text:
Equilibrium between barium carbonate, and barium oxide and carbon dioxide
The USNCO 2017 Question 33 is as follows:

Barium carbonate, $\ce{BaCO3}$, is stable at ambient temperatures, but decomposes to barium oxide and carbon dioxide at higher temperatures.
$$\ce{BaCO3(s) <--> BaO(s) + CO2(g)}$$
At a certain temperature, this system in in equilibrium in a closed system and contains appreciable amounts of all three compound. Which changes will lead to an increase in the pressure of $\ce{CO2}$ present at equilibrium?
I. Adding more $\ce{BaCO3(s)}$
II. Increasing the volume of the container
(A) I only
(B) II only
(C) Both I and II
(D) Neither I nor II

What I understand so far:

Concentrations of reactants and products at equilibrium are affected by:

Increasing/decreasing concentrations
Change in pressure/volume
Change in temperature

Temperature remains unchanged, so that factor can be ruled out

(I) does not change concentration, as $\ce{BaCO3(s)}$ is considered to have a constant concentration, and it could be assumed that change in volume due to the addition of $\ce{BaCO3(s)}$ is negligible. Hence, (A) nor (C) are correct

(II) If $\ce{CO2}$ was in a closed system itself, increasing the volume of its container will lead to a decrease of pressure and decrease of concentration.

However, according to Le Châtelier’s Principle, the system should react by favouring the forward reaction, leading me to believe (D) is the correct answer

How do I judge whether or not the increase in volume of the container can be balanced by the increased rate of the forward reaction, such that the pressure of $\ce{CO2}$ remains unchanged at the new equilibrium?

Your conclusion is correct.
In the expression for equilibrium, we consider the activities for the solids to be 1. This ultimately leads to:$$K_p=p_{\ce{CO2}}$$
[Note that $K_P=f_{\ce{CO2}}$, where $f_{\ce{CO2}}$ is the fugacity of $\ce{CO2}$. Under "ideal" conditions, $f_{\ce{CO2}}=p_{\ce{CO2}}$, where $p_{\ce{CO_2}}$ is the partial pressure due to $\ce{CO2}$.]
In other words, the equilibrium partial pressure of $\ce{CO2}$ will be constant. One may compare this to the aqueous tension in case of water VLE.
Increasing the volume would cause more $\ce{BaCO3}$ to dissociate, producing more $\ce{CO2}$. The ideal gas law tells us that: $$n_{\ce{CO2}}\propto p_{\ce{CO2}}\cdot V$$
Your equilibrium pressure due to $\ce{CO2}$ does not change.


text:
Do acids react with neutral salts?
I am new to the concept of neutralization.
My teacher told me that acids do not react with neutral salts.
I want to know why not.
If both are dissolved in water and both completely dissociate, why can't the ions of acid and salt exchange and perform a double displacement reaction.
Moreover I have seen the following reaction in a book: $$\ce{2 NaCl + H2SO4 -> 2 HCl + Na2SO4}$$

Indeed, they can and sometimes do react.

That last equation you have there isn't a neutralization reaction ;-)

This examples shows the danger of uncontrolled generalisations. Whether a reaction can 'happen' or not depends on its thermodynamics and kinetics, and that in turn depends on the species involved. You can't say in general that 'neutral' salts don't react with acids. Take for instance $\ce{Ag2SO4 + 2 HCl <=> 2 AgCl + H2SO4}$. This reaction does happen and has a very high equilibrium constant, because $\ce{AgCl}$ is highly insoluble. Substitute $\ce{Ag}$ with $\ce{Na}$ and then it's the opposite of your example.

https://chemistry.stackexchange.com/questions/5048/hcl-synthesis-from-sulfuric-acid-and-sodium-salt

https://chemistry.stackexchange.com/questions/76147/what-are-some-ways-to-predict-that-a-double-replacement-reaction-will-occur?noredirect=1&lq=1

Neutralization is a reaction between acid and base which results in the formation of a salt and water.
When you have any ionic compound and dissolve it into water, dissociation occurs.
This means that you get a solution of ions.
So if you dissolve acids, bases or salts into water, you end up with a mixture of ions.
Then, more things may happen or not.
In the case of the given reaction, you will get a solution with:
$\ce{Na+ + Cl- + H+ + SO4^{2-}}$
If the system is closed, nothing will happen.
But if the system is open, then $\ce{HCl}$ will evolve because it is a gas.
Since the sulphuric acid is very polar and a liquid, water is not even required.
This is, in fact, the standard procedure to prepare anhydrous HCL in the lab. Just drop $\ce{H2SO4}$ on $\ce{NaCl}$. It must be said that $\ce{NH4Cl}$ works better.

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text:
Does deuterium exhibit hydrogen bonding?
I searched for this on Google but I only got research articles in the SERP. I know of the concept of hydrogen bonds, that they are a special class of van der Waals forces between highly electronegative atoms ($\ce{F/O/N/Cl})$ and the hydrogen atom. I am asking if deuterium also exhibits the same hydrogen bonding, and how does its "$\ce{D}$ bond" compare to the hydrogen's original "$\ce{H}$ bond"?

Boiling point of D2O: 101.4°C. Does that answer your question? Also: https://en.wikipedia.org/wiki/Heavy_water

@Karl Do you mean to say that D2O possesses H-bonding, and that the "D bond" is in fact stronger than "H bond"? EDIT: Ok, I found this passage in Wikipedia "As a hydrogen bond with deuterium is slightly stronger than one involving ordinary hydrogen, in a highly deuterated environment, some normal reactions in cells are disrupted." I wonder why Google didn't catch it :/

Why does water have a Bp of 100°C? In comparision CH4: -161, NH3: -33, SH2: -60°C.

@Karl Well I don't know. I always thought it was the opposite. We arbitrarily assigned 100degC to the boiling point of water. Water doesn't know what the temperature is anyway.

That's not the point. The Bp of water is extremely high, and that has one reason: hydrogen bonding.

@Karl And D2O has an even higher BP because of its stronger D-bond. Is that what you intend to say?

Btw. hydrogen bonding is NOT a vdW interaction but an exeptionally strong form of polar interaction. And nitrogen and chlorine don't do hydrogen bonding, only fluorine and oxygen. Bp HF: +20°C, HCl: -85

@Karl I added Chlorine because of its exceptional H-bonding in stabilizing chloral hydrate. And ammonia does have H-bonds, so I am not really sure about what you just said.

Chlorine is NOT taking part in hydrogen bonds in chloral hydate! CCl3 is a strong electron widthdrawing group, and that stabilises the hydrate. Which makes regular O-H--O hydrogen bonds.

What do you mean by "SERP"?

@paracetamol Search Engine Results Page  - one that's displayed when you use a search engine.

@Karl Oh! didn't know that before...

To say hydrogen bonds only form with oxygen and fluorine is a gross oversimplification. Many common elements can be both hydrogen bond donors and acceptors: carbon, selenium, sulfur, chlorine, iodine, phosphorus, cobalt, iron, manganese, even hydrogen itself in dihydrogen bonds. Also a misconception: the hydrogen bonds with oxygen and fluorine must always be stronger than others which is not true: see, e.g., doi: 10.1039/C7CP05265K.

@LinearChristmas Ok, I understand.  Thank you for pointing out!

@GaurangTandon It was actually more directed at Karl's comment. Futhermore, I don't think that the slight difference in boiling points of $\ce{H2O}$ and $\ce{D2O}$ can immediately be translated into a stronger hydrogen bond. Helium-3 and helium-4 have boiling points of $\pu{3.19 K}$ and $\pu{4.23 K}$, for instance. In other words: boiling point does immediately signify a stronger intermolecular bond but whether it comes from a hydrogen bond needs to be proven separately.

@LinearChristmas "I don't think that the slight difference in H2OHX2O and D2ODX2O can immediately be translated into a stronger hydrogen bond." To be honest I was slightly wary of that myself at first. I didn't go deep enough to get that helium isotopic data, but yes, I agree with you.

@Linear That last comment of yours is a good point :-) However, in polar compounds (something helium is not), I'm still firm in my belief that dipole moments are ultimately what primarily governs boiling points. In light of your comment, I've revised my answer to accommodate your point. Thanks!

@LinearChristmas I just checked the link you give above (an article from last year!), and in the first sentence i read ..recent re-definition of  hydrogen bonding... Could you point out what exactly they have re-defined? I find the idea of hydogen bonds to carbon very ridiculous, but perhaps I'm getting old. Or are we talking about funny ultravacuum gas-phase and argon-matrix species?

@Karl 1. Yes, it was redefined and improved upon, I think, in 2011. It contains a list of criteria. 2. The previous definition did not limit hydrogen bonds to $\ce{NOF}$ either, see 'usually (but not necessarily)'. 3. Indeed, I see that referencing one relatively new paper may have been confusing; it was the first to come to mind. There are more common examples, if you like, for example chloroform (1955) has carbon as a hydrogen-bond donor.

see also: IUPAC techical report which accompanied the redefinition, and explains the process in more detail

@LinearChristmas Thanks! So the gist is, in daily life, hydrogen bonds are FH-F,OH-O,NH-N (and permutations), but if you search for a proper definition, you come to realise that it can only be "there is some directed bonding" between the participants, and then you find there is a huge zoo of species that fit this description. For starters, every reaction intermediate where a proton is transfered.

Comments here are getting a little excessive, but they still seem to be productive.  I noticed that some have been incorporated into answers, so can we either clean those up or shift this over to chat?

@jonsca I've thought about it but haven't really reached a conclusion. The comments (although I am obviously biased) do deal with common misconceptions, and should thus stay. However, the volume is now indeed excessive. Maybe they can be transferred to chat with a disclaimer left by a moderator that the here-deleted exchange had some value? Not sure what the best strategy is; what do you think?

@LinearChristmas We can move to chat without deleting and people can remove what they see fit from here.  Looks like things have petered off anyway.

@jonsca One of you can summarize everything in 2-3 mega comments (giving credits where due) and move everything else to chat. Could be useful imho?

 (Based on ron's answer here on the inductive effect accorded by deuterium - do give that a read, and consider upvoting it)

Since deuterium has a higher mass than protium, simple Bohr theory tells us that the deuterium 1s electron will have a smaller orbital radius than the 1s electron orbiting the protium nucleus.
The smaller orbital radius for the deuterium electron translates into a shorter (and stronger) $\ce{C-D}$ bond length.
 See the bottom half of ron's answer for the proof

In your case, that would mean a shorter $\ce{N/O/F-D}$ bond.

A shorter bond has less volume over which to spread the electron density (of the 1 electron contributed by $\ce{H}$ or $\ce{D}$) resulting in a higher electron density throughout the bond, and, consequently, more electron density at the carbon end of the bond. Therefore, the shorter $\ce{C-D}$ bond will have more electron density around the carbon end of the bond, than the longer $\ce{C-H}$ bond.
The net effect is that the shorter bond with deuterium increases the electron density at carbon, e.g. deuterium is inductively more electron donating than protium towards carbon.

So we can expect the $\ce{N/O/F-D}$ bond to be smaller and more polar than the corresponding bond with protium.
Karl very astutely pointed out that the boiling point of heavy water is higher than that of normal water. In light of the inferences drawn earlier, I took the liberty of interpreting this fact as being (at least somewhat) indicative of deuterium permitting a stronger hydrogen bond than protium.
As Linear Christmas pointed out, the variations in mass across different isotopes is known to affect boiling point (as in the case of helium). This is likely to contribute to some degree, to the increase in boiling point in heavy water. However, I'm still of the opinion that the higher dipole moment of $\ce{D2O}$ over $\ce{H2O}$ is the main cause for the higher boiling point.

Finis
Hydrogen bonding with deutrium does occur and should be, in theory (drawn from ron's answer), stronger than that with protium. The fact that heavy water $\ce{D-O-D}$ is more polar (and hence has a higher boiling point) than regular water $\ce{H-O-H}$, appears to be a consequence of this.

Everything in your answer seems great, except that ron's answer is about inductive effect in covalently bonded atoms. I am talking about H bonding in non-bonded atoms. How can we say that the factors influencing them will be the same?

I actually only wanted to point out that heavy water has practically the same high Bp. ;-)

@GaurangTandon I used ron's answer to explain why the $\ce{N/O/F-D}$ bond is more polar than its protium counterpart. H-bonding is largely an electrostatic dipole-dipole interaction. So if the covalent bonds (that effect H-bonding) are more polar, then the H-bonding is stronger :-)

@Karl I figured (from your follow-up comments under the OP) O:)

Perfect! Thanks :D I'll accept your answer in a day or two .

Do you know what's the reasoning for counting the much weaker N-H--N bond among the much stronger O-H--O or F-H--F hydrogen bonds?

@Karl Nope. Earlier I was under the impression  that there was some (arbitrary) threshold M-H bond moment value that had to be crossed before you could call it a "hydrogen bond", but apparently that's not the case. So yeah, I'm curious...why is it? :-D

Well, I'm guessing that a hydrogen bond should, at least principally, be able to transfer the proton, while a merely polar interaction means that all covalent bonds are definitely permanent. But is that the definition?

@Karl Ah, that's insightful! But then $\ce{HCl}$ should also able to form "hydrogen bonds", but apparently, this gig only includes $\ce{N/O/F}$ :-(

Not necessarily. There's no autoprotolysis in HCl, i think. Or? Does $\ce{H2Cl+}$ exist?

@Karl Whelp! You're right ($\ce{HCl}$ protolysis happens in an aqueous solution) :-D

What about masses ; ? I was given this question/exercise many years ago. I didn't solve it nor I did it ast a follow up. Qualitatively it should raise the boiling point even further (extent might be very small).

@Karl H2Cl+ does exist and autoprotonation of HCl is small but not nonexistant.

@Mithoron But under any relevant conditions? I'm not sure if liquid HCl is relevant. ;-)  Apart from this nitpicking, what do you say to this "definition" of a hydrogen bond i gave above?

@Karl "nitrogen and chlorine don't do hydrogen bonding" facepalm vide Linear's comment.

@Mithoron I meant " a hydrogen bond should, at least principally, be able to transfer the proton, while a merely polar interaction means that all covalent bonds are definitely permanent. "

In short:
Chemistry is about the last shell of electrons in the atoms.
Nuclear differences are in the field of Physics.
Deuterium is only a nuclear isotope of hydrogen, so no fundamental differences are to be expected in the chemical sphere.
Hydrogen bonding is a chemical property.

Yes, that's also another perspective to look at it!

Thank you. I do not want to discredit @paracetamol 's answer. Just to remember the fundamentals. Something my father, a Chemical Engineer, used to tell me.

Yes, I understand :D

I think is not the best perspective tough. Not for fine details.  Else we have to eliminate boiling point arguments as well and stop at rough level, such as D2O does react wit sodium and so on.  (Of course it does).

Aha, this is something my teachers repeatedly sought to hammer into my brain :-) That nuclear differences don't affect the chemistry of substances is a general statement; it is usually (but not always) true. For instance, ron's answer (which is linked in my post), explains how hydrogen isotopes (which differ in the composition of their nucleus) produce different inductive effects, and the inductive effect is indeed a chemical phenomenon. O:)

Actually it is the good answer to the Q. My previous comment is on another line brought in by the discussion about boiling point. Mass differences count too:, even assuming the same H bond strength water and heavy water will be different, chemically (like in biochemistry fine details) and physicochemically ( b.p. , vibrations, etc...).

This is not at all an anwer, but the discussion above pointed to the 2011 IUPAC definition of hydrogen bonds, which I did not know about:
The hydrogen bond is an attractive interaction between a hydrogen atom from a molecule or a molecular fragment X–H in which X is more electronegative than H, and an atom or a group of atoms in the same or a different molecule, in which there is evidence of bond formation.
That definition paper from Pure and Applied Chemistry is accompanied by a technical report that gives history, examples, etc.
So, while hydrogen bonds are most well known for fluorine (FH-F), oxygen (OH-O) and nitrogen (NH-N) (and permutations), there is a huge zoo of other examples (e.g. reaction intermediates for proton transfers, etc) where this definition applies, and they include most of the periodic table except the  electropositive metals.
(Any hydrogen bond of course also works with deuterium, just the stability and reaction dynamics are in most cases slightly different.)

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text:
Redox - concentration cell
I'm studying galvanic cells and I've found a demonstration that lead me to an impossible result, so there must be a mistake, but I can't find what it is.
My problem is about concentration cells made by two standard hydrogen electrodes (SHE; $\pu{1 bar}$, $\approx\pu{1 atm}$, $\pu{298 K}$). I want to show that the reduction always happens in the most acid solution.
Demonstration
Consider the half-cell with $x > \pu{1 M}$:
$$\ce{Pt | H+ (aq, $x$\,\pu{M}) | H2 (\pu{1 atm})}$$
$$
\begin{align}
\Delta G &= \Delta G^\circ + RT\ln{Q}\\
E &= E^\circ - \frac{RT}{nF}\ln{Q}\\
&= -\frac{0.0592}{2}\log{Q}
\end{align}
$$
The reaction is
$$\ce{2 H^+ + 2 e- -> H2}$$
$$
\begin{align}
Q &= \frac{P(\ce{H2})}{[\ce{H+}]^2}\\
E &= \frac{0.0592}{2}\log{[\ce{H+}]^2}\\
&= 0.0592\cdot\log{[\ce{H+}]}\\
&= -0.0592\cdot\mathrm{pH}
\end{align}
$$
where $E$ is the reduction potential, and therefore the reaction is shown as a reduction.
So, this means that since $E<0$, $\Delta G>0$, this reaction is not spontaneous, consequently it never reduces when linked to a SHE. Can you find the mistake?

The 'standard' in 'Standard Hydrogen Electrode' also indicates the concentration of the electrolyte must be 1mol/l (pH 0).
Therefore, if you link two Standard Hydrogen Electrodes, you won't be able to measure a voltage.
If you don't intend to have standard conditions however, the reduction will indeed take place in the more acedic half-cell (simply spoken). This is also indicated by the electrode potential you calculated: -0,059V * pH. The reduction takes place in the half-cell with the higher potential and the potential you calculated increases with lower pH (more acedic solution).

Yes you're right, they are not two SHE. Anyway thanks for the answer because i didn't notice that if x > 1 then pH < $0$ so E > $0$ and G < $0$


text:
Why are polysaccharides not sweet in taste?
Polysaccharides are defined as polyhydroxy aldehyde or ketone which on hydrolysis yield many units of monosaccharides.
I got one answer(to my question above) as:

On our tongue, we have things called taste receptors. These receptors are loosely categorised into sweet, sour, bitter and salty. Our sweet-receptors bind to specific types of molecules, namely monosaccharides and disaccharides. Polysaccharides are not as sweet because they do not readily bind to the sweet-receptors on our tongue, as the other smaller molecules do!

My question is that in our body (due to the presence of water),polysaccharides and oligosaccharides are hydrolysed to simpler units of monosaccarides...
So why does the latter respond to the receptors(indicating sweetness) while the former does not(while both yield monosaccarides on hydrolysis)?

They don't form monosaccharides on hydrolysis in your mouth. Monosaccharides are formed in a later stage when the polysaccharides get to your intestines.

Most polysugars do not hydrolyse at all on a reasonable timescale. You need enzymes (e.g. amylose for starch) !

Well, while indeed polysacharides are metabolised to simpler units, even in mouth (saliva in the mouth can account for 30% of initial starch digestion), it usually doesn't happen fast enough, so full digestion follows later in you digestive system.
You can`t sense the sweetness of polysaccharides as they do not fit the receptors.

If you chew on wheat for five minutes, you indeed get some sweet taste, proving your above statement.

@A.Fleming Your point does make sense,but just before i accept your answer,I would be glad to have a valid reference regarding your views.

@Divyanshu  I am under the impression that this answer is factually correct. If you're seeking a reference for the 30% claim, here it is.


text:
Carbocation stability order 3
I have come to an example which suggests that +M effect of -OH is better than =O . How is that possible?
Here is the picture with their Hydride affinities (in Kcal/mol)

I think -OH should have stronger +M , as it's lone pair's hybridization is sp2 which will be closer in energy of sp2 orbital of carbon  centre as compare to sp hybridization of =O .
Source: www.cpp.edu/~psbeauchamp/pdf/315_SN_E_answers.pdf
Problem no. 21 , page no. 10 .

I'm voting to leave this open, but to prevent it from getting closed as homework, you should edit your post and include your thoughts and why the source sounds weird.

Ok I've added my thoughts too and this is from  California State Polytechnic University-Pomona
.


text:
Why is cysteine more acidic than serine?

How is that possible that the -I effect of $\ce{-SH}$ is greater than $\ce{-OH}$ group?
In case of cysteine $\mathrm{p}K_\mathrm{a}$ value of $\ce{-COOH}$ is $1.96$, whereas in case of serine it is $2.1$; a similar trend can be observed in case of $\mathrm{p}K_\mathrm{a}$'s of $\ce{-NH3+}$ group.
$$
\begin{array}{lccc} \hline
\text{Amino acid} & \mathrm pK_{\mathrm{a}1} & \mathrm pK_{\mathrm{a}2} & \mathrm pK_{\mathrm{a}3} & \mathrm{pI} \\
\hline
\text{Serine} & \color{red}{2.21} & \color{red}{9.15} & - & 5.68 \\
\text{Cysteine} & \color{red}{1.96} & \color{red}{8.18} & - & 5.07 \\ \hline
\end{array}
$$

The pKa2 value for cysteine here is probably that of the SH group.

Good question. First let's recognize that thiols (-SH) are more acidic functional groups than the corresponding alcohols (-OH) (some rationale here). So if we were talking about pKr we would see around 8 for cysteine's thiol side chain and around 15 for serine's alcohol side chain.
Now, to address your question about the carboxyl acidity, pK1, of these compounds. Serine's OH side chain has the more electronegative atom than cysteine's SH group, so we would indeed expect a stronger electron-withdrawing inductive effect in serine, possibly increasing the carboxyl acidity. However, the O atom is two carbons away from the carboxyl group, so this effect may be minimal to negligible... Not to mention of course the empirical data you provided implies that cysteine has the more acidic hydroxyl, not serine.
So I'm just speculating without looking at the actual experiments used to establish the pK1s, but it might be possible that when attempting to measure the pK1 of cysteine's carboxyl, the acidity of its thiol may have slightly skewed the measurement to make the carboxyl appear more acidic. I'm not sure if this was controlled for in the pK1 determination, but either chemically or computationally removing the contribution of the thiol to cysteine's acidity might result in a slight discrepancy compared to what it "really" is.
Additionally, one might imagine that the acidic thiol of could intramolecularly protonate cysteine's amino group, creating a proton-labile positive charge (-NH3^+) quite close to the carboxyl, and this might serve to stabilize the negative charge of the carboxylate conjugate base (-COO^-) thus increasing the carboxyl group's acidity. This effect would be more prominent in cysteine (due to the acidity of the thiol) than in serine (whose hydroxyl would be less likely to protonate the amine to stabilize the carboxylate). So this is a possible (but perhaps oversimplified) mechanistic rationale of why cysteine's carboxyl is more acidic than serine's.
As for the amino group, being more often protonated by the side chain in cysteine, it may also appear to be more acidic in serine.
This is just conjecture, for a complete understanding I would recommend looking at some experiments in the literature to see how they handle determining pK1 and pK2 when an acidic (or basic) side chain is present. It's possible that common practice is to eliminate the side chain's effect by computational or chemical methods, so without examining the primary sources one cannot be certain exactly what the reported pKa values actually represent.

Just to add to answer by electronpusher, also just speculating: it might be also possible that the thiol would coordinate and stabilize the protonated amine group better than the alcohol (due to lower electronegativity and larger radius of sulphur). Lower pKa of the amine would also make the self protonation have greater effect on carboxylic acid pKa.
Also, I wonder if the more pronounced effect on pKa of amine group can be any evidence for any of the mechanisms.

But for that, it has to make 4 membered ring. Can it be accounted for such large effect?


text:
Isopropanol in contact lens solution plastic bottle
I have a rather large plastic bottle of isopropanol ($99.7\%$), and I wanted to store some of it in a smaller, more handy container. I have two empty plastic bottles from contact lens solution that I wanted to use, but I'm not sure if it is safe for storing them in there for longer periods of time.
One of them has a 01 symbol inside a triangle, from which I concluded that this means PET and it also has a number 3 inside a circle (I'm not sure what that means).
The other one is HDPE (2 in a triangle) and has a number 5 elsewhere. Are both of those safe for long term storage of isopropanol?
I do intend to mark them very clearly.

Based on your description, it looks like you've determined both plastics correctly. In general, they both would work fine for storing isopropanol; however, according to the Thermo Fisher's Labware Chemical Resistance Table, HDPE is preferred:
$\ce{i-PrOH}$ – PET

$\pu{20 ^\circ C}$: E Little or no damage after 30 days of constant exposure;
$\pu{50 ^\circ C}$: N Immediate damage may occur. Not recommended for continuous use.

$\ce{i-PrOH}$ – HDPE

$\pu{20 ^\circ C}$: E Little or no damage after 30 days of constant exposure;
$\pu{50 ^\circ C}$: E Little or no damage after 30 days of constant exposure.


text:
Kinetic energy of molecules in liquid state?
My book (book link) has this question:

The kinetic energy of molecules at constant temperature in gaseous state is:

more than those in the liquid state
less than those in the liquid state
equal to those in the liquid state
None of these

The answer given is (3). It gives the reason as KE per molecule is still $=3/2\cdot k\cdot T$. So, the KE are both same!
But, I wonder, how can the kinetic energy of atoms in gaseous state be equal to that in the liquid state? Being in the gaseous state implies rapid and random motion of molecules, with little to no intermolecular attraction, spreading in all directions of any container. A liquid, however, is bound by intermolecular forces and is not so free to move.
In fact, I even wonder whether the formula $\text{KE per molecule }=3/2\cdot k\cdot T$ is valid in the liquid state, as it's derived from the kinetic theory of gases.

What exactly is correct in this situation here?

Comments are not for extended discussion; this conversation has been moved to chat.

The average translational kinetic energy of a molecule is $3kT/2$ irrespective of whether the molecule is in the gas, liquid, or solid phase.  In the liquid the motion giving rise to kinetic energy is restricted to a narrower range about the potential energy minimum  than it is in the gas phase. The equipartition theorem is quite general. A derivation is given here;  Derivation of mean kinetic energy
For ideal gases, the result may also be derived via kinetic theory. However, the most general derivation comes from the equipartition theorem.

Woah! That math passed over my head too quick! Though I'll accept the  truth that (I still don't know too much physical chemistry :P) average KE only depends on temperature and is independent of phase of the substance. And that this is based on the equipartition theorem.

I don't see how particles in a solid could have translational kinetic energy. All their kinetic energy would be vibrational, no?  There are exceptions, such as the free electrons in metals—but those are exceptions.

Lattice vibrations

A vibration is not the same degree of freedom as translation

Of course molecules in a solid can have translational kinetic energy! Ever been hit by a flying brick? Translational motion, in essence, is when the centre of mass (CoM) of the whole object moves but otherwise leaves the relative disposition of the constituents unchanged, and you can certainly have energy associated with that. Rotational motion is when neither the CoM nor the relative disposition changes. Vibrational is when the CoM does not change, but the relative disposition does. Energy can be associated with all three.

A deep dive for the sceptics
Since this is a question about which there seems to be a lot of misunderstanding, it seems worthwhile to take a deeper look, even though the given answers are perfectly satisfactory.
First, some terminology clarification. When people say something like “the kinetic energy of molecules in a liquid is the same as in a gas at constant temperature”, the “kinetic energy” in question is very specifically the average molecular translational kinetic energy. Mathematically, this is
$$K.E. = \frac12m\langle v_x^2\rangle + \frac12m\langle v_y^2\rangle + \frac12m\langle v_z^2\rangle$$
Where $\langle v_x^2\rangle$ is the average of the square of the x-component of the velocity of the molecule’s center of mass for a large population of molecules.
Note that rotations and vibrations in which the molecular center of mass does not move do not contribute to this total, even though there is atomic kinetic energy in those movements. Those kinetic energies are part of the total energy of the molecule, but are not part of the kinetic energy that we are referring to in the context of thermal equilibrium between phases.
So the claim that is being made is that this specific sum of translational kinetic energies is the same for molecules at the same temperature regardless of state. As I’ll explain below, this is true for systems that behave classically, which is a good approximation for most cases, but at low temperatures the quantum behavior of actual molecular systems causes deviations from the classical result.
Equipartition
You can find much more thorough information on equipartition elsewhere. What we need to know here is that the equipartition theorem tells us that in a classical system at thermal equilibrium, energy is evenly distributed between all “quadratic” energy terms. Our kinetic energy terms represent some of these quadratic energies, since the velocity terms are squared. The other type of quadratic term that is of interest is the potential energy of a harmonic oscillator, which has the form $\frac12 k \langle\Delta x^2\rangle$, where $ \Delta x$ is the deviation from the equilibrium position.
In this classical framework, if a gas and a solid are in thermal equilibrium, then the three molecular translational KE terms of the molecules in the gas must have the same amount of energy as the three molecular translational KE terms of the molecules in the solid. Thus the sum of the three terms is the same for a gas as for a solid. This sum is $\frac32 k_B T$ per molecule or $\frac32 RT$ for a mole of molecules.
Equipartition is directly observable in the heat capacity of substances. Because all of the quadratic energies must remain equal to each other, one cannot increase the molecular translational energies without simultaneously increasing the energies of all of the other quadratics. Since the temperature is proportional only to the molecular translational kinetic energy, the total number of these quadratic terms determines the amount of total heat that has to be added for a given increase in the molecular translational KE, ie the heat capacity of a substance.
Quantum effects
Of course, molecules are not purely classical systems, so we need to consider how quantum effects change this result.  A main difference between classical and quantum systems is the quantization of energy levels. These energy levels can be spaced far enough apart that there is not enough energy at a given temperature to excite the system from the ground state to the next higher energy level. This results in some quadratic energy modes not containing as much energy as would be predicted by equipartition.  A commonly given example is the intramolecular vibration of dihydrogen, which is not observed to contribute to the heat capacity except at high temperatures. We say that this vibrational mode is “frozen out” at lower temperatures.
In the gas phase, translational modes can be treated as classical continuous energies as long as the container in which the gas is held is sufficiently large. We need now to determine if the same is true for solids.
So the question we must now answer is whether molecular translations can be frozen out in solids, and if so, at what temperatures?
The nature of translations in solids
Within a molecular solid crystal, each molecule typically cannot move very far from its equilibrium position. When it does move from that position, it is pulled and pushed back towards the equilibrium position by interactions with adjacent molecules. Because the center of mass of the molecule moves a short distance and then moves back, these movements are typically described as vibrations rather than translations. It is for this reason that some textbooks will (confusingly) say that solids do not have translations, only vibrations. Because the center of mass of the molecule is in motion, however, these motions contribute to the molecular translational kinetic energy terms as we have defined them above. So with respect to equipartiation, they provide the three KE terms that sum to $\frac32 k_B T$.  Mathematically, though, they can be modeled quite well as quantum harmonic oscillators just like intramolecular vibrations (with stationary COM) can. To distinguish between the two, the translational vibrations are often called “external vibrations” or “lattice vibrations”, while the intramolecular vibrations are “internal vibrations.” [That means that these translations also provide potential energy terms to the equipartition, but those aren’t of concern here.]
Because the energy levels of quantum harmonic oscillators can be widely spaced (as in the $\ce{H2}$ vibration example above), it is entirely possible that these translational vibrations in solids can be frozen out at sufficiently low temperature, so we need to know more about the frequencies involved.
Conceptually, we can anticipate that since intermolecular forces in molecular solids are weaker than intramolecular forces and the molecular masses are greater than atomic masses, the frequencies (and therefore vibrational temperatures) of the external (translational) vibrations are expected to be lower than those of intramolecular vibrations, so they would require lower temperatures in order to be frozen out.  This is indeed the case.
Example
For crystals with Z molecules in each unit cell, there will necessarily be a total of 3Z orthogonal external vibrational modes spread across a range of frequencies. The highest of these tend to be in the far infrared region with wavenumbers around 100 cm$^{-1}$ or less. For example, solid carbon disulfide has two IR-active modes with wavenumbers around 65 cm$^{-1}$[1].  Solid iodine has three modes between 40 and 60 cm$^{-1}$ and three at even lower wavenumbers.
Using carbon disulfide as a test case, let’s look at the contribution of these modes to the heat capacity at the melting temp, when the solid and liquid phases are in thermal equilibrium.  Carbon disulfide has two molecules per unit cell, so there are six external vibration modes, and the two at 65 cm$^{-1}$ are the highest frequency.  The melting temperature is 161 K.  Using the standard formula, we find that a vibration with wavenumber of 65 cm$^{-1}$ contributes 0.97 R to the molar heat capacity at constant volume at 161 K, only 3% less than the value of 1R predicted by equipartition in a classical system. [Only $\frac12 R$ is translational for each mode. The other half is in the mentioned potential energy of the "vibration". So the contribution to $C_v$ is 3% less than a gas phase translational mode.]
In order to reduce this contribution to less than 90 % of the classical value, we need to go down to nearly 80 K, and this, remember, is for the highest frequency translational vibrations. The contribution of the lower frequencies will remain higher, and so the overall translational KE will be higher than 90% of the predicted value. In this example, therefore, we can conclude that the statement that the translational KE is equal between two phases at thermal equilibrium is a good approximation except at very low temperature.
The same analysis can be conducted with other molecular solids with similar results.
[1] Ishi, K. and Takahashi, S. (1985) Temperature dependence of the infrared active translational modes of solid carbon disulphide. Journal of Chemical Physics 82, 1476. doi: 10.1063/1.448422

When you perform a vibrational analysis you remove modes that are associated with translation of the CM or rotation of the inertial axes from consideration of the vibrational normal modes.

Then in the harmonic treatment half of the energy is potential, the other half of the vibrational energy is kinetic (expectation value), with motion about the CM. At least that's how I remember it.

@BuckThorn You're right about removing translation from vibrational analysis in the case of the gas phase normal mode analysis, but not for modeling of solid lattices. And you are correct that for any harmonic oscillator, half of the energy is potential and half kinetic (hence R per mode rather than 1/2R), but that KE is not counted towards the KE in the kinetic temperature. Hence the specific definition in the beginning of molecular translational KE.

@BuckThorn - here's a good summary reference on modeling motions in solid lattices: https://aip.scitation.org/doi/10.1063/1.1678617

I can't see why it would be any different for a crystal than a "gas phase molecule", but thanks for the ref. Any CM translation will move the CM of the crystal at constant momentum, otherwise (in the absence of an external field) it would violate some conservation law, no?

@BuckThorn - in the gas phase, there is no restoring force opposing translation, so there is no associated PE term, and the translations do not model as harmonic oscillators. The normal vibrational modes that do not involve translation that one calculates for the gas phase are still present in the crystal. They are the "internal vibrations" I mentioned. For the conservation, the CM of the whole crystal does not move, just the CM's of the individual molecules, so there is no conservation violation.

Ok, I grasp you might seek a different treatment of vibrations in solids and liquids, for mathematical reasons. Still, in the harmonic oscillator treatment the translations associated with vibrations are often called displacements, no doubt to avoid confusion. Also, stating that equipartition applies to solids is equivalent to stating that the law of Dulong and Petit holds, which is not true below a critical temperature (eg Debye temperature).

@BuckThorn Which is why I concluded with a discussion of the temperature at which the statement holds and where it fails. The Debye temp (and law of Dulong and Petit) results from essentially the same analysis as I described, though theirs is a more generalized treatment mostly applicable to atomic solids with assumptions that aren't true for molecular solids.

Ok, thanks for the clarifications (I think I will delete my extensive comments in a bit). The take-home message is then that in molecular solids the intermolecular potential is soft, and therefore a classical treatment of the vibrational partition function is reasonable already at relatively low T.

@BuckThorn - Right. And I'd add that although the Debye temp for atomic solids is often quite high, so are the melting and boiling points, so if we are considering two phases of the same substance in thermal equilibrium, the temperatures required to have two phases are high enough that the assumption of approximate equivalence of kinetic energy between phases is often still valid.

But, I wonder, how can the kinetic energy of atoms in gaseous state be equal to that in the liquid state? Being in the gaseous state implies rapid and random motion of molecules, with little to no intermolecular attraction, spreading in all directions of any container. A liquid, however, is bound by intermolecular forces and is not so free to move.

You would probably have no objection to saying that the temperature of a gas can be equal to the temperature of a liquid. In fact, over time (when there is nothing else going on), two phases in contact will approach temperature equilibrium, i.e. will have the same temperature.
In general, this happens through heat exchange, which could be through radiation, convection or conductance. The latter happens through collisions between particles of the two phases at their interface (let's say there is no particle exchange to make it simple). What happens in a collision depends on the speed and mass of the particles, and on what kind of intermolecular interactions they have. No matter what the details, though, after many collisions between particles of two phases, the kinetic energy will be distributed evenly (equi-partition theory mentioned in answer by porphyrin).
The reason that the same kinetic energy "looks" different in a gas, liquid and solid are the different intermolecular interactions. In a solid, there are strong and persistent intermolecular interactions. A given amount of kinetic energy allows the particle to "wiggle" a bit, against a strong intermolecular force toward the equilibrium position. In a liquid, the particle will diffuse a bit fast at higher temperature, bumping against other particles and making and breaking non-covalent bonds. In a gas, the particle moves pretty fast. Some of the kinetic energy is also present in bond vibrations (all phases) and in molecule rotation (gas phase).

In fact, I even wonder whether the formula KE per molecule =3/2⋅k⋅T is valid in the liquid state, as it's derived from the kinetic theory of gases.

Because of the equipartition theorem, answer (3) of the exercise posted in the question is correct in a rigorous way. In fact, the new 2019 SI definition of the Kelvin temperature scale is based on it, relating temperature to kinetic energy and the Boltzmann constant, which is now fixed to a constant value.

I've been wondering about this recently:  Does the same T for different phases of the same substance mean the same KE?  We know this isn't generally the case for different substances—e.g., at the same T, a diatomic ideal gas will have more KE than a monatomic ideal gas, because of the former's rotational dof's. But what about two phases of the same substance?  Consider my answer here:  https://chemistry.stackexchange.com/questions/129290/what-exactly-is-temperature/129366#129366  For a liquid and gas, say, to have the same KE at the same T, they would need to have the same total dof's..

...weighted by their fractional accessibility.  Would this always be the case?

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


text:
How is the cavity of a cyclodextrin chiral?

I am studying the host-guest chemistry of cyclodextrins with achiral aromatic molecules, and I found out that a useful way of studying them is by induced circular dichroism spectroscopy. This technique is used by taking an achiral molecule and placing it in a chiral environment to induce a CD signal. Apparently, the cavity of a cyclodextrin is chiral, but for the life of me I cannot see HOW it is chiral. I know the glucose molecules the ring is made of are D-glucose, but I still see mirror planes in the structure, and I cannot figure out how the cavity is chiral! Help??

related https://chemistry.stackexchange.com/questions/153/how-do-you-separate-enantiomers

You don't look close enough. From pictures it would seem there's center of symmetry no planes, but rotation is blocked - inside of molecule can't become outside so there's no center and cyclodextrin is chiral.

It is because the constituent glucose molecules are chiral.
In the image you can see that all the $\ce{CH2OH}$ groups in the center are out of the plane (towards you). These groups are on the upper side of the cavity. If they were below the plane, the cyclodextrin would have other interactions with the host molecule.

Not necessarily because there are more chiral centers, but probably the interactions would have much less chiral nature. That cyclodextrin would serve still for non-chiral purposes like solubilizing hydrophobic molecules.

So regardless of the mirror planes in the molecule as a whole, it is still chiral, simply because the glucose molecules provide the chiral information individually?

Yes, as long as there are chiral centers, the whole is chiral (except if it is a meso form). If you could invert all the chiral centers of glucose and then make a cyclodextrin of them, you would have the specular form; exactly the same properties except for the optical activity.  This would be, however, very complicated since cyclodextrin is made from starch, and starch comes from biological sources and there the chiralty is as it is.

There's no mirror planes, there could be only one plane making it meso, but half of glucose units would need to be "backwards".

@Mithoron I meant it in general. Like in tartaric acid. In the case of glucose I do not think that it is possible to have a meso form since both ends of the molecule are not the same ($\ce{CHO}$ and $\ce{CH2OH}$).

IF cyclodextrin were comprised of alternate D- and L- glucoses, it would be meso even though the both ends of the monomers are different. Cf with nonactin.


text:
Optical Activity of Organic Compounds
If an organic compound has only one asymmetric carbon in it, is it an Optically Active compound for sure?
Similarly, If an organic compound has No asymmetric Carbon atom in it, then is it Optically Inactive for sure?
(without considering any other specification of  its structure in both the queries.)
Thanks.

What are your thoughts on this? This site is NOT a homework answering service.

I'm sorry, but that's not my homework question. But why do you think so? (No grudges, asking genuinely)

https://chem.libretexts.org/Core/Organic_Chemistry/Chirality/Chirality_and_Stereoisomers   https://chemistry.stackexchange.com/questions/87869/how-to-find-the-plane-of-symmetry-in-a-molecule

One question too much. Also both are kinda broad - this isn't a place for "yes or no" and there are lots of examples why answer to second question is no.

If we limit ourselves to hydrocarbons there are 1,3-disubstituted allenes.  Build a molecular model of any such compound, and you see that no single carbon atom is asymmetrically bonded yet you can't rotate the molecule to match its mirror image, or find any mirror planes.  Ypu find that even if both substituents are the same like 1,3-dimethylallene.  You have a chiral molecule with no chiral center.
A molecule with just one chiral center will be chiral,  but with two they could have mirror image configurations so that a mirror plane passes between the chiral centers.  See here.


text:
Boiling point of alkynes from n = 2 to 10
From a video on organic chemistry, I found this table that shows the boiling point temperatures of alkynes $\ce{C_{n}H_{2n-2}}$ with $\,n=2...10$.
$$\begin{array}{|c|c|c|c|}
\hline n & t_\mathrm{boil}/^\circ\mathrm{C} & m & T_\mathrm{boil}/\mathrm{K}\\
\hline 2 & -83.6 & 26 & 189.6\\
\hline 3 & -23.2 & 40 & 250.0\\
\hline 4 & 8.1 & 54 & 281.3\\
\hline 5 & 39.7 & 68 & 312.9\\
\hline 6 & 71 & 82 & 344.2\\
\hline 7 & 99.7 & 96 & 372.9\\
\hline 8 & 125.2 & 110 & 398.4\\
\hline 9 & 150.8 & 124 & 424.0\\
\hline 10 & 174 & 138 & 447.2\\
\hline\end{array}$$
Here $\,m=14n-2\,$ is the molecular mass and $\,T_\mathrm{boil}\,$ is the boiling point temperature converted to kelvins. Surprisingly, when I plot $T_\mathrm{boil}$ v.s. $m$, I find a power law $T_\mathrm{boil}\propto \sqrt{m}$. The melting point temperatures were also shown in the video but were not so regular. Can someone explain why the boiling point satisfies such a simple power law?

Which isomers are these?

related https://chemistry.stackexchange.com/questions/28576/alkane-alkene-alkyne-boiling-point

@Mithoron, based on the boiling point of $1$-butyne ($8.1,^\circ\mathrm{C}$) and $2$-butyne ($27,^\circ\mathrm{C}$), these should all be $1$-alkynes to exhibit such consistent pattern.

For the curious, I plotted the first ten boiling points, melting points, and their square roots in kelvin of alkanes, and there's no discernable pattern in them. Here's the graphs. Here's the sheet to fiddle with

@GaurangTandon, the melting points are not so regular as I observed for alkynes as well. For the boiling points, if you plot $T_{\mathrm{boil}}$ v.s. $\sqrt{m}$ with $m=14n+2$, you get a straight line. I'm not sure if there's a simple theory behind this empirical $\sqrt{m}$-law.

@ZhuoranHe oh, you're right. I should have plotted against molar mass instead (facepalm). Editing

@ZhuoranHe Ok, here's the new sheet. You're right. There's a straight line relation between $Tb$ and $\sqrt{m}$.

I did linear regression analysis (fancy name for data entry into calculator :P) and obtained $T_b=0.431287764+0.024704463\sqrt{m}$ for alkynes. Not a very good linear relation though :( Let's check if it fits higher alkynes.

I would like to try to give a more detailed answer, but this is all I can give right now. I am fairly confident this is what is going on.
The dominant intermolecular force which determines the boiling point is dispersion. Because dispersion is the result of instantaneous dipoles, the strength of the interaction is proportional to the "surface area" of the molecule. There is no well-defined measure of the surface area of a molecule, but the most common way to describe this would be by using a van der Waals' radius for the atoms/groups.
The important point, however, is that when we add another carbon with its hydrogens, to a hydrocarbon, the mass increases linearly, while the surface area increases proportional to the square of the van der Waals' radius.
So, if we believe that dispersion forces are truly proportional to surface area and hence truly proportional to the boiling temperature, then a change of mass proportional to $\sqrt m$ should make a linear change in the surface area of the molecule, which should make a linear change the boiling temperature.
Notice, however, that this argument should be exactly the same for a linear alkane, and not just an alkyne. I made this same plot for the first fifteen n-alkanes, and the trend is very linear.

As to why there isn't a similar trend for the melting points, melting temperature is much more sensitive to symmetry than the boiling point, so how the molecules stack will make quite a difference. I would expect that hydrocarbons do not form very good crystals, so the trends may be highly dependent on the specific molecule.
Also, notice that the more branched these molecules are the less this argument should apply to the boiling point because each groups added at branching points will add less surface area than for the straight alkane because some of the surface area will be "redundant" so to speak.
Great observation! I hope my answer is right. I'm sure this has been noticed before, so it would be interesting if someone could find references.

Another way of thinking about this which I like is this:
Imagine that we can add mass to the carbon chain continuously rather than in lumps of atoms. Because adding mass also adds length to the chain, adding mass adds surface area. From this perspective, the relationship between $\sqrt m$ and surface area is the same as the relationship between length and surface area. This follows because these molecules are shaped a lot like cylinders.
So, the only jump left is from surface area to boiling point, which is already understood to be related through dispersion forces.

The empirical $\sqrt{m}$-law is found to work for small molecules. For example, if you extrapolate the alkane data to $n=0$, the boiling point of $\ce{H2}$ is reproduced with good accuracy. For very big $n$ I found a paper that uses a logarithmic function to fit the boiling point temperature (no explanation). To understand the $\sqrt{m}$-law, we would need a model for small molecules that allows to do calculations.

The geometrical explanation is very attractive, but the outer surface of a cylinder scales linearly with the cylinder length $S = 2\pi r * L$, so I would expect a linear scaling with the number of atoms rather than a square root ?


text:
Silk Amino Acids and Fruit
Can I mix silk amino acids, bamboo extracts and keratin extracts with fresh organic fruits for a hair treatment?

You probably can.
So-called silk amino acids are also known as sericin, which are small proteins with a high serine content. There is a review in Scientific & Industrial Research, 2004, 643, 323-329 (PDF) for some background information.
Just like the sericin, keratin extracts are commonly used in hair care products.
Bamboo extracts seem to be used as well, probably due to the flavonoids contained and their antioxidant properties.
All this plus the fruit does not sound harmful in general, however:

I can not know whether your skin will show an allergic reaction on one of the ingredients or their combination!
You will probably prepare your hair care mixture without any preservatives. It might therefore be a good idea to prepare and use it freshly. You certainly do not want microbial growth in your cosmetics!

Given that you have beautiful hair while my preferred hair care product is a sharp blade, you might not want to base your experiments on my opinion alone.


text:
Order of Acidic strength: Phosphorus oxoacids
I was asked to compare the acidic strength of the following phosphorus oxoacids,

I thought the order (of strength) would be $\ce{H3PO4 > H3PO3 > H3PO2}$, because they have three, two and one ionizable $\ce{O-H}$ bonds (that furnish protons / $\ce{H^+}$ ions).
However, looking up the $\pu{pK_a}$ values on the respective Wikipedia pages of each of those acids, suggests the exact opposite order of acidic strength:
$\ce{H3PO2 > H3PO3 > H3PO4}$. Apparently the acid with one $\ce{O-H}$ bond is a far stronger than the acid with three $\ce{O-H}$ bonds!
This is horribly counter-intuitive :-(
So, why is the correct order of acidic strength  $\ce{H3PO2 > H3PO3 > H3PO4}$, and not the other way around as I believed? What was wrong with my analysisi ( acidity in oxoacids $\propto$ number of $\ce{O-H}$ bonds)?

(+1) It's possible that this question should be separated into two parts. The $\pu{p}K_\pu{a}$ values are measured in water where tautomerisation is significant. To directly compare the acidities of the depicted compounds we should move into the gas phase, and compare the $\Delta G_\pu{r}^\circ$ values of deprotonation. Sadly, I only found data for $\ce{H3PO4}$ as shown in OP, written as  $\Delta_\pu{r} G^\circ = 1351\ \pu{kJ mol^{-1}}$, doi: 10.1063/1.473465. Further information might be available from computational studies.

It was asked maybe half a dozen times already, check out the site before asking.

There was also https://chemistry.stackexchange.com/questions/38034/why-is-h3po2-more-acidic-than-h3po4 and terrible https://chemistry.stackexchange.com/questions/65146/why-is-phosphorous-acid-more-acidic-than-phosphoric-acid

@Mith Ah, I see :-(

There was a better duplicate I think. If I find it mod can add second one. It's recurring question, because it's defying common "rules".

https://chemistry.stackexchange.com/questions/77309/can-the-effect-of-resonance-on-acidity-be-generalised needs duping probably

@Mith Gee, thanks ;-) But the answer to the duplicate you proposed first seems less controversial than the answer to this post. I'm not sure myself, so I'd like the Mods (or anyone with a sound knowledge of the matter) to compare the answers (with existing literature). Let the post with the better answer be the "original" O:)

It's better then this one but not fully correct I think; there was still better one IIRC

Acidic strength of the acids is determined on the relative tendency to donate one proton.
The source where you have seen then $\mathrm{p}K_\mathrm{a}$ value has considered the tendency of first deprotonation of the acids ($\ce{H3PO2}$ deprotonates once, $\ce{H3PO3}$ can deprotonate twice and $\ce{H3PO4}$ can deprotonate thrice).
The double bonded oxygen has negative inductive effect (-I). And this electron withdrawing effect is experienced by one $\ce{H}$ of the one $\ce{-OH}$ group in $\ce{H3PO2}$. Hence it has greatest acidic strength.
Similarly, the -I effect of $\ce{O}$ is experienced by two $\ce{H}$ of the two $\ce{-OH}$ groups in $\ce{H3PO3}$, hence electron withdrawing effect is reduced because it is getting distributed between two groups. Hence, it has lesser acidic strength. Similarly, in $\ce{H3PO4}$, it is distributed between 3 groups and so the acidic strength is least.
Thus the acidic strength, is:
$$\ce{H3PO2 > H3PO3 > H3PO4}$$
(I hope it's clear.)

I'm not convinced by this explanation. Do you have a source?

@orthocresol I have read this explanation for the oxyacids of P in my textbooks. And whether you are convinced or not, it's a fact. But, if you have any other considerations and explanations, feel free to comment and answer this question.

I second what @ortho said. It would be helpful if you could list a source that mentions this. Thanks! :-)

@paracetamol Actually, I studied this in my textbooks and also my teacher told this while studying various applications of inductive effect, Mesomeric effect, hyperconjugation, etc. This is a basic concept. But, I think I have the name of a popular book in which I read it.... I'm checking out in google books. I'll upload a screenshot, if possible, shortly.

Why can't this explanation be applied to the oxyacids of sulfur, namely sulfuric acid and sulfurous acid?

@TanYongBoon see... In sulphuric acid there are two double bonded O and in sulphurous acid there is one double bonded O with the central atom. But, there are same number of -OH groups in both. So, in H2SO4, the negative inductive effect is much more due to two double bonded O and hence it is more acidic. You can practically explain several oxyacids by this method. For example take..HOCl, HClO2, HClO3 and HClO4. This can also be explained way. Just considering the number of double bonded O and -OH groups present.

@paracetamol And while answering, we have some knowledge from which we answer and it is not always possible to cite a source. But if possible, if usually do that. This is based on a basic concept of chemistry. Suppose, you are said to compare some carbocation stability. So will you cite a source?

Please do upvote my comments if you really agree with me.

While your answer seems convincing to me (I'm not a chemist though), it's always appreciated if a source is included when talking about not-so-familiar cases. As for carbocation stability, I would cite Wikipedia  (which cites other articles) or some standard Org. Chem textbook :-)

This answer makes no sense, it's just not working like this and this question can't be answered on basis of inductive/mesomeric effects.

I’m not convinced because inductive effects simply aren’t “shared”. As the writer of an answer, the burden of proof lies on you. I can claim that the Sun orbits around the Earth and I’m sure you would agree that the responsibility for proving it would be entirely mine.

@orthocresol See....I have read this as an explanation and hence I wrote this here. Even you can google it and you can find that in many sites such as Quora, libretexts, quantumstudy, etc.

Also Pauling’s rules are based on resonance not inductive effects. Inductive effects won’t explain a pKa difference of ~5 between HClO3 and HClO4

@orthocresol see... I'm not a chemist and not a researcher..... I'm a student and since during my chemistry course I studied this I wrote it here.

@orthocresol I know that they are not trustworthy sources. for that reason only I did never cite those. But, this concept regarding acidic strength of oxidised of phosphorus is a common and basic concept. I studied this while studying inductive effect in my textbooks.

This is wrong, so very wrong. Wherever you got this information from: throw it away, it's wrong. And if that is wrong, I wouldn't trust anything else from that source either.


text:
For what reasons, is isopropyl alcohol widely used as a cleaning solvent?
I have worked in electronics and aeronautics industry 30 years ago, and I have always wondered, what were the specific properties of isopropyl alcohol, which make it used instead of other alcohols like common methyl or ethyl alcohol ?
Edit:
I know that it's less toxic than other solvent and that it is easier to separate from water (by adding salt) than other alcohols (reduce producing costs ?), but some reasons might be less pertinent than others.

From this Wikipedia article

Isopropyl alcohol dissolves a wide range of non-polar compounds. It also evaporates quickly, leaves nearly zero oil traces, compared to ethanol, and is relatively non-toxic, compared to alternative solvents. Thus, it is used widely as a solvent and as a cleaning fluid, especially for dissolving oils.

 You should try Googling first :-)
Also, the antiseptic properties of isopropyl alcohol are an added benefit (over the other lower alcohols).
As you pointed out, methanol is notoriously toxic, so using that as an industrial cleaning agent is a no-no.
Ethanol for industrial purposes is often available as denatured alcohol; which has added impurities to render it toxic and/or alter its taste and smell.
These impurities may interfere with the equipment (you mention the electronics and aeronautics fields) you clean it with.
However commonly available isopropanol isn't denatured (then again, there's no real need to denature it). Since it is lacking in added impurities, it presents itself as a better choice over ethanol for cleaning such equipment.

I did it, but knowing some difference doesn't tell me which are the most pertinent for the choice.
btw, I wonder why ethanol would leave oil traces.

@Camion I've edited my answer. As for the bit on oil traces, both ethanol and isopropanol leave oil trace, but the latter less so (from the Wikipedia wording). I took the liberty of interpreting this as being indicative of  most oils being more soluble in isopropanol than in ethanol (though I'm unsure why exactly this would be so)

At the lab bench scale (like cleaning a microscope slide eg) one can use ethanol as well. https://www.berkshire.com/learning-center/dyk_1408_why_ipa/

there's no real need to denature it || you are underestimating capabilities of human liver and needs of human drunkards. While not very common, the use of isopropanol to achieve drunkenness is known.

@permeakra Do you have a source on this? I've never heard of anyone trying to consume isopropanol, and I've always assumed the bitterants added to it would have made it too difficult to do.

@Hashim I heard it repeatedly and talked to people who claimed to use it. Official data suggests that it is possible to drink small (<10 g) dose of isopropanol with no severe effects for health, and isopropanol does produce narcotic effects https://www.ncbi.nlm.nih.gov/books/NBK208299/ (read carefully, it is well hidden in "effects on animals"). I have no idea how widespread this practice is.

It depends on what impurities are being removed. Methanol, ethanol and isopropanol can be purchased or purified to equivalent degrees of purity. If they are used in a condensing vapor degreaser neither will leave an "oily film". Methanol removes ionics better, isopropanol removes nonpolar materials better.  Alcohols aren't the best to remove greases [methylene chloride or trichloroethylene are best here], and nothing removes fluorocarbon greases. Safety and environmental concerns are addressed by good lab techniques. There is serious trend towards using water based detergent systems. Their inception can be problematical, and I sometimes wonder if their use is better for the environment.


text:
Order of dipole moments of chlorine substituted methane (CH3Cl, CH2Cl2, CHCl3, CCl4)
In my book, the order of dipole moment for CH3Cl, CH2Cl2, CHCl3 and CCl4 is given as:
CH3Cl>CH2Cl2>CHCl3>CCl4
The way I understand this is—

In case of CCl4, tetrahedral symmetry exists. Therefore, the dipole moment of three C-Cl bonds on one side, gives a resultant moment that is equal and opposite to the dipole moment of the single C-Cl bond on the opposite side. So, $\ce{\mu=0}$.

This explains CHCl3>CCl4.

Again, in case of CHCl3, the three C-Cl bonds point to one side, so a moment equal to C-Cl bond is generated as a result. A moment of C-H bond is added to it. In CH2Cl2, the moment of two C-Cl bond is added at $\ce{109^\circ}$(approx.) so the resultant is greater than on C-Cl bond.

This explains CH2Cl2>CCl4

In case of CH3Cl, the dipole moment of one C-Cl bond is added with moment of one C-H bond (resultant of three C-H bonds).

But this should mean moment of CHCl3$\ce{\approx}$ moment of CH3Cl, but clearly this is not so.
Where did I go wrong?

"But this should mean moment of CHCl3≈≈ moment of CH3Cl" In making this assumption, did you take into account that the direction of the dipole moment in a $\ce{C-H}$ bond is opposite to that in $\ce{C-Cl}$ bond?

@Gaurang Direction of dipole moment for C-H bond is in the same direction as the resultant due to 3 C-Cl bonds. C attracts electron from H, and Cl attracts electrons form C. So, the moments are added.

https://chemistry.stackexchange.com/questions/48066/logical-way-to-compare-dipole-moment


text:
Potentiometric determination of fluoride
We did an experiment of potentiometric determination of fluoride ions. In several beakers we added several concentrations of NaF solutions and then we added water. We calculated with the electrode the potentials.
Then in different beakers we added again NaF solutions, but now we added total ionic strength adjustment buffer (TISAB). We calculated again the potentials. I see that the potential values in the beakers with TISAB are bigger than those with water. Why?
I know that TISAB works in order to adjust the total ionic strength and to work against the interfering ions, but at this point why is there a difference? Does the hydroxide ion play such a big interference role such as the ferric ions, for example?


text:
Identifying luminescent transitions for lanthanides
Using the ground-state of Eu(III), how can I identify the possible (number of) luminescent transitions from the $^{5}D_{0}$ excited state?
I identified the ground-state term symbol for Eu(III) using Russel-Saunders coupling as $^{7}F_{0}$. I understand the selection rules and allowed transitions, but I'm not sure where to start on listing all of the possible transitions. I know that using RS coupling the $J$ states are $J=6,5,4,3,2,1,0$ but the transition from the $^{5}D_{0}$ state to the $^{7}F_{0}$ ground-state has $0\rightarrow 0$ with respect to $J$.
The only thing I could think of was to take the possible combinations of $\Delta S = 0,1$ and $\Delta L = 0,1$ and identify which of these transitions were allowed; is this along the right lines?


text:
Varying data about solubility of sodium acetate trihydrate
I recently looked up the solubility of sodium acetate trihydrate in water at 20°C and found surprisingly diverse values such as:
613g/l (GESTIS-Stoffdatenbank)
762g/l (TU Freiberg)
1250g/l (applichem, a vendor)
612g/l (German Wikipedia)
1233g/l (English Wikipedia)
I also found 408g/l somewhere but I can't find the source anymore.
As those values differ by a huge amount, I wanted to ask if someone knows how it comes to such a big difference.
Furthermore, I would be very grateful if someone could point out the correct value.

I suspect it has to do with the ability of sodium acetate to easily form supersaturated solutions, making experimental determination of solubility tricky. -1 for not comparing Wikipedia articles in all the languages (just kidding:) ).

The ability to form supersaturated solutions was my first assumption too but shouldn't it be possible to get more exact results than "Somewhere between 408 and 1250 g/l"? That's a pretty wide range. By the way, I also searched the Latin Wikipedia which (sadly) didn't have an article about sodium acetate.

@GreenSmurf I briefly checked some other resources and it seems like there is also a confusion between the the solubilities for the anhydrous substance and its hydrate. But yeah, this is indeed an interesting question.

CRC Handbook 2017 lists same solubility values for sodium acetate and sodium acetate trihydrate: $\pu{50.4 g/100 g \ce{H2O}}$ at $\pu{25 ^\circ C}$.

seems like mixture of two factors. 1) solubility of the salt changes with temperature rather significantly 2) some values actually are for anhydrous salt.

My own experiments have given me a value of 620g/l which seems reasonably close to 613g/l and 612g/l.


text:
Electronic strucure of cyclopropene radical cation

Suppose I have a cyclopropene molecule with a + charge on the singly bonded carbon.I think the singly bonded carbon is $\mathrm{sp^2}$ hybridised, isn't it? To obtain this species from cyclopropene, one can consider removing two hydrogen atoms, one homolytically (the one bonded to pure p orbital of carbon) and the other heterolytically (the one bonded with $\mathrm{sp^2}$ orbital of carbon).
So, finally, there is one electron in the pure p orbital and one unit positive charge on one $\mathrm{sp^2}$ hybridised orbital, shouldn't the electron in the pure p shift to the $\mathrm{sp^2}$ orbital as $\mathrm{sp^2}$ has more s character (so it is near the nucleus), does this shift happens?
I am unclear about this scenario, can someone please explain where the electron and the positive charge reside in the C atom.

Correction: "I removed one H atom from the singly bonded Carbon heterolytically" Yes, it's heterolytically, but then it's the "hydride ion" that you have removed and not the hydrogen atom.

https://chemistry.stackexchange.com/questions/69925/how-to-compare-stability-in-tropylium-and-cyclopropenyl-carbocation

You do not really have a singly bonded carbon atom. When you remove the hydrogens from the originally saturated carbon the pi electrons delocalize to make an aromatic ring.  In effect you have a cyclopropenyl cation with one of the hydrogen atoms homolytically removed.  The positive charge is thus delocalized in the ring but the odd electron is localized to the carbon with no hydrogen, in the outward pointing $sp^2$ hybridized orbital.

Thanks, what is the hybridisation of carbons here, is it sp2?

All carbons are sp2.


text:
How are poisons discovered? Does someone have to die/be poisoned from it first?
How are poisons discovered? Does someone have to die/be poisoned from it first?
Or are there other ways of discovering the harmfulness of a substance? Perhaps everything is tested on other animals prior to testing them on humans?

Related question on WorldBuilding SE - A poison that only works if you know it is there

Yeah, could be :P Except that it's quantum mechanics while we are on a macroscopic scale, so, no chance of it ever existing :P

Most poisons will also make you ill at lower doses than the lethal dose. So, a very careful person can discover many poisons by being very cautious about the dose of anything they eat that has unknown qualities. This is presumably how ancient societies got to know so many poisons.

Alle Dinge sind Gift, und nichts ist ohne Gift, allein die Dosis macht dass ein Ding kein Gift ist (The dose makes the poison) - Paracelsus

Poisons (I'm going to use this as an umbrella term for "toxins" and "venom" as well. Bear in mind though, they are not the same thing) have been known since antiquity.
Back in the good old days, you figured out if something was poisonous or not by eating/touching it (or getting someone else to do it), i.e- "discovering" a poison was simply a matter of chance.
These chance encounters alone, led to the discovery of numerous poisons.
With the advent of Chemistry, the gents in white lab-coats figured out that compounds that bear resemblance to the already well-known poisons, in terms of their functional groups and structures, are also toxic (albeit, to vastly different degrees). Poisons could now be identified a priori (you could tell it would end badly if you were exposed to such a substance, but not how badly).
And no, you can't really quantify a poison's effects without testing it on something. Nor can you tell how much of something would be needed to kill or severely maim.
This is where we bring in the idea of a median lethal dose $\pu{LD_{50}}$ which is, simply put, the minimum amount of a substance required to kill off half of all the animals in a particular test group.
$\pu{LD_{50}}$ values for a particular substance depend on the animal used.
Of course, the only way to get an accurate $\pu{LD_{50}}$ (which itself, is really a "mean/average" value of sorts) for a human is to actually poison someone, which doesn't sound very nice. So you do the next best thing, you measure it for a rat/chimp, couple it with your knowledge of the poison's mechanism of action, and extrapolate the value to something that would kill a person.
Another way to establish degree of toxicity, which doesn't involve killing animals, would be by exposing a cell/tissue culture (and not the entire animal) to the potential poison .

Great answer! But Google translate gives "All things are poison, and nothing is without poison, but the dose makes that a thing is not a poison" for your quotation. I don't know German though so I can't verify it in that case I leave the discretion upto you :)

@GaurangTandon That's correct, but the version I put up is the better known one  ;-)

Alright! Also, Re: "These chance encounters alone, led to the discovery of numerous poisons."  slightly off-topic, but this reminded me that I remember watching a show on Nat Geo where they showed how a scientist had drank a poison knowingly in order to record what effects it had on his body, because till then its effects on human bodies were unrecorded. To his last minute he kept writing of the negative effects of the poison until he finally collapsed and died. Tragic! Did you ever come across anything about this scientist, the poison, or the story?

@GaurangTandon Since NatGeo hasn't been taken down, I'll assume they just mentioned that incident (if it ever happened), and not demonstrated on live T.V. That said, if that incident did occur, I suspect the chap never did intend to die, it's just that the path of Science is a long and dangerous one (1) (2) (3) O:)

@GaurangTandon did you know that a scientist volunteered to test vx (spoiler: he didn’t die)

@paracetamol it can also be noted that the Chinese legendary toxicologist Shen Nong allegedly tasted different herbs to see if they were toxic and died this way (just a note that your link only includes European people)

+1 for the quote. As my tutor used to say (we did synthesis back then) the compounds we synthesize or are poisons or do nothing. If they do nothing they have no interest. If a poison, it can be a medicine when used with  the correct dose.

Note that culture testing is prone to false positives.  For example, a culture test would conclude that theobromine and ethanol are deadly poisons, because a tissue culture doesn't have a liver.  Further, it would conclude that a number of substances are deadly neurotoxins, when in fact they're harmless to humans because they can't cross the blood-brain barrier.

Of course, the only way to get an accurate LD50 ...  for a human is to actually poison someone. Actually rather a lot of people, as it's a statistical measure

@Chris True. I did say you need a test group to compute the $\pu{LD_{50}}$. I suppose a less ambiguous way to put it would be "Of course, the only way to get an accurate $\pu{LD_{50}}$ for a human would involve actually poisoning someone"  ;-)

I don't know about chemistry, but in nuclear physics and radiation health science some of the dosage information was determined from either unwitting exposure or exposure before harmfulness was known. Of course, this has the problem of not being repeatable.

As you already correctly deduced, the discovery of poisons was in former times quite accidental, but once its potency was discovered, the (mis)use of it was predictable.
It must also be said that our ancestors were very careless with poisonous substances. The old houses used wallpapers and paint which were spiked with arsenic, lead and antimony. Copper arsenite or Scheele's Green, Copper (II)-acetate and White lead were used for the resulting luscious colors.
The alchemists and later the chemists did test substances on themselves. While the alchemists were often reckless by trying their concoctions undiluted, the chemists were more careful and used strong dilutions which were carefully increased in potency. Still, taste and smell were methodically used until the first part of the 20th century. Christian Wilhelm Posselt and Karl Ludwig Reimann e.g. tested the extracted nicotine on the tip of the tongue and remarked the burning feeling (Nicotine is by the way very poisonous and is used as insecticide). Carl Wilhelm Scheele was not so lucky, his continued testing of every discovered substance finally overpowered his body and killed him. LSD was experienced by Albert Hofmann despite precautions: He chose a dilution which was considered safe, but LSD was so powerful that he still got a 20 times overdose.
Since the advent of science, the information about the potency of poisons has been recorded independent of the cause: accident, suicide or homicide. Animal tests were used, but they were never really completely reliable. LSD seemed to be noneffective in animal tests, but is one of the most potent human drugs in relation to its quantity. The poison of the funnel-web spider is ineffective against dogs or cats, but could be fatal for humans. In contrast, wonderful tasting chocolate is deadly for dogs and cats. This can be mitigated a bit by using primates as test animals, but they are still different enough that deductions from results must be taken with care.
What we learned is that the effect of most poisons is unreliable. There are many, many recorded instances of humans dying on harmless doses caused by an overreaction and also people surviving incredible overdoses of e.g. cyanide poisoning. So the value of $\pu{LD_{50}}$ is a rule of thumb how poisonous a substance really is. Chemistry is now advanced enough that people have a pretty good idea how poisonous the result is, but there will always be uncertainty about the exact effects.
The only way to get an impression of the exact effects would be (apart from heroic self-tests) human experimentation, and as repugnant as it is, this has been in fact done deliberately. Paracetamol's link to the old Romans mentioned that they not only tested animals, but also slaves. During WWII the infamous Nazi and the Japanese Unit 731 experiments tested poisons on prisoners.

Sometimes poisons are discovered by chance.
At least that is what happened to me.
We were researching on products made with malonodinitrile and enones.
Since I was interested in the mechanism, I used milder conditions in the reaction so I could isolate the Michael and Knoevenagel intermediates and in one case a much more complicated product (see here).
The Knoevenagel adduct between malonodinitrile and cinnamaldehyde yields a beautiful yellow product.
At that time there were no gloves in the lab (and we were very young and a bit careless), so sometimes some mother liquor spilled on your hands when the product was filtered.
The first time I made the product my hands got red.
I did not pay attention.
The second time, my hands got red and my arms burnt.
I started to think that there might be a relation.
The third time I made it, I was red from head to toes and all burnt like hell.
I had to go to the hospital.
It seems to be an allergic reaction, which, as usual, become worse every time you come in contact with the product that causes it.
Not exactly a poison, but maybe in a higher dose it can be fatal.
I have not made the product anymore.
So in this particular case, it was discovered by the effect on a human.
A note for the younger ones: everything you synthesize can be a poison, so use gloves, work in the fume cupboard and do not get in contact with it.

"The third time"???

If you want to get an idea if a substance could be a venom prior to animal or human testing, you can use computational chemistry to model possible interactions with human proteins. It is primarily used to study possible cures or better understand how a disease works, but if you can not find any information about the substance elsewhere, it will help you narrow the options and lower the number of experiments needed to be done, especially if you not only want to know if it can harm your body, but also how and what a possible treatment might look like. It is a far shot, but it lowers the costs rapidly.


text:
Why does 1,2-dimethylcyclohexane only possess three stereoisomers?

Why does 1,2-dimethylcyclohexane in chair form have only three stereoisomers?

I understand that the trans isomer doesn't possess plane of symmetry and thus it has two stereoisomers. But my book says that the cis isomer has does not have two  stereoisomers. Why is this so?

The structure of cis-1,2-dimethylcyclohexane is shown above. Its mirror image is non-superimposable and cannot be obtained by rotation about single single bonds or by a ring flip. Can anyone explain?Thanks

Related: Enantiomers of 1,2-dichlorocyclohexane

Draw the other conformation...

Related, bordering on duplicate? Is cis-1,2-dimethylcyclohexane a meso compound?

related https://chemistry.stackexchange.com/questions/87847/why-is-trans-1-2-dimethylcyclohexane-an-enantiomer

Quoting , ORGANIC CHEMISTRY second edition by JOSEPH M. HORNBACK, UNIVERSITY OF DENVER
1,2-dimethylcyclohexane has two chirality centers.

The cis-diastereomer has a plane of symmetry bisecting the ring bond between
the two methyl groups  and therefore a meso compound.It is not chiral and does not rotate plane-polarized light.The trans-diastereomer exists as a pair of enantiomers.
The chair conformation of cis-1,2-dimethylcyclohexane  shows that it is chiral.
But due to  ring-flip , another  conformation ,its  enantiomer of the original conformation is also formed.These conformers interconvert rapidly at room temperature, therefore cis-1,2-dimethylcyclohexane is not chiral.
Further I quote
"stereochemical analyses such as the preceding one can be done with drawings using planar rings" pg.235.
A molecule cannot be optically active if its chiral conformations are in
equilibrium with their mirror images. We consider such a molecule to
be achiral. PG- 193 - ORGANIC CHEMISTRY  .EIGHTH EDITION
L.G. WADE , JR . WHITMAN COLLEGE


text:
How is Speed Mathematically Transformed into a Reaction Rate?
The basic physical definition of speed is "distance/time", and the rate of a reaction can be defined as "concentration/time".
However, how is distance (d) mathematically transformed into concentration (C) to go from the first definition to the second (reaction rate)?
(Edit):
I've been messing around with variables and might have found the answer. If speed (units: m/s) is multiplied by area (m^2), then multiplied by density (g/m^3), then divided by molar mass (g/mol), then divided by volume (m^3) we get to reaction rate (mol/L*s). However, I'm not sure if this is what was expected of me by my professor.

I'm not really sure what you are looking for here. Speed doesn't really have anything to do with reaction rate.

According to my Chemical Equilibrium Professor, it is possible to start from the first definition, do math, and obtain the second. He said it has to do with Fugacity and Activity.

You either grossly misunderstood you teacher, or he is fooling you. If he said what you say he did, i suspect the latter. Or you (not the teacher) are just months behind on the curriculum. Catch up!

The speed (or rate) of a chemical reaction is by definition how frequently it happens. Since it is happening everywhere in our reaction container, we describe it using how many times it happens per unit volume per unit time. Since the number of times it happens to molecules is enormous, we use the unit mole to do the counting. So if a chemical reaction is happening at $1\,\mathrm{mol/(L\cdot min)}$, it means in $1$ minute, $6.022\times 10^{23}$ reactions happened in every liter of reaction space.
For example, if we have a reaction
$$\ce{N2 + 3H2 <=> 2NH3},$$
when every $\ce{N2}$ molecule combines with $3$ $\ce{H2}$ molecules to form $2$ $\ce{NH3}$ molecules, we write down one such reaction equation. Then if we find ourselves writing down $1\,\mathrm{mol}$ of such chemical reaction equations for all reactions that happend in $1$ liter of volume and $1$ minute of time, we say the speed of the reaction is $1\,\mathrm{mol/(L\cdot min)}$. In a fixed volume this means the concentration of $\ce{N2}$ is reducing at the speed of $1\,\mathrm{mol/(L\cdot min)}$, the concentration of $\ce{H2}$ is reducing at the speed $3\,\mathrm{mol/(L\cdot min)}$, and the concentration of $\ce{NH3}$ is increasing at the speed of $2\,\mathrm{mol/(L\cdot min)}$. If the volume is not fixed, or if there is extra source of these materials, then the change of concentration is not only due to reactions, so the speed of reaction is not measured by how quickly the concentrations change, but is still measured by how many reactions happen per unit volume per unit time.
Another point to make is that the overall coefficients matter. For example, if the above reaction is happening at $1\,\mathrm{mol/(L\cdot min)}$, then the reaction below,
$$\ce{\frac{1}{2}N2 + \frac{3}{2}H2 <=> NH3},$$
is happening at twice the speed, i.e., $2\,\mathrm{mol/(L\cdot min)}$. This is because $1$ reaction above is equivalent to $2$ reactions below. The same goes for thermochemical equations.

My Chemical Equilibrium Professor says there's a way to start with the first definition (d/t), do math, and end up with the second definition (C/t). He also mentioned it is related with Fugacity and Activity.

@Sam202, then perhaps he's trying to calculate how many successful collisions there are based on thermodynamic properties of the gases and statistical physics, to link the "speed" in physics (how quickly molecules move) to the "speed" in chemistry (how frequently reactions happen). But these are different concepts with different definitions. The former determines the latter given some model.

It very well could be something along those lines; however, this question was given as the first assignment of the course, and he said it wasn't too complicated. I attempted to use dimensional analysis to try to find factors of units that would lead from meters/seconds to moles/(Liters * seconds), but that would require a factor of (moles/(Liters*meters)), which doesn't translate to other variables (or at least none of the ones I can think about).


text:
solve steady state of simple product reaction
consider these biochemical reactions (simpler than usual michaelis-menten setup):
$E + S \rightarrow^{k_{f}} ES$
$ES \rightarrow^{k_{r}} E + S$
$E$ and $S$ reversibly make $ES$. the forward/reverse rate constants are equal: $k_{f} = k_{r} = k$. how can you solve for steady state concentration of $[ES]$ given initial amounts $[E]_{0}, [S]_{0}$ and assuming $[ES]_0 = 0$?
$d[ES]/dt = k[E][S] - k[ES] = 0$
rate constant cancels:
$[ES] = [E][S]$
then $[E] = E_{t} - [ES]$ where $E_{t}$ is total enzyme concentration. by substitution:
$[ES] = [S]([E]_{t} - [ES])$
$[ES] = [S][E]_{t} - [ES][S]$
$[ES] + [ES][S] = [S][E]_{t}$
$[ES] = \frac{[S][E]_{t}}{1 + [S]}$
but if $E_{0} = 10, S_{0} = 5$ then $[ES] = (5 * 10) / (5 + 1) \approx 8.3$ but the right answer is closer to 4. what's wrong with this?

You considered $[E]+[ES]=E_0$. I think you also need $[S]+[ES]=S_0$? Another weird thing is the enzyme doesn't convert the substrate to any product.

it's not weird not to have a product. there is no product in this system. can you explain how substituting [S]_0 = [S] + [ES] helps?

There are $3$ equations for $3$ unknowns $[E],[S],[ES]$ given by
$$\left\{\begin{array}{l}
k_f[E][S]=k_r[ES],\\
[E]+[ES]=E_0,\\
[S]+[ES]=S_0.
\end{array}\right.$$
We can use the last two equations to cancel $[E]=E_0-[ES]$ and $[S]=S_0-[ES]$ and plug them into the first equation to get a quadratic equation for $[ES]$ given by
$$k_f(E_0-[ES])(S_0-[ES])=k_r[ES].$$
Then we plug in numbers $k_r/k_f=1$, $E_0=10$, $S_0=5$ to obtain $[ES]=8\pm\sqrt{14}$. Since we have $[ES]\leq\min\{E_0,S_0\}=5$, the solution is $[ES]=8-\sqrt{14}\approx 4.26$.


text:
What style clamp holder is this?
I cannot for the life of me find this style of clamp holder with beaker base:

I can find beaker bases on sites like Fisher Scientific/etc., and flimsy alligator clamp soldering stands on Mouser/etc., but not this specific alligator (or other wire holding) clamp with sliding connector, on solid rod, attached to a beaker base.
Does anyone know what those sliding clamp holders are called, or where I can get this as a kit? I use this for e-chem and love it but need more.
Edit, 15-Jan This is an older version of my actual setup, which I would typically have on top of a magnetic stirring plate:

I would then be taking 2 of the 3 probes and passing them through a number of other poured solutions. My current clamping setup is great because I can simply release the counter electrode and just move my working and reference to different baths; really, I just need more of what I already have. As these experiments are limited in scope and do not require it, I do not make provisions for salt bridges, nitrogen purging, hydrogen gas aeration, etc.

Isn't this off topic as the question itself is not about chemistry and the mentioned equipment is not specifically designed for chemistry?

I doubt there is a special name for those. Alligator clips are fine for keeping wires in place, but are unsuitable for holding the glassware. The setup on the photo looks a lot like a DIY project to me, probably someone came up with this design and used standard "crocodiles" to simplify the process of mounting and adjusting electrodes in the beaker.

I think you would have more success on the Electrical Engineering Stackexchange or something similar

@GreenSmurf This setup is used specifically for 3-electrode cyclic voltammetry experiments.
@ andselisk I have standing beakers that work just fine, but the electrode holding is quite helpful. I was hoping to avoid cheezy soldering setups.

Cyclic voltammetry (CV) is often performed in an enclosed vessel with the purge gas (argon/nitrogen) constantly bubbling through to prevent the solution from aeration and unnecessary side processes, mainly oxidation by the atmospheric oxygen. You'd rarely see alligator clamps in any CV apparatus also because they are not suited for holding moderately thick round objects (such as electrodes) in place, they can damage the inlet and outlet piping (e.g. the one for the purge gas, or chemicals if one studies kinetics), plus they make the experiments less reproducible since the position of the electrodes is changed each time you change the beaker.
Typically a teflon lid with 4 (or more) holes is used, where all 3 electrodes and the purging line are adjusted in place with rubber rings or plugs so that they are all firmly attached to the lid. This is very convenient since you don't have to manually adjust the apparatus every time the solution needs to be changed. All that's required is take the lid off, rinse the vessel and electrodes, fill the new portion in and put the lid back in place.
Alternatively, one can use a 3-neck round bottom flask with the electrodes attached through the rubber stoppers in the side necks. The downside is problematic cleaning, so this setup is more appropriate for the continuous measurements when one doesn't have to switch solutions too often. This is the simplest setup; a more sophisticated one can as well look like this:

Image source

Either way, I would discourage from using alligator clips for CV. Even though they might work for simple measurements, in my opinion there are too many downsides.

I added an edit to my original post showing my setup. I really do just need the clamps that I have, not the more traditional CV setup, as I'm doing rapid device fab and testing in a bunch of different baths. If I find what I'm looking for, I'll post it here. If not, I'll just mark this as answered and use a soldering setup.

@PoGaMi Thanks for the update; I think that beaker on the second photo looks too big for the amounts you work with. I'd just extend the electrodes (e.g. by soldering wires), put them in the lid, make the compartment smaller, and let the wired alligator clips lie somewhere aside so that they don't mess with the setup.


text:
Dissociation constants of leucine
Leucine is a type of amino acid. Let me write its chemical formula as $\ce{NH2-C5H10-COOH}$. On the wikipedia page, https://en.wikipedia.org/wiki/Leucine, there are two dissociation constants, one for $\ce{-COOH}\,$ given by $\mathrm{pK}_{a1}=2.36$, and one for $\ce{-NH3+}$ given by $\mathrm{pK}_{a2}=9.60$. But I also know that once $\ce{-COOH}\,$ loses its proton $\ce{H+}$ into the solution, the proton may combine with the $\ce{-NH2}\,$ group of the same molecule to form the zwitterion $\ce{NH3+-C5H10-COO-}$ so that the solution now has $4$ species with $4$ dissociation equilibria
$$\begin{array}{rclll}
\ce{NH3+-C5H10-COOH}\!\!&\!\!\ce{<=>}\!\!&\!\!\ce{NH2-C5H10-COOH +H+},& \ &(1)\\
\ce{NH3+-C5H10-COOH}\!\!&\!\!\ce{<=>}\!\!&\!\!\ce{NH3+-C5H10-COO- +H+},& \ &(2)\\
\ce{NH2-C5H10-COOH}\!\!&\!\!\ce{<=>}\!\!&\!\!\ce{NH2-C5H10-COO- +H+},& \ &(3)\\
\ce{NH3+-C5H10-COO-}\!\!&\!\!\ce{<=>}\!\!&\!\!\ce{NH2-C5H10-COO- +H+}.& \ &(4)
\end{array}$$
So there should be $4$ equilibrium constants. Which two do $K_{a1}$ and $K_{a2}$ refer to and how do I know the other two? Or is it that we can assume that $\ce{-COOH}\,$ and $\ce{-NH3+}$ dissociate independently so that the equilibrium constants of the $4$ equations are $K_{a2},K_{a1},K_{a1},K_{a2}$ respectively? I've also seen in some calculation that the zwitterion $\ce{NH3+-C5H10-COO-}$ and the original molecule $\ce{NH2-C5H10-COOH}$ are treated as the same species. But I don't quite understand why this is allowed. They clearly have different charge distributions and should be treated as distinguishable particles in entropy (or Gibbs free energy) calculations.

related https://chemistry.stackexchange.com/questions/32645/why-amino-acids-zwitterion-become-either-negative-or-positive-at-low-and-high

Well, I guess 1 and 3 are "included" into effective pKa values or neglected as very minor.

@Mithoron, I think I finally figured this out after spending the whole afternoon on it. Let me post an answer so you can check if it's correct.

Among the $4$ equilibria, only $3$ are linearly independent, because we have $(1)+(3)=(2)+(4)$. Let's then choose only three independent reactions
\begin{align}
\ce{NH3+-C5H10-COOH}&\ce{<=> NH3+-C5H10-COO- + H+},\\
\ce{NH3+-C5H10-COO-}&\ce{<=> NH2-C5H10-COO- + H+},\\
\ce{NH3+-C5H10-COO-}&\ce{<=> NH2-C5H10-COOH},
\end{align}
as the basis equations, which are obtained from $(2)$, $(4)$, and $(1)-(2)$. The equilibrium constants of these equations are $K_{a1}$, $K_{a2}$, and $K_{a2}/K_{a1}=10^{-7.24}\ll 1$. Therefore most of the neutral amino acid molecules would exist in the zwitterionic form $\,\ce{NH3+-C5H10-COO-}$. Neglecting the last equilibrium, the equilibrium constants $K_{a1}$ and $K_{a2}$ are those of the $\ce{-COOH}\,$ and $\ce{-NH3+}$ groups in the zwitterions $\,\ce{NH3+-C5H10-COO-}$, which is much more abundant than the original molecules $\,\ce{NH2-C5H10-COOH}\,$ making the latter negligible.


text:
Does alum react with zinc?
I want to remove a hardened steel bolt in a block of zinc (there is a through-hole, so I have two sides to work with). I have done this previously with aluminum and steel before, using a solution of alum to loosen the steel bolt.
Will I have as smooth a time using the alum method in the present case, or will the alum react with the zinc?

Related - but not duplicate - post here.