text:
Why is cyclopropane planar?
What would cyclopropane look like if it weren't planar?
How and why does non-planar cyclopropane have higher torsional strain than planar cyclopropane? can you please show me visually via image that non-planar cyclopropane has higher torisional strain and maybe more angle strain than planar cyclopropane?
On the otherhand, why and how is cyclobutane bulged or puckering, a.k.a non-planar?

Considering that three points define a plane, geometrically, it's not really possible for cyclopropane to be non-planar...

Why is cyclopropane planar?

This is like asking why $\ce{HCl}$ is linear–just as two points define a line, three points define a plane. There isn't any way to orient the three methylene groups in space that they aren't coplanar.
$\hspace{8cm}$
Torsional strain in cyclopropane is also significant for this reason (the molecule has a $\mathrm{C_3}$ axis through the center of the ring, meaning every hydrogen is overlapping). Any reduction in torsional strain, however, decreases the already weak carbon-carbon $\ce{sp^3}$ orbital overlap. Though the $\ce{C-C-C}$ bond angles are $60º$, the orbitals are still oriented roughly $109.5º$ from one another, resulting in very weak and highly strained banana bonds.
$\hspace{8cm}$

Why is cyclobutane nonplanar?

Cyclobutane has one additional methylene, and can reduce its torsional strain by orienting one methylene out of the plane of the other three.
$\hspace{7.3cm}$
This is in part possible due to the much greater carbon-carbon $\ce{sp^3}$ orbital overlap.

Due to symmetry, the carbon in cyclopropane is not $sp^3$ but $sp^2$ hybridized.

If cyclopropane did pucker then a new plane would exist. This is because any set of three points (a.k.a three carbons) is coplanar. Cyclobutane puckering can occur and make a difference as a set of four points/carbons is not necessarily coplanar.


text:
Is cis-1,2-dimethylcyclohexane a meso compound?
I was watching a video lecture today on hydrocarbons and came across this. The instructor says that, as there is a plane of symmetry, cis-1,2-dimethylcyclohexane is a meso-compound:

However, the cyclohexane ring is never planar as it adopts a chair conformation. I cannot find any plane of symmetry in chair form. Like this:

Is the compound meso or not?

cis-1,2-Dimethylcyclohexane is achiral, not because there is a plane of symmetry, but because it consists of two enantiomeric conformations which interconvert rapidly via ring flipping at normal temperatures.
This is exactly the same case as amine inversion. "Chiral nitrogens" such as that in $\ce{NHMeEt}$ do not lead to chirality or optical activity because of rapid inversion of configuration at the nitrogen atom, leading to interconversion of the two enantiomeric forms 1a and 1b.

Likewise, the ring flip in cis-1,2-dimethylcyclohexane leads to two different conformers. I have deliberately chosen to depict the ring flip in the following fashion, to make the mirror image relationship more obvious. The green methyl group, equatorial in conformer 2a, is changed into an axial methyl group in conformer 2b. Likewise, the blue methyl group goes from axial to equatorial. The 1,2-cis relationship between the two methyl groups is retained in both conformers.

Each individual conformer can be said to be chiral, but just like how the amine is considered achiral, cis-1,2-dimethylcyclohexane as a whole is considered achiral.

Is the compound meso? According to the IUPAC Gold Book, a meso-compound is defined as:

A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

1,2-Dimethylcyclohexane possesses two diastereomers, one cis and one trans form. The trans form is chiral, but the cis form is achiral, as explained above. Therefore, the cis form satisfies the above definition and is considered a meso-compound.

Can we get it cold enough to freeze the enantiomers long enough to detect them?   Can we see evidence of non-racemization in an optically active environment?

@OscarLanzi I don't know. I presume it is possible, but for all intents and purposes it's achiral.

Sir, I'd like to ask one thing, suppose we have the compound 1-(chloromethyl)-3,5-dimethylcyclohexane in which all three substituents (the 2 methyl and one chloromethyl) are coming out of the plane, is this compound meso? Due to the same reason as explained in this answer (flipping of ring)? It does have two stereogenic centres (the 3,5 positions)...

@LightYagami Yes, it is the same thing.

Is it wrong then to see "plane of symmetry" at all in the rings? Because that is not how in reality it exists, but is it a conicidence that if we answer according to the 2D picture (using POS idea), the answer still comes out to be the same. I am also not able to see why the same argument for the trans form cannot be applied in this case.

Is the answer to my second question that, the same argument cannot be applied to the trans form because in cis,due to ring flip we were getting both $(R,S)$ and $(S,R)$ but in trans, if the isomer is $(R,R)$, then it will remain $(R,R)$ only even after flipping and hence the other conformer is not obtained, making it chiral?

@LightYagami I think you should ask a new question, this is getting too substantial for the comments. To be honest, I also don't understand what you mean, so you will need to explain it a bit more clearly. Your compound has three different groups on the cyclohexane, you can't call it cis or trans. (R) and (S) descriptors are to do with configuration and not conformation. A ring flip doesn't interconvert (R) and (S) under any circumstances whatsoever.

@orthocresol cis-1,2-dimethyl cyclohexane at room temperature is optically inactive due ring flipping right? It means in mixture we have equal percentages of enantiomers. So why should we call it as a meso reather than a racemic mixture.

And also I have a doubt, if we write absolute configurations at two chiral carbons in compound 2a of your post we R and S . This means that the compounds 2a ,2b possess internal compensation. So none of the compounds 2a ,2b are optical active. Then how both are enatiomers?

@Infinite (1) It's a bit of a grey area, but the key point is that the enantiomers are not isolable at RT. "Racemic" implies that you can separate and isolate the two enantiomers. (2) It being (R,S) means that the compound is optically inactive, which completely agrees with what I wrote. This, however, does not rule out that a specific conformation may be chiral. On balance, when all conformations are considered, it will be optically inactive. Note that the same is true of, say, meso-tartaric acid, for which one can easily construct a chiral conformation.

@Infinite Note, further, that if you are describing one particular species or conformation, then it is better to use the term "chiral" or "achiral". On the other hand, "optically active" or "optically inactive" should only be used to refer to a bulk sample of such a compound, because optical activity is something you can only measure on a bulk sample. You can't go into the lab and determine how a single molecule rotates plane-polarised light. You can, however, use symmetry arguments etc. to show that the molecule is or isn't superimposable on its mirror image: hence chiral or achiral.

I'd like to add a few points definition wise from the IUPAC goldbook.
(Because recently the original website seems to be broken I am using links via the internet archive.)

meso
  A prefix to indicate the presence of a symmetry element of the second kind (see chirality) in a chemical species which is usually one of a set of diastereoisomers that also includes a chiral member.

However, there is a more general definition, too:

meso-compound
  A term for the achiral member(s) of a set of diastereoisomers which also includes one or more chiral members.

The first part clearly does not apply to the cis-isomer, (1R,2S)-1,2-dimethylcyclohexane, the second does because of the definition of chirality.

chirality
  The geometric property of a rigid object (or spatial arrangement of points or atoms) of being non-superposable on its mirror image; such an object has no symmetry elements of the second kind (a mirror plane,  σ = S1, a centre of inversion,  i = S2, a rotation-reflection axis, S2n). If the object is superposable on its mirror image the object is described as being achiral.

I think the key point of this definition is in this case, that the object needs to be rigid, which (1R,2S)-1,2-dimethylcyclohexane is not at room temperature. If you cool down the solution, inhibiting the ring flip, you would probably be able to find both configurations; that is a racemic mixture of them.
Since (1R,2R)-1,2-dimethylcyclohexane and (1S,2S)-1,2-dimethylcyclohexane, the trans-conformers, are both chiral (and enantiomers), the remaining member of the set (1R,2S)-1,2-dimethylcyclohexane can be considered a meso-compound, since it is achiral.

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text:
Phosphoglucose isomerase mechanism
The mechanism for phosphoglucose isomerase changing glucose  6 phosphate to fructose 6 phosphate shows histidine protonating the c5 oxygen and lysine deprotonation the c1 oxygen to form an open chain aldolase. Glutamic acid then deprotonates c2 to create an enediol intermediate and donates the proton to c1. For the proton to be transfered from c1 to c2 the c1 oxygen needs to be protonated and the c2 oxygen deprotonated. Does anyone know of a source that specifies the amino acid residue or residues that facilitate the deprotonation and protonation of the c2 and c1 oxygens?

tl;dr: The active site catalytic residues (BH+ and B’) are thought to be Lys and a His–Glu dyad, respectively. These facilitate the abstraction of proton from C2 and subsequent protonation on C-1.

Some important points to note:

"shows histidine protonating c5 oxygen and lysine deprotonating c1
  oxygen to form an open chain aldolase
  ”.

The opening of the ring, is facilitated by Lysine and not by Histidine.

“Glutamic acid then deprotonates c3 to create an enediol intermediate
  and donates the proton to c1”

It is actually the His–Glu  (B’) dyad (see image below) that abstract this proton from C2. See how this process occurs at the same time as the exchange of H+ (protons) with the solvent or medium. In this step C2 loses a proton while C1 gains a proton (Step 3-4)
Finally a proton is added  from the medium and the ring closes

A proposed reaction mechanism for the Phosphoglucose Isomerase reaction involves general acid–base catalysis by the enzyme:
Step 1

Substrate binding.

Step 2

An acid, presumably the Lys ε-amino group, catalyzes the opening of the pyranose ring.

Step 3

A base, presumably the imidazole portion of the His–Glu dyad, abstracts the acidic proton from C2 to form a cis-enediolate intermediate (this proton is acidic because it is α to a carbonyl group).

Step 4

The proton is replaced on C1 in an overall proton transfer. Protons abstracted by bases are labile and exchange rapidly with solvent protons. This step was confirmed by Irwin Rose by demonstrating that $\ce{[2-^3H]G6P}$ is occasionally converted to $\ce{[1-^3H]F6P}$ by intramolecular proton transfer before the $\ce{^3H}$ has had a chance to exchange with the medium.

Step 5

Ring closure to form the product, which is subsequently released to yield free enzyme, thereby completing the catalytic cycle.

Reaction mechanism of phosphoglucose isomerase

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text:
Why was the definition of chiral carbon changed?
Before 1984, the definition of chiral carbon was,

carbon atom that is attached to four different types of atoms or
  groups of atoms

In 1984, the definition was changed to

Any carbon molecule that is not superimposable on its mirror image.

Why was this change made? Don't both definitions mean the same or is/are there cases where the former is false. Preferably, please provide an example

The second definition is much broader. An object which is not superimposable with its mirror image does not have to be carbon, or any atom, for that matter. Your left hand is an example of such object.

how is it broader? what do you mean by the second sentence @IvanNeretin

Now that you've changed the definition, your left hand is no longer an example. Still, the second definition is broader (there are chiral molecules in which no carbon is attached to four different groups). Related: http://chemistry.stackexchange.com/questions/39830/what-kind-of-optical-activity-requires-chirality/39833#39833

No! The first statement is entirely incorrect as 1. there are carbons with four different substituents that are achiral and 2. there are chiral molecules with less than four different substituents. Meaning that the first definition is entirely wrong in both directions. By the way, the second definition is not really wrong, but it is missing the point that it is the existance of any $S_n$ element that renders a molecule achiral. Please be precise here.

Martin and I both spent some time discussing the definition of chirality here: http://chemistry.stackexchange.com/questions/59124/what-is-the-perfect-definition-for-chirality/59144#59144

The IUPAC definition of asymmetric carbon is still:

The traditional name (van't Hoff) for a carbon atom that is attached to four different entities (atoms or groups), e.g. Cabcd.

The definition of chirality centre is:

An atom holding a set of ligands in a spatial arrangement which is not superposable
  on its mirror image. A chirality centre is thus a generalized extension of the concept
  of the asymmetric carbon atom to central atoms of any element, for example N+abcd,
  Pabc as well as Cabcd.

Asymmetric carbon is a complete misnomer because the carbon atom is indeed very symmetric (spherical symmetry). And again, chirality centre is bad for the same reasons (see my answer). Does it make sense to define a chirality centre in a way that it can also be contained in non-chiral molecules? And even so that a molecule can be chiral without having a chirality centre? How can that ever be a sensible definition! The IUPAC gold book also gives those definitions as "traditional names" (see http://goldbook.iupac.org/S05980.html).

@ketbra yes it makes sense to me.  The local environment at the carbon is asymmetrical, yet the molecule may or may not be chiral.  See Le Bel-van't Hoff rule https://en.wikipedia.org/wiki/Le_Bel-van%27t_Hoff_rule It made sense in 1874.  It still makes sense.

I just don't get it, sorry. The 'local environment at the carbon is asymmetrical' don't cause even more confusion because that is already covered by the term chirotopic.

By the way if you say it like that, then most of the carbon atoms would be asymmetrical carbons, as all atoms in a chiral molecule are in a chiral environment, so then all atoms of a chiral molecule would be chiral centers, isnt that kind of confusing. Why not just stick with the clearcut $S_n$ definition then?

@ketbra "The crux of the matter is that chirotopicity and stereogenicity
are conceptually distinct; consider, for example, the halogen atoms
in CHBrClF (chirotopic but nonstereogenic) and the carbon atoms
in the CHCl groups of 1,2-dichloroethene (achirotopic but stereogenic).
Nevertheless, stereogenicity and local chirality appear
to be inseparably linked in the practice of organic chemistry, as
epitomized by the very expression “asymmetric carbon atom”." http://pubs.acs.org/doi/abs/10.1021/ja00323a043

Taking cis-1,2-Dimethylcyclohexane as an example. You think it is reasonable that this molecule contains two chiral centers even though it is achiral? And please explain to me why it is reasonable to have a chiral molecule and always achiral atoms and then to speak of a chiral carbon atom even though it is clearly achiral and only (in some cases) chirotopic.

Sure, stereogenicity and chirotopicity are completely different. I mean a double bond for example may have stereocenters as well. Still it makes much more sense to have sometimes achiral molecules with stereogenic centers than to have achiral molecules with chiral centers. I just prefer the short and correct definition (existance of $S_n$) over the long clumsy and wrong one with 'four substituents'.

@ketbra like the "Stereoisomerism and Local Chirality" article explains, the point of the term "asymmetric carbon" is describe local chirality.

But if asymmetric carbon describes local chirality how is it different from chirotopicity?

@ketbra chirotopic refers to an atom or other entity in a chiral situation http://old.iupac.org/goldbook/C01065.pdf so it is global. See also the quote a few comments above "...halogen atoms in CHBrClF..."

This discussion leads nowhere. My point is, if the carbon is not asymmetric than it is wrong to call it asymmetric carbon. If I say "the kitchen is painted red" than I mean that the kitchen is red and not the house surrounding it. If a carbon atom is not asymmetric and also not chiral and it is termed either asymmetric carbon or chiral center even though it is the surrounding that is asymmetric/chiral than this is a clear case of a misnomer to me, and that's it. If you don't agree with that logic please tell me why but please stop given references that have nothing to do with the discussion.

The first definition is entirely wrong. Take cis-1,2-dimethylcyclohexane as an example. While any of the two tertiary carbons bears four different substituents the molecule is achiral. Then there are molecules (like helicene, BINAP, ...) that have no atoms with four different substituents and are chiral nonethless. So the definition of chirality based on four different substituents is entirely wrong. Then there is the problem that the expression 'chiral carbon' implies that the carbon itself is somehow chiral, which it is of course not, but it is the molecule which is chiral instead. This is why use of the terms 'chiral carbon' and 'chirality center' is discouraged. The correct term is stereogenic center, avoiding those issues alltogether. The second definition is also not good, as it isn't clear what superimposable means. Just use the clearcut definition that an achiral molecules point group contains $S_n$ with $n \geq 1$ elements and that's it!

So basically, if you still use the terms 'chirality center' or 'achiral carbon' or whatever you might end up in the paradoxical (but funny) situation of having an achiral molecule that contains one or even multiple 'chirality centers'. How is that a sensible definition?

your first point ...helicene, BINAP, ...) that have no atoms with four different substituents and are chiral nonethless is unjustified because if 1-> 2 in not necessary equal to 2->1 EDIT: Im sorry, i understood what you meant


text:
Are there any polyprotic acids where Ka2 is greater than Ka1?
My chemistry textbook says that in the ionisation of a polyprotic acid with Ka1, Ka2 and so on, 'the first Ka is usually the largest.' I took the word 'usually' to mean that there are exceptions to the rule, but I haven't been able to find any cases where Ka2 is greater than Ka1.

That's a really odd statement. In my experience, they're always numbered in decreasing order of numerical value (increasing order of $\mathrm p K_\mathrm a$). I'm curious to see if anyone comes up with an exception to this rule.

How do you know which proton is the first? No, it goes the other way around: the largest Ka is the first, because that proton is naturally the first to react. It can't be otherwise.

-sigh- the statement doesn't make much sense to me either, but I'm hoping someone will come up with a case where the second dissociation constant is greater. I recently lost a percentage in an exam for answering according to the textbook and saying Ka1 is 'usually' greater than Ka2, rather than 'always.'

@IvanNeretin What about a hypothetical molecule where the loss of the first proton facilitates the loss of the second proton?

@DHMO I don't see how's that possible, but let's assume it is. Well, then these two protons would react together and effectively have the same pKa.

I'd have to agree with Ivan. Ka1 and Ka2 are just labels. They don't carry any information. Which one is most acidic? The one that is larger. It just intuitively makes sense let the largest one be Ka1. It's an exam question that doesn't really require any understanding of the subject.

One cannot dissociate the second proton without dissociating the first proton therefore, ipso facto, the first proton must dissociate first and therefore, the Ka1 for the first dissociation must be larger than that for the second. If the second proton were to have a Ka greater than that of the first then it would be the first proton to dissociate and not the second.  So, by definition, the first Ka is the largest. Strike "usually"  and substitute "by definition". There are no exceptions to a rule which defines itself.

The only possible exception that I could think of is some compound where losing a second proton after losing a first is easier. This could possibly happen due to the first deprotonation causing the second one to participate in resonance. I don't know if the extra stability from resonance could be more important than the formal charge on the compound, though.

@DHMO You're right! It works exactly like that and I read about an example, long time ago but maybe I'll find it.


text:
Probability of finding a molecule in the ground vibrational level
I wanted to estimate the probability of finding a molecule in the ground vibrational level using the Boltzmann distribution:
$$
p_i = \frac{e^{-\epsilon_i/kT}}{\sum_{i=0}^{N}e^{-\epsilon_i/kT}}
$$
Using the quantum harmonic oscillator as a model for the energy
$$
\epsilon_i = h\nu (i+1/2) =/i=0/=\frac{h\nu}{2}
$$
In the Boltzmann distribution, we have the state of interest divided by the sum of all possible states. But how should I treat the denominator?

Searching a bit I found that the analytical expression for this geometric series is ($i$ not imaginary number)
$$
\sum_{i=0}^{N}e^{-i h\nu/kT} = \frac{1}{1 - e^{h\nu/kT}}
$$
However, is this using a shifted energy scale for the harmonic potential? In that the vibrational energies are $0$, $h\nu$, $2h\nu$, ..., and not $\frac{1}{2}h\nu$, $\frac{3}{2}h\nu$, $\frac{5}{2}h\nu$, ...? Should I make sure I use the same energy scale for the nominator and denominator in the Boltzmann distribution?

Doing what porphyrin suggested, I get
$$
\sum_{i=0}^{\infty} e^{-h\nu(i+\frac{1}{2})/kT} = e^{-h\nu/kT} \sum_{i=0}^{\infty} e^{- ih\nu/kT}
$$
Expanding the four first terms
$$
e^{-h\nu/kT} \sum_{i=0}^{\infty} e^{- ih\nu/kT} = (e^{-h\nu/kT} \cdot 1) + (e^{-h\nu/kT} \cdot e^{-h\nu/kT}) + (e^{-h\nu/kT} \cdot e^{-2h\nu/kT}) + (e^{-h\nu/kT} \cdot e^{-3h\nu/kT}) \\
= e^{-h\nu/kT} + e^{-2h\nu/kT} + e^{-3h\nu/kT} + e^{-4h\nu/kT} = \sum_{1}^{\infty}e^{-n\cdot h\nu/kT}
$$
which has an analytical expression for the converged value, right?

The denominator should be calculated exactly. Ever heard of a geometric progression and its sum?

Ah, yes. Looking at it, it should converge.

The summation is called the partition function and it always extends over all possible levels $n=0 ..,\infty$.  You are assuming a harmonic oscillator which has an infinite number of levels. Its easier if the 1/2  is separated out first and treated separately, this just adds a constant to the energy as you realise. Expand the summation as $1+e^{-a}+e^{-2a}+....$ ($a=h\nu /(kT)$ which converges to the result you quote.

You're on the right track.
Also, using $i$ as an index can be confusing some times because it can be confused with the imaginary number; however, here it should not present a problem. As a matter of habbit however, I like to use $j$ or $n$ or something else..there are only so many letters in the alphabet.
The sum in the denominator is called the partition function, and has the form
$$Z = \sum_{j}e^{-\frac{\epsilon_j}{kT}}$$
For the harmonic, oscillator $\epsilon_j = (\frac{1}{2}+j)\hbar \omega$ for $j \in \{ 0,1,2.. \}$
Note that $\epsilon_0 \neq 0$ there exists a zero point energy.
Let's write out a few terms $$Z = e^{-\frac{\hbar \omega/2}{kT}} + e^{-\frac{\hbar \omega3/2}{kT}} + e^{-\frac{\hbar \omega5/2}{kT}} +..... $$
factoring out $e^{-\frac{\hbar \omega/2}{kT}}$
$$Z = e^{-\frac{\hbar \omega/2}{kT}} \left( 1+ e^{-\frac{\hbar \omega}{kT}} + e^{-\frac{2\hbar \omega}{kT}} +.....\right) $$
The sum in the bracket takes the form of a geometric series whose sum converges as shown below $$1+x+x^2+... = \frac{1}{(1-x)} $$
 herein, $ x \equiv e^{-\frac{\hbar\omega}{kT}} $
Putting all of this together
$$Z = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}$$
Now, $$p_0 = \frac{e^{-\epsilon_0/kT}}{Z} = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{Z} = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{\frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}} = (1-e^{-\frac{\hbar\omega}{kT}}) $$

Great answer. But I strongly disagree with not using i as an index variable. The possible confusion with the imaginary unit only arises from typographic sloppiness. Variables should be italic and mathematical constants should be upright set. This includes e.g. the euler number, but no one would think that you exponented the variable e. From context it is clear that you exponented the euler number. The same holds true for sum symbols, it would make no sense to write the imaginary unit underneath them and even with typographic sloppiness it is clear from context that you mean the variable i.


text:
Why does caffeine powder cause apple cider (or any carbonated drink?) to foam?
Something I noticed today, when adding some pure caffeine powder to apple cider.
Why does caffeine powder cause apple cider (or any carbonated drink?) to foam?
It seems to foam (with around 150mg caffeine added) slowly, but constantly so that it almost seems that it will eventually make all the carbon dioxide fizz to air.
Is this really related to caffeine or the powdered form in general?

Almost all powders will work that way. Each crystal/particle will act as a nucleation site and cause oversaturated solutions of carbon dioxide to fizz out.

The solid grains act as nucleation sites to form bubbles of $\ce{CO2 }$. The $\ce{CO2 }$ is of course unstable in the solution. If you leave the bottle open for a day then it will go flat.

http://chemistry.stackexchange.com/questions/3958/does-citric-acid-react-with-carbonated-water?s=3|0.3475

I don't need to know why you're mixing uppers (caffeine) with downers (alcohol) but if the powdered caffeine is removing the fizz from your cider then you might try dissolving it in hot water first and then adding the hot caffeine solution to your cider.  In dissolved form, the caffeine cannot provide the nucleation sites for the formation of CO2 bubbles that MaxW and vapid pointed out.
The solubility of caffeine in water is 200mg/ml at 80 oC so you should be able to dissolve an adequate dose of caffeine in a small enough quantity of hot water to avoid both excessive warming of your cider and precipitation of the caffeine upon cooling and dilution.
It behooves me to mention that recommended caffeine administration should not exceed 200mg every 3-4 hours in adults, but since you're adding it to alcoholic beverages, I doubt you give a monkey's.  Here's the drugs.com link anyway:
https://www.drugs.com/monograph/caffeine-caffeine-and-sodium-benzoate-injection-caffeine-citrate.html
And solubility data here:
http://www.sigmaaldrich.com/content/dam/sigma-aldrich/docs/Sigma-Aldrich/Product_Information_Sheet/c0750pis.pdf

Cider in North America isn't necessarily alcoholic.


text:
Why adding salt to kneading make it wetter?
Last week I tried a new recipe of pizza dough:

65mL of Water;
140g of Wheat Flour (typo 3);
7g of Yeast;
1 big spoon of Olive Oil;
25g of cooked potatoes;
1g of common salt;

I think the recipe uses cooked potatoes as direct source of strach for Yeast. I have mixed all ingredients and kneaded 20 minutes until having a homogeneous dough that is not sticky. Then I added salt (about 0.5g) to the dough and I mixed it. Then the dough became wetter and stickier, I had to add flour in order to recover from the previous state.
It was an amazing experience, because it is somehow unexpected and counter intuitive. Adding salt to preparation seems to lead to more free water in dough.
I have the feeling that water is bound into macro-structure and adding salt may release some water in exchange of ion caging.
Anyway I would like to deeper explain this phenomenon. I am looking for concrete, scientific and reliable explanation.
Update
I read comments about osmosis phenomenon but I do not think it is the good explanation, it happens but does not explain the final state. Osmotic pressure is a transient phenomenon and really need a system boundary to be defined (after all this is a Fick force applying over a surface). I agree that a gradient exists at the beginning, and when mixing but it does not explain why afterwards there is more free water in the whole system. I also agree that there are local gradient, because of macro structure (proteins, etc.) but I am not asking about the transient phenomenon, I am asking about the overall equilibrium.
I express the question in another way:
Why adding salt to a finite mass of dough (of fixed recipe) releases some water and force us to add flour to keep dough constancy as before?

Osmosis.

@DHMO have you read the question, your comment was really fast (quite instantaneous)? Anyway, could you develop what you mean by osmosis in this particular case (where are boundaries and how fluxes are defined), and make an answer then I could accept it if relevant.

The wheat proteins trap water. You're pulling out that water via osmosis by adding more salt.

I doubt it's just the proteins. Starch is a polymeric sugar, and the salt ions are too big (hydrate shell) to freely diffuse into them. So, they create an osmotic pressure.

updated question

maybe the salt killed the yeast and made it excrete it's water content

@Riftcaster nice idea, just for information the process is really quick, about 10 seconds, do you think it holds?


text:
Bromination of acetanilide
Why does acetanilide gives exclusively para isomer. I know that -I of nitrogen must decrease the yield of ortho product, but still it should be made in accountable amounts. Where am I going wrong.
I am inserting the question and solution screenshots for reference. The question is from IIT JEE 2016 exam, chemistry section of paper 1
Question

Solution

It probably does give some of the ortho product. Do you have a link or a specific example in mind?

Agree - I can certainly imagine that there is less ortho product compared to, say, aniline, but 0% ortho doesn't seem plausible.

@Zhe I uploaded the source. Here you go!

@orthocresol I agree. That should be true. I have uploaded the source.

I am not familiar with this bromination reaction. Usually, you see something like $\mathrm{Br_{2}}$ and $\mathrm{FeBr_{3}}$. Bromate and hydrobromic acid are just going to give bromine, and I'm a bit suspicious that bromine without a catalyst will perform EAS.

Well, apparently this does work: http://websites.rcc.edu/grey/files/2012/02/Bromination-of-Acetanilide.pdf
In retrospect, steric hindrance had to be the right answer... :/

How comes that the amide is not cleaved in HBr and only later when treated with $\mathrm{H_3O^\oplus}$? : D

@ketbra You need water to hydrolyse an amide.

@orthocresol Ok thats kind of an argument. But I suspect HBr means HBr/H2O right? At least thats how I understand it. Or do you think they do a gas phase reaction there?

@ketbra Yeah, HBr probably means hydrogen bromide dissolved in water. I suppose there is just not enough water for hydrolysis to compete with the bromination.

But one serious problem with those highschool questions remains: Not giving equivalents, temperatures, solvents, reaction times etc. really makes it impossible to answer those questions without knowing the solution already : )
They could have at least provided the number of equivalents during the bromination.

@ketbra Yeah, indeed. The truth though is that most people only care about such things when they actually have to go into the lab and do it.

One reason: 1,3-allylic strain of the amide group hindering attack at the ortho position. However, I still find it amazing that there is exclusively no byproduct formed. Most aromatic brominations (mostly with NBS) I did required cooling to -78°C and then slowly warming up to room temperature to prevent extensive formation of multiply brominated side products. By the way, the $\mathrm{BrO_3}/\mathrm{HBr}$ combination is used to generate $\mathrm{Br_2}$ in situ thereby leaving the concentration of elemental bromine at a bare minimum. This prevents the formation of polybrominated site products. A similar protocol is $\mathrm{KBr}$/oxone. However, the question is clearly ill-defined as one should have provided the fact that 1 eq of $\mathrm{BrO3}/\mathrm{HBr}$ is used. If you would have used more equivalents, chances are good you would have gotten (d) as well.
Also, I find the description given in (http://websites.rcc.edu/grey/files/2012/02/Bromination-of-Acetanilide.pdf) questionable, I think that the main point for the high regioselectivity is the use of $\mathrm{BrO3}/\mathrm{HBr}$ and not the steric hindrance of the amide. I did bromination on aromatic amides as well and regioselectivities were not that much higher at all, in my experience.

Please explain how did the use of KBrO3+HBr enhanced regioselectivity.

KBrO3 oxidizes HBr -> Br2 is formed. Br2 does electrophilic aromatic substitution. But because the oxidation is relatively slow, low concentration of Br2 is formed at a time. Hence, there is not too much bromine there at a given moment and multiple bromination is prevented.

Turning the amino group into an amide serves two purposes:

We are introducing a sterically bulky group. We expect the amide bond to be as complanar as possible with respect to the benzene ring to maximise electronic interactions. That should block one of the two ortho positions sterically.
We are turning an electron-rich, $+M$ aromatic system into an electron-poor, $-I$ one.

Especially the second transformation will greatly reduce the reaction rate. Furthermore, a bromine atom is mildly deactivating, too, due to its $-I$ effect (the weak $+M$ effect is neglegible and only responsible for the ortho/para directing abilities). Thus, it is very likely that we can selectively stop the reaction after monobromination.
The question still arises why we only obtain the para-product. Obviously the amide behaves much like a bromide and displays a weak $+M$ effect in spite of the amide resonance. Yet we can imagine the the proximity to the bulky amide group inhibits substitution at both ortho-protons sufficiently to allow isolation of the para-product.


text:
How can I remove a PEG shampoo from a natural horn-comb without damaging it?
I washed by accident a horn-comb with a shampoo that contains PEGs, now how can I remove these types of chemicals from the horn-comb without damaging its surface?

Soap, water and a dash of elbow grease. PEGs are all surfactants which will wash away with water and if they don't then an alcohol-water mix (vodka) will certainly remove any PEG residue, probably along with some of the natural oils in the horn comb that has built up over time and usage.
This may leave the comb with whitened patches or streaks on it.  These are not a PEG residue but fine cracks in the horn, interstices in the layers of keratin that the horn is made of, that used to be filled with oil which the PEG shampoo has removed.  Any number of natural oils will improve the appearance of the horn, including, but not limited to, natural scalp oils (i.e. use the comb for its intended purpose), linseed oil (a favourite of cricketers for their bats), castor oil, beeswax, olive oil (a bit sticky) or a light application of any proprietary furniture polish.  Choose something unscented if you don't want Mom to find out. And start with the least viscous options you have, these will achieve greater penetration of the fine cracks in the horn and "seal" them with a wax if possible.
I've seen antique tortoiseshell respond very badly to cleaning with detergents, but the damage is really illusory and can be restored with the gentle application of oils and waxes.

Isopropyl alcohol is a lot cheaper than vodka for basically the same effect.

Thanks, guys... @Isobutane, it's actually the comb of my girlfriend. :)))


text:
Which side of an aluminum foil should I put in contact with food?
Aluminum foil has two sides: a matte one, and a shiny one.
My question is: which one should I put in contact with food?
Criteria are the lowest risk of toxicity and best qualities for food preservation (e.g. do not favor oxidizing, or react with acid nor alkaline ingredients)

For information, I've come across these several explanations; they are however contradictory (and might not be accurate):

Both sides have the same properties. The difference in terms of appearance is only due to the manufacturing process. (Two foils are laminated at the same time, so each foil is in contact with the rolls on one side only — hence the asymmetrical look). In this case, we can use both sides of the foil.
Aluminum foil is actually made of two layers: one of aluminum, another that is coating. The surface of the aluminum layer that is in contact with the atmosphere oxidizes and turns into aluminum oxide — hence the matte side.In that case, we should use the shiny side, since aluminum oxide is not that healthy.

Aluminum oxide is very inert, so I'm not sure how one can imply that it is not healthy. Not that I'd suggest eating sapphires, but still...

@JonCuster : I'm not implying these statements are true: I actually don't know. I just listed them for highlighting I looked for answers by myself, and as potential explanations (that can be falsified because you explain why they are false as you did). || I've also found this question, but I am looking for an answer rather based on chemical reaction (hence here asked).

Both sides will be coated with aluminium oxide which is inert and safe. Shiny or matt has nothing to do with it.

In this post (in french), Prof. Fenster of McGill Univ. states that the difference is due to the manufacturing process (and so, both side have same chemical properties). This is an argument from authority (which I wanted to avoid), but since no source have been cited in given answers, it's the best I've found.

https://universityhealthnews.com/daily/memory/aluminum-linked-to-alzheimers-disease/

After some thought, I guess I'll go with the first argument, that seems plausible (though I don't think it's accidental). As for coating one side with a layer, that seems like a bit too much.
Assuming the second argument is correct, then it's most probable that it doesn't matter which side you use. Because if one side is a 'coated', that 'coat' has got to be safe. Since well, aluminium foil is used pretty much only for packaging food, I don't think people could go about selling Al foil coated with some toxic film for the past couple of decades or so without someone pointing out something very wrong.
Now if it's the first argument, then it still doesn't matter which side you use. Aluminium oxide isn't really as dangerous as out make it out to be, especially if you're dealing with an oxide film that's still on the foil (Hogging a tablespoon of powdered aluminium oxide is another story altogether...). We sometimes call that process (aluminum developing an oxide coat) Passivation, because the foil is rendered 'passive' or 'inert'. Why else do you think it takes so long to get aluminium to react with acids or alkalies?
In a nutshell, regardless of which argument is correct, it'd be perfectly safe to use either side of the foil.
The reflective surface will reflect heat and the matte side will absorb heat. If you're baking or defrosting, the matte side will absorb more radiant heat and reflect less infrared heat while the shiny side will reflect more of both, so it makes more sense to bake and defrost with the matte side facing up. Conversely, during freezing, it makes more sense to reverse the sides as the shiny side face up would reflect heat away from the food.
I've always felt that one of the more practical reasons for creating two visually dissimilar sides on the foil, is to make it easier for you to identify which side you had initially used to wrap some food in. Say, you had just hurriedly unwrapped a cheeseburger and you don't feel like finishing it off. If you re-wrapped using the wrong side, you just get greasy fingers, which is quite annoying.
EDIT: I've read @Vedant's answer and he seems to have provided a good explanation as to why the two surfaces are dissimilar. But I wouldn't entirely dismiss the reason I've mentioned (identification), since, from my personal experiences...I'm pretty sure my keyboard would be covered in grease by now.
But all the same, I've included @Vedant's explanation as well, to give you the complete package.

Thanks for citing my answer.  If i have a paper and I rub one side with sand paper it become rough, but that does not change its chemical properties, it still remains paper. The same way one side of the aluminum is optically reflective and the other is not.

Well, I didn't know about the advantage it holds when you're baking/freezing before I saw your answer, so kudos to you!

Surprisingly the comments did not yield the correct answer
The reflective surface will reflect heat and the matte side will reflect less heat .
If you're baking or defrosting, the matte side will absorb more radiant heat and reflect less infrared heat while the shiny side will reflect more of both, so it makes more sense to bake and defrost with the matte side facing up. Conversely, during freezing, it makes more sense to reverse the sides as the shiny side face up would reflect heat away from the food.
Thought it is true that this happens because of the manufacturing process, it is also true that both have different uses.
EDIT:
Also note, that which side facing is out causes very little difference because aluminum is a very good conductor of heat.
EDIT 2:
From Aaron Abraham's answer

In a nutshell, regardless of which argument is correct, it'd be
  perfectly safe to use either side of the foil.

In a nutshell, regardless of which argument is correct, it'd be perfectly safe to use either side of the foil.

I don't think it is accurate to say the matt side will absorb heat. It might be slightly less good at reflecting it, but that is a minor issue in practice.

It makes no difference; but, if the food is wet and has salt or is acidic it will corrode through in a matter of hours (at ambient temperature, not if immediately frozen). The different appearance is because the final rolling is done with 2 layers of foil. One side of each sheet contacts the polished steel roller, the other side faces the second sheet of foil.

I have noticed in my years of use, that when you use aluminum foil to cover anything tomato based, if you use the matte side, the acid from the tomatoes eats through the foil. So, I have always put the shiny side down or inward, toward the food.

Neither side of the aluminum foil is coated. (Source.) Tomatoes are acidic. It's recommended not to let salty or acidic foods touch either side of aluminum foil. See here.

Back when I was in my final year of high school, the subject of our major chemistry assignment was Aluminium.  I remember the class being taken on a tour of an enormous aluminium factory, where we stood overlooking the massive rolls of aluminium and the successive stages it went through during production.  I distinctly recall the foreman/guide telling us that one side of the aluminum foil was coated during the rolling process with an extremely thin layer of another substance, and this is why it had a duller appearance.  I don't remember what the substance was (I believe he used the words 'plastic substance' but I doubt he was referring to the usual concept of 'plastic'), or any reason given for this.  This was in the late 80's.  I never forgot it, & from that day forward I always used the shiny side against any food.
Some people don't believe me when I tell them this, but it's absolutely true, and I see no reason why the man would have made something like that up.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How can one tell from the MO diagram of the cyanide ion that the HOMO is carbon-centred?
The cyanide ion, when acting as a nucleophile, typically attacks via carbon as its HOMO has a greater contribution from carbon. But how can we deduce this from the MO diagram?
To me it appears that the nitrogen $\mathrm{2p}$ orbitals are energetically closest to the $3\sigma$ HOMO, so its coefficient should be larger in the HOMO. What is wrong with this line of reasoning?
.

related http://chemistry.stackexchange.com/q/18449/16683

can u please @martin-マーチン answer the question ..as in the linked question it is mentioned that We can clearly see that the highest coefficient of the HOMO is at the carbon end, this is consistent with the fact, that it will most likely bond with this end to a more electropositive element. i dont know how to tag , can someone pls tag him.

@akso854 There are 4 MOs with the same $\mathrm{\Sigma}$ symmetry. So the exact energies are going to matter here. I suspect that you're getting a much larger contribution from the carbon $\mathrm{2s}$ orbital, and that seems consistent with the shape of the boundary surface in the diagrams.

thanks @Zhe ..i also think the same. 2s orbital is behind greater lobe on carbon in HOMO.

Whilst I don't think this post in unclear in it's meaning (and so should not be closed), it could definitely do with some tidying up to use better grammar and formatting. You are much more likely to get an answer if you present your posts in a nice way.

thanks @orthocresol for editing. I am new to stackexchange so not much aware about posting rules , format and editing methods.

You can’t, just from the orbital diagram alone.

The 2s and 2pz atomic orbitals are of the same symmetry in the molecule. That means that the 2 sigma* and 3 sigma are actually hybrids. The former is a hybrid were most of the electron density points to the bond (making it partially bonding), while in the latter most of the electron density points outside, i.e. a lone pair in the carbon.


text:
How does boiling water extract dyes from vegetables?
Why does boiling water always seem to release and become the colors of whatever vegetables I am boiling?
For example, beet root and red cabbage both vividly color hot water.
I'd assume it's something to do with collisions with water particles breaking down cell walls and releasing a pigment, but how would this work on a physical level?

I believe this is a Chemistry question.

@QuantumBrick I'm very interested in analysing it from a physics perspective, specifically that of how the kinetic energies of the particles in the water succeed in breaking down the solid cell wall.

cross post at: http://biology.stackexchange.com/questions/52533/how-does-boiling-water-release-dyes-from-vegetables/52542?noredirect=1#comment92063_52542

These dyes (NOT pigments$^*$) are highly water soluble. As you pointed out, boiling causes cell walls to rupture, releasing the cell plasma into the water. Boiling of the plasma fluid causes pressure build up, something the cell walls aren't designed to withstand.
You can also extract them with cold water by mashing up the raw material in a food blender or grinding it in a mortar and pestle. This has the same effect of rupturing the cell walls.
An interesting fact is that many of these dyes are pH sensitive: their colour is different in acid or alkaline conditions. Notably red cabbage extract changes from wine red to greenish blue when you add some sodium bicarbonate (baking powder) to it, because 'Bicar' is a (weak) alkali. Then adding a lot of vinegar (a weak acid) will change it back to its original colour.
$^*$ pigments are inorganic substances (many are refined minerals) and mostly insoluble in water.

But how does the water actually cause the cell walls to rupture?

Just edited it: it's due to pressure.


text:
Why don't heavy and light gases separate in the atmosphere?
Everyone must have heard that balloons are filled with helium, and the fact associated with it that helium gas is light and light gases always go upward.
There comes a question to mind: if the molar mass of $\ce{CO2}$ is greater than that of $\ce{O2}$ and $\ce{N2}$, then why doesn't $\ce{CO2}$ occupy the lower layer of the atmosphere, since it is heavier than $\ce{O2}$ and $\ce{N2}$, as in the case of balloons, where helium being light rise upwards.

Proportionally, CO2 is slighter more prevalent at the surface than high up. But, as you may have noticed outside, there is a lot of activity in our atmosphere, from winds to large scale updrafts (and downdrafts) that keep the atmosphere pretty well mixed over long length scales.

Related: Why does ozone, being heavier than air, not settle down?

Related: Chlorofluorocarbons rising

That's because of two reasons. One is entropy, the ultimate force of chaos and disorder. Sure, gases would like to be arranged according to their density, but even above that, they would like to be mixed, because mixing creates a great deal of entropy. If you prevent the mixing, then they would behave just as you expected. Indeed, a balloon filled with $\ce{CO2}$ would drop right to the floor and remain there.
On the other hand, if you allow mixing, light gases wouldn't necessarily go upward. Just pierce that balloon with helium, and... Well, you won't actually see much, but in fact, the helium will disperse in the atmosphere and remain there. True, its contents in the upper layers would be somewhat higher, but only somewhat. It is not like a layer of pure helium floating atop all those $\ce{O2}$ and $\ce{N2}$.
The importance of entropy is by no means limited to gases. Think of all that salt in the oceans. Salt is much more dense than water; wouldn't it just drop to the ocean floor? Well, no, it rather wouldn't.
The other reason is the constant action of winds and currents mentioned by Zhe. They stir the atmosphere (or sea water, for that matter) and make it even more uniform than it might have been otherwise.

Here's how you can get a quantitative handle on this.  Suppose you have a tall column of two gases with molecular weights $M_1$ and $M_2$ in a cylindrical container of height $L$ and cross sectional area $A$, and you have the amount of substances $n_1$ and $n_2$ of each gas, respectively, in the container.  Suppose that the gases can be treated as ideal gases, and that the container is maintained at constant temperature.  When the system reaches equilibrium, what is the extent of the gravitational separation of the two gases?
Since you are dealing with an ideal gas mixture, each gas can be treated as if it is the only gas present.  Each gas must satisfy the local static equilibrium condition:
$$\frac{\mathrm{d}p}{\mathrm{d}z}=-\rho g, $$
where $z$ is the elevation from the bottom of the container, $p$ is the partial pressure of a given species at elevation $z$, and $\rho$ is the local density of the species at elevation $z$.  From the ideal gas law:
$$\rho=\frac{pM}{RT},$$
where $M$ is the molecular weight.  If we combine these two equations, we obtain:
$$\frac{\mathrm{d}p}{\mathrm{d}z}=-\frac{pMg}{RT}$$
Integrating this equation between $z = 0$ and $z = L$ yields:
$$\frac{p(L)}{p(0)}
=\frac{\rho(L)}{\rho(0)}=\exp{\left(-\frac{MgL}{RT}\right)}$$
So, comparing species $1$ and $2$,
$$\frac{(\rho_1(L)/\rho_1(0))}{(\rho_2(L)/\rho_2(0))}=\exp{\left(-\frac{(M_1-M_2)gL}{RT}\right)}$$
As an example, for the case of $L = \pu{1 km}$, $M_1 = \pu{44 g/mol}$, $M_2 = \pu{28 g/mol}$, and $T = \pu{298 K}$, the ratio of the densities at $z = 0$ compared to $\pu{1 km}$ differ by only about $6\%$.

That all looked very convincing until the final calculation. Should not the units of L be metres? 169.81000/8.314/298 > 60, implying diffusion doesn't prevent complete separation over 1000 metres of still air.
Suggests CO2 mixing requires wind: https://scienceblogs.com/illconsidered/2010/10/is-co2-well-mixed
Another discussion: https://www.researchgate.net/post/Is_gas_gas_gravity_separation_possible_What_conditions_and_restrictions_are_needed
Reliable sources would be better...

@CedricKnight. I never implied that convective transport was not involved.

OK. I was hoping you would clear up my 'factor of 1000' confusion by clarifying the answer. The units of g are metres, so why is L in km? Your answer derives the scale height of the atmosphere (about 8 km), but M₁ and M₂ are described as 'molecular weights' (shouldn't that be molecular masses?) in g/mol. To use entirely SI units, these would be expressed in kg/mol and then L would be in metres. So the 6% appears correct, but seemed doubtful in context of the second link I posted (where sulfur hexafluoride flows like an invisible liquid initially, before diffusing).

The primary reason why you don't get separation of gases in the atmosphere by density is that the atmosphere (especially the lower atmosphere) is turbulent. The second reason, but one that is unimportant in the lower atmosphere, is diffusion.
If you mix a bunch of gases in a jar, diffusion will ensure that they are thoroughly mixed. So diffusion can cause a lot of mixing. When the atmosphere is thin, then diffusion is competing with gravity and some (small) degree of separation by density does occur (but this is trivial in the lower atmosphere: the air at the top of the tallest mountain has essentially the same mix of gases as the air at sea level). But in the lower atmosphere this is unimportant and the dominant force is turbulent mixing (or as it is more commonly know to non-chemists, wind). Given the amount of wind and the constant turbulent motion of air, things get mixed quickly and stay well mixed. Only in the very rarefied upper atmosphere does the battle between diffusion and gravity lead to any separation between light and dense gases.
If you are in a large closed space where wind doesn't exist, then things will still mix, but a lot more slowly. Some industrial accidents, for example, are caused because some gases or vapours will accumulate in the lower parts of a vessel pushing out the oxygen and air. If you don't follow good safety procedure this can rapidly asphyxiate you (good practice would involve fully ventilating the space allowing turbulent mixing to do the job of mixing up the dense gases with everything else). Equally, if you filled a closed vessel with helium, it could accumulate in the upper parts of the vessel and have the same effect at least in the short term before diffusion evened out the mix). But if you partly filled the vessel with air and another gas and left it long enough, the two would be mixed eventually.

There's certainly a preference for carbon dioxide to be lower, but I think you're forgetting that the outside world isn't a static mass of air. There are a lots of convection currents. Just think how much a major storm like would have mixed up the gases.


text:
Work to overcome adhesion of a liquid to a solid
I'm interested in the van Der Waals force of a liquid adhered to a solid. For example, if you spray some water on a vertical piece of glass, the little droplets with adhere to the window rather than be pulled down by gravity.
How does one calculate the energy of that bond, or re-stated for my specific interest, the work required to de-adhere the liquid?
I haven't been able to find any good resources for macroscale liquid films on solid surfaces. My first guess was surface tension of the liquid multiplied by the interface area; the units work out and it makes some intuitive sense, but I don't like that its independent of the solid material.

Every kind of phase boundary has its own specific surface energy. The value listed in the books for water actually refers to the water-air boundary. You need the value for other interfaces, namely water-glass and glass-air.

@IvanNeretin A google search for "water glass surface tension" doesn't reveal alternative values. Based on google results, "surface tension seems to be only defined for liquid-gas interfaces. Is there a better term? I suspect something like "surface energy" is better but that doesn't really get me a value either.


text:
Uranium enrichment via effusion through a porous membrane
The following problem is from Principles of General Chemistry, Silberberg, 1st edition:

One way to utilize naturally occurring uranium ($0.72\%$ $\ce{^235U}$ and $99.27\%$ $\ce{^238U}$) as a nuclear fuel is to enrich it (increase its $\ce{^235U}$ content) by allowing gaseous $\ce{UF6}$ to effuse through a porous membrane. From the relative rates of effusion of $\ce{^235UF6}$ and $\ce{^238UF6}$, find the number of steps needed to produce uranium that is $\ce{3.0 mol}\ \%$ $\ce{^235U}$, the enriched fuel used in many nuclear reactors.

Elsewhere in the text the following ratio is given:

$$\frac{\text{Rate}(\ce{^235UF6})}{\text{Rate}(\ce{^238UF6})} = \sqrt\frac{M(\ce{^238UF6})}{M(\ce{^235UF6})} = 1.0043$$

I've found two different ways to solve this that give different answers.

Solution 1 based on similar problem in LibreTexts Chemistry:

$$\begin{align}
1.0043^n &= \frac{0.030}{0.0072} \\[5pt]
n &= 333
\end{align}$$

Solution 2 based on similar problem in Principles of Modern Chemistry by Oxtoby, Gillis, Butler:

$$\begin{align}
1.0043^n &= \left(\frac{3.0}{97.0}\right) \left(\frac{99.27}{0.72}\right) \\[5pt]
n &= 338
\end{align}$$
Which of these approaches is correct? 2 makes more sense to me but 1 is closer to the answer given in the book (332).

The second approach looks more correct to me. The first approach is using raw percentages without converting to a ratio between isotopes, and the ratio rates is going to be directly related to the ratio of isotopes, not raw percentages.

From the comments, ed:
Well, $1.0043$ is the ratio between rates, so each stage will cause the ratio between isotopes to change by this factor.
I simple set up:
$$\frac{0.72}{99.27} \times 1.0043^{n} = \frac{3}{97}$$ which is the same equation as method 2.
$\frac{0.72}{99.27}$ is the initial ratio (before any enrichment). You need to increase it successively by $1.0043$ until you get to $\frac{3}{97}$, the desired mol%.

Can you expand on this, perhaps with numerical examples that justify your answer? As it reads now, it's a comment and is primarily opinion-based with no quantitative information illustrating/justifying your reasoning.

Right that's where solution 2 came from. Should have been more explicit.


text:
Origin of "ortho", "meta", "para"
Definition
The ortho position refers to the two adjacent positions on a benzene ring.
The meta position refers to the positions separated by one carbon atom on a benzene ring.
The para position refers to the opposite position (separated by two carbon atoms) on a benzene ring.
Background
The dictionary.com pages for ortho, meta, and para state that they are independent from the prefixes ortho- (correct), meta- (after; beyond; a higher level), para- (alongside) respectively.
However, I suspect that the meta position is just after the ortho position, so it is named meta ("after" in Greek).
Question
How did these words become affiliated with the positions on a benzene ring?

lost in translation, apparently ... https://en.wikipedia.org/wiki/Arene_substitution_pattern

Buttonwood the info on that link would be a great basis for an answer!

Related: http://chemistry.stackexchange.com/q/40506

@CurtF. Your suggestion is convincing, especially after recognizing how some of them (ortho) are used in inorganic chemistry (the indicator set by @Jan).

The designation of an ortho-, meta-, or para-substitution pattern was coined when spectroscopic analysis was just starting (Robert Bunsen, for example).  Some of the spectroscopic techniques to assign substitution patterns (IR-, and more importantly, NMR spectroscopy) were simply not available.  To quote wikipedia's entry about origines of arene substitution pattern:

The prefixes ortho, meta, and para are all derived from Greek, meaning
  correct, following, and beside, respectively. The relationship to the
  current meaning is perhaps not obvious. The ortho description was
  historically used to designate the original compound, and an isomer was
  often called the meta compound. For instance, the trivial names
  orthophosphoric acid and trimetaphosphoric acid have nothing to do with
  aromatics at all. Likewise, the description para was reserved for just
  closely related compounds. Thus Berzelius originally called the racemic
  form of aspartic acid paraaspartic acid (another obsolete term: racemic
  acid) in 1830. The use of the prefixes ortho, meta and para to distinguish
  isomers of di-substituted aromatic rings starts with Wilhelm Körner in
  1867, although he applied ortho prefix to a 1,4 isomer and the meta prefix
  to a 1,2-isomer. It was the German chemist Karl Gräbe who, in 1869,
  first used the prefixes ortho-, meta-, para- to denote specific relative
  locations of the substituents on a di-substituted aromatic ring (viz,
  naphthalene). In 1870, the German chemist Viktor Meyer first applied
  Gräbe's nomenclature to benzene. The current nomenclature was
  introduced by the Chemical Society in 1879.

Besides many among the chemists are familiar of this designation, it is limited to describe the relationship of two substituents of a benzene.  Hence frequently chemical suppliers and data bases prefer to describe the position of them by numbers.  IUPAC's Blue Book still allows the use of o, m, and p (rule A-12.3), yet the consent moves into the direction of a consistent numbering that is more adaptable in terms of number of (potentially different) substituents and their relative arrangement to (hetero)aromatic nuclei.
To illustrate this, think of examples like lutidine (systematically 2,6-dimethylpyridine), a frequently used mild, sterically hindered base that does not fit into the other system.

In the end, we've to learn understand both dialects + some of non-systematic names, too.


text:
What does the prime symbol mean in organic nomenclature (e.g 2')?
I encountered the following compound on Wikipedia:
$\hspace{6cm}$
Its systematic IUPAC name is:

4,4',5,5',6,6'-hexahydroxy-2,2'-biphenyldicarboxylic acid

What do the prime symbols in the locants 4', 5', 6', and 2' signify?

Primes, double primes, triple primes, etc., are added to differentiate between the same locant in different parts of the structure. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), they are used

to differentiate the nitrogen atoms of hydrazides, imidamides (amidines), amidrazones, and hydrazidines;

N'-acetyl-N'-ethyl-N-methylpropanehydrazide

with superscripted numerical locants indicating the position of the suffix on the parent structure, to differentiate the nitrogen atoms in other cases, especially when symmetry conditions are not fulfilled;*

N3-ethyl-N'3-methylhexane-3,3-diamine

in multiplicative nomenclature to denote multiplied units and modify locants accordingy;

2,2',2''-nitrilotri(ethan-1-ol)

in spiro-fused compounds, to denote positions in polycyclic systems

7,7'-spirobi[bicyclo[4.1.0]heptane]

in ring assemblies, to number identical ring components;

1,1'-biphenyl

in fusion nomenclature, to identify first and higher attached components, identical attached components, and multiparent names;

difuro[3,2-b:2',3'-e]pyridine

in fullerenes ortho-fused to organic ring or ring systems, to identify positions in the nonfullerene component;

3'H-cyclopropa[1,9](C60-Ih)[5,6]fullerene

in natural product nomenclature, to identify positions in ring(s) fused to a fundamental parent hydride;

naphtho[2',1':2,3]-5α-androstane

The example that is given in the question is a ring assembly. The numbering of locants corresponds to the numbering of 1,1'-biphenyl, where primes are used to differentiate between the same locant in different phenyl groups. Therefore, the correct systematic name is 4,4',5,5',6,6'-hexahydroxy-[1,1'-biphenyl]-2,2'-dicarboxylic acid.

* Note that, according to current IUPAC recommendations, primes are no longer used to differentiate the nitrogen atoms of symmetrical geminal diamines, diimines, diamides, di(imidamides), (diamidines), di(hydrazonamides), di(amidrazones), di(hydrazonohydrazides), imidohydrazides, and poylimidopolycarbonic and inorganic oxoacids.

The locants are "primed" and are used for ring systems not fused together.
For example, this is the locants for biphenyl (courtesy ACD/Labs):

Relevant recommendations (Nomenclature of Organic Chemistry, Sections A, B, C, D, E, F, and H. Pergamon Press, Oxford, 1979. Copyright 1979 IUPAC.):

$52.2$ - If there is a choice in numbering, unprimed numbers are assigned to the system which has the lower-numbered point of attachment.
$52.3$ - If two identical hydrocarbon systems have the same point of attachment and contain substituents at different positions, the locants of these substituents are assigned according to Rule A-$2.2$; for this purpose an unprimed number is considered lower than the same number when primed. Assemblies of primed and unprimed numbers are arranged in ascending numerical order.

The name follows from the usual rules.
More information
For more information about how non-identical ring systems or ring systems with three or more rings are numbered, see Rule A-$53$ and Rule A-$54$.


text:
Is it possible to have a dispersion interaction with no inductive interactions?
In symmetry-adapted perturbation theory, the interaction energy of a non-covalently bound pair of molecules (it could be done to arbitrary order but is usually just pairs) is built up from a perturbative expansion of the interaction potential between two monomers.
The first-order perturbation drops out a term which we call induction. This has components associated with the induction on monomer B by monomer A and vice versa. The second-order perturbation drops out a term which we call dispersion (this is what people usually refer to when they say van der Waal's force (although that's not technically correct...)).
My question is, can that second-order perturbation term--dispersion-- have a non-zero contribution to the total energy while the first-order induction terms do not contribute?
When I imagine this situation physically, it doesn't make much sense because it means that the presence of monomer A near monomer B does not perturb the eigenstates of monomer B (beyond electrostatics, exchange, etc.) and vice versa, but somehow there is a mutual dispersion effect.
So basically what I'm asking is if it is mathematically possible to have a non-zero dispersion term while simultaneously having zero contributions from the first-order induction terms in the interaction potential expansion?
Also, if it is not possible, it seems as if we ought to be able to write the dispersion energy in terms of the induction energies, but clearly we can't because dispersion is a second-order perturbation. If the effects are related, however, can we approximate dispersion using induction? I imagine this to be similar to how in Moller-Plesset perturbation theory you can approximate the fourth-order perturbation using the second-order perturbation.
I don't know too much about the details of what I'm asking about, so if I've said something quite stupid please let me know.

This is an excellent question but I don't have time to think about it properly right now. In case you didn't know about it already, here is a review containing all the terms you mention.

Thanks for the link! I was actually just reading that before you linked it. It's very interesting stuff.

The first-order correction in the interaction operator is an integral over the frozen reference densities, so in a product basis $|i_A,j_B\rangle$ that would read
$$ E^{(1)} = \langle 00|V|00 \rangle $$
and as such is the electrostatic interaction. This is the same as the Coulomb interaction of the two non-interacting reference densities.
Induction is defined as the  polarization of one monomer due to the reference density of the other monomer, so must involve integrals over the virtual orbitals of the monomer being polarized in order to capture its polarization. So induction is a second-order correction and has the form
$$ E^{(2)}_{ind} = \sum_{j\neq 0} \frac{|\langle 00|V|0j \rangle|^2}{E_{00}^{(0)}-E_{0j}^{(0)}} + \sum_{i\neq 0}\frac{|\langle 00|V|i0 \rangle|^2}{E_{00}^{(0)}-E_{i0}^{(0)}}$$
where the first term represents the polarization of monomer $B$ due to the reference density of monomer $A$ and the second term is the reverse process.
What sets the dispersion energy apart from the induction energy is that it is the simultaneous polarization of both monomers, and so the integrals run over the virtual orbitals of both monomers as in
$$
E^{(2)}_{disp} = \sum_{i\neq 0} \sum_{j\neq 0}\frac{|\langle 00|V|ij \rangle|^2}{E_{00}^{(0)}-E_{ij}^{(0)}}
$$
Mathematically, the dispersion and induction terms are both second-order correction terms in the energy expansion, but they have been grouped in a physically-intuitive way to give two different types of interactions. There are systems where dispersion is the dominant interaction, such as rare gas dimers.
EDIT:
An example of a dispersion-dominated bound system is the neon dimer. Below is the SAPT decomposition of the binding curve at the HF-SAPT(2)/aug-cc-pVDZ level.

At the equilibrium distance, dispersion dominates the binding interaction but there is a non-zero induction contribution. I am inclined to say that although the ratio of dispersion and induction may become large, you can never truly have zero induction. Some naive mathematical reasoning follows.
We need $ \langle 00|\frac{1}{r_{12}}|0i\rangle, i\neq 0$ to be zero in a Gaussian product basis. At a long-enough distance we can expand the Coulomb operator in a multipole expansion to give something like
$$
\frac{1}{r_{12}} = \alpha x_1x_2+\beta(x_1x_2^2-x_1^2x_2)+...
$$
in a one-dimensional model picture. I've also folded all the coefficients into the constants $\alpha,\beta,...,etc.$ The important thing here is the symmetry of the integrals. So, lets take $i=1$ and see what happens...
$$
\langle 00|\alpha x_1x_2+\beta(x_1x_2^2-x_1^2x_2)|01\rangle = \langle 00|\alpha x_1x_2|01\rangle + \langle 00|\beta x_1x_2^2|01\rangle-\langle 00|\beta x_1^2x_2|01\rangle+ ...
$$
The first two integrals are zero because $\langle 0|x_1|0\rangle=0$. The third integral is non-zero and survives in the sum. This can happen any time the operator in monomer $A$ is an even function. As $i$ gets larger, higher-order terms in the expansion contribute to the induction energy. So mathematically, there is a tiny but non-zero induction contribution.

Mathematically that helps me to understand the distinction between dispersion and induction better so thanks for that, but I don't think you have said yet if it is possible for the induction integrals to evaluate to zero but have the dispersion integral be non-zero. It feels to me like there should be a way of proving it one or the other, but I don't know if I can quite work it out myself.

@jheindel, I've expanded my answer to include why induction may be small but not zero.

Thanks! That's a very helpful contribution. Thanks for the answer.


text:
What are the products of the double displacement reaction of oxalic acid and cesium hydroxide?
A solution of oxalic acid is mixed with a solution of cesium hydroxide. This chemical reaction requires double displacement. I got this as my chemical equation:
$$\ce{C2H2O4 + CsOH -> H3O + CsC2O4}$$
My teacher says it is wrong, is it really incorrect?
Is the one below the right one?
$$\ce{C2H2O4 + CsOH -> CsHC2O4 + H2O}$$

I would start by considering what the products are. This is a special type of double displacement; it would probably also help for you to identify which one.

You have a dicarboxylic acid (Oxalic Acid) reacting with a base (Cesium Hydroxide). This reaction will give you a carboxylate salt (Cesium Oxalate) and water. This is basically a neutralization reaction.
$$\ce{C2H2O4 + 2CsOH -> C2Cs2O4 + 2H2O}$$
You need 2 Cesium cations to neutralize the -2 charge of the dicarboxylic acid.
Here is an image of how Cesium oxalate looks like:


text:
Effect of pressure on adsorption of a gas
I am currently studying about the basic terms of surface chemistry, and am facing some conceptual doubt regarding the following:
What effect does pressure have on the adsorption of a gas (adsorbate) on a adsorbent? I don't understand the fact that the adsorption of a gas increases very rapidly as the pressure is increased from small values at low temperature.
I have the idea that increasing the pressure brings the molecules of the gas close together so that they have more intermolecular forces of attraction, but how do I correlate this with adsorption?

Short Answer
The following equilibrium roughly represents what goes on during physisorption of a gases onto a a surface.
$$\ce{Gas + Surface <=>[adsorption][desorption] Gas--Surface}  $$
The effect of pressure can be understood on the basis of Le Chatelier's principle, which states that if a stress (such as a change in pressure) is applied to a system in equilibrium, the equilibrium will shift so as to tend to counteract the effect of the constraint.
In this case, we are increasing the pressure, and the system would respond by wanting to reduce the number of gas phase molecules, and in this case it can do so by favouring the forward reaction wherein the gas phase molecules are immobilised on the surface.
Adsorption is usually described through isotherms, that is, the amount of adsorbate on the adsorbent as a function of its pressure (if gas) or concentration (if liquid) at constant temperature. Several isotherm models exist to describe this phenomenon.

Bonus Round: Additional Commentary
One of the earliest mathematical fits for experimental data was the Freundlich adsorption isotherm. Experimentally it was determined that extent of adsorption varies directly with pressure and then it directly varies with pressure raised to the power 1/n until saturation pressure $P_s$ is reached. Beyond that point rate of adsorption saturates even after applying higher pressure. Thus Freundlich adsorption isotherm failed at higher pressure.
$$\frac{x}{m} = Kp^{\frac{1}{n}}$$
where $x$ is the quantity of adsorbate adsorbed in moles, $m$ is the mass of the adsorbent (this is basically a normalisation to allow for comparison between different materials), $P$ is the pressure of adsorbate and $K$ and $n$ are empirical constants at the given temperature.
The function is not adequate at very high pressure because in reality, as I have already mentioned, because $\frac{x}{m}$ has an asymptotic maximum as pressure increases without bound. As the temperature increases, the constants $k$ and $n$ change to reflect the empirical observation that the quantity adsorbed rises more slowly and higher pressures are required to saturate the surface.
Irving Langmuir was the first to derive a semi-empirical model with a kinetic basis and was derived based on statistical thermodynamics.
It is quite simple, (and elegant in my opinion), and finds extensive use. it is based on the following four assumptions

The surface containing the adsorbing sites is a perfectly flat plane with no corrugations (assume the surface is homogeneous) .
The adsorbing gas adsorbs into an immobile state.
All sites are equivalent.
Each site can hold at most one molecule of A (mono-layer coverage only).
There are no interactions between adsorbate molecules on adjacent sites.
There are no phase transitions.

Unfortunately,these four assumptions are seldom all true: there are always imperfections on the surface, adsorbed molecules are not necessarily inert, and the mechanism is clearly not the same for the very first molecules to adsorb to a surface as for the last. The fourth condition is the most troublesome, as frequently more molecules will adsorb to the monolayer; this problem is addressed by the BET isotherm for relatively flat (non-microporous) surfaces.
Anyway, the surface coverage is given as follows $$\theta = \frac{K_{eq}p}{1+ K_{eq}p}$$


text:
What does IUPAC mean by these rules?
I don't understand IUPAC's rules/recommendations to name the compound.

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(e) saturation/unsaturation:
 (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
 (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Does hydro or dehydro refer to the $\ce{OH}$ group?
The rule first says that low locants are given to ene and yne endings.
The next part of the rule says locants are first given to multiple bonds and then to double bonds. Aren't double bonds multiple bonds?
Do multiple bonds refer to triple bonds?

Hydro/dehydro means having added or removed hydrogen, usually a pair. For example: tetrahydrofuran.
I think they mean that triple bonds have higher "seniority" than double bonds, but not both of those are lower than double and triple bonds indicated by "ene" or "yne" in the name.

When there is a choice for numbering, the criteria given in Subsection P-14.4 are applied in order until a decision can be made.
For example, since Criterion (e) takes precedence over Criterion (f), the systematic name of allyl chloride is 3-chloroprop-1-ene, not 1-chloroprop-2-ene, because low locants are given first to the double bond (the ‘ene’ ending) and then to the ‘chloro’ prefix.
The term ‘multiple bonds’ includes double bonds and triple bonds. According to Criterion (e) (ii), low locants are given first to multiple bonds as a set, even though this may at times give triple bonds (‘yne’ endings) lower locants than double bonds (‘ene’ endings).
For example: a correct systematic name is pent-3-en-1-yne, not pent-2-en-4-yne, because the locant set 1,3 is lower than 2,4.
Only if a choice remains, preference for low locants is given to the double bonds in accordance with Criterion (e) (ii) (“… and then to double bonds …”).
For example: a correct systematic name is pent-1-en-4-yne, not pent-4-en-1-yne, because the locant 1 for the double bond (the ‘ene’ ending) is lower than 4.
(See also Subsection P-31.1.1.1.)
The prefixes ‘hydro’ and ‘dehydro’ are not related to hydroxy groupds ($\ce{-OH}$). They are used to indicate addition and subtraction, respectively, of hydrogen atoms (see Subsection P-31.2.1).
For example: a systematic name for benzyne is 1,2-didehydrobenzene, which indicates the removal of two hydrogen atoms.


text:
Reaction of H3BO3 and NaOH
I have been searching around and have found varying answers to the question. What exactly would reaction of $\ce{H3BO3}$ and $\ce{NaOH}$ give? Shouldn't it be a neutralisation reaction as $\ce{H3BO3}$ is an acid and $\ce{NaOH}$ is a base?

This isn't something you could know except by experiment, but boric acid doesn't exist as solely $\ce{H3BO3}$ in solution. In addition to being an Arrhenius acid, $\ce{H3BO3}$ is also a Lewis acid. This means it is capable of accepting a lone pair, and will undergo the following reactions in aqueous solution:
$$\ce{B(OH)3 + H2O <=> B(OH)4- +  H+}\tag1\label{Eq1}\\$$
$$\ce{4B(OH)4- + 2H+ <=> B4O7^{2−} + 9H2O}\tag2\label{Eq2}\\$$
By combining \eqref{Eq1} and \eqref{Eq2}, we get the full reaction:
$$\ce{4B(OH)3 <=> B4O7^{2−} + 2H+ + 5H2O }\tag3\label{Eq3}\\$$
Boric acid does not produce tetraborate exclusively, however, and many other species species are also present in solution. The system is actually so complex that it remains difficult to characterize, one paper mentioning that there are some ten different equilibrium reactions in solution$^{[1]}$. In performing an acid-base calculation, however, it is okay to simply treat it as a solution of $\ce{H2B4O7}$, neutralizing according to the equation:
$$\ce{H2B4O7 + 2NaOH <=> Na2B4O7 + 2H2O}\tag4\label{Eq4}\\$$
Titration confirms that $\ce{H3BO3}$ reacts with $\ce{NaOH}$ in a $\text{2:1}$ ratio, and crystallization of the product confirms $\ce{Na2B4O7}$ is produced.

$[1]$ Trejo G., Frausto-Reyes C., Gama S.C., Meas Y., Orozco G. Raman Study of Benzylideneacetone on Silver. Int. J. Electrochem. Sci. 2012, 7, 8436–8443.

Please note that the sodium tetraborate solution actually contains the $\ce{[B4O5(OH)4]^2-}$ anion.

There are conflicting interpretations for the origin of the acidity of aqueous boric acid solutions. Raman spectroscopy of strongly alkaline solutions has shown the presence of B(OH)−
4 ion,[3] leading some to conclude that the acidity is exclusively due to the abstraction of OH− from water:[3][4][5][6]
or more properly expressed in the aqueous solution:
This may be characterized as Lewis acidity of boron toward OH−, rather than as Brønsted acidity.
However other sources say that boric acid is also a tribasic Brønsted acid, with successive ionization steps:
Since the value of Ka1 is comparable to that of the reaction with OH−, the concentrations of BO(OH)−
2 and B(OH)−
4 are similar.

could someone help format my answer

I think you'll find this meta post quite useful.


text:
Explain why the noble gas Kr does follow the general trend for covalent radii
Looking at covalent radii I know the trend states that as you move from LEFT to RIGHT the radii decreases due to increased nuclear charge. Up until you hit the noble gasses like neon. Neon does not want to form a covalent bond with itself so we have to look at the van der Waal forces to determine its radii (which is bigger than Fluorine, disproving the trend).
Up until you reach Krypton, the noble gasses tend to be slightly larger than the halogen group.
I am having a hard time understanding why the noble gas Kr does follow the general trend for covalent radii in the fourth period, even though the noble gas Ne does not.
Would it be because of the greater shielding effect?

Covalent Radii of 2nd Period and 4th Period Halogen and Noble Gase.
F
 - 0.064 nm
Ne
 - 0.070 nm
Br
 - 0.114 nm
Kr
 - 0.109 nm

Could you cite a source containing the values?

Added covalent radii data.

Since the increased atomic radius causes a decrease in the electric interaction between electrons and nucleus, the heavier noble gas elements, such as Kr and Xe can form compounds with other elements. (KrF2, XeF4, ... &c.)
The covalent radius is quite meaningless in for elements such as Ar and Ne since its compounds seldom exist, but Kr and Xe compounds can occur with relatively high stability, so its 'covalent radius' can be meaningful.
That 0.109nm is probably a covalent radius.
Something like measuring a Kr-F bold length in KrF2 and then subtracting the covalent radius of a fluorine atom.
Trivially, the Van der Waals radius tends to be bigger than the covalent one, the noble gas atom radius seems as if it does not follow the trend, but from Krypton which the covalent radius is measurable, the data follows the general trend in periodic table.


text:
How to find the equivalence point on a graph of an acid-base titration curve?
Here is the data taken from the lab:
Acid pH, Vol (cm3)[
(9.67,  4cm3)
(9.08,  8cm3)
(8.70,  10cm3)
(8.06,  12cm3)
(7.93,  13cm3)
(4.76,  14cm3)
(2.45,  15cm3)
(2.23,  16cm3)
(2.10,  17cm3)
(2.01,  18cm3)
(1.94,  19cm3)
(1.88,  20cm3)]
Thanks

Not the answer per se, however:

If you were to titrate an acid in aqueous solution, why would be the pH value in the beginning superior to 7?  Because reading the data provided suggests you added the acid (to be characterized, so your question) instead.
The data interval is too coarse, especially in the region "something is happening", at about 13 mL to 16 mL.   Repeat the experiment (with correct addition of the reagents), and while adding the base to the acid, keep a log running.  Like:
\begin{array}{|c|c|c|c|}\hline
\text{pH value}& \text{total added volume of base} & \Delta\ce{pH} &\Delta V & \frac{\Delta \text{pH}}{\Delta V}\\
\hline
\ldots&\ldots&\ldots&\ldots&\ldots\\
\ldots&\ldots&\ldots&\ldots&\ldots\\
\end{array}

Where $\Delta\ce{pH}$ is the change of the pH-value since the last addition of the reagent (conversely, $\Delta V$ the volume of reagent added since the last addition.
From the poor (because very little) data here, the plateau-like data provided by you (esp. pH 4.76) that is away from 7 might be indicative for a weak acid; like acetic acid, for example.  So read about equivalence points of weak acids (and bases), and how this differs from strong acids.  When you reach the equivalent point(s), small quantities of bases / acids added may yield a significant change in the pH value; a volumetric analysis reducing the volume of added reagent to 0.1 mL in this particular region is not uncommon.  And dervivate from the reading (the lower section of the reference provides some links).

Here's the result in R
titration <- data.frame(Vol=c(4,8,10,12:20),
                        pH=c(9.67, 9.08, 8.70, 8.06, 7.93,4.76,2.45,2.23,2.10,2.01,1.94,1.88))

plot(titration$Vol,titration$pH, main="Acid Base Titration",
      xlab="Volume /ml", ylab="pH")

If you copy and paste the code into r-fiddle.org you'll get your graph and result.

I would have used matplotlib on a dataframe from a list of tuples in Python. Admittedly, the short in R notation for the volume range 12:20 is pretty neat!

I used R because the OP can simply post the code into the web page I gave and get the result.  No software downloads needed etc...

Good point! If I remember correctly, wakari.io requires (free) registration.

As to whether or not the OP does anything with the answer is a different matter.

I can't speak for the OP, but I found your answer useful and upvoted. More often than not, it is helpful to plot the data before one starst to fiddle around with them.

I don't find this answer helpful at all, therefore I downvoted. I don't even consider this a valid answer. To be fair, I don't find the question helpful at all, too, so I downvoted that also.