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How can 2 molecular orbitals form simultaneously?
I understand (for example in the case of H2) that two s orbitals produce through constructive interference a bonding orbital and that the out of phase combination of two s orbitals leads to destructive interference, followed by the apparition of an antibonding orbital, but, as far as I know, these two orbitals form at the same time (the number of orbitals is conserved, although only the lower energy one is occupied in the first instance). How can this be possible? Are specific regions of the two s orbitals in-phase and others out of phase so as to produce two different outcomes at the same time? Do same-phase regions interact selectively with other same-phase regions or does the s orbital coexist in two phases at the same time, thereby simultaneously producing two different bonding orbitals (having to do with superpositions). Please help me, I am really confused and similar questions contain answers that rely on mathematical proofs - I would like to grasp it conceptually: are the two phases of the s orbital simultaneously occurring?

Hello. I'm afraid that deleting and reposting your question doesn't make it any less of a duplicate of this. You want something "conceptual", but in fact if you read it, the previous OP already asked for something similar. If the answers there don't satisfy you, then you should seek a new answer to that question, not a new question.

I (quite substantially) expanded on my answer there, with an analogy which you might hopefully find more helpful. You might want to check it out again. That's as far as I can help, though; this doesn't affect the fact that this is still a duplicate, so I'm not inclined to reopen, unless there is a convincing argument (and I don't see one now).

I will try to answer without using any mathematical formula, as your question is just made of words, without any mathematical symbols.
You should know that the wave function of an electron in the $\ce{H}$ atom is normalized. It means that the integral of the wave function over the whole space is equal to $1$. And it must be equal to $1$.
If you add two of these wave functions from two $\ce{H}$ atoms, the integral of this sum will be $2$. This is forbidden. It must be $1$.
So you cannot simply add two normalized wave functions, like the $\mathrm{1s}$ in the $\ce{H}$ atom. You have to take one-half of the wave function of the first atom and to add it to one-half of the wave function of the second atom. This would produce a new molecular wave function whose integral is equal to $1$. It is nice. This new molecular wave function describes a sigma bond, designed as $\ce{H-H}$ in the Lewis model.
Now you have to use and combine the other half of these wave functions, without adding them. The only thing to do is to subtract them. But such a subtraction produces a point between the two nuclei where the obtained wave function is equal to zero. As a consequence, this molecular orbital is a antibonding orbital. And the integral of this antibonding is also equal to $1$.
Hopefully the pure theorists will pardon me for having simplified the description of the problem. I should have spoken of the factor $√2/2$ instead of $1/2$.

You basically answered exactly the same question as this. You might wish to consider answering there, instead of, or in addition to, this question.


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Nomenclature of Colors of precipitates and solutions
I have just started learning salt analysis in inorganic chemistry. In it i came across the names of colors of various precipitates and solutions. Why do they have prefixes such as "scarlet" in "scarlet red"? Why does it need to be called "red litharge" and not just "red", "Prussian blue" and not just "blue", "crimson yellow" and not just "yellow", "apple green" and not just "green"?
I wanted to know if there are any rules for naming such colors, or they are just named randomly.

Crimson yellow is something I don't think exists. And it's Prussian blue, not Persian blue. Properly phrased questions are more favourably seen. 2. This is a trivial question of how colours are designated, and does not have anything to do with actual chemistry. 3. Colour is not something scientific that should have a systematic nomenclature. Since many compounds appear red, making it a bit more specific as to which shade of red it is, helps, specially while reporting the experiment. It all depends on what shade you see in the test tube. There's no systematic naming.

@TRC Persian blue is also a color but most likely the author meant Prussian.

They are just descriptors. Unless one is affected by color vision problems, the typical examples in text books have been observed so many times to be reliable. Then what is pale yellow can be different between you and me.... But I don't understand the questions really. It is not really different in normal day life. An accurate description of "colours" is something else, as said in other comments and answers.

There are many shades of "red", so for specificity, prefix are used. "Scarlet-red" sound very specific.

It is an awful practice to describe the colors of precipitates with object-colors, like apple green, strawberry red, olive green, etc. I have seen these weird descriptions of hydrogen lines in its spectrum in first-year lab reports. You will not encounter this language in professional work. Avoid such usage while writing lab reports.
You can undoubtedly use simpler expressions without using an object like an apple or a strawberry, e.g., deep blue, bright yellow type expressions. There is no hard and fast rule as stated in the comments.
If you were to discover a new precipitate whose color is hard to describe, then a picture is worth a thousand words for a formal report (assuming the camera won't distort the colors).
Those who work in the textile industry and deal with fancy dyes use more sophisticated instruments to "describe" color. Color measurement science is a whole field of science, and there you do not need the crutches of such words. See, for example, this essay on Wikipedia Colorimetry

+1 For the hint to just take a photo, which in some publications became part of the documentation (e.g., on OrganicSynthesis, example, instructions to the authors revised in 2019). And put a reference for scale and color next to your object of interest, too, the cap of your ball pen isn't the best yet better than nothing (because it is standardized) since cameras capture withe and colors differently in daylight/flashlight, etc.

This goes too far. If I see apple green I note in my lab journal apple green. All the rest is true. But has nothing to do with the actual question. If my crude is olive, that is the best description. Just we know that is the pictorial description of a single operator. To make a wise use of that is up to us reading that journal.

source: Stack Exchange

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

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What does the following diagram represent?
Suppose, the following is a diagram of a protein's polypeptide chain:

What does this diagram represent?

What are the letters A, E, M, W, L, N, S, etc. represent?  (I suppose these are amino acid codes. Right?)
Which one is the c-alpha atom here?
What are the sequence numbers i-2, i-1, i, i+1, i+2, etc. represent? Why there are negative and positive increments from i?

The following is a related diagram:

Note the schematic is from https://chemistry.stackexchange.com/questions/153935/what-is-the-meaning-of-this-data-from-a-protein-data-bank-pdb-file

What is the origin of the diagram?

@BuckThorn, some book. i don't know the exact source.

Yes, those are 1-letter amino acid codes
The image just shows each amino acid as a single point, so it doesn't the individual atoms, including the C−α.
This may require knowing where this figure came from to be more specific, but they seem to be interested in the distance between amino acids before and after some particular point in the chain.

i don't know the exact source. all I know is that it's from a book.

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


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Do decomposition potentials change during electrolysis?
I was just wondering if decomposition potentials actually change during electrolysis. For example, if we reversed the Daniell cell reaction so that copper is oxidized, the half cell reactions would be:
$$\ce{Cu <=> Cu^2+ + 2e-} \qquad \pu{+0.34 V}$$
$$\ce{Zn^2+ + 2e- <=> Zn} \qquad \pu{-0.76 V}$$
if we assume both of the solutions are at $\pu{1 M}$ at $\pu{298.15 K}$, the decomposition voltage should be $\pu{1.1 V}$.
If we applied the voltage until half of the zinc cation solution reacted, we would have $\pu{1.5 M}$ $\ce{Cu^2+}$-solution and $\pu{0.5 M}$ $\ce{Zn^2+}$-solution, so that the resulting decomposition voltage should be higher:
$$E_\ce{Cu} = 0.34 + \pu{0.0295 V}\times \log (1.5) = \pu{0.345 V},$$
$$E_\ce{Zn} = -0.76 + \pu{0.0295 V}\times \log (0.5) = \pu{-0.769 V}$$
This should result in a decomposition voltage of $\pu{1.115 V}$ approx. Am I correct on that matter? I haven't found a single article that talks about the actual potential changes during electrolysis, even in my exam I only had to calculate the initial voltage, not what happens afterwards.

You can simply use Nernst equation for the cell reaction, $\ce{Zn + Cu^2+ <=> Zn^2+ + Cu}$ with $n = 2$.

Hah! I was just having the same question a few days ago. Your conclusion is correct.


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Water density in the calculation of the ionic product of water
In 2019 the International Association for the Properties of Water and Steam released this document for the semi-empirical calculation of the ionization constant of water: http://www.iapws.org/relguide/Ionization.pdf
My idea was to insert this formula in my spreadsheet for calculating the pH of an acid base mixture, where until now $K_w$ I entered it manually.
On the other hand, this formula, in addition to requiring the inlet temperature (between $\pu{0°C}$ and $\pu{800°C}$), also requires the density of the water (between $0$ and $\pu{1.25 g/cm3}$).
Since the density of water in turn depends on temperature and pressure, I wondered if in the literature a sufficiently compact formulation could be found that would give me the density of the water at the input temperature and pressure.

Before asking here of course I searched far and wide:

in MapleSoft here they show their procedure, which in turn refers to this other routine which is apparently secret;

in WolfraAlpha I found this widget which in turn refers to a long line of books in which I have not found anything that can do for me;

this is the only page where without many turns of words they show a simple algorithm, too bad it is only valid for temperatures between 0°C and 40°C, if it had been valid at least up to 100°C it would have been enough.

This is why I was wondering if anyone was aware of some more general formulation of the latter.

The problem with equations for a wide range is that they are either not  accurate enough either not simple enough.  There is always possible to involve numerical interpolation of tabelized values that are available, e.g. by a cubic spline.

Here is the scanned PDF on nvlpubs.nist.gov with the formula from 1975 for 0-150 deg C range.


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Meaning of $^2[\,1/2]$ terms in atomic spectroscopy
I was trying to analyze some atomic emission spectroscopy data and I encountered unfamiliar notation of the terms for me.
Commonly, the electronic states of the atoms can be described by quantum numbers using the term notation ("L-S coupling"). The terms are given as:  $^{2S+1}L_{J}$, where $S$ is the total spin quantum number, $J$ is the total angular momentum quantum number and $L$ is the total orbital quantum number in spectroscopic notation.
However, in this table there are also terms having notations without letters ($S$,$P$,...), like $^{2}[\,1/2]^\mathrm{o}$ (black box). What does that mean and why are there separate cases for $J = 0$ and $J = 1$?
[Edit] The data has been taken from NIST Atomic Spectral Database for Ne I (neutral neon in spectroscopic notation).
NIST database
The notation, asked in this posted has already been explained here Notation key by NIST

The notation of coupling schemes are explained here https://www.nist.gov/pml/atomic-spectroscopy-compendium-basic-ideas-notation-data-and-formulas/atomic-spectroscopy-2

I am not familiar with this type of coupling but it is called LK couping. The number in square brackets is K. It is always good to a give a reference when you paste an image.

I dont understand that notation. But my lecturer told me to study up L-S coupling so this kind of notation confused me.

But the transition looks like this? For example: I have a wavelength 352 nm so the transition is $^2[1/2]$ to $^2[1/2]^0$?

This notation is certainly not LS coupling. Ask your professor if he expects you to learn LK coupling.

It is probably, just not written with letter because L is not in the form of integer because the databese uses classic LS form for other elements so I guess the notation must remain the same

In addition to previous comments, you may find helpful to have a look at Herzberg's Atomic Spectra and Atomic Structure,

May also be helpful to read this original paper by Cowan and Andrew in 1965 "Coupling Considerations in Two-Electron Spectra" https://opg.optica.org/DirectPDFAccess/42879AAE-F5CE-452A-82787AE00DCEA0A0_52651/josa-55-5-502.pdf?da=1&id=52651&shib=578648&seq=0&mobile=yes


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What does it mean by spatial neighbor?
What does it mean by a spatial neighbor (SN) in the case of a protein residue or protein chain?
Why is it different or special as opposed to other neighbors?

A neighbor in the space? // Amino acids that are not neighbors in protein primary structure, can be spatial neighbors in secondary/tertiary/quarternary structure. // Just guess, due lack of  context.

There are chain neighbors and then there are spatial neighbors.

Proteins are long chains of amino acids joined together. Atoms or amino acid residues that are a few bonds away from each other can be considered "neighbours".
But the long chains of atoms/amino acids usually show complicated folding in 3D space. Indeed, the specific shape of that folding is critical for the protein to function. The implication of the folding in 3D is that residues or atoms that are far apart if we count bonds can be very close in 3D space. The last amino acid in the chain might be hundreds of bonds away from the residue at the other end of the chain, but might sit right next to it in 3D space. That would be a spatial neighbour.
The point is that, in complex 3D structures, we can count "neighbours" by either the number of bonds between them or by how close they are in space when the protein is folded.


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Adsorption on solids - homogeneous/heterogeneous surface definition
How could we define homogeneous and heterogeneous surfaces (rigorously)?
It is among my exam questions (but I could not find any definition in my textbook or on the internet).
I would say the homogeneous surface is a surface made of one type of molecule (opposite of the heterogeneous surface). But my knowledge about surfaces is very surface (and I am surprised it is listed as an exam question) so are there any other conditions or is it enough?

Homogeneous surface is not a thing at all.

Why? Not even in an ideal case?

We talk of homogeneous mixtures and reactions when there is one phase and no surfaces at all. Otherwise we use the word heterogeneous. That's all.

Yet it is literally one of the questions which my professor gave me

@IvanNeretin All depends on context.  We can talk about homogenous/heterogenous surface in context of the surface itself, not the whole system.  Like an object made of solid pure compound can have homogenous surface, but if made of a metallic alloy, it can have heterogenous surface, with properties location dependent.

Yeah, I guess we can.

Homogeneous surface is the surface where we can (in the context of the scenario, like the purpose and  required/available resolution and sensitivity) neglect any surface variability. Otherwise, the surface is heterogeneous.
It is important to say it is relative evaluation. For a set of surfaces, it can reach both extremes like "everything is homogeneous" to "everything is heterogeneous".
A visually homogeneous surface can be considered heterogeneous when evaluated by the enough zooming-in optical or electron microscope.
A visually heterogeneous surface can be considered homogeneous when evaluated from far enough or with low resolution or sensitivity for differences.


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Product of reaction between cyclohexene and bromine in methanol at 273 K
What product could we expect when we let cyclohexene react with an equivalent of bromine in methanol at 0 °C? Is it just the trans-addition of bromine (rac.) or also the SN1-susbtitution on a carbon due to MeO? Maybe twice? See below:

The first product (bromoether) is correct but the mechanism is wrong. Consequently,  the second and third structures do not form. Have you tried googling "bromohydrin formation"?

Thank you. I just looked it up, so the bromoaddition gives us an intermediate where meoh can attack, but I wasn’t sure if it’ll really deprotonate

The oxygen of MeOH attacks the intermediate bromonium ion from the rear (SN2) then the proton is lost. With no strong base, no prior deprotonation of MeOH occurs.

Note that at 273K MeOH is not a strong enough nucleophile to attack a secondary alkyl bromide so the third product, the di-ether, will not be formed.

It starts with the formation of brominium ion, as in direct bromination of alkenes. If you do not know the mechanism, you can refer to it here, at MasterOrganicChemistry.

But here, the $\ce{MeOH}$ can itself act as a potential nucleophile. It is similar to if $\ce{H2O}$ is used as a solvent - refer the formation of halohydrin here.
Since $\ce{MeOH}$ is available in larger concentrations, it more readily attacks the carbon. Markovnikov rule is followed for determining which carbon is attacked, but in this case, both are symmetrical.
The attack proceeds through $\mathrm{S_N2}$ mechanism, forming two enantiomers. After the $\ce{C-O}$ bond is formed, the $\ce{MeOH}$ loses $\ce{H+}$ to form ether and $\ce{HBr}$.


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Mechanism of acid-catalyzed ring opening of a cyclopropane ring
I have a question regarding following synthesis which apparently only needs a sour medium to occur:

What exactly happens as a mechanism? What I thought was: The alcohol group gets protonated and forms a stable carbocation at it's tertiary position. This charge can now wander to the tertiary carbon next to it; but an H-Shift should not be possible. What is the mechanism here?

Ring opens instead of shift.

Because of the carbocation in this position? Will it destabilize the ring because it is sp2?

While acids are indeed sour, the medium is acidic. ;)

I believe the mechanism should be somewhat like this (after the formation of carbocation) -

It would account for the product shown, but I am not sure if it's correct. I feel it's more favourable for a dehydration to occur with the hydrogen from the ring itself (shown on the dash at the top) to form this -

This double bond is conjugated and more substituted, so I should have thought that this would be the major product.

If my mechanism is wrong, please let me know.

Probably elimination is concerted with ring opening, so "less favorable" product is created.

Did the OP's question state that the observed product was the major product? No.

@Mithoron Thanks for the insight. Yes, if it's a single step then to form my proposed product it would need to start making a double bond before the ring was completely broken - which of course would be difficult owing to angle strain.

@user55119 Great point! It shows how reading things closely matters a lot :)


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Grignard-type reaction of 3-bromocyclohexene, or rather recombination?
I have a question regarding a proposed Grignard-type reaction which involves the conversion of 3-bromocyclohexene to 1-(cyclohex-2-en-1-yl)propan-1-ol, in diethylether. As the bromine atom lies in allyl position to the double bond, it should form the [1,1'-bi(cyclohexane)]-2,2'-diene, instead of the alcohol, shouldn't it? See below:

Allyl Magnesium Bromide is commercially available in ethereal solution https://www.fishersci.co.uk/shop/products/allylmagnesium-bromide-1m-solution-diethyl-ether-acroseal-acros-organics-2/p-3771057. It does all the usual Grignard reactions and does not dimerise. I see no reason why the cyclohexenyl analogue should be any different.

Some preparations of allyl magnesium bromide suggest lower temperatures while others suggest 30 mesh magnesium over the traditional magnesium turnings The 30 mesh increases the surface area of the Mg to maximize Grignard formation over coupling to form 1,5-hexadiene.

The dimerization under Grignard conditions is described here: https://www.sciencedirect.com/science/article/pii/S004040200700316X

I am completely ignorant as to why an allylic halide should be more susceptible to coupling with Grignard than a normal alkyl halide (which generally doesn't couple). Is it because the allylic halide is more reactive to substitution reactions?

Like so much organic chemistry the answer is it depends.
As I said in the comments you can buy Allyl Grignard in ether solution, but you can also dimerise it by allowing it to reach room temperature Organic Syntheses prep here.
Assuming the cyclohexenyl Grignard behaves similarly (and I think it will) then you can get the expected Grignard addition product with propanal if you add you bromide to the magnesium suspension slowly, keep the Grignard solution cold and add the aldehyde promptly (basically using standard conditions). Let reach rt and take a long lunch break before adding the aldehyde and you will get dimer.

My understanding is that dimerization occurs during Grignard formation. As bromide is added to the pot, it either reacts with already formed reagent to give dimer or with Mg to give more Grignard. Dimerization would be more prevalent later in the addition rather than in the early stages of addition. Your Org. Syn. link uses rapid addition of allyl chloride which, intentionally,  maximizes biallyl formation.  How does a long lunch break after the reagent is formed lead to dimer? ;)

I was under the impression that the dimerisation is a slow process that occurs at higher temperatures. You may be correct.

Dimerization of the Grignard reagent would require an oxidation.

There is no obvious oxidant present which leads me to ask is this actual dimerisation or nucleophilic displacement of Cl- by allyl anion?

Exactly! Not a true dimerization. Because the halide is a good electrophile, substitution competes with Grignard formation unless steps are taken to minimize substitution.

So likely to be somewhat less favoured in the cyclohexenyl case due to steric factors.

See my link in the Comment to the OP for the cyclohexenyl coupling.

Can only see the abstract, the rest is behind a paywall

I just came across this paper. Allegedly solves the coupling problem. https://www.tandfonline.com/doi/pdf/10.1080/00397919209409258?needAccess=true

@user55119  Paywalled again. I know Keith Jones too


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Synthesis of allylic ether, starting with phenalene
I stumbled across a retro-synthesis question, for which I don't know the exact answer; I prepared two synthesis ways, but I don't know if they could be realized in a lab. The question is the following:

Here are my two ways:

Are those ways feasible or should I take another approach?

The method of installing the allyl ether in your top method will not work. NaH + allyl bromide would be better.

Ah, so basically, the alcohol gets deprotonated, forms an alkoxide ion, and attacks the allyl bromide in an Sn2-reaction to form the ether?

Exactly so, NaH is easier to handle than sodium metal for creating the alkoxide

You could also epoxidise the double bond and do a reductive opening to give the same alcohol as in method 1. There are a number of methods for this, catalytic transfer hydrogenation with ammonium formate and Pd/C should work as described here:  https://www.researchgate.net/publication/232995596_Pd-Catalysed_Regiospecific_Reductive_Ring_Opening_of_Epoxides_and_Glycidic_Esters

In your second solution, the allylic alkoxide is likely to effect E2 elimination on the bromide. Go with @Waylander.

@Waylander is right. For the top method you need a stronger base and one hydroxyl group to deprotonate, plus another halide where the deprotonated alcohol can attack (hence his suggestion of allyl bromide). NaH should be good enough for deprotonation. In the second method (Williamson synthesis), alkoxide can act as a base and will give significant amounts of elimination product, so though the reaction works, the yield will be low.


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LEED - determining surface structures
How can one describe surface structures (before and after adsorption)? I understand the diffraction theory but I have never really tried to determine the structures of samples by myself.
LEED (low electron energy diffraction images of Pd(1,1,0) surface (clean and with adsorbed CO):

All I can see is a rectangular diffraction pattern (which is regular so we are dealing here with a single crystal) and that the pattern has changed a bit after the adsorption. Is there any other information I can get from the images? I searched online for some tutorials but all I could find were scientific articles and I really could not grasp it fully. I need to work out some examples.

EDITED:

An incomplete list of suggestions:

From the position of the spots, did you determine their mutual distances (e.g., to discern a square from a rectangle), and angles if you connect «the dots»?  Beside symmetry, this eases to relate between direct and reciprocal space.

What about the intensity of the spots (in either one of the frames) relative to each other, is there a regular (symmetry related) pattern?  Depending on the space group (in 3D crystallography), there are systematic extinctions, so I think 2D and 2.5D crystallography may have an equivalent to this, too.  For the clean substrate, match the spots the relative anticipated position (e.g., $d$-spacing) and intensity pattern?

Do you recognize a second, weaker pattern if you extend the time of exposure for the sample (substrate + deposit)?  (Again in analogy from 3D crystallography, where an additional set of diffraction spots may be observed in presence of a sample's superstructure).  If the experiment anyway is recorded with a CCD detector, this may be easier to recognize visually if intensity $I$ is not scaled linearly, but follows e.g., $\sqrt(I)$ (this pushes a bit the weaker signals relative to the stronger ones) and good color map (e.g., viridis, cross-reference) for visualization.  For this, a good correction vs. background and empty substrate is helpful.

I think it is a FCC structure, I finally found a nice database.

Nice answer @Buttonwood! We've had a bunch of LEED or diffraction-related questions unanswered on MMSE for quite a while, do you happen to have any insight on any of them? For example: https://mattermodeling.stackexchange.com/q/2380/5, https://mattermodeling.stackexchange.com/q/4459/5, https://mattermodeling.stackexchange.com/q/6501/5,
https://mattermodeling.stackexchange.com/q/6171/5 https://mattermodeling.stackexchange.com/q/3865/5, https://mattermodeling.stackexchange.com/q/6095/5, https://mattermodeling.stackexchange.com/q/6269/5,
https://mattermodeling.stackexchange.com/q/6574/5

@user1271772 While I recall passing answered questions which mention Vesta, CCDC Mercury, .cif files, etc., I wasn't aware MMSE would touch this much diffraction experiments.  Your comment expands nicely my perception of MMSE, that not only holds a tag x-ray-diffraction (like the one I knew from chemistry.se), but equally about (what small-molecule crystallography may consider as «on occasion, work on much smaller crystals») electron-diffraction.  I'm going to read these other questions, too.

Palladium is fcc so you are looking at the fcc(110) surface.
If the 3D lattice constant is $a$ then the (110) surface has a rectangular 2D lattice with lattice constants $a$ and $a/\sqrt{2}$.
From https://www.ibiblio.org/e-notes/Cryst/FCC.html
 click for larger
The longer direction in space is the shorter direction in reciprocal space as shown by LEED.
In the drawings below the added red spots could represent added features to the real and reciprocal lattices, perhaps CO adsorption on the Pd(110) surface. We don't know if they are on top of Pd atoms or in the bridge or hollow points and can't tell from normal LEED images.
A: fcc(111) surface
B: one explanation for the LEED image seen
C: another explanation for the LEED image seen
But I don't have those extra spots
With only one image at one beam energy, it's impossible to tell if those spots are missing due to interference at 76 eV only, or if they never appear at any energy.
If a spot appears, you know you have some structure with the periodicity it suggests, but the mere absence of a spot in one image does not by itself guarantee that there is no corresponding periodicity present.
The electrons pass through several atomic layers before attenuating and so the returning reflections can interfere constructively and destructively in complicated ways.
If you have many images at many energies and you never see the spots appearing in C but not B, then you can *tentatively conclude your structure is most likely B and probably not C, but without more advanced studies like IV-LEED (plots of spot intensity versus beam energy for a long series of closely spaced energies, fit by a theoretical calculation or surface XRD or electron holography, it's sometimes hard to differentiate definitively between possible solutions.
From Matter Modeling SE:

How do I know if the reconstruction in the picture is a 2x2 reconstruction? Highly Relevant!
Overview of how self-consistent dynamical low-energy electron diffraction simulations are performed
Simulated low energy electron diffraction (LEED) patterns
Have FDTD methods made inroads into dynamical simulation of electron and/or X-ray scattering by crystals?
Straightforward formalism to get four sets of 2D hexagonal lattice vectors of fcc(111) planes that I can also cite?

@uhoh  And I actually do have other images, less spots there, i Edited my question.

@Leif I see you r edit. Those are different beam energies for the sample before adding CO, right? The 76 eV image in the new group of images doesn't show the extra spots. But it's a beautiful demonstration of how each spot changes intensity differently as a function of energy. Very nice!

+1 I found this because you posted it in the "What's going on elsewhere" chat room. Since I advertised one of your questions on the other answer to this question, perhaps I can also ask if you could help on some of our related questions at MMSE? https://mattermodeling.stackexchange.com/q/4459/5  https://mattermodeling.stackexchange.com/q/6501/5
https://mattermodeling.stackexchange.com/q/6171/5 https://mattermodeling.stackexchange.com/q/3865/5 https://mattermodeling.stackexchange.com/q/6095/5 https://mattermodeling.stackexchange.com/q/6269/5
https://mattermodeling.stackexchange.com/q/6574/5

@user1271772 It seems that I need to wean myself off of Space SE and move over there, my "day job" MM-light so it would behoove me to dig in and start learning from that trove of treasure. I'll take a look at these. Thanks!


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How can we synthesize this cycloester, starting with methyl 4-phenylbutanoate?
I was wondering how to realize following synthesis:
I am struggling with this one; for sure, this has something to do with the benzyl position which is particularly reactive, and something nucleophilic which can close the ring and form the cycloester. But how to do this? Thanks in advance.

Step 1 - benzylic bromination with NBS, Benzoyl peroxide
Step 2 - displace bromine with cyanide using KCN/DMSO
Step 3 - ester hydrolysis to acid with LiOH/THF/water
Step 4 - acid reduction to alcohol with BH3
Step 5 - nitrile hydrolysis with KOH
Step 6 - lactone formation with cat. PTSA/toluene reflux.

An alternative approach to the proposal of @Waylander for the synthesis of lactone 7 from ester 1 depends on the efficacy of the Barton nitrite photolysis, 2$\rightarrow$3. Examples of this reaction are typically conducted with secondary nitrite esters in rigid, steroidal frameworks with few degrees of freedom, an advantage that the transformation 2$\rightarrow$3 lacks. The generation of an alkoxy radical (8$\rightarrow$9) allows for 1,6-H transfer. In this case, the C-H bond has the advantage of being benzylic. Hydrolysis of the oxime 3 affords the hydroxyketone 4 that may be in inconsequential equilibrium with its hemiketal 5. A Wittig reaction with (methoxymethylene)triphenylphosphane produces an enol ether that is a masked aldehyde group. Acid hydrolysis of the enol ether affords the hemiacetal 6, which can be readily oxidized by Jones reagent. In the event that a keto acid is produced upon oxidation, $\ce{NaBH4}$ reduction of the ketone and acid-catalyzed cyclization may be effected to form lactone 7.

Neat! Silver Carbonate/Celite will do the lactol to lactone selectively if there is any problem with Jones reagent

@Waylander: I couldn't find 2-->3 in Chem. Absts. Who knows? Another Tet. Lett. for that ambitious young chemist? ;)

if only I still had a lab

@Waylander: Unfortunately, what we are doing currently is our default position.


text:
Reaction of 1-Bromo-1-phenylethene with two equivalents of LDA
I was confronted with the reaction of 1-bromo-1-phenylethene with two equivalents of LDA, then the product is hydrolyzed.
As it is a bulky base, it will surely deprotonate and form the phenylacetylene, and form phenylacetyl-lithium in a second step, but what happens when it is hydrolyzed?
Will the product stay as it is or what happens?

The phenyllithium acetylide is a reasonably stable species (pKa mid 20s). Protonation will give phenylacetylene.


text:
How could we synthesize tert-butyl-alcohol, starting with just methane as only carbon source?
would it be possible to synthesize tert-butyl-alcohol with methane as the only carbon source? I mean, I almost figured it out, so basically methane is radically substituted by bromine under UV, then it is given to a suspension of Li in dimethyl ether, and in the last step, it is combined with CO2 (dry ice or by burning CH4) to yield the lithium propane-2,2-bis(olate), but I don't get the last step where we would get the tert-butyl-alcohol.

How would only adding water give the tert-butyl alcohol?

Carbon dioxide is also a carbon source in this scheme. Presumably the last step is to add more MeLi to the acetone, and something got skipped in the scheme.

If my memory serves me correctly, the addition of MeLi to lithium carboxylate to form the dilithioalkoxide, requires carboxylic acid salts of 6 carbons or greater. There is solubility problem. As an academic exercise, this example is fine. Follow @orthocresol’s lead. If you have access to Organic Reactions, you can read Margaret Jorgenson’s review here: https://onlinelibrary.wiley.com/doi/full/10.1002/0471264180.or018.01


text:
Reaction of 3-chloro-2,2-dimethylbutane with sodium ethanolate in EtOH, which mechanism?
When we let 3-chloro-2,2-dimethylbutane react with sodium ethanolate in EtOH, which reaction do we expect? SN2, E2, E1 or SN1?
I would say it's an SN2 reaction because ethanolate is not that strong of a base, but is that correct? A carbocation would also be stabilized insufficiently, because it's a secondary one, so either E2 or SN2 would be probable, but I don't know if [OEt]- is sufficiently basic to favor the E2 reaction. I would be glad over any tips on this question, thanks in advance.

How different are ethoxide and hydroxide in basicity? What are the pKa's of their conjugate acids?

This is a very hindered system so you are not going to get very much SN2. Neo-pentyl halides are very poor electrophiles and this is yet more hindered. E2 is the most likely event.


text:
Elucidating an unknown compound using 1H- and 13C-NMR spectral data
I am currently struggling with the following NMR spectrum for this compound:

I know for sure that we should multiply the integrals of the hydrogen atoms by four, and that the compound contains at least one ethoxy group. Furthermore, the molar mass of the compound is estimated to be $\pu{258g/mol}$. We tried our best, but could not find the solution. We also figured that we should multiply the $\ce{^{13}C}$-$\mathrm{NMR}$ by two, but aren't sure about it. Any tips would be helpful.

Did you ascertain the sample is uniform (only one molecule of interest), e.g., by chromatography? How many different signals you count for the 13C/for 1H NMR spectrum?  What does the chemical shift and multiplicity of the signals tell?  Instead of  «we should multiply the 13C-NMR by two», likely your intent was in lines of «there are (at least) six types of carbon atoms». Did you record an IR spectrum, e.g. to check for the presence of $\ce{-NH2}$ or $\ce{-OH}$?  Is the estimate of $\pu{258 g/mol}$ what aim for in your synthesis, or result of an analysis by MS?

Typically, you know the reagents used in a synthesis, and anticipate a transformation; this knowledge may render some of the characterizations to combine less important.

I'm no NMR specialist, and I'm not sure pending closing reason as HW is justified, but it appears that the question lacks details. Please add as much data as possible, such as sample preparation, experiment conditions, and, in case you also performed analysis with another instrumental method, provide supplementary data (FTIR/UV-Vis/HPLC/MS/ICP-AES/…).

I do not think the homework chose reason is justified either. Is personally prefer tabulated data though; especially in this case it'll also help us understand how you interpret the signals. Like when you write that there is an ethoxy group, it would be good to know how you have deduced that.

It was in my previous exam. I was given no other information than the two diagrams, and I was supposed to deduce the structure with just the information available, which I already gave. As 81% failed this question, the professor told us we were supposed to look for an ethoxy group, no other information

@冰淇淋 If you perceive the need to train how to resolve such puzzles, have a look at the spectroscopy section of the page's resources. The Silverstein book taps more into teaching theory and reference, the ones by Field more into the practical training to combine UV-Vis, MS, IR, NMR.  Depending on the level you want to apply them later on your own, the 2D NMR book by Field may go beyond your needs, which you might check with your instructor.

I appreciate Karsten Theis for his effort to solve this problem, but I have to say that I completely disagree with the structure he has come with as the answer (as a matter of fact, I came up with the same structure but it didn't fit the given $\ce{^1H}$-$\mathrm{NMR}$ spectrum; vide infra). Followings are my reasons:
As correctly pointed out by Karsten Theis, the resonance at about $\pu{174 ppm}$ of the $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum indicates a presence of an ester carbonyl carbon. The quartet at about $\pu{4.05 ppm}$ and triplet at about $\pu{1.15 ppm}$ in $\ce{^{1}H}$-$\mathrm{NMR}$ spectrum indicates a presence of $\ce{-O-CH2-CH3}$
group, as hinted by OP's teacher. The presence of $\ce{-O-CH2\!-}$ fragment is confirmed by the resonance at about $\pu{60.0 ppm}$ in $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum.
The second triplet at about $\pu{2.20 ppm}$ in $\ce{^{1}H}$-$\mathrm{NMR}$ spectrum also indicates a presence of $\ce{-CH2-C(=O)-}$ fragment. Thus, Karsten Theis' suggestion of the presence of $\ce{-CH2-CO-O-CH2-CH3}$ fragment is correct. But here is the problem. The proton signal of $\ce{-CH2-C(=O)-}$ fragment displays a triplet, not a doublet as Karsten Theis' suggested structure would have given in reality. Consequently, most possible fragment should have been $\ce{-CH2-CH2-CO-O-CH2-CH3}$ (total of 5 carbons).
Since the $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum shows only 6 carbon resonances and the molar mass of the compound is suggested to be about $\pu{258 g mol-1}$, the compound should have at least one plane of symmetry as suggested by OP. It is also noteworthy that the intensity of the third carbon from the right in $\ce{^{13}C}$-$\mathrm{NMR}$ spectrum is almost double compared to other possible $\ce{-CH2 \!-}$ carbon signals. Therefore, arguably, the compound should have at least two $\ce{-CH2-CH2-CO-O-CH2-CH3}$ fragments plus two $\ce{-CH2 \!-}$ parts, giving that $\ce{C14H26O4}$ molecular formula (molar mass of which come up as $\pu{258.36 g mol-1}$ that agrees with the given value). Thus the structure should be:

The molecule is diethyl decanedioate, with molecular formula of $\ce{C14H26O4}$  and molar mass of $\pu{258.36 g mol-1}$. To confirm the structure, let's look at the integrations given in $\ce{^{1}H}$-$\mathrm{NMR}$. The proton signal of $\ce{-CH2-C(=O)-}$ at around $\pu{2.2 ppm}$ is integrated as $1.00$, meaning one proton is equivalent to $0.25$ integration value. The integration value of the triplet and the multiplet coincided with it at the range of $\pu{1.1-1.3 ppm}$ is marked as $3.5$, thus is equivalent to  14 protons $(\frac{3.5}{0.25} = 14)$. Also, the integrations of multiplet at $\pu{1.55 ppm}$ and quartet at $\pu{4.05 ppm}$ are marked as $\approx 1$, and hence each of them is equivalent to  4 protons. Thus, the total proton count is $3 \times 4 + 14 = 26$ protons, which agrees with the suggested molecular formula.
Further, the ChemDraw stimulated $\ce{^{1}H}$- and $\ce{^{13}C}$-$\mathrm{NMR}$ spectra for the suggested molecule are shown below:

Both spectra are in good agreement with actual spectra, specifically see the overlapped carbon signals at around $\pu{29 ppm}$. However, the triplet at around $\pu{1.07 ppm}$ has moved downfield to $\pu{1.17 ppm}$ in real time spectrum, showing more deshielded nature, probably due to the presence of oxygen at $\beta$-position.

Edit:
As Buttonwood has suggested, it is always better to compare your data with experimentally recorded data. The SDBS Database, which provides the spectral data of compound in interest under SDBS 2666 has given following $\ce{^{1}H}$- and $\ce{^{13}C}$-$\mathrm{NMR}$ spectra to compare:

Both spectra match the given spectra of the compound.

Suggestion: While there are programs to predict NMR spectra (predict as in a more or less reasonable/educated guess) sometimes plot cleaner/simpler than eventually met in the lab, I think the additional comparison with experimentally recorded data were beneficial for this question, as well as a habit to adopt in general for the training of spectroscopic identification of molecules. Diethyl decanedioate/diethyl sebacate is filed as entry #2666 in the SDBS database, thankfully, with NMR recorded in $\ce{CDCl3}$, too.

@Buttonwood: Thanks for the suggestion and the database. I have edited my answer accordingly. Thanks again.

Taking the hint from your instructor and the 13-C signal at 175 ppm fits nicely to a $\ce{-CH2-CO-O-CH2-CH3}$ fragment, i.e an ester. The 13-C spectrum shows only 6 peaks and the molar mass is about 258 g/mol, so there is probably some symmetry in the molecule.
There is something that puzzles me. The signals around 1.2 ppm in the proton NMR integrate to 3.5, so after doubling (to explain the other peaks as $\ce{CH2}$ groups), we get 7. There is a $\ce{CH3}$ group already accounted for (from the ethoxy group), so there would be another two $\ce{CH2}$ groups (or even more $\ce{XH}$ groups). That means there are at least 6 different signals in the proton NMR, but just 6 different peaks in the 13-C NMR. With one coming from the carbonyl, that's a mismatch and we need another carbon. One of the peaks in the 13-C NMR is about twice as high, so maybe that's the solution to this puzzle.
A possible molecular formula would be $\ce{C14H26O4}$. That would mean all the signals in the proton NMR are "doubled up" because of symmetry, and all of the ones in the 13-C NMR as well, except for one that corresponds to four carbons.
I did not find a structure that satisfies all of these constraints. The closest I could find is this:

This has the correct number of 13-C peaks with pretty good matching of chemical shifts, and explains much of the proton NMR as well. Unfortunately, the molar mass is off by 2 (degree of unsaturation is 3 instead of 2), and one of the signals has an integral that is too low, and the signal at 2.2 ppm in the proton NMR would be a doublet instead of the observed triplet. This incorrect structure, however, does illustrate how having different degree of symmetry in different parts of the molecule can explain the number of signals found in the 13-C spectrum.


text:
How rate of absorption remains same throughout the process?
While reading surface chemistry, I got

Rate of adsorption is high at begining and then decreases till equilibrium is attained . On the other hand , rate of absorption remains same throughout the process .

Regarding adsorption I have no doubts as logically after some time there will be no surface area  for upcoming molecules.
But how the rate can be same in case of absorption ?
Just if we  take an example of cotton ball , it absorbs water and after some time being fully wet by water how further the rate of absorbing be same ?
Please help if I am lacking somewhere in the concept ...

Your presumption is reasonable, but not complete. The text states, "remains same throughout the process." In your example, when the cotton ball is fully saturated, the process ceases, so the text appears to be correct. You've described the endpoint, where equilibrium has been reached, after the process ends.
However, you might test absorption to see if it behaves according to that general rule:

Obtain some agar gel, phenolphthalein indicator, dilute $\ce{HCl}$ and dilute $\ce{NaOH}$ solution.
Dissolve the agar in warm water, mixed with a little phenolphthalein indicator and a drop of $\ce{HCl}$, and pour into a vertical test-tube, leaving a few cm empty at top. Wait for the gel to cool to room temperature.
Add a few ml of $\ce{NaOH}$ solution at the top.
Every minute (or every ten minutes, if this is slow), mark the lower limit of the pink color.

Does the absorption, and diffusion, proceed uniformly, or does it slow with time?


text:
Why can't we add ozone to the ozone layer?
So that it becomes the ozone layer?

Check how ozone is destroyed by free radical chain reactions by chlorine containing compounds. The agents which destroy ozone, keep on destroying more and more ozone molecules.

Suppose we solve this problem, then what? The real problem always is "where is the money". See, Earth is YUGE (about as big as Donald Trump's ego, only bigger). You'll need all the money in the world, and then some. Sad!

This is a chemist's version of "if you're homeless, just buy a house" meme.

@IvanNeretin the earth IS kinda YUGE, but the atmosphere is really just a thin little sleeve of gas separating us from the vacuum... It is a sham! We need a bigger wall!

Ozone is generated at the much faster rate ( by UV C ) that we could deploy artificially. (Using Hg UV lamp, you can sniff ozone yourself ).  And at the same time destroyed by compounds with significant Ozone depletion potential.

@StianYttervik Not so thin - it has mass of the 10 m thick water layer.

@Poutnik while the earth has a ~6.370.000 m thick layer of water, stone and iron. In comparison it is quite slim.

@StianYttervik In context of required  selective passage of EM radiation, it is not slim.

The ozone layer is stabilized and recovering on it's own.
There is not much reason TO add ozone to it, though I guess we could - at great expense either inject or create it in situ.
The reason it was a problem in the 80s was the widespread use and release of certain gases to the atmosphere. These gases had a terminating effect on the ozone radical mechanism, that is they end the chain reaction. The gases that were worst were refrigerator gases, used for all kinds of heat exchange pumping. Freon, for instance. It was subsequently capped and banned. It is now only used sparingly, it is to my knowledge in most developed countries illegal to use it IF a different type can suffice.
Following that the release of such gases are only a fraction of what they were and the half lifes of these gases in the atmosphere has been passed several times, and the ozone layer has stabilized and is recovering.
That being said, good old Freon was a fantastic heat exchange gas. My office still has a fridge from the 60s that just keeps on running.

I wasn't asking what caused the ozone layer to be what it is today, so I guess your answer caused confusion and got this question closed. Sigh.

@NealConroy or maybe you can edit your question to clarify your doubts. As it stands, your question is not clear, hence it's closed.

@NealConroy Questions get closed for the question itself.  Answers are being dealt independently. //  Always include enough context and background of your question to make the question clear,  preventing  misunderstanding and clarification. // When you ask, it is expected you have thoroughly searched and thought about the topic, providing explicit summary of partial answers/ideas/thoughts you have got until then. It is quite common requirement across all StackExchange sites.

I think my question is ambiguous, and that should matter the most. I don't know of how it can be interpreted another way.

@NealConroy Your question now contains 3 questions which are different from each other, and the title is different from the body. You are also adding one anecdotal and non-relevant  statement. If you cannot see how the question is ambiguous then it is hard to help you.

Chlorofluorocarbons are worth looking at. In summary, a chlorine free radical is constantly regenerated and constantly took out ozone molecules. That is why the ozone layer was depleting quickly.

Yes, I had more than 1 question. And somewhat attempting to answer a question.


text:
Enthalpy/Temperature change in a chemical reaction
I was wondering how exactly absolute temperature change in a chemical reaction could be calculated.
For example, consider this reaction:
3H₂+N₂→2NH₃, ΔH=-46kJ/mol (298.15K)
Assuming the given reaction goes to 100% completion (which it doesn't, it's the Fritz-Haber reaction, but this is a theroretical question), and is conducted at 298.15K and 1 atm of pressure, so that the enthalpy change is -46kJ/mol, how can we calculate the temperature change?
My thought was that we divide the reaction into two steps of thought; first we let three moles of hydrogen (at 298.15K) and one mole of nitrogen (also at 298.15K) react and form two moles of ammonia gas. The temperature stays at 298.15K, so the two moles of ammonia also are at 298.15K. (Is that correct?)
Then, in a second step (assuming we have a closed container and the released heat is again absorbed by the products) we just add 2 times 46kJ to the products,(correction: 1 times 46kJ) so this problem reduces to a simple heat capacity problem. So Cp(NH3)=35.06kJ/molK roughly, so the temperature of the 2 moles of NH3 should increase from 298.15K to 299.46K.
As I explained, I am absolutely unsure if we can even calculate it like that, so any help would really be appreciated, maybe other examples and sources, too.

This is correct if the reaction is carried out at constant pressure, except that, in step 2, you only add 1*46 kJ to the products.  And also, that's 46 kJ, not 46 J.

Ah thank you. That is because it is 46 kJ per mole of reaction right? Not mole of products

Yes, that is correct.


text:
Can an elementary reaction have zero order?
I am very confused right now about my approach to this question could somebody help me out with this. I reasoned my answer as given below
Let's assume if a complex reaction is of zero order. It implies its R.D.S. has order 0 ,but R.D.S. is an elementary reaction.
So, that particular elementary step of reaction is 0 order.(for the obvious reasons)
So if we take that particular (elementary) step as individual reaction
then it will be elementary and will have 0 order.
R.D.S. here implies Rate determining step
But many people whom I shared my doubt with, explained it on basis of Molecularity = Order Rule for elementary step,but it does not sound logical to me and also bit confusing as if we see pseudo first order reaction there molecularity is 2 but order is 1.

The concept of rate determining step is outdated, but unfortunately still used. However, if one would use its implications and the order of the RDS is in fact zero, all steps must be of order zero since the RDS is the slowest. Physically, this doesn't make sense.

Photolytic reactions $\ce{A ->[light]B}$ have typically molecularity 1 and order 0.

Aren't they first order in photons, the forgotten reagent? What happens at very low pressures? I encountered one zero order reaction that I attributed to product catalysis but never figured out the details.

The combustion of wax in a burning candle is zeroth-order. Has it anything to do with a rate determining step ?

The rate of burning is related to the diffusion of wax in the wick definitely rate determining. It is similar to turning the gas valve on your stove.

As far as I understand, the assumption that the RDS has a zero-order is false. The RDS has its order equal to its respective molecularity, however, in zero-order reactions, the rate-determining step will not have the initial reactants as parts of the elementary step. Thus, when you calculate the rate constant in terms of the initial reactants using the other steps in the reaction, you will get their orders to be zero. Another thing, zero-order reactions are ALWAYS complex reactions. There are no elementary zero-order reactions.

I presume RDS refers to rate-determining step. What does ALWAYS refer to, then?

When a zero order rate is observed it means that the reaction is not a single step reaction and either two mechanisms are involved, or a reactant involved in the rate determining step is being throttled, or possibly there is product catalysis. Such an event invites further investigation. Perhaps someone can tell me what the rate determining step has been replaced by?


text:
What is "tail" in case of secondary structure of protein?
What is a "tail" in the case of the secondary structure of a protein?
How does it influence the "protein assignment" task?

Look at a model of a protein and see for yourself.

Please provide context, otherwise the answer is guesswork.

«Tail» probably is just one word to describe both N-terminus (amino/ammonium group) and C-terminus (carboxylic acid/carboxylate) of a sequence of amino acids/proteins.
The shape of a sequence of amino acids is conformationally constrained depending on the specific amino acids covalently bound together.  Empirically, this is described by $\varphi$ and $\psi$ in the Ramanchandran plot.
If the sequence growths, its interaction with e.g., cytoplasma becomes more and more important; thus curling up is one form of energetic optimization.  In the form of $\beta$-sheets, coils, etc., some of the conformations eventually lock-in because of the formation of intra- and intermolecular hydrogen bonds.  Whenever such ordering interactions are less many, the conformational flexibility of the protein chain increases again, i.e. multiple slightly different folding patterns differ little in energy and thus are «accessible».  If the structure of a large protein is resolved by NMR spectroscopy, the description of these regions then is more «more fuzzy». This may happen for the termini.  It equally may happen well within the sequence if the consecution of amino acid is not prone to the formation of e.g., $\beta$-sheets.


text:
Chain transfer in polystyrene synthesis
while I was studying the many side reactions of radical polymerization, I stumbled across this source which illustrates the chain transfer side reaction in polystyrene synthesis:

Of course, there is a typo in this picture, it shouldn’t be -CH2-H- but -CH-H-, but you get the idea. A monomer radical can abstract a hydrogen atom from an already existing chain in the first step, so that it’s radical function is forwarded to the chain.
The chain then, in a second step attacks another styrene molecule from the back side of the double bond so that the more stable, secondary radical forms, so the molecule can also form branches.
Everything alright with me, but I don’t get the regiochemistry here. In the first step,a secondary carbon atom is attacked, which also forms a secondary radical intermediate; but wouldn’t an attack on the tertiary -CH group be more favorable thermodynamically as it is a tertiary radical intermediate which is also stabilized by the phenyl group? Or is it a kinetically controlled process, where more hydrogens can be subtracted on the bridging carbon atoms, so that after the chain transfer, the bridging carbons are tertiary as a result?
Any tips would really be appreciated!

Not sure but steric hindrance might play a role. Because if we attack on CH to form tertiary radical it will be having two phenyl group that might be unstable.

So bascially, if it comes to hindrance, that would indicate a kintetically controlled process, because it is more likely to attack on the bridge, and form sterically stable bridges, if I am not mistaken.

See if we talk about intermediate that intermediate will be more stable in ur case like removing H from tertiary carbon containing phenyl group but as final product it won't be much stable to have two phenyl group on single carbon. So intermediate is stable but product is not so I am not sure whether it will come under TCP or KCP.

The image shown in the question can be found with some context at https://www.open.edu/openlearn/science-maths-technology/science/chemistry/introduction-polymers/content-section-4.3.3
In that document, they say of chain transfer:

A similar mechanism accounts for the side branches in LDPE where it is a more important mode of termination than in polystyrene

This indicates that chain transfer is a rare event in a typical synthesis of polystyrene.

[...] but wouldn’t an attack on the tertiary -CH group be more favorable thermodynamically as it is a tertiary radical intermediate which is also stabilized by the phenyl group?

As this is a rare event, you don't expect any of the intermediates to be the most favorable ones.

Or is it a kinetically controlled process [...]

If the chain transfer is rare, you would expect faster processes to react with the radical intermediate faster than it forms (because these steps are less rare). This means it is kinetically controlled.
Moreover, all the steps with two radicals combining to form a bond would be expected to be kinetically controlled because once formed, the covalent bond would not fall apart into radicals (that is why you need initiator molecules in the first place).
References

Synthesis of Polystyrene and Molecular Weight Determination by 1H NMR End-Group Analysis, Jay Wm. Wackerly and James F. Dunn, J. Chem. Educ. 2017, 94, 11, 1790–1793
Preparation and Properties of Branched Polystyrene through Radical Suspension Polymerization, Wenyan Huang et al., Polymers (Basel). 2017 Jan; 9(1): 14
Self-Branching in the Polymerization of Styrene
J. C. Bevington, G. M. Guzman and H. W. Melville
Proceedings of the Royal Society of London. Series A, Mathematical and Physical Sciences
Vol. 221, No. 1147 (Feb. 9, 1954),


text:
Which atom is the smallest atom?
Is hydrogen or helium the smallest atom? My teacher said that the smallest atom is the helium atom, but I think that the smallest atom is the hydrogen atom. It has a single electron and a single proton, whereas the helium atom has a pair of electrons, a pair of neutrons, and a pair of protons, which obviously occupy more space than the hydrogen atom.
My question is, which atom is the smallest atom of all?

There is generally applicable and mostly successful way called searching, usually providing bigger number of  more comprehensive answers in shorter time than interactive asking. The latter way is recommended only if the former one failed, as it spends resources of others by answering, instead of your own by searching.

SE and Q/A sites are very small part of internet. I do not discourage asking, I discourage asking without proper searching. // When you ask, it is expected you have thoroughly searched and thought about the topic, providing explicit summary of partial answers/ideas/thoughts you have got until then. It is quite common requirement across all StackExchange sites. // Resources are wasted, when answering what is already written on multiple places. // Atomic radius + Data page

Be aware the proper terms to search do not always match your question, see e.g. google: atomic radius data page

@Poutnik The data page has two answers depending whether you look at the empirical radius or the theoretical radius. I was surprised that hydrogen is not even the second smallest accourding to theory.

@huzaifaabedeen It is quite well possible, evaluating electron probability densities. OTOH, the Bohr model does not work for multielectron atoms.

The behaviour of electrons in atoms is not well describes by the classical points and orbits model of the atom. a more accurate modle for electrons is using quantum physics, in the is system the electron aren't at a specific spot, instead they are described by a quantity called the wave function which gives the probability of the electron being in a specific position.
from statistics you can then  work out the average position (the expectation value), this is the product of the position and how likely it is to be there (the probability)
if you do this for hydrogen and helium you get the figure below, where the solid lines are the probability distribution and the dashed lines are the average radius, which show that on average the electrons in helium are closer to the nucleus than in hydrogen.

hopefully that helps
Also if you want to break from the periodic table, there are atoms where the electrons are replaced by their subatomic cousins, muons. muons are heavier so are a lot closer to the nucleus so the atoms are a lot smaller, they are called Muonic atoms

I think something went wrong with the integration: the (average) theoretical radius of Helium should be larger than that of He+ (0.500).

The volumes of the nuclei are negligible compared to the atomic radii (like a "pea in a soccer/football/cricket stadium").
The higher the nuclear charge, the closer the electrons to the nucleus in a given shell.

Source: https://www.mdpi.com/1422-0067/3/2/87/htm

Helium has the smallest atomic radius.

@huzaifaabedeen No, electrons are considered to be point particles, with no internal structure we know of. Protons are made of quarks, etc., and are extremely tiny, but not point particles.

Maybe I should edit to say space visited by the electrons. @EdV

No problem: I was just trying to head off the OP’s notion that electrons were bigger than protons and neutrons. I upvoted, by the way!

Just the fact that a small bird is flying in larger area than an elegant is walking does not mean a bird is bigger than an elefant.

Please add the link to the source of the image.

@KarstenTheis I've initially upvoted your comment but "space visited by the electrons" is too orbit-oriented. Actually we shall speak of orbitals etc although it can go to far for OP. I would stick to the idea that the key is in the atomic radius. I also note that the question is someone surprising from a logical point of view.

@alchimista How would you say it, given the level of the question?

@huzaifaabedeen There is a quick-ish and exact answer compare hydrogen and the helium +1 cation using quantum mechanics, but the picture from the textbook suggests that the textbook is not up to quantum mechanics yet.

@KarstenTheis Good question.... But your answer now gives at least the facts.. Combining this answer with that of Nyra, the OP should get the picture.

Rutherford's gold foil experiment taught us that the protons and neutrons fill only a tiny portion of the atom, which we call the nucleus. The size of the atom is therefore only defined by the electron shells (orbitals), which in turn depend on the nuclear charge.
Your intuition about more electrons leading to a larger atom is correct, but more protons actually lead to a smaller atom. Which effect dominates depends on whether or not an additional electron finds a place in an existing electron shell.
To gain some intuition as to how this works, it may be helpful to do a Gedankenexperiment wherein we build atoms by adding protons and electrons one at a time, neglecting the neutrons which are not important here.

Imagine you start with one proton and one electron; this is the $\ce{H}$ atom. Let's say its radius is 1 (the Bohr radius).

Add one proton to the nucleus to obtain the $\ce{He+}$ ion. Because the central charge of the nucleus has increased by a factor of two, but the number of electrons hasn't changed, the radius of this ion is halved compared to $\ce{H}$, i.e. 0.500.

Now add an electron to this system to obtain neutral $\ce{He}$. This second electron joins the first one in the 1s shell, expanding the atom only a bit to a radius of 0.567. This is just a sliver larger than $\ce{He+}$ and considerably smaller than $\ce{H}$.

If you were to move on to $\ce{Li+}$, the additional proton would contract the radius to about 2/3 of $\ce{He}$, but when you add a third  electron to obtain $\ce{Li}$, it spawns a new 2s shell at a radius of 2.556, making the $\ce{Li}$ atom much larger than $\ce{H}.$

You can in principle continue this procedure, whereby each additional proton contracts the radius, and each additional electron expands it slightly. Whenever an electron shell is filled, adding another electron will cause a step-change in the atomic radius as can be seen in the jagged graph in one of the answers above.
Helium turns out to be the smallest atom on that graph, which is one of the reasons it was discovered in the sun before it was discovered on earth.


text:
Approximation for pH calculation with diluted acids in weak-strong titration curve
I was wondering how one could calculate small pH changes in titrations with dilute acids.
As we know, usually we approximate the pH by saying that the conversion is quantitative. For example, when we titrate i.e 100ml 0.1M sodium acetate solution with hydrchloric acid, the start point would be just pOH=0.5(pKb-log(c)), with pKb=9.25 we get 5.125. So pH=14-5.125=8.875.
If we add 25ml of 0.1M HCl we say that HCl consumes acetate directly:
CH₃COO⁻ + HCl → CH₃COOH + Cl⁻
So n(HCl)=0.0025mol consumes equal amounts of acetate and forms 0.0025mol of acetic acid. 0.0075 moles of acetate remain. We get 125ml of solution, so c(Ac-)=0.06M and c(HAc)=0.02M, so Hendersson-Hasselbalch gives us pH=pKs+log(0.06M/0.02M)=4.75+log(3)=5.227.
That in itself makes sense, but it doesn't if we, for example, only titrate with 10⁻⁶M HCl? If we added 0.5L of that, according to this approximation we would have n(HAc)=0.5*10⁻⁶mol and n(Ac-)=0.0099995mol both in 0.6L of solution, and subsequently a higher pH value than if we just let the acetate dissociate in pure water. So how can we replace this approximation in this, admittedly, extreme case so that the titration curve can predict the real result?

In such cases, you cannot afford to involve implicit simplifications.
In this particular case, the equilibrium ratio $\frac {[A-]}{[\ce{HA}]}$ is not the same as the nominal $\frac {[A-]}{[\ce{HA}]}$ ratio, but reaching acido-basic equilibrium is to be considered. One has to count also with water auto-ionization equilibrium.
If the added  $\ce{HCl}$ converted the full equivalent amount of acetate to acetic acid, $\ce{pH}$ would not change. A small portion of acetate will not convert to acid and the respective portion of remaining $\ce{H+}$ will add itself to acidity of the solution, all together honoring the acidity constant.
Full analysis, involving mass+charge balance/conservation and ongoing equilibria is to be evaluated and only those simplifications that are justified can be accepted.(1)

Charge balance: $\ce{[H+]} + \ce{[Na+]}= \ce{[OH-]} + \ce{[A-]} +\ce{[Cl-]}$
Weak acid mass balance : $c = \ce{[HA]} + \ce{[A-]}$
Weak acid mass balance 2 : $c = \ce{[Na+]}$
Acidity constant: $K_\mathrm{a} = \frac{\ce{[H+][A-]}}{\ce{[HA]}}$
Water auto-ionization: $K_\mathrm{w} = \ce{[H+][OH-]}$

By solving 5 equations with 5 variables, we get $\ce{[H+]}$. Justified simplifications are always welcome.
Substituting  for $\ce{[OH-]}$ and $\ce{[Na+]}$:

Charge balance: $\ce{[H+]} + c = \frac {K_\mathrm{w}}{\ce{[H+]}} + \ce{[A-]} +\ce{[Cl-]}$
Weak acid mass balance : $c = \ce{[HA]} + \ce{[A-]}$
Acidity constant: $K_\mathrm{a} = \frac{\ce{[H+][A-]}}{\ce{[HA]}}$

Substitution with last 2 equations:

Charge balance: $\ce{[H+]} + c = \frac {K_\mathrm{w}}{\ce{[H+]}} + \ce{[A-]} +\ce{[Cl-]}$
Weak acid mass balance : $c = \frac{\ce{[H+][A-]}}{K_\mathrm{a}} + \ce{[A-]} = \ce{[A-]} \cdot (\frac{\ce{[H+]}}{K_\mathrm{a}} + 1) \implies \ce{[A-]} = \frac {c}{\frac{\ce{[H+]}}{K_\mathrm{a}}+1} = \frac {c \cdot K_\mathrm{a}}{\ce{[H+] + K_\mathrm{a}}}$

and finally:

Charge balance: $\ce{[H+]} + c = \frac {K_\mathrm{w}}{\ce{[H+]}} + \frac {c \cdot K_\mathrm{a}}{\ce{[H+] + K_\mathrm{a}}} +\ce{[Cl-]}$

Multiplied by denominators:
$\left(\ce{[H+]} + c \right)\ce{[H+]} \cdot \left(\ce{[H+]} + K_\mathrm{a}\right) = K_\mathrm{w}\left(\ce{[H+]} + K_\mathrm{a}\right) + c \cdot K_\mathrm{a} \cdot \ce{[H+]} +\ce{[Cl-]}\ce{[H+]} \cdot \left(\ce{[H+]} + K_\mathrm{a}\right)$
Separated by the power order:
${\ce{[H+]}}^{3} + {\ce{[H+]}}^{2} \cdot (c + K_\mathrm{a} - \ce{[Cl-]}) - \ce{[H+]} \cdot ( K_\mathrm{w} + \ce{[Cl-]} \cdot K_\mathrm{a}) - K_\mathrm{w} \cdot K_\mathrm{a} = 0$

(1)- The classical related case is
$\mathrm{pH} = \frac 12 ( \mathrm{p}K_\mathrm{a} - \log{c} )$, implying $c \gg [\ce{H+}] \gg [\ce{OH-}]$


text:
Displacement of salts due to higher solubility
I was wondering why mixtures of salts in the same solution precipitate excactly in order of their solubilities. Imagine a solution with NaCl, NaBr, NaI in equal moles. If AgNO₃ solution is now added in excess, first silver iodide, then bromide, then chloride will precipitate out of solution in that order due to their different solubilities (Ksp=1.8x10⁻¹⁰, 5x10⁻¹³, 8.3x10⁻¹⁷ respectively).
First of all, what is the thermodynamic background of this phenomenon? It's obvious that for example the iodide is less soluble than let's say the silver chloride, but under the given conditions, all of them are basically insoluble, so I don't really get why the silver iodide selectively precipitates first. Secondly, would it matter if the sodium salts wouldn't be there in equal moles?
And on another note; would it be possible to dissolve silver chloride just by adding another, more insoluble salt? For example:
$$\ce{AgCl(s) + NaBr(aq) -> AgBr(s) + NaCl(aq)}$$
A saturated solution of AgCl with already precipitated, solid AgCl is treated with sodium bromide (NaBr) salt . Given the explanation above, would it be possible that the silver(I) ions form a precipitate with Br- ions giving NaBr, and let the AgCl precipitate dissolve and form NaCl? Thanks in advance.

For better site experience, you can find useful How can I format math expressions or chem. equations on Chemistry SE.

You can write $\pu{1.23E-12}$ to get $\pu{1.23E-12}$, or $\ce{H2SO4}$ to get $\ce{H2SO4}$

https://chemistry.stackexchange.com/questions/124923/calculating-equilibrium-concentrations-of-silver-and-halide-ions?rq=1

would it be possible to dissolve silver chloride just by adding another, more insoluble salt? For example:
$$\ce{AgCl(s) + NaBr(aq) -> AgBr(s) + NaCl(aq)}$$

It is possible to get the chloride ion back into solution by adding NaBr. The effect is to lower the silver ion concentration so drastically that chloride goes back into solution. This kind of process is used in analytical chemistry to get ions of insoluble salts into solution. Here is a related example:
$$\ce{Mg(OH)2 (s) + 2HCl(aq) -> Mg^2+ + 2H2O + 2Cl-(aq)}$$
In this case, you lower the hydroxide concentration by neutralizing it with hydrochloric acid.

Imagine a solution with NaCl, NaBr, NaI in equal moles. If AgNO₃ solution is now added in excess, first silver iodide, then bromide, then chloride will precipitate out of solution in that order due to their different solubilities (Ksp=1.8x10⁻¹⁰, 5x10⁻¹³, 8.3x10⁻¹⁷ respectively).

If you add the silver nitrate dropwise and wait a bit inbetween, you will observe solid silver iodide after the first drop. You should not make any statement about what forms first (question of kinetics) but rather what is present at equilibrium. Once the iodide concentration is 360-fold lower than the bromide concentration, you will also form a silver bromide precipitate when adding more silver ions. When most of the bromide is precipitated, silver chloride precipitate will also be observed.
So in that sense, they form stable precipitate in that order, with barely noticable overlap.


text:
Photochemical cycloaddition of ethene
I've got a question regarding the (2pi+2pi)-photocycloaddition of ethene to form cyclobutane. If I understood correctly, the reaction doesn't take place under normal conditions, even if we increase the temperature because the entropy term of gibbs energy will favor more molecules at higher temperatures, so we need to use energy in form of UV light.
Additionally, as the molecular orbitals don't have matching symmetries, one of the ethene molecules needs to be excited by UV light, so one electron can enter the pi* orbital. So basically the HOMO of excited ethene A (pi*) and the LUMO of ethene B (also pi*)can form two sigma bonds needed for the cyclobutane, in addition to the already existing two sigma bonds, at the same time (concerted mechanism).
But why is it like this? Why can one single excited electron form (or basically induce) the formation of two sigma bonds only through orbitals (even though the LUMO is empty and the HOMO only has one electron)? And why is only one ethene excited and not two at the same time? Any tips would be of great help, thanks in advance.

Have you heard of the Woodward–Hoffmann rules? The explanation you suggest (about entropy / temperature) is a little bit superficial, to be honest; I'd be a bit skeptical of the source of that information.

Not yet, I just started taking organic chemistry 2, but I'm sure it will be covered sometime in the semester :D

OK, I'll try to answer in an appropriate way, and hopefully it will make more sense when you cover these things in the near future.

For any pericyclic reaction (including cycloadditions), the key set of rules are the Woodward–Hoffmann rules. The rules are actually quite general, but people often first learn about them in the specific context of cycloadditions*:

Cycloadditions involving $4n$ π-electrons ($n$ being an integer) proceed photochemically.
Cycloadditions involving $4n + 2$ π-electrons proceed thermally.

The [2+2] cycloaddition has four π-electrons and so falls into the first category, i.e. photochemical. On the other hand, the Diels–Alder reaction is a [4+2] cycloaddition and falls into the second category, i.e. thermal.
The full rules are way too long to explain here, so I'd suggest to look up a good textbook on it if you're interested. Clayden's Organic Chemistry is a good starting place, IMO, and if you want something more in-depth there's Fleming's Molecular Orbitals and Organic Chemistry Reactions.
Note that there is also Fukui's frontier molecular orbital theory, which in fact more closely matches the HOMO/LUMO-based analysis in your post. However, this generally sees less use than the W–H rules. You can already see that in the length of the respective Wikipedia articles. I think the Woodward–Hoffmann rules are perhaps a bit more complicated, but are more easily applied and generalised.

If I understood correctly, the reaction doesn't take place under normal conditions, even if we increase the temperature because the entropy term of gibbs energy will favor more molecules at higher temperatures, so we need to use energy in form of UV light.

I mentioned this in a comment, but this explanation is a little bit superficial. The key consideration isn't entropy, because if that were the case, then no cycloadditions should ever happen, not even thermal ones like the Diels–Alder reaction. (Basically, you can't selectively apply this argument to [2+2] cycloadditions and then suddenly forget it when moving to [4+2] cycloadditions.)

Why can one single excited electron form (or basically induce) the formation of two sigma bonds only through orbitals (even though the LUMO is empty and the HOMO only has one electron)?

The part of the system that reacts isn't just the single excited electron (that would just be a plain old radical reaction). Pericyclic reactions are quite special: they involve the cyclic movement of quite a number of electrons. So, in fact, it is not the single excited electron that reacts: it is the four electrons, three unexcited and one excited, which collectively react.
So, the followup question is: why does the "3 unexcited + 1 excited" configuration react (I'll abbreviate that to 3u1e), whereas the "4 unexcited + 0 excited" configuration (4u0e) doesn't react?
To truly understand this requires one to delve into MO theory. In particular, you need to understand molecular symmetry and electronic states: it's not the individual electron's states which matter (i.e. which electron is in which orbital†) but rather the total state of the molecule (loosely speaking, the product of all the orbitals). But I will try to give a flavour of the explanation here.
The first part of this explanation is given in the form of correlation diagrams, and you may likely find it in the textbooks mentioned above. Essentially, what it says is this: the electronic state of the reactants must "map to", or "transform into", the electronic state of the products.
There are many possible states for the reactants: for example, we have 4u0e, 3u1e, 2u2e, ..., and likewise, the electronic states for the product are similar: we have 4u0e, 3u1e, ...‡ It turns out that, based on symmetry arguments, the 4u0e reactant configuration maps to the 2u2e product configuration, which is very much uphill in terms of energy.
On the other hand, the 3u1e reactant configuration maps to the 3u1e product configuration. Both are high in energy compared to the ground state, but relatively speaking, they are fairly close, so the activation energy for the forward reaction is not too high, and the reaction can proceed. The tradeoff is that you need to start from an excited state of the reactant, and that's where the light comes in.
If you want to go even deeper, then there is this question from a while ago which I answered: Theoretical basis behind orbital correlation diagram for pericyclic reactions. It has diagrams, symmetry labels, etc. and also introduces the non-crossing rule, which doesn't change the conclusion but makes the discussion more technically accurate.

And why is only one ethene excited and not two at the same time?

The simple answer is that you don't need to excite two of them at the same time. Basically, the combination of 3 unexcited + 1 excited is sufficient for the reaction to proceed; you don't need to go to 2 unexcited + 2 excited. (Indeed, the reaction might not even be possible with that combination, for similar reasons as why the 4 unexcited + 0 excited combination doesn't work; I'm not sure, though.)
A secondary answer is that only a fraction of the molecules are excited at any one point in time. Getting two of the excited molecules to collide and react before they return to the ground state, is not likely.

Footnotes
* Technically speaking: cycloadditions where both components react in a suprafacial sense. But let's not go there (for now). Refer to the textbooks above for more info.
† Electrons are indistinguishable, so it's not possible to say which electron is in which orbital. But that's more of a technicality in the present context.
‡ These can be enumerated and labelled more precisely using symmetry labels.


text:
Mixing hydrogen peroxide and sodium hydroxide doesn't lead to sodium peroxide anymore
I've been using same bottle of hydrogen peroxide and same bottle of 50 % of sodium hydroxide to make sodium peroxide for a month now. You could see the presence of sodium peroxide in the solution because it turned yellow, but it won't turn yellow now and I don't know why, but hydrogen peroxide is not efficient oxidizer for my current use. Where is the problem?

If the result is suddenly different the conditions are suddenly different. We should ask you what is different and not vice versa. Perhaps slow decay of H2O2, or absorption of CO2 by hydroxide.

Poutnik is right, what likely happened is one or both of the reagents lost its potency through chemical reactions to which they are susceptible.
In the case of the sodium hydroxide, if it reacted with carbon dioxide from the air (it is best stored and handled under a CO2-free atmosphere), the resulting carbonate is no longer a strong enough base to react with the hydrogen peroxide (this is an acid-base reaction, not a redox one). If the sodium hydroxide is/was as strong as 50%, you may be able to see a sign of this contamination: sodium carbonate is not as water-soluble and might be seen as cloudiness or a white precipitate.
If the hydrogen peroxide decomposed, that is that as far as forming sodium peroxide goes. Decomposition of hydrogen peroxide, unlike carbonation of sodium hydroxide, cannot be prevented by using an inert atmosphere, and it has no visible effect on the solution until, as you describe, it's called on to perform. If there was no apparent outward change in your reagents, I would therefore suspect hydrogen peroxide decomposition first.


text:
Why doesn't NaCl work for electrolysis?
So I found out online that to increase the conductivity of water for electrolysis, NaCl or NaOH is used.
However, upon using a low amount of NaCl in water and passing 12V through the solution, I noticed something weird, the cathode was bubbling a lot more than the anode (as expected in electrolysis - Hydrogen)  but after 20 mins, there was no collection of gas on the anode. Why could this be?
When repeating the experiment with NaOH, the desired result was obtained with hydrogen and oxygen on the cathode and anode respectively.

What was it that you expected, really?

Important info is missing: Electrode materials!

@Ivan Neretin I expect to see oxygen on the anode, which was not happening

@M. Farooq I used stainless steel

Steel could be reacting with chlorine. Did you see it corroded?

Yes the anode was a bit corroded

After running the setup for sometime with NaCl, the solution had turned a slight yellow

Oxygen? Why not chlorine?

At the cathode, $\ce{H2O}$ is decomposed according to $$\ce{2 H2O + 2 e- -> H2 + 2 OH-}$$ So some $\ce{H2}$ gas is produced, and the solution becomes basic. At the anode, two reactions occur simultaneously and independently : $$\ce{2 Cl- -> Cl2 + 2e-}$$ $$\ce{Fe -> Fe^{2+} + 2 e-}$$ So the anode is corroded. Now the chlorine $\ce{Cl2}$ produced by the first reaction is a gas relatively soluble in water, which reacts with the ions $\ce{Fe^{2+}}$ produced by the second reaction, according to :$$\ce{2 Fe^{2+} + Cl2 -> 2 Fe^{3+} + 2 Cl^-}$$ The ion $\ce{Fe^{3+}}$ is usually hydrated in aqueous solution taking a orange-brown color due to the ion $\ce{[Fe(OH)]^{2+}}$ produced by the reaction $$\ce{Fe^{3+} + 2 H2O -> [Fe(OH)]^{2+} + H3O^+}$$ And the solution becomes brownish and acidic around the anode


text:
Understanding weak acid-strong base titration
Today I stumbled across a website which defined the titration of weak acid and strong base as the following:
"The titration of a weak acid with a strong base involves the direct transfer of protons from the weak acid to the hydoxide ion. The reaction of the weak acid, acetic acid, with a strong base, NaOH, can be seen below. In the reaction the acid and base react in a one to one ratio." (https://chem.libretexts.org/Ancillary_Materials/Demos_Techniques_and_Experiments/General_Lab_Techniques/Titration/Titration_of_a_Weak_Acid_with_a_Strong_Base)
CH₃COOH + OH⁻→ CH₃COO⁻ + H₂O
But, I clearly remember my professor for analytical chemistry saying that the neutralization of a weak acid and a strong base occurs due to the shift of equilibrium:
CH₃COOH + H₂O → CH₃COO⁻ + H₃O⁺
Where the strong base (NaOH) would continue to remove hydronium ions from the equilibrium, so that more and more of conjugated base, acetate, would form.
So the first part of my question is; what definition is correct?
Furthermore, indeed we usually formulate following equation to approximately quantify the change in concentrations of acid and base:
CH₃COOH + NaOH → CH₃COO⁻ + H₂O  +  Na+
This way, we find the moles of acid/base and divide by the total volume to get the pH using Hendersson-Hasselbalch equation. But is it correct this way? Are there more accurate methods? The weak acid should dissociate first before being consumed, is that correct?
Anyway, this is not a question about how to calculate the titration curve of this example, rather it is about understanding the nature of the conversion. Thanks in advance!

For better site experience, you can find useful How can I format math expressions or chem. equations on Chemistry SE. ( Not to be applied to titles )

The neutralisation happens by both reaction paths. Hydroxide ions react with both hydronium and acetic acid.

Your second equation is an equilibrium. Both acetic acid and acetate ions are always present in a solution, although their proportion may vary. When $\ce{CH3COOH}$ is dissolved into water, a fraction of the molecules are transformed by reaction with water, producing some $\ce{H3O+}$ But if these ions are destroyed by adding $\ce{OH-}$ ions, more molecules $\ce{CH3COOH}$ have to react with water. Acetic acid is transformed into acetate ions to compensate for the consumption of the first $\ce{H3O+}$ ions. So more and more molecules of acetic acid are transformed into ions. The final result is similar to the first equation : more and more $\ce{CH3COOH}$ is consumed by adding $\ce{NaOH}$ and more and more acetate ion and water is produced from this process.

I would 100% agree. But that would be hard to quantify using Hendersson-Hasselbalch wouldn't it? The same websites just uses ICE and quantitative neutralisation like equation 2.

Hendersson-Hasselbach's equation is valid for a solution that does not change in time. It does not apply in the present problem.

How so? Why doesn't it. I learned it this way in university and don't see why

H.-H. equation is just rewritten equation for the acid dissociation constant $K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}}$, which is obviously true if and only if the reaction is in equilibrium.

@Maurice Sir when we reach the point where there is very very less acetic acid in the container , then how will the equilibrium be there ? In next few moments of dropping more naoh

I am sorry. I do not understand your question. The equilibrium constant is a constant. it will not change with the amount of acetic acid, or of NaOH.


text:
Reversible compression and proving relation with state function
I am currently working on a question I am having problems with. I tried to solve it, but to no avail; it states that this expression is always valid for any three-variable dependent function:

It states that I need to prove that by considering the done volume work while doing a reversible compression on an ideal gas, expressed by:

and we set (x,y,z)=(p,V,T). The last derivative can be solved by using the ideal gas law.
Honestly, I don't know where to start. I don't understand how I should put up the differentials, because I don't see how w is dependent on p, the only thing that I thought of would be that the work is defined as -pdV, but I would not know how to work with the dV on itself that would be left.
I know that those are all partial derivatives and that the variable on the outside of the bracket is kept constant and that I need to formulate the function in such a way that both variables in the derivative expression appear on the right side, but I don't know the way to combine all three needed parameters. Any help would really be appreciated.

For better site experience, you can find useful how-can-i-format-math-chemistry-expressions-here. ( Not to be applied to titles ).

$$pdV=\frac{nRT}{V}dV$$If temperature is constant, this is an exact differential.

For inspiration, use  $$\left(\frac{ \partial f} {\partial x}\right)_z=\left(\frac{ \partial f} {\partial x}\right)_y + \left(\frac{ \partial f} {\partial y}\right)_x \cdot \left(\frac{ \partial y} {\partial x}\right)_z$$ as $$\left(\frac{ \partial f} {\partial x}\right)_z=\left(\frac{ \partial f} {\partial x}\right)_y + \left(\frac{ \partial f} {\partial y}\right)_x \cdot \left(\frac{ \partial y} {\partial x}\right)_z$$

How should I start then?

@冰淇淋 You have been a contributor here for some time now, please follow the link that  Poutnik gave you. Also note you can copy-paste Poutnik's second comment content. Making an effort to improve readability will pay dividends.

https://chemistry.meta.stackexchange.com/questions/86/faq-how-can-i-format-math-chemistry-expressions-on-chemistry-stack-exchange


text:
Finding the percent of light absorbed by a sample
I have an absorbance of $0.266$ and need to find the percent of light absorbed by the sample. Is it found by just turning it back into transmittence% and subtracting 1? So by that, would it be $(1 - 0.542)$ making the percent absorbed $45.8\%$?

Your calculation is correct. The percent absorbed is $45.8$%.

I wonder why have you needed confirmation?

You need Beer-Lambert law
$$\frac{I}{I_0}=10^{-aLC}$$
where $I_0$ is initial intensity $I$ that transmitted by a sample of concentration $C$ with extinction coefficient $a$ in a cell of path length $L$. The ratio $I/I_0$ is called the transmittance and $\log_{10}(I_0/I)$ is called the absorbance or optical density (OD) which is  $aLC$.
As a rule of thumb an OD of $1$ corresponds to $10^{-1} \times 100 = 10\%$ transmittance, and OD of $2$ to $10^{-2} \times 100 = 1\%$ transmittance, etc.
(This law applies at each wavelength used, and assumes that the wavelength spread is tiny compared to the natural shape of a line in the spectrum)
The extinction coefficient is a measure of how strongly a molecule absorbs at a given wavelength and usually has units of $\pu{dm^3 mol-1 cm-1}$ if the cell length is in $\pu{cm}$ and concentration in $\pu{mol dm-3}$.

I made some edits to clear it for OP. That way, the percent transmission is clearly to be $10^{-0.266} \times 100 = 54.2%$.


text:
Synthetic Testosterone Laboratory
I can't seem to find any research articles stating how testosterone is made synthetically in a lab. I am looking for research papers on how anabolic agents (synthetic testosterone) are made in a lab?

Maybe a quick “don’t try this at home” or “don’t use your own product” disclaimer would be nice

Interesting is, the question is not about pharmaceutical production and hormonal treatment, but it is about lab production of anabolic agents. You do not ask about a doping cookbook, do you?

From Wikipedia:

The chemical synthesis of testosterone from cholesterol was achieved in August [1935] by Butenandt and Hanisch. Only a week later, the Ciba group in Zurich, Leopold Ruzicka (1887–1976) and A. Wettstein, published their synthesis of testosterone. These independent partial syntheses of testosterone from a cholesterol base earned both Butenandt and Ruzicka the joint 1939 Nobel Prize in Chemistry.

Both original papers are in German. I don't know if these have industrial relevance nowadays. Might be better to ask on Chemistry SE that.
There's a somewhat fully 2009 article on Pfizer's "2nd/green generation" process. They use androstenedione as the immediate precursor. (The "green" claim seems mainly because of no longer producing heavy metal byproducts.) As they mention both "bioprocess" and the Kalamazoo, Michigan plant in there, which was acquired from Upjohn way back, the androstenedione might be derived from plant sterols (which was Upjohn's big innovation in the 70's), but that's not made explicit in that Pfizer article. You can read about the bioprocess route e.g. in a recent review. It's called a bioprocess because conversion from plant sterols is rather inefficient done by chemical means, so microbes are used. There's another paper/review which talks about that:

For example, the steroid hormone testosterone (TS) is chemically synthesized from the steroidal intermediate 4-androstene-3,17-dione (AD), which is previously obtained from natural sterols by microbial biotransformation (Fernández-Cabezón et al., 2017a).

For details on the chemical conversion from AD, you could probably look at Ercoli and de Ruggieri (1953). More recent papers (like Fernández-Cabezón et al.) propose to use biosynthesis in this step as well. Pfizer's article only speaks of a "proprietary intermediate" being used in this step.
Note that due to extremely short half-life of testosterone in the blood (like 10 mins), medical products use esterified testosterone.

A google search on "synthesis of testosterone" yields this pdf document which describes the steps for making testosterone from cholesterol without experimental conditions (did you want conditions?)  https://www.lkouniv.ac.in/site/writereaddata/siteContent/202003291608409347arun_sethi_steroid.pdf

@Waylander: why are you asking me? I didn't ask the Q. I tried to answer mostly with how it's made in industry nowadays because the Q was posted to Medical Sciences when I answered it. (My answer was migrated together with the Q.)


text:
Acidity and structure of EDTA at different pH
I was conducting complexometric titrations to measure the concentration of magnesium ions in my solution.
But of course, as EDTA is a chelating agent, knowing it’s structure and ionization form at different pH would be relevant.
Wikipedia lists 6 different pKa values for EDTA. What I was wondering is, as it has 4 carboxy groups, which all have pKa around 0-2.6, is how the other two pKa values come into effect (~6 and ~10 pKa).
Am I right in the assumption, that when EDTA is dissolved in water, the four carboxy groups release oxonium ions, which then could protonate the amino groups in intramolecular fashion? Or do the amino groups react themselves with water as a base? Basically, the EDTA structure after dissolving in water (pH ~3) would rather look like:

Then, if the pH is gradually increased to ~6, the first, then at ~10 the second amino group will be deprotonated?
As I said, I am quite unsure about the facts and any help or explanation of this structure at different pH levels would be appreciated.

The premise seems false. As of November 2012 WP lists only the expected four pKa values.

@OscarLanzi Curiously, the sidebar (which is presumably what you read) lists 4 pKa values, but the main text lists 6 (see the section on "coordination chemistry principles"). The main text also states which pKa belongs to which group, which seems like it should address OP's question. But I'm not convinced that the amino group has a lower pKa than the carboxyl groups, as it claims...

... indeed, it seems like the Supporting Information of Belle-Oudry, D. J. Chem. Educ. 2008, 85 (9), 1269. (PDF link) has it the right way round. The first four pKa values (0 to 2) are the carboxyl groups; the 6 and 10 are the amino groups (technically they're pKa values of the protonated ammonium groups). Seems like somebody should edit that Wikipedia page.

@orthocresol IP user introduced this error about a month ago, fixed now.

So is what I said correct? Are the amino groups protonated due to the carboxy l groups, in addition to reacting with water as a base?


text:
Chemical analysis of campholene nitrile
I have a problem. I am synthesizing an organic compound called campholene nitrile (2,2,3-trimethylcyclopent-3-en-1-ylacetonitrile).

I need to analyze it. But there's a problem. I do not work for a university or company, so I do not have access to many of the modern analytical methods. I do have access to:

Boiling point (it's a liquid)
Basic chemical tests

So you see, I don't have that much at my disposal. In case it helps, I'm synthesizing it by the isomerization of camphor oxime with 30% sulfuric acid.
Does anyone have any suggestions for how I could do this analysis using my resources and be reasonably sure of its identity?

It very much depends on what are your purity requisites. If you just need to know if the reaction went well, just follow the literature including the isolation step (likely distillation, if you have that setup). Explain what "I have to analyse it" means. If that is true, likely you have to contact an university service or a private lab.

Can you get TLC plates?