text:
Are pi bond like like interactions possible when substituents in a compound are not directly connected?

identify the more acidic compound (and the answer is the III is more acidic than II ).

My reasoning: Consider II compound, we have oxygen attached to the carbon forming a sigma and a $\pi$ bond, this oxygen has two filled p orbitals. The oxygen in the ortho position to it (in II), also has two filled p orbitals. My theory is that as we have filled p orbitals, they can form a $\pi$ bond type interactions (*). After forming the two $\pi$ bonds, the compound is stable.
Now discussing III, applying the same idea as above, there is a larger distance for the  $\pi$ overlap and hence I think $\pi$ overlap is less stable compared to the case in II.
Hence, we can say that III is more acidic (**) than II, as the pi overlap of the p orbitals of the two oxygen atoms are farther and long in III, and II is more stable (less acidic) as the pi overlap is over a small distance.
I reached the correct answer but I am not sure if my reasoning for it is correct because I haven't seen directly similar explanations in textbooks, however, I felt this idea was hinted at.

*: Not a $ \pi$ bond in the strong sense of the word
**: Less stable

Well for one thing, ketones are more acidic than esters (lactones).

I am not asking about which compound is more acidic or basic, its already known to us. I just want to verify if my reasoning for it is a correct one. @user55119

It's not OK. In esters second oxygen donates electrons to the one in carbonyl group competing with similar donation from carbon in the conjugate base - you need to compare conjugate bases rather then acids.

So, you're saying the idea completely false/ inapplicable? I feel like it has some truth in it. For example consider how the p orbitals become a donut sort of shape in benzene, pretty sure there is some mixing and mashing of p's somewhere.

Have you never heard of ester resonance?

First, let's get to the root of this problem, then I'll double back to your question. We are asked which of the two compounds is most acidic. As a side note there are various definitions of acidity, but note that broadly these theories describe acidity as linked to propensity to accept electron pairs in some form rather than general instability.
Considering neither II nor III have any immediately acidic functionalities (alcohol, carboxylic acid, sulfonic acid), we identify the alpha carbons of the carbonyl functionalities in each molecules as the most acidic hydrogens with pka ~20-25 (since all other hydrogens are alkyl (pKa ~50) and cannot form a stable conjugate base upon deprotonation).
With that in mind, the real difference between II and III is the identity of the carbonyl group. Comparing II to III we see II is an ester, with the carbonyl center bonded to an alkoxyl substituent and an alkyl substituent. Comparatively, III is a ketone, with its carbonyl center bound to to alkyl substituents. Recognizing that O lone pairs can hybridize to conjugate with the carbonyl center were C atoms cannot, we predict III to be more acidic.
The reasoning for this can be approached from a couple of angles, so here's two:

O lone pairs present in II destabilize the enol formed by deprotonation, making it more likely to donate electrons (I.E has more basic character than III), thus making its conjugate acid form less acidic (since weaker acids have conjugate bases that are stronger bases and vice versa)

O lone pairs can form an additional bond to the carbonyl carbon in a octet-satisfied resonance contributor, which greatly reduces the contribution of the analogous hyperconjugative resonance contributor with dissociation of H+. Note that this hyperconjugative contributor is a rather poor description in a thermodynamic sense, but it does play an effective stand in for a mechanism with a basic species.

Now to your actual question: You are correct with your analysis of the pi system of II, but I would note that your description is using atomic orbitals which can become very confusing in the world of ochem. In the hand-wavy hybridization terms, each oxygen would adopt an sp2 bonding orbital hybridization with unhybridized p orbitals participating in the pi-system. However, these pi-interactions can only occur if sequential atoms have unhybridized p orbitals.
This is due to the fact that the stability of pi systems arises from overlap of p orbitals. When two p orbitals are close enough to overlap, their wavefunctions can recombine into  relatively lower energy (bonding) and relatively higher energy (antibonding) wavefunctions which become orbitals. This same recombination can occur for any number of p-orbitals, with the sole condition being that overlap actually occurs.
This can only have a fully stabilizing effect when atoms are ~1.20-1.5 angstroms apart as indicated by spectroscopic bond lengths. Overlap is feasible between single-bonded substituents (Like C-O in II), but impossible once the continuous system of unhybridized p orbitals is broken.
Thus while the O in II can directly conjugate its electrons with the carbonyl pi system by "rehybridizing" from the general expected sp3 hybridization to stabilize the system, in III there are no adjacent unhybridized p orbitals to interact with the carbonyl pi system. The only electronic interactions present between the substituent (cyclic ether) and the carbonyl system would be inductive/field effects.
While one could argue that a conformer of III could possibly involve delocalization of O p electrons into the carbonyl system, I would respond by noting that such a conformer

Would have to have an unhybridized O p-orbital in order to be involved with pi interactions
Would have to have a torsional angle Θ between the oxygen and the carbonyl group of Θ=/=0 in order for overlap to actually occur (Distance between orbitals r<~1.5).

However, this would mean that the oxygen's unhybridized p orbital and the carbonyl pi system would have different symmetry elements, meaning such a pi-interaction would be symmetry-forbidden
That last bit might be jumping the shark a little, but I hope this helps!

Could you give a reference for this re-hybridisation concept?

"When two p orbitals are close enough to overlap, their wavefunctions can recombine into relatively lower energy (bonding) and relatively higher energy (antibonding) wavefunctions which become orbitals. " are you combining both molecular orbital theory and hybridisation here?

" conformer of III" what kind of conformed are you talking about here?

alright, your answer is really nice. But you are talking about pi interactions between the C and O, whereas I am talking about pi interactions being possible between the two Oxygen atoms themselves (as they have 2 filled p orbitals). Your reasoning for III being more acidic is right, but you never talked about my question which said "is pi interactions possible between the Oxygens themselves. And in the middle you say that the Oxygen is $sp^2$ hybridized, but isn't it $sp$ hybridized? Please do let me know what is wrong with my thinking. Thank you for your answer. @bigboss87

" unhybridized p orbitals participating in the pi-system. However, these pi-interactions can only occur if sequential atoms have unhybridized p orbitals." Do you mean like a chain of atoms with unhybrized p-orbitals?
Why must they be sequential for such an interaction?

Oh I misunderstood it, we can't have pi bonding because all of them are filled, and filled p orbitals can't form pi bond as they already are full, so we need only unhybridized p orbitals. Is this right?

@Buraian In organic, descriptions of molecular orbitals tend to be a little hand-wavy, so there can be a little mixing of theories that describe the same thing. Here, I'm discussing LCAO using hybridization to explain the presence of atomic orbitals on a molecular species. As I noted in my post, for favorable recombination to occur, the p-orbitals have to be able to interact through space, which is not possible if the probability density regions are not overlapping. This overlap becomes infinitesimal once you are >1 single bond away.

@buraian Here's a paper discussing rehybridization from a pretty rigorous computational perspective https://onlinelibrary.wiley.com/doi/pdf/10.1002/jcc.20524

@FinalBOSS I did discuss both possible pi interactions, but my point may have been obscured by my presentation. When we discuss "bonds" or "pi interactions", what we are really describing is a recombination process of individual electron wavefunctions that leads higher and lower energy states; when there are 2 electrons and 2 wavefunctions involved we can put both in the lower energy state, stabilizing the system. Similarly, when n consecutive p orbitals overlap, n possible recombinations are possible; but this is only possible when these p orbitals are actually interacting (distance<2 A)

@FinalBOSS this recombination is possible because we describe electrons as waves and not particles. The p orbitals represent the probability density of the electron if it was a particle, which is generated from the wavefunction by Ψ*Ψ. When these wavefunctions overlap, they add as waves; thus new wavefunctions for the interaction are generated as a result of constructive (lower energy) and destructive (higher energy) interference. However, an additional stipulation of shared symmetry elements must be imposed in order for a stabilizing interaction to occur.

@FinalBOSS this symmetry stipulation ensures that overlap actually generates stabilizing interactions by making these wavefunctions either symmetric or antisymmetric in phase (constructive and destructive interferences). If two nonsymmetric orbitals interact, their possible recombinations would have an equal amount of bonding and antibonding character. Considering that quantum physics tells us that this antibonding interaction is more destabilizing than the bonding interaction is stabilizing, thus a symmetry forbidden interaction would raise the overall energy rather than stabilizing it.


text:
Can someone help me figure out what the noble gas configurations for zirconium and holmium are?
I am currently working on noble gas configurations in chemistry and I am having a hard time understanding why I got two questions wrong on a practice.
The instructions for the practice are here:
Write out the noble gas configuration for the element. Enter your answer with all values without any spaces, hyper scripts or subscripts
Example: Sulfur S - $\ce{[Ne] 3s^2 3p^4}$   -->  [Ne]3s23p4
My questions are:

Write the noble gas configuration for Zr:

Write the noble gas configuration for Ho:

My answers were:

$\ce{[Kr] 5s^2 4p^6 4d^2}$ -->  [Kr]5s24p64d2

$\ce{[Xe] 6s^2 5p^6 4f^{11}}$  -->  [Xe]6s25p64f11

I am still getting the answers wrong. Where did I go wrong in my configurations?

Check the prefixes for s and p orbitals.

What does that mean? Can you show me what that looks like? Are my letters not in the correct order?

4p is part of [Kr], so does 5p of [Xe]. Check the periodic table.

I'm still not understanding.

As a steering wheel is a part of a car, we do not say there is a car and a steering wheel in the garage, do we ?

I don't know wether Zr and Ho have a "noble gas configuration" (sounds ridiculous to me), but the answers expected are the same as for the electron configurations of, say, iodide and astatide ions?

Are you saying the electron configurations of iodide and astatide are the same electron configurations for both Zr and Ho?

No, I'm saying your teacher is stupid. This question makes no sense. Neither Zr nor Ho are noble gases. Does he mean ground state configuration? Or should you strip off or fill in electrons until you get a noble gas configuration? $\ce{Zr^{4+}}$ is OK, with Ho that doesn't work in reality. This question is unchemical. The example points to the former, i.e. ground state. But what have you been taught if you thought this was called a noble gas electron configuration?

I add to Karl comment otherwise OP might get even more confused. It is not a real chemistry mistake but rather language, or viceversa. Or both. What you are actually asked to do is to write down the electronic configuration of an element and render it in a more compact, short form by using the preceding noble gas one instead of carrying a long string of character. Nothing about chemistry in a way. But at beginning we all have to do it. Hope this helps. For the actual question see other comments or answers I am not going to check this exercise.

"noble gas configuration for zirconium" is a terrible slang to be taken "configuration of zirconium written using the previous nearest noble gas and adding the remaining electrons".

Zirconium is $\ce{[Kr] 4d^2 5s^2}$. Holmium is $\ce{[Xe] 4f^{11} 6s^2}$. But they don't have the configuration of a rare gas.

Is this the answer?

@AudreyW.- No that is not correct. Zirconium is $\ce{[Kr] 4d^2 5s^2}$

@ MaxW. Thank you. My mistake was ridiculous.

source: Stack Exchange

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

sort: votes


text:
A question about ligand binding

I have done part a and part b but I keep getting a negative concentration for part c.

For part a I got $\pu{2 \mu M}$
For part b I got $\pu{5.051 \mu M}$

For part c, I get  $2=3.2 + \text{[bound ligand]}$, but obviously that is wrong as you cannot have negative concentration for the bound ligand.

I roughly calculate it as $\pu{0.793 \mu M}$. Is that closed to your answer (if given)?

Hey there thanks for your reply. Did you use the same values I calculated in part a and part b for your answer of 0.793 cause I keep getting a negative answer when I try.

No, I used the graph. The answers A and B are correct, but you must use the graph to answer Q3.

You know the total amount of ligand, 5 micromole. You also know that 3.2 micromole are unbound. Isn't the remaining 1.8 micromole bound? And it is in 10 mL, so the concentration is 0.18 mM? Which is weird because the protein concentration is much lower.

You have calculated correctly for parts $\bf{a}$ and $\bf{b}$. However, to answer part $\bf{c}$, you need to use the given plot, the interception and slope of which can be calculated as follows:

The slope is $\tan \alpha = - \frac{1}{K_d}$:
$$\tan \alpha = \frac{1-0.2}{0.1-0.8} = -1.14 \ \Rightarrow \ \therefore \ \frac{1}{K_d} = 1.14$$
According to the given Scatchard equation (where  each receptor has a single ligand binding site), interception of the plot is also equal to $\frac{1}{K_d} = 1.14$. Therefore, the Scatchard equation can be rewritten as:
$$\frac{r}{[A]} = 1.14(1-r) \tag1$$
where $[A]$ is actually free ligand concentration (Wikipedia). Therefore, when $[A]= \pu{3.2 \mu M}$, find the relevant $r$ using the equation $(1)$, which is the average amount of $A$ per $\pu{mol}$ of $P$ $(r = \pu{0.785 \mu M})$.
This means: Total number of binding site per receptor is $1$ and only $0.789$ is bound by $A$. Protein concentration in the dialysis tube $= \pu{2.0 \mu M}$
Therefore, $A$ bound concentration of $P$ $ = \pu{2.0 \mu M} \times 0.785 = \pu{1.57 \mu M}$

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


text:
Symmetry Operation on Product of Functions
In regards to direct product representations, I'm trying to find a proof for:
$R(X_iY_j)=R(X_i)R(Y_j)$
Where $R$ is a symmetry operation of a group and $X_i$ and $Y_j$ are members of different basis sets for the group. See e.g. Chemical Applications of Group Theory by F.A. Cotton, 3rd ed., p.105 or Atkins' Molecular Quantum Mechanics, 5th ed.,p.152.
A lot of books I've seen covering direct product representaions start with this equality but I haven't seen any justification for it yet.
Any help is greatly appreciated!

You could try D. Bishop 'Group Theory & Chemistry' publ Dover 1973, chapter 8 p 156 where he expands the product of operators as two matrices. (its not easy)


text:
Feasibility of a chemistry experiment involving combustion vegetable oil and fuel
As a part of a chemistry experiment I would like to see the effect on the enthalpy of combustion of a mixture of vegetable oil and gasoline/diesel fuel when its percentage composition is changed (i.e 90% vegetable oil, 10 % fuel, and then 80% vegetable oil, 20% fuel etc.).
I am a bit confused about whether to use gasoline or diesel fuel with vegetable oil. I researched this and usually found that waste vegetable oil is used with diesel fuel for combustion in diesel engine cars. I will also be using straight vegetable oil. Is this fine?
I am also confused as to how I should conduct the experiment, what should I use to burn the fuels ( a spirit burner) or would this not be sufficient. I am thinking about using a spirit burner and placing the mixture with the spirit burner below a stand with a metal calorimeter with a certain volume of water. Taking the initial mass of the mixture and temperature of the water, I would then light the spirit burner up and wait for the maximum temp. of water to be reached and record it and then measure the final mass of the spirit burner. I would then repeat this process with a different percentage composition and calculate the enthalpy change each time.
Is this experiment is doable? What other materials might I need?

Note that vegetable oil based diesel fuels do not use the original oil as mixture of esters of glycerol and fatty acids, but mostly methyl esters of these acids. The methyl esters are prepared by trans-esterification by methanol, using sodium hydroxide as an alkaline catalyzer, with glycerol as a side product. This  way is the plant-origin fuel closer in composition and properties to mineral-based diesel fuels, compared to the original oil.
$$\ce{CH2(-O-CO-R)-CH(-O-CO-R)-CH2(-O-CO-R) + 3 CH3OH \\->[NaOH] 3 R-CO-O-CH3 + CH2(OH)-CH(OH)-CH2(OH)}$$
where $\ce{R}$ are generally different alkyl or alkenyl chains.
The carbon chain length in these esters ( typically $\ce{C14}$ or $\ce{C16}$) is much closer to diesel fuel  than to gasoline, so do their physical properties like volatility and specific energy or energy density.
Some oil burning devices or similar is preferred, as spirit burners expect high volatility at relatively low temperature. They usually burn previously evaporated spirit vapour, like if they were gas burners ( e.g. portable outdoor spirit burners ).
But you probably mean a low power,school grade, wick based spirit burner, that should be applicable for the oil as well. But even here, preferred may be wick burners for scented oils or similar. Or their wicks at least, as they may potentially better address lower viscosity and hydrophobic property of oils.
Little preparation experiments beat everything else.
Note that low power wick burners have  high heat dissipation due long time of burning, that may affect measurement.

A comment to the comment:
I have not expected you would make any synthesis, just making aware the plant based diesel oil is not really the plant oil. I have already expressed the concern gasoline is too different to both diesel fuel and edible oil.
I suppose the school burner may be good, but I have concerns rather about the calorimetric measurement itself.

There will be needed enough oxygen. Either you burn just a little, or you need big calorimeter, or calorimeter with pressured air.
A school spirit burner may be too heavy and contains a lot of fuel and stuff.
Be aware the difference between the mineral diesel fuel and edible oil will be minimal.

We can simplify the burning by the equation:
$$\ce{CH2 + 3/2O2 -> CO2 + H2O}$$
For $\pu{14 g}$ of fuel ( $\ce{-CH2^{}-}$ ) is needed $\pu{48 g}$ of oxygen ( 3/2 $\ce{O2}$). $\pu{1 L}$ of air with oxygen partial pressure $\pu{p \simeq 21000 kPa}$ and $25 ^{\circ}C$ has oxygen density $\rho=\dfrac {pM}{RT} \simeq \pu{0.271 g/L}$.
Stuff has generally trouble to burn below 16% of oxygen, so consider rather about 5 times less oxygen than there really is: $\pu{\dfrac{0.271}{5} g/L} \simeq \pu{0.054 g/L}$.
So $\pu{1 g}$ of fuel (diesel, edible oil ) would need $\dfrac {\pu{\dfrac {48}{14} g}}{\pu{0.054 g/L}} \simeq \pu{64 L}$ of air.
With pressuring, the needed volume of calorimeter is much smaller because of higher percentage of oxygen depletion.
About the temperature measurement procedure itself:

Measure temperature at regular intervals also some time before and some time after the action to get a priori and a posteriori temperature trends.
Draw the temperature versus time plot.
Extend both respective a priori and a posteriori trend lines across the "action time interval".
Note the time with the half step of the temperature rise.
Note the temperature difference of both trend lines at this time.
The difference is your measured warming up of the calorimeter.

Be cautious not to burn Diesel oil in an open basin or bowl. When cold, diesel fuel does not burn like alcohol lit by a match. It needs a wick, and then it burns around the wick like a candle. But suddenly as the temperature gets high enough, the whole liquid suddenly catches fire. This can be dangerous. Take care !

@Maurice He already plans to use a wick based spirit burner, frequent in school in my young days. So it would be rather like scented oil esoteric burner, just not so well scented.

This is, I'm sorry, a pointless experiment. If you know the composition, and the enthapies of the components, you can perfectly predict the result.
I can only see one motivation to do it: You have made a home-built calorimeter that is quite a lot better than people would expect, and want to demonstrate that.
Your proposed calorimeter will however loose a lot of heat to the outside, also produce smoke and char, and probably in a different way for all your different fuels.
Not gonna work, sorry.


text:
Which textbook is this image on protein tertiary structure from?
I am trying to find the original textbook image that is the inspiration for many schematic figures showing tertiary interactions (hydrogen bonding, ionic interactions, hydrophobic interactions, disulfide bridges). The earliest I found was in a blog, without source attribution, and is reproduced here:

It must be from Chapter 8 of some textbook. The same blog shows other figures and tables from the same textbook, I think, for example:

The quantitative nature of the table points at a upper-level text (rather than the last chapter of organic chemistry, or a biology introductory text). The black and white figure (intended as such as judged by the rastering of the gray shade) suggests 1960s or early 1970s. There is a picture of the structure of hemoglobin in another figure (a schematic of a wood slab layer model, and a low resolution model next to it), so it cannot be written earlier than 1959.
I don't have access to a library with old textbooks at the moment, so I am asking here.
I did find a newer colorized version (with at least one mistake) here:

Google books and images do not bring up any similar result. It more or less looks like an Indian textbook. Have you seen the identical figure elsewhere besides this blog?

I checked my 1977 copy of Metzler's biochemistry, and it's not that. By 77 they seem to have been using more advanced models for their hemoglobin drawings, so that could help set the limits on the book's age.

Possibly the original 1975 edition of Stryer's Biochemistry? I don't have a copy on hand to check.

After more thinking: not getting to basic protein structure info until chapter 8 suggests that it's a general biology textbook rather than specific to biochemistry?

@Andrew Stryer (foreword Oct 1974 / print 1975 by WH Freeman, San Francisco) already was polychrome in representing formulae / diagrams. Chapter 4 «Hemoglobin. An allosteric protein» (p. 21) shows it with yellow $\alpha$-chains, blue $\beta$ chains, and red heme groups as copy from a publication by Perutz himself (1964, Scientific American) from a very different perspective than the one used in the blog.  Thus, Stryker1975 is unlikely.  (I can't confirm the elder books by Lehninger edited, but then, await chapter 8 to go show proteins is unlikely for the Principles, too.)

@KarstenTheis Is writing to the maintainers (at https://www.notesonzoology.com/contact-us) an option?

@Buttonwood I did, but anyone can post, so they probably don't know.

@M.Farooq I have seen many newer versions, so I think it was read by folks in many countries, including UK and US (because those are the source of the newer versions).

Probably Morrison-Boyd?

The illustration is likely taken from a later edition of Baum's Introduction to organic and biological chemistry.
The first edition [1, p. 292] presents nearly identical illustration under a different number alongside with a note that the image was adapted from Bennett's Graphic Biochemistry [2]:

Unfortunately, at the moment (2020-10-16) [1] is available on The Internet Archive only for a limited amount of time for borrowing, and I was unable to find a full-text source for [2].
References

Stuart J. Baum. Introduction to Organic and Biological Chemistry, 1st ed.; Macmillan: London, 1970.
Bennett T. P. Graphic Biochemistry. The Chemistry of Biological Molecules; Macmillan, 1968; Vol. 1.

@karsten theis My local library has the Bennett book. I could check it out and post images if you can’t get a copy yourself

@Andrew Not necessary, but thanks for the generous offer. I am happy to have the source confirmed. I was using the colorized version (with the mistakes) in my slides (from Ball, Hill, Scott, The Basics of General, Organic, and Biological Chemistry, which also has a Flatworld, a Saylor and a Libretext version).

@andselisk, (+1): Could you share how you did an image search?

@M.Farooq Image search was of no help for me. I used Google Books to search for the quoted parts of the caption (such as "bonds that stabilize the tertiary structure of proteins"), and then sorted the results from older to newer. This helped to discover Baum's Introduction to organic and biological chemistry.

@andselisk This demonstrates nicely how having multiple strategies helps to solve problems! And I learned about a new resource (Internet Archive).

@Andrew I am telling this story at OpenEd20 next month. A draft of the talk is at https://www.youtube.com/watch?v=AqQN7ypisjo, and the relevant 30 seconds are from 01:52 to 02:22 of the 10 minute lightning talk.

@KarstenTheis Great stuff, and all the best with the talk! The draft video is insightful and I found several tips and tricks for myself. Kudos for sticking with free and open-source tools as well as freely available web resources and apps. Finally, the "pretzel instead of an α-helix" made my day:)


text:
The charge on given atom cannot be dispersed on the other atom in

The answer given is (3)but I am not quite sure about it. I feel it's (2)
Can anyone please explain this ? Thank you in advance kind human.

why do you 'feel' it's 2? please state your reasoning/working as well in the question

Count the available orbitals and eventual electrons that can occupy them. All 3 kinds of atoms have the same orbital count, variably occupied.

Feelings are worthless in science. Instead of "I feel", write "I think" followed by arguments.

Structures 1 and 2 can be draws in alternate forms, specifically, in 1, a double bond between the C and the N atoms is an accepted alternate configuration.  In 2, similarly, a double bond between the C and B atoms gives a different configuration. In both cases, the electron charge distribution is now "dispersed" differently relative to the original structure as drawn.  Remember, these are all constructs, or our way of understanding such molecules.
In structure 3 however, aromatic stability of the six member ring can not be changed without creating a 5-valent carbon atom and and 5-valent nitrogen atom.  So there are no alternate structures for 3 as there are for 1 and 2.


text:
Units to use in Morse potential Equation
I am to graph the Energy potential curve (in $\text{cm}^{-1}$) for a molecule (HI),in the fundamental electronic state, according to the Morse oscillator. I believe I am to use this equation
$$E(R)=D_e\left[1-e^{-a(R-R_e)}\right]^2$$
However when I look at the units they don't seem to make sense to me. Parameters $a$ and $D_e$ should be $\text{cm}^{-1}$, so that the energy can be In $\text{cm}^{-1}$ as well. But that doesn't coincide with the formula given for the parameter $a$:
$$a=\left(\frac{\mu}{2D_e}\right)^{\frac{1}{2}}2\pi\nu_e$$
I know $\mu$ has atomic mass units, and $D_e$ supposedly has $\text{cm}^{-1}$. $\nu_e$ should be Hertz. So the units really don't add up between equations, and I don't know where I am going wrong.

Welcome on the Chemistry SE! I had happily converted your image into Latex formulas, but my eyes are not enough good for the task.

The exponential is dimensionless so $a$ has units m$^{-1}$ which it does as $D$ is an energy , joules, and $\nu$ a frequency, Hz, $a \equiv ( kg/(m^2 kg s^{-2}) )^{1/2}s^{-1} $. Convert $D_e$ from wave numbers into Joules and then plot $R$ in something like femtometers to get sensible numbers. You could plot $E/D_e$ vs $R$.


text:
Performing an SN1 reaction in a polar aprotic solvent
Straightforward question
What would be the outcome of having an SN1 reaction carried out in a polar aprotic solvent? I have found this question approached scarcely and with unsatisfactory answers.
What I know so far
I know the SN1 reaction is carried out in a polar protic solvent since it helps stabilize the carbocation that gets formed (thanks to the lone pairs adjacent to the electronegative atoms of the solvent itself) and also the leaving group (stabilized by hydrogen bonding).

Also, many such reactions are performed as solvolysis and it is preferred - and is much more efficient - to have the solvent act as the nucleophile when this is possible. Based on this we can ascertain the fact that a polar protic solvent would be the best choice for such a reaction and its polar aprotic homologue would not do as well.

All of the reaction rates comparisons I have seen so far take into account a polar protic solvent and a nonpolar one. But what about the protic-aprotic dichotomy?
Final note
However, my question is - can we not use a polar aprotic solvent at all? I have no problem imagining it would perform in a less efficient manner. However, stabilization is not completely absent - at least it has the capacity to stabilize the carbocation. Would the sheer lack of stabilization of the leaving group (and of the nucleophile even though many SN1 reactions do not use charged nucleophiles) render this kind of solvent completely useless for this reaction? Or would it work to some extent?

The rate of an $S_N1$ reaction may be increased by increasing the polarity of the solvent of reaction.  Because, the more polar a solvent (as measured by the dielectric constant / relative permitivity, the easier to separate ions from each other; here: carbocation and its counter anion.
Contrary to polar-aprotic solvents (think DMF, DMSO) eventually leading to «naked ions», polar-protic solvents (like ethanol, methanol, water) may weakly interact with these ions; similar to a Lewis acid/base reaction, they stabilize the ions to stay in proximity to each other as an ion pair.  Eventually, it is by experiment to figure out which effect is dominating at the reaction conditions (structure of the substrate and reagents, concentration of starting material and reagents, temperature, etc.).

Can you offer more information on the "naked ions" part? Also, what you're saying is that basically a polar aprotic solvent would work for the SN1 reaction, but much slower and with lower efficiency, as I have stated previously, right?

Re «naked ions»: these are ions lacking the typical sphere of coordinated solvent molecules.  Thus a «naked $\ce{Cl-}$» in DMF is more reactive than in MeOH along the $S_N2$ reaction during the short blink of the transition state.  Re rate of a reaction in polar-aprotic vs. polar-protic solvent: I have no kinetic data about a $S_N1$ reaction at hand, say in water and formamide of about similar DC (DMF and methanol are already way lower than water ...) to substantiate if, for a $S_N1$ in water, the high polarity does outrun the hindrance to remove the solvent molecules. But I speculate, yes,

.../... a $S_N1$ reaction in a mixture of EtOH/water  would accelerate by increase of the relative amount of water vs. EtOH.  Hopefully, there is a second answer by someone else, accounting in favour or disfavour for this speculation. (E.g., by a bounty).


text:
Why does the gas constant R appear in heat capacities?
Often, expressions for the heat capacity of a gas include a term $R = \pu{8.314 J K-1 mol-1}$. For example, the constant-volume molar heat capacity of a monoatomic gas is
$$C_{v,m} = \frac{3}{2}R$$
Why is this factor of $R$ so ubiquitous?

What do you think, based on the derivation of kinetic gas theory?

Temperature is just a form of energy and it would be more naturally expressed in energy units than in degrees. But because temperature is an older concept than (thermal) energy and very familiar to us, we keep using temperature. Just like with the exchange of foreign currency, there is a simple conversion factor to go from temperature to energy and that is the Boltzmann constant $k_\text{B}$, that is, if we consider the thermal energy of a single atom or molecule. When you consider a mole of particles, you need to multiply the Boltzmann constant with Avogadro's number $N_\text{A}$ to get the conversion factor. The product of $k_\text{B}$ and $N_\text{A}$ is the gas constant $R$.
A more thorough derivation comes from the equipartition theorem, which states that every degree of freedom contributes $\tfrac{1}{2}k_\text{B}T$ to the energy of a particle. Taking the derivative with respect to temperature of the resulting expression gives then a constant term containing the Boltzmann constant for a single particle (or the gas constant for a mole of particles).


text:
Which allotropes of sulfur exist naturally?
I'm writing a paper on chemistry and I would like to know which allotropes of sulfur exist naturally. I've searched on the Internet but I haven't found anything about natural sulfur allotropes except this book(Ref. 1). Does anyone know the answer?
Reference

Steudel R., Eckert B. () Solid Sulfur Allotropes. In: Steudel R. (eds) Elemental Sulfur and Sulfur-Rich Compounds I. Topics in Current Chemistry, vol 230. Springer, Berlin, Heidelberg, DOI: doi.org/10.1007/b12110

Even the elemental sulfur in its most common yellow powder form is rare. I would be surprised if any other would exist. Maybe temporarily, in active vulcans, a very little bit of.

Some browsing in Wikipedia, e.g. here suggests only $\alpha$ and $\gamma$-form are found naturally in significant amounts at ambient conditions.  The $\beta{}$ form probably only if the molten sulfur drops into nearby water, but over time, it is not a stable form.  But varying pressure and temperature changes the typical $\ce{S8}$ molecule.

@Buttonwood Thanks for the answer! However, from the reference that I provided above, I already know that $\alpha{}$ and $\gamma{}$-$\ce{S8}$ naturally occur.

@NilayGhosh Oh, okay, I'll add a self-answer. Thanks!

I just answered: How do native sulfur crystals form?

The alpha and gamma form of cyclo-S8 are the only forms to occur naturally. Besides the reference provided in the question, you can also find more information in the wikipedia article of "sulfur allotropes".


text:
How would someone in 1917 describe elements 43, 61, 72, 75, 84, 85, 87, 89, and 91?
https://books.google.com/books?id=aaELAAAAYAAJ&printsec=frontcover#v=onepage&q&f=false Notice the list of elements on pages 62–63. It includes a short description of each element's "Nature under Ordinary Conditions." I am curious to know what the author might have written for elements 43, 61, 72, 75, 84, 85, 87, 89, and 91 (based on 1917 knowledge; for those which hadn't yet been discovered, some properties could be told from the element's position on the periodic table). Would people be willing to suggest period-appropriate descriptions for them?

Please don't underestimate 19th and 20th century scientists, they were as intelligent (perhaps more creative) as today's scientists. The scientists who discovered most of your listed elements were trained by them.

I'm asking mostly because I sometimes have trouble wording things. While I'm sure the scientists of the time were as intelligent as today's scientists, I haven't seen much about which aspects of the listed elements had or hadn't been discovered yet in 1917.

What you have to do, is to look up the original papers or documents which described the elements. See how do they describe it. Have you explored the Internet Archive?

"I sometimes have trouble wording things" Could you provide an example. This might explain the context of your question which is not very clear yet.

See this: https://en.wikipedia.org/wiki/Timeline_of_chemical_element_discoveries

Unknown elements were discussed according to Mendeleiev's nomenclature. The prefix eka- was used for the element just under a known element, and it was dvi- for the following one, two lines under. For example, the element 31 under aluminum was celled eka-aluminum, just before it was discovered as gallium. With this same technique, the element 43 was named eka-manganese before it was discovered as technetium, and element 75 was dvi-manganese before its discovery as rhenium.
In your list, the element 61 was not expected. The 72 was eka-zirconium. The radioactive elements (Z > 83) were not expected.
Ref. : M. Fontani, M. Costa, M. Orna, The Lost Elements, Oxford University Press 2015, 530 p., ISBN 978-0-19-938334-4.

Perhaps the most interesting aspect of this answer is that it presumes a periodic relationship, whereas the modern notation (where they make Latinized names from the digits of the atomic number) does not.  Are we subtly assuming that the periodic relationships are breaking down with the new crop of much heavier unknown elements?

@OscarLanzi They do increasingly break down and that people didn't expect actinides, when lanthanides were known, perhaps shown a thing about predictions.


text:
Could there ever be a way to safely handle visible amounts of elements like astatine, francium, or protactinium?
This is something I'd actually be interested in doing if possible. I've asked about artificial stability here https://physics.stackexchange.com/questions/584595/could-there-at-least-theoretically-ever-be-a-way-to-prevent-radioactive-atoms because that would be ideal (it would be nice if these rare elements could stick around), but I'm not sure if that could ever be practical. Note that I don't mean with current technology; I'm well aware that it's currently impossible (at least in the case of At and Fr). I mean with any technology that could potentially be developed in the future.

The UKAEA has extracted 125g of Protactinium back in the sixties from old fuel rods. As for Fr and As, of course not.

You cannot physically handle those elements because there isn't sufficient amount of those element available to handle in the first place. There is only 25g and 30g of astatine and francium available at a time in the whole of Earth respectively. Those are really rare elements. Don't forget, they are also radioactive.

Title says it all - Titles should never say all, such questions are not good questions. Titles should be just a top of an iceberg.

In a sense, the real safety problem is not with us but with the samples of these elements.  We can don protective equipment to fend off the effects of radioactive decay, but the samples cannot.

They have been clickbait videos of throwing francium into water ,saying that it would cause this insanely large explosion.This is probably not true and there's no visible amount of francium as it has a half-life of 22 minutes. And the largest amount of francium atoms ever captured was about 300,000 atoms which is invisible to the naked eye.

Sometimes it is possible in very special cases to affect the half time.
The nucleus could be denied
the electron capture as the isotope only ( or major ) mode of beta decay, keeping fully ionized atoms in cyclotron or similar device.
Or, some nuclei form nucleus isomers with different decay rates. A nucleus can be excited to the metastable isomer  which may have counter-intuitively longer half-time than the base state. See as rather rare such an exception $\ce{^{180}Ta}$ with $t_{1/2}=\pu{8.125 h}$ and naturally occurring metastable
$\ce{^{180m}Ta}$ without observed decay.
But such elements would not be in a visible form.
Even if some visible amount of francium or astatine had been obtained, it would have been fast evaporated and ionized to plasma by enormous produced heat and flow of ionizating particles, evaporating also it's container.


text:
Which is more stable: Ortho or Para Hydrogen?
It is known that Para-Hydrogen has opposite spin and Ortho-Hydrogen has same spin.
Therefore, the energy of Para-Hydrogen is lower than that of Ortho-Hydrogen.
Therefore, the stability order should be: Para-Hydrogen > Ortho-Hydrogen.
Is this true or is the stability order the opposite of what I predicted?. What is the reason for this stability order?

https://en.wikipedia.org/wiki/Spin_isomers_of_hydrogen

It depends what you mean by 'stability' see also the answer here https://chemistry.stackexchange.com/questions/67316/why-do-spin-isomers-of-hydrogen-ortho-and-para-hydrogen-change-their-nuclear-s

At room temperature, about $75$% of $\ce{H2}$ is $\ce{o-H2}$ and $25$% is $\ce{p-H2}$. As the temperature drops, the relative amounts of $\ce{p-H2}$ increases. The two forms are in a temperature-dependent equilibrium $\ce{o-H2<=> p-H2}$, $\Delta H = \pu{-1.66 kJ/mol}$.
Ref. M. Eagleson, Concise Encyclopedia Chemistry, Walter de Gruyter Berlin, New York $1994$, 1202 p. ISBN 0-89925-4547-8.

This fails to answer OP's question: why?

So, which one is more stable?

Both are more stable. It is dependent on temperature and their ratio.


text:
Helmholtz Free Energy Minimization Proof
In the proof that Helmholtz free energy is minimized at equilibrium, we use the fact that $$\mathrm dS_\text{total}=\mathrm dS_\text{system}+\mathrm dS_\text{bath}\geq 0$$
But we also use the fact that
$$\mathrm dS_\text{bath}=\frac{\mathrm dU_\text{bath}}T=-\frac{\mathrm dU_\text{system}}T$$
Why does this not imply that $\mathrm dS_\text{system}=\frac{\mathrm dU_\text{system}}T$ and therefore $\mathrm dS_\text{total}=0$?
Is this because only the bath is always at temperature $T$, so we can imagine a reversible path, whereas the system may have some temperature fluctuations during the process of heat exchange with the bath?

I went through this for dG here:   https://chemistry.stackexchange.com/questions/124412/what-is-wrong-in-this-argument-that-dg-must-always-be-zero/124470#124470 The equiv. argument applies for dA.   But the essence of it is we treat the bath as infinite, so all heat flow into it is rev., and thus dS for the bath can be calculated directly from the actual heat flow. But that's not the case for the sys.  There, to calculate dS, you need to find a rev. path (where both T and q can be different from the actual T and q).  Hence, for a spontaneous process, dS_sys + dS_bath > 0.

Who says that $dS_{bath}=\frac{dU_{bath}}{T}=-\frac{dU_{system}}{T}$?  The equation should correctly read $\Delta S_{bath}=\frac{\Delta U_{bath}}{T_{bath}}=\frac{Q_{bath}}{T_{bath}}=-\frac{Q_{syst}}{T_{bath}}$, where the Q's are for the irreversible path.  The Q of the system for the reversible path is different from that for the irreversible path because of both temperature gradients and viscous dissipation of mechanical energy to internal energy during the irreversible path (both of which generate entropy).  For the reversible path of the system, these entropy generation effects are not present.


text:
Misconception in valence bond theory
Since the overlap increases with directional properties of orbital,
$$\ce{p - p > s - s > s - p}$$

However it is also observed that the bond strength of
$$\ce{H-F > H-H > F-F}$$
$$\ce{\{s - p > s - s > p - p\}}$$
Why does this happen? I've done some research and found this which says that it may depend on the p orbital's alignment with inter-nuclear axis.
The reference doesn't seem to apply here, since all the above orbitals are aligned along the inter-nuclear axis.
What am I missing here?

Bond dissociation enthalpies:
$$\begin{array}\\
\ce{H-F} &&& \pu{568 kJ/mol} \\
\ce{H-H} &&& \pu{436 kJ/mol} \\
\ce{F-F} &&& \pu{157 kJ/mol} \\
\end{array}
$$
Source

It may have to do with the fact that HF bonds are partly covalent and ionic, increasing the bond strength by introducing more ionic character then the latter two bonds

@dval98 Quite possibly. But this was specifically mentioned as an application of VBT in my textbook.

I think if you look at HBr and HCl then the order is fine. It is just that the bond in HF has too high ionic nature. You don't need overlap between of anything between two opposite point charges, to bring it to an extreme. In other words these kind of rules are qualitative and should be taken within contexts. Note that even the graphics rendering is totally qualitative, a real overlap should consider non only the lobes but also the probability density within them, which brings Electronegativity into the discussion as well. Finally the argument is also used differently but to discuss the imp..

... the impact of hybridisation, like in a part of the thread you have linked.

@Alchimista so what you are saying is $\ce{ p - p > s - s > s - p }$ is not universal and the bond strength depends on other factors as well. So is it safe to say that VBT even fails to achieve what it was devised for?

@newbie105 I would say that I do not need valence bond theory to justify why a Cation and an Anion stick to each other. Newtownian mechanics works well when not out of scope. I do not see the reason for both generalisations you made. In other words you certainly need a covalent bond in H2 and F2. For HF that's not straight forward

You should read about hybridisation, if you're so hung on VBT. What you think about is total basics and somehow you think you can get with comparing letters stuff that needs a computer - yes, actual VBT is for computational chemistry.

The images often used for the 'phenomenological interpretation' of VBT are crap, and even if these orbitals existed and had such a shape they would be wrong. Also, with more overlap doesn't only come more 'attraction', but also more 'repulsion'. VBT as often applied in OC may provide a good starting point for descriptions, but it'll never be enough to shed actual light on the bonding patterns and is most likely not yielding the right description. Additionally $\ce{F2}$ is a molecule where there is still quite some debate about the bonding, some going as far as introducing new concepts.


text:
Will equilibrium ever establish if one of the products starts to disassociate?

When $\ce{N_2O_5}$ is heated at a certain temperature, it disassociates as $$\ce{N_2O_5 (g)<=>N_2O_3 (g) + O_2 (g); K_c = 2.5}$$ At the same time $\ce{N_2O_3}$ also decomposes as $$\ce{N_2O_3 (g)<=> N_2O (g) + O_2 (g)}$$If initially 4 moles of $\ce{N_2O_5}$ were taken in a $\pu{1 L}$ flask and allowed to disassociate, and at equilibrium $\ce{[O_2]=2.5 M}$ then find equilibrium concentration of $\ce{N_2O_5}$

Lets say the first reaction reached equilibrium and then $\ce{N_2O_3}$ started decomposing. This means more amount of $\ce{O_2}$ will be produced and the equilibrium for the first reaction will get disturbed.
The first reaction will then proceed backwards as predicted by Le Chatelier's principle (since the concentration of $\ce{O_2}$ is more than the equilibrim conc.).
Due to this the oxygen level will drop and equilibrium for the second reaction will get disturbed and $\ce{N_2O_3}$ will decompose further to form oxygen.
But this process will go on forever. There is definitely wrong with this. Will this really happen or is there some other mechanism at work?

Where did you see that $\ce{N2O3}$ gets decomposed into $\ce{N2O}$ and $\ce{O2}$ ?  This is improbable. As far as I know it gets decomposed into $\ce{NO + NO2}$. A good proof is the fact that $\ce{N2O3}$ reacts with $\ce{NaOH}$ to produce a mixture of nitrite and nitrate through $$\ce{N2O3 + 2 NaOH -> NaNO3 + NaNO2 + H2O}$$ See Mary Eagleson, Concise Encyclopedia Chemistry, Walter de Gruyter Berlin, NewYork 1994, p. 698.

This is a question in my textbook. But let's just suppose hypothetically that it got decomposed into $\ce{N_2O and O_2}$. Would equilibrium ever be established in this case?

@Eyyboss this might have some useful information. (ChemLibreTexts on multistep reactions).

Because N2O3 is at equilibrium both reactions come to equilibrium, i.e.  the N2O5 is not completely used up. You can add both equations together, eliminate N2O3 and carry on.

In fact, N2O3  is the adduct $\ce{NO + NO2 <=> ON-NO2}$ with N-N bond, formed at very low temperature, similar to N2O4..

Let's make this an abstract problem with species A, B, C, and F:
$$\ce{A <=> B + F}\tag{1}$$
$$\ce{B <=> C + F}\tag{2}$$
We can also write down the sum of the two reactions, which will also be at equilibrium:
$$\ce{A <=> C + 2 F}\tag{3 = 1+2}$$
And let's use a for the concentration of A and so on.
Finding the solution using reactions 1 and 3
The concentration of F is equal to the equilibrium constant for reaction (1), so the concentrations of A and B have to be equal:
$$a = b$$
For every mole of B made, we also get a mole of F. For every mole of C made, we get two moles of F. We know the concentration of F is 2.5 M, so we get:
$$b + 2c = 2.5 M$$
Finally, we started with 4 M of A, which makes equimolar B and makes equimolar C, so the sum of all three has to be 4 M:
$$a + b + c = 4 M$$
Now we have three equations and three unknowns. We can combine the first and third equation to eliminate $a$:
$$ 2b + c = 4M$$
We can multiply the second equation in preparation for solving for $c$:
$$ 2b + 4c = 5 M$$
Subtracting the latter equation gives the solution for $c$:
$$ c = \frac{1}{3} M$$
$a$ and $b$ come out as 1.833 M.
Le Chatelier

Lets say the first reaction reached equilibrium and then $\ce{N_2O_3}$ started decomposing. This means more amount of $\ce{O_2}$ will be produced and the equilibrium for the first reaction will get disturbed

Without the second reaction, the first will attain equilibrium when $a$ is about 1.85 M. The second reaction makes F and uses up B, so the first reaction almost stays at equilibrium, and the concentration of A does not change much.
Will it ever stop?

Due to this the oxygen level will drop and equilibrium for the second reaction will get disturbed and $\ce{N_2O_3}$ will decompose further to form oxygen.
But this process will go on forever. There is definitely wrong with this. Will this really happen or is there some other mechanism at work?

If you look at this step-wise (figure out how much one reaction would go, disregarding the other), you are never done. This is similar to Zeno's paradox. However, the corrections get smaller and smaller, and your approximation gets better and better.
In general, reactions approach equilibrium, they don’t reach it. Nothing special about this system.

My doubt is not regarding the solution to the problem, but the mechanism of equilibrium establishment. Please read my question again.

@Eyy boss: There is no solvent here, all are gases. Volume is $\pu{1.0 L}$ and it's okay to consider concentrations as given moles.

@Eyyboss You now have my completed answer.

The equation $\ce{N_2O_5 <=> N_2O_3 + O_2}$ is really two equations $\ce{N_2O_5 \to N_2O_3 + O_2}$ and the reverse $\ce{ N_2O_3 + O_2 \to N_2O_5}$. Suppose we start with pure $\ce{N_2O5}$ frozen at 77K so no reaction can occur then suddenly heat it up to room temperature so that reaction starts. As soon as $\ce{N_2O3}$ is formed there is the possibility that it will react with oxygen and reform $\ce{N2O5}$ and so on back and forth until equilibrium is reached where the rates of forwards and backwards reaction are equal and all three species co-exist in proportions depending on the equilibrium constant.
In you case there is a second equilibrium $\ce{N2O3 <=>N2O + O2}$ so as soon as any $\ce{N2O3}$ is formed there is a chance that it will form $\ce{N2O}$ and a chance that it will form $\ce{ N2O5}$ depending on the rate constants. However, the $\ce{N2O3}$ can reform by the second equilibrium as well as by the first so some of it is present at equilibrium as are all the other species. See first figure.
Now consider the case that the second reaction is not at equilibrium $\ce{N2O3 \to N2O + O2}$ then once $\ce{N2O3}$ is formed it reacts irreversibly and is lost. This will also deplete all the $\ce{N2O5}$. Eventually only $\ce{N2O}$ and oxygen will remain.
It is possible to solve the differential rate equations for these reactions and technically equilibrium is never reached, but this is an artefact of using this approach and not real. Instead, using what is called a Monte-Carlo approach uses the fact that at any given time a molecule may react or may not and by counting the number of molecules present at any time the kinetics can be determined. As we use only a few molecules this shows that there is some random fluctuation in the numbers of reactants and products, this fluctuation represents what happens with small numbers of molecules and means that the population properly reached equilibrium but fluctuates about this. This may seem silly when there are $10^{20}$ molecules reacting but if a small region is examined, say a sphere of some nm radius then this fluctuation will be apparent. This is, of course, the same throughout the sample.
The first figure shows the equilibrium calculation with both reactions, I have arbitrarily chosen rate constants to show the populations clearly. The second figure (with a longer time scale) shows how the species are depleted when the $\ce{N2O3}$ equilibrium is made 'one-way' by making the reverse rate constant zero.

Above, the log populations vs time under equilibrium conditions.

Log populations when the second equilibrium is made irreversible, i.e. $\ce{N2O3 \to N_2O + O2}$. Only $\ce{N2O}$ and $\ce{O2}$ remain as there is no further reaction possible for them. Notice that the time scales are different in the two plots.


text:
Calculations of concentrations in two different equilibria
I'm a high school student, and I'm learning about acid dissociation equilibrium.
I wanted to make the idea of Le Chatelier's principle rigorous in math, so I decided to take to it with some calculus. It was all fine with just the acid equilibrium:
$$\begin{align} K_a = \frac{\ce{[H3O+][A-]}}{\ce{[HA]}} \end{align}$$
My thought process was to call $\ce{[H3O+]}$ $c_1$, to call $\ce{[A-]}$ $c_2$, and to let $c_1$ increase by some small $dc$. By Le Chatelier's principle the backwards reaction will be favoured, and both $c_1$ and $c_2$ will change by some other $dc_2$. In this light, the total change in $c_1$ would be $dc_1 = dc + dc_2$.
Plugging into the equation for $K_a$, noting that $K_a$ is constant throughout the changes in each concentration, and observing that $\ce{[HA]} = c_0 - c_2$ (where $c_0$ is the concentration of $\ce{HA}$ before any dissociation), you can get an expression for the change in $c_2$ in terms of the initial change in $\ce{[H3O+]}$, namely:
$$\begin{align} dc_2 = -\frac{c_2 ^2}{c_0 K_a + c_2 ^2}dc \end{align}$$
This can be used to find the final change in $c_1$:
$$\begin{align} dc_1 = -\frac{c_0K_a}{c_0 K_a + c_2 ^2}dc \end{align}$$
This is (pretty much) a system of differential equations. Now I'm not that experienced with systems of differential equations, but I knew enough to solve this particular system.
However, I then tried the same thing with the ionic product also taken into account:
$$\begin{align} K_w = \ce{[H3O+][OH-]}. \end{align}$$
Call $\ce{[OH-]}$ $c_3$, and change it by $dc$. Long story short, this makes $c_2$ change by some $dc_2$, $c_3$ change by some $dc_3 = dc + dc_3'$ (where $dc_3'$ is the amount the concentration changes strictly after $dc$ has been added) and it makes $c_1$ change by some $dc_1 = dc_2 + dc_3'$. Plugging all of this into the two equilibrium expressions as before gets a system of three differential equations:
\begin{align} \frac{dc_1}{dc} &= -\frac{ac_1/c_2^2}{(c_1+c_3)(1+a/c_2^2)-c_3}; \\
\frac{dc_2}{dc} &= \frac{c_1}{(c_1+c_3)(1+a/c_2^2)-c_3}; \\
\frac{dc_3}{dc} &= \frac{ac_3/c_2^2}{(c_1+c_3)(1+a/c_2^2)-c_3},
\end{align}
where $a = c_0K_a.$
Problem is, this system of differential equations looks... ahem viciously unsolvable. So much so that I doubt that the process I took to get to it was right. So is the way I'm doing this correctly, or is there a more efficient and less calculus-y route? Also, what is this field of study even called? I couldn't find any resources on these types of calculations, which is why I'm on here.

I think you overcomplicate it by differential equations. First step is calculation of equilibrium conditions of the weak acid + water system. The second step is adding some strong acid and calculation new equilibrium conditions, using dissociation constants of the acid and water, mass and charge  balance in ordinary algebraic equations. The practical evaluation  of acidobasic systems are often  a trade off between accuracy and simplicity, but this is a simple enough system.

I agree with Poutnik, you are overcomplicating this. Differential equations are usually required to solve chemical systems that evolve over time (e.g. reaction kinetics). Equilibria by definition describe systems that have stopped changing (macroscopically at least). And you can handle the latter using only ordinary algebraic equations, namely equilibrium constants, mass balances and charge balances. Then you can put as many components as you want, and it will always work out. In your case, initially you have a weak acid $HA$ in water; then you add a strong acid $HX$. Try it out.

@Poutnik Ahhh, you're right!  I'm too used to differential equations from physics, so I assumed the way to do it was to add a little bit of $\ce{OH-}$, see the result and sum all of those additions. Of course, you can add all of it at once. Are there any benefits you know of to viewing the process quasistatically, or was this all a bit of a waste of time? In any case, thank you for pointing this out.


text:
Is there an 'easy' way to determine if a ligand is a pi acceptor, donor or neutral?
First of all I am confused how something can be pi neutral? something either does or doesn't have filled p orbitals for example.
given something I have never seen acting as a ligand before like trimethylphosphate how can I work out whether this is pi accepting, donating or neutral? I am thinking the donor atom would be the oxygen bonded to the Me group and P but do I only consider the orbital of the donor atom?
For a tetragonal Cu(II) complex how can a ligand even act as pi donating if since the t2g orbitals on the metal are filled?
Sorry about this messy question, I am very confused about this in general, I would be so grateful if someone could help me understand this

If you ask for an "easy" way to determine if a ligand is a pi acceptor/donor or none, look at the spectrochemical series. A quick-and-dirty rule (which means that it is correct most of the time but not always) is that weak ligands (iodide, bromide, hydroxide etc) are pi-donor ligands. The medium ligands (water, ammonia etc) are pi-neutral, and the strong ligands (cyanide, carbonyl, bipyridine, etc) are pi-acceptor.
But this method is some kind of reversing here. Because we normally use a ligand's pi character to guess if it is a strong or weak ligand.
If you can, ligand field theory is your best friend in this problem.
Generally, a pi-donor ligand has an extra electron pair that lies in an orbital that has approximately the same symmetry as the metal center's orbital (usually t2g). This allows the two orbitals to interact with each other and creating two new sets of orbitals: t2g and t2g* (as in the molecular orbital theory). The ligand's pi electron will reside in the  lower t2g orbital, whereas the metal's original d-electrons will reside in the t2g* one. That explains your question of the copper complexes.
In the other hand, a pi-acceptor ligand has an empty molecular orbital of low energy and approximately the same symmetry as the orbital of the metal center. In this case, it will interact with the metal's orbitals and the d-electrons of the metal will be delocalized to the ligand. These ligands usually have a highly conjugated system that greatly decreases the energy of the LUMO (bipyridine, triphenylphosphine etc). In the cases of carbon monoxide (called carbonyl as a ligand) and cyanide, they have low-energy LUMO for other reasons.
A pi-neutral ligand (also called sigma-donor ligand) does not process any of the above characteristics. For example, ammonia has only one electron pair on the nitrogen, and that pair is already used in the formation of the sigma bond, so no more pi electrons to donate. The LUMO of the ammonia is also high in energy, so no pi accepting here.
I think trimethyl phosphate is mostly a pi-neutral with the donor atom is the oxygen double-bonded to phosphorus. The molecule is not conjugated at all, and the oxygen can only donate one pair of electron since the other's orbital does not have the same symmetry.

Is the ethylene ligand in Zeise's Salt also a π-donor?


text:
Why is work and heat energy always viewed from the perspective of the system?

So my answer was 3 and it was wrong, And i really would like to know what is the correct answer, after giving it some thought, I think the answer is 4 because indeed it is easier to control the system than surroundings, can you please assure it that it's D.
I appreciate any help

In my judgment, this question is totally idiotic.  Who cares?  will knowing the answer to this question, whatever it is, in any way improve your ability to correctly analyze the behavior of any specific situation?

I agree, but still :)

@ChetMiller Curiously enough, I would not entirely agree with the answer (D). For instance, in a typical adiabatic bomb calorimetry experiment we monitor the change in temperature of the jacket surrounding the bomb in which the reaction takes place. The temperature change of the jacket is due to exchange of heat with the bomb. One can argue about semantics, but the jacket could be regarded as the surroundings in this case.

Why would it - how could it - be viewed from any other perspective?

The basis of this question seems silly to me, because you're going to focus on the heat and work exchange between the system and the surroundings, which means it is all a matter of sign convention (i.e. this simply defines the physical meaning of $q>0$ and that of $w>0$). Somehow the question seems to be saying that, say, defining $q$ to be the heat change in the surroundings would necessitate measuring the entire heat change in the surroundings from all factors, which is absurd.

Does this answer your question? Why do chemistry and physics have different sign convention in thermodynamics?

The problem is in the word "always". The statement "Work and heat energy are always viewed from the perspective of the system?" is wrong. Please inform your teacher that is true in the chemist's world only- and chemists did not invent thermodynamics.
Chemists have adopted a different convention than physicists. All we have to do is to be aware of the context.
Recall the First Law of Thermodynamics, which has 2 different equivalent statements.
In some books it is
$$∆U=q-w$$
and in some it is
$$∆U=q+w$$
Why, because the word "always" is not correct.

"Please inform your teacher that is true in the chemist's world only".  This statement is incorrect. While both chemists and physicists usually reference heat and work relative to the system, both chemists and physicists can switch to referencing them relative to the surroundings if that makes sense for the particular set-up they are analyzing. Ironically, you're making the same mistake the problem did: making a broad, incorrect generalization.

By way of comparison, according to the international standard ISO 80000 Quantities and units – Part 5: Thermodynamics, $\Delta U=Q+W$ where $Q$ is amount of heat transferred to the system and $W$ is work done on the system provided that no chemical reactions occur. This applies to physicists as well as chemists.


text:
Permanent ink from urethane, 99% IPA, and pigment?
I’m an artist and I started working with recycled plastic bottles as a material recently. I want to paint the plastic and retain its transparency, but acrylic paint won’t stick without priming and generally looks bad. Permanent markers are good, but they leave streaks over large swathes of color and are expensive.
I want to make alcohol ink, which is basically bulk highly pigmented permanent ink that you can paint with or do whatever. I could buy permanent markers and make my own (again, expensive), but I have
solvents (99% iso, mineral spirits, naphtha--no toluene or xylene, though) and lots of dyes and pigments, as well as experience with epoxy resins.
I’ve read that permanent markers are simply a resin (urethane) dissolved in a solvent and pigment is added. I have resin-friendly pigments and was looking at spar (has more resin) or poly urethane as possibly a good choice for the resin. Before I invest in another experiment I can't afford to fail, I was wondering if this would even work or if the two-part epoxy resin I have would work, too.
So my questions are:

What resin should I use?
What solvent for that particular resin?
What ratio of solvent:resin:pigment?

I know thinning epoxy resin can weaken it, but my work is purely decorative. I just need it to stick to the plastic and be ok with low heat for shaping purposes.


text:
Why is potassium permanganate used to kill mollusks?
I know that potassium permanganate is used as a disinfectant for bacteria. Permanganate acts as a strong oxidizing agent that can rip the electrons off of bacterial cell membranes, resulting in a disruption that can lead to death. I've also read that potassium permanganate can be used as an oxidizing agent to kill mollusks (source: http://www.chemistry.wustl.edu/~edudev/LabTutorials/Water/PublicWaterSupply/PublicWaterSupply.html). I've tried looking into it and I can't find anything helpful. Does permanganate kill mollusks in the same way? If that is the case, will the disruption of the outer cells of the mollusks really kill the whole organism? Any insights would be helpful, thanks in advance!

An army flamethrower can kill bugs. It will kill frogs, too. It will kill humans (that's what it was designed for, BTW). Same thing here.

The provided link (Lab Tutorial at Washington University in St. Louis)  states the following:

Another problem for water-treatment plants in the Great Lakes and Mississippi Valley is zebra mussels, which can "congregate" at the water intake and clog the screen, eventually sealing it off. Potassium permanganate, KMnO4, can be used to kill these mollusks.

Zebra mussels are a problematic invasive species in the midwestern USA. Permanganate has been found to inhibit their proliferation. The main effect is apparently oxidation. From a report released by the US military (ref [1.]) on use of oxidizing molluscicides including permanganates:

In general they have similar modes of action based on the oxidation of organic matter,which leads to toxic and lethal effects. They are suitable for use in preventative treatment, where they are added to a system throughout the breeding season at from 0.1 to 0.5 mg/L (ppm) total residual oxidant (TRO) to prevent settling. For reactive treatments,continuous application of 0.5 to 1.0 mg/L TRO for 2 to 4 weeks can eliminate established adult colonies, but concentration and contact time required depends on temperature, water chemistry, and physiological state of the zebra mussels. Mussels do detect oxidants, and shell closure for up to 2 weeks may reduce efficacy in adults(Claudi and Mackie1994)

See also ref [2] cited in ref [1.].
Ref [3] compares the toxicity of various molluscicides, noting that

An  LC50 value  could  not  be  estimated  for antimycin,  Bulab  6002,  Bultab  6009.  Calgon DM -DACC,  KML  V2.  KML  V54.  potassium  chloride, and  potassium  permanganate  due  to  insufficient mortality  of  the  test  organisms.

In addition, potassium can also have an effect on mollusks. For instance, from a thesis (Ref [4]):

Potassium can reach increased levels that are toxic to aquatic organisms, such as mollusks, in the environment (Romano and Zeng 2007). For example, Fisher et al. (1991) reported that increased potassium concentrations were toxic to Dreissena polymorpha, causing asphyxiation by damaging gill epithelium. In addition, Imlay (1973) suggested that high potassium concentrations could be the cause of absent mussel populations or lower diversity within Upper Midwest mussel assemblages. Any river or stream with a potassium concentration of 7 mg/L or higher lacked mussels, but mussels could be found in rivers with a concentration lower than 4 mg/L.

Note responses vary across organisms. Discussion of that is bound to take us increasingly away from chemistry.
References

Zebra mussel chemical control guide (ERDC/ELTR-00-1) by Susan L Sprecher, Kurt D. Getsinger; prepared for U.S. Army Corps of Engineers (2000).

Klerks,P.L., Fraleigh, P.C,and Stevenson, R.C.(1993)."Controlling zebra mussel {Dreissena polymorpha) veligers with three oxidizing chemicals: Chlorine, permanganate, and peroxide+iron."Zebra mussels: Biology, impacts,  and control. T.F.Nalepa and D.W.Schloesser, eds., Lewis Publishers, Boca Raton,FL,621-641.

Diane L. Waller, Jeffrey J. Rach, W. Gregory Cope, and Leif L. Marking, Susan W. Fisher, Henrycka Dabrowska. Toxicity of Candidate Molluscicides to Zebra Mussels (Dreissena polymorpha) and Selected Nontarget Organisms. J.  Great  Lakes  Res.  19(4):695-702 Internat.  Assoc.  Great  Lakes  Res.,  1993.

Gibson, Kesley. (2015). ACUTE TOXICITY TESTING ON FRESHWATER MUSSELS (BIVALVIA: UNIONIDAE) AND FRESHWATER SNAILS (GASTROPODA: CAENOGASTROPODA).


text:
Combustion of methane and ethane
I'm having trouble with the following problem:

A gas fuel of Methane ($\ce{CH4}$) and Ethane ($\ce{C2H6}$) is burned
with atmospheric air, and the volumetric analysis of the products of
combustion yields the following: 5.32% $\ce{CO2}$, 1.60% $\ce{CO}$,
7.32% $\ce{O2}$, 12.24% $\ce{H2O}$ and the rest $\ce{N2}$.
Calculate the percentage of excess air and the percentage of Methane and Ethane in the fuel.

The balance equation from the data given above should be:
$x \ce{CH4}+y \ce{C2H6} +a(\ce{O2}+3.76 \ce{N2})\rightarrow 5.32\ce{CO2}+ 1.60\ce{CO}+7.32 \ce{O2}+12.24 \ce{H2O}+73.52\ce{N2}$
Balancing each element:
$\ce{N2}: \quad 3.76 a =73.52 \\
\ce{C}: \quad x+2y=5.32+1.60 \\
\ce{H}: \quad 4x+6y=24.48$
And solving the system I get: $a=19.55, \quad x=3.72, \quad y=1.60$
How can I get the percentage of excess air and types of fuel from there? Thanks!

You can get the fuel composition from the total H/C ratio of  products.

The excess of $\ce{O2}$ is ${7.32~ mol}$. So the excess air is : ${7.32 ~mol· (1 + 3.76) = 34.84~ mol}$.
The total amount of air is : ${a·4.76 = 19.53~ mol ·4.76 = 92.96~ mol}$.
The percentage of excess air is : $34.84/92.96 = 0.3748 = 37.48$%


text:
how is the esterification of amino acid carried on in this reaction
i was asked to synthesis this compound from this amino acid, i first alkylated the alpha carbon using enolate alkylating method then i esterficated the amino acid but i'm doubtful that my steps are right, do i need to protect the amino group in the esterification step? or even before ?

I don't think your alkene hydration will work on the unprotected amino acid

I think you need a protection strategy and there are several choices. This one uses O'Donnell's benzophenone imine strategy details here.
First you prepare glycine ethyl ester hydrochloride from glycine with EtOH/HCl, a method is here patent.
From the ethyl ester hydrochloride prepare the benzophenone imine by O'Donnell's method as described in ref 1. This gives a protected species that can be easily alkylated under phase-transfer conditions with allyl bromide as described in ref 1
The allyl product is then epoxidised (MCPBA) and the epoxide opened with iodide procedure here
Deprotection with acid hydrolysis gives your target molecule.


text:
Thermodynamics heat, enthalpy, Gibbs
We know the basic Thermodynamic equations are:
$$ dU=-PdV+ TdS + \mu dN$$
$$ dH=dU+d(PV)= TdS+ \mu dN + VdP$$
we typically assume constant Pressure for a reaction carried out in air and neglect the dP term
$$dG=dH-d(TS)=  \mu dN - SdT+ VdP$$
once again we neglect the dT and dP terms
but for heat of reaction we use the enthalpy not just the TdS term. When we have a battery though the electrical energy is the Gibbs free energy with the heat of reaction being TdS and other energy losses also contributing to the heat. When the reaction is not separated (not an electrochemical cell) then the Gibbs is part of the heat.
For a typical reaction where the chemicals are in contact why do they use the enthalpy to determine the heat of the reaction produced?

fixed the equation. Interesting how a non spontaneous reaction can become spontaneous with temperature

I'm not entirely clear on what your question is, but the reason for using $\Delta H$ in a lot of equations rather than $q$ is because enthalpy is a state function, whereas heat is not.

When a reaction occurs at constant pressure and no work other than expansion work  (such as electrochemical) is performed, then it can be shown, from the definition of the enthalpy and the first law of thermodynamics applied to a closed system, that the change in enthalpy is equal to the heat:
$$\begin{align}dH&=dU+d(pV)\\&=dw_{pV}+dw_{other}+dq+pdV+Vdp\\&=-p_{ext}dV+dw_{other}+dq+pdV+Vdp\\&=dq+dw_{other} \end{align}$$
The last equality follows when the pressure is constant and equal to the external pressure. When only pV work is done then clearly heat and enthalpy change are equal. Readily measured heats of reaction provide a very convenient way to determine values of the enthalpy state function which can be tabulated and employed in further thermodynamic computations.

Yes this is a great explanation. Once again at constant pressure and temperature whether a reaction occurs in the forward or reverse is still dependent on the temperature and pressure which means the chemical potential is a function of both $$ \mu= RTlna $$ where the activity (effective concentration) is also a function of both

@ChemEng It's worh recalling that $(\partial \mu/\partial T)_p=\bar S$

from the multivariable chain rule we know $$ \mu=\frac{\partial{G}}{\partial{N}}$$ and $$S=\frac{\partial{G}}{\partial{T}} $$ so the second partials are equal

Sorry, I miss-typed. Should have written $(\partial \mu_i/\partial T)_p=-\bar S_i$. This is analogous to $(\partial G/\partial T)_p= -S$ except true for the partial molar quantities (indicated by the bar). You can integrate this relation to obtain the potential at some other T.

yes this would be an interesting way to measure entropy as gibbs is the free energy or    -nFE for electrochemistry and enthalpy-reversibleheat = heat of reaction so we can measure those


text:
Why is the monomer for this polymer drawn in this way?
I am a bit confused about why the monomer for a particular polymer is drawn in a particular way.
Shown below is the polymer:

Shown below is the monomer for the polymer which the answer scheme gives:

I am a bit confused about why the monomer is drawn in this way. For each CH in the polymer, there is a CH3 molecule which is either above or below it. Why is it that, in the monomer drawn above, the CH3 is to the side of the CH. Isn't it above, so should be drawn above?

The "real" molecule is neither planar nor rectangular. So it is just a 2D projection of a 3D molecule.

Thanks, but wouldn't it be more accurate to put the NH part on the side?

I agree that placing the amino group to the left, as is usually done in drawing amino acids like the monomer you've given (which is alanine), is more helpful to the reader, but it isn't more or less accurate - the 2 drawings are equivalent. As MaxW says, in reality the bond angles around the central carbon would be approximately 109.5 degrees, rather than the 90 degrees shown.

Schematic structural formulas are just formal notations, following structures of chemical bonds, but not actual 3D arrangement of molecules.
They can have multiple equally right and equivalent forms, as they all describe the same molecule, with focus on different aspects..
The one drawn above, with methyl on the left and amino group below, follows general organic chemistry approach of alanine being formally an aminoderivate of propanoic acid $\ce{CH3-CH2-COOH}$, i.e. 2-aminopropanoic acid.
The approach with amino group on the left and methyl group below is the biochemical approach, following the line of the peptidic structure. It takes aminoacids as formal derivates of the most simple aminoacid glycine $\ce{NH2-CH2-COOH}$, in our case 2-methylglycine.
So the latter approach is more suitable/illustrative in context of polypeptides, but the former one is not wrong.


text:
Why do neutron scattering lengths vary seemingly randomly with atomic number?
I'm a bit unclear as to what determines neutron scattering lengths, and hence why they can vary so much even between adjacent elements like Fe and Co. I gather it's related to resonant scattering, but I would be very grateful for an account (ideally not purely mathematical!) of how and why.

Isn't it surprising that chemical properties vary so much between adjacent elements like Xe and Cs?

Honestly, I would say no, because I can readily explain that by referring to atomic orbitals, whereas neutron scattering is a nuclear rather than electronic property. I don't think saying "Gosh, all chemical properties can vary significantly" is terrifically meaningful. I am aware that the variation is not random, which is why I asked the question.

Well, nuclear properties also vary pretty significantly between neighboring elements. Indeed, these changes are unrelated to chemical properties.


text:
What does the order of a class mean in a character table?
I am not from a chemistry background. I am trying to see how the symmetry operations affect the stiffness tensor which led me in the path of understanding character tables. I am looking at the character table for C6v group and I cannot understand what the factor 2 for C6(z) means. Is it representative of the fact that I can rotate in clockwise and anticlockwise direction?
Please let me know if you need any further information.

Yes.$\mathstrut$

Shouldn't C2(z) also have a factor of 2 then?

No, because rotating for $180^\circ$ clockwise and anticlockwise is the same thing.

This answer might help in general https://chemistry.stackexchange.com/questions/58609/understanding-group-theory-easily-and-quickly/58613#58613

@IvanNeretin That makes sense. Thank you

@porphyrin Thank you. This is very helpful.


text:
Why does diethylmethylamine have such a low melting point?
According to PubChem, diethylmethylamine has a remarkably low melting point of $-196.0\ \mathrm{^\circ C}$. This is substantially lower than the melting points of dimethylethylamine ($-140.0\ \mathrm{^\circ C}$), trimethylamine ($-117.1\ \mathrm{^\circ C}$), triethylamine ($-114.7\ \mathrm{^\circ C}$), and 3-methylpentane (given as both $-162.9\ \mathrm{^\circ C}$ and $-118\ \mathrm{^\circ C}$, not sure which is right). Why is that? Is diethylmethylamine uniquely ill-suited to forming a stable crystal lattice for some reason (maybe its highly asymmetric structure)?

As a small aside, diethylmethylamine is potentially the substance with the lowest melting point that remains liquid in ambient conditions. The value -196 °C is a little suspicious in that it happens to equal the boiling point for liquid nitrogen, so I hope it's not actually some kind of data misinterpretation.

The true melting point for diethylmethylamine is yet unknown; the original reference for this "melting point" value is J. Org. Chem. 22, 840-843 (1957), which states that the material became a glass at liquid nitrogen temperature. Also of interest is that the same exact data points are given for isopropyldimethylamine and diisopropylmethylamine, but have apparently never been misconstrued as melting points for those compounds!


text:
Why can't iodide be oxidised to iodate in acidic medium?
Even if we use a strong oxidising agent like 'permanganate ion in acidic medium' it doesn't help oxidise iodide to iodate, but a relatively weaker oxidising agent 'permanganate ion in alkaline medium' helps to form iodate from iodide.
I have found one discussion similar to this but it's not convincing.
Even in alkaline medium the potential is negative although significantly lesser than in acidic medium.
Then how does the reaction even proceed in alkaline but not in acidic?

Draw yourself a chart of standard redox potentials for eventually involved redox systems as functions of pH, involving formally and implicitly [H+] or [OH-] to the redox potentials pH dependent values. See https://en.wikipedia.org/wiki/Standard_electrode_potential_%28data_page%29

Note that there can be kinetic reasons, as $\ce{IO-}$ is more stable in alkaline solutins than $\ce{HIO}$ in acidic one.

Permanganate oxidizes iodide to iodine $\ce{I2}$ at all pH values. In acidic conditions, the reaction is finished here. But in basic solution, the iodine $\ce{I2}$ is transformed into iodide and iodate according to $$\ce{3I2 + 6 OH^- -> 5 I- + IO3^- + 3 H2O}$$The mixture  iodide + iodate reacts the other way as soon as the solution becomes acidic : $$\ce{5I^- + IO3^- + 6 H+ -> 3 I2 + 3 H2O}$$

So its because of the property of iodine & not permanganate that we see a larger difference of oxidation state in the reactions in the basic pH mediums?


text:
Maximum absorption wavelength directly proportional to the donor number of the ligand for a copper(II) coordination complex
I am looking at a set of results showing a directly proportional relationship for an increasing maximum absorption wavelength (UV-Vis spectra) with increasing donor number of solvent ligands. The complex has a $\ce{Cu^2+}$ centre with $\mathrm{d^9}$ electronic configuration. Assuming a tetragonal configuration, how does this tie in with ligand strengths?
I am not sure what I can deduce here as I thought the more available a lone pair (basicity) would be able to interact more strongly, a higher wavelength indicates the energy gap between the $\mathrm d_{x^2-y^2},$ and $\mathrm d_{z^2}$ orbital has decreased. So, has there been more significant Jahn–Teller distortion, or is the splitting parameter somehow smaller?
Where does my reasoning fall down?


text:
"Nivea Sun protect & refresh" melted my tub
I stupidly applied spray sunscreen on my children inside our white, acrylic bathtub and the floor ended up with a white, very sticky residue that rubs off on things, but will not wash away with detergent. The sunscreen turns out to be based on denatured alcohol and contains Benzyl Alcohol, so I suspect it melted the top layer of the tub.
Is there anything to do to re solidify the surface or remove the sticky layer? I've seen refinishing kits at the hardware store, but I assume this requires a clean surface beforehand.
Other ingredients:
Denatured Alcohol,
C12-15 Alkyl Benzoate,
Homosalate,
Octocrylene,
Cyclomethicone,
Glycerin,
Butyl Methoxydibenzoylmethane,
Ethylhexyl Salicylate,
Acrylates/Octylacrylamide Copolymer
Butylphenyl Methylpropional,
Benzyl Alcohol
Product and ingredient list:
https://www.nivea-me.com/en-me/products/protect-and-refresh-invisible-cooling-sun-spray-spf30-40058082837740214.html

You might want to give this question at home improvement (diy) SE if you don't get an answer here.

Common sense chemistry says water with household detergent should get the organics out of the acrylic, so it becomes rigid again. Give it a day or two. Afterwards try whatever your hardware store offers to polish acrylic surfaces.


text:
Unusual Titration Curve Explanation

From Costanzo's Physiology book in Acid-Base Physiology chapter, this titration curve of a weak acid ($\ce{HA}$), in the context of discussing buffers is provided:

What confused me is that in this diagram, the $\mathrm{pH}$ is on the $x$-axis rather than the $y$-axis. Also, why does the peek of the curve has ($\ce{HA}$) only and the trough has ($\ce{A-}$) only?

Thanks for giving me the chance to think about it, @Poutnik. I flipped it but don't seem to understand it. As for the second question, I think that's because acids and bases occur in pairs, with an acid having a conjugate base??

Acidic environment turns conjugate pairs toward acid side and vice versa.

Yes, but allow me to ask what does this do for our question @Poutnik ?

Do not write questions at the moment they appear, but think about them. It gives you more, if you resolve it yourself, including textbook and online resources. Only if you really failed, then ask.

Acid (Brønsted–Lowry Acid) is a substance that gives away its $\ce{H+}$, but the environment should be able to accept that $\ce{H+}$. After it accepted it - the acid (which is now its conjugate base) can accept it back. There are 2 extremes:

The environment already has plenty of $\ce{H+}$ already. So even if acid gives its $\ce{H+}$ to the environment, it immediately receives $\ce{H+}$ back. Hence the acid always has its $\ce{H}$ and you see $\ce{HA}$ on the picture.
The environment is basic - it can take lots of $\ce{H+}$ and won't give them back to the acid. Hence acid stops being acid - it gives away $\ce{H+}$ and becomes $\ce{A-}$.

$\mathrm{pH}$ is a measure of how many $\ce{H+}$ there are in the solution. Low $\mathrm{pH}$ means there's plenty of $\ce{H+}$ (our 1st case), high $\mathrm{pH}$ means there's a "shortage" of $\ce{H+}$ (2nd case).
In the middle (where $\ce{H+}$ equals acid's $\mathrm{p}K_\mathrm{a}$ which in your case is 6.5) you'll have a case when 50% of $\ce{H+}$ is taken by the environment and the other 50% of acid molecules are $\ce{HA}$.
When you titrate you do exactly that - you change the $\mathrm{pH}$ of your environment which then changes how many molecules are ionized. And then at some point your indicator changes the color which means you reached some $\mathrm{pH}$ you were looking for.
Titration curves that I've seen have the volume of solvent added (your graph probably shouldn't be called Titration Curve?) - not $\mathrm{pH}$. But these are related since if you're adding a basic titrant to acidic solution you're changing its $\mathrm{pH}$.

Thanks for the answer! But could you eleborate more on how does this fit on the big picture of titration, i.e how is the acid titrated this way? @Stanislav Bashkyrtsev

@Positron12, I thought I did that.. Anyway, I've added more info there.

@Positron12  Remember that $\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log{\left(\frac {\ce{[A-]}}{\ce{[HA]}}\right)}$ and the concentration ratio depends on the titration progress.

For a weak acid $\ce{HA}$ with acid constant $K_\mathrm{a}$:
$$\ce{HA + H2O <=> H3O+ + A-} \tag1$$
$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]} \ \Rightarrow \ [\ce{H3O+}] = K_\mathrm{a} \cdot \frac{[\ce{HA}]}{[\ce{A-}]} \tag2$$
Taking $-\log$ of both side of equation $(2)$:
$$-\log [\ce{H3O+}] = -\log K_\mathrm{a} -\log \frac{[\ce{HA}]}{[\ce{A-}]} $$
Or
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{A-}]}{[\ce{HA}]} \tag3$$
The equation $(3)$ is called Henderson-Hackleback equation (see Poutnik's comment elsewhere). According to the Le Chatelier principle, if you remove $\ce{H3O+}$ from the equilibrium $(1)$ (Note: you can do this by adding appropriate amount of strong base as a titrant, e.g., $\ce{NaOH}$ solution; $\ce{H3O+ + OH- -> 2H2O}$), more of $\ce{HA}$ ionizes to form $\ce{H3O+}$, thus, decreasing $[\ce{HA}]$ as a result. As a consequence, $[\ce{A-}]$ also increases at the same time. Therefore, $\mathrm{pH}$ of the solution increases according to the equation $(3)$.
To answer your question about "why does the peek of the curve has $\ce{HA}$ only and the trough has $\ce{A-}$ only?": Again, if $\ce{HA}$ is in acid medium (increasing amount of $\ce{H3O+}$), according to the Le Chatelier principle, the equilibrium $(1)$ favored the backward reaction so that more $\ce{HA}$ produces. At one point, if $[\ce{H3O+}]$ high enough (say $\mathrm{pH} \lt 4.5$ in your case), there won't be sufficient $[\ce{A-}]$, because all $\ce{A-}$ would be protonated to give only $\ce{HA}$.
During the titration (removing $\ce{H3O+}$), when you reached the endpoint by removing all $\ce{H3O+}$, you have only $\ce{A-}$ remaining in the solution. Yet, resulting $\ce{A-}$ would established following equilibrium:
$$\ce{A- + H2O <=> HA + OH-} \tag4$$
Suppose you add one extra drop of $\ce{NaOH}$ solution. That increase the $[\ce{OH-}]$ in the mixture and again, according to the Le Chatelier principle, the equilibrium $(3)$ favored the backward reaction so that more $\ce{A-}$ produces. Thus, as further addition of the titrant, only $\ce{A-}$ remains in the solution (of course, with $\ce{OH-}$, thus further increasing $\mathrm{pH}$).


text:
Why is chlorpromazine not characterized as tricyclic?
Chlorpromazine is an old ("typical") anti-psychotic medication with a chemical structure which I as a non chemist find almost identical to that of all "ine" tricyclic anti depressants such as imipramine, amitryptiline, nortryptiline, maprottyline, and alike.
If I understood correctly, imipramine itself and later its counterparts were all originally synthesized from chlorpromazine. That said, in any occasion I've read about chlorpromazine, never have I found it being described as a "tricyclic", but rather just as "antipsychotic" (instead "tricyclic antipsychotic" compared with "tricyclic antidepressant" as with its counterparts), so I ask: Why does chlorpromazine not characterized as tricyclic?
By the way, both chlorpromazine and tricyclic antidepressants tend to cause weight gain if it matters to the reader.

Apparently anti-psychotic phenothiazines and thioxanthenes got separate categories.  Why they did when there were still "tricyclics" with no more specific designation? Dunno and I don't see importance of that. BTW I think it's borderline off-topic when one asks 'bout some lame pharmaceutical categorisation and not for example biochemical aspects.


text:
Question about reading NIST datasheets & Gibbs energy of reaction
I'm peeping the NIST thermo tables for $\text{H}$ and $\text{H}_2$, and I'm perplexed.
Seems to me that using
$$G=[\Delta H^\circ_f+H-H^\circ(T_r)]-T\cdot S^\circ$$
(well, actually correcting the entropy for partial pressures in a mixture of ideal gases) obtains some bewildering values for simple reactions.
Consider $\text{H}_2 \rightarrow 2\text{H}$ at standard pressure and standard temperature. If I'm calculating it correctly (which I am sure I am not), using the values at $298.15^\circ K$ in columns 2 ($S^\circ$), 4 ($H-H^\circ(T_r)$), and 5 ($\Delta H^\circ_f$) from the links above, I find that
$$G_{\text{H}_2}=0\frac{\text{kJ}}{\text{mol}}−298.15^\circ \text{K}\cdot130.680\frac{\text{kJ}}{^\circ \text{K}\;\text{mol}}\approx-38962\frac{\text{kJ}}{\text{mol}}$$
And
$$G_\text{H}=217.999\frac{\text{kJ}}{\text{mol}}−298.15^\circ \text{K}\cdot114.716\frac{\text{kJ}}{^\circ \text{K}\;\text{mol}}\approx-33984\frac{\text{kJ}}{\text{mol}}$$
Which is great, the reaction isn't spontaneous. Except... if I recall correctly,
$$G_\text{mixture} = \sum^\text{species}_j \mu_j n_j$$
Where $n_j$ is the number of mols of that species and for ideal gases at constant pressure & temperature,
$$\mu_j = H_j-TS_j = G_j$$
That would imply the mixture of one mol of $\text{H}_2$ has a total gibbs of the aformentioned $-38962\frac{\text{kJ}}{\text{mol}}$, but the mixture of two mols of $\text{H}$ (pressure neglected) would be $2\times(-33984\frac{\text{kJ}}{\text{mol}})=-67969\frac{\text{kJ}}{\text{mol}}$, and now supposedly I've found that hydrogen gas spontaneously decomposes at standard temperature. That's clearly wrong.
(I've even done the calculations for a reaction considering pressure, and the doubling of mols of mixture post-reaction isn't enough to drive it into non-spontaneity except at absurdly high pressures)
So my question is: do I misunderstand how the Gibbs works for mixture of gases, how to read values off of the NIST themo tables, or both?
Thank you!

I've had experience doing calculations like this, but I have no idea what you are doing.  Are you trying to determine the. equilibrium constant?

See this thread, and get back to me with any questions you might have:  https://chemistry.stackexchange.com/questions/141516/why-isnt-standard-gibbs-energy-zero-at-equilibrium/141537#141537

@ChetMiller I'm writing a numerical solver to find the equilibrium composition of a mixture of species at a set T, P. The numerical solver constrains the system to preserve element quantities using the method of lagrange multipliers and then seeks a root for dG/dn_j by Newton's method. It does this all perfectly... it's the same method that Gordon & McBride use for NASA's CEA. The only problem is that it seems to want to disassociate species, and I think it's because I'm either pulling data from the NIST datasheets incorrectly or I am wrong about the basics--how to sum up system Gibbs.

Perhaps I should reframe the problem. From NIST datasets, how would I calculate the Gibbs of 1 mol of H2 at 1 bar & 298.15 K? Then how would I calculate the Gibbs of 2 mols of H at 1 bar & 298.15 K?

Your calculation for H doesn’t look right to me.  The heat of formation should be 217999, not 217.999.

@ChetMiller I can't believe it, the issue was inconsistent units this whole time. Thank you! From limited testing, the reaction solver is operational now.


text:
How would you examine samples of rocks using a handheld spectrometer?
My friend has an old spectrometer set that his father used to own and we are trying to figure out how to use it. Asking his father about it is not an option unfortunately.
The box contains a handheld spectrometer. It is cylindrical in shape, length around 10 cm and diameter around 1 cm. There is a lens at both ends. We can't see through it.
In the box there are around 10 samples of different rocks. They are around the size of a golf ball but irregular in shape.
There is nothing else in the box. We presume there are components missing.
Is there some way to analyse these rocks with the spectroscope ?
Maybe I need to shine a laser or light on the rocks in some particular way, or grind a sample of the rock into dust and put it through a flame.
All suggestions welcome.

You may add a photo of the device to substantiate your question.  And don't shine with a laser on the samples in front of the spectrometer, your eyes may be damaged beyond repair by this.

You might have a gem's testing spectroscope. https://lotusgemology.com/index.php/library/articles/282-gem-testing-with-the-spectroscope-lotus-gemology


text:
Sign of work done by gas during isothermal expansion

$\pu{10 mol}$ of an ideal gas confined to a volume of $\pu{10 L}$ is released into atmosphere at $\pu{300 K}$ where the pressure is $\pu{1 bar}.$ The work done by the gas is:

Work done can be found as
$$W = -p_\mathrm{ext}(V_\mathrm{f} - V_\mathrm{i}).\tag{1}$$
$$V_\mathrm{f} = \frac{nRT}{p} = \frac{\pu{10 mol}\times\pu{0.083 L bar K^-1 mol^-1}\times\pu{300 K}}{\pu{1 bar}} = \pu{249 L}\tag{2},$$
$$W = -\pu{1 bar}\cdot(\pu{249 L} - \pu{10 L}) = \pu{-239 bar L}\tag{3}$$
As the question is asking us to find the work done by the gas, so shouldn't it be $\pu{+239 bar L},$ as work done by the gas is negative of the word done by external pressure?
But someone told me that the work done calculated is the work done by the gas and $\pu{-239 bar L}$ should be the answer. So, what mistake am I making and how do I determine if work done by the gas (and external pressure) is positive or negative under different conditions (expansion and compression)?
P.S.$\pu{-239 bar L}$ is the answer according to the answer key.

It is the usual confusion. During invention of thermodynamics within physics, it was defined Delta U = Q - W, i.e. positive work is work done by the system. Chemistry later came with the opposite convention, Delta U = Q + W, with positive work done on system. Currently, physics is abandoning historical convention, accepting the one from chemistry. 6th

Your answer for the work done by the gas is correct, and your friend and the answer key are wrong.

If we imagine our proverbial, frictionless piston containing some ideal gas, when $P_{surr} > P_{syst}$, the system is compressed by the piston and work is done on the system by the surroundings (i.e. work is positive). If the reverse is true ($P_{syst} > P_{surr}$), the gas in the piston will expand and work is done on the surroundings by the system (work is negative)
If you think of it in terms of internal energy, it might help. When work is done on the system, the internal energy increases for the ideal gas in the system (work is positive, internal energy increases). When the system does work on the surroundings (expansion), its ability to do work decreases (internal energy decreases, work is negative).
Hope this helps.

During compression of a gas , surroundings does positive work on the gas and during expansion of gas system does negative work on the surroundings.
Work done by the gas on surroundings against zero external pressure is zero in magnitude. Since external pressure can't be less than zero so work done by the gas can't be positive.


text:
Does silicone hard as plastic exist?
I saw rubber-like silicones. Are there types of silicone that can be used in injection molding and once the process is finished - become as rigid as, let's say PVC? At the same time they should have all(/majority) of silicone's properties (citing Wikipedia):

Low thermal conductivity
Low chemical reactivity
Low toxicity
Thermal stability (constancy of properties over a wide temperature range of −100 to 250 °C)
Does not support microbiological growth
Resistance to oxygen, ozone, and ultraviolet (UV) light
High gas permeability

If such silicones exist - what are their names/formulas?

Homework. Hint: What generally limits the application temperature range of polymer types?

The answer is unlikely to be as simple as you might think. Most properties of material involving polymers are determined by the additives and means of production which allow a very wide range of possible properties. While, for example, PVC is usually a rigid, stiff material, with the right plasticisers it can be made into a flexible, soft tube. The same is probably true for silicones.

If there are so many options - name at least/just one. I need something as rigid as a regular PVC pipe.