text:
Why do some biofuels have a maximal blending ratio?
ICAO has several conversion processes for sustainable aviation fuel (SAF) with various maximal blending ratios. What would happen if we tried to increase the blending ratio with these processes?

According to the snapshot taken in July by the WayBack Machine on archive.org (here), ICAO possibly reorganized how to access the content.

Many pairs of liquids just won't mix in arbitrary ratios.

Many engines are tuned for specific fuels. Many biofuels have very different characteristics to the fuels engines are tuned for. So too much of them upsets the performance of the engine.

Petrol/gas already is a blend of multiple different hydrocarbons when harvested from soil.
After processing (e.g., washing/drying, passing the cracker to split larger molecules into smaller ones, passing the reformer to interconvert some of the molecules into other isomers, and large scale distillations to separate lighter fractions of hydrocarbons (e.g., butane for lighters) from heavier ones, these fractions are intentionally mixed again to adjust some desirable properties like the octane rating.  To quote:

«[This] is a standard measure of a fuel's ability to withstand compression in an internal combustion engine without detonating. The higher the octane number, the more compression the fuel can withstand before detonating. Octane rating does not relate directly to the power output or the energy content of the fuel per unit mass or volume, but simply indicates gasoline's capability against compression. »
(quote from Wikipedia)

In this respect, biofuel is nothing particular on its own; again, it is a mixture which, for the purpose of running an engine, has to pass certain criteria.  Some engines are more tolerant about the power fuel used by design (e.g., the turbine in M1 Abrams battle tanks), than others.  This is one reason (in states where you gas up your car by yourself) why the nozzles for petrol and diesel differ in diameter.
As aviation in general is regulated internationally in minute details, the use of aviation biofuel as such as well as the relative quantity to conventional fuel however extends beyond what technical is possible (e.g., engine design) without sacrifice of performance/longevity of the engines, but accounts for infrastructure, security, etc.


text:
Just looking for some help with finding relative ratios of components in a mixture using proton NMR
G'day,
I am used to identifying molecular structure in isolation using proton and carbon NMR, but have never worked with a mixture before. The mixture relates to the following spectrum:

The spectra for compounds A and E are as follows:

The spectra for the pure compounds were easy to delineate the structure, however it's the mixture that gave me the most trouble. I did some digging, and found a youtube video that gave me the following approach, in order to determine mole ratios:
mol(A)/mol(E) = integration(A)/integration(E)*N(E)/N(A)
The above relates to taking two signals, one from the pure spectrum of A, and one from the pure spectrum of E, and taking N, the number of hydrogens represented by those signals. Then looking at the NMR spectrum of the mixture, one can find the corresponding signal taken from A and E, and use the integration reported in the mixture NMR for the above equation.
Inspecting the pure spectra: there is a triplet in the spectra for unknown A at about 0.95 ppm, which represents three protons, and has a corresponding signal in the mixture at about the same chemical shift, with an integration of 3.056; and for the spectra of unknown E, there is doublet at about 1.32 ppm, which represents six protons, and has an integration in the mixed spectra of 9.272, with the signal shifted upfield to about 1.15 ppm. I suspect this could possibly be due to interactions between the components, perhaps van der Waal's forces.
Finally, returning these to the above equation:
mol(A)/mol(E) = (3.056/9.272)*(6/3) = 0.6592
Given the above equation can be related to the mass and molar mass of A and E as follows:
mol(A)/mol(E) = m(A)/m(E)*M(E)/M(A)
where M(A) = 148.08881 g/mol and M(E) = 134.10955 g/mol (taken from Mass Spectroscopy data)
and in the mixture, if the total mass m(T) = m(A)+m(E), then letting m(T)=100 g, one can solve for the percentage weight of each component, or %w/w as follows:
0.6592 = (100-m(E))/m(E)*(134.10955/148.08881)
Rearranging:
m(E) = (100 g)/1.7279 = 57.87 g
m(A) = 100 g - 57.87 g = 42.13 g
This the %w/w are 57.87% and 42.13% respectively.
This is where I hit a snag however, because if I were to chose other signals from the pure spectra and plug them into the above equation, I don't get the same %w/w. I know this to be a limitation, given no internal standard was used, thus %w/w is not absolute but relative. But what are other limitations, besides NMR being inherently insensitive and requiring large concentrations? How could I fix these limitations?
Appreciate the help in advance.

There should be a kind of averaged MW somewhere, in reference to weight. Is the mole ratio alone more independent of the peaks you choose?

Integration is never going to be perfect, there’s some error. Make sure you use a long recovery delay if you want quantitative integrals. Speak to someone if you dont know how to set that up

The mole ratio was just to be taken from any two signals, one for each compound, that corresponded to the pure spectra, but had different integration and even chemical shift in the mixture spectrum. We didn't have a standard, so I suppose that would be why using different signals will give a different mole ratio. But again, it's a relative and not an absolute measurement.

On that note, why does the integration and chemical shift change when you have a mixture? I am unfamiliar with the recovery delay part too. If anyone wouldn't mind helping me address possible deficiencies, especially the concept of why NMR is an "insensitive" technique, I'd really appreciate it.

@Archimedes_Eureka My real suggestion re: your duplicated accounts, would be to repost your question with the account which you have a login for. There's not much information you're losing by doing so. I can delete this one. Re: the actual question: speak to somebody who does NMR at your institution about the recovery delay. You will end up needing their help anyway, if you want to rerun the spectrum. You are correct that the sensitivity of NMR is intrinsically low, but that is not an issue for the problem you're solving, it's not relevant.

The polarization of the different protons in your molecules will relax back to equilibrium at different rates. The recovery time has to be long enough so that even the slowest relaxing signals have equilibrated to the desired degree. This typically means 5 to 10 times the longest T1. There are are several other precautions needed to get accurate integral ratios. Here is a brief guide to qNMR: https://enfsi.eu/wp-content/uploads/2017/06/qNMR-Guideline_version001.pdf

If you want the best estimate from the existing spectra, add up the signal areas from the non-overlapping peaks (so not the signals near 2.8 ppm), add up the expected number of protons for those peaks, and use the formula you have. This will average out some of the noise.
If you want to know how many significant digits would be reasonable (less than 4), look at how much the ratios of integrals between peaks from the same component differ from the expected ratios. For example, the ratios for compound E (excluding the 2.8 ppm signal) are 3.88:3.09:6 instead of the expected 4:3:6, so they are off by more than 1%, but less than 10%. This means two significant figures might be reasonable.
If you want an estimate of the product distribution with more certainty, you should probably do experiments directly aimed at that (and probably do multiple runs of the synthesis to see how much variation there is from batch to batch).

If you resort to $\ce{^1H}$-NMR for an estimation of concentration you should look out for unique signals (here: either belonging to the butenone (A), or the cymene (E)) simple in shape (this prefers a singlet over a triplet), easy to recognize, which do not overlap with signals by the other compound, and still refer to similar nuclei in terms of their relaxation time in the NMR experiment.
As a suggestion:

the butanone (A, $\ce{C10H12O}$, $M_r  = \pu{148.21 g/mol}$) exhibits a triplet around $\pu{0.96 ppm}$ to account for the terminal $\ce{-CH3}$ group (i.e., 3H)

the cymene(E, $\ce{C10H14}$, $M_r = \pu{134.22 g/mol}$) exhibts a singlet at about $\pu{2.4 ppm}$ for the toluene $\ce{-CH3}$ group (i.e., equally 3H).

for the spectrum of the mixture, you integrated $\pu{4.77 a.u.}$ ( $\approx \pu{2.2 ppm}$, E; possibly the small peak should not be included here), and $\pu{3.06 a.u.}$ ($\approx \pu{0.9 ppm}$, A).

since both signals report for 3H in their corresponding structure, the molecular concentrations of the analyte (excluding the solvent for the NMR and any other compound not recognizable by this technique) compute for
$$c(\textbf{E}) = \frac{4.77}{4.77 + 3.06} \cdot 100\% = \pu{60.92 mol\%} $$
$$c(\textbf{A}) = \frac{3.06}{7.83} \cdot 100\% = \pu{39.08 mol\%}$$

if interested in the mass concentration, you compute e.g.,
$$c(\mathbf{E}) = \frac{0.6092 \cdot \pu{134.22 g/mol}}{(0.6092 \cdot \pu{134.22 g/mol}) + (0.3908 \cdot \pu{148.21 g/mol})} \cdot \pu{100\%}$$
which yields $\pu{58.53 m\%}$.

If you want obtain data reliable enough, you should get in touch with your NMR group because the typical setup for the routine $\ce{^1H}$-NMR experiment favours high throughput of samples with a rather short $t_1$ time. However, the relaxation of the nuclei depends both on the chemical identity ($\ce{H}$, $\ce{C}$, $\ce{N}$) as well as its neighbors.  Determine these $t_1$ times experimentally for your signals of interest ($0.5 \dots \approx \pu{5 s}$ for $\ce{^1H}$) and then adjust the NMR experiment to accommodate $3 \dots 5 \cdot t_1$ per individual FID recorded.  Thus overall, recording a good spectrum to quantify components of a mixture takes a bit more time.
(One would record $\ce{^1H}$-NMR spectra of the pure components and a few samples within the molar range $0 \le x \le 1$ to ascertain that the peaks monitored do not move (too much) along the chemical shift while varying the concentration.  For the two signals suggested, the variation ($\Delta\delta = f(c)$) and danger they overlap with signals of the other component however seems small.  For reference samples used over a long time again and again, flame sealing an NMR tube is better than using the plastic cup.)

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text:
Multiplicity prefixes for complex compounds with organic ligand
Consider the compound $[\ce{Ni(gly)_2}]$, with $\ce{gly}$ denoting glycinate ligand.
Should the name be

bis-(glycinato)nickel(II)
or
diglycinatonickel(II)

In my book, it's written that the prefixes "bis-", "tris-", "tetrakis-", etc. should be used with ligands which already have "di-", "tri-", etc. in their name, or with organic ligands. So, it mentions that the first name is correct.
But my teacher says that these are just used with the ligands having the prefixes in their name, doesn't matter if organic or not. So, he says the second name is correct.
Which one is correct?

See: https://chemistry.stackexchange.com/questions/152640/using-bis-over-di-as-multiplicative-prefix-in-the-nomenclature-of-urea-for-coo

According to PubChem, the IUPAC name of nickel bis(glycinate) is carboxymethylazanide;nickel(2+), which has a given structure of:

However, according to suggestions in Ref.1, the correct IUPAC name should be bis[glycinato(2-)-$\kappa^2$N,O]nickel.
Since the complex is simply given as $\ce{[Ni(gly)2]}$, it is no need to go for the compllicated name since I consider OP as a nivice to coodination chemistry. Let's look at what we can find in Ref.1. The Ref.1 specifies that:

The number of each entity present has to be specified in order to reflect the composition of the compound. For this purpose multiplicative prefixes (Table 2) are added to the name of each entity. The prefixes are ‘di’, ‘tri’, ‘tetra’, etc., for use with names for simple entities, or ‘bis()’, ‘tris()’, ‘tetrakis()’, etc., for names for most entities which themselves contain multiplicative prefixes or locants. Care must also be taken in situations when use of a simple multiplicative prefix may be misinterpreted, e.g., tris(iodide) must be used for $\ce{3I–}$ rather than triiodide (which is used for $\ce{I3–}$), and bis(phosphate) rather than diphosphate (which is used for $\ce{P2O7^4–}$).

Note: Table 2 in above concept is table of multiplicative prefixes for simple and complicated entities upto 20.
Ref.1 also states that the names of complexes are constructed by adding the names of the ligands before those of the central atoms, using appropriate multiplicative prefixes. Formulii are constructed by adding the symbols or abbreviations of the ligands after the symbols of the central atoms. For xample:

The first structure is relative to OP's question while the second structure is beyond OP's knowledge of the field. The name of the first structure is pentaammineaquocobolt(III) chloride. Note that in complexes, ammonia is named as ammine (with two $m$s) and water is aqua. All other amines would be named ith one $m$ as usual. Also note that five ammonia molecules are named "pentaammine," not "pentakisammine." It is because ammonia is a simple molecule (the molecular formula of this molecule is written as $\ce{Co[(NH3)5H2O]Cl3}$).
On the other hand, if the the ligand has a complex name or multiple bonded metal with more than one ligand or etc., the complex would be named with complicated  multiplicative prefixes. For example, look at the second structure of the above image. It is a triply bonded two rhenium atoms with each having four chloride ligands. Thus, it is a bis(tetrachloridorhenate).
According to above example, it is safe to assume that since glycine is simply given as "gly," the name diglycinatonickel(II) is correct. However, if glycine is shown as  2-aminoethanic acid, the the name should be bis[glycinato(2-)-$\kappa^2$N,O]nickel (A $\kappa$-term is required for ligands where more than one coordination mode is possible). Also, note that there is no hypen between bis term and parentesis, technically making the first name given by OP is incorrect! :-)
Note: OP's textbook and what teacher has said are both technically correct.

References:

Richard M. Hartshorn, Karl-Heinz Hellwich, Andrey Yerin, Ture Damhus, and Alan T. Hutton, "Brief guide to the nomenclature of inorganic chemistry," Pure and Applied Chemistry 2015, 87(9-10), 1039-1049 (DOI: https://doi.org/10.1515/pac-2014-0718).

What do you mean by "shown as 2-aminoethanoic acid"? As in being given the structure of the complex or something else?

@ultralegend5385: Yes, it is the structure in the first image. glycine is aminoacetic acid.

Okay, so I'm given $\ce{[Ni(gly)_2]}$, so I will just write diglycinatonickel(II)?

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text:
Can methoxide ion be used as a base in an E1 elimination reaction?
This one problem given in my textbook is confusing because all I know is we can use a weak base for an E1 Reaction. But the reagent given is a methoxide ion. I thought the answer would B but it is C. Compound C is a result of ring expansion which is a characteristic of an E1 Reaction. I still firmly believe that the pathway should be an E2 reaction.
It's the third problem.

The agent given is not methoxide, it is methanol and heat which makes E1 far more likely.

Think about ionization to get C.

Please add a complete citation for "my textbook".


text:
Is there a way to experimentally verify the shapes of atomic orbitals?
The shapes of the orbitals as we know, is the interpretation of the Schrödinger equation that calculates the probability distributions of where to find the electrons with a significant probability. But have there been any experiments to verify these shapes practically?

No, there have been many claims about the experimental verification shapes in very big journals, but they later turned out to be non-sense.

https://physics.stackexchange.com/questions/200143/is-there-experimental-verification-of-the-s-p-d-f-orbital-shapes


text:
What are the necessary conditions for formation of a Hydrogen Bond?
Usually Hydrogen bonds are associated with N,O,F as the atoms involved. In some cases OH---Cl bond is observed as in case of o-chlorophenol.
According to the updated IUPAC definition a Hydrogen bond can be formed between molecules having EN greater than hydrogen (which explains the H-bond in DCM(aq) and chloroform(aq)).
My first question is -How is the EN decided for concluding the formation of H-bond between two General atoms?
Is it the Pauling scale(defined for sp3 hybrids), or an application of Bent's Rule for determining s% in a bonded orbital.By General I mean not N,O, or F as the only participants and a definition that can be extended for C, Cl etc.
My second question is- Is it possible to quantitively determine the strength of a H bond with minimal data like EN between atoms , or at least sufficient to conclude the formation of a H-bond between them?

Verification of presence of hydrogen bond needs serious experimental and/or computational tools not some lame ballpark parameters like electronegativity.

The full definition https://doi.org/10.1351/goldbook.H02899 it doesn't really matter what value you take, the elements are the same. 2. No. Structure is key.


text:
Molybdenum assay (phosphorus vs inorganic phosphate)?
Recently, I have been getting blood tests done for phosphorus. I see some laboratories report phosphorus and others report inorganic phosphate. The normal range for phosphorus seems to be about 0.1 mmol/l higher than that of inorganic phosphate. I expected that, because I expect that there are some phosphorus-containing molecules that are not H$_2$PO$_4$.
My expectation is also that the laboratories are doing something similar to the Molybdenum-method to test the blood serum. But does the Molybdenum-method provide estimates for inorganic phosphate or phosphorus? And how do the laboratories that report inorganic phosphate perform their assay differently than those that report phosphorus?

You use phosphate and phosphorus interchangeably in the context of inorganic compounds. Is that correct? Is the it true in general that every inorganic phosphorus compound is a salt from phosphoric acid? Is that why we can use these words interchangeably in the context of inorganic compounds? I am learning

I have posted that as the answer.

Direct molybdenum method detects inorganic phosphorus/phosphates(1). To determine total phosphorus, a sample gets first mineralised, forming phosphates, e.g. at high temperature by peroxodisulphate oxidation(2).

(1) I have used it interchangeably depending on if phosphate content is expressed as phosphorus or phosphate. It is advantageous to express phosphates as inorganic phosphorus, if compared with organic or total phosphorus.
It is still possible there is inorganic P in other form than phosphate, not reacting directly. But in context of biology, it is unlikely.
It is common practice that concentration of some analytes may be expressed in terms of equivalent components that may not be present in the sample at all. Recalculation is trivial.

(2) This is done during total phosphorus analysis in waste water, where samples are heated with peroxodisulphate solution in pressurised sealed vials. Samples with colloid parts usually become clear when oxidation is done.


text:
Intesive Properties - Temperature
Why is the temperature of a sample of matter intensive?
I don't think it would be extensive, but I also don't see why it would be intensive, either.
It's a very basic and easy question, but I can't seem to rationalize why temperature is independent on the sample size. Nor is it dependent on the sample size. In fact, to me it seems that temp would be unrelated to the type of material, amount of material, and I don't even know if I would necessarily through temp into a "property of matter" so to speak.
This all started with this basic quiz question:
Which of the following are correctly referred to as “intensive properties”?
(i) density of a substance
(ii) mass of a substance
(iii) temperature of a substance
(iv) boiling point of a substance
(v) volume of a substance

(i) and (iv) - my answer

(ii) and (v)

(i) (iii) and (iv) - correct answer

(ii), (iii), and (v)

(i), (ii) and (iv)

I understand how Density and Boiling point are intensive, and I understand how Volume and Mass are both extensive, but what I am confused about is how exactly Temperature is intensive.
Because if the definition of an intensive property is simply that said property does not rely on the amount of the material, then I have a hard time understanding how temp is independent of the amount of material.
The way I always think of it is that if I had, for example, 30 grams of iron in a closed system, and then added 30 more grams of iron to it, the mass would increase, the volume would increase, the weight would increase, the number of mols would increase, etc (all extensive properties). Whereas the density, melting point, etc would remain constant, as they are just a property of the material itself, and can’t be altered unless the substance itself is changed (say it oxidized to iron oxide, then those properties would change).
;TL/DR
How would temperature be intensive, as the temperature can be changed by a plethora of factors, and has little to nothing to do with the quantity of mass of the substance itself (aside from the the  specific heat, for example). Temperature is just the average kinetic energy of the particles in the material, so if 30 grams of iron had x amount of energy put into it to, which set the iron's temp to be 20 degrees C, and then 30 more grams of iron were added to the system spontaneously, the kinetic energy of the first 30 grams of iron would transfer to the second 30 grams, resulting an overall lower temperature, correct?
Alternatively, if the 30 grams of Iron were 20 C, and then 30 grams of iron at 50 C were added, the temperature would change and adjust to be 40 C on average, correct? Thus it would technically depend on the sample size, as when the sample increased, the temp increased also.
Not to mention that the temperature of any material isn’t an intrinsic property of the material itself, but is based on the system/circumstances surrounding the material, whereas intensive properties rely on nothing but the properties of the matter itself (such as the bonds between atoms, structure of any lattice formed when solid, polarity, and even how many subatomic particles make up the atom), while temperature is just a temporary state of the energy applied to the atoms/particles, causing them to move and creating a temperature.
Thanks in advance.

Your major confusion may be thinking intensive property must be property of material.
There are 3 types of quantities, depending how they are related to system scalability:

Extensive properties are additive wrt(with respect to) the system scaling. If there is twice as big system, there is twice as big value of an extensive property. Examples are mass, volume, various forms of system energy.

Intensive properties are invariant wrt the system scaling. They may be intrinsic properties of system components, or may be state variables of the system conditions. Examples are density, concentration, refraction index, chemical potential, specific and molar heat capacity, temperature.  Two times $\pu{1 L}$ of $\pu{20 ^{\circ}C}$ water gives $\pu{2 L}$ of $\pu{20 ^{\circ}C}$ water.

Properties, that are wrt system scaling neither additive neither invariant. Examples are system surface, heat conductance or ohmic conductance, as these depend on the system scaling geometry.

The question was poorly designed with incorrect choices. Density is also intensive property. Density of 1 kg of water is same 0.001 gram of water so your answer is also correct.
I feel the source of your confusion is between heat and temperature. This concept has bewildered big physicists and philosophers in the 18- and 19-century, so no need to feel bad about this confusion.
Let us say you have 1 kg of water at and 5 gram of water in a container. You insert a thermometer, keep heating and you measure the temperature. When all water has converted into steam, you measure the temperature  again. The themrometers registers 400 K in both containers. Which container has higher heat content, yet the temperature is the same? Temperature is the manifestation of the amount of heat, it is not heat itself.
Temperature indeed is a peculiar intensive quantity which is experimentally measurable. In thermodynamics, it is defined rather circulary. For example, Wikipedia say temperature is an intensive property because it is ratio of partial derivative of internal energy and partial derivative of entropy.
At a deeper level, what causes objects to have mass to begin with and what is charge itself? What causes molecules of continually move and never rest? There are certain quantities and observations which are not fully understood.
Your second example,

30 grams of Iron were 20 C, and then 30 grams of iron at 50 C were
added, the temperature would change and adjust to be 40 C on average,
correct? Thus it would technically depend on the sample size, as when
the sample increased, the temp increased also.

This is again an example transfer of heat, this is a special case. If the masses of iron were different you have to do proper calculations, as discussed in calorimetry examples.
Example


text:
Will all aldehydes and ketones give 2,4-DNP test?
I've recently gone through 2,4-DNP test for aldheydes and ketones on Byjus:

An aqueous solution of 2,4-dinitrophenyl hydrazine (DNP) is known as Brady’s reagent. It reacts with carbonyl compounds (aldehydes and ketone) to give a colored precipitate.

From the above statement, I think only water-soluble aldehydes and ketones give the test, but my teacher said all aldehydes and ketones (carbohydrates excluded) give 2,4-DNP test.
Who is correct? My teacher or the website?

Glucose doesn't give 2,4-DNP test.

@NilayGhosh Yes, of course, but I actually wanted to ask about simple aldehydes and ketones.

2,4-DNP test can be based on solvents other than water e.g. alcohols


text:
Why do isosteric molecules have similar shapes?
Isosteres (Definition): Isosteres are atoms, molecules, or ions of similar size containing the same number of atoms and valence electrons.[1]
Questions:
A. Why should they have same molecular structure (or geometry)?[2][3]
B. Are there exceptions where isosteres don't have same structure? If yes, please give one.
1. Chemcool Dictionary, Definition of Isosteres  2. Wikipedia: Isostere  3. Chemcool Dictionary, Definition of Isosteres

Isosteric molecules have similar shapes because that's what the word "isosteric" really means.

@IvanNeretin My teacher just gave the definition as "isosteres are those molecules which are isoelectronic (have the same number of total electrons) and have the same number of atoms", for example CO₂ and N₂O. Interestingly, in Wikipedia, I found the definition as "Classical Isosteres are molecules or ions with similar shape and often electronic properties." Are these two definitions equivalent? Does the fact that they have same number of electrons and atoms imply the fact that they will have same structural arrangement? And are there exceptions?

OK, let's get this straight. "Iso" means "same" and "stereos" means "shape", so "isosteric" means "having the same shape". That's what it means because that's what it says, as clearly as the day. There is nothing more to it.
This is the correct definition, and the other one is wrong. See, butane and isobutane contain the same number of atoms (even atoms of the same elements) and of valence electrons, so what? Do they have the same shape? Definitely no.
So it goes.


text:
Orbital angular momentum of an electron in s-orbital
I'm a beginner in quantum mechanics, and I'm pretty much confused with the orbital angular momentum of electron in s-orbital of hydrogen atom. I've read that average angular momentum of an electron in $1s$ orbital of $\ce{H}$-atom is $0 \pu{kgm^2s^-1}$, but when I took the wave function of $1s$ orbital of $\ce{H}$-atom and solving by the following way is giving me instantaneous angular momentum is $0\pu{kgm^2s^-1}$.
My way of solving:
The momentum operator is $\hat p = -i\hbar\nabla$
and the $\psi_{1s}= k \times e^{\frac{-r}{a_0}}$, where $k$ is normalizing constant
Applying momentum operator for $1s$ orbital,
$$\hat p \psi_{1s} = p \psi_{1s}$$
I'm getting some thing like $\hat p \psi_{1s} = -(x \hat i+ y\hat j+z\hat k)k^{'}e^\frac{-r}{a_0}$, which implies that the momentum $\overrightarrow p$ of electron in $1s$  orbital is
$$\overrightarrow p = k^{''}(x \hat i + y \hat j + z \hat k)$$
So, angular momentum can now be given by $$\overrightarrow L = \overrightarrow r \times \overrightarrow p$$
The position vector $\overrightarrow r = x\hat i + y\hat j + z\hat k$
As the position vector, $\overrightarrow r$ and the momentum vector, $\overrightarrow p$ are in anti-parallel directors at every point in space, I have concluded that $\overrightarrow L = \overrightarrow 0$ at every instant, which contradicts to what I've read.
(NOTE: The source from where I've read is not very much reliable, but I didn't find any thing related to this any where else)

So you've read that it should be $0$ and also found it to be $0$. What's the problem?

@IvanNeretin  I've read average angular momentum is zero but I got instantaneous also zero as well. Is it correct?

Does it make the average different from $0$? No. So, there is no contradiction.

@IvanNeretin It doesn't make any thing different, but is the instantaneous value also $0$?

Yes.$\mathstrut$

@IvanNeretin Oh, okay. Thanks for clarifying.

Electron orbitals in atoms are eigenstates of both energy and total orbital angular momentum. These are therefore constant in any orbital unless the orbital is changed, as it would be for instance when forming a bond to another atom. So whatever "average" you have for these variables applies at every instant as well.


text:
Predicting time taken to reach equilibrium
Consider the following reaction, $\ce{A <=>[k_f][k_b] B}$, $k_f$ and $k_b$ are the rate constants of forward and backward reactions respectively.
Let the initial concentration of $\ce{A}$ be $a \pu{M}$ and the final equilibrium concentrations would be $(a-x_e)\pu{M}$ and $x_e\pu{M}$ respectively, where $$\frac{k_f}{k_b} = \frac{x_e}{a-x_e}$$i.e,$$x_e = \frac{k_fa}{k_f+k_b}$$
Let $x\pu{M}$ $\ce{A}$ be consumed at any time, $t$, then the net rate of formation of $\ce{B}$ is
$$\frac{dx}{dt} = k_f(a-x) - k_b(x)$$
Solving for $x$, with limits from $x=0$ to $x=x$ and $t=0$ to $t=t$
$$\frac{1}{k_f + k_b} ln(\frac{k_fa}{k_fa - (k_f+k_b)x})=t$$
Substituting $x=x_e$, $t \to \infty$, but in many books, it is shown that reactions reach equilibrium after a certian time like the following

Image taken from NCERT Text book for Chemistry, Class XI, Part-1.

Image taken from Chemistry Libre Texts.
A google search result:

All the above images show that equilibrium state is attained at a finite time but mathematically it takes $\infty$ time to reach equilibrium.
Is there any flaw in the equation or in the graphs?

I wouldn't even call that a flaw. The graphs are made for kids who don't have the concept of $\infty$ yet.

Also, related: https://chemistry.stackexchange.com/questions/57075/is-there-no-actual-equilibrium/57135#57135

The graphs are made by kids who don’t know about asymptotic attainment of equilibrium yet.

The problem, is anything, is with the mathematics. The equations–if interpreted strictly–only reach equilibrium at infinity. But they implicitly assume that matter is continuous and not made of discrete molecules. In a real-world reaction happening quickly real reactions will reach equilibrium give or take a few molecules difference quickly: in a continuous math world this will not be true. If you wanted a better equation you would need to include a precision level for the math that defined equilibrium as being reached when the difference was a few molecules not exactly zero.

The flaw is in the graph if you take its labels in their literal sense.

So technically, the label "at equilibrium" is incorrect for the left panel. You can either weaken the label language (center) or change the graph (right) to have a watertight graph and labels.
Source of original graph: https://chem.libretexts.org/Courses/University_of_Kentucky/UK%3A_CHE_103_-_Chemistry_for_Allied_Health_(Soult)/Chapters/Chapter_8%3A_Properties_of_Solutions/8.2%3A_Chemical_Equilibrium

Don't you think your third equation has a wrong sign ? You write it : $$\frac{dx}{dt} = k_f(a-x) + k_b(x)$$ In my opinion, it should be : $$\frac{dx}{dt} = k_f(a-x) - k_b(x)$$ The same mistake is reported later on, in the next equation.

I see that you have edited your question, with my suggestion. Nice ! Thank you !


text:
How is nitrogen gas passing through condensed water in the U-Tube while reducing metal oxides with ammonia gas?
In my school chemistry book, the experiment about the reduction of metal oxides with ammonia has a diagram of the set up as follows:

$$\ce{NH₃ + CuO → 3Cu + 3H₂O + N₂}$$
The tube entering the U-tube from combustion tube A is not dipped into the condensed water.
So doesn't that mean that nitrogen (density: $\pu{1.165 kg/m^3}$), being lighter than water(density: $\pu{1000 kg/m³}$) will push the water up on reaching a sufficiently high pressure and the water will flow up through Tube B?
I believe nitrogen gas won't pass through as bubbles as the tube is not under the water level. Nitrogen is also almost totally insoluble in water, hence the nitrogen doesn't dissolve in the water either(which can also be seen by nitrogen gas obtained in the gas jar by downward displacement of water) at the right side of the setup.
So is there a flaw in the diagram? Or am I missing something?

It is rather a problem of mechanics and hydrostatics. With sufficient pressure, gas will push liquid down until it bubbles through. But I agree the experiment is not well designed, they should avoid water accumulation in the gas path.

I suspect the purpose of the U-tube is to trap water (the label says "condensed water" and the tube is surrounded by a coolant). If so, the picture is very badly drawn as the volume condensing would not block the tube and prevent gas flow.

@matt_black Additionally, there is not addressed "sucking up" gas volume because of gaseous ammonia dissolution in water. The reportedly  condensed water could as well be remnant of sucked up water from the end cylinder. :-)

No, the experiment as depicted is basically correct, see my comment below, however, on what is likely understated.
Remember ammonia is highly soluble in water, so without Tube 'B', the water in Tube 'C' could backflow.
In reality, the volume of Tube 'B' is much larger than depicted to address backflow issues. I would personally replace it with a large heavy jar that sits in the cold water basin (while a light vessel could problematically float, hence the original design of secured tubing).
Don't believe, see what happens upon removing the backflow trap (aka, Tube 'B') when all the CuO is consumed leaving just highly soluble ammonia gas in the system.

So is the diagram drawn considering the unreacted ammonia proceeding further into the system? Because the explanation provided in our book avoids any mention of unreacted ammonia and assumes that a perfect mass of ammonia is being passed into the setup, just enough to complete the reaction without any residual ammonia left.
Also, can you just elaborate slightly more why will the backflow take place if tube B is removed?

This would be a bad setup if the purpose if to avoid backflow. Moreover the U-tube is labelled "condensed water" and is surrounded by what seem to be coolant. This implies it is designed to condense out the surplus water from the reaction not to avoid backflow. But it incorrectly shows the tube to be overfull with the condensate which would be unlikely in this reaction.


text:
What is the mechanism of paracetamol synthesis from N‐acetylsulfanilic acid?
C can be converted to D with NaOH and heating at 300 °C. What is the mechanism of the reaction?

Actually I am fairly curious. Google searching reveals sources such as Quora saying it's nucleophilic addition-elimination, but there are few details in those answers and I am not finding a primary source. Vote to reopen.

I am frankly surprised that the N-acetyl group survives the conditions.

@waylander that group probably deprotonates at N and hunkers down. Would not be as good a leaving group as sulfite.

Due to apparent interest from community I voted to reopen upfront after adding chemical names and correcting units. It would be nice if OP could add their research to prevent the question from being closed again.

Finally I find a primary reference by Oae et al.[1] They react benzenesulfonic acid with fused potassium hydroxide using isotopically labelled chemicals. The isotopic labelling results favor a direct nucleophilic substitution of hydroxide for sulfite, so that the only ring position involved in the reaction is that originally holding the sulfonate function. This is called $\ce{S_N2}$ in the paper; today we use addition-elimination since with an aromatic substrate, a full-fledged addition intermediate is formed in this mechanism. A benzyne mechanism and attack on sulfur (followed by rearrangement of the oxygen atoms) are rejected. The abstract is quoted below.

Benzenesulfonic acid-1-¹⁴C was fused with potassium hydroxide at 320-350℃ for five minutes. The phenol obtained was then degraded as usual, and the distribution of carbon-14 in the phenol was determined. Practically all the ¹⁴C activity was found at the 1 position of the phenol, which means that there is no migration of the ¹⁴C originally labeled at the position 1 or the benzenesulfonic acid. Benzenesulfonic acid was also treated with potassium hydroxide enriched with ¹⁸O. The resulting phenol was found to be incorporated, with the same concentration of ¹⁸O as that of the ¹⁸O enriched water used . The results show that the mechanism of the reaction is of a simple SN2
type and not of the elimination-addition type involving the benzyne intermediate, while there was no oxygen migration in the reaction.

Reference
1.
Shigeru Oae, Naomichi Furukawa, Masahiro Kise and Mitsuyoshi Kawanishi, "The Mechanism of the Alkaline Fusion of Benzenesulfonic Acid". Bulletin of the Chemical  Society of Japan 39, 1212-1216 (1966).

Old paper. Isn't this mechanism called addition-elimination now, because it's two step, unlike "ordinary" SN2?

@mithoron Rephrased my summary to reflect this. The quotation still has to use the authors' SN2 term.


text:
How does addition of side chains affect solubility and processibility of polyphenylene vinylene?
I am studying electrical engineering and physics. In a recent lecture I was told the $\ce{OC1C10}\text{-PPV}$  molecule has the chains added to the oxygen atom down below because it allows for solubility and processibility:

How does the addition of these chains affect the solubility and chemical processibility of the molecule?

Conjugated chains, especially if delocalisation does indeed take place, tend to be flat or at least stiff.
This results in a tendency to adhere to each others. In addition, their stiffness also means that not enough conformational space open up upon solubilisation.
In order to enhance their solubility, or at least their processability^, flexible chains are linked to the main polymer chains.
The presence of the side flexible chains can result in

a more exothermic solubilisation process, the entity of this being dictated by the exact nature of the long substituent as well the specific solvent counterpart (enthalpy variation of the process);

and perhaps most important and surely general as solvent independent, an extra entropic term (entropy variation of the process).

The impact of the above on the free energy of solubilisation can combine and lead to a soluble material, or at least one that dissolves in aggregates small enough to allow processability from "solution".
The solubility of flat stacking molecules can be enhanced by the same "trick".
Finally, the entropic effect can lower the temperature of evaporation (vacuum usually nevertheless required) of organic materials for opto-electronics. So that at least a kind of processability is attained.
^For sake of clarity as OP used the words "chemical processability". There is no much chemistry involved, the meaning is that the material must be easily processed from melts, solutions, evaporation, etc. Literally "intractable" materials (no melting point nor glass transition, insoluble) were indeed the first conjugated polymers - willingly (polyacethylene) or not - prepared.
Another note specific for the discussed conjugated polymer. The particular branched chain was introduced to further reduce interchain interaction and so enhance fluorescence. In fact, one of the first envisioned application (nowadays mature) of conjugated polymers & molecules was electroluminescence for LEDs.


text:
Cobalt glass colour of potassium as
Question:

Cobalt glass colour of potassium as:
A) Crimson red
B) Lilac
C) Apple green
D) Brick red

I thought option B as true because on flame test, potassium gives lilac colour. But the given answer is option A. I can't understand, why they have given the answer as crimson red colour. Does cobalt glass changes the color of potassium flame?

Hint: What wavelengths does potassium emit in flame? Find this out and modify your post.

Cobalt glass has its Wikipedia page which is supposed to be already read by you.

I have searched in internet and there is no such type of question.And I have visited Wikipedia before posting this question but I couldn't able to get the answer.So that's why I have asked this question.

Cobalt glass is used for removing the intense yellow lines of sodium that is present everywhere in small amount, and which prevents to see the color of potassium flames. Seen through a cobalt glass, the sodium flame color is invisible (entirely absorbed). And the potassium flame may be dim, but it is visible.

..and I would say it is lilac, not crimson red( even if who am I as a man to know colors? :-)  ) // Search also youtube for potassium flame cobalt glass videos.

Yeah that's what my doubt.I thought lilac but given red. Thanks for answering.

https://www.youtube.com/watch?v=3ScNyl943C4, flame color with potassium.

https://www.youtube.com/watch?v=5kqaTNGFL9U All flame colors with and without cobalt glass

Color is somewhat subjective, that is why people study wavelengths with atomic spectra rather than color. For elementary classes, this is fine. Certainly A is a wrong choice because I checked the transmission characteristics of cobalt glass,* the blue glass transmits both the violet and the deep red doublets of potassium. So, the flame color appears like lilac as you can see in various Youtube videos of flame tests with cobalt glass.
In low temperature flames like that of the Bunsen burner, K atoms emit 404.4, 404.7 (pure violet), and 766, 769 nm (really deep red, pretty hard to see in a pocket spectroscope). The resulting mix of emission color looks like the classical lilac potassium flame shown in textbooks.
*Ref: Frederick C. Strong III, Improving the Potassium Flame, J. Chem. Educ. 1969, 46, 3, 178 (behind paywall)


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:
The necessary energy required to ionize 1g of potasium in its fundamental state if the ionization energy is given

The energy of ionization of the potassium is $\pu{6.94\cdot 10^{-19} J/atom}.$ Find the necessary energy to ionize $\pu{1 g}$ of potasium that is found on its fundamental state.

Attempt. I really didn't want to overthink this so I tried with conversion factors:
$$\pu{6.94\cdot 10^{-19} J/atom} \cdot \frac{\pu{1 atom}}{\pu{1 molecule}} \cdot \frac{\pu{1 molecule}}{\pu{6.022\cdot 10^{23} moles}} \cdot \frac{\pu{1 mol} \ \text{of } \ce{K}}{\pu{39 g} \ \ce{K}}\\ \approx \pu{2.96775\cdot 10^{-44} J/g},$$
but that didn't give me the correct answer since the correct answer is $\pu{10.7 kJ/g}$. I'm clueless on what else to try. May anyone shine a light?

6.022e23 mol of potassium is about 0.4% of the Earth mass.

Houston, we have a "Problem" in question titles. Please use real titles. Note that the symbols of physical quantities are always upright (use \pu{…} macro). Your formula is meaningless. First, neither "atom" nor "molecule" nor "moles" (it's mol) nor "of K" (or symbol of any other chemical element and prepositions) are physical quantities or their proper symbols. Second, where is the derivation with symbolic algebra and the numeric answer you've obtained?

numeric answer is now edited, what do you mean by "symbolic algebra"?

@Acyex A symbolic algebraic expression typically implies writing an equation in generic form using symbols (not numbers or physical quantities) for all constants and variables. It's a universally recommended initial step of calculation prior to plugging in the numbers so that you tackle the problem conceptually and don't have to deal with units, dimensions and significant figures at the same time as you are figuring out the solution.

but what equation should I write about? There's no equation that I can know of related to this problem? I figured that since we have $\pu{J/atom}$ we could get $\pu{J/g}$ easily like that way. Can only think of the $\frac{1}{\lambda}=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$ formula? Since ionization means that $n_2\to \infty$ but don't know how that'll help

Don't overcomplicate. You know the energy per atom. You know the mass of potassium. How many potassium atoms are in 1g? Problem solved.

How on earth is that not showing enough effort? @matt_black isn't what you just mentioned the exact method I mentioned?

You have the correct method, but one of your ratios is flipped. $\pu{1 mol}$ is $\pu{6.022E23}$ molecules, you have that 1 molecule is $\pu{6.022E23 mol}.$ As a sanity check, even if you didn't know the right result, a gram should have more energy than a single atom, but your answer was smaller.

@Acyex The symbolic algebra is $\frac {E_\mathrm{I} \cdot N_\mathrm{A}}{M_\mathrm{K}}$.

@Acyex I didn't say "not enough effort" but the opposite. You overcomplicated the calculation and, as a result, made an error. The simple approach I suggested would have avoided the error by making the logic far more clear.

@matt_black oh no i wasn't referring to you, I was referring to the mod that deleted my question because apparently "it didn't show enough effort"... is there a way of like reporting or flagging the mod or something?

@Poutnik, Tyberius, matt_black yes it's solved now, thanks!

@Poutnik Again, neither the word "atom" nor chemical symbol of potassium are proper unit symbols, and should not be used in your expression. It's especially bad for potassium since now it looks like you are using kelvins for some reason. Also, the cross "×" should be used for the product of the numbers with the decimal sign denoted with a dot, not middle dot "·".

@Andselisk I just tweaked the OP expression. I was not going to give more effort to it. // BTW the central dot is put there by \pu{}

@Poutnik You should've used E, not e: \pu{6.022E23} $(\pu{6.022E23}),$ not \pu{6.022e23} $(\pu{6.022e23}).$ mhchem currently doesn't keep a track of what kind of decimal separator is used, unfortunately. Either way, thank you for going back and fixing things.

@andselisk $$\pu{6.94E-19 J} \cdot \frac{\pu{6.022E23} }{\pu{1 mol}} \cdot \frac {\pu{1 mol}}{\pu{39 g}}$$
Thanks, E vs e is the very new info for me, I was not aware it is case sensitive, nor what is the difference.


text:
Why does the same type of hybridization appear in both Valence Bond Theory and Molecular Orbital Theory?
I was reading about Molecular Orbital Theory, and saw that you can treat a central atom's atomic orbitals as hybrid orbitals in order to simplify the process of creating molecular orbitals by no longer having to look at the symmetries of the AOs. I was interested by the fact that the same type of hybridization (sp, sp2, sp3) that one uses in VBT is used when simplifying MO Theory with hybrid orbitals. Is this because the observations of bond angles and bond equivalency that led to hybridization in VBT could be explained by looking at the molecular orbitals of the molecule? In a way, by using hybrid orbitals to simplify the creation of molecular orbitals, are you just reverse engineering observations about molecular shape to determine what the orbitals must be without using symmetry arguments?

Molecular orbital theory and Valence bond theory are equivalent at their respective limits using the same basis; they will produce exactly the same result. They just approach the solution from different angles. Therefore it is natural that you can express characteristics of the one within the framework of the other. However, this is mathematically a bit easier going from molecular orbitals to Valence bond (or localised) orbitals. Please also note that hybridisation in VB is not discrete; look up Bent's rule.

Would Bent's rule apply to the hybrid orbitals you create to simplify MO creation?


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:
Endothermic reaction with dV < 0
I am looking for an endothermic reaction dH > 0 for which the net volume change is negative, e.g. gases reacting to form a liquid. Does anybody have an illustrative example?

$\Delta H^o <0$ would be an exothermic reaction (thermal energy is released) (at constant p, with pV-work only). Are you looking for an endothermic reaction with $\Delta V <0$, or an exothermic reaction with  $\Delta V <0$?

@theorist Right, of course, I was sloppy… I am looking for an endothermic reaction.

An example of an endothermic reaction with $\Delta V < 0$ is:
$$\ce{C(graphite)->C(diamond)}, \,\,\,\,\Delta H^o_r (\pu{298.15 K}) = 1.871 ± \pu{0.074 kJ//mol}$$
Source:
https://atct.anl.gov/Thermochemical%20Data/version%201.118/species/?species_number=951
And $\Delta V < 0$ because:
$$\rho_{diamond} = \pu{3.515 g//cm^3}; \,\,\ \rho_{graphite} = \pu{2.16 g//cm^3}$$
You'd expect this with diamond, since its formation is favored by high temperatures and pressures.
Source: Wolfram Mathematica Chemical Database  (not the primary source, but Wolfram doesn't give that in their answer).
An entire class of reactions that would satisfy your requirements would be most* combustion reactions, when run in reverse (making them endothermic), and at a high enough temperature that water would be gaseous, e.g.:
$$\ce{ 12CO2(g) + 11 H2O (g)->C12H22O11(s) + 12 O2(g)}$$
$$\ce{3 CO2(g) + 4 H2O(g)-> C3H8(g) + 5 O2(g)}$$
[*This wouldn't work with the combustion of methane, which has equal numbers of reactant and product molecules (that's of course not considering small volume changes due to non-ideality, which could go either way).]
For three other examples of endothermic reactions with $\Delta V<0$, see my answer here:
Any examples of reactions where simpler molecules are combined into more complex ones and are endothermic at the same time?

Thank you, these are very helpful examples!


text:
What is "gram atom of an element"?
I've seen so many answers on Quora that got me lost :
1st answer :
The three are similar.
Gram atom.... Take the atomic mass of an element and express it in grams (sodium for example atomic mass is 22.990 so 22.990 grams of sodium is a gram atom of sodium.
Gram molecule... for this you use the molecular mass of any molecule and express it in grams (water for example two hydrogen atoms plus one oxygen atom atomic mass of 18.015 so 18.015 grams of water is a gram molecule) the gram molecule is also referred to as a mole or mol.
Gram ion... virtually the same as gram molecule but referring to ion.
2nd answer :
Don't be confused, these are the old names of moles of these things.
1 gram atom means 1 mole of atoms
1 gram molecule means 1 mole of molecules
1 gram ion means 1 mole of ions.
Similarly,
2 gram atom means 2 moles of atoms.
So, this is the name of mole.
3rd answer :
Gram atoms are exactly Total number of atoms present in one mole of a given covalent substance.
Gram molecules are total number of molecules in one mole of given substance.
Gram ions are total number of ions present in one mole of given ionic substance.
Eg - 1 mole of $\ce{NaCl}$ is 1 gram of molecules of $\ce{NaCl}$ and 2 grams of ions of $\ce{NaCl}$.
mole of $\ce{MgBr2}$ is 1 gram molecules of $\ce{MgBr2}$ and 3 grams of atoms of $\ce{MgBr2}$.

The gram-atom is a very old terminology (mainly historical now). When you express the atomic weight of an element or a molecular weight in grams, it was called gram-atom or gram-molecule. Note the hyphen. Your second and third quoted answers are quite wrong.
As a corollary, 1 gram-atom or 1 gram-molecule contain the same number of particles.
Please note that Quora, Wikipedia or even this site are not a gold standard for scientific facts and research. For serious historical questions, you should consult multiple older textbooks, which are readily available from the Internet Archive.

thanks, i'll try to ask my professor :(  And yessss, i remembered something now : my professor said : "0.3 gram atom of iron, so it's 0.3 mole of iron", so that means that 1 gram atom is 1 mole?

No, 0.3 gram-atom of iron is 0.3 x (Atomic weight of iron). 1 gram-atom is the weight of 1 mole of atoms, it is NOT 1 mole. Hope that clarifies your query.


text:
Mechanism for alcohol synthesis via terminal acetylene reaction with sodium amide followed with epoxide and protonation

If anyone can give me some guidance here, it'd be much appreciated. I know the NaNH2 would dissociate and the NH2- base would deprotonate the terminal acetylene. I'm assuming the resultant product would act as some sort of nucleophile which would open the ring of the epoxide. The Na+ would form an ionic bond with the negatively polar oxygen after which the acid would protonate said oxygen, but I'm not entirely sure.

The final structure is wrong. Count carbons. Missing a CH2.

Basically, you are on the right track and answered the question. First, NaNH$_2$ is used as a base to deprotonate the terminal alkyne (see this Reagent Friday). Second, the generated nucleophile opens the epoxide in an S$_\mathrm{N}$2 fashion, usually adding to the lesser substituted carbon which in your case yields a quaternary alcohol after acidic workup (see also Opening of Epoxides With Acid).
(Note: Indeed you are missing the CH$_\mathrm{2}$-group where the nucleophile attacks in the final structure)


text:
SN1 mechanism on 3,5-dimethylbenzyl bromide
I wanted to see how to make the product on the right so my thought was to have an SN1 reaction to remove water and have the bromine ion attack making the product on the right hand side.
Is this theoretically possible?

yes, the benzyl cation is very stable which facilitates the process

Yes, this can be done readily facilitated by the stable benzyl radical. This patent  here shows an example.

The reaction would theoretically occur through an SN2 mechanism because the hydroxyl (OH) group is bound to a primary carbon, and SN1 will usually only occur under tertiary (steric) conditions with a weak nucleophile.
What you are trying to do is referred to as "activation of alcohols". The OH- is not a good leaving group because OH- is a stronger base compared to Br-. There are several ways to activate an alcohol group (such as the formation of tosylates, protonation with strong acids, and reacting with PBr3 etc.) Since your goal is to substitute the alcohol with a bromide ion, theoretically H-Br could be used, where alcohol will accept a hydrogen atom from H-Br and leave as H20 as you have noted. PBR3 would be a better choice compared to H-Br.
This is because you need to watch out for side products from any potential carbocation rearrangements. There is some steric bulk with the presence of the benzene ring which could arguably allow the mechanism to form a stable carbocation through SN1, in which case the carbocation could rearrange with the adjacent carbon in the benzene ring. Using PBr3 also avoids "harsh conditions".
Look up activation of alcohols with hydrogen halides and activation with PBr3 for more information.

yes PBr3 and CBr4 would have been another option but I just wanted to see if the proposed reaction was also a theoretical possibility. Thanks

Welcome to Chemistry! While we appreciate your contribution to the site, it would be preferable to not answer questions, that do not comply with our homework policy.


text:
How do I balance redox equations with three reactants?
Could anybody please explain me the concept of balancing redox reactions with three reactants? Either some solving steps or even the solution would be very appreciated.
For example, I don't know how to balance this equation:
KBr + $\ce{KMnO4}$ + $\ce{H2SO4}$ $\ce{->}$ $\ce{Br2}$ + $\ce{MnSO4}$ + $\ce{K2SO4}$ + $\ce{H2O}$.
I split it like this:
$\ce{K^+ + Br^- + K^+ + MnO4^- + 2H^+ + SO4^2- -> Br2 + Mn^2+ + SO4^2- + 2K + SO4^2- + H2O}$
And then I am not sure what to do, so I did this:
$\ce{Br^- + e^- -> Br2}$
$\ce{MnO4 + 8H^+ - 2e^- -> Mn^2+ + 4H2O}$
And it does not make sense, because I do not know what to do with $\ce{SO4^2-}$.

In contrary to some other Q/A or forum sites, answers on CH SE site are figuratively paid by the user's own effort. 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. Effort not shown may be considered as effort not done and such a question may get closed.
How do I ask a good question.

Using photos/screenshots of text alone instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, nor can be reused nor referred in answers.  Consider  copy/pasting or rewriting of at least essential parts and eventually using  MathJax for formatting of mathematical/chemical formulas or equations.

See chem.libretexts.org: Balancing_Redox_reactions and many more, following Google: How do I balance redox equations. Always search before asking.

You can find useful  Notation basics and How can I format math/chemistry expressions on Chemistry SE.

I will provide you with a comprehensive account of what to do with $\ce{SO4^2-}$. Here is what: nothing. It just sits there and does nothing. It is not important. Just write the half-reactions (this you already did, only the electrons are going the wrong way), then balance each half-reaction, then put them together.

Related

You first focus on the groups that undergo oxidation or reduction. The sulfate does neither, so you can set it aside, just like the potassium cations.
The half reactions you wrote, however, need some work. The oxidation state of manganese in permanganate is +7, so there is a transfer of 5 electrons. For bromine to bromide, the oxidation state changes by one, but there are two bromine atoms, so you are transferring 2 electrons, not 1.
Some reactions have more than one species undergoing oxidation (or reduction). In those cases, it is impossible to have a unique balanced equation. This is not one of these cases, though, so it should be possible to write a unique balanced equation.


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Mesylation of cyanohydrins
Is it possible to add methane sulfonyl chloride to benzaldehyde cyanohydrin, or add toluene sulfonyl chloride to form the tosylate group on the alcohol and then perfom a grignard reaction on the nitrile to form an imine? Or more succinctly do mesylate/tosylate sulphonyl chlorides react with nitriles as they do with amines.

Yes, according to Kiritsis et al. [1] it is possible to mesylate cyanohydrins.
The paper by Bracco et al. [2] has a number of relevant examples such as protection of the cyanohydrin by O-silylation followed by Grignard addition and O-triflation followed by Suzuki coupling of the triflate with arylboronic acid. It also shows that the −OH group may also be acetylated and the tetrahydropyranyl or t-butyl ether can be formed.
References

Kiritsis, C.; Manta, S.; Dimopoulou, A.; Parmenopoulou, V.; Gkizis, P.; Balzarini, J.; Komiotis, D. Stereoselective Facile Synthesis of 2′-Spiro Pyrimidine Pyranonucleosides via Their Key Intermediate 2′-C-Cyano Analogues. Evaluation of Their Bioactivity. Carbohydrate Research 2014, 383, 50–57. DOI: 10.1016/j.carres.2013.11.001.
Bracco, P.; Busch, H.; von Langermann, J.; Hanefeld, U. Enantioselective Synthesis of Cyanohydrins Catalysed by Hydroxynitrile Lyases – a Review. Org. Biomol. Chem. 2016, 14 (27), 6375–6389. DOI: 10.1039/C6OB00934D.


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Environmental properties of chemicals
I am trying to collect the environmental properties (GWP and ODP) of several chemicals. However, I am experiencing some problems with retrieving these properties for some fluids.
For example, I cannot find the GWP nor ODP of:

cyclobutene
Methanol (Or is the GWP and ODP negligible as Methanol is liquid in ambient conditions?)

I would think this information is widely documented, however I am struggling to find these properties for some chemicals. Is there a database available of these properties? Also, is it safe to assume a negligible ODP and GWP of fluids which are liquid in ambient conditions?

It is adviced to always explain abbreviations not used widely and generally, in text or link. // GWP = Global warming potential // ODP = Ozone depletion potential

Methanol is not much relevant, as it is not released in big amounts and it is hydrophilic, effectively purged out of atmosphere by water circulation.

Many molecules have no ODP and many reactive molecules will have such short lifetimes in the atmosphere that their GWP will be, effectively, zero. Both are probably true for your two examples.

cyclobutene GWP is likely <2, methanol is likely <15.
Reasons: closest examples, methane 25, propane 3, most hydrocarbons 0 to 5. Hydroxyl and C=C link reduce GWP by a lot.
https://refrigeranthq.com/refrigerant-2/hydrocarbons/
Their both ODP is likely 0.
Reason: closest examples, propane 0.
https://en.m.wikipedia.org/wiki/Ozone_depletion_potential
You cant assume liquids under normal conditions to have zero ODP or GWP.
Reason: water is a liquid, and yet we have clouds and it is the main global warming gas.


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Lysozyme crystals versus NaCl
If someone has some experience with lysozyme crystals I would be interested in knowing whether any of the images below seem likely to be lysozyme. Solutions were made with powdered lysozyme ($\pu{50 mg/mL}$) and $\ce{NaCl}$ ($\pu{3 M}$), except the lower left which is $\ce{NaCl}$ alone.
The top two images were made from a lysozyme solution with a small amount of $\ce{NaCl}$, then adding a piece of tissue paper. The square/cuboid crystals came out of solution immediately onto the fibers. That they are cubes suggest to me $\ce{NaCl}$ but the appearance is odd.
The middle two images have a higher ratio of $\ce{NaCl}$ to lysozyme--maybe close to 2:1 or 1:1. The crystals look OK to me but as a complete novice I am not sure. The crystals are all on the order of $\pu{0.5 mm}$.

A better image of the square crystals from lysozyme/$\ce{NaCl}$ + tissue paper.

If someone has some experience with lysozyme crystals I would be interested in knowing whether any of the images below seem likely to be lysozyme.

The bottom two have the typical shape of well-grown lysozyme crystals.

Source: Jena Biosciences (https://www.jenabioscience.com/crystallography-cryo-em/get-started/crystallization-of-model-proteins/co-401-lysozyme-solution)
Buttonwood's answer about polarized light is spot on.

You can also add a dye. If it diffuses into the crystal, it can't be NaCl.
There is also a destructive method. If you try to squeeze the crystals with a hard object (like a needle), NaCl crystals will stay intact or break with smooth faces while the lysozyme crystals will look like smashing a watermelon with a baseball bat.
Here are some more ideas: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2864706/
And here a paper about dye penetration with some great photos of lysozyme in polarized light: https://www.semanticscholar.org/paper/Penetration-of-dyes-into-protein-crystals.-McPherson/5d79f2f257f85163d5c14ad0967de1d4354bd1c8

So protein crystals don´t cleave well if tapped gently with a sharp blade of a safety razor? Based on own experience, it may take some training of the eye (comparison with other «broken» crystals in the vial), manual dexterity (angle along one of the cleavage planes) and check (Nicols for a hopefully still sharp extinction, later recording an orientation matrix or/and an early and quick $\phi$ scan on the diffractometer to interpret non-sharp diffraction peaks as an indicator for physical damage). For me, this worked well enough for some organic crystals, once too much in needle shape.

But it may be well a question of scale, too, with protein crystals often smaller than $\approx \pu{0.2 mm^3}$.

I was more or less shocked at how much the (Jena) crystals resembled mine. This is a very helpful answer. I left a comment yesterday but somehow today it's gone.

@daniel That’s great, you grew protein crystals! I did read your comment yesterday.

The one in the top right and those large in the very last photo pretty much look like $\ce{NaCl}$.  Those in the second row (left) and third row (both left and right) look different by the crystal habit's symmetry;  and what could be birefringence (especially third row, right hand side) -- a plausible first evidence for protein crystals which typically crystallize in a lower than cubic space group symmetry.
However:
The habit of a crystal (outer shape) may vary on the crystal's condition of growth.  Only the angles enclosed by the crystal's surfaces remain constant for the same polymorph of a compound (Steno's law).  The presence of impurities the purification did not remove, intentionally added detergents, dyes and buffers, the applied technique (e.g., sitting drop vs. hanging drop), the age of the crystals while growing may influence which faces may be seen or not (more).
Remedy:
A quick and easy check is to observe the sample put between two linear polarizers/Nicols.  Check if there is a polarization microscope nearby and put the dish on its table (it very well may contain the solution including the crystals).  Since there is there is no birefigence for cubic $\ce{NaCl}$, you may sort these out visually. If Nicols are crossed, crystals of $\ce{NaCl}$ appear uniformly black.
It may happen that accidentally such a salt crystal still ends on the X-ray diffractometer instead of your wanted protein crystal.  But these will be recognized quickly by symmetry, $d$-spacing and brilliance (compared to a protein sample) of the diffraction pattern.

The polarizer trick for NaCl looks very useful and I will be trying that. Very helpful answer, thanks.


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How can all amino acids point outwards from an alpha helix, if peptides are mostly trans?
My textbook in biochemistry ("Biochemistry" - Berg,Tymoczom Gatto, Stryer) tells me that:

"Almost all peptide bonds in proteins are trans" (p. 40)

At the same time, it states the following regarding the structure of the $\alpha$-helix:

"A tightly coiled backbone forms the inner part of the rod and the side chains point outwards" (p.42)

How do I reconcile these two statements?
If the peptide sequence in an $\alpha$-helix is predominantly trans, that would imply that every other amino acid residue would point inwards, towards the center of the helix.
Does this imply that peptide sequences involved in $\alpha$-helixes are predominantly cis, or that   the amino acid residues pointing towards the center of the helix must be very short, like for instance glycine?

It is neither. See, the whole thing is 3D. Take a paper sheet and draw an all-trans peptide chain along a straight line. Now roll up the paper into a cylinder and bend all residues outwards.

At this moment there are nine editions of your textbook. When citing a source please make sure your citation is complete and unambiguous. On Chemistry.SE the preferred citation style is ACS.

[Ivan Neretin in the comments] It is neither. See, the whole thing is 3D. Take a paper sheet and draw an all-trans peptide chain along a straight line. Now roll up the paper into a cylinder and bend all residues outwards.

That is so true. Here is an animation to illustrate:

The peptide planes are still planar, and wrap around an imaginary cylinder. Consecutive planes are at an angle with respect to one another because there is a tetrahedral carbon, the alpha carbon, in every amino acid. The side chain (shown as gray methyl group here) stick out away from the axis of the cylinder.

[OPs cited statements] "Almost all peptide bonds in proteins are trans"

All peptide bonds in an alpha helix are trans.

[OPs cited statements] "A tightly coiled backbone forms the inner part of the rod and the side chains point outwards" (p.42)

You could call it a rod if you show a spacefilling model, or a toilet paper roll (or paper towel roll for longer helices) if you show an all-bonds model like above. Notice it says the side chains point outwards, not the amino acids.

[OP] How can all amino acids point outwards from an alpha helix, if peptides are mostly trans?

They don't. The side chains point outwards, and that is compatible with trans-peptide bonds. In fact, different secondary structures (turn, strand, helix) are possible by changing the torsion angles on the tetrahedral alpha carbon while keeping the peptide bonds all-trans.


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:
Why is the chirality center in (1R, 2S)-2-fluorocyclopentane-1-carboxylic acid 1R and not 1S?
I'm confused why the chirality center in  (1R, 2S)-2-fluorocyclopentane-1-carboxylic acid, is 1R and not 1S.

I'm personally think that the carboxyl carbon at C1 should have a higher priority than the fluoride group at C2 because

C1 = (C, O, O)
C2 = (C, F, H)
C3 = (C, H, H)
C4 = (H)

So, the priority should go like this C1 -> C2 -> C3 -> C4. Therefore, I think it should be 1S. Where do I get wrong?

F has higher priority than O.

There are two substituted carbon atoms.  Discounting sterochemistry, in the substitutive nomenclature, the carboxy group has higher priority than the fluorine group. Thus, the framework is 2-fluorocyclopentane-1-carboxylic acid. Then, there is
a) a carbon bound to two other carbons, an hydrogen, and a fluorine.

Among the atoms around the stereogenic centre, fluorine, with an higher atom number than carbon, has highest priority within the CIP-rules now applied (labeled 1). Carbon 2 is of higher priority than the one labeled with 3 because (in the next layer), (one carbon and an hydrogen atom) is of higher priority than (two hydrogen atoms).  We observe the substituents from the side opposite to convention, though; thus, the stereogenic centre is (S) configured.
b) a carbon bound to three other carbons, and an hydrogen.

Again, it is the second layer which offers decision about the priorities.  Here, we already observe the stereogenic centre from the convened perspective.  And if we think of the four atoms like a steering wheel, with the atom of lowest priority (here: hydrogen) in the back, it turns to the right hand side; the centre is (R).
Thus (1R,2S)-2-fluorocyclopentane-1-carboxylic acid is the name for the compound.
Note: The site's  compilation resources for learning chemistry has a section dedicated to nomenclature, too.  It equally mentions programs which you may use to draw and assign the CIP labels, too.  Don´t worry, these assignments demand training, training, and training.

@Buttonwood has provided a well-reasoned explanation for the CIP assignments of configuration to fluoro carboxylic acid A. While the novice may appreciate why the configuration at C2 in amino carboxylic acid B remains the same as in A, the configuration at C1 in each compound is different even though the two compounds have the same absolute configurations. The reason lies in the fact that the CIP is a construct unrelated to absolute configurations but rather to what is attached to the stereogenic center under consideration.
The digraph illustrates that the first shell around C1 in A is bound to three carbons (C2, C5 and C6). No decision can be made here! Proceeding to shell 2, C5 is attached to C4 and two hydrogens, C6 is connected to three oxygens and C2 is attached to a fluorine, carbon and hydrogen. Now these connected atoms are to be ranked in descending order of atomic number, {C,H,H}, {O,O,O} and {F,C,H}, respectively. Here is where the student usually  becomes confused. Now compare the first member in each triad. Because F>O>C, C1 in A has the (R)-configuration. Your C1 (C,O,O) should read (O,O,O). Even so, it would come in second to the triad bearing fluorine. In compound B, where nitrogen replaces fluorine, C1 has the (S)-configuration because O>N>C.
You may want to consider this alternative to the "steering wheel" method for assigning configuration. This link will help assigning CIP configurations.


text:
Nova 2.1 stop command (graphical user interface for electrochemical measurement)
We are doing an electrochemical polymerization of a semiconducting polymer and would like to try it as a pulsed square wave with defined frequency (i.e. 100 Hz) and duty cycle (i.e. 50%), but fixing a control over the total deposited charge (stopping the deposition at a target charge value, i.e. 100 μC). How can I do to set the stop command?
I manage to do it easily with the "record signal" block in a fixed voltage electropolimerization (using the "Apply" block to set a voltage potential), where the block can also integrate the current value and measure the charge in real-time.
However, with the "chrono methods" block for the square wave development, it is not easy to link a "record signal" block to the "chrono methods" block. They seems always to be connected in series, occurring one after another. Thus, the deposition always stops at the end of the total selected cycles, overcoming the charge threshold I selected.
Using the "repeat measurement" block, I have seen that there are time-intervals between the calling of the "apply" block, thus the duty cycle changes and the measurement is slower than the selected frequency.
Do you have any suggestion to fix this?

Can you please edit to say what ‘Nova’ is and what the ‘blocks’ are? I would guess you have some electrochemical instrument that has a graphical user interface (GUI) software system, but it would be best to just give a bit more information.

Check this instruction manual maybe?

Thank you for your answer.
Yes, it's Nova 2.1 and the blocks are block diagram for programming the measurements, similar to the LabView software system.


text:
Why is my Briggs-Rauscher reaction oscillating but failing to turn blue?
As the title says, I've constructed a Briggs-Rauscher reaction according to the ever-popular "Chemical Demonstration: A Handbook for Teachers of Chemistry" by B.Z. Shakhashiri (the same procedure is available on MIT OCW here: https://ocw.mit.edu/high-school/chemistry/demonstrations/videos/briggs-rauscher-reaction/briggs_rauscher.pdf and can be seen below)
Putting the three solutions together in the correct 1:1:1 ratio and stirring it vigorously turns the solution dark yellow as expected, but it does not turn dark blue, merely "resetting" to transparency and oscillating again from there.
What could be causing this? Since the blue is caused by the starch-triiodide complex, I assume my starch solution is too dilute, and that adding some more starch to the third solution may help, but I will not be able to test any solutions until Friday, since it is part of a class. Has anyone else had the same issues?

Questions on CH SE should be rather self-contained, using links just as the further reference, not needed for understanding the question.

Thank you! I didn't know, I've edited the post now to include the MIT OCW procedure and materials.

Ehm, for future reference, it is preferred to copy/paste or retype essential quotes as text, in the scope needed to understand the question. // The image text content cannot be indexed nor searched for, nor can be reused in answers.

Try with another source of starch. I have seen starch samples for potatoes that do not react with iodine.

Does MIT offer these materials altogether with a contact address (e.g.; to receive comments/suggestions for improvement of the material, the addition of experiments)?  Possibly they could provide additional information, too.


text:
Struggling to make SMCC
I have attempted to synthesize SMCC (succinimidyl 4-(N-maleimidomethyl)cyclohexane-1-carboxylate) following the lit protocol in J. Org. Chem. 2011, 76, 21, 9169–9174. Except for the scale, I followed the protocol exactly and the purity of starting materials is confirmed with NMR analysis.

For the raw product, I recorded a $\ce{^1H}$-NMR spectrum.  Since this analysis was not conclusive, the sample was characterized by  ESI-MS spectrum before moving on to the purification step (trituration with diethyl ether). However, what I saw on the ESI Ms was 353.24 (relative abundance 100%), 375.18 (65%), 705.04 (40%), and 726.99 (70%). I am concerned with the complete absence of the desired mass peak (Exact Mass: 334.12, and no M+1 or M+23 peaks were found), and unable to figure out a potential side-reaction that gave the mass peak at 353.24.

Too few information. Since the publication describes multiple esters, it would be beneficial if you summarize your synthesis in a brief reaction scheme which includes a structure formula of your starting reagent and intended product. Contrary to IR and NMR, the shape of MS spectra very much depends on the experimental conditions to record the spectrum. Did you monitor advancement of the reaction, e.g., by TLC (e.g. by cross-spotting starting material(s) and reaction)? What if you have your product in one of the washing solutions from (back) extraction you likely still have under the fume hood?

@Buttonwood Ok, will do now. I should have outlined it better. I was making the compound 3i following the Representative Procedure B from pg 9172-9173.

Also would be good to know conditions for ESI. I would assume 353 peak is M + H2O +1.

Is there strong evidence (e.g., quenched aliquote, then MS (direct injection)) the esterification between maleic anhydride and amine proceeded well? By own experience, I'm hesitant to telescope (no isolation/workup) a new sequence right on first attempt. It won't travel well on the normal phase plate (perhaps better on neutral silica) and anyway requires a derivatisation in a stain, after elution and drying).

If you followed the protocol "exactly" and were unsuccessful, that would imply that the protocol is incorrect. Alternatively, you didn't follow the protocol "exactly" ;).

@user55119, Organic synthesis in research journals is not that reproducible because authors often fail/forget to write sufficient details. This is why only reliable protocols are those which have been reproduced in another lab e.g. the database of Organic Synthesis.

@user55119 There is a good number of publications where (even if the SI put less constrain on page numbers, than the body of publication [e.g. Baran's 11 step Maoecrystal V approach]) where experimental details are left out for whatever reason. Validation of protocols (like by Org.Syn.) is not as widespread as it could be. It is not only about «reproducible science» if lack of data shared puts additional difficulty (both about time and $$$ for chemicals/experiments) to your work when colleagues in the same group do this intentionally.

In lab classes you might be pushed to write detailed reports (depends on the TAs and on the specialty, though, analytical chemistry and physical chemistry tend to be more aware about things like standard deviation, etc. than organic chemistry).  And once the students become profs, well, other things enter the focus.

@M.Farooq,@Buttonwood. I'm well-aware of the truth of both of your comments. I was in the business for decades. Did you notice the little ;) at the end of my comment? My issue was the term exactly. ;)

@user55119, You will also agree that some synthetic chemists also hide the details to prevent their competitors from catching up. There was one famous case, where the author mentioned a wrong metals in superconductors work (?) in Nature or some other big journal. When other researcher tried to reproduce, they could not. Later he published a correction. The exact details escape, it was mentioned by a teacher in class a decade ago.

@M. Farooq: Deviousness occurring among chemists? I am as shocked as Captain Renault in Casablanca.

@user55119 I admit that the claim that 'followed the protocol exactly' can be subjective.


text:
Why does the trend of the enthalpy of reaction per atom of carbon in hydrocarbons approach a near-constant value?
I was looking at trends for enthalpies of hydrocarbons for a high school project and found an interesting paper regarding it, stating that the enthalpy per carbon atom approaches a constant. I'm thinking it has to do with the limited arrangement of bonds in hydrocarbons for bond dissociation energies, but I am not certain. Does anyone know a chemical explanation for why?
https://www.sciencedirect.com/science/article/pii/B9780444535634100057

The limit increment value would be the sum of molar enthalpies for 2 C-H bonds and 1 C-C bond.

Thank you! I should've thought about it that way.

More significant than maths alone. If you take a chain long enough, ideally infinite, you realise that atom arrangement is the same at each position, and the sides are identical.