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Are alkenes or alkynes reduced first with diimide?
I know that $\ce{NH2NH2}/\ce{H2O2}$ is used as an reducing agent for selective reduction of alkenes and alkynes. $\ce{NH2NH2}$ reacts with $\ce{H2O2}$ to form diimide ($\ce{N2H2}$), which then undergoes a percicylic mechanism with the alkene/alkyne to give the reduced product.
My textbook had this question: what does the reaction of but-1-en-3-yne (vinylacetylene) with 1 equivalent of $\ce{N2H2}/\ce{H2O2}$ give?
Now my question is which reacts first? I know that in electrophillic reactions the alkene has greater reactivity than the alkyne but this is a pericyclic mechanism so i am not sure about it. The given answer is that alkene part reacts, not the alkyne.

Your given answer to the textbook question is seemingly correct. I said seemingly, because this reaction is not well understood to date, even though its first introduction was given at time as far as 1929 (Ref.1). The reaction was mostly hidden without getting attention in literature until to the middle of 20th century. This is mostly because the reactive species of the reaction, the unstable molecule, diimide ($\ce{HN=NH}$) was not well understood.
By 1960's, the unstable $\ce{HN=NH}$ was generated in a variety of ways such as oxydation of hydrazene (by $\ce{H2O2}$ or atmospheric $\ce{O2}$), or decarboxylation of azodecarboxylic acid, or thermal decomposition of p-toluenesulfonylhydrazene. It reacts with alkynes (e.g., diagram A below; Ref.2) and unhindered alkenes (e.g., diagram B below; Ref.3) to transfer two hydrogens in a cyclic process (diagram C below; Ref.4), resulting in a strictly cis reduction. This is the mildest way of performing such reductions. As shown in the diagram A below, one generally useful application of this reduction by diimide is it did not cleave the weak $\ce{C-I}$ bond during the reduction process. Most other reducing agents will cause competing or complete cleavage of the $\ce{C-I}$. The most important advantage in this case is in the conversion of iodoalkyne to vinyl iodide, it gives predominantly the cis-isomer. In addition, it is note worthy that this method of reduction with diimide does not lead to the cleavage of other sensitive functions such as $\ce{O-O}$ and $\ce{N-O}$ bonds.

To my understanding, the type of functions reduced by diimide are depend on the way it was generated. Besides, the rate determining step is always the rate of diimide generation. In general, trans-alkenes react more rapidly than cis-alkenes. During the mechanistic study of reaction, it was reported that diimide reduction of symmetrical multiple bonds proceeds readily. Those bonds included $\ce{C=C}$, $\ce{C#C}$,$\ce{N=N}$, and $\ce{O=O}$. On the other hand, more polar functions such as $\ce{C=O}$, $\ce{C#N}$,$\ce{C=N}$, and $\ce{O-N=O}$ are either very slowly reduced or not reduced at all (Ref.4).
The table 1 has given some uniques alkene substrates used in this process (Ref.5). In this research, 2-nitrobenzenesulfonylhydrazene has been used to generate diimide in situ.

Abstract of Ref.5: A one-pot protocol for the formation of 2-nitrobenzenesulfonylhydrazide (NBSH) from commercial reagents and subsequent alkene reduction is presented. The transformation is operationally simple and generally efficient for effecting diimide alkene reductions. A range of 16 substrates have been reduced, highlighting the unique chemoselectivity of diimide as a reduction system.

A recent publication has shown that diimide can be generated by refluxing haydrazene in air (with solvent). The reduction of diimide generated by this method gives somewhat different results. For example, a norbornene derivative gives 100% conversion in Table 1. However, similar substrate used in Ref.6 has given only 42% yield (Entry 15; Ref.6). Further, a styrene derivative (cis-1-phenylpropene) gives 61% conversion in Table 1. Yet, styrene substrate used in Ref.6 (should be better since it is a terminal alkene) has given only 16% yield (Entry 17; Ref.6). Therefore, it is very convincing that the outcome is highly depend on the metheod of generating diimide,
Based on all, the result of the reaction in the question is uncertain.

References:

J. Hanus, J. Voříšek, "Action de L'hydrazine sur Quelques Acides Incomplets $\ce{C_nH_{2n-2}O_2, \ C_nH_{2n-4}O_2,\ C_nH_{2n-6}O_2}$," Collect. Czech. Chem. Commun. 1929, 1, 223-227 (https://doi.org/10.1135/cccc19290223).
Scott E. Denmark, Shyh-Ming Yang, “Total Synthesis of (+)-Brasilenyne. Application of an Intramolecular Silicon-Assisted Cross-Coupling Reaction,”  J. Am. Chem. Soc. 2004, 126(39), 12432-12440 (https://doi.org/10.1021/ja0466863).
James D. White, Rich G. Carter, Kurt F. Sundermann, Markus Wartmann, “Total Synthesis of Epothilone B, Epothilone D, and cis- and trans-9,10-Dehydroepothilone D,” : J. Am. Chem. Soc. 2001, 123(23), 5407-5413 (https://doi.org/10.1021/jo9710137).
E. E. van Tamelen, R. S. Dewey, M. F. Lease, W. H. Pirkle, “Selectivity and Mechanism of Diimide Reductions,” J. Am. Chem. Soc. 1961, 83(20), 4302-4302 (https://doi.org/10.1021/ja01481a057).
Barrie J. Marsh, David R. Carbery, “One-Pot o-Nitrobenzenesulfonylhydrazide (NBSH) Formation−Diimide Alkene Reduction Protocol,” J. Org. Chem. 2009, 74(8), 3186-3188 (https://doi.org/10.1021/jo900237y).
Hao Chen, Jianmin Wang, Xuechuan Hong, Hai-Bing Zhou, Chune Dong, "A simple and straightforward approach toward selective C=C bond reduction by hydrazine," Canadian Journal of Chemistry 2012, 90(9), 758-761 (https://doi.org/10.1139/v2012-057). Here is the available PDF of the article through Science Gate.

Just to be clear: The final conclusion is that we cannot decide whether the alkene or alkyne reacts first? Is that right? Thx!

To my knowledge, yes. Also, both could be partially reduced.Wikipedia mentioned that order of reactivity is alkyne > alkene, but doesn't  give any example.


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What is the final product?

According to the first step,nbs performs allylic bromination,replacing one H from -Ch3 group. Now,as iodine is strongly bonded to the ring due to resonance,CH3SNa replaces Br and forms NaBr as an additional product
My final answer comes out to be A,but the given answer is None of the above. What will then be the final product?

It would be good if the NBS equivalents and conditions were given, but I think you are correct that A is the expected product. Query this with your tutor

Usually, Such MCQs provided by Indian authors don't have much details of conditions or solvents or any such extra but important details. In this question, it isn't mentioned that with addition of NBS, heat or light is provided or not, so it may or may not be a radical reaction.

I think you are meant to think it is not a radical reaction as there is no initiator mentioned, either chemical or light. Hence, the assumption would be that NBS is acting as an electrophile.  Not ideal without conditions being supplied but there you go.

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Calculating Ka using Nernst Equation
I've heard multiple times that the Nernst Equation could be used to calculate K$_a$. How would one calculate the acidity constant, K$_a$, of a specific acid, using the Nernst Equation? The equilibrium itself is a proton transfer, not an electron one, so how would one use the equation in this case?

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What makes some chemicals form steep density gradients in solution?
Caesium chloride and sucrose are both suitable for isopycnic centrifugation, and presumably not much else is. I'm unclear as to the mechanism by which they form useful density gradients. Wikipedia says,

The $\ce{CsCl}$ molecules become densely packed toward the bottom, so even
  layers of different densities form.

(No doubt $\ce{CsCl}$ is molecular, despite being the chloride of a group I element, because $\ce{Cs}$ has such low electropositivity, but I digress.) But what is it about these molecules, not to mention those of sucrose, that causes their packing to be so strongly isopycnic in solution?

Wow, I've never heard of this technique, it sounds really interesting. I thought these kinds of gradients only happened appreciably in the gas phase.

@NicolauSakerNeko It's probably better-known among biologists; biology is where I learned it.

A determination was made that coordination number of cesium ions is 8 only in diluted solutions; in concentrated solutions it is near 4. The form of cation hydrate complexes depends largely on concentration and effect of counterions. https://inis.iaea.org/search/search.aspx?orig_q=RN:27004247
If the cesium hydrates are fairly stable, their volumes, and hence densities would vary with the amount of water bonded. It would seem that after centrifugation and separation, each portion with a uniform density would be a separate phase, with a different composition. Not a liquid crystal, or a crystalline liquid. Maybe a liquid glass? It should be possible to extract some of each layer and determine the total solids, or water content.
Sucrose presents a more complicated situation, but studies indicate degrees of hydration with relaxation times of a few picoseconds to 30 or so (https://www.sciencedirect.com/science/article/abs/pii/S0008621596910198), and so this could indicate hydrates of different stability that might be forced into a certain and constant composition - another liquid glass.
Another separation from a mixture that I have seen occasionally is when a mixture of different polymers in the same solvent evaporates. If the polymers are of sufficiently low molecular weight, or high mobility, or slow enough evaporation rate, they might separate into two layers. I suppose if the mobility is not high enough, the mixed polymer will evaporate with obvious inhomogeneity, or haze.

But how is this used to create the density gradient? Are solutions of different concentrations separately created, so that when they're mixed they're like water and oil? Or does the centrifuge cause a concentration gradient, and hence a density gradient upon hydration?

We expect that waters of hydration exchange fairly rapidly with bulk water. But even if a cesium 4- or 8-hydrate is extremely stable, they would seem to be too small to be centrifuged out unless, perhaps the higher gravity forces a few together to coalesce into larger moieties which also include the anions. The Cs hydrates might be a special case, not held together by long chains, but by a structural fit not quite enough for solidity. These phases ought to have definite - and different - melting points. But the entropy of mixing prevents their existence at 1g.

Gravity will make the density of a column of a gas greater at the bottom that the top. In the centrifuge, gravity is replaced by a much, much stronger centrifugal force by rapidly spinning the sample about a vertical axis. The centrifugal forces on the solute will slowly create a density gradient in the solution, the more massive the solute the better. For this type of separation the species to be separated (say a protein or DNA) must be less dense than the solution and eventually will come to equilibrium at some point along the tube when the densities are equal.

All you've described is how the process works, not what makes certain solutions especially helpful for it.

You asked how density gradients are formed.

No, I didn't. My title was, "What makes some chemicals form steep density gradients in solution?" (emphasis added). My post's last sentence was, "But what is it about these molecules, not to mention those of sucrose, that causes their packing to be so strongly isopycnic in solution?"

All solutes form a density gradient to some extent when centrifuged so it is quite general.   Packing is not the correct word to use, the density gradient gradually varies along the centrifuge tube.  Look up 'sedimentation' in a text book for the equations to use to calculate this.

"To some extent"? The question is clearly about why a few substances do it to an especial extent. If caesium chloride & sucrose form especially gradated sediments, your answer should explain why.

As I explained they behave like all other molecules will. The density gradient increases exponentially with mass but mass may not be the only consideration, solubility in the solvent to be used is clearly important as is being inert towards the species being separated.


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Why is platinum denser than gold?
The atomic masses of gold and platinum are 196.96657 u and 195.084 u respectively, meaning that (on average) an individual gold atom is heavier than an individual platinum atom.
At the same time, the van der Waals radius of gold is 166 pm, whereas that of platinum is 175 pm. So an individual gold atom is smaller than an individual platinum atom.
Given these data, how is it possible that platinum metal can actually be denser than gold (21.5 g/cm3 vs 19.3 g/cm3)?

At the atomic scale, "size" is a concept that requires careful consideration of definitions and context.

Also check how gold and Pt crystallize. The packing (arrangement) may not be the same.

They crystallize in the same face-centered cubic packing, with a unit cell length of 407 pm for gold and 392 pm for platinum. This length goes into the density calculation as cubed, so gold uses 12% more space per atom, but is only 1% heavier (196.97 g/mol vs 195.08 g/mol).

What is important is not the atomic size, but the geometry of the metallic lattice.

To add just a bit of additional information, the tabulated atomic radii is an "average" size. So the radii of Na in each of the salts NaF, NaCl, NaBr, and NaI is not exactly the same.

Shubhang Walavalkar - I don't disagree, but the gold and platinum atoms in the metal aren't free atoms either, even though the formal charge on them would be 0.

Since OP is still in the high school, I'll try to explain it simply as possible using mathematical manipulation (hoping OP is more familiar with mathematics than chemistry).
Both gold and platinum consist of same crystal packing called body-centered cubic, which is illustrated in following diagram:

Crystal studies of both gold and platinum has revealed that the lengths of their cubic unit cells ($a$) are $\pu{407.82 pm}$ and $\pu{392.31 pm}$, respectively. If you inspect closely, you would realize that each of eight corner atoms shares with eight unit cells while each of six face-centered atoms shares with only two unit cells. Thus, total atoms per unit cell  $= 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$. Keep in mind that $a$ is depend on the atomic radius of each metal and how closely they packed against each other against some concerning forces (e.g., van der Waal's). For example, the experimental finding has been reported that the closest $\ce{Au-Au}$ separation is $\pu{288.4 pm}$ while that of $\ce{Pt-Pt}$ is $\pu{277.5 pm}$. These values imply that the metallic radius of gold is $\pu{144.2 pm}$ and that of platinum is $\pu{139 pm}$. This would cover the quote "the radii of the atoms and the bond lengths within the crystal lattice."
Based on the value of $a$ and number of atoms per unit cell for each metal, you can calculate the molar volume of each metal ($V_\mathrm{Metal}$):
$$V_\ce{Au} = a_\ce{Au}^3 \times \frac{1}{\pu{4 atoms}} \times N_A \\
= (\pu{407.82 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{4 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} = \pu{10.21 cm3mol-1}$$
and similarly,
$$V_\ce{Pt} = a_\ce{Pt}^3 \times \frac{1}{\pu{4 atoms}} \times N_A \\
= (\pu{392.31 pm})^3 \times \left(\frac{\pu{1 cm}}{\pu{10^10 pm}}\right)^3 \times \frac{1}{\pu{4 atoms}} \times \pu{6.022 \times 10^{23} atoms\:mol-1} = \pu{9.09 cm3mol-1}$$
Once we know the molar volume of each metal, we can calculate the density ($d$) by using $d=\frac{\text {molar mass}}{\text {molar volume}}$:
$$d_\ce{Au} = \frac{\pu{196.97 gmol-1}}{\pu{10.21 cm3mol-1}} = \pu{19.29 gcm-3}$$
$$d_\ce{Pt} = \frac{\pu{195.08 gmol-1}}{\pu{9.09 cm3mol-1}} = \pu{21.46 gcm-3}$$
Therefore, it is true that $d_\ce{Pt} \gt d_\ce{Au}$.

Relevant resources:

John W. Arblaster, In Selected Values of the Crystallographic Properties of Elements; ASM International: Material Park, OH, 2018 (ISBN-13: 978-1627081542).
Arnulf Maeland, Ted B. Flanagan, “Lattice Spacings of Gold–Palladium Alloys,” Canadian Journal of Physics 1964, 42(11), 2364-2366 (https://doi.org/10.1139/p64-213).
https://www.webelements.com/gold/crystal_structure.html
https://www.webelements.com/platinum/crystal_structure.html

Thank you Sir. I understood everything. But one thing I couldn't  grasp is on what basis does the length of cubic unit cell depends upon?

I can't give a direct answer to that, but one of our crystallography persons would do. I'm pure organic chemist. :-)

@ShubhangWalavalkar In a nutshell, unless you are looking at crystalline noble gases the atoms will be in some way bonded to each other. How exactly these bonds are formed in different elements is a rather complex topic but as a rule of thumb if all other things (such as number of shells) are equal a shorter bond length means stronger or more bonds. Comparing gold and platinum, the shorter $\ce{Pt-Pt}$ distance can be rationalised as a stronger or more effective bond – in an extremely simplified manner. Once you know the bond length, the length of the unit cell is simple trigonometry.

The stronger bonding in metallic platinum is  confirmed by its higher melting point (1768 C for Pt, 1064 C for Au).


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Industrial production of alkali metals
My professor says that $\ce{Li}$ and $\ce{Na}$ are produced by electrolysis, while $\ce{K, Rb}$, and $\ce{Cs}$ are made via the reaction $\ce{Na + MCl -> NaCl + M}$, where $\ce{M = K, Cs, Rb}$.
I don't understand how the reaction $\ce{Na + MCl -> NaCl + M}$ can take place, given the reduction potentials of the alkali metals ($-3.04, -2.71, -2.93, -2.98,$ and $-3.03$ V for $\ce{Li, Na, K, Rb, Cs},$ respectively). From the data, we see that sodium is the weakest reducing agent and the least electrochemically active among these alkali metals. Why then, would it prefer to displace a more electrochemically active species?

Note that reduction potentials apply only to water solutions, where ion hydration energy take part in the overall Delta G and E°.

Don't forget the very high temperatures. The listed electrode potentials in ordinary textbook tables should be restricted to solution phase and ordinary temperatures only.

Things are somewhat different at elevated temperatures. For example, look up “Potassium” in wiki and see the “Griesheimer process” for reacting KF with calcium carbide.

@Poutnik do you happen to know what phase the reaction $\ce{Na + MCl -> NaCl + M}$ happens to be? Since it is molten, the standard reduction potentials do not apply?

@DrPepper Do you see there liquid water at 25 °C ? Standard elchem. potentials includes 3 formal processes: 1/ metal atomization 2/atomic ionization 3/ ion hydration. 3/ is missing above,being replaced by ionic lattice formation.

The potentials don´t really matter. It is an equillibrium reaction, and the heavier alkalis are removed via distillation, as the have a lower Bp. Afaik, the reduction is done commercially using calcium or barium.

@DrPepper The standard potentials are only valid for separated electrodes. In a one-pot, especially in a non-aqueous slovent (melt), things can be a bit different.

One aspect to consider is that the boiling points of alkali metals go down as they get heavier.
From The Encyclopedia Britannica:
Lithium:  bp = 1342°C
Sodium:  883°C
Potassium:  759°C
Rubidium:  688°C
Cesium:  671°C
Francium:  677°C (probably this one is merely predicted)
Thus the heavier alkali metals could be volatile enough to effectively escape as vapors while sodium and the chloride salts remain largely liquid.  Thereby the equilibrium that would otherwise limit the displacement of the heavier metal is pulled forward in favor of that metal being lost to the gas phase.  The vapor would later be condensed to get the metal as a solid.


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Is it possible to measure pH without physically coming into contact with a solution?
I've been researching ocean acidification, and the way that ocean pH is typically measured is by using sensors placed throughout all of the oceans. This is costly and time-consuming, and being able to measure ocean pH from space would be much more efficient as a global map could be created in a relatively short time and updated regularly.
However, when I researched pH measurement techniques, they all involve an instrument physically touching a solution, which would obviously not be possible from space. Ocean pH maps that have been created using satellite data measure pH indirectly by measuring ocean salinity, temperature, and other conditions, and they point out areas where it is "likely" that pH levels are low enough to harm ocean life. Is is possible to directly measure pH from a large distance? Would it be necessary to do so, that is to say, is measuring ocean pH levels indirectly accurate enough for environmental monitoring?

Interesting analytical chemistry problem. Currently, it is not possible to accurately determine pH in a remote fashion. Spectroscopy is the technique of choice for remote sensing, but the H$^+$ is not that spectroscopy friendly i.e., it is a quiet ion. It will respond to radiofrequency in a powerful magnetic field but the sensitivity is quite poor in an NMR experiment by current standards. You can Google "measurement of pH by NMR". In general what appears to be pure to an organic chemist by NMR experiment, analytical chemists call it a dirty sample because there a lot of minor impurities.
A quick literature search shows that that satellite data can be used to make an only an educated guess about the ocean pH. It not accurate but good enough for a ball-park estimate.
Remote sensing of ocean pH by R. Sabia, D. Fernández-Prieto, J. Shutler, C. Donlon, P. Land and N. Reul, "Remote sensing of surface ocean pH exploiting sea surface salinity satellite observations," 2015 IEEE International Geoscience and Remote Sensing Symposium (IGARSS), Milan, 2015, pp. 106-109.

Within this context, the purpose of the ESA “Pathfinders-OA” project is to quantitatively and routinely estimate surface ocean pH by means of satellite observations in several ocean regions. Satellite Ocean Colour, Sea Surface Temperature and Sea Surface Salinity data (with an emphasis on the latter) will be exploited. A proper merging of these different datasets will allow to compute at least two independent proxies among the seawater carbonate system parameters and therefore obtain the best educated guess of the surface ocean pH.

I'd remark that “educated guess” makes it sound worse than it perhaps is. Lots of good science relies on similar methods. The important thing is that the proxy's validity can be experimentally verified independently of the result the proxy is used to obtain. In this case, the problem seems to be not that proxies are involved but that that the error bars on the final result are huge – but as long as they're known, those results can still be useful.

Right, Feynman was famous for making good educated guesses in physics. I agree that educated guesses are good and even necessary for doing science but in analytical chemistry the typically "permitted" error is less than 5%. When we develop new chemical analysis methods, we would like to have the error as low as possible. It is just a narrow perspective of analytical chemistry where guesses are not very welcome.

Yeah. In physics there are some methods with error 0.000001% and some methods with error 70% or so...


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Conversion factor for chemical to physical scale of atomic masses
Before the 1960s chemists and physicists used to have different atomic mass tables. The main difference was that the physicists assigned the $^{16}$O as the O=16 from mass spectrometry. Chemists on the other hand were using the ordinary oxygen which is a mixture of $^{16}$O, $^{17}$O and $^{18}$O as O=16. Of course, this confused everyone before carbon-12 was suggested as a compromise as both parties were not willing to budge. The % abundance of $^{16}$O is 99.76, $^{17}$O is 0.04, $^{18}$O is 0.2.
People were able to develop a conversion factor between a chemist's atomic mass and a physicist's atomic mass tables. A book by Guggenheim "Physicochemical Calculations" shows the example how the conversion factor was obtained. He doesn't explain anything clearly.
Could anyone shed some light, how the author is calculating this factor? I assume by $$\frac{O}{^{16}O}$$ the author mean how heavy is ordinary oxygen with respect to true $^{16}$O . I don't know where 1 is coming from and why there is a 2 besides O-18 isotope.

$$\frac{\ce{O}}{\ce{^16O}}=\frac{(.9976\times\ce{^16O})+(.0004\times\ce{^17O})+(.002\times\ce{^18O})}{\ce{^16O}}$$
$$=\frac{(.9976\times\ce{^16O})+(.0004\times(\ce{^16O}+1))+(.002\times(\ce{^16O}+2))}{\ce{^16O}}$$
$$=\frac{(\ce{^16O})+(.0004\times1)+(.002\times2)}{\ce{^16O}}$$
Essentially, it comes from collecting factors of $\ce{^16O}$.

Thanks for solving this mystery. One part which is still troublesome is that O-16 is $exactly$ 16 by definition (let us assume we are in 1960s). However, O-17 is not 17.0000000 and O-18 is not 18.000000. The conversion factor value is correct, but this assumption that O-17 = 16+1, is not correct even by 1960s, because mass spectrometers could tell us the exact mass of O-17, if we set O-16 as exactly 16. So why do you think Guggenheim and the rest calculated would have calculated it that way. Am I missing something?

I guess the mix between the low abundance and how close they are ($\ce{^17O}=16.99913$ and $\ce{^18O}=17.99913$ from Wikipedia) made the difference small enough for whatever they were applying it to just call them 17 and 18

This is what it seems like.


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Entropy of Irreversible Isothermal Compression
If I want to find the entropy change of universe in isothermal irreversible compression, I use 2 approaches:

I use q/T. Which gives me positive entropy change for surroundings positive and negative entropy change for system. So, using clausius inequality to say that net entropy change is positive.
I find entropy change for irreversible isothermal expansion which comes out to be positive (again using clausius inequality) and say that the compression process is just the reverse of the expansion. So, entropy change being a state function must be negative(as it is just the reverse process).

What am I doing wrong here?

Let’s see the details of your analyses of the two situations.

I used the formula for change in entropy for ideal gas dS=d qrev/T  which implies dS= (dU + Pdv)/T which implies dS = n×Cvm×dt/T + nRT×dv/V and then integrating from V1 to V2 to get deltaS for system. And using Wsystem/T (Isothermal process) for deltaS of surroundings. In the expansion, for system ln(V1/V2)  is positive hence deltaS is positive and for surroundings, Wsys is negative so, deltaS is negative. Now using clausius inequality. And then similarly for compression from V2 to V1, entropy is positive according to clausius inequality but if we take it to be reverse of expansion, it is -ve.

You are aware that the work and heat for the reversible path (from which you calculated W and Q) are different from the work and heat for the irreversible path, right (in both scenarios)?

Yes, but change in entropy is a state function so as long as I integrate from V1 to V2, it should give the same value as that of an irreversible path.

That's correct for the entropy change, but not for the total heat Q. For the irreversible path, Q is different from the Q for the reversible path.  And Q for the irreversible path is what determines the entropy change of the reservoir.

The formula for entropy is given by
S= nCvln(TF/Ti) +nRln(Vf/Vi) . I hoping u got it

There is no reversible process that can take both the system and the surroundings to the same final states (simultaneously) that they reached in the irreversible process.  So, to get their changes in entropy for the irreversible process, we need to first separate the system from the surroundings and then subject each of them separately to new reversible processes (perhaps even involving a second and third "surroundings").
For the case of the original surroundings, we know that $Q_{irrev, surr}=-Q_{irrev, syst}$.  So, from the first law applied to the irreversible process $$\Delta U_{irrev, surr}=-Q_{irrev, syst}$$The alternate reversible process for the surroundings must take it between the same two states.  This means that the internal energy for the alternate reversible process must suffer the same change as for the irreversible process.  The only way this can happen is if the heat flow for the alternate reversible process is the same as for the irreversible process.  So, the entropy change for the surroundings in the irreversible process is given by $$\Delta S=-\frac{Q_{irrev, syst}}{T}$$
Here is an example of how this plays out for a specific irreversible isothermal compression or expansion.  The example works equally well for both expansion and compression.
The initial state of the system is specified by $P_1V_1=nRT$.  The irreversible process is specified by suddenly changing the external pressure on the piston from $P_1$ to $P_2$, and then allowing the gas to re-equilibrate irreversibly to its new equilibrium state with volume $V_2$ given by $V_2=P_1V_1/P_2$.  Since, during this irreversible process, the external pressure is constant, the work done by the gas on the surroundings is given by $$W=P_2(V_2-V_1)=nRT\left(1-\frac{V_1}{V_2}\right)$$This is also equal to the heat Q transferred from the surroundings to the system:  $$Q=nRT\left(1-\frac{V_1}{V_2}\right)$$So the entropy change of the system during this irreversible process is given by:  $$\Delta S_{syst}=nR\ln{(V_2/V_1)}$$And from the analysis I presented above, the change in entropy of the surroundings during this irreversible process is given by
$$\Delta S_{surr}=-nR\left(1-\frac{V_1}{V_2}\right)$$The overall change in entropy for the system plus surroundings for this irreversible process is thus given by
$$\Delta S=nR\left[\ln{(V_2/V_1)}-\left(1-\frac{V_1}{V_2}\right)\right]$$The term in brackets in this equation is positive for all values of the ratio of the volumes, whether the ratio greater than unity or less than unity.  Therefore, the Clausius inequality ($\Delta S>0$) is satisfied for both expansion and compression in this irreversible process.

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How to test for and remove lead(II) sulfate from a sample of 1 M sulfuric acid?
If you had a clear, colourless sample of dilute sulfuric acid and suspected lead contamination, probably lead sulfate, how would you test for this? It's solubility is low in sulfuric acid solutions, and lower near freezing, so I thought cooling might produce a precipitate, but I'm not sure what level if any this would be able to detect. Raising the concentration by open boiling or distilling would, from what I can gather, produce a filterable precipitate, but that is not something I want to test.
Would there be any easy way to remove any detected contamination? By easy, I mainly mean not distilling or boiling and filtering conc. sulfuric. Ideally, I would be looking for a method that does not involve raising the concentration at all. Chemical conversion to something even less soluble? Electrolysis?
EDTA can be used to combat lead intoxication. Would there be a way of using it to remove lead compounds?

PbSO4 is slightly soluble in concentrated sulfuric acid. But by diluting it to 30% H2SO4, PbSO4 is not soluble any more. Its solubility is 43 mg per liter. Is this concentration acceptable for you ?

Thanks for your answer. I based my comment about solubility on an old paper, D. Craig & G. Vinal 1939, which shows solubility of about 6.5mg/l in 1M sulfuric and actually decreasing at higher and slightly lower concentrations. I don't really know what a reasonable limit is though. I gather it's a serious accumulative toxin, so I'm really looking to get it as low as possible with my limited apparatus.

Without knowing what equipment, chemicals, and laboratory setup you have this question is impossible to answer. All in all you need not only a method to purify the 1 molar sulfuric acid, but also a way to analyze for lead before and afterwards.

Quick and dirty method: Take a few mL of your sulfuric acid and neutralize it with ordinary household ammonia solution. Use litmus paper to get approximate neutrality: red means acidic and blue means basic. Then test the neutralized solution with lead testing strips. Hardware store often have them. If anything precipitates during the neutralization step, then that, in itself, indicates that the 1 M sulfuric acid was contaminated with whatever.

What I suggested has nothing to do with possible purification of the acid, assuming it needed it. If it did have lead it it, I would take it to a recycling facility that can handle used lead-acid battery acid and then buy new battery acid.

Interesting. I would be interested to know the chemistry of lead test strips. I'll see what I can find out.

I have various bases and indicators, so I can definitely neutralise it. The only thing I'm ruling out for now, @MaxW, as I think I stated in the question, is substantially increasing the concentration of the acid, e.g. boiling and distilling. All other suggestions would be interesting, and I will happily accept a practical answer that's beyond my budget or apparatus if it is not that.

If I neutralise the acid, maybe I could get a precipitate by boiling and cooling. Would that be a sensible approach, or should I perhaps try to convert it to a less soluble salt?

Once the acid in a small test volume is neutralized, you could add a small volume of potassium iodide solution and see if you get the brightly colored lead iodide precipitate. Potassium iodide tablets are available at pharmacies (to protect the thyroid in the event of exposure to radioactive iodine). So maybe is about the best this simple qualitative test can do, unless you really have lead contamination.

That sounds inexpensive and worth a try. Thanks. There seem to be quite sensitive test kits available for applications worth the cost. I'm not going to be drinking this, but it has a workspace exposure limit, according to UK EH40 WEL, of 15ppb 8hrs or 45ppb 15min, so the potential based just on solubility that a clear solution still has 43mg/l or even 6.5mg/l seems way too much.

Suppose you use EDTA. Now how do you decide whether using it, or not, has reduced the lead concentration to what you would accept? And what is that acceptable concentration, anyway? If there is any significant lead present, in your judgement, then how much time and money do you want to throw at this? This is very fuzzy and basically becoming opinion, to be frank.

Yes, it's fuzzy, unless someone can also suggest a test, and some have been suggested. I gather EDTA is also used for quantitative tests for transition metals, but I'm not sure whether it could be used in a sulfuric acid solution. For a test, a sample could be neutralised, but using it for removal would require keeping it in the 1M/2N acid. The questions that are very much a matter of opinion are "what's acceptable" and "is it worth it". But, though I did comment on them when asked, those are not my questions. I just want to know the best methods. I will not be asking whether I should do it.

To answer your question on how to test for Lead sulfate, I refer you to this paper, to quote:

In 1929 Fischer [1] introduced a sensitive reagent for small quantities of lead which has found interesting applications in widely diverse fields. Diphenylthiocarbazone, or dithizone as it is commonly called...

And further:

We have made no study of this reagent, except to find means of using it in our problem of measuring the solubility of lead sulfate. Dissolved in chloroform, dithizone imparts a noticeable green tint to the solution, provided the solution is not too concentrated. The lead-dithizone complex, however, has a bright cherry-red color.

Also of import;

Dithizone is known to react with metals other than lead. The precautions taken to avoid interference by other metals are discussed on page 58. These precautions, together with the spectroscopic analysis of the lead salt and the redistillation of water and acid, make it highly improbable that the results were affected by other metals. Confirmation of this is found in the analysis of solutions of known lead content.

If Pb is present, dilution appears to be the best path due to solubility.
[EDIT] Here is an theoretical path to the removal of heavy metals that does not require dilution. I am suggesting UV light therapy employing TiO2 photocatalyst to act on cold H2SO4 containing heavy metal impurities. Heavy metals are expected to precipitate.
With respect to the mechanics of the photolysis, I referred, for example, to this source. Per cited Equation [R1]:
$\ce{ hv -> e- + h+ }$
Per Equation [R3], expected reaction with an electron hole:
$\ce{H2O + h+ -> .OH + H+ }$
$\ce{H2SO4 = H+ + HSO4- }$
$\ce{H+ + e- = .H }$
where the hydrogen radical behaves as a e-/H+ pair, see Hydrometallurgy 2008: Proceedings of the Sixth International Symposium, p. 818, a commercial reductive leaching equation, to quote:

$\ce{ PbS + 2 •H = Pb + H2S }$ (5)

In the current case:
$\ce{ .H + PbSO4 -> Pb (s) + HSO4- }$
However, the associated transient sulfate radical is also created:
$\ce{ .OH + HSO4- -> H2O + .SO4- }$
Leading to the persulfate ion:
$\ce{ .SO4- + .SO4- -> S2O8(2-) }$
So, the sulfate ions is not alone. Possibly more advantageous, further  pump air/oxygen into the solution during the UV light treatment. Then per Equation [R3], expect the superoxide radical anion formation:
$\ce{O2 + e- = .O2- }$
which becomes, in acidic conditions, the $\ce{.HO2}$ radical, which may similarly react with Lead sulfate as follows:
$\ce{ .HO2 + PbSO4 -> PbO2 (s) + HSO4- }$
or, just oxidize any liberated Pb.
I would even suggest you try this process out by creating a test solution (with at least added Lead) and heat in H2SO4 to form a heavy metal impurity and then apply the UV light/TiO2 (and also added oxygen) therapy.

Seems to be marginally more obtainable than Potassium Rhodizonate and reasonably safe to store and use, so this is definitely a helpful answer to the testing part of the question. If I can test before and after, that gives me a good basis for trying out purification methods, but I still need some ideas for that.


text:
Find the maximum number of tripeptides can be synthesized by using three different amino acids
I was asked the following question in a myPAT take-home test,

Find the maximum number of tripeptides can be synthesized by using three different amino acids.

Let those amino-acids (Z) be A, B, C.
So, taking three Z's at a time, we've only one set, ZZZ. So, total 3 tripeptides.
Next, we can take two same and one distinct (eg. AAZ),
For AAZ type, we've four sets, AAB, ABA, AAC, ACA. So, total 12 tripeptides.
Finally, take 3 distinct Z, for that we get total of 6 (i.e., 3!) tripeptides.

I am unsure if I've done it in a valid way, as answer given is 18. Where am I wrong?

This question is phrased very poorly. Without reading your working, I would have answered $20 \times 19 \times 18 = 6840$. The question wording isn't your fault, but it's very hard to answer a question that doesn't have a clear meaning. Even the given answer, 18, isn't helpful unless there's some explanation of how they arrived at this magic number. 2. This is more of a maths question, not chemistry. Maybe post it there with some extra clarification.

With two amino-acids, A and B, there is already 8 tripeptides : AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB. It makes 8 tripeptides with A and C, and also 8 tripeptides with B and C. Plus the tripeptides with ABC, which are : ABC, ACB, BAC, BCA, CAB, CBA. In total : 8 + 8 + 8 + 6 = 30 tripeptides.

Okay @orthocresol, I agree that the question is poorly phrased, and I'll cast my vote to close it! But, I have some question: Are the synthesized amino acids valid? If one of the amino acid (say A) is optically active, then can BAB be two tripeptides? I'm very unsure about the possibilities I've found. Once I'll get some corrections, I'll cast my vote :)

@Maurice: I see that you've mentioned AAB and BAA, but are those two distinct? Bcoz, seeing the second one from behind looks exactly like the first one. No?

The list is aaa ,  aab ,  aac ,  aba ,  abb ,  abc ,  aca ,  acb ,  acc ,  baa ,  bab ,  bac ,  bba ,  bbb ,  bbc ,  bca ,  bcb ,  bcc ,  caa ,  cab ,  cac ,  cba ,  cbb ,  cbc ,  cca ,  ccb ,  ccc   which has 27 entries and removing 9 palindromic arrangements (except aaa, bbb, ccc) this leaves 18.

@porphyrin - I'd agree that is how the book answer is found. However that seems wrong to me. For a three chain peptide, let's say that the carboxylic acid is on the third group as written. Then bcc isn't the same as ccb. In bcc the c has a unbonded carboxylic acid, but in ccb the b has a unbonded carboxylic acid. So they are two different peptides. The correct answer should be 27.

@porphyrin: I think I misunderstood your comment. I thought palindromic entries are of type Z(Z')Z, but after reading MaxW's comment, I got to know that it was of the type ZZ(Z') and (Z')ZZ :P

(1) because polypeptides have C- and N-termini, AAX and XAA are not the same thing. (2) I can't answer the question about optical activity without knowing the context in which the question is asked. I would suggest that you can't count different enantiomers (they would be different amino acids), but I am not the author of this question. (3) if you guys can't reverse engineer the book solution, then that really suggests that there's no sensible answer. I could accept $3^3 = 27$, but I would still find it extremely arbitrary because it's based on a specific interpretation of the question.

@Rahul Verma: AAB and BAA are two distinct tripepdtides. In AAB, terminal A has free $\ce{-NH2}$ group while in BAA, terminal A has free $\ce{-CO2H}$ group, Vice versa for B.

I should also be clear that it is not your Stack Exchange question that is poorly phrased, but rather the question given to you that was poorly phrased. Sorry if that was not clear the first time. This whole debate in the comment section (which I am also part of) is not great, though. @Maurice your approach double-counts AAA, BBB, and CCC. 30 - 3 = 27 bringing you in line with everybody else.

@orthocresol: I understood what you've said, by the first comment itself. You're always vivid in your words :). It seems that, with the help of everyone, I've got the point which I was missing. By the next time, I'll try to avoid asking these kind of questions. Thanks @all!

Rahul Verma -- In mathematics this whole sort of thing falls into permutations and combinations.

@MaxW I would agree with you the C and N termini do make a difference in practice, and I think that we all agree that the question should be more exact.

There is a misconception that runs through the questions, the answer key and the comments: For tripeptides, the peptide linkage is directional (i.e. the tripeptide has an amino terminal and a carboxy terminal), so ABC is different from CBA.
To get the number of tripeptides combined from three types of amino acids, you have three choices for the first, three independent choices for the second and another three independent choices for the third position, for a total of 3 x 3 x 3 = 27 possibilities.

[OP] I am unsure if I've done it in a valid way, as answer given is 18.

The answer key is incorrect. I have no way of reading the question that 18 would be the answer, even considering the poor wording of the question.

[OP] For AAZ type, we've three set, AAB, ABA, AAC, ACA. So, total 12 tripeptides.

There are three choices for Z, two remaining choices for A, and three positions for Z (AAZ, AZA, ZAA). So, total of 3 x 2 x 3 = 18 tripeptides containing two different amino acid types of the initial set of three types.

[Orthocresol] This question is phrased very poorly. Without reading your working, I would have answered 20×19×18=6840.

This answers how many different tripeptides made from (20) standard amino acids contain no amino acid twice. The question, certainly poorly stated, says "synthesized by using three different amino acids". So your reagents are a set of three amino acids types, and you are making tripeptides from them.

[Maurice] With two amino-acids, A and B, there is already 8 tripeptides : AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB. It makes 8 tripeptides with A and C, and also 8 tripeptides with B and C. Plus the tripeptides with ABC, which are : ABC, ACB, BAC, BCA, CAB, CBA. In total : 8 + 8 + 8 + 6 = 30 tripeptides.

That is indeed a math problem, not a biochemistry problem. You can make AAA from A and B, but also from A and C, so you counted AAA twice (and BBB and CCC). So your answer overshot by 3.

[Rahul Verma] Are the synthesized amino acids valid? If one of the amino acid (say A) is optically active, then can BAB be two tripeptides?

The linkage does not introduce new chiral centers (different from e.g. combining sugars), so each sequence XYZ designates a single molecule (given that you start with enatiomerically pure amino acids).

[porphyrin] The list is aaa , aab , aac , aba , abb , abc , aca , acb , acc , baa , bab , bac , bba , bbb , bbc , bca , bcb , bcc , caa , cab , cac , cba , cbb , cbc , cca , ccb , ccc which has 27 entries and removing 9 palindromic arrangements (except aaa, bbb, ccc) this leaves 18.

The list is correct, but there is no reason to remove palindromes (9 in the list) or divide non-palindromes by two (18 in the list).
Update
I missed all the comments that came in while I was writing this. They now reflect the misconceptions and how these can be resolved; what is not resolved is the poorly worded question and its incorrect answer key - maybe someone who has the original question can take on the task of sending an erratum.

very nice summary.

@MaxW Sorry I did not mention your comment from an hour ago. I started writing the answer, then had to step out, and did not check for more comments when I came back to post it.

Doesn't bother me at all. I really like the fact that you gave the answer, and then pointed out mistakes in some of the comments. The mathematical use of permutations and combinations can get tricky. Knowing why some of the other methods didn't work helps in understanding. // LOL -- "Results! Why, man, I have gotten a lot of results. I know several thousand things that won't work."  Thomas A. Edison

Ugh. I missed it. Re:"9 x 9 x 9 = 27 possibilities." It should be $ 3 \times 3 \times 3 = \pu{27 possibilities}$.


text:
Why change on pressure affects the boiling point much more than the freezing point?
I do see the mathematical solution given by the classius clapyeron equation but is there any real life and intuitive expalanation?

Because there's much greater difference in volume.

Don't make it complicated. Gases are very compressible, fluids are not.

Forget compressible. Gases are big; liquids and solids are not.


text:
Does voltage or current matter most when setting up an efficient water electrolyzer?
I am setting up a homemade water electrolyzer for producing hydrogen gas, which will solely be used by myself for experimental purposes, and have heard the wattage used has a large part to play in determining efficiency. However, seeing as watts are determined by Voltage * Current, would high-voltage-low-current or low-voltage-high-current lead to the production of more gas?

The amount of Hydrogen gas collected at the electrode depends only upon the current, in A. The voltage has no direct effect. It may be used to increase the current, that is all.

You can't necessarily take a setup that runs at 10V @ 2A and get it to run at 2V @ 10A. Things such as electrode size, distance between electrodes, electrolyte, and stirring have an effect on the operation of a real cell. // I'll point out that the cell will effectively have an internal resistance. The more current, the more power is lost to heat.

@MaxW I would strike "not necessarily " and say definitely not.

The cell run at 10V / 2A will definitely produce much less hydrogen. Exactly a fifth, if the numbers were true. The amount of gas produced is directly proportional to the current. Current x time / Faraday´s constant / 2 gives the molar amount of hydrogen. All voltage applied that is more than the nominal voltage necessary to run the electrolysis produces just heat.
Of course at two volts, the electrolysis will in reality hardly run at all. A few milliamperes with the same cell. Highly efficient, but extremely slow. There is some optimum in between. Where it is depends your special setup and your own preferences.
If you want as much hydrogen generation capacity as possible, you can choose between buying more (or larger) cells, or turning a lot of electricity into heat in the long run.

I am using a 12V 10AH LiFePO4 battery in conjunction with a 100W monocrystalline solar cell as the energy source; I apologize for neglecting to mention that. Do you think you could explain what calculations might be necessary to find such an optimum range as you mentioned?

The math is not exactly difficult. The unloaded cell voltage is around 2V, the rest of the cell resistance is more or less ohmic. Measure the current at e.g. 5V, and you have all numbers that are necessary for a rough approximation.

There are few things to consider:

The rate of hydrogen production is proportional to the passing current.
For voltage below cca 1.8-2.0 V, the current is very low.
Above that value, the current quickly grows ( in first like exponentially, later more linearly ).
The current for high enough voltage depends on the electrolyzer geometry, electrolyte conductivity and temperature. For electrodes with larger areas, closer each other, the current for given voltage is higher. The same for higher conductivity or temperature.
Set electrode current density low enough to prolong their life.
The electrolysis efficiency decreases  for higher voltages, because part of the voltage is wasted on warming up the solution.
For production, it is good to manage an electronic high power constant current source, the best with the electronic max voltage control, as keeping the  given reasonable current  is more important than a needed particular voltage, that would vary. The alternative is the obvious, more common voltage source with the electronic max current control.

I object to your point 7: If the current drops noticeably for a given voltage, that means something is changing in your cell, and most likely not for the better. In which case increasing the voltage is the last thing you should do.

I mean an  intelligent current source with the voltage check, similarly as voltage sources have the current check. I had in mind the voltage drifts depending on temperature or concentration.

Depends. A temperature increase should lower the electrolyte resistance, hence the voltage would drop for a constant current, which is good. If that initial assumption is true. If not, you blow up your cell. If the electrolyte concentration changes, something is obviously wrong. It´s not directly obvious to me wether keeping the current and thereby hydrogen generation constant would help or make the situation worse. So yea, you need some "intelligent" control system. Which should also check the temperature, perhaps conductivity of inflowing/outflowing electrolyte, etc.

You may did not read the answer update...

@Poutnik An excellent point. I am wondering now if using an Arduino board with a thermal resistor of some sort may be an adequate temperature control system as they can be used for, among other things, the regulation of both temperature and voltage. The part about inflowing/outflowing electrolyte may not be necessary, however; my former chemistry teacher recommended Na2SO4, stating that once introduced, it may not need to be replenished due to both the sodium and sulfate ions being good electrical conductors.


text:
Nucleophile in a SN2 reaction
In a $\mathrm{S_N2}$ reaction, a good nucleophile is considered to be willing to donate electrons. So how do we generalize this fact across the periodic table? Does that mean the more electronegative an atom is, the better the nucleophile?

One should exercise caution in making generalisations across the periodic table. Although related to periodic trends, the extent of an element's nucleophilic character is more complex to determine than using its position in the periodic table. Also keep in mind that as well as atoms, there are other species that are nucleophilic (eg. NH3) and their nucleophilic character is not only based on the properties of the electron donor atom (nitrogen in NH3) but also on the nature of bonding in the species.

Consider how good a nucleophile F- is

As electronegativity increases the holding capacity of electron by the atom increases. So it is less likely to donate electrons. So we can generalise that access a period in periodic table nucleophilicity decreases.


text:
What is the relationship between dipole moment and solubility?
I thought a large dipole moment meant the substance was polar, so I'm confused by the following statement (from Wikipedia) that "With a dipole moment of 3.92 D, acetonitrile dissolves a wide range of ionic and nonpolar compounds...." Why would it dissolve many nonpolar compounds?

Just because something has a dipole doesn't mean that it can't interact with other, less polar, compounds. The dipole might help get some ionic compounds into solution but acetonitrile has plenty of scope to mix with non polar compounds via dispersion forces which are still there in acetonitrile.

It's all about the energetics (assuming an isolated system--happy to elaborate in the comments): as acetonitrile has a large dipole moment, it can readily stabilize the individual ions. See this discussion of solvation energy.
The more subtle case is in the case of nonpolar compounds. In this case, the large dipole can create an induced dipole in the nonpolar molecule. Then the dipole-induced dipole attraction would be larger in magnitude than the spontaneous dipole-spontaneous dipole interactions present in nonpolar-nonpolar interactions (which are also still present as small fluctuations in the polar-nonpolar case). See this discussion of how charges/dipoles/multipoles induce dipoles and then subsequently attractively interact with said dipoles.


text:
Cis, trans diene reacting with trans dienophile in D-A reaction

I wonder if I predicted the products correctly, because it seems to be messier than I thought - there's a pair of diastereomers... Also, I feel that the endo/exo rule does not apply here. Am I correct in my understanding? I have never seen a reaction like the one I drawn mentioned anywhere before. Hopefully someone can help me out.Thanks!
Note, there's an example in Clayden reacting cis, trans diene with DEAD, but since there's no stereocenter at nitrogen we end up just getting trans product.

With fumaronitrile, when one cyano group is endo the other is exo. When the E end of the diene is endo in the TS, the Z end is exo. Then there is also Z-endo and E-exo. So two racemic diastereomers may be expected.

Macrophage: I thought you did a fine job in analyzing the Diels-Alder reaction of the E,Z-pentadiene with fumaronitrile. To complement my comment above, I offer an illustration of the transition states of the two reactions leading to their respective products. Because fumaronitrile may add with equal facility (mirror image transition states) to either face of the achiral diene, both 2 and 4 are necessarily racemic. "Racemic" transition states represented by 1 are E-endo/Z-exo that afford racemic bis-nitrile 2. Alternatively, "racemic" transition state 3 is E-exo/Z-endo, which leads to bis-nitrile 4 as a racemate.


text:
Two monoprotic acids vs. one diprotic acid
My intuition says that the titration curve for titrating a $1:1$ mixture of the  monoprotic acids $\ce{HA_1}$ and $\ce{HA_2}$ (having ionization constants $K_a = x$ and $K_a = y$ respectively) is identical to the titration curve obtained by titrating a diprotic acid $\ce{H2X}$, with successive ionization constants $Ka_1 = x$ and $Ka_2 = y$.
I feel that this should be correct because the "concentration" and "strength" of the acids are the same, but I am not sure how to prove if it is correct.

What is diuretic acid? You meant diprotic acid.

Your intuition is right, because a potentiometric sensor (a $\mathrm{pH}$ probe in your case) does not know the source of hydrogren ions, it only senses the presence of hydrogen ions. The same is true for any other detection mode. All we care is $\ce{H+ + OH- -> H2O}$.
It can come from two monoprotic acids or it can come from a diprotic acid. In order to make your wish true, $\mathrm{p}K_\mathrm{a}$ must be less than $\mathrm{p}K_\mathrm{b}$. They cannot vary randomly and they cannot be equal.
However intuition is not enough to prove a point. You have to pick numbers and simulate. Here is useful online tool to calculate titration curve and see how well they match.
pH Titration Curve Generator
Choose to add $\ce{HA}$ or $\ce{H2A}$ and $\ce{NaOH}$ as a base.


text:
How to measure the speed in a catalized reaction?
What is cheaper?

Set up a constant time of reaction and measure the concentration of the products or;
Set up a constant quantity of reactants and, with an indicator, measure the time until the reaction stops.

If you know another protocol, Can you explain it, please?

It depends on scenario.
If time is cheap, while measurement is expensive ( e.g.high fixed price for each value ), then constant time method is probably cheaper.
If time is expensive, while measurement cheap ( e.g. continuos measurement ), the constant products method may be cheaper.


text:
Will calcium concentrations be reduced when water is boiled at lower atmospheric pressure?
It's well documented that one can reduce calcium concentrations in water by boiling the water. This is caused by CO2 being driven out of the water and the calcium precipitating out of solution. Ok, but, would calcium drop out of solution if the water was to reach its boiling point at a lower pressure atmosphere? The goal being energy-savings.
Anyone with some insight on this or ... wielding a low-pressure apparatus and a TDS meter?

Could you explain the context or background of this problem?

"This is caused by CO2 being driven out..." Would CO2 also evolve at low pressure? What happens when you open a bottle of soda water? You have enough information to answer your question.

It will be less efficient due better CO2 solubility at low temperatures, what would require more boiling to keep partial CO2 pressure low enough. It may be even more energy demanding, unless T is kept below Tenv and part of heat comes from the surrounding. Additionally,  boiling at lower temperature requires more energy.

@Poutnik Thanks for your comment! So CO2 solubility is temp-dependent so I suppose I won't get around boiling the water.

Apologies for the typo-ridden previous comment.
Additional background for this question: My tap water is very hard, with just over 100 ppm of calcium dissolved in it. This is an issue for aquarists with fish that require soft water. For a moment, I was considering using an RO filter for the water but then an friend/civil engineer recently suggested I could reduce the calcium concentration in my water as much as 50% if I boil the water. That got me thinking if there was a way to have calcium precipitate with less energy input.

Malte, it makes more sense now. No you will have to provide enough heat to decompose calcium bicarbonate. Generating vacuum will also cost energy. Is there a possibility of a solar still? https://en.wikipedia.org/wiki/Solar_still


text:
Algebraic treatment of equilibrium
This is from Laitinen and Harris, "Chemical Analysis, An Advanced Text & Reference" which describes a simple acid-base equilibrium. They derive a relation in the end directly. I haven't seen this expression in many books which teach solution equilibria. The authors cite an article from the Journal of Chemical Education for the expressions (4 & 5). I didn't find article very useful. How should we derive the equations 4 and 5 in a better way than the author's explanation in italics?
The authors start from very simple equations:

$\ce{HA + H2O <=> H3O^+ + A-}\tag{1}$
Conventionally,
$\ce{HA <=> H+ + A^-}\tag{2}$
and the ionization of water,
  $\ce{H2O <=> H+ + OH-}\tag{3}$
If $C_{HA}$ and $C_{A^-}$ are the analytical concentrations of $\ce{HA}$ and $\ce{A^-}$, and if $\ce{[HA]}$ and $\ce{[A^-]}$ are the equilibrium concentrations, the relations...
$\ce{[HA]} = C_{HA} - \ce{([H^+] - [OH^-])}\tag{4}$
$\ce{[A^-]} = C_{A^-} + \ce{([H^+] - [OH^-])}\tag{5}$

The authors explain ...
"result because the analytical concentration of HA is diminished by the amount of hydrogen ion produced in Reaction (2), which in turn is the total hydrogen ion concentration minus the hydroxyl ion concentration."

..in reaction (3) is wrong imho, it should be 1 or 2.

There is only one source of hydroxyl ions, ionization of water (Eq. 3 in the OP). Ionized water contributes equal amounts of hydroxyl and hydronium ion. Similarly, dissociation of acid $\ce{HA}$ results in equal amounts of hydronium and of conjugate base $\ce{A-}$. Unless a salt of the conjugate base is explicitly added then the acid is the only source of $\ce{A-}$. In absence of salt the total hydronium concentration in solution is therefore
$$\ce{[H+]}=\ce{[OH-]}+\ce{[A-]}$$
Note that this equation reflects the charge balance condition.
Solving for $\ce{[A-]}$
$$\ce{[A-]} = \ce{[H+]}-\ce{[OH-]}\tag{1}$$
We also know that the original amount of acid equals the sum of undissociated acid and conjugate base:
$$C_\ce{HA}=\ce{[HA]}+\ce{[A-]}$$
Note that this equation reflects the mass balance condition.
Solving for $\ce{[HA]}$
$$\ce{[HA]}=C_\ce{HA}-\ce{[A-]} \tag{2}$$
Substituting Eq. (1) into Eq. (2) we obtain
$$\ce{[HA]}=C_\ce{HA}-(\ce{[H+]}-\ce{[OH-]}) \tag{3}$$
The derivation that lead to this expression is not general. It was assumed that no conjugate salt was added to the solution ($C_\ce{A-}=0$). If a completely dissociating salt $\ce{MA}$ is added at concentration $C_\ce{A-}$ then it is necessary to modify the mass balance expression:
$$C_\ce{HA}+C_\ce{A-}=\ce{[HA]}+\ce{[A-]}$$
and the charge balance expression:
$$\ce{[H+]}+\ce{[M+]}=\ce{[OH-]}+\ce{[A-]}$$
or since $\ce{[M+]}=C_\ce{A-}$
$$\ce{[H+]}+C_\ce{A-} =\ce{[OH-]}+\ce{[A-]}$$
Solving for $\ce{[A-]}$
$$\ce{[A-]} = C_\ce{A-} + \ce{[H+]}-\ce{[OH-]}$$
Combining the two balance expressions leads once again to Eq. (3):
$$\ce{[HA]}=C_\ce{HA}-(\ce{[H+]}-\ce{[OH-]})$$
By the way, the quote you cite should read:
"result because the analytical concentration of HA is diminished by the amount of hydrogen ion produced in Reaction (2), which in turn is the total hydrogen ion concentration minus the hydroxyl ion concentration."
It might be that "typo" ("3" written instead of "2") that led to the confusion.

I'm confused. Shouldn't the second charge expression be $$\ce{[H+] + [M+]} = \ce{[OH-]} + C_\ce{A-}$$

@MaxW Maybe I should edit the equation as written but it is correct. The key point is that $[M^+] = C_{A-}$

I can't figure out $\ce{[H+] +[M+] =[H+]}+C_\ce{A-}$ since $\ce{[H+]}$ is on both sides.

@MaxW I edited in an attempt to clarify

Thanks for detailed response. Why do we need to invoke another dummy variable [M$^+$] to derive  equation (5) in the original post.

There might be other ways to think about this that lead to a derivation without dummy variables (as in the OP). But if you understood such a derivation you would probably have the required background, ie you'd know the answer anyway. The point of the question seemed largely to come up with a clearer (more explicit) derivation.

@M.Farooq The analytical concentration of conjugate base is referred to as $c_{\ce{A−}}$, but for the purpose of keeping track of charge balance you have to consider both ions formed during conjugate base dissociation (A− and counterion), resulting in the "dummy" variable. In a "real" problem (creating a buffer say) you would specify the salt of the conjugate base that was used and things would be more explicit. In the OP it is ignored, which imho creates confusion.

@M.Farooq Note I made a small amedment to the answer which I think catches an error in the textbook (or in your citation)

It must be my own typo.

@M.Farooq Repeating myself somewhat, the explanation in the text is straightforward (my opinion of course) if you have a solution prepared with only acid (no salt added). My intuition fails me when mixing salt and acid, but then if I strain my noggin a little I might reach clarity. And sometimes it's ok to let the math show you the way and not worry too much about intuition.

@BuckThorn"or since $\ce{[M+]}=C_\ce{A-}$" Is it mean :
$$\ce{[M+]}=C_\ce{A-} = [\ce{MA}]_0 $$

@AdnanAL-Amleh Yes, this is correct.


text:
Something is wrong with converting structure to name in ChemDraw 18.0
I am using the tool "convert structure to name" to name 2 organic compounds below. But, it seems the names are not wright. Can someone explain to me whether I am wrong or the software made a mistake?

ChemDraw's generated names are good enough for most organic chemists, but are not usually in accordance with latest IUPAC guidance. But that is not a chemistry issue. It is a software issue, which cannot really be thoroughly answered here, because we do not have insight into the algorithm they use. You may have better luck finding more info on their forums: http://forums.cambridgesoft.com/

Can you explain why you think they are wrong?

I think they are wrong because when choosing the parent of an alkene, we chose the longest chain that includes the pi bond so that the parents of these should be octadiene and heptene, respectively.

The names given by ChemOffice are not wrong since they describe the correct structures and are in accordance with IUPAC nomenclature rules. Note that a compound may be named correctly in two or more ways in accordance with the several methods recommended by IUPAC. Also note that the IUPAC recommendations have changed several times and the current (2013) recommendations are not included in ChemOffice yet.
The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) includes the definition of a preferred IUPAC name (PIN), which is the name preferred among two or more names generated from two or more IUPAC recommendations including also the many retained names.
In particular, the name vinyl for $\ce{CH2=CH-{}}$ is retained for general nomenclature only. The systematic name ethenyl is the preferred IUPAC name.
Also in preferred IUPAC names, stereodescriptors are preceded by a numerical or letter locant to describe the position of the stereogenic unit when such locants are present.
Therefore, the preferred IUPAC names for the compounds that are given in the question are (2Z)-6-ethenyl-3,4-dimethylnon-2-ene and 4-ethenyloctane, respectively.

I am sorry because I did not make clear why I found they are wrong. Actually, I want to mention about the parent. I think we have to choose the longest chain that includes the pi bond so that the parents of these should be octadiene and heptene, respectively.

@HungNguyen No, see How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

I got it, thank you very much. However, I wonder why some popular books, like David Klein's Organic Chemistry (2017), still use the previous rules?

(2E)-6-ethenyl-3,4-dimethylnon-2-ene
using ACD chem

can I know where is Structure to Name option on ChemDraw I have chemDraw 19.0 , but I can't find that option,


text:
Biphenyl esterification

I was wondering that at optimum temperature can 6,6'-bis(hydroxymethyl)[1,1'-biphenyl]-2,2'-dicarboxylic acid show esterification reaction.
I know that the two phenyl groups aren't in the same plane so is esterification reaction still possible?

What conditions do you intend to use for the esterification? Acid catalysed or with a coupling agent?

There's a base present ie NaOH and the temperature is of 100°C

You are not going to get esterification with NaOH present!

these type of things are possible,

esterification is possible, check the reference.

Can the carboxylic acid and alcohol react to give an ester at 100°C of temperature??

is this for research purpose ? or just exercise?

It's for my knowledge and a part of my excercise


text:
Reduction of nitrobenzene in alkaline medium
My teacher listed out the following set of reactions under the aforementioned topic:
He explained the formation of azoxybenzene as follows.
Reduction of nitrobenzene produces 2 intermediates:

nitrosobenzene
phenylhydroxylamine

He then went on to say that condensation between nitrosobenzene and phenylhydroxylamine yields azoxybenzene. This takes place by the nucleophilic attack of the lone pair on the nitrogen atom of phenylhydroxylamine (as opposed to that of the lone pair on the oxygen atom of the same) on the nitrosyl nitrogen. He reasoned that nitrogen, being less electronegative, is more nucleophilic than oxygen (see figure). While I do understand that the explanation is generally true, I am not sure that it is the case here as the lone pair on the nitrogen atom is in conjugation with the benzene ring here and is thus less available on nitrogen. Thus, shouldn't it be the lone pair of the oxygen atom that takes part in the reaction instead?
He followed this discussion up with a question where we had to figure out the product of a similar condensation reaction in an alkaline medium between p-Nitrosotoluene and 4-Nitrophenylhydroxylamine which basically reduces to deciding which benzene ring will be directly attached to the nitro part of the azoxy moeity. He said that it is the ring with the nitro substituent that finally gets attached directly to the
nitro part of the azoxy moeity as the nitro substituent stabilises the negative charge on the oxygen of the nitro part of the azoxy moeity better (see figure).  But I can't see how that should come to be because the internal attack of the lone pair on the other nitrogen atom which establishes the N-N double bond of the azoxy moeity and removes the -OH group in the intermediate should be better available in case of the benzene ring with the methyl substituent in the first place. So, shouldn't it be the other way round?


text:
Color change in indicators
Methyl orange indicator changes color to red and yellow with an acid and base respectively.
I am unable to understand why this color change takes place.
Searched on Google about the same with litmus paper and some answers said that litmus paper contains an acid an a base with color red and blue respectively.
But what about phenolphthalein it does not show any color change with acid.

They have different structures in different media, phenolphthalein exists in quinonoid form in basic media (which is pink), in phenolic form in acidic media, which is colorless. Methyl orange exists in quinonoid form in acidic media (red), whereas in basic media remains in original form (yellow).

Red mixed with yellow looks orange, while pink mixed with colorless looks pink-ish.

see section on Methyl orange on the following webpage https://www.chemguide.co.uk/physical/acidbaseeqia/indicators.html

Phenolphtalein is formed of three phenyl rings which are attached to the same Carbon atom. In the basic form, this central Carbon atom is part of a double bond joining it to one of these phenyl rings, which has a quinonic structure. The delocalization of the pi electrons is possible over all rings simultaneously. The rings are coplanar, and define is a huge domain. This is a huge "box" for the electrons, and the electronic transitions are low energy, that is in the visible.
In acidic solution, the central Carbon atom is related to four simple bonds. The phenyl rings are not coplanar. Their pi systems are not even delocalized over the whole molecule. The domain available to the pi electrons is limited to the phenyl rings, which are about three times smaller than in basic solutions. So the electronic transitions are high energy, in the UV region. And the phenolphtalein is colorless.

That's basically correct, but at more extreme pH conditions phenolphthalein can undergo still more changes.  Check it out.


text:
Why isn't cyclopropane considered unsaturated?
Looking up "cycloalkanes" on wikipedia (link), I found this:

In organic chemistry, the cycloalkanes (also called naphthenes, but distinct from naphthalene) are monocyclic saturated hydrocarbons.

Of course, that implies that cyclopropane is saturated. Later I looked up "saturated and unsaturated compounds" on wikipedia (link). Here's the definition they mentioned:

In chemistry, a saturated compound is a chemical compound (or ion) that resists the addition reactions, such as hydrogenation, oxidative addition, and binding of a Lewis base.

Now the definition mentions that a saturated compound "resists addition reactions". I am aware that cyclopropane does undergo addition reactions, as mentioned here (link):

So finally, to me it seems these two wikipedia pages are contradicting each other. My question is - why is cyclopropane considered saturated if the above definitions are correct?

Saturated simply means that all the bonds are single bonds. Unsaturated would mean that the compound contains at least one double or triple bond. // The quirk here is that the molecular formula is $\ce{C4H8}$ for both cyclopropane (saturated) and propene(unsaturated).

@MaxW So does that mean the wikipedia definition is wrong?

Addition is meant without breaking the structure.

@Poutnik Hmm, I'm still a little skeptical about this. It still is an addition reaction, isn't it? (of course, it involves breaking a bond, but then all addition reactions are required to involve that)

Breaking a bond is not the same as breaking a structure.

With your approach, even ethane would be unsaturated, as you can break C-C bond to make from it 2 CH4 by H2 addition.

@Poutnik True, but ideally ethane does "resist" such addition reactions, so it would be called saturated as per wikipedia's definition.

With high enough pressure, it would not. The key is not breaking C-C bonds by addition. It does not matter if it is in alkane or cycloalkane.

"Saturated" / "Unsaturated" can refer to a bond (i.e. an alkene or alkyne), or to a whole compound. The latter is far less specific.

@Poutnik Alright then, that makes sense. I suppose the wikipedia definition is a little ambiguous as it doesn't really mention the conditions under which addition should take place.

If you are interested in the bonding in cyclopropane and how this leads to its behaving like an unsaturated molecule, then see this earlier Q&A, "Why does cyclopropane react with bromine?".

In short, there are two definitions with separate purpose.
IUPAC defines alkanes as

"Acyclic branched or unbranched hydrocarbons having the general formula $\ce{C_nH_{2n+2}}$, and therefore consisting entirely of hydrogen atoms and saturated carbon atoms."

and notes notes about cycloalkanes, contrasting e.g., to unsaturated cycloalkenes:

"Saturated monocyclic hydrocarbons (with or without side chains), e.g. cyclobutane. Unsaturated monocyclic hydrocarbons having one endocyclic double or one triple bond are called cycloalkenes and cycloalkynes, respectively. Those having more than one such multiple bond are cycloalkadienes, cycloalkatrienes, etc. The inclusive terms for any cyclic hydrocarbons having any number of such multiple bonds are cyclic olefins or cyclic acetylenes."

Thus, as in the earlier comment by @MaxW, cyclopropane equally is a saturated alkane because each carbon atom is connected with the next carbon atom only by a $\ce{C-C}$ single bond.  Or, as MSU condenses,

"Alkanes and cycloalkanes are termed saturated, because they incorporate the maximum number of hydrogens possible without breaking any carbon-carbon bonds."

This set of definitions is not the one serving to determine organic structures from spectra, e.g. in Field's training book, where

"[...] the molecular formular is reduced to $\ce{C_nH_m}$ and the degree of unsaturation is given by $$\mbox{Degree of Unsaturation} = n - \frac{m}{2} + 1$$ The degree of unsaturation indicates the number of $\pi$ bonds or rings that the compound contains."

(4th edition, p. 4; emphasis added by mine.)

Thanks, I see where I went wrong (the quote from MSU clears it).

Saturation has its IUPAC definition, but it also has its historical roots.
You can add hydrogen to some hydrocarbon up to some point where it is considered "saturated".
If you start from benzene, you can add 3 molecules of hydrogen and consider the resulting substance saturated, but if you try harder, you can add 2 more hydrogen atoms. What is saturated, then? (This observation predates the knowledge of the benzene structure.)

I don't think it has something to do with cyclopropane in particular. Indeed, cycloalkanes are naturally unsaturated, but they are associated with alkanes, which are saturated. The fact that the article specifically mentioned "monocyclic" offered enough details about the degree of unsaturation. However, just for the fact that they are the result of cyclization of saturated compounds, many sources assign cycloalkanes a somewhat saturated character.
TL;DR Cycloalkanes are the saturated hydrocarbons from the family of monocyclic compounds (which may contain other unsaturation-causing factors)

I think you might be confusing the hydrogen deficiency index (for which a monocyclic alkane would have a value of one) with the degree of unsaturation. The terms are often used interchangeably in NMR calculations; however, I make sure to only use the term hydrogen deficiency index to avoid confusion with the term unsaturation, which is in fact different.


text:
What is the magnitude of the current in an electrolyte?
if a battery is connected to an electrolytic cell, what is the magnitude of the current in the electrolyte compared to the the magnitude of the current in the wire? This doesn't make sense to me because there are cations flowing in one direction and anions flowing in the other in electrolytes, so is the current twice the magnitude of the current in the wire? does ohm's law still hold? Is current not uniform in the circuit? I hope somebody can help. Thanks!
Edit: Sorry for asking an off-topic question. Thank you so much for the answers and comments!

For every 100 electrons transported, 100 charges are transported via ions (e. g. 50 sodium and 50 chloride ions, or just 100 chloride ions, or just 50 magnesium ions, or any other combination).

If you label anode and cathode species in a chemical equation that includes ion transfer, not only does the entire equation have to be charge-balanced, but the sum of the charges at the anode has to be the same for reactants and products (same goes for cathode species, but this is ensured by the two other charge-balance conditions).

Electrolytes are non-ohmic, but your reasoning is entirely wrong..

Ok, but what is the magnitude of the current? Is it the same in the metal wire and in the electrolyte? By the way, thank you for your comments. They helped me get a better understanding of the subject.

The total current is the same in all parts of the circuit. It could be in the electric circuit made of a wire joining the anode and the cathode. It could also be the movement of the ions in the electrolyte, well described by Karsten Theis. And these ions may be negative charges travelling though the electrolyte, or positive charges traveling in the opposite sense, or both simultaneously.

My understanding is that the charge balance is accomplished with time, and if not the reaction grinds to a halt.

By Kirchoff's Current Law ("what goes in must come out") the current flowing in total in all the branches of a circuit must be equal at all points. Hence if A amperes are flowing in the wire, as no parallel branches are present we conclude that A amperes must also be flowing in the electrolyte.

For each 100 electrons coming to the cathode, 100 electrons are passed at the cathode to coming cations or neutral molecules.
Similarly, for each 100 electrons leaving the anode, 100 electrons are taken by the anode from coming anions or neutral molecules.
It is similiar within the electrolyte itself. For a given volume $\mathrm{d}V$, there is a steady state $\frac{\mathrm{d}q}{\mathrm{d}t}=0$. There are only small, initial and temporary deviations to manage the ion distribution as the reaction to the externally forced potential gradient.
If the overall current was not consistently the same across all the circuit, including the electrolyte, there would happen accumulation of the charge somewhere, leading to enormous electrostatic forces and coulombic explosion.
For an ilustration, the charge passed in 20 s at the current 1 A is comparable to the charge of GC+ thunderstorm ligtning of voltage up to 1 GV, long up to 10 km.


text:
How is 1,4,6-trichlorodiamantane optically inactive?

How this compound is optically inactive I can't analyse any element of symmetry in this compound.

The picture quality is low, but maybe one of the C-C bonds is drawn incorrectly, too.  Did you build a molecular model with a model kit?  Try, and give the flow scheme next to https://en.wikipedia.org/wiki/Molecular_symmetry#Point_groups_and_permutation-inversion_groups a chance.

Where did you invent such a monstrosity?

It could be an adamantane derivative. I may be mistaken, but it should then have 3 mirror planes. I strongly advise building a model, either in a computer or with a kit.

The sketch is wrong. https://de.wikipedia.org/wiki/Diamantan

In a correct sketch of this molecule, you immediately see that all chloride atoms lie in the mirror plane that is spanned by the two CC bonds that are furthest apart.

Just out of curiousity, where does that picture come from?

@Karl tis' certainly a JEE practice questionXD

@HappyFeetMe Potentially inspired on claim #4 of this patent, and the NMR spectra included https://patents.google.com/patent/EP1760057A1/en

M. C. Escher would have been proud of this picture.

Yes I am pretty sure this is a question from one of the many question banks available for the Indian competitive exam, the JEE. Unfortunately, in my experience, the authors for such books more often than not make erroneous questions in their attempt to make them complicated

From this indeed Escheresque' screen photo,  I assume the following molecular structure:

which reveals to be and example of

 point group $C_s$ or a meso-compound

Note:  There is a nice interactive symmetry gallery at Otterbein.edu, accessible by a web browser.


text:
Mechanism of alkene dihydroxylation with potassium permanganate
I have scavenged the internet for different mechanisms for permanganate-mediated alkene oxidation and have found satisfactory mechanisms for oxidative cleavage and dihydroxylation under basic conditions.
However,I found out that $\ce{KMnO4}$ can still dihydroxylate alkenes under acidic conditions, when it is cold and dilute. I am aware that it is not efficient as it is prone to further oxidation, but the mechanism still makes me curious. I know the overall formula:
$$\ce{5alkene + 2MnO4- + 2H2O + 6H+ -> 5diol/glycol + Mn^2+}$$
I can't really understand this, does the manganese oxidise 5 alkenes at the same time? Does it get oxidised 5 times, with a different byproduct for each one? I think that the manganese-bearing compound is $\ce{MnO2}$ at first, then $\ce{Mn^2+}$, but I'm not really sure. If anyone can shed some light on the mechanism, it'll be a great help.
I link below my best attempt at the mechanism. The byproducts are incorrect as it's supposed to be $\ce{MnO2}$ or $\ce{Mn^2+}$
Here are some papers I've found that helped me, maybe someone can get more use out of them than me:

https://www.semanticscholar.org/paper/Oxidation-by-permanganate%3A-synthetic-and-aspects-Dash-Patel/fe91f5161ca1ab481ae111f1abc82d37d336f021
https://www.researchgate.net/publication/271107223_Permanganate_Oxidation_mechanisms_of_Alkylarenes

(1) I'm not sure that the stoichiometry of the reaction is so well-defined; I wouldn't be surprised if there are multiple Mn-containing products in this reaction, so I'm not sure how much we should read into it. (2) I don't know enough about this specific reaction to write an answer (indeed I don't know if the full mechanism has ever been studied), but I would hazard a guess that there are other important redox steps purely involving Mn species (i.e. dis- or comproportionation). You can search the internet for alcohol oxidations with TPAP. The mechanisms of those have been studied.

Thanks for the input! Yes, I have considered disproportionation (never heard of comproportionation before today though), but couldn't find anything regarding it. I have some papers that I've found that might be helpful, so maybe I'll add them to my post.

Mechanism has been explored: https://pubs.acs.org/doi/abs/10.1021/ja01568a042?cf_chl_jschl_tk=04306b3b0c96ca99fcc28f5d402ff4ca7a1d00b7-1586703493-0-AUkzm2A00AzYtV6yXjjmAusmaudZZINA8gHign3FLbuvVW2_rOOuD-B8FMN2OLgmKV_tlhfpTffaSmLVhW5Z08jbuOfT5sr3SLNZT2UWE8BTmeQuzF5o62AKGt33a0RmwqVYqsIb6hs11454ZypXm0QFdiC_6H5ehOBFvS67jqda80dprtSUPS5mr_CDh3Qh2LtxH_5ivREePqBwqMOmQMqMQTKeq3cRM8hiEZMTkt3g6-jITq76R1g0T00kF7eDGHRkOlcaqygt03TOy3OW5LV59xY6fToiPW_zBNmTPliQCcbzyGxJ92TWw-GNzgb5DQ

@user55119 that mechanism is for basic conditions, though (article specified a high pH). I studied the mechanism for basic conditions already, since it is similar to that of osmium tetroxide.

Alkene ---> 1,2-diol is a 2-electron oxidation while you show Mn going from +7 to +4. In the last step of the mechanism have water attack Mn to get the desired +5 Mn. There is a lot of oxidation states of Mn involved but the electrons gained and lost must balance.

@user55119 If you can detail how the water will need to attack the Mn, I'll greatly appreciate it.

Your structure in the lower right of your diagram has Mn at +5. Add water to Mn +5 as a "nucleophile" and have glycol oxygen be the leaving group. Just exchanging one oxygen for another.


text:
Temperature Programmed Disorption

For second part of the question I'm wondering if the peaks on the right denote the chemisorbed layer, while the peaks on the left denote the physisorbed layers. Since chemisorption results in a stronger covalent bond, the energy to remove adsorbates than the physisorbed layers

What is plotted on the vertical scale ? It is 1000 or 2000 what ? What units ?

The Y axis is rate


text:
What is the structure-activity relationship of cocaine?
Cocaine is a widely known drug for both recreational and pharmacological uses. What exactly is the relationship of its structure to its drug activity?

This may be helpful:  https://www.drugabuse.gov/publications/research-reports/cocaine/how-does-cocaine-produce-its-effects

What you already know about this topic? Without a bit of knowledge about receptors answer may be not all that useful...


text:
Sensitivity of Carbon NMR Compared to Proton NMR
I am trying to figure out exactly how much less sensitive carbon NMR is from proton NMR. In a lecture, my professor said that carbon NMR is 1/5800th as sensitive as proton NMR for the following reasons:

Abundance of Carbon-13 is 1.1% so sensitivity is decreased by a factor of (0.011)
The precession frequency (difference in energy between the alpha and beta states) is about 1/4 of the precession frequency of a proton. Because this is equivalent to electrical output on the oscillator coil, this will be 1/4 as powerful (0.25)
The magnetogyric ratio of Carbon-13 is 1/4 that of Hydrogen-1 (0.25)
The dipole generated by Carbon-13 is 1/4 that of Hydrogen-1 (0.25)

Thus: 0.011 * 0.25 * 0.25 * .25 = 1/5800 (roughly)
My only objection is that the magnetogyric ratio is proportional to the magnetic moment and the spin number. Thus, I would not expect the 1/4 factor to be counted twice since it is assumed based on the definition of magnetogyric ratio. For this reason, I would expect the sensitivity of carbon NMR to only by 1/1600th of proton NMR.
i.e. 0.011 (abundance) * 0.25 (lower precession frequency) * 0.25 (magnetogyric ratio / dipole moment) = 1/1600
I am not sure who is right. I cannot find any sources that will detail the exact sensitivy differences between proton and carbon NMR and why it is so

Hint: https://en.wikipedia.org/wiki/Carbon-13_nuclear_magnetic_resonance#Sensitivity

I've already read that article; however, it does not specify the effect of the precession frequency or the relationship between the dipole moment and the gyromagnetic ratio

The receptivity ratio is ca. $5800:1$, as listed in the Bruker tables (under the column Receptivity > Natural rel. $\ce{^13C}$). Receptivity is a product of the natural abundance and an "inherent NMR sensitivity", which scales as $\gamma^3 I(I+1)$. For proton and carbon, $I = 1/2$, so the $I(I+1)$ term cancels out and we are left with
$$\begin{align}
\frac{\text{receptivity }\ce{^1H}}{\text{receptivity }\ce{^13C}} &= \frac{\text{abundance }\ce{^1H}}{\text{abundance }\ce{^13C}}\cdot \left(\frac{\gamma_\ce{H}}{\gamma_{C}}\right)^3 \\
&= \frac{100\%}{1.1\%} \cdot (4)^3 \\
&\approx 5800.
\end{align}$$
Why $\gamma^3$? One must bear in mind that at the end of the day, the signal that is detected is not the magnetisation directly. The precessing magnetisation that is excited in an NMR experiment induces a voltage in the coil that surrounds the sample. This voltage is the actual quantity that is detected.
By Faraday's Law, the voltage generated is proportional to the rate of precession, i.e. the Larmor frequency $\omega_0$, as well as the actual magnitude of the magnetisation $M$. The Larmor frequency is itself proportional to $\gamma$ (it is defined by $\omega_0 = -\gamma B_0$), so that accounts for one out of three.
Any magnetisation that we can measure must be derived from the equilibrium magnetisation $M_0$. Since the equilibrium magnetisation points along the $z$-axis, $M_0$ is the expectation value of $z$-magnetisation $\langle M_z \rangle$ at equilibrium. $\langle M_z \rangle$ itself can be related to the populations of the spin-up and spin-down states, conventionally denoted as $n_\alpha$ and $n_\beta$:*
$$\begin{align}
\langle M_z \rangle &= \gamma \langle I_z \rangle \\
&= \gamma \left[ \left(\frac{\hbar}{2}\right)n_\alpha + \left(-\frac{\hbar}{2}\right)n_\beta \right] \\
&= \frac{\hbar\gamma}{2}(n_\alpha - n_\beta)
\end{align}$$
So that accounts for one more occurrence of $\gamma$. The third and final one comes from the population difference $(n_\alpha - n_\beta)$. According to the Boltzmann distribution, we have
$$\begin{align}
\frac{n_\alpha}{n_\beta} &= \exp\left(\frac{\Delta E}{k_\mathrm{B}T}\right) \\
&\approx 1 + \frac{\Delta E}{k_\mathrm{B}T}
\end{align}$$
where $\Delta E = E_\beta - E_\alpha$ is the (positive) energy difference between the $\alpha$ and $\beta$ states. In the second step we have expanded the exponential as a Taylor series and truncated at first-order, making the (in this context, very valid) assumption that $\Delta E \ll k_\mathrm{B}T$. So now:
$$\begin{align}
n_\alpha - n_\beta &= \left(1 + \frac{\Delta E}{k_\mathrm{B}T}\right) n_\beta - n_\beta \\
&= \frac{\Delta E}{k_\mathrm{B}T} \cdot n_\beta
\end{align}$$
The energy of the spin states in an external magnetic field (the Zeeman splitting) will depend on the gyromagnetic ratio:
$$\Delta E = \gamma\hbar B$$
So, working backwards: a nuclide with a smaller gyromagnetic ratio, when placed in an external magnetic field,

has a smaller difference between the population of spin-up and spin-down states (by a factor of $\gamma$);
which leads to a smaller equilibrium magnetisation (by a factor of $\gamma$);
which, when excited, generates a smaller voltage in the detection coil (by a factor of $\gamma$).

Finally, it should be noted that this does not take into account any noise in the spectrum; it is just a measure of the signal that is available. As far as I can tell, the (root-mean-square) noise also scales with the frequency, and this leads to a theoretical signal-to-noise ratio that scales as either $\gamma^{3/2}$ or $\gamma^{11/4}$ depending on the analysis made (essentially, the exponent is slightly reduced from the nominal value of 3). I don't have the ability to explain this properly, but I can provide references that are probably helpful:

Abragam, A. The Principles of Nuclear Magnetism; Clarendon Press: 1961, especially pp 78 onwards.
Hill, H. D. W.; Richards, R. E. Limits of measurement in magnetic resonance. J. Phys. E: Sci. Instrum. 1968, 1 (10), 977–983
Hoult, D. I.; Richards, R. E. The signal-to-noise ratio of the nuclear magnetic resonance experiment. J. Magn. Reson. 1976, 24 (1), 71–85

* Populations should not be interpreted as the actual number of particles in a specific eigenstate, but rather the ensemble average of the coefficient corresponding to the specific eigenstate in the wavefunction. In other words, if each spin has a state $|\psi\rangle = c_\alpha|\alpha\rangle + c_\beta|\beta\rangle$, then we have $n_\alpha = \overline{c_\alpha^* c_\alpha}$ where the bar indicates an average over all spins in the ensemble. For more information please consult a suitable resource on density matrices.

orthocresol has explained why the signal is proportional to  $\gamma^3$. I’ll add here an argument for why the SNR should vary as $\gamma^{11/4}$. Under favorable conditions, the noise in an NMR FID is dominated by the equilibrium thermal voltage noise of the coil, $v_n=\sqrt{4kTR}$, where $k$ is Boltzmann’s  constant, $T$ is the absolute temperature, and $R$ is the coil resistance at the NMR signal frequency $f$. The RF resistance is difficult to calculate exactly, but it is often a good first approximation to suppose that the signal current flows uniformly over the surface of the coil wire to a depth given by the RF “skin depth.” The skin depth varies as $1/\sqrt{f}$, so the RF resistance varies as $\sqrt{f}$, and thus the voltage noise varies as $f^{1/4}$. This implies that the SNR varies as $\gamma^{11/4}$.

Thanks for the excellent addition! I do wish I understood more about NMR hardware, but I mainly work on pulse sequences, and don't have a great physics / engineering background (except for perhaps QM)...

I am perhaps the opposite, physics and engineering background, not so much chemistry. I have used high-field spectrometers but never had the chance to get deeply involved in pulse programming a high-field system. You are much more useful here!


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How to graph spherical harmonics in the Mac "Grapher" application?
I was wondering if it was possible to graph the spherical harmonics in the Mac grapher app. In their manual, there is a section that talks about graphing the spherical harmonics (page 54):

This is my attempt at graphing the spherical harmonic for l = 1 and m = 0:

There is clearly something wrong. My suspicion is that (for some reason) the application may be using a rectangular coordinate system, since this is the exact graph in rectangular coordinates. However, as you can see in the image, I am supposed to be in spherical coordinates. Any help would be greatly appreciated.

You should probably convert your coordinates to x, y, z from $r,\theta, \phi$ as there does not seem to be another option in this programme. Alternatively use something else such as python/matplotlib but compared to just using a plotting programme there is a learning curve.

When I start Grapher, I get to select 3D plot and which coordinate system to use. I chose the spherical one, and plotted your function. I got a sphere (s-type function).

@porphyrin Ok, so I took your suggestion and just used cartesian coordinates. When I plot the same graph, I get a sphere. However, shouldn't this particular spherical harmonic have 2 lobes (approximate "shape" of a p-orbital)? (interestingly, when I square the entire function, I get 2 lobes)