text:
Would the carbocation intermediate undergo rearrangement?
I have my concerns regarding option (a) in this question.

During the dehydration step, wouldn't the resulting carbocation intermediate formed in option (a) rearrange?
That is, the ring should contract from a $7$-carbon to a $6$-carbon one because the resulting carbocation would be more stable. But the resulting product would have to be an addition product since elimination now becomes impossible.
I know that this situation can arise because I have tackled it before and the answer was an addition product in that case, but that reasoning contradicts with the official answer key in this question. (The answers are a,b and d).
The shift I called for was the bond between carbon (endocylic with nitrogen) and the carbocation.

@Chemstack The issue here is that the carbocation after the shift is not stabilized. The nitrogen is fully protonated and is thus just an electron withdrawing group attached to a carbocation.

@Zhe can you please explain it a bit more perhaps write an answer.Thanks.

@Pink I'm not sure there's anything else to say. If the nitrogen is fully methylated, it's just a de-stabilized group to have next to a cation. If you're confused, it's probably more educational for you to draw out the mechanism for the rearrangement and look at the intermediate instead of my doing it.

Unfortunately, I cannot back this with references at the moment but in my opinion there should be no carbocation formed in the first place and therefore no tendency to undergo any Wagner-Meerwein rearrangements.
Secondary carbocations, while being less unstable than primary ones, are still rather unstable. There are no stabilisation mechanisms that suggest themselves in the reaction conditions (concentrated sulphuric acid) in which the nitrogen will be protonated and therefore unable to stabilise a carbocation in α position. I would assume that the elimination proceeds much more $\mathrm{E2}$-like: likely with no intermediate carbocation at all but if there is one it should be extremely short lived and the neighbouring proton should dissociate practically instantly.

After 3 years perhaps this post requires a fresh look. An Olympiad question is unlikely to want to bog down a participant hunting for myriad rearrangements. Rather than obsess over rearrangements, perhaps one should start at the end of the problem where 1,3,5-cycloheptatriene (6) is the final product. The presence of a cycloheptane ring is not available in structures c and e. However, the remaining structures a, b and d do contain cycloheptane rings and are all candidates as structure C. What is required is the introduction of three double bonds, one by dehydration and two by successive Hofmann eliminations.
Achiral 1 is formed from a upon dehydration while rac-2 arises from b. Compound d leads to two possible alkenes, achiral 3 and rac-4. Bicyclic unsaturated amines 1 - 4 (structures J) are all candidates for successive Hofmann exhaustive eliminations. continued

Historical Note: One of the epimers of alcohol b is tropine, one of the components of atropine. The dehydration of b to alkene 2 and it subsequent oxidation leads to tropinic acid 6 (R=H). Its ester proved to be critical to the young career of Richard Wilstätter. When he performed a single exhaustive methylation on the ester, an ether soluble amine was obtained leading to the conclusion that tropine is a bridged amine. The text below is from his autobiography, From My Life, available through Amazon for the Kindle or online at the Hathi Trust, if you have access. A must read for both aspiring and accomplished chemists.

This does not address OP's concern..

True! A learning experience.

There are multiple possibilities for this reaction, elimination may not come first,  Wagner-Meerwein rearrangements may occur before this.  For both A and B, rearrangement occurs before elimination.  For D it is straightforward, the -OH will be eliminated and the corresponding alkene will be formed.  For B, the -OH will first be protonated, the carbon nitrogen bond of the bridgehead carbon farther away will be broken.  The nitrogen atom will then form a bond with the carbon having the -OH2 substituent, the loss of water will follow.  This generates D, but a carbocation is in place of the -OH group.  This compound can undergo elimination to generate the same alkene as D generated.  Your question is pertaining to the lack of rearrangement of the carbocation.  It cannot rearrange to one of the tertiary bridgehead carbons, because of Bredt's rule.  So when heating in concentrated sulfuric acid, the major elimination product will be the alkene mentioned previously.  While B and D proceed through this alkene, A proceeds through a Wagner-Meerwein rearrangement similar to B.  This yields a carbocation.  This carbocation will rearrange (but not in the context you had) because of angle strain on the adjacent carbon.  The rearrangement increases stability and elimination then occurs to yield the same product as B and D.  This then undergoes transformation to cycloheptatriene under the conditions provided.

This is not correct because compound (a) does not undergo rearrangement. It also does not really answer the OP's question...

Straightforward elimination could occur, but that does not rule out the possibility of Wagner-Meerwein rearrangements.  In more complex orgo problems, there are numerous possibilities.  There is no way to prove either of us right or wrong without experimental evidence.

It absolutely rules out rearrangement because it would put the carbocation next to a fully substituted nitrogen.

You do not understand the mechanism of a Wagner-Meerwein rearrangement, the resulting carbocation is beta to the nitrogen, and the degree of substitution on the nitrogen wouldn't rule out the possibility of a rearrangement.

Yes, that's the whole point. The nitrogen would ordinarily stabilize a cation to form an iminium but it can't if it's fully protonated from sulfuric acid. You do not place an fully substituted, electron withdrawing nitrogen next to carbocation for a rearrangement. That rearrangement will not occur. And if you disagree, then we're not talking about the same thing. I suggest reading the OP's question again.

In this scenario, there are two different Meerwein rearrangements (contracting the ring vs. moving the bridgehead) that can take place, he mentioned one of them in the question, I mentioned the other in my response.  Both are possible, and due to the answers being what they are on the answer key, I conclude that his rearrangement won't occur and rather the one I suggested will.

@AS_1000 I agree with zhe wagner would have occured before elimination only if nirogen was not already protonated by conc. H2SO4. Infact it get's protonated before alcohol group


text:
What is the origin of the names P-1 and P-2 of nickel boride catalysts?
From Chapter 2 of Catalytic Hydrogenation in Organic Syntheses:

Highly stereospecific hydrogenations of acetylenes to cis-olefins can be achieved with a nickel (P2) catalyst, prepared by sodium borohydride reduction of nickel acetate in ethanol, and ethylenediamine as a promoter.

The Wikipedia article on nickel boride describes two forms, P-1 and P-2, differing in their ratio of $\ce{Ni2B}$ to $\ce{NaBO2}$.
What is the origin of the names "P1" and "P2"? Does it have to do with whether the catalysts add hydrogen in a syn or anti fashion?

Have a look at Wikipedia, it's right there: https://en.wikipedia.org/wiki/Nickel_boride#Preparation

The meaning of ''p -2" , the reason behind this terminology is what I wanted to know.

I'm speculating things: the only google books reference I'm able to find is this (it's not letting me preview the entire book). Notice closely, Raney Nickel has been called W2, and a footnote at this wikipedia article tells me that's a grade of nickel. So, perhaps P2 is also a grade?

The actual differences between the two forms are well-documented on Wikipedia. As for the names: as far as I can tell, the choice of "1" and "2" seems to be arbitrary. The choice of the letter "P", though, is described by C. A. Brown (who carried out investigations into the synthesis and reactivity of these catalysts).1,2 In footnote 13 of reference 1, Brown writes:

The designations are derived from Purdue University, where this work was initiated.

So, it is within reason to assume that the P stands for Purdue.

References

Brown, C. A. Catalytic hydrogenation. V. Reaction of sodium borohydride with aqueous nickel salts. P-1 nickel boride, a convenient, highly active nickel hydrogenation catalyst. J. Org. Chem. 1970, 35 (6), 1900–1904. DOI: 10.1021/jo00831a039.
Brown, C. A.; Ahuja, V. K. Catalytic hydrogenation. VI. Reaction of sodium borohydride with nickel salts in ethanol solution. P-2 Nickel, a highly convenient, new, selective hydrogenation catalyst with great sensitivity to substrate structure. J. Org. Chem. 1973, 38 (12), 2226–2230. DOI: 10.1021/jo00952a024.

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text:
Are glucose and galactose cis-trans isomers of each other?
Glucose and galactose are diastereomers of each other. However, is it correct to say that they are cis-trans isomers of each other? Does it make a difference with regard to terminology if glucose and galactose are in cyclic form i.e. are glucopyranose and galactopyranose cis-trans-isomers of each other? And is it correct to say that glucopyranose has $2^5-1=31$ cis-trans isomers?
I would also appreciate some references to verify the correct use of terminology in this situation.
NOTE: I have edited this question to clarify that I'm interested in the cyclic form as well.

Since the question was revised, I have revised the answer as well to include more information:

Part A
Glucose and galactose are not cis isomers, but diastereomers called epimers,
Two sugars that differ only in the configuration around one carbon atom are called epimers; D-glucose and D-mannose, which differ only in the stereochemistry at C-2, are epimers, as are D-glucose and D-galactose (which differ at C-4):

In other words they are nonsuperimposable, nonmirror-image stereoisomers.
The explanation below gives some insight on what forms of isomerism such sugars exhibit:
The predominant form of stereo-isomerism in such sugars as monosaccharides is optical isomerism.
Simple substances which show optical isomerism exist as two isomers known as enantiomers.
A solution of one enantiomer rotates the plane of polarisation in a clockwise direction. This enantiomer is known as the (+) form.
A solution of the other enantiomer rotates the plane of polarisation in an anti-clockwise direction. This enantiomer is known as the (-) form.
For monosaccharides with two or more asymmetric carbons, the prefix D or L refers to the configuration of the highest numbered asymmetric carbon (the asymmetric carbon farthest from the carbonyl carbon). A monosaccharide is designated D if the hydroxyl group on the highest numbered asymmetric carbon is drawn to the right in a Fischer projection, as in D-glyceraldehyde.
The stereochemistry and configurations for the aldohexoses  (monosaccharides) was put forward by Emil Fischer  in 1896. According to the Fischer convention,

D sugars have the same absolute configuration at the asymmetric centre
  farthest removed from their carbonyl group as does D-glyceraldehyde. The L
  sugars, in accordance with this convention, are mirror images of their D counterparts

Note that the designation D or L merely relates the configuration of a given molecule to that of glyceraldehyde and does not specify the sign of rotation of plane-polarized light.

If the sign of optical rotation is to be specified in the name, the
  convention of D or L designations may be used along with a + (plus) or
  - (minus) sign. Thus, D-glucose may also be called D (+)-glucose because it is dextrorotatory, whereas D-fructose , which is
  levorotatory, can also be named D(-)-fructose.

The relationship between the enantiomers
One of the enantiomers is simply a non-superimposable mirror image of the other one.
A chiral (asymmetric) carbon atom is the usual source of optical isomerism, as was the case with amino acids. The simplest carbohydrate that contains a chiral carbon is glyceraldehyde, which can exist in two isomeric forms that are mirror images of each other.

Part B
In general, a molecule with n chiral centers can have $2^1$ stereoisomers. Glyceraldehyde has $2^1=2$; the aldohexoses, with four chiral centers, have $2^4=16$ stereoisomers. Thus
In general, n-carbon aldoses have  $2^{n-2}$ stereoisomers.
Certain conventions are used for two-dimensional drawings of the three-dimensional
structures of stereoisomers. The dashed wedges represent bonds directed away
from the viewer, below the plane of the paper, and the solid wedges represent
bonds directed oppositely, toward the viewer and out of the plane of the paper.

Part C
The same is true for cyclic forms of sugars (not cis-trans isomerism):
The cyclization of a monosaccharide renders the former carbonyl carbon asymmetric. The resulting pair of diastereomers are known as anomers and the hemiacetal or hemiketal carbon is referred to as the
anomeric carbon. In the α anomer, the OH substituent to the anomeric carbon is on the opposite side of the sugar ring from the CH2OH group at the chiral center that designates the D or L configuration (C5 in hexoses). The other anomer is known as the β form;

References:

Lehinnger biochemistry
Voet and Voet biochemistry
Biochemistry Grisham
Biochemistry Campbell and Farrel

Glucose and Galactose are best described as epimers of one another.
The three structures below all have the same molecular formula and the same connectivity, but differ in the absolute configuration at a single stereocentre (this is the definition of an epimer).
You can see that glucose and galactose are epimeric at the $\mathrm{C4}$ position, whilst glucose and mannose are epimeric at the $\mathrm{C2}$ position.

Image taken from Principles of Biochemistry, Lehninger et. al. (5 ed.)

To comment on the use of cis and trans.
The cis/trans nomenclature is used to describe the geometric isomerism of alkenes - it is not correct nor appropriate to use it in this context.  What you may be referring to is whether two adjacent stereo centres are syn or anti to one another.

What about ring-structured glucose and galactose? Would it be correct to say that they are cis-trans isomers of each other. I have understood that if a ring has two substituents on the same side of the ring plane, it's called a trans isomer (eg. trans-1,2-dichlorocyclohexane) and if they are on different sides, it's called a cis isomer (eg. cis-1,2-dichlorocyclohexane). Isn't that exactly the difference between glucopyranose and galactopyranose?

They're still just epimers. You can only use CIS TRANS for rings when you're talking about precisely two groups and their relationships

@NotBaran It's probably worth putting a discussion of the (valid) ring usage of cis/trans in the body of the answer, especially as sugars do have a ring form. (And cis/trans is used in connection with carbohydrates, though not for the glucose/galactose distinction.)

A slight misconception. According to http://www.chemguide.co.uk/basicorg/isomerism/geometric.html, the definition of geometrical isomers is (emphasis mine):

These isomers occur where you have restricted rotation somewhere in a molecule. At an introductory level in organic chemistry, examples usually just involve the carbon-carbon double bond - $\cdots$

So basically classifying glucose and galactose as geometrical stereoisomers by using terms such as cis and trans is wrong in this context.
They're stereroisomers no doubt, but calling them diastereomers would be a better choice (i.e., try taking a mirror image of glucose and you're not going to get a structure which matches with that of galactose - this makes them non-mirror image isomers, better called as diastereomers; see also Addendum at the bottom of this answer). Terminologies such as R/S nomenclatures would be more appropriate in this case.

And is it correct to say that glucose has $2^5-1=31$ cis-trans isomers?

Nope, firstly because of the usage of the word cis-trans.

Considering the chain structure of glucose:
Another thing is that when you look closely at the structure of glucose which is, you'll notice that it has only $4$ chiral carbon atoms. Chiral atoms are atoms which have four different groups/atoms attached to them. I'm marking the chiral atoms with red stars for you:

The reason I've left out the lowermost $\ce{C}$ atom is that I've two same groups (namely $\ce{H}$). Now calculating, the number of isomers you've:
$$2^4=16$$
You might want to exclude glucose itself with gives you the answer as $15$.
Considering the ring form:
If you're talking about the ring (glucapyranose structures) then it's true that $\alpha$ and $\beta$ forms will double the number and therefore you'll end up with $32$ isomers. Excluding glucose, again the number lessens by unit to give us the final answer as $31$

Addendum:
Quoting a wikipedia article (https://en.wikipedia.org/wiki/Stereoisomerism):

Diastereomers are stereoisomers not related through a reflection operation. They are not mirror images of each other. These include meso compounds, cis–trans (E-Z) isomers, and non-enantiomeric optical isomers.

The case we're dealing with (in context of glucose and galactose) is of non-enantiomeric optical isomers.

When talking about $2^5-1$ isomers, I was actually thinking about the glucopyranose. Ring-structured glucose has alpha and beta forms which doubles the number of isomers.

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text:
Potassium hydroxide neutralization with acetic acid
I put 9% acetic acid to 60% potassium hydroxide solution to neutralize it and was hoping to see some fizzing, but didn't see any indication of them reacting. Why is that?

People often neutralise acetic acid with a weak base such as sodium hydrogen carbonate, which liberates carbon dioxide gas as the reaction proceeds in addition to the other neutralisation products.

The reaction for this would be: $$\ce{CH3COOH + KOH -> CH3COO^{-}K+ + H2O}$$ There are no gaseous products, so you can't really expect any fizzing.

As stated in Pritt's comment, the reaction you are performing is:
$$\ce{CH3COOH + KOH -> CH3COO^−K+ + H2O}$$
Note that none of the products are gases, so there is no effervescence (fizzing). If you want a neutralization reaction that will effervesce, then you need to form a gas. A simple candidate for this would be sodium bicarbonate (baking soda), which would give the following reaction:
$$\ce{CH3COOH + NaHCO3 -> CH3COO^−Na+ + H2O + CO2 ^}$$
With this we are forming $\ce{CO2}$ gas, which will cause the solution to effervesce. Note that this is likely to be stinky and messy: a good experiment to perform outdoors.

Acetic acid is a too weak acid to neutralise KOH.


text:
Which calcium salt is good enough to make calcium benzoate?
This question is inspired from the following questions:

Reaction of citric acid and calcium chloride
Precipitation of citric acid by CaCl2 and Ca(OH)2?

The answer to the 1st question says that calcium citrate is highly unlikely to form from calcium chloride and citric acid because of its inability to be present in deprotonated form and will not precipitate in acidic medium. 2nd question also addresses a similar kind of problem. But what about calcium benzoate?
Which calcium salt can succesfully precipitate calcium benzoate from benzoic acid? Would calcium hydroxide work for that? I am pretty sure calcium chloride won't work because it is a good dehydrating agent and would rip off water molecules to form something else. Calcium benzoate is also somewhat more soluble in water than calcium citrate. So, what should be the necessary measure to successfully precipitate calcium benzoate?

I think calcium hydroxide would be sufficient to form calcium benzoate solution, but the solution must be heated to increase the solubility of calcium hydroxide and benzoic acid. Another method is to dissolve benzoic acid in NaOH solution(in equimolar amounts) and then react Ca(NO3)2 with it.

Thanks to @Shoubhik who pointed out that calcium hydroxide is good enough to make calcium benzoate. To support this claim, I found a patent which indeed describes that calcium hydroxide was used to make calcium benzoate (See comments). Here is the summary of the patent:

The invention relates to a processing method for synthesizing calcium
  benzoate at low temperature. The method comprises the following steps:
  heating and stirring 65-85kg of industrial-grade benzoic acid, 28-32kg
  of calcium hydroxide, 1-3kg of catalysts comprising triethanol
  distearate, ethylene distearamide and ethylene bislauramide and
  300-500kg of clear water in a reaction kettle at 333-363 K to react
  for 2-4 hours; and then adding 5-15kg of separant, stirring for half
  an hour, filtering and drying. The method provided by the invention
  can be used for directly synthesizing calcium benzoate, thereby
  lowering the equipment investment, reducing the working procedures,
  lowering the labor intensity of workers, increasing the output,
  enhancing the quality, lowering the cost and preventing the
  environmental pollution.


text:
How do I measure out these salts using the proper method?
I'm planning on preparing a water solution using the following salts of reagent grade with approximately these concentrations:
\begin{array}{lr}
  \ce{CaCl2}  & \pu{10 mg/L}\\
  \ce{Na2SO4} & \pu{1 mg/L}\\
  \ce{NaHCO3} & \pu{20 mg/L}\\
  \ce{KHCO3}  & \pu{5 mg/L}\\
  \ce{NaF}    & \pu{0.25 mg/L}\\
  \ce{NaNO3}  & \pu{1 mg/L}\\
\end{array}
I suspect some of these salts are hygroscopic. Do any of them need to be oven dried prior to weighing? What is an appropriate procedure for this (i.e. what temperature, for how long, etc.). Are there any other requirements in order to correctly produce a solution of the above composition?
It's for an experiment where I'm needing to replicate a natural water source so I'm needing to be as accurate as possible.

@airhuff, I'd disagree with your statement. Chemicals like calcium chloride are almost certainly not anhydrous unless explicitly stated, and even then once opened they'd absorb any moisture in contact.

@NotBaran , somehow I read into this statement that the "artificial water" was for an aquarium, though that's clearly not stated...don't know why I assumed that, and I will delete the comment. But, with that in mind I figured worrying about drying, or other special handling, of the chemicals was overkill. Regarding the labeling of the degree of hydration, we are still in disagreement. If I buy a chemical reagent, I expect the bottle to be accurately labeled with it's contents, including whether a salt is anhydrous or not. I guess an exception would be a bucket of CaCl2 ice melt or something.

Often labs have a sealed nitrogen/argon environment, with gloves on the glass so you can manipulate what's inside. There's usually an airlock to move things in and out. Put a balance inside one (Or ask to use someone who's already set one up), weigh your material in the safe environment, and take it out, or set up your reaction also inside the safe area. You of course need anhydrous material, or it'll kind of defeat the purpose.

@airhuff The best attempt I have ever seen from a vendor to acknowledge that a hygroscopic salt was not anhydrous was the following: $\ce{CaCl2 --X H2O}$ I don't know if this meets your standard for "clearly labeled" or not.

@J.Ari , this is how I expect the bottle to be labled. Sigma-Aldrich has another CaCl2 dihydrate product labled "Calcium Chloride", sans the dihydrate part, but it is clearly shown as a dihydrate in the chemical formula on the label.

Well when I purchased from Sigma-Aldrich maybe 5 years ago, the chemical formula had the indeterminate hydrate noted. I suppose this at least should make one think about oven drying or determining moisture content and correct the experiment accordingly. I am leaning to NotBaran's point of view regarding the labeling for such details, better to verify by oneself.

You are dealing with very low concentrations and for those you will need very accurate equipment and a very clean lab. I am not sure that reagent purity is enough. In some parts of the world ordinary tap water might already have higher concentrations.
Unfortunately you have not given enough context for what you are doing, so I will outline some procedures which should guarantee quite good accuracy.
First thing you will need is bidistilled water probably lots of it.
For very accurate (and reproducible) results I would recommend purchasing stock solutions and the doing serial dilutions.
If the above is no option you will have to do some more research by yourself, because it depends on the quality and variation of the chemicals you buy. (And preparing a list of all that for your list of chemicals is beyond the scope of this summary.)
For example calcium chloride comes on merck in at least four variations: $\ce{CaCl2}$, $\ce{CaCl2.xH2O}$, $\ce{CaCl2.2H2O}$, $\ce{CaCl2.6H2O}$, and solutions. Depending on your starting material you will need to look up the decomposition temperatures. For calcium chloride you may find them on Wikipedia.
You first need to dry your container with which you will measure the mass of your reagent. Make sure you use gloves as fats and skin can change the result at these low concentrations. Measure it multiple times until you have a constant value (within the limitations of your scale). You need to dry the container at the same temperature which is appropriate for drying your chemical.
Then you follow the same procedure with the chemical. Obviously you will not hit the concentration exactly, but in science that hardly matters, as long as you know the correct value. Once you started the drying process you should keep the container closed until you are ready to transfer it into the solution.
You shouldn't store your prepared stock solutions for a long time and at below room temperature in appropriate flasks.
The other way would be checking the concentration against a primary standard. For calcium chloride you can probably use silver nitrate, which you have to dry like explained above. (It doesn't have the nasty crystal water, which is a plus as $\pu{50 - 90 ^\circ C}$ should be sufficient.)
For setting up the solutions you will need a good assortment of volumetric flasks, pipettes, micro pipettes, thermometers, and probably a bit more I just can't think of right now.
You might want to check with this open access publisher what is common in the field and adjust appropriately. It very much depends on your goal; approximately can cover a pretty large range of desired accuracy.

Addendum cross-link: Is MDPI a reputable academic publisher?


text:
Number of unpaired electron and total spin angular moment
IIT-JEE advanced 2013 has a question:

Give the correct order of the complex ions given below according to their spin-only magnetic moment:
  A. $\ce{[FeF6]^{3-}}$
  B. $\ce{[V(H2O)6]^{2+}}$
  C. $\ce{[Fe(H2O)6]^{2+}}$

So, my reasoning is more unpaired electron results in greater spin-only magnetic moment. So, the order should be B>A>C
But the answer is given A>C>B.
I don't know where I'm wrong.

The first choice lists $\ce{Fe^3+}$ as the central metal atom (CMA), so you have the configuration: $\mathrm{3d^5 4s^0}$. Remember that fluoride is a weak field ligand and would not cause pairing of electrons. This gives you 5 unpaired electrons.
The second choice has $\ce{V^2+}$ as the CMA, the configuration being $\mathrm{3d^3 4s^0}$. This gives you 3 unpaired electrons.
Finally, the third choice has $\ce{Fe^2+}$ as the CMA. Yet again water is a weak field ligand and does not cause pairing of electrons. The isolated configuration although is $\mathrm{3d^6 4s^0}$, but now you'd have 4 electrons in the $\mathrm{t_2g}$ set and 2 electrons in the $\mathrm{e_g}$ set. So now you've 4 unpaired electrons.

more unpaired electron results in greater spin-only magnetic moment.

This is true and I can see it being followed as $\mathrm{A>C>B}$.

Also see my comment for the strength of ligands.

The answer A>C>B is correct. I think you might have considered Fluoride to be a strong field ligand but it's actually not that strong so there won't be any pairing in the first compound as a result it will have five unpaired electrons. The remaining compounds have lesser unpaired electrons so yeah the order is justified.


text:
Synthesis Golf I: Sodium Fluvastatin
A full FAQ exists on meta.chem.SE explaining the premise of synthesis golf, why we're doing this, and the ground rules. Please read this before posting an answer.
The target for this first round of synthesis golf is sodium fluvastatin, an artificial statin drug used to treat hypercholesterolemia and in the prevention of cardiovascular disease. You must provide a synthetic route to the target molecule.

In addition, to narrow the scope of this question:

the synthesis must include a way of making the indole (i.e. no buying of the indole core and carrying out multiple functionalisation reactions).
all of the stereo centres must be set. Relative stereochemistry only is fine (the 1,3-syn relationship between the hydroxyls), but if you can think of a way to set them with absolute configuration, then great.

This is a neat idea - hopefully there are enough experts in organic synthesis on the site to make for a robust set of answers.

(may/may not be helpful to anyone but this doesn't cost me much) SMILES for target molecule O=C(C[C@H](C[C@H](/C=C/C(N1C(C)C)=C(C2=CC=CC=C21)C3=CC=C(F)C=C3)O)O)O and InChI InChI=1S/C24H26FNO4/c1-15(2)26-21-6-4-3-5-20(21)24(16-7-9-17(25)10-8-16)22(26)12-11-18(27)13-19(28)14-23(29)30/h3-12,15,18-19,27-28H,13-14H2,1-2H3,(H,29,30)/b12-11+/t18-,19-/m0/s1

This is so interesting! Can't we fully adopt this idea and create a new stack exchange like the code golf?

@KartikeyaBadola you would need to have sufficient interest, if these become popular then it might be a consideration. See: http://area51.stackexchange.com/faq

I don't understand this question.  Is the idea to find a route with fewer steps than previously published routes? Or are people supposed to answer without consulting previous works?

@DavePhD Was intended to be a back-of-the-envelope type of thing, else people won't answer anyway because it takes too much time/effort, but SciFinder etc is fair game (i.e. its not necessarily intended to be done blind).  The idea was really just the shortest/most efficient synthesis from users on here (i.e. not comparing to a process route which has undoubtedly had many rounds of optimisation).

What are the precise criteria for "shortest route"? The number of nominal steps in the mechanism? The number of experimental steps involved?

Are we back to arguing about what a step is? (@hBy2Py: you may be interested http://openflask.blogspot.co.uk/2016/12/11-step-synthesis-of-thapsigargin.html)

Maybe put a limit on the adjusted cost per kilo or cost per mole to eliminate some advanced intermediates that are readily available? I.e. To make it more commercially realistic?

@orthocresol Well, I've never argued about what a step is, so this particular party's all new to me. :-P

This is quite a good idea. Even this question https://chemistry.stackexchange.com/questions/75283/how-to-convert-carboxylic-acid-into-an-alkene could be a potential synthesis golf.

@PrittBalagopal - theres definitely some refining to do to make the problem easy enough to be answered and hard enough to be interesting. The linked question possibly borders on the tedious, a SciFinder search would instantly reveal every published way of doing that transformation, there isn't really much room for creativity.        I encourage you to take a shot at the current problem though!

@NotBaran I wanted to, unfortunately I don't know how to create an Indole ring .

@PrittBalagopal https://en.wikipedia.org/wiki/Indole#Synthetic_routes   Wikipedia has articles on the most common if you're interested :)

@NotBaran Hmm, this looks good. I'll give a shot after doing some scrap workouts.

I recommend a creating #synthesis-golf tag.

@zhe. I was planning on posting in meta to ask people for thoughts/opinions first. I think it's been interesting and kind of successful but if it was to become a thing it possibly has to be done in an organised way

This is a great idea. I am sad I've missed this. When will the next problem be posted?

@Marko... next Monday, and then one a month as per https://chemistry.meta.stackexchange.com/questions/3779/synthesis-golf-faq/3789#3789

I don't like the frequency. A problem per two weeks would be much better.

@Marko, there has to be a balance. If the challenges get a sensible number of responses in two weeks then it can perhaps happen more frequently. I think the concern is that too frequent and they go unanswered...

I am certain that they wouldn't be unanswered. After all, this question had excellent reception.

Edit: step count is 15, longest linear sequence of 9 steps
Just to (hopefully) get the ball rolling, here's something I scribbled down. I am sure people here will have better answers. And I have not done a lot of literature research to see whether the steps are plausible, so there might be fatal flaws... but hopefully not; as far as I know/can tell, everything here is relatively sensible.
Retrosynthesis first. I chose a Horner–Wadsworth–Emmons reaction to make the (E)-double bond, and something similar to the Reissert synthesis for the indole. It's not exactly the same because I didn't acylate the benzylic carbon, but the initial disconnection is similar. This leads to the key building blocks:

I chose the C(sp2)-C(sp3) bonds in 2 to disconnect. One can be made by SNAr with the amazing SNAr substrate 1-fluoro-2-nitrobenzene (5), and the other by Pd cross-coupling (I'm guessing cheaper ways probably exist). I've never actually seen SNAr reactions with enolates as nucleophiles, but I guess that if it doesn't work, SRN1-type conditions should still make that step possible:

As for 3, I figured it could be made from a symmetrical intermediate:

Forward synthesis
Synthesis of 3, starting from cyclopentadiene:

(i) $\ce{BH3.THF}$, then $\ce{H2O2, NaOH}$; (ii) TBSCl, imidazole; (iii) $\ce{O3}$, then $\ce{H2O2}$; (iv) $\ce{TMSCHN2}$; (v) enzymatic desymmetrisation (I did not look up specifically which, but I think it should be possible); (vi) 7, NaH

Next, the synthesis of 2, starting from methylglyoxal:

(vii) ethylene glycol, TsOH; (viii) Pd cat., ligand, $\ce{NaO^{t}Bu}$, 6; (ix) NaH, 5

Forming the indole:

(x) Zn, AcOH; (xi) $\ce{NaH, ^{i}PrI}$; (xii) aq. HCl

and finally the HWE, the TBS deprotection, and the last redox step, a diastereoselective boron reduction. I wasn't entirely sure if it would work on an α,β-unsaturated ketone, but Comp Org Synth II, section 8.01 has several examples of it being used on an α,β-unsaturated ketones.

(xiii) LiHMDS, 3; (xiv) $\ce{TBAF, H2O}$; (xv) $\ce{Et2B(OMe), NaBH4}$

Would the conversion from cyclopentadiene to the other compound work? I think cyclopentadiene is quite reactive, and undergoes a Diels-Alder reaction with itself.

You buy the Diels-Alder adduct (http://www.sigmaaldrich.com/catalog/substance/dicyclopentadiene132207773611?lang=en&region=GB), then heat it to get the monomer (e.g. http://www.orgsyn.org/demo.aspx?prep=CV4P0238); you can use the monomer in further reactions. Wikipedia article has more info.

I'm a bit concerned about your synthesis of compound 4. Methylglyoxal, being rather unstable in its pure form, is typically sold as a 40% solution in water, which would make acetal formation (and concomitant removal of water) rather difficult. Additionally, the formation of the acetal with TosOH and ethylene glycol would probably not proceed smoothly, likely forming the diacetal and some polymeric material. Methylglyoxal 1,1-dimethyl acetal is commercially available from Sigma-Aldrich, and avoids this problem (and saves a step).

@JSK thanks! That's much more tractable. I've not done a lot of stuff in the lab yet - so my practical knowledge will be rather limited too. On paper I suppose it would be possible to selectively protect the aldehyde over the ketone, but I realise that in practice it's going to be an entirely different matter.

@orthocresol no problem! I did like your use of hydroboration on cyclopentadiene; I wasn't sure of the regioselectivity on cyclic dienes, but it appears to be quite regioselective in modest yields. It seems like most syntheses of cyclopent-3-en-1-ol use either epoxidation of cyclopentadiene, followed by lithium aluminium hydride reduction, or use a metathesis reaction on hepta-1,6-dien-4-ol, but your method works nicely as a single step (in my opinion, but at least single pot) reaction from a simple starting material.

Strategy
The indole core can be made with one of numerous methods; having had classes taught by Dr. Richard Larock, my first inclination was to use the Larock indole synthesis. The Larock indole synthesis uses an N-alkylated/acylated/tosylated 2-iodoanilines and a di-substituted alkyne to produce 1,2,3-trisubstituted indoles in one step. However, care must be taken to ensure appropriate regioselectivity of the product. The use of a TMS-substituted phenylacetylene produces a 2-trimethylsilyl-3-phenylindole (Larock, JOC, 1998, 63 (22), 7652).
The main complexity of the molecule is found in the side-chain, containing a syn-1,3-diol with alkene and carboxylate functionality; installation of a chiral alcohol $\beta$ to a ketone would allow diastereoselective formation of the diol.
Synthesis
The side chain may be made in a few steps from the cheap and commercially available 2,2,6-trimethyl-4H-1,3-dioxin-4-one (1), which is first deprotonated and reacted with TMSCl to produce compound 2. A Mukaiyama aldol-type reaction, mediated by a titanium (IV) compound, e.g. titanium tetrachloride or Ti(OiPr)$_3$Cl, between 2 and acrolein produces the racemic alcohol 3.
Refluxing 3 in methanol and toluene with catalytic pTSA removes the acetal-like protecting group to form the methyl $\beta$-ketoester 4. The Narasaka–Prasad reduction on 4 uses the boron reagent diethylmethoxyborane (available from Sigma-Aldrich) as a chelating reagent to permit the diastereoselective reduction with sodium borohydride to the alkenyldiol 5 (relative stereochemistry is shown).
It should be noted that the aldol reaction to produce 3 may also be performed enantioselectively, as in Singer and Carreira, JACS 1995, 117, 12360, to give the desired enantiomeric alcohol; compound 4 retains this stereochemistry, and the diol 5 would then be the single diastereomer shown. Alternatively, chiral resolution could be performed at this point, or the racemic compound could be used to produce racemic Fluvastatin.

As the Larock indole synthesis allows for the use of a 2-iodoaniline and a trimethylsilyl phenylacetylene derivative, the main core is synthesised using this reaction. A palladium- and copper-catalyzed Sonogashira coupling between 1-bromo-4-fluorobenzene (6) and ethynyltrimethylsilane gives the TMS-substituted 4-fluorophenylacetylene 7. Meanwhile, N-isopropyl-2-iodoaniline (9) is synthesized from 2-iodoaniline (8, from Sigma-Aldrich) and acetone via formation of the imine in acetic acid, followed by the in situ reduction with sodium triacetoxyborohydride.
The palladium-catalyzed Larock indole synthesis is then used to produce indole 10 from alkyne 7 and aniline 9. The 2-TMS-substituted indole is capable of undergoing a Heck reaction (see Larock, JOC, 1998, 63 (22), 7652) with alkenyldiol 5 to form 11, the methyl ester of Fluvastatin. Ester hydrolysis with sodium hydroxide in water gives the sodium carboxylate, Fluvastatin, 12, in nine steps.

I thought given the challenge, it was only fair that I 'wasted' some time thinking about this myself. The following route is hopefully plausible, even if not 100% thought through. Comments welcomed, and hopefully provides some competition to Ortho's great attempt.

Strategy and retrosynthesis
The synthesis proposed here disconnects fluvastatin into two main components: an indole core, and a side chain containing the stereocentres.
Fragment union is proposed to take place via selective cross metathesis (mainly because @Orthocresol made use of a HWE and I thought some variety would be nice [though seriously, cross metathesis is great for not having to use pre-functionalised precursors]).
The indole core itself is proposed to be brought together using a modern variant of a Bischler Indole synthesis by cyclisation of a linear precursor under L.A. conditions, avoiding the potential regioselectivity issues of many indole syntheses involving carbonyl condensations.
For the side-chain, a vinylogous Mukaiyama aldol reaction should provide the required stereochemistry, using a copper-bisoxazoline catalyst to control the absolute facial selectivity.

Synthesis of the Indole core
Starting from commercially available fluorobenzene 1 (Aldrich: 500G/16.30GBP) a Friedel Crafts acylation using the ethylene glycol (Aldrich: 1000mL/52.60GBP) derived acyl chloride in the presence of a Lewis acid catalyst could be carried out to give the acylated ring 2. Regioselectivity of the Friedel-Crafts is well precedented based on directing effects of the fluorine, with the 4-position both activated electronically and sterically most accessible. Due to the availability of the starting materials, poor regioselectivity at this stage not crucial, and the isomers could be separated.
Subsequent cleavage of the PMB under oxidative conditions furnishes the free alcohol 3 which may undergo an Appel-type iodination (could also do a Finkelstein via formation of the tosylate), displacement, and reductive amination with acetone to provide the indole precursor 6.
Upon treatment with a Lewis acid catalyst and heat, 6 should undergo (what I think I can call) a Bischler indole synthesis to furnish the indole core of fluvastatin 7, via this method no regioselectivity issues arise, since the two reacting partners are tethered together, with only the 5 membered ring being favoured.
Failing the milder Lewis acidic conditions, polyphosphoric acid (PPA) is common, and given the lack of delicate functionality in the molecule, this should pose no issue.

Completion of the synthesis
With the indole core 7 in hand, functionality must be installed at the 2-position. Standard Vilsmeyer-Haack conditions afford aldehyde 8 which may undergo Tebbe olefination (or similar) to provide a terminal olefin 9 capable of undergoing (E) selective cross metathesis with side chain 18 (synthesis described below). Final ester saponification using NaOH provides the sodium salt of Fluvastatin 11.

Stereoselective synthesis of the side-chain

The crucial step of this synthesis is the setting of the 1,3-syn diol with complete absolute and relative stereochemistry.
Chan's diene 12 has been shown to undergo enantio- and diastereo-selective Mukaiyama aldol reaction to furnish the 1,3-syn diol 13 using a copper-bisoxazoline catalyst. Protection of the diol as the acetonide, reductive debenzylation, DMP oxidation and Tebbe olefination then affords the required terminal olefin 18 for the planned cross metathesis.

Conclusion
Based on the proposed synthesis, Fluvastatin Sodium 11 would be synthesised in 19 steps overall, with 13 steps in the longest linear sequence.

Why not do the acylation of fluorobenzene with chloroacetyl chloride?

I wasn't confident which chloribe would get ripped off first by the Lewis acid - I'm sure there's precedence, I just would take my chances

Org. Syn. Coll. Vol 3, 191        4-F-phenacyl chloride is commercially available

I admittedly forgot about the existence of Vilsmeier formylation. Regarding the Mukaiyama, does that produce the diol in one step? All I can find is Evans' report on it, where he used $\ce{Me4NBH(OAc)3}$ to reduce the beta-hydroxyketone to the anti diol: JACS 1996, 118, 5814 P/S Did you use ChemDraw for this?

@orthocresol. The Vilsmeyer is an old'un but has its uses.  I think I might have mis-read a step for the aldol (will correct shortly)- it likely just needs a Narasaka afterwards. It is ChemDraw, the settings have just been significantly changed (thicker lines, shorter bond lengths)

Ah, ok, I'm just curious as to what font that is?

Frutiger (Frutiger Neue to be precise IIRC).

So the conversion of 1 to 4 is one step depending on how much of a purist one wants to be vs the phenacyl bromide

As a compromise to being a synthetic purist and buying the phenacyl bromide, you could also start with 4'-fluoroacetophenone and perform a bromination to compound 4 as per Org. Synth. 1939, 19, 24.

Reference through step 2: Synthesis of carbon–14 labeled fluvastatin, Journal of Labeled Compounds and Radiopharmaceuticals volume 41, pages 1-7.
Step 0 (product is commercially available):
Fluorobenzene + Bromoacetylchloride $\ce{->}$ 2-Bromo-4′-fluoroacetophenone
(Aluminum chloride, 75 °C)
Step 1:
2-Bromo-4′-fluoroacetophenone + N-Isopropylaniline $\ce{->}$ 1-(4-fluorophenyl)-2-[(1-methylethyl)phenylamino]-ethanone $\ce{->}$ 3-(4-Fluorophenyl)-1-isopropyl-1H-indole
(intermediate created under nitrogen atmosphere, ethanol solvent, 78 °C, 1 hour.  Intermediate then reacted with zinc chloride, ethanol solvent).
Step 2:
3-(4-Fluorophenyl)-1-isopropyl-1H-indole +  N-methyl-N-phenyl-3-amino acrolein $\ce{->}$ 3-[3-(4-Fluorophenyl)-1-isopropyl-1H-indol-2-yl]-propenal
(nitrogen atmosphere, phosphorous oxychloride, acetonitrile solvent)
Then see Facile and Highly Enantioselective Synthesis of (+)- and (−)-Fluvastatin and Their Analogues Journal of Organic Chemistry, volume 75, pages 7514–7518.
Step 3:
3-[3-(4-Fluorophenyl)-1-isopropyl-1H-indol-2-yl]-propenal + diketene $\ce{->}$ 3-[3-(4-Fluorophenyl)-1-(1-methylethyl)-1H-indol-2-yl]-5-hydroxy-3-oxo-(6E)-heptenoic acid-1-isopropyl ester
(in the presence of Ti(O-i-Pr)4 and chiral Schiff base, -40 °C, methylene chloride)
Step 4: reduction of ketone with $\ce{NaBH4}$ to yield fluvastatin isopropyl ester
Step 5: saponification to yield fluvastatin product.
(Step 6: preparation of the chiral Schiff base from 3-tert-Butyl-2-hydroxybenzaldehyde and valinol)

For alternatives to steps 3-6, see An Improved Manufacturing Process for Fluvastatin Organic Process Research & Development volume 11, pages 13-18.
And for alternatives generally see:
Section 9.1 of Fluvastatin in Pubchem
US patents  4739073 and 5354772
Cardiovascular Drugs in Ullmann's Encyclopedia of Industrial Chemistry.

I'm stuck with the lateral chain, but here are my two cents for the formation of the indole with just a few steps/reagents (I admit I'm quite a newbie, yet an enthusiast in organic chemistry).
Maybe help is not admitted, but I hope my idea could be integrated by others:
Sigma-Aldrich has readily available the p-fluorophenylacetic acid.
A typical Fischer synthesis for the indole might be performed with said compound and (the diazonium salt of) aniline.
The product is brominated with $\ce{PBr3}$. This results in 2-bromo-3-(4-fluorophenyl)-1H-indole
The amine might be alkylated by reaction with a 2-halopropane.
A Grignard reagent could then be formed, and by reaction with a haloalkene, a Kumada reaction could attach the side chain. Yet, I admit that I didn't come up with a decent way of preserving the stereochemistry of the side chain.
"Ai posteri l'ardua sentenza"

A synthesis which involves an Evans auxiliary to induce asymmetry, and a Narasaka-Prasad reduction to control diastereoselectivity.

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text:
Rate equation for a reaction involving a polymer having absorbed small proteins according to langmuir's model
First, let me apologize for the awful question title..
I have a solution with polymers with $N$ binding sites and small proteins, which can attach to a binding site. The polymers will absorb the proteins till $N^*$ out of $N$ sites are occupied. From that moment the polymer can undergo a certain structural deformation and will drop some small proteins in the process.
My problem is this: I would like to write down the rate equations of the problem and I thought that langmuir's model would be a good way to describe the absorption of the proteins to a polymer. However, this theory gives the rate of the surface decomposition $\theta$, but I would like to be able to let the polymer undergo the deformation reaction (involving the concentration of polymers having absorbed $N^*$ proteins). How can I incorporate the langmuir model in the "ordinary" set of first order rate equations? How can I get to a numerically solvable system of equations that describe the kinetics of this problem having so many absorbing surfaces (i.e. polymers)?
In other words, is there a way to obtain the concentration of polymers having a certain $\theta$?


text:
Chemicals for electrochromic reaction ?
I would like to detect the electric current going through the water solution using the colour reaction in this water solution. What chemicals can be used for this solution to make it electrochromic.

Hello and welcome to Chemistry.SE. If you have any questions about how this site works, please feel free to visit the [help]. Regarding your question, as it is I think it may be too broad to get a good answer. I think your chances of getting a good answer would be greatly increased if you would add several sentences outlining exactly what you are trying to accomplish, what reagents and equipment you are using, etc. Again, welcome and best of luck!

Easiest way is with an iodide, e.g. $\ce{NaI}$ or $\ce{KI}$, and starch.
An old trick was to sprinkle a bit of $\ce{KI}$ on a (damp) slice of potato, stick a couple of wires into the potato and attach a small cell to the wires. Iodine is liberated at the positive terminal and it reacts with starch in the potato to make a deep blue color.
There are more suggestions from Flinn Scientific, such as using an indicator to show acid or alkali liberated at electrodes.


text:
Could you charge a recharable battery with a voltaic cell?
In class we were learning about voltaic cells and my chemistry teacher joked that after modern society falls, we'll be the only ones who can recharge batteries with simple voltaic cells. And this got me thinking, is this possible? Could I create a salt bridge cell, attach wires to a store bought rechargeable battery, and charge it? I'm really curious and I can't ask my chem teacher as he's in the hospital recovering from an emergency appendix removal

Any source of electricity of the required voltage and current could be used to charge a rechargeable ("secondary") cell. A home-made "primary" cell would work (though you might need a few in series to provide the needed voltage).
For that matter, you can generate DC to charge the cell many other ways:

Using a thermopile, e.g. from copper and iron wire
Magnetohydrodynamically, e.g. with flowing salt water and a magnet
Using a photovoltaic panel...


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The history of ethylene
According to the history section in Ethylene, it is written that:

the name ethylene was used in this sense as early as 1852

However, the earliest mention of it I could find was in 1859 (Wurtz, "Sur l'oxyde d'ethylene"; Comptus Rendus 48, p.101). I also found mentions of "ethylene" in 1862 (Wurtz, "[...]On oxide of ethylene, considered as a link between organic and mineral chemistry[...]"), and in 1878 (Frankland and Thorne, "[...]On the Luminosity of Benzol when burnt with non-luminous Combustible Gases[...]"). Can anyone find any earlier publications that mention "ethylene"?

I considered ethane, and got the fact that the prefix eth comes from ether. https://i.stack.imgur.com/1GbRO.png

Thanks for responding, @PrittBalagopal. However, what I essentially want to know when the word "ethylene" was first used. This may have been unclear in my title; perhaps "etymology" is a slightly inaccurate word.

For what it's worth, Merriam-Webster says first usage is 1849, but reference is not provided online. Have you checked the Oxford English Dictionary? It is not available free online, but is often the best source for first usages.

According to the Oxford English Dictionary entry on "ethylene", the first English usage was in 1849 in the Chemical Gazette (alternate name Journal of Practical Chemistry), published in London. The specific citation is Vol. 7 p. 116 (Mar. 15, 1849). The sentence in which it is used is in an article by "A. Wurtz" titled "On a Series of Organic Alkalies Homologous with Ammonia.", and the relevant sentence reads "By adding to ammonia the elements of ethylene, $\ce{C4H4}$, ethylic ammonia, $\ce{C4H7N}$, is obtained."
The full text of the journal is available digitally via Google Books.

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Going from La to Ce, why is the extra electron added into the 4f orbital?
So, I am studying electronic configuration but the elements of the series of the lanthanides confused me. The electronic configuration of $\ce{La}$ is $\mathrm{6s^2\,5d^1}$ and that of cerium is $\mathrm{6s^2\,5d^1\,4f^1}$.
So, why does the extra electron that was added make the $\mathrm{5d}$ orbital have a higher energy?

Marginally related: Why is the 5d orbital instead of the 4f orbital filled in lanthanum?

Due to two contradictory factors, one is the effective nuclear charge which is greater in $\mathrm{5d}$ making it of lesser energy and the second one is a nodal factor. Nodal factor depicts localization of electrons, which indicates of higher energy. Hence to balance these two effects out, for Ce the second electron goes to $\mathrm{4f}$ and its electronic configuration stands as $[\ce{Xe}]~\mathrm{(4f)^1 (5d)^1 (6s)^2}.$


text:
Physical interpretation of the Gibbs-Duhem equation
For a binary system at constant temperature and pressure, the Gibbs-Duhem equation can be reduced to
$$\mathrm d\mu_B=-\frac{n_A}{n_B}\mathrm d\mu_A$$
How does one make sense of the Gibbs-Duhem equation in terms of intermolecular forces and lattice packing? Why does the chemical potential of one component decrease (relative to the other) by a factor of $n_A/n_B$?

@Mithoron for a two component system under constant T and p, Gibbs-Duhem reduces to the formula given above. Of course, it can be made slightly clearer.

@Mithoron Apologies, I should've made it clearer - I have corrected it now.

Why don't you write it out for the case of an ideal gas, as see what it looks like?  That should answer your question.

@Chester Miller I've gone through the derivation, and I see that the equation takes the form $\sum_{i} n_id\mu_i=0$. My question, however, is how to make sense of this equation chemically?

You can that equation, but it's for reaction not forces, not packing and this question still looks like nonsense.

It means that, if your system is already at chemical reaction equilibrium, if you slightly perturb the number of moles of the reactant and product species in such a way as to achieve a new reaction equilibrium (at a slightly perturbed equilibrium point), the chemical potentials must undergo certain changes to make this happen.  This will be the case if your equation is satisfied.  Otherwise, the proposed changes in the moles of reactant and product species will not correspond to a new equilibrium mixture.

If you write down the equation in terms of the very definition of $\mu_i=\frac{\rm dG}{{\rm d}n_i}$ and impose ${\rm d}n_\text{totoal}=0$, it's just the conservation of the energy potential ${\rm d}G=0$.

Say you own a theme park where, every hour, each adult has to pay ${\rm d}\mu_a=\$7$ (money=energy) and each kid has to pay ${\rm d}\mu_k=\$3$.
We assign positive signs to when they put money out of their pockets.
If there are $n_a=10$ adults and $n_k=5$ kids in the park, and if this number of people stays fixed (${\rm d}n_\text{totoal}=0$ over some time), then they together have to spend
$$n_a\cdot {\rm d}\mu_a + n_k\cdot {\rm d}\mu_k = \$70 + \$15 = \$85$$
Now say you, the park owner, are not allowed to actually make any money (${\rm d}G=0$) and you relax the condition that the kids have to pay any money at all. That is, now just the adult have to pay ... and since the money has to go somewhere, the kids are on the receiving end. Then the ten parents still spend $\$70$ per hour and now the five kids will split that money among each other. Each kids receives
$$-{\rm d}\mu_k = \frac{1}{n_k}\cdot n_a\cdot {\rm d}\mu_a = \frac{1}{5}\cdot \$70 = \$14$$

The comparison with "money per time" lacks in that here people don't bring in money just from coming to the park (as it's the case with particles coming into the system and energy). But I hope this clear up the meaning of factor $\frac{1}{n_k}$ clear.


text:
How do you predict isomers for certain hydrocarbons?
Hexane $\ce{C6H14}$ has 5 isomers, but I'm able to draw at least 6 different structures, which initially led me to think that there were 6 isomers.
So how can I tell which structure will qualify as an isomer?
The only rule my textbook suggests is connectivity exemplified as a $\ce{CH3}$ (for example) that runs downward at the end of the straight chain, which is no different than the straight chain itself.

is your question rather like How do I recognize if the two structures are the same?

No. I can tell when structures are the same. It's just that, during a textbook question, I'm give hexane and told to draw the 5 isomers of it. One of my structures was different and yet, I still had the 6 carbons and 16 hydrogens, which got me wondering why this odd T-shaped isomer wasn't in the textbook.

Hexane indeed has five isomers, which means that your sixth isomer is identical to one of the other five. There is no going around it.

Yes I'm beginning to see it now. I think you're right.

So I believe that mykhal is right with this comment and I was wrong. I do indeed need help telling whether two structures are the same.

Would getting those organic chemistry models that are sold on amazon help me understand if two structures are the same? https://www.amazon.com/Duluth-Labs-Organic-Chemistry-Molecular/dp/B016ZAWAIS/ref=sr_1_1?ie=UTF8&qid=1495321955&sr=8-1&keywords=organic+chemistry+models


text:
Why is it difficult to measure the pH of environmental water samples accurately (near 7)?
I've read in several places (edit: 1, 2, 3, 4, 5, 6, 7, 8. 9, 10, 11, 12) that there are many practical factors that make accurate pH measurements of roughly neutral environmental water samples more challenging than significantly basic or acidic samples.
Since the glass probes are sensitive fairly specifically to hydrogen or hydronium activity, why would concentrations around $\ce10^{-7}$ be more challenging than a much higher or especially much lower concentration, say $\ce10^{-10}$?
Discussions of the problem seem to center around issues related "solutions of low ionic strength", so a basic solution near pH 10 and a buffered solution near ph 7 would both have less practical measurement issues than a non-ionic solution near pH 7.
Is there a way to explain the practical issues clearly, but in fairly simple, somewhat physical rather than purely mathematical terms?

If you carry the pH electrode to the water source it would be easy to measure the pH. The problem would come with trying to transport the water back to the lab. The water could get CO2 from the atmosphere, and the water would have active biological organisms which could cause the pH to change. // Remember too that at a pH 7 water itself has little buffer capacity. So a very small amount of acid or base can change the pH a lot.

@MaxW variation in the dissolved CO2 is a very important consideration, since the sample would be non-buffering, but those pH measurements would not be incorrect. They would reflect the pH of the solution, which is no longer a valid sample. I'm really asking about effects that would lead to an incorrect measurement - the number on the meter not reflecting the hydrogen activity of the solution currently under test.

My major professor (chemistry) at the Univ of Ga was on the committee for a PhD student in the agriculture dept. The kid was sudying pH effect on some crop. In his PhD defense my professor threw the kid what he thought was a softball  question. "How did you calibrate your pH meter?" The kid responded "Calibrate?!?"

@MaxW This line of enquiry started when helping a friend work with a third party who's anecdotal response was "Keep it wet?!?" There are quite a lot of interesting things going on with the measurement or even definition of pH, much more than I'd ever realized. I thought I'd carved out a tidy little bit of science to learn, but wow!

Can you link any of those sources? Since pH-electrodes measure the concentration of "H+" it doesn't make sense that they wouldn't work well at pH 7 but good at higher AND lower concentrations.

@DSVA check carefully what I've said. It's related to low ionic concentrations. "solutions of low ionic strength."  There are other ions besides H${}^+$.

@uhoh again: can you link any source on this. Afaik low ion concentration is no problem, high ion solution can be a problem (that's why you will get problems above pH=12 and below 0). You are claiming the exact opposite.

@DSVA a half-dozen "problem measuring pH of solutions of low ionic strength" 1, 2, 3, 4, 5, 6.

@DSVA There are all kinds of different explanations there. I am not qualified to search through these or another half dozen after this. I believe those more knowledgeable than me probably know which of all the possible things are likely to be the most likely, most common, and which are - while theoretically possible - not really the main sources of trouble.

@DSVA 7, 8, 9, 10, 11, 12

@uhoh I've looked into several of those and all pretty much describe the same problems and have similar explanations. The main Problem is that you will only get stable and correct readings once your system is in an equilibrated state. Such pH-electrodes are calibrated in high ionic strength solutions and the reference solution is of high ionic strength too. In addition the glass and especially membranes have a high ion concentration. If you now use low ionic strength analytes the whole system (especially at membranes and at the liquid junktion) won't be at equilibrium

@uhoh you'll get similar problems when doing measurements in D2O with electrodes calibrated in H2O

@DSVA would you post that as a proper answer, which can then be up/down voted? I don't need a "pretty much"-type comment, I need a quality stackexchange answer. I've looked at these and others and I disagree that the reality of the problem is as simple as you make it sound. If you post it as an answer rather than a comment, we can see how well the rest of the community agrees with your statements.

@uhoh well there's no point in repeating the content of your reference 3 and  6 (which explain the problems pretty well without any mathematics) as an answer if you seem to not accept these sources anyways.

@DSVA the point would be to select from the large variety of explanations which ones are the best and most likely. This would be for my benefit and the benefit of future readers who would like to learn about the problems of measuring solutions of low ionic concentration.  Indeed there is certainly a point to it. This is what stackexchange is for, this is how it works. It's not a repository of knowledgeable-sounding comments, it's a place for questions and answers.


text:
What is the mechanism by which reactants in endothermic reactions absorb energy from their environments?
My understanding of exothermic reactions is that, in the aggregate, the sum of the bond energies of the products is lower than that of the reactants, and definitely lower than that of the sum of any excited state intermediates, and that the resulting difference in energy is accounted for by a decrease in potential energy for the bonding electrons involved, resulting in a release of photons of equivalent energy.
Seeing as how exothermic reactions generally raise the temperature of the environment they are in, it seems logical to me that those photons are rapidly absorbed (inelastically) by atoms/molecules immediately neighboring the reaction products and thus their kinetic energies are raised, although I honestly don't truly understand the mechanism for that, either.
Similarly, for reactions that are isothermal but which are chemiluminescent, photons of a wavelength not easily absorbed by neighboring atoms/molecules simply leave the environment.
However, does that mean that the inverse is true? Is IR radiation from the environment the direct mediator of energy transfer that raises bonding electrons in reactants to anti-bonding MOs so that new MOs can form among different configurations of the original reactant atoms, thus allowing electrons to occupy the bonding MOs of the products?
I ask because the basic collision theory proposed by Trautz and Lewis doesn't seem to account for quantum mechanical models of energy, but I admit my own understanding of the topic is about as limited as my question would demonstrate.

Within the BO approximation, collisions can transfer energy into vibrational modes (and vice versa). At thermal equilibrium a small percentage of molecules will have enough vibrational energy in a certain bond to break it.

In your first paragraph you state at the end that photons are released. In all reactions, except the rather few chemiluminescent ones, this is not the case.
The energy difference is made up of vibrational and rotational energy of the products, which by collisions with solvent (or inert buffer gas) removes this energy.  This energy transfer occurs as the product is initially 'hot' with respect to the surroundings and thus energy flows out of this into the surroundings, heating it a little, until equilibrium is reached. In femtosecond laser experiments it is possible to observe the flow of energy when in solution and it is very rapid, not much longer than a few picoseconds. Ir photons could be emitted but the radiative lifetime of levels is much, much longer than collision lifetime so that i.r. emission yield will be tiny.
In the low pressure gas phase (where collisions with buffer gas are absent) infra-red can be observed (called infra-red chemiluminescence as it comes from excited vibrational levels) in reactions such as $\ce{H + H2 \rightarrow  HF(v'J') + H} $ where $\ce{v'J'}$ are excited vibrational and rotational levels. Many experiments have been reported by Polanyi and co workers for which Polanyi received the Chemistry Nobel prize. This reference https://www.nobelprize.org/nobel_prizes/chemistry/laureates/1986/polanyi-lecture.pdf gives a clear account of this important work
In the reverse case it is the few molecules that by the Boltzmann distribution have enough energy to surmount the activation barrier that are important not promotion by ir photons.


text:
How to calculate atoms of *each* element in a compound
I am trying to do the following exercise in a chemistry workbook. I have searched over internet but I couldn't find or understand it.

How many atoms of each element are found in:

5 moles $\ce{H2SO4}$

Now the only confusing thing for me is the 5 moles.
For example lets take $\ce{H2}$ element

so n=N/Na --> n × Na = N (number of atoms)
2 × 1gr/mol H = 2 moles H x 5 moles in the compound = 10 moles
10 moles $\ce{H2}$ × 6.02 × $10^{23}$ = 60.2 × $10^{23}$ atoms

Or do I do like this?

5 moles $\ce{H2}$ × 6.02 × $10^{23}$ = 30.1 × $10^{23}$ atoms

There is no such element as H2.

Think of the mole as a SI prefix, like 'kilo' or 'mega'. If you have 5k $\ce{H2SO4}$ molecules, you will have 10k hydrogen atoms, not 5k. Just as 'kilo' means '1000', a mole means $6.02 \cdot 10^{23}$ of something.
Therefore, the first approach gives the correct answer, but I don't understand what you mean by 1gr/mol. Grams are out of the picture here.


text:
Is suckback possible after removing stopper?
I am making hydrogen gas to fill a balloon. I will run the gas to a washing bottle with $\ce{NaOH}$ and then connect the balloon. I was wondering what would happen if I removed the stopper of my main reaction flask (e.g. to change wasted reagents) when the balloon is already half inflated.

Would $\ce{NaOH}$ get sucked out of the washing bottle somehow?
Would gas that's already in the balloon come out? Or the washing solution would prevent it?

What are you mixing with NaOH to give you hydrogen gas? This wouldn't be the most common way of generating small amounts of hydrogen in the lab

I am mixing HCl with aluminium and bubbling the gas through NaOH.

A picture (or drawing) of your apparatus would be worth a thousand words.

@airhuff I have made the edit.

Well, yes, now the suckback looks quite possible.

What if I clamp the balloon and then remove the stopper?

It looks like you'd be pretty safe to clamp the balloon then remove the stopper. You probably know this, but absolutely wear safety goggles.


text:
Enantiomers in anisotropic achiral chemical environments
Consider a functionalised molecular box or cage which has an electrostatic gradient due to positively charged groups one one of its faces. This environment is achiral as there are reflection symmetries

Now place a chiral molecule which for example, has formal charges as shown

Based on heuristic arguments on coloumbic repulsion, we expect the torsional potential energy surface of it and its enantiomer will look like this, where they will be mirror images of each other due to the rotational symmetry:
Now suppose the gradient is something more general such as pH, concentration of a certain reactive species temperature and so on. Then we will suspect that despite they are enantiomers, the anisotropy of the environment will caused them to behave chemically different, with e.g. one of the more positively charged groups will became more reactive due to being positioned near the positive end of the electrostatic gradient of the environment.

What is the terminology that refers to the difference in reactivity due to relative geometric arrangements of moeities between the molecule and its surrounding anisotropic achiral environment?

Any good literature examples where this "rotational selectivity" is investigated or used in the context of chemical synthesis or molecular machines?


text:
What are the microstates of the triplet oxygen molecule?
I am trying to understand molecular term symbols of oxygen molecule. Everywhere I looked, it is claimed that the molecule has 6 microstates, three of them are singlet, and three of them are triplet. Now I can draw the diagrams for the singlet states with no problem, but I am struggling to find the tree triplet configurations. According to my microstate matrix there should be a triplet state with a total spin multiplicity of zero - how is this possible?

Are you able to find the term symbols? What are the degeneracies of the terms? Note that the triplet state has $S = 1$ which allows $M_S = 1, 0, -1$.

I can find the term symbols, I have uploaded a diagram here: http://imgur.com/a/quLxo . The singlet state and the triplet state are both triply degenerate, and the triplet state can have a total ang. mom. of 1, 0, and -1, exactly as you have pointed out. The question is: how is the triplet state having zero total ang. mom. possible? On the table I've linked, why is the triplet state in the middle of the matrix considered triplet? Why isn't it singlet?

$M_S$ doesn't refer to a total spin angular momentum, but rather to a projection of the spin angular momentum onto the z-axis. The triplet state still possesses nonzero angular momentum ($= \sqrt{S(S+1)}\hbar$) but the projection onto the axis ($= M_S\hbar$) can be zero. In fact, there has to be one component of the triplet state ($S = 1$) with a zero projection $(M_S = 0)$.

(By the way the symbol $M_L$ is inappropriate for a diatomic molecule; it should be $\Lambda$ as the projection of orbital angular momentum taken is not onto the z-axis but rather the internuclear axis. But that is a relatively minor point.)

In a $(\pi2p)^2$ configuration the orbital $\lambda$ and spin s quantum numbers have to be tabulated and combinations removed which break the Pauli principle. The total angular momentum quantum number is $\Lambda$ and spin $\Sigma$, for example the combination
$\lambda_1 = 1, \lambda_2 =1 , s_1=1/2, s_2 =-1/2$
generates total angular momentum $\Lambda = 2, \Sigma =0$ and this is part of the $^1\Delta$ state.
There are only 6 valid combinations leading to $^1\Delta, ^3\Sigma^-,^1\Sigma^+ $ states of which $^3\Sigma^-$ is the ground state.
A vector spin diagram shows how a triplet can have spin multiplicity of zero. There is a picture of vector spin for singlets and triplets in my answer to this question; How does spin flipping of triplet carbenes occur?


text:
Literature values for NaCl Eutectic - Calculating the eutectic
For the NaCl eutectic I've seen two values widely used:
23.3wt% NaCl - James G. Speight: Lange's Handbook of Chemistry 2017)
5.1 M NaCl (Marion + Kargel 2008 - Cold Aqueous Planetary Geochemistry with FREZCHEM)
both of which are at about -21C.
Am I being silly - these values aren't the same (they're both at atmospheric pressure)?
Is my maths right?
[23.3wt%=3.99M and 5.1M= 29.8wt%]

What density are you using for the comparison?

I've just been reading off quote values, so would a 5.1M solution, corrected for density at -21C, be 23.3wt%?t

I guess it would.

I've found that the specific gravity of a 23.3 wt% solution is 1.179. How would I use this to work out the molarity of the solution

What is molarity, really?

So in this text, it says 5.175 Molal is the eutectic, (which should be equal to 23.3wt %). So this requires a 23.3wt% solution to have a density of 1.298 at -21.3 C, which is way off 1.179.


text:
Correctness of DSC curve for polyamid or polyester
I did a DSC on a part of an alternator car belt. We already found out that it should by a polyester or a polyamid. But then we got the following DSC curve:

How is it possible that the glass transition temperature is around 39°C for a polyamid or a polyester?

Did the recording really start at $\pu{-80 ^\circ{}C}$, as the abscissa indicates? Routine DSC I saw so far (Mettler) start their scan at about $\pu{+20 ^\circ{}C}$. And then, your characterisation already stops at $\pu{100 ^\circ{}C}$, a bit early.


text:
Is OH + CO → CO2 + H an elementary reaction?
Is
$$\rm OH + CO \longrightarrow CO_2 + H$$
an elementary reaction? How could I tell?
I could break the reaction into
$$\rm OH\ (+ M) \longrightarrow O + H\ (+ M)$$
$$\rm CO + O \longrightarrow CO_2$$
but I don't know if the former is witnessed.

$\ce{OH}$ and $\ce{H}$ arent really stable species that actually exist

The only way is to perform some experiments.  Just writing down the stoichiometric equation is not really informative as to the mechanism; for example $\ce{H2 + Cl2\rightarrow 2HCl}$ proceeds by a radical chain reaction and not as implied by this equation. Incidentally, the species H and OH are well known; OH is stable enough to be measure in the atmosphere and is important in atmospheric chemistry.

Another note on OH: while OH can be measured in the atmosphere by very sensitive techniques, it is indeed extremely unstable, and exists at levels of around 0.1 ppt, or 10^6 molecules per cc! It's atmospheric lifetime is on the order of 0.1 seconds, and as it is produced by sunlight, levels drop to very close to zero at night.

@ airhuff re OH, nevertheless at $0.1$ sec OH is a hugely long lived radical :)


text:
How did ancient chemists know the constitution of various compounds
During times when there was no spectroscopy of any kind, people did make guesses on the molecular constituency and structure.
For example, John Dalton's original atomic hypothesis was that all elements were mono-atomic and that the atoms in compounds have the simplest possible ratios. However, this was proved to be wrong and later it was recognized that oxygen and various other gases were actually diatomic.

How? How was it deduced that these gases are diatomic. Also how were the constitution of various compounds, in terms of both number and structure discerned?

This may perhaps be a better fit for/more 'on-topic' at history of science and mathematics rather than chemistry.SE.

How does migration work.

I'm sure the whole story is convoluted. But in a nutshell early chemists used reactions for a lot of the simple inorganic compounds to figure out the elemental ratio.

The book "Crucibles: the story of Chemistry" by Jaffe explains this in thorough and interesting detail. I don't remember the details, but generally they measured volumes of gases evolved from different reactions and found correlations/commonalities between the reactions. They found that the volume of the gases was related to the amount of substance reacted in whole-number ratios, etc.

For migration flag your post for moderator attention and explain that you want it migrated. I have not checked thoroughly, but I think it's likely that this is also a duplicate over at HSM, you probably aren't the first person to have been curious about this.


text:
Solvent for recrystallization of benzoic acid?

Which is best solvent for recrystallization of benzoic acid?

Water
$\ce{CH3Cl}$
$\ce{CCl3COOH}$
$\pu{0.1 M}\ \ce{NaOH}$

I understand the solute/solvent should be soluble at high temperature, and insoluble at room temperature. But how to make the specific arguments for eliminating the alternatives?

It can be difficult to guess at what a given compound will crystallise out from, and often there is a lot of trial and error involved.
Generally, the most 'common' type of recrystallisation involves finding a solvent in which the compound of interest is sparingly soluble at room temperature, but highly soluble at elevated temperature such that slow cooling of a saturated solution affords recrystallised material.
If you look up the entry for benzoic acid in Purification of Laboratory Chemicals, Armagero and Perrin, several solvents are suggested:

Benzoic acid has been crystallised from boiling water (charcoal), aqueous acetic acid, glacial acetic acid, benzene, aqueous EtOH, petroleum ether (b 60-80C), and from EtOH solution by adding water.


text:
Ketones and nucleophilicity
Is there anyone who can give me a convincing argument about the ketone, or the carbonyl oxygen in the ketone being nucleophile or not? . I really don't know what to write anymore. Everyone says something different. My instructor said that the Ketone oxygen can be protonated, thus reacting as a nucleophile. Others says that  Protonating the ketone in fact makes it more electrophilic.
Here is the molecule I'm working with (attached) and I have to mention all the nucleophilic sites - I've done that. This is the only one that really is confusing me. Hoping for some clarification.
https://i.stack.imgur.com/hQ9qD.jpg

Everyone says something different

Well, actually they don't. A ketone has a nucleophilic and an electrophilic center. The carbonyl carbon is electrophilic and can be attacked by nucleophiles, the oxygen itself can act as a nucleophile and can, for example, be protonated. And yes, if we protonate the oxygen the "ketone", or better the carbonyl carbon, get's more electrophilic.
An example would be the Luche reduction. In this case the carbonyl oxygen coordinates to cerium chloride, thus acting as nucleophile, and makes the carbonyl carbon even more reactive towards nucleophiles.

Thank you the explaining this. It makes more sense now. However, I don't have this reaction the my textbook (The Luche reduction). Do you know any other more common kind of reactions to be used in this situation?

As I have to indicate a reaction regarding this nucleophilic site.

Well, I've linked the wikipedia article so you can look it up there. Another example would be the acid catalyzed synthesis of ketals. Or the acid calayzed formation of Cyanhydrine ( https://de.wikipedia.org/wiki/Cyanhydrine ) , which is actally quite interesting since you can use lithium salts instead of acids to catalyze that. Li+ acts quite similar to H+.

Thank you so much for your great answers, DSVA! Now it makes more sense and I can carry on with my assignment.


text:
How to assign peaks to mannose using H-NMR and COSY?
I did a H-NMR experiment with D-mannose in deuterated DMSO. I have difficulty assigning the peaks. I know doublet a is the anomeric centre C1 and that triplet f is the OH attached to C6. Doublets b,c,d and e also correspond with the OH groups. Looking at the COSY you see that a and b are next to each other which suggests that b is the OH on the anomeric carbon C1. However, COSY also shows that b and h are next to each other. How is this possible if with COSY you are only able to look 3 bonds further?

What resolution did you run the 1H NMR at ? Theres often a lot to be obtained from analysis of the peak splittings, but they're essentially non-existent here.

It's also way too concentrated which makes the peaks look ugly.

In addition to the above comments, in addition to the plot provided here (an overview, and already helpful for the signals labelled with a, and b) an expansion of 3--5 ppm (both along $f1$ and $f2$) with both the projection along the two axes would be beneficial for the attribution of the remaining cross signals.  And, you likely have access to the integration of the signals and their coupling constants, too; at least from a 1D $\ce{^1H}$-NMR spectrum of the same sample, which equally helps to re-attribute the signals properly.


text:
How did early chemists determine the major product of a solution?
For example, how did they determine using evidence that when 2-chloropropane reacted with ethoxide, a strong base, formed propene as the major product and 2-ethoxypropane as the minor product?
Could someone explain how did chemists determine the major products without NMR Spectroscopy? I have no lab experience and do not have access to them either.

there's also https://chemistry.stackexchange.com/questions/54381/how-to-determine-the-structure-of-organic-molecules-without-spectroscopy

I am not familiar with how a degradation reaction is done. I know this is a forum for professional chemists and those who have studied it at a university, but for me the only chemistry course I've taken was in high school. Elaborate if you can

After 11 questions you still dunno it's not forum for chemists but Q&A site for everyone? BTW what degradation?

In absence of currently available (or, in principle available) spectroscopies like NMR, IR, Raman, UV-Vis; and mass spectrometry our ancestors tried first to separate the products obtained by sequential recrystallisation.  Or by extraction under acidic, neutral, basic conditions (including back extraction).
They then recorded other physical properties of the purified products, if accessible, which then were compared with reference data.  For example, melting point and crystal shape if the products were solid, otherwise boiling point, refractive indices of the products obtained.  Then, they knew a whole array of pretests to check for the presence / absence of elements in the products formed, like Beilstein test for halogens and Lasaigne test for nitrogen.  They would determine the molecular weight by means of cryoscopy or ebullioscopy.
From a chemical point of view, it was common to subject the products to combustion analysis to determine the molecular formula.  (One of the instruments for this, the  Kaliapparat is still present in the logo of The American Chemical Society.)  Till today, there are journals preferring this over a mass spectroscopic determination of the molecular weight.
They knew about reactions that would yield derivatives, for example from the reaction with picric acid (affording pircrates), or with toluenesulfonic acid (yielding amides); often crystalline with a sharp melting point, easing to compare with reference data.  If not, other reactions like the haloform reaction would sequentially degrade the products into fragments, and then these products obtained would be characterised by physical or chemical means.
And obviously, the balance was much more "sacred" than today, where often we "just" weight-in reactants, and weight-out products.


text:
How to solve for concentration of two equilibrium reactions with finite equilibrium constants?
I have a small problem of equilibrium that I counldn't solve. I want the exact answer. No approximations please.
$$
\begin{array}{ll}
\ce{H2O + H2O <=> H3O+ + OH-} &\quad\left(K_w = 1\cdot 10^{-14}\right) \\
\ce{HNO2 + H2O <=> H3O+ + NO2-} &\quad\left(K_a = 4\cdot10^{-4}\right) \\
\end{array}
$$
Objective: Find the concentration every substance, given that the initial value of $[\ce{HNO2}]_0$ is known. Feel free to insert any numerical value for it, if you wish. I have no idea how to solve for the concentrations, when I have two finite ionization constants. How to solve it?
Of course, in this situation, most people would ignore the contribution of the auto-ionization constant of water (that is, ignore the first reaction), leaving only the second one, which is fairly easy to calculate the concentrations. I do not want that.

The reading of this part is optional. Feel free to skip everything down below if you wish. This is an example that I understood how to solve:
$$
\begin{array}{ll}
\ce{H2O + H2O <=> H3O+ + OH-} &\quad\left(K_w = 1\cdot 10^{-14}\right) \\
\ce{HCl + H2O <=> H3O+ + Cl-} &\quad\left(\mbox{Full Dissociation}\right) \\
\end{array}
$$
Hereby, lets say we had an initial concentration of $[\ce{HCl}]_0$. It will then fully dissociate, thus we shall have: $[\ce{HCl}]_0 = [\ce{H3O+}]_0$. Now we correct considering the water autoionization:
$$
\begin{array}{ll}
[\ce{H3O+}] = [\ce{HCl}]_0 + x \\
[\ce{OH-}] = x \\
\end{array}
$$
We solve for $x$ such that $[\ce{H3O+}][\ce{OH-}] = K_w$. This will give me a quadratic equation, which using bhaskara becomes easy to solve.
Again, most people here would ignore the first reaction. And, in certain circumstances, they could, for instance, find a potentially wrong value of $\text{pH}$. Classical example: Just let $[\ce{HCl}]_0 = 10^{-8}$, and then ignoring water autoionization, it will fully dissociate thus $[\ce{H3O+}]_0 = 10^{-8}$, giving $\mathrm{pH = 8}$. So, we just got a basic $\text{pH}$ after just inserting hydrochloric acid in water!

https://chemistry.stackexchange.com/questions/10249/determine-the-ph-of-mixture-of-two-weak-acids

https://chemistry.stackexchange.com/questions/7648/does-the-number-of-h-ions-in-solution-go-up-on-dilution-of-a-weak-acid

https://chemistry.stackexchange.com/questions/69307/ph-of-mixture-of-strong-and-weak-acid

Completely accurate answer :
The main reactions which take place are
$$
\begin{array}{ll}
\ce{ H2O <=> H+ + OH-} &\quad\left(K_\mathrm{w}= 1\times 10^{-14}\right) \\
\ce{HNO2 <=> H+ + NO2-} &\quad\left(K_\mathrm{a}= 4\times10^{-4}\right) \\
\end{array}
$$
$\ce{HNO2}$ acid is a weak acid, which does not dissociate completely in water. This means it does not give all its hydrogen ions into the water, the resulting solution contains four species ${\ce{H+}, \ce{OH−},\ce{NO2−}}$, and undissociated acid $\ce{HNO2}$. In order to specify the concentrations of the four species
present in an aqueous solution of $\ce{HNO2}$, we need four independent equations
 between them. These are :
1- Equilibria. We have two equilibrium equations:
The dissociation equilibrium of water
$$\ce{ K_\mathrm{w}=[H+][OH−]}\tag{1}$$
The dissociation equilibrium of acid
$$\mathrm{ K_\mathrm{a}}=\frac{[H^+]_e[NO_2^−]_e}{ [HNO_2]_e}\tag{2}$$
2- Mass balance. The mass balance equation that relates the concentrations of the various dissociation products of the substance to its initial concentration, which we designate here as $\ce[HNO_2]_0$. For the solution of $\ce{HNO_2}$, this equation would be
$$\ce{[HNO2]_0  = [HNO2]_e + [NO2−]_e}\tag{3}$$
3-Charge balance. In any ionic solution, the sum of the positive and negative electric charges must be zero; in other words, all solutions are electrically neutral. This is known as the electroneutrality principle.
$$\ce{[H+]_e = [NO2−]_e + OH−}_e\tag{4}$$
We can use ${\text{Eq}\mathrm{3}}$ to get an expression for $\ce{[HNO2]_e}$, and this can be solved for$\ce{ [NO2−]_e}$; these are substituted into $\text{ Eq}\mathrm{2}$  to yield
$$ K_\mathrm{a}=\frac{\ce{ [H+]_e} {([H^+]_e - [OH^-]_e)} }{[HNO_2]_0 - [H^+]_e + [OH^-]_e}  \tag{5}$$
This equation is cubic in $\ce{[H+]_e}$ when $\ce{[OH−]}$ is expressed as $\frac{K_\mathrm{w}}{\ce{[H+]_e}}$; it becomes as
$$ K_\mathrm{a}=\frac{\ce{[H+]_e} {([H^+]_e – K_\mathrm{w}/{[H^+]_e})} }{[HNO_2]_0 - [H^+]_e + K_\mathrm {w}/{[H^+]_e}}  \tag{6a}$$
Or
$$\
 [\ce{H+}]_e^3
  + K_\mathrm{a} [\ce{H+}]_e^2
  -(K_\mathrm{w} + [\ce{HNO2}]_0K_\mathrm{a})[\ce{H+}]_e
   - K_\mathrm{w} K_\mathrm{a}
  \tag{6b}
$$
 For most practical applications, we can make approximations that eliminate the need to solve a cubic equation.
a- Unless the acid is extremely weak or the solution is very dilute, the concentration of $\ce{OH−}$ can be neglected in comparison to that of $\ce {[H+]_e}$. If we assume that $\ce{[OH−]_e<<[H+]_e}$, then $\text{Eq}\mathrm{5}$ can be simplified to
$$K_\mathrm{a }=\frac{[H^+]_e^2}{[HNO_2]_0 –[H^+]_e}\tag{7}$$
which is a quadratic equation:
$$\ce{[H^+]_e^2 + K_\mathrm{a}[H+]_e – K_\mathrm{a}[HNO_2]_0 = 0}\tag{8}$$
b-If the acid is fairly concentrated (usually more than $\mathrm{10^{−3}}\text{M}$), a further simplification can frequently be achieved by making the assumption that$\ce{ [H^+]_e << [HNO_2]_0}$. This is justified when most of the acid remains in its protonated form$\ce{ [HNO_2]_e}$, so that relatively little $\ce{H+}$ is produced. In this event,$\text{Eq}\mathrm{7}$ reduces to
$$K_\mathrm{a}=\frac{[H^+]_e^2}{[HNO_2]_0}\tag{9}$$
Or
$$\ce{[H^+]}  \approx  \sqrt{K_\mathrm{a}\ce{[HNO_2]_0}}\tag{10}$$
-   Comparing of the$\ce{ [H+]_e }$values  resulted from solution the three
 derived equations when the initial acid concentration equal $\mathrm{0.1}\text{M}$
1-  The completely accurate answer solution using cubic equation $\mathrm{(6b)}$by wolframalpha
$\ce{[H+]_e} = 0.00612772\text{M}$
2- The approximate  answer solution using quadratic equation $\mathrm{(8)}$
$\ce{[H+]_e} = 0.00612771680\text{M}$
3- The further approximate  answer solution using the reduced equation $\mathrm{(10)}$
$\ce{[H+]_e} = 0.00632455532\text{M}$

I'd run a solid black line after the equation 6b and say something to the effect "This is exact. Now to the approximations."


text:
Calculate the molarity of Na+ ions in 20 mL of a .130 M solution of Na2Cr2O7?
The answer is $\pu{0.260 M}$, but I keep getting $\pu{0.0052 M}$ because I thought the right way to do it was by multiplying $\pu{0.130 M}$ by $\pu{0.02 L}$, then multiplying by $2$ moles of $\ce{Na+}$. What would be the right way?

$0.0052$ is the number of moles of $\ce{Na+}$ ions in solution, but to find the molarity, you need to divide by $\pu{0.02 L}$. Therefore, the final answer is $\ce{2\cdot 0.130 \cdot 0.02 \cdot \frac{1}{0.02} = 0.260 M}$


text:
Elephant Toothpaste demo: How to concentrate hydrogen peroxide (freezer?)
I would love to do an 'elephant toothpaste' demo in my Chemistry classes this week, but I do not have enough time to order 30% $\ce{H2O2}$ from a supply company.  I have been reading (and seeing on YouTube), however, that some people choose to create their own concentrated $\ce{H2O2}$ (concentration allegedly in the vicinity of 20-35%) by putting 3% $\ce{H2O2}$ in the freezer overnight, then separating the concentrated $\ce{H2O2}$ from the then-frozen water.
My ideal plan would be to use all of the $\ce{H2O2}$ that I produce during my classes throughout the day.  I've never worked with concentrated $\ce{H2O2}$ and so would like to ask.

Would anyone here actually recommend this method to me, given that I would be using this in class for the Elephant Toothpaste demo?  Are there any safety-related deal-breakers that I should be aware of?
What is the highest possible concentration of $\ce{H2O2}$ I could feasibly achieve by storing a bottle in a school's lunch-room freezer overnight?  I don't need anything above ~30% and would just love to know if this method has the potential to make >35% concentrations.
If I do happen to have extra $\ce{H2O2}$, what is the simplest way for me to dispose of it responsibly?  I know that $\ce{H2O2}$ will eventually decompose into $\ce{H2O}$ and $\ce{O2}$, and would rather not have to find a way to store home-made $\ce{H2O2}$ of an unknown concentration, but want to be safe and responsible with any disposal methods I use.

It is a very very bad practice to put chemicals in a fridge used for food storage. It is a safety consideration that you shouldn't violate. Especially if there are others that don't know about the "experiment."

@MaxW I appreciate your concern, but I am not sure what you are trying to warn me of.  Some of the substances found in home refrigerators are what we might think of as "chemicals" (not specifically intended for consumption), such as baking soda (an open box is often used to enhance "freshness"), vinegar (effectively diluted acetic acid), and even gel-based cold packs generally kept in freezers.  If we really want to go into detail, we could accurately say that everything in a refrigerator is a "chemical" product.  Hydrogen Peroxide decomposes into water and oxygen.  Is there a specific risk?

@MaxW I'm sorry, the rest of your comment was initially hidden, and I didn't get past the first sentence.  If I were to tell my colleagues not to touch the freezer bottle (I know they would not), what are the specific potential safety risks that I would be taking?  I definitely understand that if someone were to take the bottle out and start fiddling with it, there could be problems, but the same would go for the rest of my chemical supply.  If everyone were to heed my request, should I have any reason to think that the H2O2 would damage or contaminate anything in the fridge or freezer?

My point is simply that you shouldn't try to rationalize that putting chemicals in a school lunch fridge is safe. Anytime that you're sure that you've made something idiot proof, you've vastly underestimated the intelligence of idiots. // I'm a chemist, so I fully understand that a strawberry is a bag of chemical;s. ;-)

Thanks, I do definitely see your point.  I probably won't end up doing this in the school fridge then.  In theory, though, are there any specific risks I should consider, beyond those associated with working with a relatively strong oxidizer and specific to the case of a plastic bottle in the freezer?  My students see so many DIY chemistry "experiments" on YouTube, and I really want to be able to give them good, chemistry-based explanations of how safe or not the author was (since my students are curious about that).  This understanding would generally add to my ability to answer their Q's.

I have seen a number of youtube videos that are extremely reckless. Often a basic safety requirement like safety glasses isn't met. What happens if you dump the acid on yourself? Where is the safety shower and eyewash?

I feel like you are assuming that I am a reckless chemist, and I do not appreciate that.  I am trying to be a voice of reason for my students who ARE often reckless/forgetful but who are also trying to be proud individuals, and I find that saying "don't do it, it's not safe" generally makes them feel belittled, and they don't listen.  They might do sthg just because I said not to.  If I can provide them with an explanation of chemistry involved, they listen.  I came to this forum with a legitimate request for help with my own chemical understanding.  Please let me know if you can provide that.

Here is a link using 12% H2O2 which can be bought at a beauty supply store. https://www.stevespanglerscience.com/lab/experiments/elephants-toothpaste/  // beauty supply link -- http://www.using-hydrogen-peroxide.com/hairdressing-peroxide.html

The first consideration of any chemical experiment is understanding what safety procedures are needed. Also how to safely dispose of chemical waste. // "Small" quantities of hydrogen peroxide solutions ( a quart, not a railroad tanker car...) can be disposed of with no special treatment. You can safely pour them down the drain of a sink while flushing with water.

Thank you, I appreciate this.  My original, ideal situation was to purchase 30% from an educational chemical distributor, since that was called for in the curriculum I'm currently using.  I'll probably purchase the 12% you linked me to for this year, but in the future, if I were to use 30% in the demo and have left over solution, can I still drain and flush it?  If I buy a 500mL bottle, for example, but can only use 50 mL each in 6 class demos, can I store the remainder for a year (in an expandable accordion bottle), dilute it then drain it, just drain it, etc.?  Just want to be clear.

H2O2 degrades with time. If you use 50ml*6classes = 300 ml then you won't have enough for another year. Why keep it then?

That makes sense, I just don't know how quickly it would degrade, and so was thinking in extreme hypothetical terms so that I could push the limits of my understanding.  I suppose if I kept 200 mL and it kept well enough for a year, the following year I may only need to purchase a 100 mL container to come back to 300 mL total, but I won't bank on it.  If I buy a 500 mL bottle next year and need to dispose of 200 leftover mL safely, how can I do it?

I have to ask the question of why do the elephant's toothpaste experiment, what part of chemistry will it teach the students ?


text:
Positive charge on a zwitter ion at pH=7?

I faced this question. But I am having some problems understanding the solution to this:

How can the negative charge on the peptide vanish on $\pu{pH = 7}$? Suppose, we have some $\ce{H+}$ donor on either $\ce{R1}$ or $\ce{R2}$. Then the $\ce{H+}$ donor can make the carboxyl radical neutral. But by donating a $\ce{H+}$ ion, the donor becomes negatively charged. Maybe, the donor can take the $\ce{H+}$ from the $\ce{NH3+}$ radical. But the result, a completely neutral tetrapeptide, doesn't satisfy the condition of the question.

Need a hint.

Basically, positive charge at pH=7 means there are more basic groups than acidic groups. This would make the peptide accept more protons than it ejects, giving a net positive charge

@Pritt_Balagopal That means I misunderstood the question.

@Pritt Balagopal Then I assume option 4,5,6,9 are correct.

In theory, yes. Although option 6 is not quite sure. Amides are soo weakly basic, I doubt if that counts too.

@pritt
option 6 has no amide. are u talking about option 5? I am confused between option 5 and option 8. -OH radical isn't that acidic. So the amine group alone can make the tetrapeptide basic?

To sum up: of the side chains listed, only the aminoalkyl ones will be positively charged at pH 7, so four peptides -- 4, 6, 8 and 9 -- will have a net positive charge at pH 7.