text:
What is the best approach to balance this equation? Why I cannot do it efficiently?
What is the best approach to balance this equation?
$$\ce{CO2 + H2O <--> C2H2 + O2}$$
I did it at first this way and I was not successful:
\begin{align}
\ce{4CO2 + H2O &<--> C2H2 + 3O2}\\
\ce{2CO2 + 6H2O &<--> 6C2H2 + 5O2}\\
\ce{12CO2 + 6H2O &<--> 6C2H2 + 5O2}\\
\end{align}
Should I start from right side or left? They say start from the most complicated molecule? If it is like that, then what is the most complicated one and why?

@airhuff Please do not add the reaction tag any more. We are in the process of blacklisting it. See meta for more information.

Related: http://chemistry.stackexchange.com/questions/68924/what-is-the-simplest-approach-to-balance-a-complex-reaction-equation/70540#70540 (Personally, I'm biased toward my own answer to this question, but I think all of them address how to balance a reaction in general without guesswork).

This may be a bit dumb method, but sometimes I do use it as it is pretty straightforward and doesn't require any guessing. I setup a few equations and try solving over integers.
 $$\ce{aCO2 + bH2O <--> cC2H2 + dO2}$$
Balancing $\ce{C}$ atoms both sides,
$$a = 2c\\$$
Balancing $\ce{H}$ atoms both sides,
$$2b=2c\\$$
Balancing $\ce{O}$ atoms both sides,
$$2a+b=2d$$
Solving gives this. Not hard to see that $a=4$ does the job.

The key for this type of equation is to balance the diatomic molecule last, and then just add a 1/2 of the diatomic molecule to finish the balance.  Then, double all the coefficients and you're done.

balance everything but the diatomic molecule:
$$\ce{2CO2 + H2O <=> C2H2 + O2}$$
Then add the right fraction of the diatomic molecule:
$$\ce{2CO2 + H2O <=> C2H2 + 2.5O2}$$

Notice that 2.5 oxygen molecules is 5 oxygen atoms.

Double all the coefficients to get whole numbers
$$\ce{4CO2 + 2H2O <=> 2C2H2 + 5O2}$$

There is a standard equation for balancing such reactions:
$$\ce{C_xH_y + [x +(y/4)]O2 -> x CO2 + y/2 H2O}$$
Hopefully, now you should be able to apply it to your reaction. It's just the reverse reaction that you're dealing with.

Please don't use shorthand or text speak.

This answer would be more helpful if it demonstrated how the stoichiometric coefficients are found in the first place.


text:
NiO and ZnO electrical conductivity with oxygen pressure
NiO is a Mott-Hubbard insulator and increasing oxygen concentration leads to anion excess and oxidation of Nickel so electron hopping is more favourable so electron repulsion is reduced and it becomes metallic. This makes sense to me but I think it can also be explained as a p-type semiconductor since you're essentially adding Ni(III). So you'd expect to get a band structure with a very narrow empty band above a filled band. However, I don't see how these ideas match up?
ZnO has ten d electrons so isn't a Mott Hubbard insulator so can only be a n-type semiconductor since it won't be oxidised. This leaves a narrow filled band below and empty band right? But what is the empty band? The s band? Is my reasoning okay?

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text:
Best Radioactive Isotopes for Energy Generation
First off, this is research for a video game storyline, I like to be thorough.
--
I am looking for the most efficient method of nuclear power generation which:

Can survive for hundreds/thousands of years (a no-brainer for isotopes)
Has some kind of non (or minimally) radioactive decay product/by-product.

The storyline is based in the 2050's, so I can use a little poetic license, I just want to get things as close to reality as possible.
This would be used to power a large city and scientific research base, for which several generators could be used, but I was wondering on which material would be best to generate such power consistently.
These facilities have consistent access to clean, flowing water so heavy water theoretically could be used in the energy generating process.
I started looking at our currently used isotopes, $\ce{^{234}U}$ and others, but I am looking for a material that would be compatible with both fusion and (what would then be legacy) fission (if possible).
$\ce{^{232}Th}$ seems like a promising start, but I thought I should ask before basing everything on extremely flawed science.
Any suggestions?

I always hoped that humanity would move beyond the realm of nuclear energy generation by 2050. Welcome to Chemistry.se though.

@Martin-マーチン Doubt it, fusion reactors can start to be useful though.

This would probably be better asked on Worldbuilding.SE.

I'm voting to close this question as off-topic because it belongs on World Building.

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text:
How to find the moles of carbonic acid I need to add to make a solution of KOH acidic
I wanted to know if it is possible to add enough carbonic acid to a $\ce{KOH}$ solution to make it an acidic solution.
The $\mathrm{pH}$ of my KOH solution was $13.2$. After adding $\pu{50 g}$ of carbon dioxide (and I do not know how much reacted), the $\mathrm{pH}$ decreased to $10.302$.
Is there a way to know how many moles of carbonic acid more to add to the $\ce{KOH}$ solution to make it acidic?
At first, I thought I could take the change in pH to calculate the additional moles of carbonic acid needed.
So I thought going from a $\mathrm{pH}$ of $10.302$ to $7$, I would calculate the moles of carbonic acid needed for a change in $\mathrm{pH}$ of $3.302$ $\mathrm{pH}$ ($10.302 - 7 = 3.302$). Thus, the calculation would be $10^{-3.302} = \pu{4.989 x 10^{-4} M}$ $\ce{H2CO3}$. There are 2 moles of $\ce{H+}$ in 1 mole of $\ce{H2CO3}$; thus, I need:
$4.989 \times 10^{-4}/2 = 2.49 \times 10^{-4}$ moles of $\ce{H2CO3}$
$\ce{H2CO3}$ is a $1$ to $1$ ratio with $\ce{CO2}$; thus, I need $2.49 \times 10^{-4}$ moles of carbonic acid in solution.
But I do not know if it is right to think it this way?
Can someone tell me if what I am doing is right, and if not, what I can do to theoretically calculate moles of carbonic acid needed to make the $\ce{KOH}$ solution as acidic as I could.

"as acidic as I could" is too vague. Can you be more specific about, say, your target $\mathrm{pH}$ for the solution?

between a pH of 5-7

Are you adding carbonic acid by bubbling pure CO2 through the solution? I think this will be tricky to calculate then assume it will all be absorbed and react (as you found in the first stage of your experiment). Can you add a pH indicator or other means of monitoring pH and just bubble CO2 in via a fine frit until it's neutralized / slightly acidic? I think your calculations represent a minimum of how much you will need in practice.

no put dry ice. And @airhuff, you think my calculations are right?

Yea, looks right to me.

okay but when I go to like to pH of 5, the difference in pH is 8, so doing the same calculations as I did in my question, I get lower moles of CO2 needed. So does my calculation break down after pH of 6?

You have not stated the volume of your $\ce{KOH}$ solution. I therefore have to make some assumptions.
I will use $\mathrm{p}K_\mathrm{a1} (\ce{H2CO3}) = 6.37$. I further assume you have $\pu{1 L}$ of the $\ce{KOH}$ solution and that this volume will not change by the addition of the carbon dioxide.
For the following, we have also to assume that all the solid carbon dioxide will react, i.e. no evaporation of any gas -- tricky, but perhaps not impossible.
You start with a solution of KOH that has the $\mathrm{pH} = 13.2$. From this we calculate $\mathrm{pOH}$:
$$\mathrm{pOH} = 0.8 \quad \to \quad [\ce{OH-}] = \pu{0.16 M}~(\text{or}~\pu{0.1584 M})$$
First, we suppose that the reaction between the solid carbon dioxide
and the solution of KOH will result in only potassium hydrocarbonate:
$$\ce{CO2 + OH- -> HCO3-}$$
OK, how much carbon dioxide is needed to transform all hydroxide ions into potassium hydrocarbonate? You will need (per liter $\ce{KOH}$): $\pu{0.16 mol}$.
After addition of $\pu{0.16 mol}$ carbon dioxide we will face the same situation as if you had dissolved $\pu{0.16 mol}$ of potassium hydrogencarbonate in $\pu{1 L}$ of water. The $\mathrm{pH}$ of such a solution will still be alkaline $(\mathrm{pH} > 8)$.
You wanted the solution to be acidic, but you did not specify the pH. I suppose you will be happy with $\mathrm{pH} = 6.37$.
Accordingly, we have to add more carbon dioxide. The charge balance is:
$$\ce{[H3O+] + [K+] = [OH-] + [HCO3-] + 2[CO3^2-]}$$
At $\mathrm{pH} = 6.37$, both $[\ce{OH-}]$ and $[\ce{CO3^2-}] << [\ce{HCO3-}]$.
Therefore, we re-write the charge balance as:
$$\ce{[H3O+] + [K+] = [HCO3-]}$$
We know the values of $[\ce{H3O+}]$ and $[\ce{K+}]$, so we calculate:
$$[\ce{HCO3-}] = 10^{-6.37} + 0.16 \approx \pu{0.16 M}$$
We can now calculate $[\ce{H2CO3}]$:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} \quad \to \quad [\ce{HCO3-}] = [\ce{H2CO3}] = \pu{0.16 M} \quad \to \quad [\ce{HCO3-}] + [\ce{H2CO3}] = \pu{0.32 M}$$
Accordingly, in total we need $\pu{0.32 mol} \approx \pu{0.3 mol}$ of carbon dioxide to reach $\mathrm{pH} = 6.37$. Per liter KOH solution this will correspond to $\pu{0.3 g} \times \pu{44 g mol-1} = \pu{13 g}$ solid carbon dioxide.

I reacted 10 grams of KOH in 200mL of water

I am not quite getting the same molar concentration of OH as you are using 13.2 as my pH value. Wouldn't the concentration be 10 raised to the negative pOH?

also, how does CO2 transform hydroxide ions to KHCO3? And I do not understand how you got the charge balance for KHCO3 ([H3O+] + [K+] = [OH-] + [HCO3-] + 2[CO32-])

It is not the charge balance for KHCO3. It is the charge balance of the system, which tells you that the total concentration of all positive ions in the solution have to be equal to the total concentration of all the negative ions.

Or better:
It is not the charge balance for KHCO3. It is the charge balance of the system, which tells you that the total concentration of positive charges in the solution have to be equal to the total concentration of the negative charges.

what about the molar concentration of OH-


text:
Realistic chemical spill accident in high school chemistry class
In a story I'm writing, a teenager spills chemicals on herself in a high school chemistry class, causing her to be put under the safety shower. The spill destroys her clothes and causes minor burns. I would like this to be realistic and since I know nothing about chemistry, I have two questions:

Is there a chemical used in high school chemistry that when spilled could cause this situation to occur?
If so, do you know of an experiment that would use that chemical?

Seems realistic to me, assuming high schools still do chemistry practicals. A large spillage of even dilute hydrochloric acid would have me putting the victim under the emergency shower just to be on the safe side. Others should be able to comment on the sort of chemicals high schools use as I'm not American and long out of education I don't know.

Generally, school chemistry avoids such chemicals. There should be regulations on what chemicals may be handled by students. Consider finding a local chemistry teacher, look at the chemicals (s)he has in the lab and post here.

I had looked into hydrochloric acid but it seems the delution (not sure if that's the right word) used in high school seems to be far from worrisome to spill. I looked in to nitric acid but I wasn't certain if that would be used in high school. I was thinking about rewriting so that her clothes are just stained (which still gets the point across) and I can discard the part about being burned. Maybe I should rephrase my question as, what chemical used in high school chemistry would cause a teacher to put a student under a safety shower? That's the most important aspect of this story.

Even dilute potassium permanganate will stain clothes as well as skin. I think the 'better safe than sorry' approach works here. Safer to have the student under the shower than risk any chemical burns. Would need to be a decent volume of solution, I think. A few hundred mls. (Fellow author)

As Waylander said, for stains KMnO4 (routinely used in redox titrations) is by far the best option. For acids it will probably vary from place to place. Personally I used H2SO4, HCl, and HNO3 in high school; of the three the most dangerous is HNO3. However, I doubt that we had anything > 1 M. Advanced students might possibly be allowed to use more concentrated acids under very strict supervision although that will obviously depend on the system (I used 6 M HCl couple of times).

Dump 200ml of 1N HCl over yourself in my lab and I will put you under the shower.

I think I would too, however as you said the volume would need to be quite large.

Bromine. CDC writeup on exposure effects and what to do with clothing here.

@ToddMinehardt That's not really a high-school-chem thing though, is it?

Unless the clothes being destroyed are an important part to the story, it might not matter too much since after a chemical spill/burn like that you typically strip down then go into the safety shower, if my memory is correct, it's to prevent further contamination of chemicals back onto your skin from your clothes.

@hBy2Py Depends on your high school chemistry lab. My teacher had a (sealed) ~250 mL flask of liquid bromine she brought out on occasion (as a demonstration piece for element properties). It always filled me with an abundance of trepidation, and she never let students handle it, but I could certainly imagine a more cavalier student with a more lax teacher accidentally breaking such a thing and spilling it on themselves.

I think you need to better define what you mean by "destroyed clothes". I've certainly rendered more than one article of clothing "unwearable", as far as I'm concerned, and even had acids or bases eat a small hole through  one shirt, but "destroyed"... enh, down to how you define it.  You should probably also define what you mean by "put under the safety shower" as well - I've been told to go use the safety shower as a result of spills in AP chemistry, but my response was always that I wasn't doing that ... maybe current attitudes about "safety" wouldn't allow a student these days to say "no".

When I was a student at an Australian university, a chemistry teacher mentioned to class an incident where a female Muslim student suffered a spill but didn't want to take off affected clothing because of modesty, and it made the situation worse.

Maybe I should rephrase my question as, what chemical used in high school chemistry would cause a teacher to put a student under a safety shower?  |||| NONE. this is the point of high school chemistry - it is as safe as possible.  The worst case scenario is eye burn from spills and in the case the eye would be rigorously rinsed with water (preferably, isotonic salt solution).

Does the setting have to be in the actual high school lab? I recall, for example, in high school (US), in the AP level courses, it was not unusual to take trips to the college campus that was next door (Carnegie-Mellon or Pitt in our case, actually some of our high school teachers were also  common guest lecturers at CMU) for various demonstrations and lab experiments and such, and a wandering student or careless placement of something in a lab could certainly expose a high school student to something not normally found in a high school lab. This held for both chemistry and bio courses.

@Tina If getting naked or partially naked is an important element in your plot, I suggest removing clothes before using the safety shower instead of getting clothes destroyed.

If so, do you know of an experiment that would use that chemical? Are you plotting a crime scene? Given this record here, you may be accountable for the expected damage. Insure yourself! We have chemistry safety quizzes, WHS guidelines to follow prior to anything in that Chem square. It's darn strict. That quiz - geez!

Uhm, at our high school, we ll had to wear lab coats and glasses all the time in the lab. But we didn't have any safety showers. So, whatever scenario you make will be completely impossible here :) As for chemicals, the most dangerous stuff I personally handled there was a small quantity of 37% HCl. 2 friends at another school were permitted to make nitrocellulose.

Collecting together the responses from the comments and Ben Norris's answer and adding one of my own, individual solutions in water of the following chemicals are realistic options for a liquid spill in a high school chemistry lab, leading to destroyed clothes and time under a safety shower:

Potassium permanganate $(\ce{KMnO4})$
Sulfuric acid $(\ce{H2SO4})$
Hydrochloric acid $(\ce{HCl})$
Nitric acid $(\ce{HNO3})$
Sodium hydroxide $(\ce{NaOH})$

As for an experiment leading to a spill, one possible example would be in the course of filling a buret in preparation for a titration. Burets are usually quite tall and narrow (click image for larger version):

      (Marketing image from Microtech Scientific)
A slip while filling one could easily result in spilling a few hundred milliliters or more of liquid on oneself.

From looking at the list, is it safe to assume this is a "pick one" list, not a "mix 'em all together" list?

@CortAmmon Yes, I intended this as a "pick one" list. See my edit to clarify.

@CortAmmon There may be a case for a mix of two in a titration experiment, for ex: Titrating a known concentration of sulfuric acid into an unknown concentration of sodium hydroxide  with an indicator to determine the concentration.

I accidently poured concentrated sulfuric acid on me. It caused itching and rashes.

conc H2SO4 does not belong in a high school lab.

@orthocresol Conc H2SO4 is sold as industrial-strength drain cleaner.

Yes, but it should not be let near the students. Actually, I have an idea, and maybe I should submit it as an answer to cash in on that sweet HNQ rep... but I am too lazy. A high school lab would probably still need a preparation room, in order to make stock solutions, etc. There will likely be concentrated acids inside there. Now, a naughty student wandering about the lab can easily find him/herself in trouble.

@orthocresol Is high-school the same as UK secondary school / sixth-form? More advanced classes use conc H2SO4. I recall practical lessons in preparation of esters and dehydration of glucose.

Likewise, @orthocresol - I recall doing experiments with concentrated sulfuric too. It's not brought out until the upper years, but it was certainly there.

@orthocresol My school do have concentrated sulfuric acid. One of the first experiments I did was with sulfuric acid.

My 10/12th grade HS chem classes used 6M acids and bases as the default concentrations; there were a handful of times we used stronger (up to 12M) for one reason or another.

The real high school accidents that I know of involved flammable liquids (ethanol or methanol) igniting.

Mishaps in school labs reveal lack of safety (PDF link)
Two high school kids burned in lab accident
6 injured in chemistry classroom fire at Woodson High in Fairfax
Safety Alert Rainbow Demonstration

If you want to be realistic, base on the above.

To be fair, the fact that these made national news probably indicates these are worst case scenarios. (Not that this means they are fake or anything.) The news has a selection bias toward the unusual.

Here's a couple more from schools near me: https://www.dnainfo.com/new-york/20161117/inwood/28-students-hazardous-material-northeastern-academy-school and http://abc7ny.com/education/6-students-teacher-hospitalized-after-nj-acid-spill/1666898/

My answer is a carbon-copy of hBy2Py's answer except to outline the specific damage that would occur in each case. And to add another candidate, potassium hydroxide$\ce{KOH}$.
Potassium permanganate ($\ce{KMnO4}$)
This is probably the most innocuous. It has antiseptic properties and I wouldn't consider it especially dangerous. That being said in high concentrations caustic burns have been reported. These sort of concentration is not likely to be encountered in a high school laboratory though.
Visually this spill would be the most spectacular. It is very strongly colored even in dilute solutions. It will stain clothes a bright purple color. The permangenate ion then reduces to a kind of gross brown colour over time.
Sulfuric acid ($\ce{H2SO4}$)
This, and the other two acids below, is substantially more dangerous than the $\ce{KMnO4}$ above. A 1 mol.L$^{-1}$ (or perhaps 2 mol.L$^{-1}$, again, I'm not sure the max allowable concentration in a high school lab) solution of this acid given enough time will slowly degrade most fabrics. It turns paper a brown color as it slowly oxidizes the cellulose in it over time. A similar reaction would occur with cotton I assume.
I'm not sure how bad any burns to the skin would be from solutions of the above concentrations. That being said, I would definitely want to wash myself if I spilled any quantity on me though.
Hydrochloric acid ($\ce{HCl}$) and Nitric acid ($\ce{HNO3}$)
Much of what was said for sulfuric acid above can be said for these two; though, these two in high enough concentration though can also produce noxious fumes. Hydrogen chloride ($\ce{HCl}$) gas in the case of hydrochloric acid, and Nitrogen dioxide ($\ce{NO2}$) for nitric acid. The fumes of these two gasses are highly toxic. To achieve these gases though you'd need to have a highly concentrated solution, again above a concentration I'd expect to have a modern high school lab.
Sodium hydroxide ($\ce{NaOH}$)
I'm not sure of the clothes destroying nature of sodium hydroxide (sometimes called lye), but it would definitely give you issues at a concentration found in a high school lab. Any hydroxides are going to be especially dangerous as they can convert the naturally occurring fats and oils in your skin in to soap. This process is called saponification. If you get some on yourself and it's left unwashed you will feel your fingers and hands become slimy from the soaps on your fingers. It feels like when you have soapy water on your fingers.
Potassium hydroxide ($\ce{KOH}$)
This is an extra compound I've added. It's a hydroxide like the sodium hydroxide above, and so everything I've said there applies to it too. Potassium hydroxide is a stronger base than sodium hydroxide and so its effects will be felt at a lower concentration. A 2 mol.L$^{-1}$ solution will definitely cause saponification. And so, if potassium hydroxide were to be found in a high school lab it might be found at a concentration that would cause chemical burns.

"Potassium hydroxide is a stronger base than sodium hydroxide." Can you provide a citation for this? I've never heard or read this before, as best I can recall.

@hBy2Py "Dissociation, Chemical Exchange, and the Proton Magnetic Resonance in Some Aqueous Electrolytes" The Journal of Chemical Physics 21, 1688 (1953) http://aip.scitation.org/doi/abs/10.1063/1.1698644

Interesting -- so the $\ce{Na+}$--$\ce{OH-}$ ion pairs are more strongly associated, essentially?

It looks like at low concentrations you wouldn't notice the difference, but at at higher concentrations the odd Na+ and OH- will form ion pairs more than KOH would. From figure 4 in the above reference it looks like you'd notice the difference from a concentration of 5% protons or higher, which would be about 5 mol/L based on my late night calculations. That's actually a bit higher than I thought it would be. Definitely higher than high school lab concentrations. Nice ref @DavePhD. I learnt something today.

Also, technically permanganate reduces to the ugly brown color, going from $\ce{Mn(VII)}$ to $\ce{Mn(IV)}$.

Sulfuric acid will do truly nasty things to organic material if you don't wash it off very quickly.  See this video for a famous example: it will suck the hydrogen and oxygen directly out of organic materials such as sugar, leaving the carbon behind.  That's something you seriously do not want happening to your skin!

Having taken high school chemistry, hydrochloric acid is by far the most likely one as it is the most used and straight out of the bottle it is plenty strong enough.

A real example from personal experience which, with slightly altered circumstances, could have had the effects you describe.
A student chemistry teacher taught us the chemistry of manufacturing oleum (by passing SO3 through sulfuric acid) then set up an apparatus to demonstrate the reaction. All went well.
After the demonstration, we all began to write up the experiment and the teacher began to clear up. When he reached the flask of oleum he had manufactured he tipped it into the sink. I remember the horror on my (and other pupil's) faces but he was too quick for us to stop him.
There was a loud explosion. Fortunately, he hadn't made very much oleum and the reaction between it, the water and rubber tubing in the sink was not overly violent. He was not hurt. We were too far away to be injured, though a lot of the equipment was damaged.
A little more oleum, a little more water in the sink and a student sitting closer and things could have been worse.

Sulfuric acid will do this. Concentrated sulfuric acid might not be common in high schools due to safety considerations. However, perhaps you could include the teacher getting in trouble for using something so dangerous.

Nitric acid would be a good example. It will dissolve certain synthetic fibers, color any organic material (including skin) exposed to it yellow/orange and cause a chemical burn wound to form, which would certainly require the use of a safety shower.
Major caveat is that the fumes are extremely toxic and thus the chemical would normally only be handled in a fume hood. However I could easily imagine the chemical being spilled during transport.

Yup, in multiple years of chemistry lab the only harm I ever suffered was from nitric acid--a drop of it got on the outside of a bottle and I picked up the bottle.  (College chem lab, not high school!)  It was handled under a fume hood but that doesn't mean somebody couldn't be careless, or the victim of other factors.  Suppose somebody else has an oops and jumps back, bumping the person handling a bottle of it.

Nitrating mixture (concentrated H2SO4 and HNO3).
Christian Friedrich Schönbein spilt this on his apron (no lab coats in them days I guess) and dried it over the stove, resulting in spectacular deflagration and leading to the invention of nitrocellulose.
We were allowed dropper bottles of those acids when I was in 6th form, for instance in one fun afternoon, trying to find an acid that dissolves aluminimum.

The funny thing about nitric acid is that it becomes less active against metals when it is at very high concentration (99% or higher). Pretty sure you were "only " allowed to handle 63% nitric acid. 99% is really hard to obtain unless you distill it yourself.

I can imagine some "teacher's pet" being allowed to help prepare (after school) reagents for class. Realistically, the only thing which would "destroy" clothing is concentrated sulfuric acid, which should not be part of the chemicals available to high school children. The problem with your question is that it assumes the teacher/school administration has allowed the classes to involve materials which are inherently unsafe, not just "unsafe" but actively dangerous. It's certainly possible, but not very likely, imho. IOW, not very realistic. Frankly, I've seen more really bad practice in high school biology labs. For instance, one class I was in had an OPEN 5 Gallon Bucket of formaldehyde solution (it may have had an animal corpse in it, I don't recall)!! They had it "by" the open windows (which was apparently the teacher's idea of "proper ventilation"!!). And while that was like 20 years ago or so, it was in an affluent suburban school. If someone were to be carrying that at chest level and collide with another student, a safety shower (if they even had that in biology labs, but why not?) would be appropriate. But while concentrated sulfuric acid will burn holes in clothing, it won't "destroy" it, nor will any of the other chemicals (at least not on time scales of minutes).

At first I had trouble with the "destroys her clothes" part, too. But a common use of the phrase implies more "renders undesirable/unfit to be worn again", rather than physical obliteration. The former is a much lower bar to reach, and I think most parents would just throw away any chemical-contaminated clothing out of an overabundance of caution.

Tina, don't listen to the naysayers :)
You absolutely can write a realistic scenario that accomplishes everything you are asking in the context of a high school chemistry class. The only thing is that we need a bit more information about your context and your aims. For example:
In what historical era and location is your scene going to play out? The reagents  available in a 2017 public high school chemistry class will be very different from what was available in a public high school in 1965.
How risky / dangerous do you want this to be? Do you want this to be a normally safe and common activity that goes wrong because of clumsiness or tomfoolery, or do you want this to be something slightly dangerous or perhaps even very dangerous, so that both teacher and students are warned and ready to respond when something goes wrong?
This guide for high school teachers comes from the American Chemical Society  and it gives a lot of specific information that may be helpful to you.
Do you want the student to put herself under the safety shower? Current published guidelines for high school chemistry classes say students should be taught to walk quickly but calmly to the shower and pull the handle immediately under certain circumstances. For example, the ACS guide says this:

Students should be taught the “stop, drop, and roll” technique to be used if their clothes catch on fire elsewhere, and in the laboratory taught to walk calmly to and use the safety shower to extinguish clothes that are on fire. A drill to practice these exercises is recommended.

If you do not want the student to be forced to undress completely, perhaps a spill resulting in a small clothing fire (with or without burned hair) might  accomplish all of your goals. If that would do, I suggest that you could have the students be performing metal salts flame tests using the petri dish technique, in which metal salts are burned in a small amount of methanol (methyl alcohol) and the material is identified by the rainbow of colors that are observed in the flame. This would be a great visual when your book is made into a movie :)
Here is a link to a middle school guide for one version of the flame test.  Having a small spill of burning alcohol cause clothing to catch fire is quite realistic. Most such events are easily handled, but the risk is real and the safety shower, while not the best approach for significant amounts of burning liquids, would absolutely work to put out a clothing fire ignited by a small amount of flaming alcohol.
I should point out that over the years there have even been a few published classroom incidents in which poor ventilation caused a large amount of methanol fumes to accumulate and a massive fireball resulted. There are some people who feel the flame test should not be part of a modern curriculum, but is is widely performed, colorful, fun for students, teaches something quite fundamentally important both in chemistry and in physics, and falls well within a safety zone appropriate to this age group. You can purchase a kit here and elsewhere.
Because links may evaporate, I will embed a description of a flame test involving methanol in a petri dish:

Flame Tests
Description: Methanol solutions of salts are burned to observe that
  different wavelengths of light are emitted by different materials
Materials:

Methanol
Petri dish
Cover (to extinguish flame)
NaCl (yellow)
SrCl2 (red)
LiCl (red)
KCl (violet)
CuCl2 (green)
CaCl2 (red-orange)
H3BO3 (green)

Procedure:
Several variations exist including the “Petri dish” method (described
  here), the squirt bottle method (ref 1), the H2 balloon method (ref.
  2), the guncotton method (ref 3) and the Magic Eraser method (ref 4).

Sprinkle a spatula full of each salt into separate Petri dishes.
Cover salt with methanol (10 mL). Turn off the lights for a better visual effect. Use the lighter to start up the flame for each sample.
To extinguish the flame, put the cover over the Petri dish and let cool before removing.

Discussion: When an element is burned, the electrons are excited. As
  the electrons from an excited state relax back to a ground state, they
  will emit photons of light. These photons will have different colors
  depending on the element and its discrete energy levels. That is,
  different wavelengths of light (colors) will be emitted when the
  electrons of different elements go down the step(s) between their
  energy level(s). Each element will have its own set of steps therefore
  each will have its own color.
Safety: Wear safety glasses and goggles while preparing and during the
  demonstration. Be sure to allow time for the cover to cool before
  removing.

And here's what it looks like:

If you prefer a chemical spill that causes the teacher to put your student under the shower, that is also easily accomplished in a high school chemistry class. Somebody has already suggested that if a chemical spill occurred the student would be undressed before entering the shower. This is not correct: the ACS high school chemistry teachers guide I referenced above says about corrosives:

If splashed on clothing, the clothing must be
  removed while under a safety shower. Do not remove the clothing and then get
  under the shower. While under the shower, remove all clothing, including
  shoes, socks, wristwatch and strap, and other jewelry if they are splashed with
  corrosives (this is no time for modesty). Stay under the shower for at least 15
  minutes while someone else calls a doctor.

Most safety showers are fully exposed, but there are times when they may be located in an alcove with a door that can be closed or a privacy curtain that can be pulled across. There are also safety showers with built-in plastic curtains that can be pulled around for privacy.
If you want the safety shower to be required for a chemical spill resulting from an experiment generally recognized as safe, then any of the mild corrosives will do but you will need to have the student be drenched with the material -- a small spill of most "safe" high school chemicals would not result in a full shower, and of course if ANY splash involves the face then the eyewash station would be the first thought. I've seen people fully drenched from the eyewash station, though, so perhaps that would fit your needs?
If you want to create real fear of injury from a minor chemical spill or splash, this can still be accomplished within a realistic high school setting. I'll pick just one example: it's not unknown for chemistry classes to have students etch mirrors using hydrofluoric acid, which in liquid form is one of the most noxious substances you can come into contact with anywhere below graduate-level chemistry. Even with immediate irrigation the stuff has a tendency to percolate into the body, and you will probably die despite treatment if a small glassful spills in your lap. For this reason it's not used at all as a reagent in high school. However, HF acid is one of the few things that will etch glass, and it is used in various forms for glass etching. The demonstration of an acid that can dissolve glass appears in chemistry courses even at the high school level. This stuff is so nasty that the slightest splash will cause immediate alarm. The correct response is immediate and prolonged irrigation and the immediate application of calcium gluconate gel as a neutralizing agent.
Here's a link to a college-level class guide in which liquid bath glass etching is performed, and if this were an AP class perhaps something like this might appear. Somebody is bound to comment that hydrofluoric acid would not be provided to regular high school students, and thankfully in a strict sense that's true. However, the stuff does still show up in high school classes in various forms. It's routinely found in high school electronics classes in the "liquid tin" solutions used for tin-plating the copper traces on printed-circuit boards, and that same reaction could easily be part of chemistry segment on metals. And here's a link to a fund-raising prospectus in which a high school chemistry teacher recently raised money so that inner-city high school students could "etch glass mirrors with hydrofluoric acid". The link includes photos of the students displaying their etched mirrors, one of which I will embed here.

Of course the material given to these students used was not a liquid form of HF even though it's billed that way -- it's actually a cream that contains fluoride salts. If washed off quickly enough it will cause only superficial injury (and ruin your clothes) -- but if there were any sort of splash you can bet the teacher would be rushing a student under the shower ASAP.
So if you want to create a real fear, the mere mention of hydrofluoric acid would do it. As a physician (board-certified in Emergency Medicine) I have treated several patients for exposure to hydrofluoric acid, including one unfortunate janitor who died a week later (despite rapid and aggressive treatment) after what initially appeared to be a minor splash from an unknown roof-cleaning chemical. It wasn't until we called the poison center to find out what was in that branded stripper/cleaner that we knew it was going to be bad, and even then it was hard to believe how bad. Believe it or not, all the area hospitals actually ran out of calcium gluconate supplies before he died, so yes, scary.
Lots of options for you, but in the end I think you'd be best served by the colorful flame test: put a Bunsen burner together with a spilled petri dish of alcohol and you get a lot of excitement, a quick trip to the shower, and the kind of damaged clothing and minor burns that people can easily understand. And no need for your student to disrobe.


text:
Which will react with gaseous bromine faster and why: C3H6 or C2H2?
I know that for both $\ce{C3H6}$ and $\ce{C2H2}$, it will be an addition reaction where bromine will be added through a free-radical mechanism. But I'm not sure which one will be faster.
At first, I thought it should be the $\ce{C2H2}$ because it has a triple bond and so it has a higher electron density ( i.e. it is more nucleophilic) and thus the reaction would proceed faster. But then I realized that the reaction for $\ce{C2H2}$ will proceed in steps so I believe it being more nucleophilic won't make much of a difference.
So that's why I'm stuck with which one will be faster?
P. S. : the C3H6 refers to Ethyne and NOT cyclopropane

While I think from context it's clear what you meant, cyclopropane also has the formula $\ce{C3H6}$.

Regardless if the addition of $\ce{Br2}$ on propene or ethyne proceeds along a electrophilic, or a radical pathway, the reaction with propene will be faster, then on ethyne.
In ethyne, the carbon atoms in question for the reaction are formally $C(sp)$ hybrids; the ones in propene $C(sp^2)$.  The higher the $s$-character, the higher the electronegativity of the carbon atoms in question.  Hence, the $\pi$-electrons in ethyne are tighter bond to the carbon atoms, than in the case of propene.  In addition, the methyl group adjacent to the initial C=C double bond offers stabilization of the intermediate carbocation (for the electrophilic addition, favouring the Markovnikov product) or radical (for the radical addition), that is absent in the instance of ethyne.
(Since your special example is about the addition of $\ce{Br2}$, Markovnikov selectivity is not an issue here.  For halogen addition, choice of solvents, presence / absence of light or initiators may favour one or the other mechanism, too.)

Hey!  Wow thanks that was really helpful!  I'm a bit confused about what you mean by "s-character" though.  Can you please explain that.

@Pink_NinjaTurtle The "$s$-character" refers to the concept of how atomic orbitals are (formally) mixed to obtain hybrid orbitals.  Hence, a $sp$ is the result of mixing one $s$ and one $p$ atomic orbital; where each contributes 50%.  In instance of $sp^2$, there are two $p$ and one $s$ AO mixed (formally), hence the contribution by $s$ is only about 33%. And so forth.

If you do not feel easy with hybridization of AO's, revise the concept with a textbook.  Robert Grossman's The Art of Writing Organic Reaction Mechanisms may be a handy companion, including plenty of exercises. Don't be reluctant to use this book just because the presentation of the material differs from "classical textbooks".


text:
If I mix carbon dioxide and sodium hydroxide in the right amounts in water, do I get baking soda?
Suppose I need a sodium bicarbonate solution (for non-food related reasons). Can I get something resembling it by mixing $\ce{NaOH}$ and $\ce{CO2}$ in water?
Of course one should not attempt to make food in a chemistry lab, so lets rule out the food safety requirements for food-grade baking soda.

This is certainly one method by which sodium bicarbonate can be produced.  According to this Wikipedia article:

$\ce{NaHCO3}$ may be obtained by the reaction of carbon dioxide with an aqueous solution of sodium hydroxide. The initial reaction produces sodium carbonate:
$$\ce{CO2 + 2 NaOH → Na2CO3 + H2O}$$
Further addition of carbon dioxide produces sodium bicarbonate, which at sufficiently high concentration will precipitate out of solution:
$$\ce{Na2CO3 + CO2 + H2O → 2 NaHCO3}$$

The net reaction for this two-step process is then simply:
$$\ce{CO2 + NaOH -> NaHCO3}$$
Note that this is not the common method by which sodium bicarbonate is produced commercially, but it is a perfectly valid method of it's synthesis.

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text:
Chemical compatibility
I'm an electrical engineer trying to understand a failure in a plastic part that I'm working with.
I am using a plastic called polysulfone- and the data sheet has a lot of words ending in "amine" that are listed as not-being compatible with said plastic.
I want to bond this plastic with an epoxy, but the very-vague datasheet states that the hardener contains 80% "Modified Cycloalipathic Amines".
With this admittedly limited information, should I look for a different adhesive?
The process temp is $5$ to $85~^\circ\mathrm{F}$ ($-15$ to $29~^\circ\mathrm{C}$).

I have a bachelor's in chemistry, but I am not an expert in materials science. All I can tell you is that looking at the structure of a polysulfone polymer, going by textbook logic I would not expect much reactivity with typical amine compounds. Perhaps the amines listed in your MSDS have functional groups, other than the amino group, that would react with your polysulfone. From a glance at structures found by googling "modified cycloaliphatic amine", it looks like such compounds (representing the epoxy) will likely behave like typical amines (which may or may not react with your polysulfone).

I suggest taking the experimental route and exposing another test piece of the plastic you are working with to the hardener. Keep an eye out for any softening, swelling, change of color and any other indicators of a reaction occuring.

Short answer: yes.
Every manufacturer of two-part, room temperature cure epoxies has their own formula. There is no way to know without either knowing the formula or testing your specific application whether the one you are talking about is or isn't compatible.
First thing you want to find out is the composition of the other 20% of the hardener. It could easily be a tertiary amino phenol (or just phenol itself), which might be a problem since it would tend to be "fugitive". I can think of at least a couple of dozen questions that might need answering (by empirical test). How is dielectric surface break-down affected by the epoxy, for instance. No way you're going to find that data - someone will have to generate it for you (I'm assuming this isn't a one time thing. If it's a one time thing, I'd advise against an amine cured epoxy if you're concerned with anything except bond strength (and even that would have to be tested).There are a number of mercaptan cure epoxies that would be my first picks - but I've no direct experience with this adhesive/substrate combination.
To keep it simple, I'd suggest you try a 2 part acrylic adhesive (they are expensive). Also, the stoichiometry (ratio of epoxy to hardener) could be important. More epoxy reduces the amount of low molecular weight amine (meaning less will diffuse out of bond and into Polysulfone). But higher epoxy means a more brittle bond, generally. So, this isn't something that can be answered without testing.
If you're a "big customer" perhaps the adhesive or polysulfone part or resin manufacture can help. I used to be an adhesive formulator, so perhaps I'm overly complicating this, but I wouldn't be confident in making a recommendation about type of adhesive without knowing a whole lot more. You've not mentioned even one requirement of this "bond", bond strength of 1 nanogram/m² or 1 Gigaton/cm² for instance, and as I said, there are several dozen more characteristics that are probably required.
Incidentally, marine supply stores may have two-part acrylics, if you're at the retail level. FWIW, talking about epoxy adhesives as if they are all one thing is like talking about the 0-to-60 acceleration of "cars". There may actually be more epoxy formulations available than car models. My experience is that the more vague the data sheet the more it is targeted for the retail market (that is, clueless consumers). This type of epoxy might work just fine, or might cause a failure. No way for us to tell you that. Heck, you didn't even tell us what the application or maximum use temperature was.


text:
chirality of conformations of cyclohexane
How is the chair form of cyclohexane achiral?
I don't think it has any plane of symmetry. I have tried very much to find the symmetry in it, but I cannot figure it out.

The figure shows the symmetry elements in the chair form of cyclohexane. (From molecule-viewer.com ). The point group is $D_{3d}$.
There is a $3$ and a $6-$fold rotation axis through the centre of the molecule, $3$ two-fold axes between opposite sided and at the mid point of the CC bond, and $3$ (dihedral) mirror planes that bisect the two-fold axes.


text:
Gibbs Free Energy and Equlibrium Constant
I have some problems understanding the relationship between $\Delta$$G°$ and $K$. For example, in the reaction of $$\ce{N2 + H2 <=> 2NH3}$$ where $\Delta$$G°= - 33.5$ kJ/mol.
$\Delta$$G°=- 33.5$ kJ/mol is obtained from the $\sum G^°_{\text{f, products}} – \sum G^°_{\text{f, reactants}}$
. This means that the complete conversion of 1 molar of $N_2$ gas and 1 molar of $H_2$ gas gives 2 molars of $NH_3$ gives -33.5 kJ/mol of free energy. This is assuming that the reaction goes fully to completion, yet in reality it doesn't.
From this formula below
$\Delta{G^°} = –RT \ln K_p \tag{5-6}$
$K_p$ can be obtained as $7*   10^5$, assuming at 298K and 1 atm.  My question is, since the complete reaction corresponds to  $\Delta$$G°= - 33.5$ kJ/mol of free energy, why should there be $K$ in the first place? $- 33.5$ kJ/mol tells me this value is for the complete reaction, so why should we insert this value into the equation just to get $K$ which tells me otherwise that the reaction doesn't go to completion? It doesn't make sense to me? In fact, shouldn't have K be infinity since the value that we had inserted is one obtained for a complete reaction?
Also, from this equation $\Delta{G} = \Delta{G^°} + RT \ln Q \tag{5-5}$
does having 1 mol of $N_2$ , 1 mol of $H_2$ and 1 mols $NH_3$ of in the reaction correspond to $- 33.5$ kJ/mol, since that makes $Q = 1$? And since $Q$ slowly reaches the value of $K$,  $\Delta{G} = 0$, then  $- 33.5$ kJ of free energy must have corresponded from the initial state to the equlibrium state. Yet again to me this doesn't make sense, because my understanding is that $- 33.5$ kJ is a value obtained only for the ideal scenario that the reaction goes to completion, and not from its initial standard states to equilibrium.
How does this all work out?

The $\Delta G$ value doesn't need to correspond to the reaction going to completion. It only tells us what the free energy released per mole of reaction is. At standard conditions, this reaction releases $-33.5$ kJ of free energy per mole of reaction, but that doesn't necessarily mean that if we have $1$ mol of $\ce{N2} \text{ and } 3\text{mols of } \ce{H2}$ that the reaction will proceed to completion.

Q is supposed to be calculated in terms of partial pressures, not moles.  $\Delta G^0$ is based on a reversible process you have devised to take the pure reactants in stoichiometric proportions and 1 atm (actually, 1 bar) and 298 K and convert them to pure products in stoichiometric proportions, also at 1 atm. and 298 K.

You can't really measure $\Delta G^\circ$ directly, rather you measure $K$ and convert it to $\Delta G^\circ$.  Put another way, $K$ is the quantity you measure to get $\Delta G^\circ$, just like heat exchanged ($q$) is the quantity you measure to get $\Delta H^\circ$.
To measure $\Delta H^\circ$ for the reaction you would, for example, measure $q$ under conditions (excess $\ce{H2}$) where a known quantity of $\ce{N2}$ is completely converted to $\ce{NH3}$.
To predict $q$ for an arbitrary set of reaction conditions you compute how much reactant/product is consumed/produced from $K$ and multiply that change in partial pressure by by $\Delta H^\circ$.

First of all, it confuses many when $\Delta$ in thermodynamics is used for both real changes (as in $\Delta U = q + w$, unit e.g. kJ) and for derivatives, i.e. infinitesimal changes (as in the $\Delta G$ in the question, unit kJ/mol). In my opinion, to reduce this confusion it is better to use $\Delta_r G$ and $\Delta_r G^0$ than $\Delta G$ and $\Delta G^0$, as (some) textbooks do.
Next one needs to understand that $\Delta_r G$ and $\Delta_r G^0$ are not changes from beginning to end of the reaction, and not at all to completion of the reaction, but they describe infinitesimal changes (change in $G$ per mole reaction: $\Delta_r G$ for the actual experimental composition and $\Delta_r G^0$ for the standard reference composition). Experimentally the mathematical term "infinitesimal" means so small a change that the molecular composition in the reaction vessel does not change significantly. Provided the change in composition is negligible, one gets the actual change by multiplying with the actual number of moles that has reacted. This number does not have to be 1 mol. Just as if you drive on the highway with, say, 120 km/h, this does not mean that you must drive for exactly 1 hour and move 120 km, in fact it may be only for one minute and you have moved 2 km. Your speed was still 120 km/h. And it is easier to keep a constant speed for 1 min than for 1 hour.
So, $\Delta G_r^0$ = -33.7 kJ/mol is not for the reaction running to completion, but for the special case where the partial pressures of $\ce{N2}$,  $\ce{H2}$, and $\ce{NH3}$ all are 1 bar (or 1 atm if old standard), both before and after the reaction. In this very special case, after 1 mol reaction, i.e. after 1 mol $\ce{N2}$ has reacted with 1 mol $\ce{H2}$ and formed additional 2 mol $\ce{NH3}$, you would then get $\Delta G = \Delta_r G^0 \times 1\ {\rm mol} = -33.7\ {\rm kJ}$. Experimentally this three digit number would only be correct to the three digits if one had a very big container with many moles of the three gases, such that adding or removing a mole would not significantly change the partial pressures of the three gases. Of course, keeping the change of composition negligible would be easier to achieve after e.g. only 1 $\mu$mol reacted instead of 1 mol; cf. my comparison of 1 hour and 1 minute driving on the highway. (Note that it is not important if you are able to measure this $\Delta G$ experimentally or not, it is still the correct number if the measurement could be done!)
In your eq. 5-5 ( $\Delta_r G = \Delta_r G^0 + RT \ln Q$ ) the term "$RT \ln Q$" describes the change in $\Delta_r G$ when changing from the reference standard composition to the actual composition. In this case, changing from 1 bar partial pressure of each of the 3 gases to the actual partial pressures in the experiment. We can then ask ourselves: when will there be equilibrium? The mathematics of thermodynamics tells us that we have equilibrium at constant pressure and temperature when $G$ has its minimum value, and minimum means that the derivative of $G$ with respect to further reaction is zero: $\Delta_r G = 0$. Inserting this in eq. 5-5 gives $0 = \Delta_r G^0 + RT \ln Q_{eq}$, which is easily rearranged to eq. 5-6. Conventionally we call the numerical value of $Q_{eq}$ the equilibrium constant $K$.


text:
Recoil velocity of atom when electron is excited
If an electron is excited from the $n = 1$ level to the $n = 2$ level in a hydrogen atom, how can we find the recoil velocity of the atom?

Not sure if chemistry.se is the right place for this, since it seems like a physics.se conservation of momentum question. Or mabye its just that i haven't ever heard of such a subject.

I have never seen the term recoil velocity in this context. Are we talking about the lifetime of an electronically excited state here?

Question makes no sense... The electron and the atomic nuclei are "bonded" together and as such the electron transition cannot change the velocity of the atom.

I am sorry it's more of physics question. You conserve momentum of photons liberated with atom

There are three things to conserve during an electronic transition: energy, momentum, and angular momentum. As always, these things are related to some extent. Let's forget about the angular momentum because it isn't relevant to this question.
The energy of a photon is given by,$$E_\gamma=h\nu.$$This energy is related to the momentum of a photon through the mass-energy equivalence,$$E^2=m^2c^4+p^2c^2.$$ Remembering that the mass of a photon is zero, we have,$$P_\gamma=\frac{E_\gamma}{c}.$$
The energy of an electronic transition between two states in a hydrogen atom must equal the energy of the photon being absorbed, which conserves the energy upon either absorption or emission. Thus, we have,$$E_\gamma=\frac{m_ee^4}{8h^2\epsilon_0^2}\left (\frac{1}{n_1^2}-\frac{1}{n_2^2}\right ),$$where we have used the solution to the energy states of a hydrogen atom coming from using a Coulomb potential to describe the potential energy of the system.
Then, we impose the conservation of momentum condition that $P_\gamma=P_{atom,f}-P_{atom,i}$. When I write $P_{atom}$ we have to note that this is given by the sum of momenta of the nucleus and electron.
Let's define the recoil velocity as, $v_{recoil}=v_f-v_i$
I don't wanna deal with the special relativistic correction to the momentum, so I'll leave that to you if you wanna figure it out.
Now, we can write,$$P_{recoil}=\frac{E_\gamma}{c},$$which can be written in terms of the recoil velocity according to,$$v_{recoil}=\frac{E_\gamma}{(m_e+m_p)c}.$$In the last expression, $m_p$ is the mass of a proton (i.e. the mass of the nucleus).
The expression would get uglier if you used $P=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$ instead of $P=mv$.
Using the above expression, the recoil velocity ends up being,$$v_{recoil}=\frac{10.2\cdot10^{-5}MeV}{938.272\frac{MeV}{c}}=3.26\ \frac{m}{s}.$$
So that's a reasonable velocity I think.
It's totally possible I made a mistake somewhere because I've never thought about this question before, but I think that is all correct. It's a little bit weird to think about whether the momentum is being transferred to the electron or the nucleus, but it seems like it must be transferred to both somehow?
One way to check the work (which I don't wanna do) is to try to get the same expression by only balancing energies and using $\frac{1}{2}mv^2$.

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Why are super absorbent polymers not made from polysaccharide-rich wheat straw and chaff?
What are the qualities in a substance that chemists look for when making a super absorbent polymer?
If I am right, polysaccharides are one of them.

Until the 1920's, polysaccharides were a prime choice for producing absorbent materials. However, their water absorbancy was limited to about 10:1 by mass (i.e. 1g of materiel could absorb 10g of water).  This metric, along with absorbancy by volume, are the properties of interest when designing a super absorbent polymer (SAP).
Subsequently, for several decades hydrogels dominated the SAP landscape. According to this Wikipedia article:

"Water-absorbing polymers, which are classified as hydrogels when
  cross-linked, absorb aqueous solutions through hydrogen bonding
  with water molecules. A SAP's ability to absorb water is a factor of
  the ionic concentration of the aqueous solution. In deionized and
  distilled water, a SAP may absorb 300 times its weight (from 30 to 60
  times its own volume) and can become up to 99.9% liquid...
  [...]
  Superabsorbent polymers are now commonly made from the polymerization
  of acrylic acid blended with sodium hydroxide in the presence of an
  initiator to form a poly-acrylic acid sodium salt (sometimes referred
  to as sodium polyacrylate). This polymer is the most common type of
  SAP made in the world today.
Other materials are also used to make a superabsorbent polymer, such
  as polyacrylamide copolymer, ethylene maleic anhydride copolymer,
  cross-linked carboxymethylcellulose, polyvinyl alcohol copolymers,
  cross-linked polyethylene oxide, and starch grafted copolymer of
  polyacrylonitrile to name a few.

There is also ongoing research into improving the ability of natural products to perform as SAP's.  According to the same Wikipedia article quoted above:

"Today´s research has proved the ability of natural materials, e.g.
  Polysaccharides and Proteins, to perform super absorbent properties in
  pure Water and saline solution (0,9%wt.) within the same range as
  synthetic polyacrylates do in current applications."

Summary, TL;DR:
Until the 1920's, polysaccharides were the prime choice for producing absorbent materials. Primarily due to their relatively low absorbancy (~10:1 by weight, and lost water when pressure was applied), they were subsequently replaced with hydrogels (absorbancies of 50:1 to 300:1, and maintained water when pressure was applied). However, by the 1960's, polysaccharide based super adsorbent polymers (SAPs) saw a resurgence in use as researchers (primarily at the US Department of Agriculture) were able to re-engineer polysacharide based SAPs to have absorbancies comparable to that of the synthetic hydrogels.
The qualities that make a SAP desirable are primarily it's absorbancy and ability to maintain absorbancy under pressure, but also cost and availability of materials. It is the latter of these that drives the research to develop better techniques for the use of polysacharide based SAP's.

Check this out: Today´s research has proved the ability of natural materials, e.g. Polysaccharides and Proteins, to perform super absorbent properties in pure Water and saline solution (0,9%wt.) within the same range as synthetic polyacrylates do in current applications. It's from the same article that you have mentioned.

Several natural hydrophilic polymers including polysaccharides (e.g. hyaluronic acid, alginate, chitosan, and cellulose), proteins (collagen, gelatin), and DNA etc. form a three-dimensional (3D) network that can retain a large amount of interstitial water, hence forming a naturally derived hydrogel.

That is interesting how they've been able to improve the SAP-like properties of the natural compounds. I haven't had any luck finding quantitative data, and the link they gave in the Wikipedia article seems odd, like it doesn't quite refer to the statement they made. I'll try to look for some other comparisons. Anyway, good find in that article I quoted that I didn't really appreciate at first glance!

@Abs , note that I appended my answer to include the very good point you made about the progress with natural products.

So, what is the conclusive answer to my question?

@Abs , note that I made a couple changes to the original text and added a new summary section. Please let me know if my answer is still unclear. Thx for the feedback.


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Why do Ca-Ca distances get shorter in ionic structures, as compared to the metal?
I saw in the book Chemistry of the Elements by Greenwood & Earnshaw, 2nd ed., p 66, that the distance between calcium atoms decrease when it's forming an ionic structure.

The closest $\ce{M-M}$ approach in these compounds [metal hydrides] is often less than for the metal itself: this should offer no surprise since this is a common feature of many compounds in which there is substantial separation of charge. For example, the shortest $\ce{Ca-Ca}$ interatomic distance is $\pu{393 pm}$ in calcium metal, $\pu{360 pm}$ in $\ce{CaH2}$, $\pu{380 pm}$ in $\ce{CaF2}$, and only $\pu{340 pm}$ in $\ce{CaO}$ (why?). [sic]

Why does this happen?

You understand the passage correctly. It is referring to Ca-Ca distances not Ca-X.

Simply put, the distance is measure from the center of the atoms. In the metal, the uncharged Ca atoms are kept at the van der waals contact distance.
In the ionic compounds the atomic radii of the Ca2+ ions have shrunk due to the loss of the outer shell electrons. Furthermore, negative ions cancel out the repelling charges and allow the Ca2+ ions to pack even closer together.


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Why does mutarotation occur only in specific media?
I was reading a book on biomolecules, and read that muta-rotation in hexoses, can happen only in media like water, which can act as an acid and as a base. Why is this so?
Essentially, it is just the formation of the open chain form, and then back to the other cyclic anomer, right? What role does the medium have in this process?

You need acid or base to convert the cyclic and open-chain forms. See formation of hemiacetal in any organic chemistry textbook, for example.

Essentially, it is just the formation of the open chain form, and then back to the other cyclic anomer, right?

Precisely, mutarotation involves interconversion of the α-and β-forms (cyclic anomers) of the monosaccharide with intermediate formation of the linear aldehyde (or ketone).

The mutarotation of glucose is a general acid- base-catalysed reaction which involves two proton transfers. Studies have proposed that these proton transfers are either carried out in a stepwise or concerted manner, with the latter alternative possibly incorporating several water molecules in a cyclic transition state.

Two possible mechanisms of mutatoration have been investigated:

The first involves an intramolecular proton transfer i.e acid-catalyzed reaction.
The second uses one solvent water molecule to assist proton transfer i.e water-catalyzed process.

There is some experimental  evidence1 for the solvent participation in the mutarotation reaction that appears from the studies of primary solvent effects on the rates of glucose and tetramethylglucose observed in $\ce{H2O + D2O}$ mixtures. In addition the absence of intramolecular proton transfer can be deduced from the small values found for the mutarotation rate of glucose in organic solvents.

Eigen1 proposed that the two proton transfers could proceed
  in aqueous solution via a pathway involving two or more water
  molecules in a cyclic hydrogen-bonded transition state, where the
  intramolecular proton transfer does not take place in aqueous
  solution.

Previous studies2 have shown that the reaction order with respect to water is zero for an acid-catalyzed reaction involving a stepwise mechanism, however in aqueous solution without added catalyst, water molecules must act as an acid or base and it is possible that the mutarotation reaction does proceed via a cyclic-concerted mechanism involving about three water molecules.

The results obtained in experimental work in the  newer
  studies1 have shown that mutarotation is a
  solvent-assisted process that obeys pseudo first-order kinetics. Besides that, the first water molecule from the solvent acts as a
  catalyst, and plays a decisive role, since it completely changes the
  mutarotation mechanism.

So a solvent is necessary, the evidence been seen  in proton transfers in  reactions that do occur via cyclic concerted mechanism as proposed in these studies.

References:

A theoretical study of glucose mutarotation in aqueous solution. A. M. Silva et al. / Carbohydrate Research 341 (2006) 1029-1040. Carbohydrate Research 341 (2006) 1029-1040.
The Mutarotation of Glucose in Dimethylsulfoxide and Water Mixtures. Ballash and Robertson CAN. J. CHEM VOL. 51. 1973.
CM Garrett, Grisham Biochemistry


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Can Isopropyl β-D-1-thiogalactopyranoside be degraded into either 1-deoxy-galactose or just galactose?
This may be a silly question, but is it feasible that Isopropyl β-D-1-thiogalactopyranoside (IPTG) when acted upon by a beta-galactosidase could produce 1-deoxy-galactose or just galactose?
The reason I am asking is because I have a crystal of a beta-galactosidase that was exposed to 5mM IPTG for ~6 months at 4 degrees C for inhibitor studies.I have X-ray crystallographic data (Beta-galactosidase + IPTG) to ~1.8 angstroms and can easily place the galactosyl ring into my electron density, however, I cannot place the sulfur atom or the isopropyl group.
This would normally indicate that the sulfur and isopropyl atoms are disordered, however, the possible conformations are sterically restricted to one direction.
As such I am wondering if IPTG could, in fact, have been turned into either 1-deoxy galactose or galactose?

I'm not sure if I understood your question correctly, and my thought is based on this assumed reaction:

Your indication of a resolution of about 1.8 Å reads like the data were collected by Cu K$\alpha$ radiation, rather than by Mo K$\alpha$.  If sulfur was split off from your molecule, than (even more with softer Cu than Mo X-rays) this should have decreased a lot the contribution of anomalous scattering in the X-ray data, as CHO's atomic scattering factors -- in comparsion -- are mainly described by $f'$, and not so much about $f''$. What about looking on value .and. esd of the Flack parameter of your model for your new product, compared to a "typical / reference" model of IPTG?
(By the way, Howard Flack passed away just at the beginning of this year.)

The data was collected at 1.000A at a synchrotron. I have succesfully phased the data by molecular replacement  and built a publication ready model. I do not have the required redundancy needed for anomalous diffraction maps. The reaction is what I had in mind.

@JeppeNielsen Your experiment was approved by the gate keepers, and now there could be a crystal-to-crystal reaction. So I assume, there is no resonable quantity of substance left (assuming Murphy) to eventually pursue further spectroscopy or GC-MS, or even to collect additional diffraction data of high brilliance again.  Ok, "true biochemists" should look at the question.  Still, with invested work & advanced X-ray data in hand, may I hint you to other experts on https://www.xrayforum.co.uk/ or Olex2's olexsys who may provide additional hints regarding "where is the sulphur gone"?  Good luck!


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Calculating purity from NMR spectrum
I just started organic chemistry in university. I have no idea what I'm doing — it's wonderful.
My assignment was to produce $\pu{10g}$ of fluorescein, the theoretical yield is $70~\%$ — I need at least $60~\%$ yield. I've made $\pu{11.07g}$ of product and my lab instructor analysed a sample of my product on the NMR spectrometer, which gave us the following result:

There are some impurities in my product (I know that f.e. $1.12$ is from the impurity and $3.42$ is from the fluorescein, same with $1.00$ and $3.46$) and I have no idea how to use that data to compute purity or actual yield.
Can somebody help?

Your question can't be answered without more information. We don't know what the side products are. The aryl region of the spectrum is not particularly helpful in any case...

@Zhe That's actually helpful. Would you mind telling me why this question can't be answered? That should help me when I'll turn my paper in tomorrow - maybe my lecturer didn't provide enough information to solve the task.

Impurities could also include reactants and intermediates. Be sure to review all the information about this exercise for clues. If you know what the impurities are and they have distinct NMR signals you can identify and integrate, you can calculate purity.

Beside proper identification and attribution of signals related to the product; remaining starting material(s), reagent(s), and intermediate(s); watch-out if there is evidence for the presence of remaining solvents from reaction / extraction / recrystallization. There should be no signal in the NMR without reasonable attribution.  Conversely, say you know ethyl acetate still is present in the sample, than all three expected signals should be consistently present in the spectrum (along $\delta$, in mulitplicity and relative integration).  The original question contains just too many unknowns.

For many substances (commonly used solvents, many starting materials and reagents), reference data are already published in journals, by chemical suppliers, data bases -- these are "ready for comparision".  If you just start in org chem, chances are that the products are literature known compounds, too.   No need for anger in the tune.

Identifying the purity of your substance is a non-trivial task, even for a seasoned chemist. Figuring out exactly what the other substances are and quantifying how much is even more difficult. You are in this latter camp, if you've provided all of the information you have. Because we don't know what impurities are present, we can't quantity the mass of material that is not your product in your final sample. Your best bet is actually to purify the sample and isolate only the desired product. Then the whole mass of the sample determines the experimental yield.

A couple of points for starters:

to say anything about impurities you need to show the entire spectrum, typically from $0$ to $\pu{10ppm}$ for $\ce{^1H}$. Merely describing the existence of peaks is not enough.
you also need to make sure your baseline is a proper baseline as you need to use the integrals to calculate purity if you even can. Here again, $n_\mathrm{t} = 8$ is absolutely too little.

However, it still needs to be said that determining the purity in percent just from NMR spectra is a very non-trivial task. It can only be done, if you know exactly which impurities you have, if they all contain the corresponding NMR-active nucleus, if the NMR-active nucleus is sufficiently predominant for the read to be quantitative and if there is no mechanism by which a non-NMR-active nucleus may be introduced into your impurity. Especially the predominance of NMR active nuclei along with another few requirements for ‘nice’ NMR spectra means that purity via NMR can in practice only be measure for $\ce{^1H, ^19F, ^31P}$ and a few much less common elements.
The main deal breaker is typically the requirement to contain the NMR-active nucleus. You can have a beautifully clean $\ce{^1H}$ NMR spectrum but the actual substance can be as little as $1~\%$ pure: most notably if the impurities are inorganic salts and do not contain any hydrogen. In most non-carbohydrate organic chemistry, this is not so much of a problem, though.
For any substances that contain the NMR-active nucleus and are impurities, the first step is to find out exactly what the substance is. To be able to calculate anything from an NMR spectrum, you need to know the impurity’s molar mass as NMR readings can only give you relative amounts.
Finally, you also need to make sure that your impurity does not lose its NMR-active nucleus. For example, if you are measuring in $\ce{CD3OD}$ and the impurity happens to be $\ce{C3F7OH}$ (1,1,1,2,3,3,3-heptafluoropropan-2-ol), the reading of the final remaining signal (if it even remains) will not be accurate due to proton/deuterium exchange:
$$\ce{C3F7OH + CD3OD <=>> C3F7OD + CD3OH}\tag{1}$$
(The equilibrium is shifted towards the products due to the abundance of methanol as the solvent.)
This is why NMR analysis is typically only used to determine relative purity, e.g. if two diastereomers form during a reaction. These will have the same molar masses allowing for an easy comparison.
To actually perform this, you need to:

Identify the peaks belonging to an impurity, the impurity’s structure and the signals’ integrals.
Normalise the impurity’s integrals to a single proton equivalent (i.e. if your impurity is cyclohexane, divide the integral by $12$ since you have $12$ magnetically equivalent protons.)
Normalise your desired compound’s integrals so that a single proton has the integral 1.
Determine a ratio of amounts in moles. (e.g. $\pu{8mol}:\pu{2mol}$)
Calculate the corresponding masses of desired product and impurity.
Add up the masses you calculated to a total mass.
Divide your desired product’s mass by the total mass to get your purity. Multiply by $100~\%$ to get a percentage.

From the information you give and the spectrum you show it is absolutely impossible to tell you anything about the purity of your compound. Signal integrals themselves are also not helpful; if your impurity is huge in molar mass but only has few protons, or if it has multiple magnetically identical protons a simple integral reading will be way off.

The practical approach for an organic chemist to determine purity is to use chromatography (either flash or HPLC) until no signals can be detected that do not belong to the compound in question and then label it ‘pure’.


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Do dipole dipole interactions need to be added explicitly to vdW forces while doing DFT simulations for organic molecules?
I  am modelling the adsorption of organic molecules on metal surfaces using plane-wave dft. While there is literature on how important van der Waals forces are for organic systems, and there are also some corrections to incorporate vdW forces into DFT (DFT-D, DFT-TS, and so on...); I was wondering if these corrections incorporate electrostatic dipole-dipole interactions within them (or do the dipole-dipole terms need to be added separately)?

(Permanent dipole)-(permanent dipole) interactions are covered by the electrostatics of DFT (and Hartree-Fock, for that matter, because no [Coulomb] electron correlation is involved). In my opinion, interactions between a permanent dipole (such as the one of HCl) and an instantaneous dipole (methane) are covered by the parametrization of dispersion-accounting methods in an averaging fashion.


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How does one compute the number of unique 2-electron integrals for a given basis set?
I have run into trouble with an exercise in Szabo and Ostlund's Modern Quantum Chemistry.

Exercise 3.14: Assume the basis functions are real and use the symmetry of the two-electron integrals [$(\mu\nu|\lambda\sigma)=(\nu\mu|\lambda\sigma)=(\nu\mu|\sigma\lambda)$, etc.] to show that for a basis set of size $K=100$ there are $12,753,775=O(K^4/8)$ unique two-electron integrals.

I have seen answers to this on the Physics SE here and here, but the case-by-case analysis used in the answers seems confusing to me and not easily generalized to other values of $K$. The symmetry of the integrals when all the basis functions are real suggests that there should be a simpler way of determining this. How can one find the number of unique two-electron integrals in this case and, ideally, in general?

To find the number of unique $2e^-$ integrals $\left<AB|CD\right>$, it is useful to first find the number of unique $1e^-$ integrals $\left<A|B\right>$.
If $n$ is the number of real basis functions, we can find the number of $1e^-$ integrals as ${n+1}\choose{2}$. This stems from the general formula for combinations with replacement: $${n+k-1}\choose{k}$$ where $n$ is the selection pool and $k$ is how many are being chosen. Combinations with replacement are needed to ensure we count the cases where the two Gaussians are the same.
As an aside, this is only true for real basis functions, since we can only say the order doesn't matter (what we do when we compute the number of combinations) because all the basis functions are explicitly real and thus the complex conjugate equals the original function. This explains one of my own questions on the site regarding why we don't commonly use complex basis functions.
For this particular case ($n=100$), we find that the number of unique $1e^-$ integrals is $${{101}\choose{2}}=\frac{(101)*(100)}{2}=5050$$
So how does this allow us to find the number of unique $2e^-$ integrals? Well, for the $1e^-$ integrals, all we did was find the number of unique pairs of real basis functions. So, for a $2e^-$ integral, we might expect that all we need to do is find the unique pairs of basis function pairs. This winds up being exactly the case (again, only for real basis functions).
To find the pairs of pairs, we need to again find the number of combinations with replacement, but instead of sampling from the $100$ basis functions, we are now sampling from the $5050$ basis function pairs. We again need replacement to account for the fact that we can pick the same pair twice. So, we obtain: $${{5051}\choose {2}} =\frac{(5051)*(5050)}{2}=12,753,775$$ as the number of unique $2e^-$ integrals. This method avoids the checking of specific cases (i.e none of the basis functions are the same, two are the same, etc.) and gives a general way of finding the number unique integrals for any amount of real basis functions.
To show explicitly why complex basis functions aren't used, we can look at the same problem, but assume all the functions are complex instead. In bra-ket notation, the bra is the complex conjugate of the function inside, so for $\left<A|B\right>$, the order would matter if $A$ and $B$ are complex. So finding the number of pairs requires finding the number of permutations rather than combinations. This only requires a subtle change of the original calculation $$2!{101\choose2}= (101)*(100)=10100$$ So, we now have twice as many unique pairs, so there are twice as many $1e^-$ integrals.
The equation for the $2e^-$ integrals remains the same, but with the new amount of unique pairs filled in. $${{10101}\choose {2}} =\frac{(10101)*(10100)}{2}=51,010,050$$ We still need to do a combination for the pairs because regardless of whether the functions are complex or real $\left<AB|CD\right>=\left<CD|AB\right>$ via exchange of electron labels. So using complex basis functions would double the number of $1e^-$ integrals and nearly quadruple the number of $2e^-$ integrals. So even if some feature of complex basis functions made them easier to integrate, it would have to make them easier enough to warrant doing and storing 2-4 times as many integrals.


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Help with idea for an optical pH sensor
I am experimenting at home with the idea of making an optical pH sensor that can read the colour of pH indicator using photodiodes.
I am wondering if there is some kind of ion permeable membrane that could be used to contain the indicator solution, so it can 'read' the pH of the solution being tested, without mixing with it.
I would like something that is fairly easily obtainable to me as a home experimenter.
The sensor could possibly be used make an automated pH corrector for my local aquaponics farm.
I looked into commercially available pH sensors. The glass bulb type and ISFET type suffer from drift and need regular calibration. There are optical types available, but they are too expensive for me.

In terms of a membrane material, did you think about something like Nafion (https://en.wikipedia.org/wiki/Nafion)?  This polymer already finds deployment as membrane material for fuel cells, for example.

Ok, I might give it a go. I will need to work out a way to make a sealed pocket to contain the ph indicator. Maybe clamp the Nafion between solid frames.

This review article might give you some ideas: D. Wencel, T. Abel, C. McDonagh, "Optical Chemical pH Sensors", Analytical Chemistry, 2014 86 15-29.

I really like the concept but the problem is to find a membrane that is permeable to water and H3O+ ions but impermeable in respect to the indicator.
I understand you need something as in the picture below.
Dialysis tubing are commonly used to prevent bigger molecules from crossing the membrane, but unfortunately the typical lower limit is for molecules of around 1 KDa so much more than a pH indicator molecule.
 The only option available seems reverse osmosis membranes, for example used in water purification filters. However, the resulted filtered water has an altered pH (ref.1) therefore not suitable.
Finding a membrane that is permeable to water and H3O+ ions but not to small polar molecules (i.e.Phenolphthalein, blue bromothymol, Anthocyanins etc.) is something difficult to achieve especially if you have also the requirement of not altering the pH.
I think that is the reason why we still don't have a pH meter that works as you described.

References:
1)Nir, Oded & Bishop, Noga & Lahav, Ori & Freger, Viatcheslav. (2015). Modeling pH Variation in Reverse Osmosis. Water Research. 87. 328–335. 10.1016/j.watres.2015.09.038.


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How to calculate the number of atoms of each element in Mg(NO3)2?
I'm working on chemistry and I'm not quite sure how to solve this question? I haven't worked with moles before and whenever I try to figure out how to do these types of question I only end up confusing myself more.

How many Mg atoms are in 5.7892 mol of $\ce{Mg(NO3)2}$?
How many N atoms does the sample contain?
How many O atoms does the sample contain?

I haven't any clue how to do this so a walkthrough on how to solve these types of problem would be nice.

$\ce{Mg(NO3)2}$ has (Figure 1):

One $\ce{Mg}$ atom
Two $\ce{NO3}$ groups, which are composed of:

One $\ce{N}$ atom
Three $\ce{O}$ atoms

Figure 1: we have one $\ce{Mg}$, two $\ce{N}$ ($2 \times 1$) and six $\ce{O}$ ($2 \times 3$) in $\ce{Mg(NO3)2}$.
Now all you need is to know that 1 mole equals $6.022\,140\,76\times 10^{23} \approx 6.022 \times 10^{23}$ particles.
Just multiply:

How many Mg atoms are in 5.7892 mol of $\ce{Mg(NO3)2}$?

$$1 \times 5.7892 \text{ mol $\ce{Mg}$} = 1 \times 5.7892 \times 6.022 \times 10^{23} \text{ atoms $\ce{Mg}$} \\= 3.48625624 \times 10^{25} \text{ atoms $\ce{Mg}$}$$
And so on.

Your equation is not correct since there are different dimensions on both sides of the equal sign (on the left amount of substance in mol, on the right a dimensionless number). It would be better if you use the Avogadro constant instead of the Avogadro number four your calculation.


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What does "oxidized form" and "reduced form" mean?
What does the phrase "oxidized and reduced forms of the reactants" mean?
For some context: The book was talking about half-reactions in a half-cell of a galvanic cell, and it said this,

Sometimes, both oxidized and reduced forms of the reactants in a half-cell are soluble and cannot be used as an electrode.

I am confused as to what the "oxidized and reduced forms of the reactants in a half-cell" might mean because in a half-cell either only oxidation or reduction takes place.

In this case, the adjective "reduced" does not imply something is actually being reduced. It simply refers to the form of the element in question which has the lower oxidation state. Let's say something is getting oxidised in a half-reaction, for example
$$\ce{Fe^2+ -> Fe^3+ + e-}$$
Iron takes two forms here, $\ce{Fe^2+}$ and $\ce{Fe^3+}$. $\ce{Fe^3+}$ is the oxidised form, that much is obvious, right? Then, $\ce{Fe^2+}$ is the reduced form.
No $\ce{Fe^3+}$ is actually being reduced to $\ce{Fe^2+}$. However, it is still correct to refer to $\ce{Fe^2+}$ as the "reduced form", since it is the form of iron with the lower oxidation state.


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Enthalpy in adiabatic process

A gas adiabatically expanded from $\pu{32atm}$ and $\pu{273K}$ to $\pu{1atm}$ and $\pu{251K}$, Calculate Joule-Thomson coefficient $\mu$ at $\pu{273K}$.

Answer:
$$\mu = {\Delta T \over \Delta p} = {-22 \over -31} = 0.7$$

I think the answer uses the definition of $\mu$ that is $\displaystyle\left({\partial T \over \partial p}\right)_H$, but this definition assumes the process to have constant enthalpy.
But in adiabatic process $dH = dU + Vdp + w_{ad}$, so for $dH = 0$ we need to have $dU + Vdp + w_{ad} = 0$ which is not possible since pressure is clearly changing and so $Vdp$ is not zero.
My questions are

Is $dH = 0$ for a adiabatic and I am missing something ?
If not then how did the author used $\displaystyle\left({\partial T \over \partial p}\right)_H$ to get the answer ?

Do you not understand how adiabatic flow through a porous plug or valve, from high pressure upstream to low pressure downstream, gives no change for the enthalpy change per mole?  Are you aware of the equation:  $$dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP$$

No I only know $\displaystyle dH = -\mu C_pdp + CpdT$.

If you combine these two equations, you get a relationship for calculating $\mu$ from knowledge of the heat capacity and the equation of state.

Ok I subtract my equation from our equation,  $$0 = \left[V-T\left(\frac{\partial V}{\partial T}\right)_P +\mu C_p\right]dP$$
How can I get rid of $C_p$ ?

You can't.  But, in any event, you did the algebra incorrectly.  What happened to the dT.

$$dH - dH=C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP -  ( -\mu C_pdp + C_pdT) =C_pdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP +\mu C_pdP - C_pdT  = \left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP +\mu C_pdP = 0$$
can't I just cancel $dT$ ?

You want to be solving for dT/dP at constant H.

Since $H$ is constant I get $\displaystyle 0 = -\mu C_pdP + C_pdT$ or  $\displaystyle {dT\over dP } = {C_p\mu\over C_p  } = \mu$. But I did not use your equation ?

$$\mu=-\frac{\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]}{C_p}$$

@ChesterMiller But how is that useful ? since I don't know $V$ and $C_p$.

If you know the equation of state of the gas (P,V,T) and its heat capacity at constant pressure (which is also often known), you can predict in advance the JT coefficient.  But, that wasn't your question, and I was trying to address your question.  Regarding how he got his answer:  He just measured $\Delta T$ and $\Delta P$, and thus determined $\mu$ experimentally.

In this problem, calling $\Delta T/\Delta P$ the Joule-Thompson coefficient really isn't proper for such a large change in P .  The Joule-Thompson coefficient $\mu$ is, strictly speaking, defined only in terms of an infinitesimal difference in pressure between the upstream and downstream sides of the porous plug (and a corresponding infinitesimal change in the real gas temperature).  A change from 32 atm to 1 atm does not really qualify.  At best, the data provided would give a very crude approximation to $\mu$.  And the equations that we have given in terms of Cp and the EOS (continued)

(continued) would not give an accurate prediction of $\Delta T/\Delta P$ for a typical real gas.

The Joule-Thomson experiments occurs with no change in enthalpy.
Suppose that at the left of a porous plug there is a pressure $p_1$ and temperature $T_1$ and $p_2,T_2$ to the right of the plug, as $p_1>p_2$ the gas moves left to right. The experimental configuration must ensure that pressures remain constant and that the experiment is performed under adiabatic conditions when $q=0$.
If a volume $V_1$ of gas moves from the left to right the work done/mole is $W=p_1V_1-p_2V_2$. This is the difference between the work of compression on the left of the plug and work recovered on expansion on the right. If the gas were ideal then $w=0$, but real gases are not. The gas expansion is also adiabatic so that no heat leaves or enters then $q=0$ and the change in internal energy $\Delta U$  is equal to the net work $$\Delta U =U_2-U_1 = p_1V_1-p_2V_2$$ therefore $$U_2+p_2V_2=U_1 + p_1V_1$$
As $H=U+pV$, then  $$\Delta H = H_2-H_1=U_2 +p_2V_2-U_1 -p_1V_1 =0$$
The Joule-Thompson  coefficient $\mu$ is defined, as you write, $\left ( \partial T/\partial P \right)_H$ and this measures how much the intermolecular interactions make the gas differ from an perfect gas. Most gases cool when passing from high to low pressures at room temperature.
Notes:
The coefficient can be rewritten in other forms using
$$ \left ( \frac{\partial T}{\partial p} \right)_H \left ( \frac{\partial H}{\partial T} \right)_p \left ( \frac{\partial p}{\partial H} \right)_T =-1$$
then
$$ \mu C_p= -\left (\frac{\partial H}{\partial p}  \right)_T $$
as
$$ \frac{dH}{dP} = C_p\frac{dT}{dp} +V-T\left (\frac{\partial V}{\partial T}  \right)_p$$
then if $\alpha=(1/V)(\partial V/\partial T)_p$ is the coefficient of expansion then at constant T, $(\partial H/\partial p)_T= V(1-\alpha T)$ which means that the Joule-Thompson coefficient can be written as $\mu=(V/C_p)(\alpha T-1)$ (This last equation has been used as a way of measuring absolute temperature because $V,\mu$ and $ \alpha $ are all measurable quantities.

Most gases cool when passing from high to low pressures at room temperature?  Not according to Wikipedia:  https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect

perhaps 'most' was loose, 'many' would be better, or best only  those with positive coefficients at the prevailing temperature will cool, i.e. $\alpha T>1$

@porphyrin What is the $p$ in the partial derivative? $p_1$ or $p_2$? Why for ideal gas $w=0$? Thanks in advance.

@adosar corrected typo in $\Delta H$  eqn. The last equations are general for any pressure.

@porphyrin So we can measure either the change of $T_1$ with respect to $p_1$ or $T_2$ with respect to $p_2$?

@ChetMiller  I believe that Poutin is correct for most real-world applications, i.e., those in which p_final ~ 1 atm, and p_initial is not more than ~ a few hundred atm.  In that regime, the only gases I know of which heat during a JT expansion at room temp. are hydrogen, helium, and neon. Do you know of any others? The interesting question would be what happens when the initital pressure is so high that it puts, say, N2 into a heating regime at room T -- then as its p decreased it would first heat and then cool; you could test or calc. how high p_initial would need to be for heating to win.

@theorist Why don’t you start a new thread so that we can analyze this in greater detail?

@ChetMiller That's a nice idea, but in posing the question I'd first have to see if I could derive an expression for the pressure-depence of the JT coefficient at v. high p for a specific gas. [Once I had that, I could integrate to get $\Delta T$.] I'd also need to account for a possible transition from gas to liquid or SC fluid at v. high p, which would in turn require a different expression be obtained. [Plus I'd need to think about the effect of a phase transition within a JT expansion upon $\Delta T$!] All of this requires time, so it will need to wait until I have a chance...

@theorist I think all this is doable, depending on the kind of data one has available, including large pressure drops (not just differential pressure drops).


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Electrolysis of AgNo3
I have a solution of $\ce{AgNO_3}$ (is diluted with water). I electrolyse it. What products from at anode and cathode.

You would get silver at the cathode, and oxygen at the anode.
Since silver is below hydrogen in the spectrochemical series,it tends to get reduced over hydrogen, similarly, since $\ce{NO3-}$ is above $\ce{OH-}$ , it tends not to get oxidised, therefore giving oxygen.
$\ce{Ag+ + e- -> Ag}$ (at cathode)
$\ce{ 4OH- -> O2 + 2H2O + 4e-}$ (at anode)

But isnt the Voltage for NO3-  (+0.96) which is above Ag (+0.80)?? @Supernova. there fore NO3- will reduce. can u give the acutal voltage of the half reactions if i got it wrong.

lets just put it like this, you go to a mall, and do some shopping with your dad. Now you see Bill gates(who is in a mood to help out anyone who asks him) there. You shop all you want, but see that you forgot your wallet, and then at the end, you approach him, and he says,"I'll take care of the bill, you can leave if you want". Now consider a different scenario, where you see a rather poor dude standing there while you wereshopping. Now if you request him to pay for you, that would put him in a much more unstable state, that what Bill gates wouldve faced

This is somewhat similar to what happens when you have both $\ce{OH-}$ and $\ce{NO3-}$ in the medium. With $\ce{Ag}$ at the other side @KierenAlbuquerque

@Supernova, marvelous analogy


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What i the reaction of an alkene with cold, dilute, neutral potassium permanganate?
How will an alkene react with dilute, cold, neutral $\ce{KMnO4}$?

@YashasSamaga We prefer to not use MathJax in the title field, see here for details.

The behaviour of $\ce{KMnO4}$ in a neutral medium is nearly the same as its behaviour in slightly alkaline medium.
In alkaline medium:
$\ce{3R-CH=CH-R + 2MnO4- + 2OH- -> R-CH(OH)-CH(OH)-R + MnO4^2-}$
In neutral medium:
$\ce{3R-CH=CH-R + 2MnO4- + 4H2O -> R-CH(OH)-CH(OH)-R + MnO2 + 2OH-}$

Wow, interesting. I wasn't expecting that. Thank you for your help!


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Geometrical Isomerism (Cis-Trans) in trans-2-fluoro-3-methylpent-2-ene
trans-2-fluoro-3-methylpent-2-ene this compound in also known as (Z)2-fluoro-3-methylpent-2-ene.
I have given my possible structure in the image but it is wrong, Can anyone explain this ?

Just a guess but since the fluro and methyl are the named sidegroups, shouldn't the trans apply to their orientation to one another?

@bpedit it might be true, but it doesn't seem convincing enough. My general perception about trans structure remains that it should be symmetrical in some way. If they actually were side groups then how many more typed of side groups are there in organic ?

The implication of my suggestion is that the "-2-fluoro" should be on the bottom in your diagram so it is trans to the "-3-methyl" group. I'm no expert on the rules but it appears to make sense,

@bpedit I got your point, but why are we doing that shouldn't same groups be on opposite side in trans structures if the two same type of groups are present, isnt that the basic definition ?

The structure you drew

is ‒ according to the CIP-rules ‒ (E)-configured.  These rules are applied on each side of the double bond; namely

C of the ethyl group has a higher priority than C of the methyl group
F has a higher priority than C
C(ethyl group) opposes F, as seen across the double bond

This is not in contradiction that the two methyl groups are opposing each other, are in (trans)-relationship with each other.
In instances with only two subsitutents around a double bond, like (E)-2-pentene, you may use (E), derived from German "entgegen", literally "opposing", as synonym to (cis).  For the opposite instance, (Z), derived from German "zusammen", like "together", to describe a (trans)-configuration.
In instances with three or four substitutions around a double bond, and instances of several conjugated double bonds, follow IUPAC and adhere to the (E/Z) notation; which is absolute and not ‒ as (cis/trans) ‒ relative and potentially ambiguous.  See the example E/Z notation, for example. (Or, as one teacher told me, "stay on the E-Z pass".)

I know that when 3-4 subsitutions around a double are present, we follow CIP rules. But suppose this compound was given to you as (Z) confirguration and asked whether it will trans or cis then how will you account for that ?

@Aviral Agarwal The system of (cis)/(trans) works differently as the more general absolute, based on CIP rules.  Hence there might be confusions if applying the former in the instance of three / four times substituted alkenes (see the example in the link).  While cis corresponds to Z for simple cases of twice substituted dienes, this is no longer every time true for three and four times substituted dienes.  IUPAC even discourages deployment of (cis)/(trans), as in the Gold book: http://goldbook.iupac.org/html/C/C01092.html


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Balancing complex haloform redox reaction
I just completed a haloform reaction experiment, using acetophenone and sodium hypochlorite to form benzoic acid and chloroform.
My lab report has a question regarding balancing the equation of the reaction.  I have been working on it for quite a while, and I can't figure out how to do it.  I've tried looking up methods but too many atoms are reliant on too many molecules!  Can someone help with this question and explain how to do it so I can really understand this process?

I need to be able to identify the limiting reactant as well, which relies on the molar ratio of the reactant to the product, so I really need to know how to do this.

This question is related, as it describes how to balance reactions in general without guesswork. I'm partial to my own answer involving matrices, but all the methods given as answers would work.

just go by the mechanism, youll have every reactant over there, s those numbers are done, then just get the product numbers, in accordance with reactants

As you noted, you need to find the limiting reactant. This is most of the time done by taking the one with a 1:1 ratio of educt:product.
So in this case, the $\ce{ROCH_3}$. So you take that once. Now you see immediately that you need 2 $\ce{Na}$ on the right side, so you can put a 2 in front of the $\ce{NaOH}$. You have two $\ce{X}$ on the left, but 4 on the right, so put a 2 in front of the left one. Now count all your atoms. Subtract them from each other:
You have $a * \ce{H}$ on the left and $b * \ce{H}$ on the right. if $a-b=0$ its balanced. Do that for every reactant and you're done.
There are of course more strictly formulated methods, but in my time I noticed that a straightforward approach with a little thinking can be applied most universally without having to memorize some formulas.

The coefficients should be
1 3 4 1 1 3 3, from left to right. The limiting reagent is acetophenone.
Used this website to find the answer.
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch18/ch18-3-2b.html
-- a fellow owl ;)

It would likely help the OP more to have a method of finding an answer rather than just being told what it is.