is a light source relevant here, or is it merely a marketing gimmick?

'High energy' means using ultra-violet light and high intensity means lots of them. The absorption spectrum of hydrogen peroxide increases significantly below 300 nm so any light here will speed up its dissociation into radicals. However, UV of these wavelengths is dangerous which is why we add sunscreens when outside in high sunlight to prevent sunburn.

@porphyrin Thanks, I wonder if the LED lights in consumer tooth whitening kits even reach the UV range.

in consumer LED's i'm sure you are correct that there is very little if any uv. In that case the effect of illuminating will be small as very few photons will be absorbed.

I hate to answer my own question, but after some digging, I believe that the use of a light source is in fact a gimmick, at any concentration of the reagent, contradicting the only other answer so far (But I'm happy to be corrected, if I'm wrong)
This meta-study https://www.ncbi.nlm.nih.gov/pubmed/22525016 concluded that at high concentrations of the reagent (typical of in-office use), the light is useless; and there was insufficient or no data for lower concentrations (as of 2012):

Light increases the risk of tooth sensitivity during in-office
  bleaching, and light may not improve the bleaching effect when high
  concentrations of HP (25-35%) are employed. Therefore, dentists should
  use the light-activated system with great caution or avoid its use
  altogether. Further rigorous studies are, however, needed to explore
  the advantages of this light-activated system when lower
  concentrations of HP (15-20%) are used.

A slightly more recent study came to the same conclusion https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4178330/ :

The in-office bleaching treatment of vital teeth did not show
  improvement with the use of light activator sources for the purpose of
  accelerating the process of the bleaching gel and achieving better
  results.

I'm inclined to think that if light were useful at all, they'd see this at higher concentrations of HP too, but I haven't seen any recent studies that address the lower concentrations specifically.

Before answering your question, we must review the mechanism of bleaching action. Hydrogen peroxide ($\ce{H2O2}$) is replacing other beaching agents as the preferred beaching agent in many applications. Although it was accepted that $\ce{H2O2}$ acts by destructively oxidizing colored organic molecules involved, its bleaching mechanism by such oxidation is still not well understood. However, recent works on the bleaching of textile by $\ce{H2O2}$ have identified that the hydroxyl radical ($\ce{HO^.}$) and perhydroxyl radical ($\ce{HO2^.}$) are the active oxygen species mostly responsible for such action [1]. The facts found by a separate study on the effects of the $\ce{HO^.}$ radical and $\ce{H2O2}$ on tooth bleaching suggest that $\ce{HO^.}$ has the main role in tooth bleaching with $\ce{H2O2}$ [2]
Suggested path for formation of $\ce{HO^.}$ and ($\ce{HO2^.}$) radicals from aqueous $\ce{H2O2}$ is as follows:
$$\ce{H2O2 + H2O <=> HO2- + H3O+}$$
$$\ce{HO2- + H2O2 -> HO2^. + HO^. + HO-}$$
Now, we look at bleaching process in Dentists' office: Stable hydrogen peroxide-urea solid is commonly used in these situations to avoid unstable aqueous $30\%~\ce{H2O2}$. It is described in Wikipedia article on hydrogen peroxide - urea as follows:

Hydrogen peroxide-urea (also called Hyperol or artizone) is a solid composed of equal amounts of hydrogen peroxide and urea. This compound is a white crystalline solid, which dissolves in water to give free hydrogen peroxide.

In the Wikipedia article on tooth whitening mentioned in your question states that:

The bleaching agent is either carbamide peroxide, which breaks down in the mouth to form hydrogen peroxide, or hydrogen peroxide itself. The bleaching gel typically contains between 10% and 44% carbamide peroxide, which is roughly equivalent to a 3% to 16% hydrogen peroxide concentration. The legal percentage of hydrogen peroxide allowed to be given is 0.1-6%. Bleaching agents are only allowed to be given via dental practitioners, dental therapists and dental hygienists.

At a dentist's office, they may use the light source to accelerate (and light-influenced catalytic radical cleavage) above spontaneous reactions to generate $\ce{HO^.}$ and ($\ce{HO2^.}$)Radicals. To support my argument, I'd like to point out the following statement from the same source you noted above in the Wikipedia article on tooth whitening:

Use of light during bleaching increases the risk of tooth sensitivity and may not be any more effective than bleaching without light when high concentrations of hydrogen peroxide are used.

Thus, use of light source is not a gimmick. It is a necessity.
References:

The mechanism of hydrogen peroxide bleaching: J. Dannacher, W. Schlenker, Textile Chemist & Colorist, 1996, 28(11), 24-28.
Effects of the Hydroxyl Radical and Hydrogen Peroxide on Tooth Bleaching: K. Kawamoto, Y. Tsujimoto, Journal of Endodontics, 2004, 30(1), 45-50 (DOI: https://doi.org/10.1097/00004770-200401000-00010).

I wonder if the second quote is accurate. 35% carbamide peroxide is available over the counter and online (in the U.S.)

I think you are right. $35%$ carbamide peroxide approximately equals to $12%$ $\ce{H2O2}$. One of my friends bought, I believe, $15%$ $\ce{H2O2}$ solution online for his son's high school project.

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A question about "post it note"
I dipped a sticky note in water and there was a clear difference between the water absorption in the area with glue and the one without glue. The area with glue was more damp than the one without it. I would like to know why.

I like your "why?" question. So I dig down to find out why Post-It notes stick to any surfaces, yet, when removed, it's able to post again in another place. This is my finding and it makes sense Adhesives:

Look at the back of a sticky note using an electron microscope and you'll see not a continuous film of adhesive but lots of microscopic glue bubbles, known as microcapsules, which are about 10–100 times bigger and much weaker than the glue particles you'd find lazing around on normal sticky tape. When you push a Post-it® onto a table, some of these relatively large sticky capsules cling to the surface, providing just enough adhesive force to hold the weight of the paper in the little note. Every time you attach and peel off the note, dust and dirt attach to the adhesive capsules, so they progressively lose their stickiness. But since there are so many capsules of all different sizes, a Post-it® note does go on sticking for quite a while.

I believe, when you dipped your sticky note in water, these microcapsules filled with water, which would be a thicker layer than the water in other areas. So, gum layer area contains more water, thus, that part is actually more damp than the rest of the paper.

yeah, i think you`re probably right.

but why did that sticky note retain its stickiness after i dried it? since we know if we poured  water on glue, the glue is dissolved in water and thus loses its stickiness.

These microcapsule have 'air-tight' kind mechanism. When you push it to surface, air in capsule escape and it stick by pressure difference. I bet it won't be efficient as before, though.


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Finding molecules that you randomly create using a modeling kit
I recently bought https://snatoms.com/ (which is just a real-life modeling kit for atoms/molecules) and was just keen on learning about chemistry in general. I often randomly join atoms to make up some molecule. Was wondering if there is an easy way to find the molecule i created?

A molecule obtained by joining random atoms from a modeling kit is most likely a nonsense and not worth looking up. That being said, you may start from searching by the molecular formula.

Meantime, you may sign in with Reaxys and  draw your structures and search for molecules or reactions.

As far as I know, databases such as Reaxys or SciFinder require institutional access, which OP may or may not have

@IvanNeretin the kit is mostly carbon, hydrogen, and oxygen, so as long as they use the model correctly, it's not likely they would make something that doesn't exist.

Chemspider allows a user to input a structure and tries to find a match, you could give that a try

Just curious, why dont folks add answers instead of comments. That way i could pick the most helpful suggestion/answer as the answer.

You can apply the rules of the IUPAC nomenclature of organic chemistry to construct the name of the compound you've created. Albeit you might find this way being not too easy, it will give you the maximum insight to chemistry you can expect from playing this kind of game. https://en.wikipedia.org/wiki/IUPAC_nomenclature_of_organic_chemistry

Here's a list of all the stuff you can try:
For you, the common theme of all these steps would be to be able to correctly identify the skeleton of the molecule that you built yourself, and then being able to draw it on the online molecular editor.

Sign into Reaxys or SciFinder, draw your structure, and search for molecules or reactions. Note that these require institutional access (or a very hefty fee).
ChemSpider Structure search: It's a free tool, and my personal goto for getting access to several matching compounds. You can search for "exact matches", or partial matches like "tautomers", "same skeleton", or even "all isomers". There's search filters as well. But again, the first few options are sufficient for your snatoms kit.
eMolecules (in case you didn't notice, we also advertise it in our right sidebar) very similar to ChemSpider above.
PubChem Structure search - again, similar to the above two. Note that though this one has an odd (and old-looking UI) which you might not find intuitive.

The main difference between all four of these is primarily the level of detail and depth of their data. Functionally, they all do the same thing. But, they do so on different databases. That means for more complex molecules, their outputs may differ. However, this hopefully won't be a problem with your simple snatoms kit, so you can go ahead and try any of them. Good luck!

Addendum: generating a 3D structure from the molecule name
You may use this free online tool based on JSMol. Type the IUPAC name of the molecule (say (1-methylcyclobutyl)benzene) into the lower right box, and press Enter. Its 3D structure is then immediately transferred into the box above. You can zoom it in or out, or rotate it by 360deg. This would help when you're trying to create the 3D molecular structure given the IUPAC name.


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What happens to a quaternary ammonium cation at a pH above its pKa?
I am working with a lipid and I have firm reasons to believe that its pKa is around 8. This lipid is a cationic lipid where the nitrogen is a quaternary amine with two lipid chains and two methyl groups.
Now, my question is that at a pH above the pKa of this lipid, the quaternary ammonium ion should get deprotonated, or basically act as an acid. But the quaternary ammonium cation doesn't have any protons to lose, so what would happen to it? My guess would be it accepts a lone pair and becomes negatively charged.

You seem to think that the pKa of a quaternary salt is the same as the pKa of any other ammonium salt (ie around 9), but that’s not true. The pKa of something is a direct measure of how easily it’s deprotonated. Ammonium salts like R3NH+ are relatively easy to deprotonate, so their pKas are relatively small. A quaternary ammonium salt having no N-H bonds is not easy to deprotonate, so its pKa is large. Things can certainly happen at high pH values, eg elimination or substitution, but a pH of 14 (for example) is not larger than the pKa of the quaternary salt, and it won’t behave as an acid.

If you have "firm reasons" that the $\ce{pK_a}$ is around 8, then acid group must be located at the lipid chains. So what lipid groups are you talking of?

there is no acid chain. the long chains are long aliphatic chains. with no double bonds or other groups present... my basic question is that in case of a quaternary ammonium ion with no H+ to deprotonate what really happens when pH exceeds its pKa?

You keep saying pKa. What is the pKa of a quaternay ammonium ion? What exactly is the deprotonation you're looking at?

hi! its me again. I am trying to find more data on quaternary ammonium ion exchanging their counter ions, like at what pH it is triggered, or in what conditions. or is it possible that above a certain pH the counterions can shield the quaternary ammonium ion from interacting with other species in the solvent.. can someone help with these questions or guide me to some literature regarding this?

Amine compounds in which the nitrogen is bound to four carbon atoms through covalent bonds are known as quaternary ammonium salts. They are stable compounds that will not converted back to their relevant amines by treatment with base. Generally, they are very soluble in water because of their ionic nature, and also soluble in commonly used organic solvents due to (long chain) organic substituents (e.g.: phase-transfer catalysts). Because they do not contain a free electron pair on nitrogen, in general, the quaternary ammonium ions are inert towards chemical reactions. Thus, pH changes course only them to change their counter ions.
In fact, not only quaternary ammonium salts are relatively unreactive, but also, they do not diffuse readily across biological membranes when used in medicine, because they exist as permanent cations.

thank you! my experiments do show that that the cationic lipid retains its charge. but i just needed some explanation as to what could happen in such a scenario.
also, how high could be the pKa of the basic nitrogen of the quaternary ammonium salt? maybe my assumption of its pka being 8 was wrong.

There is no exchangeable $\ce{H}$ on Quaternary $\ce{N}$, thus, theoretically, it doesn't have $K_\mathrm{a}$.


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Acid promoted Enol ring formation Mechanism

I know this has to be an enol mechanism but which carbon gets attacked by the $\pi$ electrons?

Why does it have to go via an enol? In fact, it doesn’t. The electron-rich arene is nucleophilic (you can surely figure out which positions) and the ketone is electrophilic.

This reaction proceeds in a few steps. In acidic media, the upper carbonyl oxygen is proton. This results, in greater electron deficiency of the upper carbonyl carbon, which strongly attracts nucleophilic attack. Due to strong activation by the methoxy substituent, the arene's $\pi$ electrons can attack the electron-deficient carbon in the carbonyl group. This nucleophilic addition to the carbonyl group gives a tetrahedral intermediate with a positive charge delocalised in the neighbouring ring.
Note that two types of reactions are happening here: Electrophilic aromatic substitution at the position para to the methoxy substituent and also, a nucleophilic addition to the upper carbonyl group.
A chloride ion now comes in to remove the proton from the para position of the arene. We then get back the aromatic ring with the methoxy group. In the next phase of the reaction, we now want to remove the $\ce {-OH}$ that we have created on the previous carbonyl carbon. This would entail us first protonating it, to make it a good leaving group. How do we create the double bond then?
Well... It is now important to note that $\alpha$ hydrogens in carbonyl compounds are relatively easily lost due to the electron-withdrawing effect of the carbonyl oxygen. Furthermore, the negative charge generated from this loss of the proton can also be delocalised into the the neighbouring aromatic ring on the right of the figure, enhancing the stability of this conjugate base, increasing the acidity of this hydrogen. What happens now is simply elimination to form the double bond. The protonated $\ce {-OH}$ first leaves as water to generate a tertiary carbocation. Then, a chloride ion acts as a base to remove the acidic $\alpha$ hydrogen, with the electrons going into forming a $\pi$ bond between the two carbons. This is essentially an $\ce {E 1}$ process. The desired product is now obtained.


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Is it possible to prevent sacrificial anode contaminate the soil?
I have a mild steel pipe (black pipe, NOT zinc plated) water well. My question: is it possible to somehow put a sacrificial anode in an electrolyte contained in a bucket instead of burying it in the ground, which allows zinc to eventually enter and contaminate ground water?

I wonder if a sacrificial anode near the surface could prevent corrosion at the deepest part of your pipe, where the water and wet soil are. Usually, sacrificial anodes are placed fairly close to the item being protected.
It would be worth testing some of the water (that comes up thru the pipe) for pH and corrosivity. If the water is alkaline (hard water), it may not be very corrosive. The oxygen level far below ground would be low, perhaps not enough for rapid corrosion.
Maybe a sacrificial anode is unnecessary.

It's also possible to use magnesium, for example in protection of pipelines.  Magnesium will still "contaminate" the soil, but the environmental effect is much less than with zinc or cadmium because magnesium compounds are already common in soil.

Why not use the standard aluminum anodes ? I can't imagine how your "bucket of electrolyte " will work.

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Which attack occurs first: nucleophilic acyl substitution or nucleophilic substitution?
If there is a compound with a site for nucleophilic acyl substitution (i.e. an ester group or the like) as well as a nucleophilic substitution (i.e. a bromine group or the like), will a nucleophile preferentially attack one of them? How do i decide which one a nucleophile would prefer attacking?

Based on the fact that the pi bond is more reactive, in the sense that it is more easily broken, nucleophilic acyl substitution may be more kinetically favourable and would thus occur first.

The answer is - it depends. On the conditions, on the solvent, on the nucleophile being used, on which ester is being used. You may get selectivity, you may get attack at both sites which generates a mixture.
In the real world of chemistry you often don't know for certain until you run the experiment unless someone has published an exact procedure for the transormation which you follow.


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Which species are oxidized and reduced when sufficient voltage is applied?
For example, you have two salts in a solution. Say one of the species half reaction requires 1V (standard electrode potential), and another requires 3V, and you have a current running through the solution at 4v.
Here are the possibilities I thought of, I don't know which or if any, are right:

Both  species react simultaneously at the same rate.
Both species react simultaneously, but the one that requires 1V reacts faster.
The species that requires 1V reacts first, and once it is completely reacted, the species that requires 3V reacts after.

Am I completely off or what is the right answer?

If you applied 4V right away, both species would begin reacting. Generally speaking, the higher you are above the half-reaction's themodynamic potential ("overpotential"), the faster the reaction goes, so the 1V would react faster. The rates are not determined solely by the overpotential, though, and at some point you'll hit the rate-limiting current. At this point, increasing the potential doesn't increase the rate any more because you can't supply reactants to the electrode surface fast enough from the solution bulk, so you could wind up with the 1V species reacting slower.
A more interesting experiment is cyclic voltammetry, where you slowly increase the potential from open-circuit up to 4 volts. Measuring the current as a function of potential tells you at what potential the reactions happen and something about their character. In this case you'd see the first species oxidized slightly above 1V and species 2 oxidize just above 3V. By the time you hit 4V there would not be much current (assuming the reaction is fast compared to the sweep rate), since many of the reactants have been consumed.


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What is the difference between an ionic compound and a salt?

What is the difference between an ionic compound and a salt? They both have a cation and an anion bonded.

On one of my homework sheets, it says that $\ce{NaCl}$ is an ionic compound (which it obviously is). How would i determine here if $\ce{NaCl}$ is an ionic compound or a salt?

A salt is an ionic compound that's formed during the neutralisation reaction between an acid and a base. So really, a salt can be described as an ionic compound where a hydrogen ion in an acid is replaced by a metal ion or the ammonium ion. An ionic compound in general has bonding where ions electrostatically attract.

The GCSE answer is.
An ionic compound is one in which the bonding between the atoms is ionic, for example sodium chloride is an ionic compound. It is formed from sodium cations and chloride anions.
There are substances which can be regarded as being ionic which are not normally viewed as salts. For example some metal oxides such as potassium oxide are probably best described as being ionic solids but in this case the compound is formed from oxygen and potassium.
A salt is the product of reacting an acid and a base together. Many salts are ionic. For example if you were to react hydrochloric acid with zinc and then to dehydrate the product (avoiding hydrolysis) you would end up with zinc chloride.
The PhD chemists view of the matter
One great problem in chemistry is that there are very few perfectly ionic compounds. I hold the view that the more chemistry you know the more one becomes aware of the fact that simple explanations do not work well.
The great problem in my mind is that a great spectrum of bonds exist from perfectly ionic to simple unpolarised covalent bonds. All cations have some ability to pull electron density towards themselves. A simple anion such as a chloride should in the idea world be a sphere of electron density.
The electrostatic attraction of the electron density by a cation on an anion pulls electron density towards the cation. This can be regarded as leading to some covalent character of the bond existing. Thus as a result I hold the view that there are very few true perfectly ionic compounds.
One example would be tetraphenylphosphonium tetraphenylborate which has two very large ions, the interaction between these ions is close to a perfect ionic attraction. Here is a diagram of the tetraphenyl phosphonium cation.

We can see here that the ion has the positive charge deeply buried inside it, here is the solvent accessible surface, I have rendered it in colours to represent the charges at points on the solvent accessible surface.

The atom locations can be seen in the following picture to help you understand what is happening.

Even the tetramethyl phosphonium cation is able to isolate the positive charge away from the solvent accessible surface, in contrast the trimethyl sulfonium cation (Me3S+) presents to the outside world (the solvent accessible surface) part of the ion which is quite positive.

We can also take a molecular orbital view of bonding, a lot of coordination chemistry of transition metals is considered with the ligand field model. The ligand field model is based on the crystal field model which treats all bonding as ionic. However the same results can be obtained for d block metals using molecular orbital theory.
It is also interesting to note for the lanthanides that they do not have strong ligand field splitting effects, this is due to the fact that their f orbitals are close to the centre of the atom and do not interact well with ligands. In many compounds often the bonding of lanthanides is viewed as being ionic.
If we consider for a moment palladium chloride then it is clear that the attraction of the chlorides to the palladium cations results in the formation of covalent bonds. If we move back from palladium to a metal which is less able to form bonds to chloride (cobalt).
Then if we have anhydrous cobalt chloride, we have in the anhydrous solid a Co-Cl distance of 2.453 Å. If we dissolve cobalt chloride in water then the water molecules will react with the cobalt(II) centres to form hexaaqua complexes of cobalt [Co(H2O)6]2+ which have a larger radius than cobalt(II) cations. These are less able to electrostatically attract chloride anions and the bonding of the chloride anions to the [Co(H2O)6]2+ cations will become closer to being a simple electrostatic (ionic) attraction / bonding than the bonding between the chlorides and the cobalt in the anhydrous cobalt chloride.
Here is a picture of the crystal structure of anhydrous cobalt chloride. (H. Grime and J.A. Santos, Zeitschrift Fur Kristallographie, 1934, 88, 136-141) You can see how the cobalt has a octahedral coordination environment in which six chloride anions form bonds to the cobalt. Keep in mind that the chloride anion is a ligand with a weaker ligand field than water. As a result the cobalt adsorbs red light in the form of anhydrous cobalt chloride.

By the way the solid is a layered solid, the solid can be thought of as being like a series of slices of bread (chlorines) with the cobalt as the jam.

By the way the distance between cobalt and chloride in the [CoCl4]2- complex is 2.276 Å, which is a bit shorter than the distance in the anhydrous solid.

Molecular orbital structures can also enhance ionic bonding.  Boron has an unassumingly low electronegativity, but it can form clusters or networks that are electron-deficient with respect to filling their bonding orbitals.  Such aggregates then tend to form predominantly ionic bonds with electropositive metals such as magnesium.

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Is there a theoretical way to predict the coordination number a metal can have in a complex?
We know that metals can form complexes with different coordination numbers, but I was told that certain metals have a greater probability to have a certain coordination number.
But I was not able to find any such list that correlates a metal with its most favorable coordination number.
As an example, I guess $\ce{Fe^3+}$ forms complexes with coordination number 6, but can I actually predict the coordination number in a reaction or not?

Summary of molecular orbital theory for coordination compounds-Ligand Field Theory
Molecular Orbital Diagram can explain what you are looking for. First, you choose the geometries you expected to have with corresponding metal cation and ligands. You need to do this since different complex structures will have different more and less energetic d-orbitals.
Second, do MO diagrams for all geometries, and compare the energies, or more properly, stabilities of compounds. Looking at antibonding interactions with ligands you can even predict which bonds of the ligands are becoming more vulnerable to attack.
Additionally, there are such lists in inorganic chemistry textbooks in general. Ferric ion; for instance, can make stable 7-ligand-complex. However, ligands are also very crucial on this aspect and there is ongoing research, like this one:
8 coordinate ferric ion

https://www.york.ac.uk/media/chemistry/research/douthwaite/Metal-ligand%20and%20Metal-Metal%20Bonding%20Notes.pdf
Have a look on this. Page 17 is kind of a list you are looking for. You may find more if you search "ligand+field+theory+number+of+possible+ligands" or such.


text:
Why do vinylic carbocations generally not undergo hydride rearrangement from neighbouring sp3 carbon to get more stability?
Why do vinylic carbocations generally not undergo hydride rearrangement from neighbouring sp3 carbon to get more stability?
According to me, a rearrangement would lead to an allylic carbocation, which is more stable than vinylic, hence the rearrangement should be favorable.

They do. https://en.wikipedia.org/wiki/Vinyl_cation#Rearrangements_in_vinyl_cations

If you think about protonating 1-propyne on the terminal acetylenic carbon, you get a vinylic cation that could apparently form an allylic cation by a hydride shift. The bond to the hydride that would shift has to be lined up with the empty orbital to which it is migrating.
So far, no problem. But after the shift, the empty $\mathrm{2p}$ orbital of the new carbocation is orthogonal (at right angles) to the $\mathrm{2p}$ orbitals of the adjacent double bond. Therefore, what looks like an allylic carbocation is actually a twisted allylic carbocation.
The $\mathrm{2p}$ orbitals of the double bond can't stabilize the new cation by resonance. Worse yet, the $\mathrm{2p^2}$ carbons of the double bond are relatively electronegative, so they actually destabilize the new carbocation. In other words, the migration is forming another unstable carbocation.
This is why rearrangements in vinylic carbocations don't occur. (There is a Russian Chemical Reviews article [1] that explains all this, but it's hard to get a copy of this, and their PDFs seem to be corrupted.)
References

Shchegolev, A. A.; Kanishchev, M. I. Rearrangements in Vinyl Cations. Russ. Chem. Rev. 1981, 50 (6), 553–564. https://doi.org/10/b7rrk8. (mathnet.ru mirror; PDF (in Russian))

Since "Russian Chemical Reviews article, vol. 50, no. 6" is an ambiguous reference (there are six articles in total) I took a liberty to pick the one you are probably talking about and supplied full bibliography alongside with alternative links from the Russian mirror mathnet.ru (the PDF link on iopscience.iop.org is indeed pointing to the wrong document). Please confirm the article is the correct one; if not, please add its title and authors.

(Wikipedia has a very extensively detailed section on vinyl cation rearrangement, so be sure to give it a read as well for a more in-depth view. I'll base my answer on that section)
1,2-hydride shifts are very uncommon in vinyl cations. "The interacting orbitals during the conversion of a linear vinyl cation to a non-linear allyl cation are orthogonal and passes through a non-planar transition state". This leads to a very high activation energy for such a rearrangement. 1,2-methyl shifts too have a very high activation energy.
Hence, even though such a rearrangement would lead to a thermodynamically favorable product, it is uncommon.


text:
Modification to improve antibacterial activity

I have difficulties with part b).
For C, it's clear to make a bulky dimethoxybenzene group in order to block the C=O group from reacting with b-lactamase.

However, in A, how does the structure of sulfonamide affect the bacterial activity in general?
Is there an alternative for C as well?

I suspect you are reffereing to B on your second paragraph. Also I think 2 structures  being refered here are A and B, they can be directly modified. C on its own has very weak antibacterial activity but serves a useful purpose when used in synergy with potent antibiotics e.g amoxicillin

Essentially the basis of antibacterical activity of sulfonamides is the inhibition of  the enzyme dihydropteroate synthase, an
important enzyme needed for the biosynthesis of folic
acid derivatives and, ultimately, the thymidine required
for DNA.
Now they do this by competing at the active
site with p -aminobenzoic acid (PABA), a normal structural
component of folic acid derivatives.

PABA

So how does the structure of sulfonamide affect the bacterial
activity in general?

The basis of the structural resemblance of sulfonamides to PABA is clear. The
functional group that differs in the two molecules is the
carboxyl of PABA and the sulfonamide moiety of sulfanilamide.
Now there are requirements that are vital for activity:

Para-amino group:

unsubstituted (i.e $\ce{NH2}$).
very important for activity because any modification of it other than to
make prodrugs results in a loss of activity. For example, all
of the N4-acetylated metabolites of sulfonamide are inactive.

Sulfonamide:

$\ce{N}$ must be primary or secondary
$\ce{R2}$ can be substituted

Aromatic ring:

Para substituted

Modifications
The pKa of the carboxyl group of PABA is
approximately 4.9 thus an optimised sulfonamide has a pKa somewhere close to this value.

It was observed that replacement of one of the $\ce{NH2}$ hydrogens by
an electron-withdrawing heteroaromatic ring enhanced the acidity of
the remaining hydrogen and dramatically enhanced potency.
With suitable groups in place, the pKa is reduced to the same range as
that of PABA itself. Not only did this markedly increase the
antibacterial potency of the product, but it also dramatically
increased the water solubility under physiologic conditions.

Structural variation among the clinically useful sulfonamides
is restricted primarily to installation of various
heterocyclic aromatic substituents on the sulfonamide
nitrogen.
An example  is Sulfisoxazole (pKa 5.0) which can also be formulated as a prodrug, by inclusion of N-acetyl groups. Another advantage is the acetyl derivative
is tasteless and, therefore, suitable for oral administration, especially
in liquid preparations.

Is there an alternative for C as well?

Clavulanic acid is a naturally occurring, mechanism based
inhibitor , which causes potent and
progressive inactivation of β-lactamases.
There  are alternatives e.g sulbactam (A synthetic derivative of penicillins undergone the oxidation of the sulfur atom to a sulfone which greatly
enhances the potency) or tazobactam.
References

The 12th Edition of Wilson and Gisvold’s Textbook of Organic Medicinal and Pharmaceutical Chemistry
Foye’s Principles of Medicinal Chemistry 7th ed.


text:
Chlorinating specifically one carbon on a symmetrical alcohol
The final target molecule is 1-chloro-2-amino propane.
The amination step would be simple via reductive amination of 1-chloro-iso-propanone.
Obtaining 1-chloro-iso-propanone would be from nucleophillic addition-elimination of 1-chloro-iso-propanol by H2O under acidic conditions (or oxidation using potassium permanganate but no need for such a reagent when a simpler route available).
However, assuming you started with only iso-propanol and you wanted to chlorinate only one of the two terminal carbons, how would you do this? I understand $\ce{SOCl2}$ would of course be the chlorinating reagent, but with no blocking group it would simply attack both carbons. Attempting to add a blocking group would block both carbons also as they are symmetrical and identical.

SOCl2 would not be the chlorinating agent of choice, all that will do is give you 2-chloropropane from iso-propanol.

Ah yes, I see that now. In that case, how could this be done? Free radical substitution? Wouldn't that also cause attack at both terminals?

If you start from iso-propanol, you can first perform dehydration using concentrated $\ce{H_2SO_4}$ and heat. You will get prop-1-ene.
Then you can use 1 equivalent of $\ce{HOCl}$ (hypochlorous acid) to facilitate chlorohydrin formation. Thus, you will have 1-chloropropan-2-ol. Then, you can simply partially oxidise the secondary alcohol group to a ketone by using pyridinium chlorochromate (PCC) or pyridinium dichromate (PDC), or by performing Swern oxidation, etc. That will be a better way to get 1-chloropropan-2-one (i.e. chloroacetone). After that you can perform reductive amination.

It might be overkill for your purposes, but there is a nice reaction that coverts secondary alcohols to $\alpha$-chloroketones in a one-pot synthesis, using trichloroisocyanuric acid (for instance, they got 75% yield of 2-chlorocyclohexanone from cyclohexanol).  The paper is 'Direct conversion of alcohols to $\alpha$-chloro aldehydes and $\alpha$-chloro ketones' (Jing, Daniliuc, Studer 2014, Org. Lett. 16 (18) pp. 4932-4935).

Direct conversion of primary and secondary alcohols into the
  corresponding α-chloro aldehydes and α-chloro ketones using
  trichloroisocyanuric acid, serving both as stoichiometric oxidant and
  α-halogenating reagent, is reported. For primary alcohols, TEMPO has
  to be added as an oxidation catalyst, and for the transformation of
  secondary alcohols (TEMPO-free protocol), MeOH as an additive is
  essential to promote chlorination of the intermediary ketones.

Trichloroisocyanuric acid has the advantages that it's inexpensive, widely available and easy to handle.

Alternatively you can oxidise iso-propanol by any of the methods mentioned above to acetone, directly chlorinate with $\ce{Cl2}$ to chloroacetone then do the reductive amination to 1-chloro-2-aminopropane. It should be noted that this material can form the aziridine fairly easily so should be kept cool. Indeed I would favour doing the reductive amination by forming the benzyl imine then hydrogenating in the presence of $\ce{HCl}$ so that the hydrochloride is formed directly.

Alternatively, you can use acetone in place of 2-propanol directly.  The selective chlorination on a single carbon can be achieve in one step by using a dichloroemethane solution of sulfuryl chloride ($\ce{SO2Cl2}$) in the presence of methanol (see reference below). Then, using a one of the methods suggested by others here, you may convert resulting 1-chloroprop-2-one (e.g., the reductive amination) to 2-amino-1-chloropropane.
For your convenience, I have put the abstract below:

Cyclohexanone reacts with sulfuryl chloride in sulfur dioxide, ether, tetrahydrofuran, dioxane, and tetraglyme to give mixtures of mono- and dichlorocyclohexanones. However, when the reaction was carried out in sulfur dioxide solution containing a slight excess of methanol, only monochlorocyclohexanone was obtained in high yield. Almost as good selectivities are obtained in methylene chloride solution in the presence of methanol. A similar procedure with acetone affords monochloroacetone cleanly. While phenols do not react with sulfuryl chloride in methylene chloride even under reflux, the chlorination can be induced by the addition of various organic “bases”, resulting in selective monochlorination.

Reference:
Sulfuryl chloride as a reagent for selective chlorination of symmetrical ketones and phenols: D. Masilamani, and M. M. Rogic, J. Org. Chem., 1981, 46(22), 4486–4489 (https://pubs.acs.org/doi/abs/10.1021/jo00335a033).


text:
Can 1,3,5-trihydroxybenzene perform haloform reaction?
As far as I know it's not possible for 1,3,5-trihydroxybenzene to give the haloform reaction.
The prerequisites(that I know) for the haloform reaction are:
Only those organic compounds which can be converted to a methyl ketone or acetaldehyde by means of oxidation (by $\ce{NaOX}$ ($\ce{NaOH + X2}$) type agents which are present in the reaction mixture) give this reaction. And obviously a methyl ketone or acetaldehyde would give this reaction too.
So the problem is I don't see how  1,3,5-trihydroxybenzene meet those prerequisites. Is there any way out because in a recent practice test I came to know that 1,3,5-trihydroxybenzene does give the haloform reaction.

Phloroglucinol(1,3,5 trihydroxybenzene) will not give haloform reaction. See this: https://chemistry.stackexchange.com/questions/50538/reactions-of-phloroglucinol

Did they give you the reason or even a hint as to why it gives the haloform reaction?

No hints were mentioned in the answer key.

Then that's probably the wrong answer. Although you may surely wish to keep this question open for a while, just in case someone has more detailed views.

There are two things here: 1) undergo actual haloform reaction 2) react,  when tested  for methyl ketone group, with iodine and sodium hydroxide mixture. First is impossible - no way you'll get haloform, second very probable.

@Mithoron alright...I couldn't get the second one /:

Actually it does give iodoform test. Its tautomer which is cyclohex-1,3,5-trione exists in quite a good amount in the solution which is why it gives the rest.

@FreakyChemist sounds great...I'd be grateful to you if you can show the mechanism because I still have a doubt as the keto form would not possess an independent methyl group as in acetaldehyde which is required for the reaction...

Could it be that instead of the usual -CX3 which is usually the leaving group in haloform reaction due to resonance stability and high -I of halogen, instead this time CX2 could form and with 2 Keto groups around it,they can provide high -I effect, resulting in favour of bond breaking and then the corresponding carbanion could be stabilised by the -I and -M effect of 2 Halogen and one double bonded Oxygen. Am I right @FreakyLearner. Or can anyone else........ kindly correct me if I'm wrong.

It can undergo tautomerism which causes it to have active methylene grp which undergo haloform test


text:
Why does Benzoin give positive Tollen's Test?
Certain alpha hydroxy ketones which can tautomerise to form an aldehyde will give a positive Tollen's Test. I was pretty clear with this idea until I came across this question and the answer given is A,B,C. I can't figure out why Benzoin gives a positive Tollen's Test as it can't give an aldehyde by tautomerisation.
What exactly does happen to Benzoin over here and why does it give a positive test? Is my idea about tautomerisation of alpha hydroxy ketones to aldehydes wrong?

As tollen's reagent has base also, we know that the most acidic hydrogen is the alpha hydrogen here. After tautomerising you will get the same product but you may notice that the hydroxy group has now become a ketone, hence being oxidised, it gives tollen's test

For benzoin, a similar mechanism of tautomerisation (of alpha hydroxy ketones tautomerising to aldehydes) happens , except for one step, where instead of the hydride ion (H-) migrating to the sp2 hydridized carbon, it is the Benzene carbanion (C6H5- or Ph-) that migrates to the sp2 hybridized carbon, since it is more stable than the hydride ion (an example of the stability of benzene carbanion over the hydride ion can be found in the pinacol-pinacalone rearrangement). Other than that, the rest of the mechanism is exactly the same.
Hope this cleared your doubt.

Alpha-hydroxy ketones do not tautomerize via hydride shift; they tautomerize via proton shift as indicated in this answer: https://chemistry.stackexchange.com/questions/73313/why-do-α-hydroxy-ketones-give-tollens-test


text:
Electronic spectra of transition metal complexes
I have been studying about the UV-Vis spectroscopy. Sometimes we are given a graph which shows the absorbance and the wavelength,the appropriate Tanabe-Sugano diagram and we are asked to assign the transitions to the bands.
I know that the bands can be d-d transitions, Ligand To Metal Charge Transfer or Metal To Ligand Charge Transfer or π→π* etc.Out of the things that I have studied, d-d transitions have low intensity whereas the LMCT phenomenon is extremely intense.
The part which gets confusing for me is I have seen some exercises for the complex [Co(H20)6]2+ for example, and there was a question that said why there is a small intensity band at ~640nm and the answer was because there is a two electron transition. Also, in another example of the complex [Ni(H2O)6]2+ where there was a question about why there is a shoulder, the answer was that it is spin forbidden and it has to do with borrowing energy from a close band.
Is there a way in order to understand if a small band is because of d-d transition or borrowing energy from another band or of a two electron transition etc?

'is there a way to understand', yes but its not so simple: you will need to understand some symmetry and quantum stuff. Have a look at a good text such as 'Molecular Symmetry and Group Theory' by R. Carter (publ Wiley), there is a whole chapter of about 60 pages on your topics. There are also sections of chapters on bonding in metal complexes etc.


text:
Quantum mechanical model of atom
While going through the quantum mechanical model of atom section in my chemistry textbook, the text states:

The existence of quantized electronic energy levels is a direct result of the wave-like properties of electrons and {the energy levels} are allowed solutions of the Schrodinger wave equation.

What does 'direct result of wave-like properties of electrons' mean?

Welcome to chemistry.stackexchange! :) If you are confused by the terminology, then try to replace "direct result of" with "consequence of". If we assume wave-like properties for electrons, then the mathematical derivations naturally lead to quantized energy levels for these electrons.

I would really appreciate if you could give an insight on how wave like properties of electrons ultimately lead to quantized electronic energy levels

The idea is based on a very simple/basic  model of an atom (such as Bohr model) which is that the only energy levels possible are those in which an electron (as a wave ) has an integer number of wavelengths in its 'orbital', i.e. the same wave exactly repeats itself in its orbital.

While waiting for an answer, perhaps try to read about "standing waves". Maybe this question can help you?

Just skim through these pages, might help with idea || https://en.wikipedia.org/wiki/String_vibration || https://en.wikipedia.org/wiki/Vibrations_of_a_circular_membrane

That said, the quote is wrong on many levels. QM states that only some states are observable and points out how to find which ones. It so happens that for atomic levels the resulting math problem might be modified to resemble some oscillation problems, and this is the only reason for similarities.

To choose a more basic example, let us consider the particle in a box with
\begin{equation}
V(x)={\begin{cases}0 & 0\lt x \lt L\\\infty & \text{otherwise}\end{cases}}
\end{equation}
To solve this we use the Schrödinger Equation
\begin{equation}
-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)=E\psi(x)
\end{equation}
subject to the boundary conditions
\begin{equation}
\psi(0)=\psi(L)=0.
\end{equation}
The origin of these boundary conditions is, that the wave function needs to be a) continuous and and b) zero outside of the potential well, since probability goes to zero for an infinite potential.
As an ansatz for the wave function we take the solution of the free particle, i.e. planes waves:
\begin{equation}
\psi(x) = \exp\left(-ikx\right)
\end{equation}
with
\begin{equation}
k^2=\frac{2mE}{\hbar^2}.
\end{equation}
What we now have is a periodic wave function, with period $L$ (or an integer multiple of that) due to the boundary condition, hence
\begin{equation}
kL = 2\pi n.
\end{equation}
At this point we have introduce the integer quantum number $n>0$. With this we can now write down wave function and the corresponding energy. Both depend on $n$ and are therefore quantized.
So the origin of the quantized nature here are the boundary conditions. The quantum number naturally show up in the solution of the Schrödinger Equation. Obviously this can happen for other system, such as the Hydrogen atom as well.
As a final side note: The Schrödinger equation does not always lead to quantized solutions, it depends on the system. There are for example continuous states when considering the dissociation of molecules.


text:
Why does a molecular sieve work for gas separation?
I was looking at membrane separation of air to nitrogen / oxygen and saw some details on the molecular sieve which on Wikipedia states you can get much higher purity from an adsorber / sieve, but when I look at these images, it doesn't make any sense.
Molecular Sieve material / Zeolite seems to be supplied in granules that are fairly large, why doesn't the compressed gas just go around all the material instead of going through it?
Without melting it / packing it into a solid column, it doesn't seem like it would be much more efficient than just glass beads since there are so many ways to get around each piece. Are you really supposed to just dump it in and have it work?
http://www.instructables.com/id/Nitrogen-gas-generator/

There are two key principles involved in gas separation using molecular sieves: pore size and strength of surface absorption of the different gases involved.
The key overall idea is that sieves are selective in some way about which gases in a mixture get stuck in the sieve and this allows separation. This is a little like chromatography where each component of a mixture partitions differently between the liquid phase flowing through the column and the solid phase. It isn't that it is impossible for the liquid to carry the dissolved compounds straight through the solid phase, it is just that the average speed of flow of a particular component is affected by the different attractions to the solid phase. The net result, if you get the setup right with appropriate flow rates, is that some compounds pass more slowly than others and the end result is some degree of separation.
Molecular sieves are like this too. The two effects combine to make one gas preferentially stick to the sieve and, with the right setup, this allows a degree of separation. One factor is simply the size of the pores in the sieve: some gas molecules pass through the pores easier than others just because of their size. The other effect (the strength of the absorption on the surfaces inside the pores may also matter: some gas molecules are more attracted to the surface than others.
In the case of this particular setup, this allows the separation of oxygen and nitrogen in air if the conditions are right. Strictly it isn't a continuous flow of gas that is being separated (that is why it is called a "pressure swing" apparatus). One vessel containing the sieve is filled with air under pressure. The oxygen is preferentially absorbed onto the pore, leaving the remaining gas very concentrated in nitrogen. When the pressure is lowered the first thing to come out of the vessel is very enriched nitrogen (which is mostly present as gas). As the pressure is lowered further, the oxygen absorbed in the sieve is also released and discarded. If you get the conditions right, this allows fairly pure nitrogen to be produced.
If you need really pure results, you can do this in multiple stages, each stage giving a better separation than the previous one.

What you need to consider for a moment is the fact that zeolites and many other solids have large surface areas. Gases can bind to the surface of a zeolite, commonly the Langmuir isotherm is used to explain or consider this binding. The amount of gas on a solid (moles of gas per gram of solid) can be expressed using the following equation.
$q = kP/(1+kP)$
Where $q$ is the moles of gas per gram of solid, $k$ is a constant and $P$ is the pressure of the gas.
This equation relates to a model which has three assumptions.

The solid has an array of identical sites
A site can either have one gas molecule on it or none
There is no communication between the sites.

If we have two gases present then for gas a we can write
$$
\begin{align}
q(a) &= \frac{kP(a)}{(1 + (kP(a)) + (k'P(b)))}\\
q(b) &= \frac{k'P(b)}{(1 + (kP(a)) + (k'P(b)))}
\end{align}
$$
If the values of k and k' are different then it is possible to use a bed of the solid to make a separation of the two gases. The plant you show uses a cyclic process in which one gas is absorbed onto a column and occasionally desorbed from the column. There are other applications of similar ideas in chemistry.
If we consider for a moment the classic Tc cow, this is Mo-99 which is adsorbed onto alumina. The binding of $\ce{Mo}$ as molybdate onto alumina is much stronger than that of pertectinate. I am sure that if one was to do experiments with very small amounts of Mo or Tc as the anions then it would be possible to measure the values of K for these anionic metal complexes onto alumina. Then we could use the same Langmuir equation to model and understand a Tc cow (Tc-99m generator).
Another use of the same maths is in soil chemistry, for the adsorption of a small amount of a metal onto the surface of soil minerals it is common to use Kd values to predict what happens. Consider for a moment a soil which contains a very small trace of uranium or zinc.
Under these conditions the concentration in the soil water is directly proportional to the concentration of the metal in the soil. This is at the low concentration set of conditions where
$[\ce{M}]_\text{aq} = k[\ce{M}]_\text{soil}$
The way to get this is to write
$[\ce{M}]_\text{soil} = k'\frac{[\ce{M}]_\text{aq}}{1 + (k'[\ce{M}]_\text{aq})}$
Where $k'[\ce{M}]_\text{aq}$ is much less than one then we can write
$[\ce{M}]_\text{soil} = k'[\ce{M}]_\text{aq}$
So $k = 1 / k'$
Under some conditions in soils (and zeolites) when the concentration of either the gas or the metal is high then we can deviate from the simple conditions. In a soil we can reach a situation where all the binding sites for a metal are occupied, under these conditions the Langmuir equation will be a better way of modelling the metal concentration in the soil rather than assuming that the metal content of the soil water is directly proportional to that metal concentration in the solid particles in the soil.
As an alternative in soil we can reach solubility limits, for example if I add lime to my garden soil then we can reach the solubility limit for calcium carbonate. Again the soil will deviate from the simple model of
$[\ce{M}]_\text{aq} = k[\ce{M}]_\text{soil}$
In a zeolite we can at high coverage of gas we can get adsorption of gas molecules onto other gas molecules (see the BET isotherm for more detail) we can also get other effects such as liquid gases condensing into pores of silicas and zeolites.
I think that to understand the zeolite system then you would need to make a lot of measurements of isotherms for the adsorption of both nitrogen and oxygen onto the zeolite. Doing this at different temperatures would also help you understand the system better as well.
Another thing you need to consider is how quickly does diffusion in and out of a zeolite particle occur, this will make the system more complex. If you want to understand this kinetics issue then I suggest that you read about the maths of GC columns. The resolution of a GC column goes down both at low and high flow rates. A good mathematical model exists for HPLC and GC columns, I recall it was published years ago by a Dutch chemist.
I will warn you that many systems do deviate from the Langmuir isotherm model.

I have seen finer zeolites but this one does not seem as coarse as you sounded it. Consider it again, likewise in fractional distillation, there will be some molecules with never being an equilibrium with another phase. However, much more of them will go through zeolites and with the passage of each unit distance and time, you will have more of the compound you want to get and less of the redundant one. If your column is enough long and zeolites are sufficiently fine and ordered, it will be a good purifier.
Try to visualize it in more detail, while some molecules try to pass through the cavities between zeolites, other molecules will be forced to collide and travel through directly zeolites.

I think what you are saying is that the gas that is filtered through the material will "fill the space" between zeolite particles and thereby will provide a push back against non filtered gases occupying the same space, provided the filtration speed doesn't exceed it's ability to occupy that space? Does that mean that initially you would have low purity separation, but getting better with time?

Sort of. There will be an optimum speed for elution, yet in every speed, there will always be retardation for the redundant species, as whenever the gas comes in contact with zeolite particles, it will stop when the other moves relatively freely. A gas fluid particle may not even go through this equilibrium through entire column but it is not likely. I wanted to stress that.


text:
Can glucose burn in chlorine?
Chlorine is an oxidizing agent, so could it replace oxygen? If so, what would be the products, and how much energy would such reactions release? Do any other carbohydrates burn in chlorine? I cannot find any information on this.
Clarification: How does glucose interact with chlorine chemically?

related https://chemistry.stackexchange.com/questions/64541/is-oxygen-necessary-for-burning https://chemistry.stackexchange.com/questions/19883/fire-resistance-of-pvc

It shouldn't work for same reason as why PVC is difficult to put on fire, IMO.

"Burn" is typically used only to described reactions with oxygen.

@Mithoron I don't quite see the analogy. ;-) The reaction yields HCl, and will definitely work. Ok, sugar even doesn't burn very well in oxygen.

@MaxW Not necessarily, you can make flame with various other oxidants, fluorine would work even in cases when oxygen isn't strong enough.

@Mithoron I wanted to say in air. In pure oxygen, sugar burns as well as practically everything else except rocks.

I have not heard of glucose being burned in chlorine, but I do know that glucose reacts with chlorine in aqueous-buffer medium. See reference 1.
The full paper is behind a paywall.
It says that oxidation of glucose by bromine water was studied previously and the reaction of glucose and aqueous chlorine also happens in a similar fashion although the reaction is slow and products yield is less. Nonetheless, major product of the reaction is D-gluconic acid in both acidic and alkaline medium but unbuffered chlorine may further react with D-gluconic acid to various products like 5-keto-D-gluconic acid and D-glucaric acid.
Reference

J. Am. Chem. Soc., 1955, 77 (7), 1875–1880. DOI: 10.1021/ja01612a050.


text:
Limit to volume change in a discretized mathematical model?
I have set up a mathematical model describing the diffusion of ozone out of a gas bubble. The bubble is surrounded by a thin gas film. So actually, the model describes the diffusion of ozone through this gas film.
The mathematical model is created by discretizing the volume from $V_0$ to $V_1$. Where $V_0$ represents the outer volume of the bubble, and $V_1$ represents the outer volume of the gas film (surrounding the bubble). The discretized scheme consists of $N$ equally spaced volume elements from $V_0$ to $V_1$ (finite difference method):
\begin{equation}
    \Delta V = \frac{V_1 - V_0}{N}
\end{equation}
The volume of the bubble changes as a function of the amount of ozone leaving the bubble, which in turn changes as a function of time.
The volume of the bubble and the amount of ozone inside the bubble is linked by the ideal gas law:
\begin{equation}
    V_0 = \frac{n_\text{total}(t) \cdot R \cdot T}{P}
\end{equation}
$n_\text{total}$: the total amount of gas inside the bubble, $T$: the temperature, $R$: the gas constant, and $P$: the pressure.
The bubble does not only contain ozone. It also contains inert gases, so that:
\begin{equation}
    n_\text{total} = n_\text{ozone}(t) + n_\text{inert}
\end{equation}
$n_\text{inert}$ will remain constant and only $n_\text{ozone}(t)$ will change over time.
There should be a limit for how much the volume can change before numerical errors will start to occur. Beyond this limit, the discretization scheme should break down and cause errors. How do I express this limit?
Is the limit given by:
\begin{equation}
    \text{Ratio} = \frac{V_{0,\text{ initial}}}{V_{1,\text{ initial}}}
\end{equation}
So that the volume change must not exceed the ratio of the two initial volumes of the bubble and the gas film?

Link to cross-posted question on CompSci.SE.

You would normally use Fick's Laws to solve problems of this sort, in addition, in a bubble you will also have to account for the change in pressure and the bubble volume changes. Numerical stability may be the least of your problems. :)

@Siglis https://scicomp.stackexchange.com/q/29426/23791 crossposting is generally discouraged. I was actually going to say the question might be a better fit on Comp Sci. I personally don't mind the cross post, but I would recommend linking to the other post in this one.

In addition to what @porphyrin comments, when you use a finite difference method the number of finite differences is crucial, as numerical errors will depend on the size of your cells ($(V_1-V_0)/N$) more than on the size of the entire system you are simulating; too many cells and you will have a very costly simulation prone to numerical noise, too few and you will be smothering the diffusion phenomenon you want to describe.
What are the exact equations that you are using for transport?


text:
How long does a wine sauce need to simmer for to reduce the alcohol to less than one percent
Perhaps you can shed some light on this topic for me.  I've recently researched the topic of boiling off alcohol while cooking--specifically how long one needs to simmer alcohol to boil it off.
I'd come across a couple of sites which showed tables of simmering times and percentage of alcohol remaining.  Since I am a math-head, I noticed that something didn't quite jive in the tables I was viewing.
I thought to ask a chemistry expert to see if there is something unusual going on chemically to explain these inconsistent values in the tables.
I don't want to make this a novel, so I will use only a few entries in the tables to illustrate my point.
Simmering time                    Alcohol remaining
15 minutes                                  40%
30 minutes                                  35%
2 hours                                     10%
So what the table suggests is that if you simmer a solution containing alcohol, after the first 15 minutes, you will have 40% of the original amount of alcohol in the solution.  If you simmer for an additional 15 minutes (total of 30 minutes) you will have 35% of the original amount of alcohol left in the solution.
So, (according to the table) if I have a solution which originally contains 100 ml of alcohol in the solution, and I simmer it for 15 minutes, I will have 40 ml of alcohol left in the solution.  Then if I simmer it another 15 minutes (30 minutes total), I will have 35 ml of alcohol remaining in the solution.
However, If I start with a solution containing 40 ml of alcohol in the solution (let's say it's the solution from above after the first 15 minutes), and I simmer it for 15 minutes, I will be left with 16 ml of alcohol in the solution (again, according to the table).  You can see my dilemma; 16 ml does NOT equal 35 ml.  The math doesn't work.
I am wondering about the nature in which they'd come up with these values or the methodology they used in their experiments.
Can you help me to understand what is going on here?

(Who deletes my comments here?) @Bob: After half an hour of boiling, your sauce is not the same as it was when you started cooking. It wouldn't even work for water, because the numbers depend on the original absolute concentration. Also: Lid or no lid? Heating power? Stirring? Amount of fat in your sauce? Humidity of surrounding air? Those numbers you found in some book are just arbitrary, anecdotal, unscientific nonsense, unless they were accompanied by some lenghty, good explanation, but then you wouldn't be asking here. Being good in math, you suspected sth is wrong here, and you're right

Thank you for your response Karl.  I appreciate your time and efforts.

Let's say, for the sake of argument, that there is no lid, 50% relative humidity, 25% fat content, 210 square cm surface area, and just enough heat to simmer, starting concentration of 10% alcohol.  How would I determine the rate at which the alcohol is evaporating at various times throughout the process?

By measuring it. :-|

The original source of the data is Alcohol retention in food preparation Journal of the American Dietetic Association vol. 92, April 1992, pages 486-488.
After 2.5 hours of simmering at 85 degrees C, pot roast Milano still retained 4-6% of the original alcohol content.

... and orange chicken burgundy contained 0.07% if cooked in a 12" pan, and 0.4% in a 10" pan. A nice example of cargo cult science, at least they didn't suppress that particulary obvious piece of evidence. :-D


text:
How do I find the theoretical pH of a buffer solution after HCl and NaOH were added, separately?
This question is about the theoretical pH of a buffer solution.

Calculate the theoretical pH values expected for a $\pu{200mL}$ buffer solution containing a 1:1 ratio of acetic acid and sodium acetate (their concentrations are $\pu{0.05M}$), following the addition of;

$\pu{10mL}$ of $\pu{0.2M}$ $\ce{HCl}$ and
$\pu{10mL}$ of $\pu{0.2M}$ $\ce{NaOH}$.

$K_\mathrm{a}$ of acetic acid is given.

So I just found the $\mathrm{p}K_\mathrm{a}$ for the Henderson equation.
Attempt for (1):
$\ce{[HCl]}$ = $\pu{0.2M}$ and $V=\pu{10mL}$/$\pu{0.01L}$ thus $n = 0.002 $
$0.002 \times \pu{0.2L} = \pu{0.01M}$
ICE table:
$$
\begin{array}{cccc}
&\ce{H3O+ + &CH3COONa <=> &CH3COOH + &H2O}\\
\text{I}&   x& 0.05&0.05\\
\text{C}& -0.01&-0.01&+0.01\\
\text{E}& x - 0.0&0.04&0.06\\
\end{array}
$$
$\mathrm{pH} = 4.74 + \log (0.06/0.04) = 4.926$
Although in our experiments the actual pH was 4.17 after $\pu{10mL}$ of $\ce{HCl}$ was added and the pH after 10 mL of NaOH was 4.98
Attempt for b:
$\ce{[NaOH]} = \pu{0.2M}$ and $V = \pu{10mL}/\pu{0.01L}$ thus $n= 0.002$
$0.002 \times \pu{0.2L} = \pu{0.01M}$
Is this how I would properly set it out?
$$
\begin{array}{cccc}
&\ce{CH3COOH + &OH- &<=> &CH3COONa + H2O}\\
\text{I}& 0.05&x&&0.05\\
\text{C}& -0.01&x-0.01&&+0.01\\
\text{E}& -&0.04&&0.06\\
\end{array}
$$
$\mathrm{pH} = 4.74 + \log(0.04/0.06)= 3.74$
So I feel like I have done these the wrong way because it makes more sense but I don't know why or how.

The Wikipedia page on buffer solutions has an example ICE calculation similar to yours towards the bottom of the page.

First I want to point out a couple things.

When you add more liquid to a solution, the concentration of the solute is going to decrease. Therefore, after the addition of 10 mL of HCl, the concentration of sodium acetate and acetic acid will no longer be 0.5 M since you diluted the solution a little bit. When calculating the concentration after dilutions, keep in mind the amount of moles is still the same, so just use the formula $\ce{M1V1 = M2V2}$.
Also, sodium acetate is a soluble salt (see solubility rules), so in the net ionic equation, don't write out the sodium as it is a spectator ion. The same applies for HCl. It is a strong acid, so it completely dissociates in water.
For this problem, using the Henderson-Hasselbalch equation is not really the right way to do it. When you are using this formula, keep in mind that it is only good for calculating pH when you already know the equilibrium concentrations of the acid/conjugate base. Here's a derivation of the formula if you're interested that will probably help clear things up a bit.

I think you mostly understand this point last point since you did made an ICE table before you applied the equation. However, in the context of this problem, making an ICE table will actually lead you to the pH value without requiring the HH equation. You'll see why.
Alright; now we can actually get to how to approach the problem. First, you're going to want to look out for a reaction. When solving these problems, being able to identify a reaction will make it 10 times easier.
How do you identify a reaction? Well, look at the reagants in this problem: $\ce{HCl}$, which is essentially just hydronium and chloride, sodium acetate, which as mentioned earlier is essentially just sodium and acetate, and acetic acid. Acids react with bases, so using that, it isn't too hard to determine what exactly happens in the solution.
You actually got the net ionic equation for the reaction correct (without the Na, however), but you didn't solve the ICE table correctly.
Actually, we're going to start with simple stoichiometry as a reaction is occurring. To simplify the process, we're going to assume the reaction reacts to completion, and then reverses until it reaches equilibrium.
For this, just use limiting reactants and such to calculate the final concentrations of $\ce{CH3COOH}$ and $\ce{CH3COO-}$.
As I said earlier, we're going to first assume that the reaction reacts to completion, and then goes back on itself and reaches equilibrium. We're at the second part, the equilibrium part. This is where an ICE table comes in.
First, write out the equilibrium reaction. This is pretty simple; it is just the acid dissociation reaction. It's going to look something like this: $$\ce{HA<->H+ +A-}$$
For your ICE table, you didn't really set it up properly. Remember these three things:

You usually won't be writing any unknowns for your initial concentrations. Everything should be known.
You usually won't be writing any numbers for your change. It's an equilibrium. You don't know the exact amount it is going to react to reach equilibrium. That's why we write out the ICE table, so we can calculate how much needs to react for the system to reach equilibrium.
Don't use the Henderson-Hasselbalch equation. Use the $K_a$ and its definition to calculate the concentration of $\ce{H+}$, which leads to your pH.

I know I pretty much just wrote a wall of text, but hopefully that helps nonetheless.

You seem to be on the right track. You have the correct number of moles of HCl added (0.002 moles H+) and the correct number of moles of acetate (0.01 moles Ac-). This reaction will result in an increase in the moles of HAc and a concomitant decrease in the moles of Ac- since H+ + Ac- => HAc. After addition, moles of Ac- = 0.01 - 0.002 = 0.008
moles HAc = 0.01 + 0.002 = 0.012
In a final volume of 210 ml (0.210 L), you can calculate final concentrations:
[Ac-] = 0.008/0.210 = 0.038 M
[HAc] = 0.012/0.210 = 0.057 M
You can then use the HH equation as you did, substituting these concentrations. Do similar calculation for addition of NaOH where HAc will decrease and Ac- will increase.
NOTE: The experimentally determined pH will not always be the same as the calculated pH. There are several reasons why this might be the case. Use your imagination.


text:
Hydroboration oxidation reaction on trans,trans,trans-1,5,9-cyclododecatriene

The given solution:

Could some one please explain the mechanism of the reaction? How it proceeds and why it proceeds that particular way? Also, what is the use of $\ce{CO}$?

The CO is the source of the central carbon

Well, I hope you're asking about Q. Because, while P does look kinda weird, it's actually very easy to understand if you know the mechanism.

@Alphanerd not a complete answer, but this should put you on the right track: http://www-oc.chemie.uni-regensburg.de/OCP/ch/chb/oc5/B_Organometallics.pdf

There are two questions here. The first is, basically, what is the mechanism for hydroboration oxidation of an alkene. That is covered in various textbooks and online resources. The second is carbonylation of organoboranes, which is significantly more complicated because there's triple migration (and it turns out the migration is condition dependent) and is definitely not covered in most introductory organic textbooks. The original reports on hydroboration by H.C. Brown cover this topic quite extensively, so if you're really interested, you should look those up.


text:
HCl/H2O with alkene: Which one reacts?

I think that the initial step forms a tertiary amine, which then in the presence of the ketone and trace acid undergoes nucleophilic addition. But then, for the $\ce{HCl/H2O}$ in the last step, which one reacts with the double bond in 1G?

Do you intend to ask whether HCl adds to the pi bond or water adds to the pi bond?

@GaurangTandon Yes, my point exactly. I'm not sure if it's a trick question or my misinterpretation.

@Carlson Miller: See my amendment to Das's solution.

I think the last step  is for the Hydrolysis.  After reaction with $\ce{LiAlH_4}$, the amide is reduced to amine and thus when it is reacted with a ketone it forms an imine($1G$) like this below,
which quickly tautomerises to form the most stable Enamine as this
After  that when it is reacted with $\ce{CH_3Br}$, the nitrogen's lone pair conjugates and the nucleophilic attack occurs from the adjacent carbon(like Michael addition) with attached methyl group. Thus you have the intermidiate substituted proiduct as,
But this imine  is hydrolysed in the presence of $\ce{H^+/H_2O}$ to give you back the original ketone but with an extra methyl group attached to it. So I think your final product $1H$ should be this,
and also you get the amine after hydrolysis which is actually the  same as $1F$.

(+1) Good answer. Though your lines looks a bit crooked. I would recommend using Marvinsketch it's online and it's free.

The product in @Soumik Das's otherwise fine solution should be 2,6-dimethylcyclohexanone (4) NOT the 2,2-isomer 7. The enamines (pyrrolidine, piperidine, morpholine) of 2-methylcyclohexanone 6 are more stable as structure 1 rather than structure 2. The methyl group in enamine 1 is undoubtedly axial to avoid the A1,3 interaction present in enamine 2. The kinetic alkylation product, imine 3, probably has trans methyl groups but acid hydrolysis will likely equilibrate the trans and cis ketones to the more stable cis isomer 4. This very reaction with a different secondary amine is discussed in Gilbert Stork's 1963 paper.

G. Stork, et al., J. Am. Chem. Soc., 1963, 85, 207.
DOI: 10.1021/ja00885a021


text:
Dealkalization of water
I want to remove Alkalinity from tap water by using an acid. The starting point is about 370pm alkalinity, and I am wondering to reduce this to about 50ppm, could I use a weak acid like citric acid? Or do I need to use a strong acid, such as HCl?

Why do you want to do this?

^ If what you want is to remove the calcium and other alkalines from your water, your best option may be reverse osmosis devices. Also common strong acids will not remove the alkalines, only make them more soluble.

It's for coffee and tea brewing, as too much alkalinity makes the brew taste flat. I can use Brita / BWT type filters to reduce calcium and alkalinity, but only to a certain extent. RO is a much larger investment, I would rather add a few drops of HCl to reduce the alkalinity to acceptable levels.

By the term "dealkalization" do you mean the removal of alkalinity ions or the adjustment of the pH value? https://en.wikipedia.org/wiki/Dealkalization_of_water

Roughly speaking, any acid is equally effective if its pKa lies below the pH 7 you want for a neutral solution.  It does not have to be HCl at all, in fact citric acid works for all three of its commonly observed deprotonations (pKa = 6.4 for the third deprotonation and of course lower for the first two; reference).  So you may use citric acid assuming three equivalents per mole and you should get where you want to go.
As a matter of taste I prefer citrate in my tea versus chloride, anyway.

Thanks! I will give this a try and let you know the result. I don't have a pH kit but only a titration kit which tests alkalinity. However this should give me a reduced reading after adding the required amount citric acid? i.e. 1 mole citric acid to match 3 moles of hydrogencarbonate molecules?

That's what I would do, good luck!


text:
Predict melting point of water CaCl2 solution
I'm doing a science project for my school. I experimentally measured the melting point of water-salt solutions, and I want to compare my results to theoretical melting points, but I encountered a problem.
I used this equation to calculate the melting point depression: $\Delta T = b \cdot i \cdot K_\mathrm{f}$. I also compared my data to a phase diagram. I expected my measured melting point would be on the liquid line. But these two methods give me a different prediction. A phase diagram tells me that $\ce{CaCl2}$ would be much more effective at lowering the freezing point of water while the above equation tells me otherwise.
Here is a comparison: Purple indicates the melting point calculated with the equation. The black line is the liquid line on the phase diagram.

Which method is correct and why is there such a big difference?
(I didn't encounter this problem with $\ce{NaCl}$ solutions, the two methods gave more or less the same predictions)
EDIT: my math was wrong on the graph i posted, here is the corrected graph:

What is extended UNIQUAC ?

@Karl It is a model from which the phase diagram is plotted. See this: https://chemistry.stackexchange.com/questions/25123/how-to-raise-the-melting-point-of-water/25127#25127

https://chemistry.stackexchange.com/questions/5023/why-calcium-chloride-is-used-to-melt-ice-over-sodium-chloride

I think there are three issues here. The first is that your purple line of calculating the freezing point doesn't appear to be correct. Let's arbitrarily select $b=1 ~ \mathrm{mol/kg}$ to compare. $i=3$ and $K_\mathrm{f}=1.86 ~ \mathrm{^{o}C/mol}$. The molecular weight of $\ce{CaCl2}$ is $111 ~ \mathrm{g/mol}$. Then, our mass concentration is $b·\mathrm{MW}/(b·\mathrm{MW}+1 ~ \mathrm{kg})=0.1 ~ \mathrm{kg}_\mathrm{salt}/\mathrm{kg}_\mathrm{total}$. The freezing point is $b·i·K_\mathrm{f}=1·3·1.86=5.58 ~ \mathrm{^{o}C}$. This is much closer to the red/black line than the purple one.
The second problem is that that relation is only valid at low concentrations. As the concentration of salt goes up, it has an increasing effect on the activity of the water. As we can see from the phase diagram, once you get to a 6:1 mol ratio of $\ce{Ca^{2+}}$ to $\ce{H2O}$, the interaction is extremely strong. I'm just speculating here, but I would guess six water molecules are forming a weak octahedral coordination complex with the $\ce{Ca^{2+}}$, with the electronegative oxygen ends oriented toward the ion. This effect becomes so strong, that the hydrate becomes the relevant element for consideration above that concentration.
Third, you're ignoring the existence of hydrates. If you find the melting points and the cryoscopic constants of the hydrates, you can actually calculate the right side of the plot the same way.

Concerning the first issue, yes, my math was wrong. I have now fixed it and the graphs seem to only diverge at about 15% mass concentration.
I don't fully understand the second and third issue you described. Should I consider the effect of hydrates at mass concentration lower than 30 percent (6:1 mol ratio as you said) or only above that? I'm only interested in the first segment of the graph (until 30%) but even after I corrected my math the graphs are still quite different (see my edit). Is that because of the hydrate issue?

Calculating the difference between the the ideal freezing point depression and the real one turns out to be a very hard problem. The way the UNIQUAC model (referenced in your graph) deals with it is by assigning each pair of ions and solvent molecules a some interaction coefficients. Some, like CaCl2, interact strongly and pull the real curve down. Others, like MgSO4 if I recall, interact weakly and push it up. Unfortunately, its not at all intuitive and your best bet is to look into more complete models.


text:
Boric Acid and Brass Alloy
Why does boric acid prevent zinc from vapourizing out of the brass alloy when it is being melted?  What is the actual chemistry involved?

It depends on the temperature, specifically the 1700 F +/- boiling point of zinc , not the flux. At 1700 F the zinc will vaporize regardless of any flux . I never saw a single component flux, likely the the flux also contains soda ash, silica, etc. and it does protect the molten metal from atmosphere by forming a physical layer. Footnote: many years age I worked with copper alloy fluxes and found that for some alloys, the boron promoted the formation of spinels in the metal ; These are exceptionally hard materials that essentially prevent machining of the bronze/ brass.

In the melting of copper alloys (as well as other alloys), various auxiliary molten metal processing steps are necessary other than melting and alloying to improve meltquality through the control of impurities. Fluxing and metal refining is one among these processing steps. One of fluxing process is neutral cover flux, which is used to reduce metal loss by providing a fluid cover. Fluxes of this type usually used are based on borax, boric acid, or glass, any of which melts at copper alloy melting temperatures to provide a fluid slag cover over the melt surface. For example, borax melts at approximately $\pu{743 ^\circ C}$ ($\pu{1369 ^\circ F}$), compared to the lowest melting yellow brass, C85800, which melts in the range of $\pu{871-899 ^\circ C}$ ($\pu{1600-1650 ^\circ F}$). These glassy fluxes are especially effective when used with zinc-containing alloys (such fluxes are also used by goldsmiths, etc.), preventing zinc flaring and reducing subsequent zinc loss by $3$ to $10\%$.  The glassy fluid cover fluxes also  help to refine alloys by accumulating and absorbing nonmetallic impurities from the melt such as oxides (e.g., $\ce{CuO}$), molding sand, machining lubricants, etc.


text:
Mathematically balancing chemical equation yields weird results
I found this equation while testing my equation balancer:
$$
\ce{a C7H6O3 + b C4H6O3 -> c C9H8O4 + d C2H4O2}
$$
Obviously, $a = b = c = d = 1$ is a solution. However, if we were to bring in some maths:
$
\begin{cases}
7 a + 4b &= 9c + 2d\\
6a + 6b &= 8c + 4d\\
3a + 3b &= 4c + 2d\\
\end{cases}\\
\begin{cases}
a &= \frac{11}{9}\alpha - \frac{2}{9}\beta\\
b &= \frac{1}{9}\alpha + \frac{8}{9}\beta\\
c &= \alpha\\
d &= \beta\\
\end{cases}
$
This leaves us with many ways of balancing the equation:

when $\alpha = 1, \beta = 1$, $a = b = c = d = 1$; $\ce{C7H6O3 + C4H6O3 -> C9H8O4 + C2H4O2}$
when $\alpha = 1, \beta = 2$, $\begin{cases}
a &= 7\\
b &= 17\\
c &= 9\\
d &= 18\\
\end{cases}$; $\ce{7 C7H6O3 + 17 C4H6O3 -> 9 C9H8O4 + 18 C2H4O2}$
when $\alpha = 2, \beta = 1$...

From what I know, this happens when the equation consists of 2 or more independent equations (as is the case here: More than one way of balancing a chemical equation). However, here the two equations are:
$$
\ce{11 C7H6O3 + C4H6O3 -> 9 C9H8O4}
\\
\ce{-2 C7H6O3 + 8 C4H6O3 -> 9 C2H4O2}
$$
What is concerning here is that the second equation has a negative stoichiometric coefficient. My question is, is a negative coefficient acceptable in an equation, and in this case which is the correct way to balance the equation?

Your equation was already balanced, so why would you try to get other solution?

@Mithoron I guess I am being more a mathematician than a chemist here, but clearly there are other solutions, which makes me wonder if they too are valid solutions.

There is nothing special about a negative coefficient. If you prefer not to have it, transfer that term to the other side.

@IvanNeretin Ohh, I didn't think about that! So a negative coefficient in the reactants is equivalent to a positive coefficient in the products, right? Would you like to elaborate in an answer so I can accept it?

You may answer your question yourself.

@Jingjie YANG you are correct, negative coefficients just flip to be positive on the other side.

As @Ivan Neretin and @Tyberius pointed out, the negative coefficient of a reactant can be flipped to be positive by moving the reactant to the products' side.
So, in my case, the equations become
$$\begin{align}
\ce{11C7H6O3 + C4H6O3 &-> 9C9H8O4}\\
\ce{8C4H6O3 &-> 2C7H6O3 + 9C2H4O2}
\end{align}$$
Now, since the original equation $\ce{C7H6O3 + C4H6O3 -> C9H8O4 + C2H4O2}$ consists of 2 independent equations which can be separately balanced and combined in an arbitrary proportion. See  More than one way of balancing a chemical equation for a more detailed explanation.


text:
Why does sugar water diffuse better than water through a gummy bear?
Our science class did a lab about osmosis, where we left a gummy bear in solutions of salt and sugar, as well as tap water. When we came back, we were all surprised to see that our gummy bears in sugar water were nearly twice the size of any of the others.
This couldn’t possibly be some sort of anomaly since it happened with every group.
Our class has been searching online for reports that reproduce this result. We haven’t found one. My question is, why does this happen?

Hint: Solute (in this case sugar from the gummy bear) diffuses from region of high concentration to low concentration.

Please describe the experiment in more detail. Exactly which solutions were you comparing (what was their composition) and in which one did you see most swelling?

The answer is osmosis. In biology, osmosis is the movement of water molecules from a solution with a high concentration of water molecules to a solution with a lower concentration of water molecules, through a cell’s partially permeable membrane.

The OP is surprised because the sugar water, not the tap water, supposedly showed the biggest increase in size. Your standard explanation explains what you would expect, not what was reported by the OP.


text:
What is the detailed mechanism for the steps of this question?

Detailed mechanism especially if the aldol condensation and Diels Alder reaction are occurring.

The hydrogenation with palladium poisoned with lead will form the cis alkene from the acetylene. This will form a 1,3-diene. This is a reaction which occurs on the surface of the palladium which has been modified with lead. What will happen is that hydrogen will chemicially adsorb onto the palladium, this forms hydrogen atoms which will be on the surface. These will react with chemically bonded acetylenes to form a vinyl group which is bonded to the surface. The vinyl group will react with a second hydrogen atom thus forming a cis alkene. If you consider the reaction of RuHCl(CO)(PPh3)3 with acetylenes which have a D atom as the terminal hydrogen then it will make it clear why the product is a cis alkene.
I think that the reaction will be a coordination of the acetylene (alkyne) followed by a migotory insertion.
It is normal to do the Diel-Alder reaction under thermal conditions, but I know that the 1,3-diene can adsorb UV light. We have the problem of knowing what will happen next.
If we convert the 1,3-diene into a diradical using the UV light then it could react with the alkene (ethylene). This could then ring close to form 7-methyl-1,2,3,4,4a,5,6,7-octahydronaphthalene.
The exact example in the book does not appear in Reaxys, so I suspect that it may been made up by the author of the book rather than being a real reaction that someone has done somewhere.
If we assume instead that the reaction is a [2+2] reaction which will form a cyclobutane then I think that the product B will not be able to form any of the final choices.
Treatment of 7-methyl-1,2,3,4,4a,5,6,7-octahydronaphthalene with ozone followed by dimethyl sulfide will form 2-methyl-4-(2-oxocyclohexyl)butanal. This reaction is a bit complex, it will a pericyclic reaction forming a five membered ring which then undergoes further reactions to form a second five membered ring. The dimethyl sulfide is a reducing agent which forms the ketoaldehyde.
If we were to react this with a base and form the more thermodynamically favoured enolate using the ketone, before using the aldehyde to react with the anolate then we would have a spiro compound (1-hydroxy-2-methylspiro[4.5]decan-6-one) which is not one of the options. But if we deprotonate the aldehyde to form an enolate and then react this enolate with the ketone then we would get 7a-hydroxy-1-methyloctahydro-1H-indene-1-carbaldehyde which is option a. This also puts the isotope lable (*) in the right location in the molecule.


text:
Would the End Point and Equivalence Point be Labelled the Same?
I created two graphs of two different titrations : One graph was the titration curve of acetic acid with sodium hydroxide (weak acid-strong base) and the other was that of nitric acid with sodium hydroxide (strong acid-strong base). I've marked the equivalence point of the acetic acid curve to be at a pH of ~8.5, while the nitric acid curve has a pH of 7.
Knowing this, where would I label the endpoint of the titration curves? In the same general location? On most of the photos I've researched, people don't tend to label the end point but it's required for this assignment. For the titration, I didn't use an color indicator like methyl red or anything but rather a pH meter. If anyone can give an idea, that would be greatly appreciated.

The endpoint and equivalence point are different.
According to this Wikipedia article , "The endpoint (related to, but not the same as the equivalence point) refers to the point at which the indicator changes colour in a colourimetric titration."
Essentially, the endpoint is when the titration visually ends in an experimental setting, but the equivalence point is when the titration actually ends, or when the moles of acid are equilalent to the moles of base.
Since you didn't use an indicator, you didn't have an endpoint.
Also, if you did use an indicator, you wouldn't really know what pH the titration ended at because you don't have a pH meter to measure the pH with.
Unless you had both a pH meter AND indicator (which seems kind of useless), you shouldn't have to label the endpoint on any titration curve.

I did use a pH meter. I recorded the amount of NaOH added in 1-2 mL increments to see the change in pH until it reached the equivalence point. Would the end point be the moment in which a huge spike was seen? For instance, in the nitric acid, the pH meter read 3.4 to 10.8 from 13 mL to 14 mL added. Would this be it? Thanks!

There is no endpoint. The endpoint is when the indicator changes color, but you didn't have an indicator.

There's still an end pont. The end point is the end of the titration. You waited until pH>7 to stop. But you didn't stop at exactly pH = 7, the equivalence point.

I wouldn't call that an endpoint. The endpoint is supposed to be near the equivalence point if anything. When plotting a titration curve with a pH meter, you usually overshoot the equivalence point by quite a bit so you can get data across the entire curve. I guess you could say that is the point where you ended, the literal ending point, but that isn't really a useful thing to measure.

@Frank It is definitely an endpoint. The equivalence point is when you the amounts are stoichiometrically equal. The end point is when your indicator, meter, etc. says that you're past the equivalence point, indicating that you have reached the end of the titration. Hence, "end point." See, for example: https://en.wikipedia.org/wiki/Titration#Measuring_the_endpoint_of_a_titration