text:
Suggestions of GHS-compliant labelling software
Having amassed a rather large collection of reagents and products from previous years of chemistry experiments, I find rather annoying that their labels aren't even close to being standardized to the GHS rules.
Each supplier has its own variation, some GHS-compliant, some not, and of course for the end products of the experiments there is no standardization at all since they weren't purchased. I try to find a pre-made label for said compound on the internet, print and use it.
If there is no pre-made template for it, the next step is to find a MSDS and fill in the data in one of many online label generators. Tedious, and prone to error. And online generators come and go offline all the time meaning after some time there are several different "formats" around.
To complicate things more I do not feel comfortable at all letting some random site know that I have this or that chemical.
That said, I ask for suggestions of software for GHS-compliant label generation.

Offline only
Windows/Linux
Can use standard inkjet/thermal/dot-matrix printers (not proprietary printers)

And bonus if:

Has internal database of chemicals. If not, well, it would be a one-time effort to input the data.
Is open-source (or is cheap)

Prior adoption of GHS symbols, our group gave open enventory a shot (open source, APGL; runs in Ubuntu) as comprising both a chemical inventory + label generator + electronic lab book. Thanks to the INSTALL folder and help by the IT it run some time. If fed by a .csv file of chemicals present, it fetched the MSDS data from the vendors and public sources. But it was not enough «ChemDraw like» (there are interfaces to an ELN and inventory) for the elder folks, and the department adopted an ERP billing (no chem) software by SAP to track the bottles.

Maybe the current version (updated February this year) adjusted the label generator to include the GHS symbols on the labels to print (which you may purchase as sticker labels a little bit smaller than 2 x 2 cm wide, too).


text:
Why does activation energy depend on temperature?
To me it makes sense that the activation energy $E_\mathrm a$, as given by the Arrhenius equation, doesn't depend on the temperature as the needed energy (potential energy) to start a reaction has nothing to do with the temperature (kinetic energy) of the molecules. And a higher temperature simply means that more molecules have sufficient energy to proceed in the reaction $E\gt E_\mathrm a$.
So I don't understand why activation energy would depend on temperature, but this equation shows that it does $E_\mathrm a=\Delta H^{\ddagger}+RT$.
Can someone explain why?

Although the question has been already answered, I will like to add a bit more of detail.
The expression you are quoting come from the thermodynamic formulation of the TST that expresses the dependence of the rate constant of temperature as
$$k(T)=\frac{k_\mathrm bT}h\exp\left(-\frac{-\Delta G^\ddagger_0}{RT}\right)$$
and can be partitioned in an entropic and enthalpic contributions
$$k(T)=\frac{k_\mathrm bT}h\exp\left(-\frac{-\Delta S^\ddagger_0}R\right)\exp \left(-\frac{-\Delta H^\ddagger_0}{RT}\right)$$
From this equation you can obtain the activation energy $E_\mathrm a$ using the relation
$$\frac{\mathrm d\ln k(t)}{\mathrm dT}=-\frac{E_\mathrm a}{RT^2}$$
Using this equation you will obtain
$$E_\mathrm a=\Delta U^\ddagger_0+RT$$
Since $H=U+pV$
$$\Delta H^\ddagger_0=\Delta U^\ddagger_0+p\,\Delta V^\ddagger_0+RT$$
For a condensed phase reaction $\Delta V^\ddagger_0\approx0$ and
$$E_\mathrm a=\Delta H^\ddagger_0+RT$$
For a gas phase reaction, $p\Delta V^\ddagger_0=\Delta n^\ddagger RT$.
For a gas phase unimolecular reaction $\Delta n^\ddagger=0$, the result is equal to that the obtained by a condensed phase reaction, while for a second-order reaction $\Delta n^\ddagger=-1$, and
$$E_\mathrm a=\Delta H^\ddagger_0+2RT$$
You can find a more detailed discussion in Keith Laidler, Chemical kinetics 3rd. ed., Harper Collins Publishers, (1987); Jeffrey I. Steinfeld, Joseph S. Francisco, William L. Hase, Chemical Kinetics and Dynamics, Prentice Hall (1989); or Paul L. Houston, Chemical Kinetics and Reaction Dynamics, McGraw-Hill (2001).

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Would Oxygen Gas and Ozone be a pure substance together?
If I have oxygen gas and ozone ($\ce{O2 + O3}$) together would it be considered a pure substance or a mixture?
And would pure substances always have the same molecular structure?

What do you think?

I am pretty new but I think it is not since the molecules have different quantities of oxygen.

Oxygen and ozone are two very different molecules. Ozone is a pungent gas. So you cannot call it a "pure substance". The problem with even calling pure dioxygen as pure oxygen is that it will be mixture of oxygen isotopes.

@M.Farooq And it also contains $\ce{O2-}$, $\ce{O2+}$, $\ce{O+}$, $\ce{O-}$ ions and monatomic oxygen. Not too much, but they are in thremodynamical equilibrium. But, I think, "pure substance" is a concept of the experimental chemistry, i.e. if there is thecompound on your desk, in a container, and not a theoretical thing.

Ozone is highly reactive and unstable, while dioxygen is stable. There do not combine to form a compound. So, clearly it is a mixture.
To answer the second part of the question, "And would pure substances always have the same molecular structure?", first a Wikipedia definition on substances, to quote:

A chemical substance is a form of matter having constant chemical composition and characteristic properties.[1][2]...
Chemical substances can be simple substances[4], chemical compounds, or alloys. Chemical elements may or may not be included in the definition, depending on expert viewpoint.[4]
Chemical substances are often called 'pure' to set them apart from mixtures. A common example of a chemical substance is pure water...
However, in practice, no substance is entirely pure, and chemical purity is specified according to the intended use of the chemical.

And further:

A chemical substance may well be defined as "any material with a definite chemical composition" in an introductory general chemistry textbook.[5] According to this definition a chemical substance can either be a pure chemical element or a pure chemical compound. But, there are exceptions to this definition; a pure substance can also be defined as a form of matter that has both definite composition and distinct properties.[6] The chemical substance index published by CAS also includes several alloys of uncertain composition.[7] Non-stoichiometric compounds are a special case (in inorganic chemistry) that violates the law of constant composition, and for them, it is sometimes difficult to draw the line between a mixture and a compound, as in the case of palladium hydride. Broader definitions of chemicals or chemical substances can be found, for example: "the term 'chemical substance' means any organic or inorganic substance of a particular molecular identity, including – (i) any combination of such substances occurring in whole or in part as a result of a chemical reaction or occurring in nature".[8]

So, in a very elementary introduction to chemistry, the law of constant composition likely applies in the 'correct' answering of your question.
Note: In the context of advanced chemistry, there are also interesting examples of polymers of length 'n', again not precisely specified.

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Hydrogen isotope effect on methyl shift
Consider this typical 1,2 methanide shift in a carbocation:

Which isotopic form of methanide ($\ce{CH3, CD3, or CT3}$) would migrate from secondary carbocation to form more stable tertiary carbocation?
This link explains the mechanism but I could not understand the abstract.
Assuming that the migration of $\ce{CH3-}$ takes place, the following facts can be considered:

The electronegativity of $\ce{CH3}$ is highest, thus, the formation of $\ce{CH3->CH3-}$ (methanide) is easiest. Consequently, the migration would be fastest in this case.

The transition state would be most unstable in this case due to highest electronegativity of $\ce{CH3}$.

The tertiary carbocation would be most stable in this case due to order of $\mathrm{+I}$ effect: $\ce{CT3 > CD3 > CH3}$

There are few exchanged comments below the second answer of this question and the proper explanation can be found in this link, which I can't access.

@Safdar So, the stability of T.S. is more important than that of final carbocation. In your comment, the numerator and denominator are same.


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Preparing aqueous solution of ethanol

You need to prepare $\pu{250.0 mL}$ of an aqueous solution of ethanol in the concentration will be $\pu{0.500 mol L-1}$. You have $\pu{100 mL}$ of a commercial solution of ethanol in water $(35.0\% \,V/V)$ available. The density of ethanol (pure) under these conditions is $\pu{0.7893 g mL-1}$ and that of water is $\pu{0.9970 g / mL}$.
1.What volume (in $\mathrm{mL}$) will you need to withdraw to prepare this solution?
2.What is the mole fraction of ethanol in the resulting solution?

Answer to (1)
We can prepare $\pu{250 mL}$ of $\pu{0.5 mol L-1}$ solution of ethanol by diluting $\pu{20.85 mL}$ of $35 \%$ $\ce{C2H5OH}$ (density $\pu{0.7893 g/mL}$) with distilled water to a volume of $\pu{250 mL}$.
The dilution is therefore made by adding $\pu{229.15 mL}$ of water.
However, the volume of a liquid obtained by mixing measured volumes of different solutions is not always precisely the sum of the component volumes.
Calculation
Source (B): $35 \%$ solution of $\ce{C2H5OH}$ (density $\pu{0.7893 g/mL}$)
The molar concentration of a solution of a chemical species A is the number of moles of that species contained in one litre of the solution (not in one litre of the solvent).
$$C(A) = \pu{0.5 mol/L}$$
$$\text{Molar concentration} = \frac{\text{Density} \cdot \text{Weight percent} }{ 100\cdot \text{Molecular weight}}$$
$$C(B) = \frac{d(B) \cdot w(B)}{100 \cdot M(B)}$$
$$C(B) = \frac{\pu{789.3 g/L} \times 35 \%}{100 \times \pu{46.0684 g/mol}}$$
$$C(B) = \pu{5.99662675499909 mol/L}$$
The number of moles of solute in the diluted solution must equal the number of moles in the concentrated reagent.
\begin{align}
V(A) \cdot C(A) &= V(B) \cdot C(B) \\
V(B) &= \frac{V(A) \cdot C(A)}{C(B)} \\
V(B) &= \frac{\pu{0.25 L} \times \pu{0.5 mol/L}}{\pu{5.99662675499909 mol/L}} \\
V(B) &= \pu{2.08450525782339 \times 10^{-2} L} \\
V(B) &= \pu{20.85 mL} \\
\end{align}
Answer to (2)
We have
\begin{align}
C(B)  &= \frac{\pu{0.125029667842 mol}}{\pu{20.85 mL}} \\
&=\pu{5.99662675499909 mol/L}
\end{align}
Now we have to add $\pu{229.15 mL}$ of water to $\pu{20.85 mL}$ of commercial solution of ethanol to make it $\pu{250 mL}$.
$\pu{229.15 mL}$ of water has a mass of $\pu{229.15 g}$. This means that there are $\pu{12.7197 mol}$ water because the molar mass of water is $\pu{18.0153 g}$
\begin{align}
\text{Mole fraction of ethanol}&= \frac{\text{Number of moles of ethanol}}{\text{Number of moles of solution}} \\
&=\frac{\pu{0.125029667842 mol}}{(\pu{0.125029667842 mol} +\pu{12.7197 mol}}\\ &
= \pu{0.009734 mol}
\end{align}

Are my answers to both questions correct? If they are incorrect, where am I going wrong?

I guess if the time spent by writing this question were rather spent on double checking your procedure, you would know the answer even before posting it, not speaking about waiting for response. Note that the purpose of Chemistry SE is not proof reading, but understanding of principles, that cannot be easily get from textbooks/online resources. Is there any principle you have difficulties to apply to solve this particular task ?

BTW, you should definitely consider reformatting the question using MathJax.

What is the meaning of the 35.0% ethanol V/V ? The volumes are not additive. 35 mL pure ethanol and 65 mL pure water do not give 100 mL, but much less, maybe 96 mL. So is it 35 mL ethanol plus enough water to make 100 mL ? Also, it is ridiculous to make such calculations with 14 significant figures, if the result is required with 3 significant figures.

@Maurice 35%(V/V) concentration is defined as 35 mL of pure substance in the 100 mL of the final solution ( i.e. not 100 mL = 35 mL of ethanol + 65 mL of water -  that would be volume fraction ) See  https://en.wikipedia.org/wiki/Volume_fraction  and https://en.wikipedia.org/wiki/Alcohol_by_volume

@ Dhamnekar. Your calculations are correct, provided your assumption (that the volumes are additive) is correct. Nevertheless note that the last value ($x = 0.0097$) has no unit : it is not in moles. And we must also admit that, due to your assumption, x is probably simply $x = 0.010$.
Anyway, next time try to trust your calculations yourself. We are not here to verify correct calculations. We are here to HELP THOSE WHO DO NOT SUCCEED in solving their problems.

Thanks for pointing out that mole fraction don't have any unit.


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Does heating potato starch destroy its resistant starch?
Resistant starch, as found in potato starch, is supposed to have health benefits. There are differing opinions on whether or not heating potato starch destroys the resistant starch in it and thereby also the supposed health benefits. I would like to know what the scientific consensus is.
That reheating potato starch destroys its resistant starch, is argued by:

Marks Daily Apple

Any idea if heating the potato starch (like using it as a thickening agent in soups/stews) negates its RS function? Yes, the RS will be completely negated.

Diabetes Sister

If potato starch is heated it converts to normal starch and you will get none of the amazing benefits.

Mark Hyman

Resistant starch is made by cooking and then cooling starches like potatoes or rice and not reheating them, transforming regular starch into resistant starch.

Healthline

It’s important to not heat the potato starch. Instead, prepare the meal and then add the potato starch once the dish has cooled.

That reheating potato starch does not destroy its resistant starch, is argued by:

John Hopkins

Reheating doesn’t decrease the amount of resistant starch.

Bernard Preston

Carbs, cooked and then cooled, and again reheated, pass through the small intestine into the colon as resistant starch where they are instead fermented by the healthy probiotic bacteria; instead of forming simple sugars with all the potential problems of obesity, diabetes and glucose intolerance.

Chris Kresser

Each heating and cooling cycle adds more resistant starch.

There also appear to be different types of resistant starches, which might complicate the discussion in ways I do not know.

The concepts are review in this paper: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3823506/

"The amylose–amylopectin ratio determines how much retrograded starch remains on reheating, with retrograded amylose being more heat stable than retrograded amylopectin. "
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7984060/
For the love of god, don't depend on sites like Hyman or Mark's Daily Apple for answers.

I agree with the general sentiment of this answer, but we usually use a less colourful language to express this. On another matter: we usually do not answer questions with a single quote, but I really think it is fine in this context.

Thanks Martin, point taken!  I'll do better if I ever leave another comment.

The original quotes don't distinguish between the 5 types of resistant starch, but from this blog community back when this was first discussed, I know they are referring to RS2 potato starch as supplied by Bob's Red Mill in their "Unmodified Potato Starch" product. Potato starch is extracted from cold potatoes and spray dried at low heat which does preserve the RS2 resistant starch. Heating destroys the RS2 starch and turns it in digestible (by humans) carbohydrate. The equivalent RS2 source from whole foods would be raw potato.
The accepted answer is referring to RS3 starch from cooked and cooled starch where starch retrogradation during cooling partially forms a new resistant starch but that is not the RS2 type found in raw starches.
People get confused between the 2 types since they can both come from potatoes. All 5 types RS1-RS5 are presented in WikiPedia. See https://en.wikipedia.org/wiki/Resistant_starch. Heating losses are mentioned under the section on isolated starch.
In the original question, the Kresser and Preston quotes are referring to RS3 so that requires heating and cooling to form retrograded starch. The Hopkins article mis-interprets their source and only is talking heating in regard to RS3. Their source ( Patterson MA, Maiya M, Stewart ML:  Resistant Starch Content in Foods Commonly Consumed in the United States: A Narrative Review. J Acad Nutr Diet 2020 Feb; 120 (2): 230-244. PMID: 32040399.) only includes data from cooked potatoes because raw potato starch was excluded. From their source "Exclusion criteria included the use of RS from flours or
starches, isolated raw starch, or uncooked foods unsuitable
for consumption (eg, raw potatoes and raw cassava)".
The Hyman quote is referring to RS3 but the linked article talks more about RS2 potato starch used in smoothies and kept cold.
Hope this helps.
Naja's source in her comment clearly states that RS2 as in raw potato starch is degraded by heat. Her answer only applies to the RS3 type.
Sorry I could post a potential answer but could not post this as a comment to the previous answer because I just signed up. Again apologies.

Formatting guides for texts and formulas/equations/expressions.


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How to make acetic acid more volatile so that it can be removed by vigorous aeration of an acetic acid-water mixture?
I want to remove acetic acid from a mixture of acetic acid and water by aeration. The concentration of acetic acid is 200 ppm (mg/L). The pH of the water is more than 5. I want to reduce the acetic acid concentration by around 80%.
I understand that acetic acid can be removed by addition of sulphuric acid to bring down the water pH to around 2 and then boiling the mixture to separate the acetic acid.
I however the water quantity I have is very huge, so it is not practically possible to boil the water. The water temperature is more than 60 deg C. Also, I do not want to put lots of sulphuric acid in the water due to various other issues and my practical limit of using sulphuric acid is around 80-100 ppm which lowers the pH to around 3.0. I can also providing aeration to the water at 0.2-0.5 vvm.
I want to know that under these conditions, can acetic acid be removed from the water?
Is there any other other method by which we can remove acetic acid by aeration at 60 deg C, by adding not more than 100 ppm of any additive?
Is there any other chance that acetic acid would be removed by aeration at 50-70 deg C temperature? If yes how to enhance the efficiency by minimum inputs?

Have you ever thought of destroying it by simply adding a few grains of sodium bicarbonate ? The acetic acid will then be transformed it into sodium acetate.

Perhaps remaining acetate can be a problem as well.

Not sure how complex a procedure you're willing to put up with, but you might be able to extract with a long-chain amine that is immiscible with water. Tri-n-octylamine has been used for industrial extraction of carboxylic acids. But with a starting concentration of only 200 ppm, any process will be inefficient and it will be very hard to get to nearly pure water.

If you employ aeration here and the acetic acid is in a tap water mix (rich in small amounts of transition metals), then O2 will introduce radical oxidizing species to chemically breakdown the acetic acid and other organics present. This is why aeration is effective for processing natural waters (it does more than just add dissolved oxygen).

The presence of sunlight during aeration (in essence, photo-Fenton and photo-Fenton-like reactions) can also create a cyclic reaction system to more effectively purify water. This topic is incorporated in so-called Advanced Oxidation Processes, on which there are hundreds on papers addressing, for example, environmental remediation, for the removal of unwanted organics (like pesticides) in rivers, lakes,....This is the only practical answer for large bodies of waters.

Maurice, sodium bicarbonate will not destroy acetic acid, just convert it into sodium acetate which doesn't solve my problem (COD).

Andrew, I know the process is complex and not obvious to many people. But I don't want to spend on heavy equipment right now and all I have is an aeration tank to which I can pump the water. I just want to know whether vigorous aeration would remove significant amounts of acetic acid at >50℃ by adding some chemicals that make the separation more easy.

AJKOER, Can you provide more details? Which transition metals can help in the reaction that you mentioned? The TDS of my water is less than 100 ppm, and I don't expect lots of metals to be present in it. How does O2 form radicals?

Aeration does not seem to be the way to go.
If there are no other ions in the water, electrolysis (Kolbe reaction) would convert the $\ce{HOAc}$ to ethane and carbon dioxide:
$$\ce{2CH3CO2^- -> 2e^- + 2CO2 + H3C-CH3}$$
This process could probably be arranged to be done in a pipe with electrodes on the sides, so that the water was treated as it was used, rather than in a huge tank, where diffusion would slow down the completion of the reaction.

This is more correctly referred to, in actual practice, as electro-fenton which produces radicals (via the application of electric current) to chemically destroy the unwanted organic. Main problem, any chloride ion presence, could be transformed into chlorine radicals and subsequent creation of very problematic chloro-organics. Unless you are very sure on the complete absence of Cl- here, I would strongly advise against this electrolysis path.

@AJKOER: I wonder if there would be an electrode that could defeat the chloride problem: e.g., a (porous?) silver electrode. Fairly noble, perhaps it would wind up precipitating any chloride ions at the anode, but still permit alkane formation. Or selecting the applied voltage?

Interesting idea. Try Silver and Activated Carbon for electrodes in water/acetic acid containing some NaCl. Do you see a white precipitate? However, while this may reduce chloride ions, I would not count on a total removal of possible very small amounts of chloro-organics.  A source: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC5480235/  as I apparently can support my comments.

The chlorides are <20 ppm in my water. There are ,however, some sugars (sucrose, glucose and fructose) below 50 ppm. The overall TDS is less than 100. Silver electrode is an interesting idea..but that would be way more costly than I can afford I guess. The water quantity I want to treat is >30 m3/h.  Such systems would require very careful design engineering.

I should have used this as an answer: a strongly basic ion-exchange resin!

@James, I know. That looks like a better solution.

Acetic acid does not form an azeotrop with water and is less volatile than water. So aeration of water would do the opposite - enriching of water by acetic acid due preferred evaporation of water.
Reverse osmosis should help, perhaps after neutralization to be mostly in acetate  form.
The question is, if it is worthy the troubles. The cheaper, easier and faster way could be replacing the water.

Given that the water is already on the warm side, I wonder how efficient distillation could be. The goal wouldn't be to boil all the water at once; instead you'd want to use a heat exchanger to reuse as much heat as possible, both from the clean water output stream and the enriched waste stream.

Boiling is very expensive, much cheaper would be water replacement.

Water will boil at 60C at about 150mm Hg pressure or 0.2 bar. That doesn't need any heat input.

@alephzero  Evaporation of 1 L would cool down 110 L of water approx by 5 deg C.

@Poutnik, We have used Reverse Osmosis after neutralization with caustic soda. The only problem is that it generates around 30% reject water which is unusable for me.

It would help if you described originally  all circumstances and relevant info, so advisors would not walk blind in darkness of what is or not acceptable for you. There is generally no magic wand to get rid of unwanted compounds in water.


text:
What are the fragments that result from taking the Mass Spectrum of o-fluronitrobenzene?
I am doing my online lab, and we are on mass spec right now. I am having a very tough time figuring out the shape of the fragments of o-fluronitrobenzene, especially the large peaks. Especially the one at 75, 83, and 111. I have been drawing fragmentation diagrams for a while now, but have come no closer to understanding what those fragments actually look like.
I am fairly confident that the peak at 95 is a phenyl ion after the loss of entire nitro group. Also, the small peak at 125 also suggests the fragmentation of simply the nitrogroup(after the loss of the oxygen radical), which results in the cation being squarely on the Nitrogen. Although I am confused why the peak is relatively small. I suspect it is to do with the unfavorable oxygen radical being produced. My understanding is that if a compound is nitrogen contaning, that it will give a odd mass on the spectrum.
However, I think that it would not apply in this case as the nitrogen is bonded to two oxygen(both even), and the benzene give us all even masses. The only other odd mass atom would be the 19 Fluorine. It is my suspicion that the 75, 83, 111 peaks all do contain fluorine to some extent. I am just confused on how the fragmentation would lead to the masses with the fluorine on it. I know it contains fluorine, and most likely not a lot of the nitro group.

Why not to write down all options for the given mass ? Community members like to see some particular OP effect before starting the effort on their own

Well, tell us what some fragments can be according to you, for example lighter ones ones if you have trouble with heavier.

H=1, C=12, N=14, O=16, F=19. Try some basic arithmetics with involved chemical knowledge. C6H4NO2F=141. 141-83=58, 141-75=66, 141-111=30.

Don't equate the chemistry from the bench 1:1 to the inside of a MS.  Low concentration of the molecules (vacuum) and high energy (if you opt for the 70 eV of electron ionization) and subsequent fragmentation may yield ions and molecules you can't store in a bottle.  Because the relative intensity of peaks in MS depends on the ionization parameters it often is acceptable to attribute only the the key and/or the most intense ones.  So play with a sketcher (to determine the masses) for a putative attribution of the signals and put them on the table for discussion.


text:
Why is water transferred from a beaker of pure water to a beaker of sugar solution when placed in the a closed container?

I know that there is something to do with the vapour pressure of pure water being higher than the vapour pressure of the sugar solution, but I still can’t really explain what happens on a molecular level. Could someone help?

Water molecules in gas phase collide and are absorbed by both solutions at the same rate, assuming the same humidity over the solutions.
OTOH, as sugar solution has lower molar fraction of water than water itself,  its rate of water evaporation is lower than pure water evaporation.
Therefore, there is the net transfer of water molecules from the water beaker to the sugar solution beaker.
It can be formulated also by other way: The saturated vapor pressure over sugar solution is lower than over pure water.  Therefore vapour saturated over water is oversaturated over sugar solution. And vice versa - vapor saturated over sugar solution is undersaturated over water.

The water goes to the sugar beaker spontaneously, be careful for this word, in that you don't make any effort to move it there except only connecting the two beakers. Any spontaneous process have a negative free energy. The negative free energy is the result of comparing the energy state of the system before the process (water only beaker) and the energy state after the process (water mixed with sugar in the second beaker). That is, systems in nature like to decrease their energy state.
What happens at the molecular level is that water molecules makes favorable hydrogen bonds with sugar molecules thus decreasign the enthalpy of the system and increase the disorder of the system of water molecules by mixing with sugar thus increasing the Entropy of the system. Both effects reults in decease in the energy state of the system and favoring the water sugar situation.


text:
opening valve separating two volatile solutions

Let there be two containers A and B containing two volatile solutions (ideal) of  pure liquids P and Q  respectively. The containers are connected by a tube such that only vapours can travel through it but it is initially closed by a valve. Now the valve is opened at $t=0$.
What will be the final equilibrium pressure over the containers?

My  Approach:
I think that final pressure will be somewhere between the pressure of pure liquid P and pure liquid Q, might their average but I can't say for sure.
vapour pressure is a function of temperature so might be there is no change ,if the liquids would have been mixed inn a single container , forming ideal solution with specified mole fraction of each component then I know how to use Raoult' s law but here I don't see the situation being an equivalent one.

@Poutnik vapour pressure is a function of temperature so might be there is no change ,if the liquids would have been mixed inn a single container , forming ideal solution with specified mole fraction of each component then I know how to use Raoult' s law but here I don;t see the situation being an equivalent one.

Remember you consider equilibrium, not the initial state. There are onviously 2 cases, for miscible and non miscible liquids.

@Poutnik    yes, I am thinking about equilibrium only

So there is nothing to solve, you already know all, what is needed, just write the expression for the final pressure.

There is not enough information to decide, and the question is not clearly worded.
As to the wording, "two volatile solutions (ideal) of pure liquids P and Q respectively," implies there is no vapor present, yet "vapours can travel through". What is the volume of each volatile liquid, and the total volume of the containers, including the connecting tubes? If the whole apparatus has no place for vapor, then the pressure is determined by uncompressible liquids.
As to insufficient information, do the two chemicals react? For example, if the liquids are $\ce{HCl(aq)}$ and $\ce{NH3(aq)}$, the resulting $\ce{NH4Cl(aq)}$ would have a far lower vapor pressure than either of the initial reactants.


text:
NaHCO3 decomposition dynamics and extent vs Temperature
I'm wondering approximately how fast and to which extent $\ce{NaHCO3}$ will decompose (to $\ce{CO2}$ and $\ce{H2O}$) at $50$, $60$, $70$, $80$, $90$, $100$, $110$, $120$, and $\pu{130 ^\circ C}$.
It seems as if J. Environ. Sci. 2019, 78, 74–80 would answer my question but unfortunately I can't access the content.
Does anyone know ?

If you notice it says "thermodynamics" and not "kinetics" . kinetics decides the speed whereas thermodynamics discusses the feasibilty.

Safdar, you are right!

The paper doesn't tell how fast will it be.
As for the rest, have this:

This doesn't seem to be an answer to the actual answer more a comment with images. Wouldn't it be better to link it to an answer in the sandbox?

True, but you can't include images in comments, and using the sandbox to this end seems somehow far-fetched.


text:
Why doesn't pure water conduct when it dissociates?
Sorry if this is an obvious question.
I am told water dissociates, one product of which is negatively charged hydroxyl ions. This makes sense, but then why doesn't pure water conduct electricity? Surely the hydroxyl ions could act as charge carriers to allow current to pass through the solution?
Thanks

Pure water does conduct but it has a very low conductivity.

Presumably due to the low ionisation of water? Is there a certain amount of conductivity below which we say something doesn't conduct? @MaxW

LOL - Well in capacitors the conductance is so low that it is called "leakage" not "conductance."

Pure water does conduct inasmuch as it dissociates.

@ Controller. You are right. In pure water, the concentration of the ions is $\ce{10^{-7}}$ M. So the currant is about $\ce{10^{-7}}$ times the value obtained with a $1$ M solution of electrolyte. It is probably some nanoamperes. It looks as if water does not conduct. Or, following MaxW, the currant may be considered as leakage currant.

Currant?  Sounds yummy, actually.  Current, with an "e", is more closely related to electricity.

Does this answer your question? Why can't pure water conduct electricity since it can be reduced at cathode and oxidised at anode?

Pure water does conduct as pointed out by the above comments. But it has very less dissociation hence there are very few ions to conduct electricity.
you may have heard that $pH$ of pure water is 7. This statement implies that the concentration of H+ ion is $10^{-7}$. Since it is a neutral solution concentration of $\ce{OH^{-}}$ ion is also $10^{-7}$. As such the conducting property of water is very less.


text:
The reason behind the alphabetical order in nomenclature
I hope my question is not too dumb but hey I am learning from the beginning
so when we have a compound such as 6-ethyl-3methylnonane, obvioysly we arrange the alkyl groups alphabetically. Then if we just follow the numbers, we can get somethign like 3-methyl-6-ethylnonane and that saves the extra step of arranging alkyl groups alphabetically. it's much easier to follow.
So I assume the extra step of arranging things alphabetically comes with a reason behind it? otherwise doesn't make sense to make things harder?

There is a thing called "alphanumerical order'. For starters, read this: Nomenclature of organic compound

We have established conventions so that compounds have one unambiguous standard name.


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What is the effect of impurities on ice?
So, I know that the salt dissolves in the thin film of water on the surface of the ice, thus resulting in freezing point depression... but how exactly do i quantify this? Is there any effect on temperature? Will perhaps water be formed? Im tryna think of a way to measure the freezing point depression by adding salt to ice. Thanks in advance.

Does this work for you? Here's another example

https://chemistry.stackexchange.com/questions/61883/effect-of-impurities-on-melting-point .... https://chemistry.stackexchange.com/questions/65929/why-is-brine-solution-more-effective-than-granular-salt-in-melting-ice-on-a-surf .... https://chemistry.stackexchange.com/questions/116302/where-does-the-energy-come-from-to-lower-the-temperature-of-a-brine-solution/116316

@NilayGhosh I dont quite understand - adding salt to ice has 2  effects: it lowers the freezing point of ice and it results in an endothermic reaction. i dont know to 2 reconcile these 2 things.

@NilayGhosh if salt causes the temp of ice to fall, how does this help prevent freezing/increase melting?


text:
Does zinc react with nitrogen?
Does zinc react with nitrogen ?
Is this reaction possible:
3Zn + N2 → Zn3N2
Or maybe there are another reaction?

Yes, it does. $\ce{}$

Wikipedia gives several synthetic routes to zinc nitride.  One of these routes is electrical discharge between zinc electrodes in a nitrogen atmosphere.  Synthesis without electrical discharge generally uses ammonia as the ultimate nitrogen source.


text:
What is the chemical formula(s) for liquid potassium-based soap?
Many places say that liquid soap is generally potassium-based, while the solid stuff is sodium-based.
But, while I have found several sites saying that solid soap is usually sodium stearate, I can't find any place that gives a specific name(s) to the chemical(s) in liquid potassium soap...

What exactly are you trying to figure out? Why they use potassium or the exact formulas for the soap? Companies tend to use a range of surfactants depending on the application, i.e. food grade surfactants, industrial cleaners, etc. additionally, co-surfactants can be added to promote micelle formation.

The so called liquid soap is made of potassium stearate, mixed with some potassium palmitate. It is more efficient than sodium based soap.

Potassium salts of fatty acids are rather liquid, compared to sodium ones, but what is commonly sold under variations of liquid soap name, is not based on that. Personally I doubt I have ever met potassium liquid soap.

You'll get an answer quickly if you search for "formulation liquid soap". There are for instance DIY guides on making liquid soap starting from natural oils (sunflower, coconut) by saponifying the oil with KOH. The fatty acid composition of sunflower oil can vary significantly depending on breed (see e.g. the Wikipedia) but generally contains mono- and polyunsaturated FAs in large proportion. Unsaturation contributes to the fluidity of the product. It is also worthwhile to check common nomenclature for FA chain length (e.g. see the Wikipedia). Sunflower oil contains palmitic (C16:0) and stearic (C18:0) acids as principal saturated FAs. Coconut oil contains shorter (lauric, C12:0) and mainly saturated FAs. Mainly because of the extent of saturation, coconut oil has a higher melting point and is used as the minor component.
In addition there are plenty of commercial vendors willing to provide recipes you can follow to make your own liquid soap. Note not all require a potassium salt. For instance, the company Stepan provides a recipe which consists of mixing the following (in decreasing order of weight %) to form one phase (which is then mixed with another lacking detergents):

Deionized Water
BIO-TERGE® AS-40 (or BIO-TERGE® AS-40 HP)  ==  SODIUM C14-16 OLEFIN SULFONATE
STEPANOL® AM  == AMMONIUM LAURYL SULFATE
STEPANOL® WAT == TEA LAURYL SULFATE
NINOL® M10  == COCAMIDE MIPA

Admittedly ammonium counterion (and TEA) is like potassium in that they are larger and softer counterions than sodium, so similar in their effect on the detergent.


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Chemistry and Plants
We're going to conduct a study to make corn husk fiber into a paper. Soda Ash is always used to extract the fibers from the husks, First question is there any chemical we can use aside from soda ash to extract fibers? Second, we want the fibers to paper white is there a chemical that could turn the yellow-greenish fiber into white? I don't have the means to do the experiment just yet so it would be helpful to have answers.

Instead of soda ash, you may use pure soda, namely sodium carbonate Na2CO3 . Soda ash contains some pure sodium carbonate. But it contains also many other substances, like phosphates, silica, etc. which do not have too much effect for extracting the fibers. Now to turn the colored fiber into white, I am not a specialist, but I would use first bleach (solution of sodium hypochlorite). Also I know that sodium chlorite $\ce{NaClO2}$  is used in the paper industry to turn the crude brownish cellulose fibers from wood into white. Please note that this is chlorite, not chloride, and not chlorate.

Suggestion: avoid bleaching by using the paper in its natural color. It would be stronger, and might be better suited for bags, insulation backing and boxes.


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Mechanism for oxidative workup of ozonide
Oxidative workup of ozonide produces ketone (in case both the substituents of carbon are alkyl groups) or carboxylic acid (in case one of the substituents is hydrogen), depending on the alkene. I have tried for the following mechanism with hydrogen peroxide but unable to find the exact one.

Use blue electrons to break C-O (green) bond. Add peroxide to cation and protonate green oxygen. Now break both O-O bonds with loss of water. Hydrolyze the anhydride.

As RCO2H begins to form, it will catalyze the opening of the ozonide.

@user55119 Thanks, but I am unable to follow or see any pattern and seems like one has to mug up this mechanism. I thought that the first step should be the cleavage of weak peroxide bond of unstable ozonide.


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Why is density of chloroform greater than water?
Hydrocarbons (i.e., molecules containing carbon and hydrogen only) are generally less dense than water.  Yet chloroform, which is a halogenated hydrocarbon, is more dense.
Why is this?  I know water is  more dense than hydrocarbons because its polar nature would cause it to pack more tightly.  But ,why  chloroform, which is less polar than water, has a higher density than water.
It can't be molecular weight, because larger hydrocarbons like hexane and octane have a higher molecular weight than water, yet water is denser.
So is there any theory or concept which will help me find why chloroform is denser than water?

Have you noticed 3 heavy chlorine atoms hanging on the central carbon ? BTW hydrocarbons do not have lower than water density because being nonpolar. It is just side attribute.

Rover your observations are very good.  In the condensed phase, it seems logical to argue, that the volume of each molecule is a nontrivial factor.  That might counteract your argument about the higher alkanes having a higher molecular weight but lower overall density.

Not all hydrocarbons are less dense than water. Then again, chloroform is not a hydrocarbon. Now what is your question, really?

@Rover I edited your question to make it clearer.  Hopefully now it won't be closed.

@Rover.  I see you removed my edits, and restored the original language.

The density of a substance depends on its molecular density and packing fraction (where you need to be consistent in definiting the molar volume in calculating both attributes).
However you are looking for a simpler explanation that gets the general trends right.   Just looking at molecular mass is too simple—larger molecules have greater mass, but take up more space.  So what would be a simple model that works reasonably well?
I suspect the main determinant of density is the average mass of a molecule's atoms, and that relative atomic sizes, as well as packing density, while important, are secondary contributors.
Given this, let's construct our very rough model for predicting relative densities by comparing just the average atomic masses of the molecule's atoms, and see if that, by itself, can explain most of the trends.
Of course, since we are ignoring these other factors, our simple model is imperfect and thus, as expected, will have many exceptions (one of which is included, and bolded, below). Nevertheless, in spite of the very simple nature of this model, it does surprisingly well in matching the general trends (at least for the substances shown here).
In each case I have calculated the average atomic mass of the atoms (u/atom) from $\frac{\text{molecular mass}}{\text{no. of atoms}}$:
Hexane ($\ce{C6H14}$): 86.18/20 = 4.3 u/atom; density = 0.661 g/$\text{cm}^3$
Octane ($\ce{C8H18}$): 114.23/26 = 4.4 u/atom; density = 0.703 g/$\text{cm}^3$
Water ($\ce{H2O}$):  18.015/3 = 6.0 u/atom; density = 0.997 g/$\text{cm}^3$ (at 25$^\circ$C)
Methyl chloride ($\ce{CH3Cl}$):  50.485/5 = 10.1 u/atom; density = 0.911 g/$\textbf{cm}^{\bf{3}}$   (note the exception to the trend; it works well in comparing like compounds—this one vs. the ones below—but fails when comparing very unlike compounds—this one vs. water—when the average atomic masses of the atoms are too close)
Methylene chloride ($\ce{CH2Cl2})$: 84.927/5 = 17.0 u/atom; density = 1.325 g/$\text{cm}^3$
Chloroform ($\ce{CHCl3})$: 119.369/5 = 23.9 u/atom; density = 1.492 g/$\text{cm}^3$
Carbon tetrachloride ($\ce{CCl4}$): 153.8118/5 = 30.8 u/atom; density = 1.594 g/$\text{cm}^3$
Carbon tetrabromide ($\ce{CBr4}$): 331.627/5 = 66.3 u/atom; density = 3.42 g/$\text{cm}^3$ (m.p. = 89$^\circ$C, so it's solid at room temperature)
Carbon tetraiodide ($\ce{CI4}$): 519.628/5 = 103.9 u/atom; density = 6.36 g/$\text{cm}^3$ (m.p. = 168$^\circ$C, so it's solid at room temperature)
Mercury ($\ce{Hg}$): 200.592/1 = 200.6 u/atom; density = 13.534 g/$\text{cm}^3$

Please note that usage of unit symbols such as "amu" or "cc" is not recommended. "amu" is obsolete since 1961 (use $\pu{u}$ instead). "cc" is an old non-standardized abbreviation that still can be found in popular literature, but is not permitted to be used in scientific writing according to the SI system (use $\pu{cm^3}$ instead).

@andselisk  You're right, cc would need to be defined or changed to $\text{cm}^3$, so I did the latter.  Often when I post I'll use convenient abbreviations (b/c having to type out the cumbersome syntax [e.g., $\text{cm}^3$] interferes with my flow of thoughts), and add the syntax later.  So I don't mind the reminder :). Also, note that SI units are not uniformly required in scientific publications. For instance,  "u" (unified atomic mass) is not an SI unit (https://goldbook.iupac.org/terms/view/U06554) [Though "cc" would not be permitted—but because it's too casual, not because it's not SI]

Both the astronomical and unified atomic mass units are non-SI units accepted for use with the SI. The issue with amu is that it was used for oxygen-based units and is now obsolete. I agree there is no need to run with SI units only, but since Chemistry.SE is a multi-national community, using up-to-date and standardized notations would be a tremendous help in scientific communication. We happen to live in a rapid era of changes and globalization, and we need to adapt no matter whether we like it or not.

Density is a macroscopic measurement so we should look for average properties spread over several molecules.
The average density of a liquid is not that much smaller than that of its solid ($\approx 0.9$, there are a v. few exceptions, water/ice, silicon, gallium, germanium, bismuth where it is greater) which means that the distance between molecules is only slighter larger than that in the solid. Just enough to allow one molecule to squeeze past another and make the liquid mobile. This means that the average distance between two molecules is not that much larger than the average distance between its atoms. Even though intermolecular forces are weak (between neutral atoms) there are many such atoms in a molecule and so the total interaction is enough to maintain the liquid. Consequently individual molecular size is not that important in determining density, it is the average mass / volume that counts, where volume means that occupied by numerous molecules.
An example is the very similar density of alcohols of differing mass and size
\begin{array}{lcc} \hline
 & \text{mass} & \mathrm{density\, (kg/m^3)} \\ \hline
\text{methanol} & 32 & 791\\
\text{butanol} & 74 & 810\\
\text{octanol} & 130 & 827\\ \hline
\end{array}
which shows that it is the average mass over several molecules that is important.
For thio- alcohols the density is greater than for alcohols reflecting the increased mass but this falls slightly as the alcohol gets larger,
\begin{array}{cc}\\ \hline
& \mathrm{density \,(kg/m^3)} \\ \hline
\ce{CH3SH} & 900 \\
\ce{CH2H5SH} & 860 \\
\ce{C4H9SH} & 836 \\ \hline
\end{array}
and this is because more lower mass $\ce{C-H}$ is present on average for larger molecules vs the increased mass of $\ce{S}$ over $\ce{O}$ atoms. The increase in size of $\ce{S}$ appears to have a smaller effect in reducing the density than the mass has in increasing it.
In the case of small molecules similarly the average  mass is v. important, for example the density of $\ce{H2O}$ is $\pu{1000 kg m-3}$ but that of $\ce{D2O}$ is $\pu{1105 kg/m^3}$ and in the case of chloroform there is a large in mass in a relatively small molecule so the average mass is far greater than for water taken over many molecules even when the increase in size of $\ce{Cl}$ atoms is accounted for; the densities are $\pu{1492 kg m-3}$ and $\pu{1000 kg/m^3}$

Yes, density is a macroscopic measurement, but the central point of my answer wasn't to average over many molecules, but to do an average over all the atoms within each molecule.  Indeed, by attempting to construct a model purely based on u/atom, I was explicitly ignoring the attribute that can only be seen macroscopically (the packing fraction), and instead attempting to see if we could see explain trends in density purely from a microscopic attribute (average atomic mass of the atoms).  Thus your characterization of what I was doing is the opposite of what I was actually doing.

@theorist, clearly I misunderstood, I will edit my answer to remove a reference to your answer.


text:
Explaining base peak for but-3-en-1-ol mass spectrum?
The mass spectrum for but-3-en-1-ol shows 42 as the base peak, but I don't know how this fragment ion is generated.
I would expect the most favorable fragmentation to occur between C1 and C2, but neither the allylic nor the hydroxymethyl fragments would have a mass of 42.

I suspect that thermal electrocyclic fragmentation occurs where the OH hydrogen is transfered to C4 and formaldehyde and propene are formed. m/z=42 is the molecular ion of propene, C3H6. See: https://sdbs.db.aist.go.jp/sdbs/cgi-bin/direct_frame_top.cgi

Either this or the other option could be standard alpha cleavage leading to an intermediate [CH2OH+, allyl radical] ion-molecule complex, followed by a proton transfer (as PA of formaldehyde 712kJ/mol is lower than that of the allyl radical 736kJ/mol). This would mean the same bond cleavage could lead to both the m/z 31 (direct complex dissociation) and m/z 42 (proton transfer in the ion-molecule intermediate) most abundant fragments. Definite answer would require ab initio (and probably RRKM) modelling.


text:
Maximum number of coplanar atoms in difluorotrimethylphosphorane

In $\ce{PF2(CH3)3},$ what is the maximum number of atoms lying in one plane?

$\ce{P}$ is the central atom with the hybridization $\mathrm{sp^3d}.$ Two $\ce{F}$ atoms will form axial bonds with it, while three $\ce{CH3}$-groups form equatorial bonds. Each of these $\ce{CH3}$-groups has tetrahedral geometry, which means that one $\ce{H}$ atom lies in the same plane as the $\ce{C}$ central atom.
Thus, two $\ce{F}$ atoms, one $\ce{P}$ atom, one $\ce{C}$ atom and one $\ce{H}$ atom with it, totalling five atoms, lie in the same plane.
Is my attempt correct?

What if you try another plane?

@Ivan Neretin Which?

For one, you state $\ce{F-P-F}$ are aligned in axial manner, without noting H, nor C.  Based on this assumption, this adds up to three.  But since the axial direction is orthogonal to the equatorial plane, count what's possible there; build a model and don't forget that methyl / $\ce{CH3}$-groups bound by $\sigma$-bonds may rotate like propellers, thus form multiple conformers.

@Buttonwood Ah, I see. There should be 7 atoms in the equatorial plane ($P$, 3 $C$ atoms and 3 $H$ atoms).

@IshanJMukherjee At maximum, true; because molecules are flexible and dynamic /  vibrate (in IR spectroscopy occasionally you occasionally hear a colloquial «breathing vibration»).

Seven is correct (the OP found this in the comments).  The methyl carbons are in the equatorial plane and each one can rotate a hydrogen atom into this plane.  In fact the hydrogen atoms fit well with this conformation so it's actually favorable.


text:
What is the mechanism for this Horner-Wadsworth-Emmons (HWE) reaction?

In this reaction the HWE reaction does not give a "normal" alkene product, but instead gives (R/S)-2. Provide a reasonable mechanism for its formation?

Interesting looking problem, but we really need you to show us your thoughts on it before it gets answered otehrwise it will be closed as homework.

First dil. aq. HCl hydrolyses the starting material to the dialdehyde.
Second the diethylphosphonoacetate anion reacts with one of the aldehyde groups to give the expected intermediate alkoxide. This is in equilibrium with the carbanion shown below. This is not normally depicted in the HWE mechanism as there is usually no pathway forward from it, however in this case there is. This carbanion can, and does, undergo a second HWE with the other aldehyde which proceeds through the usual mechanism to give the products shown in the question

Waylander: FYI, there is a skinny version of ChemDraw online. https://chemdrawdirect.perkinelmer.cloud/js/sample/index.html

@user55119  I tried drawing it in that and failed to find a way to save it so that it would post here. Clearly I am missing a trick.

@Waylander Top menu, Structure → Get Image. This would result in a popup with a PNG file on top that you can save locally. Alternatively, use screen capture software.

Did you use skinny ChemDraw on the second version? I couldn't get @andselisk's first instruction to work -- what to do with all the code--but a screenshot was fine. Fortunately, I have access to the robust version of CD.

I used the link you posted. I'm more used to ISIS draw so the scheme is not as clear as I would have liked.

@user55119 Ignore the code, grab a picture.

@andselisk: How did you get the inset that asks "save as..."?

@user55119 Right mouse button.


text:
Why is the electronegativity of potassium and rubidium same?
The electronegativity of potassium and rubidium is reckoned at 0.82 for both. Why is it same for both of them? Shouldn't it be less for rubidium as compared to potassium owing to the addition of one extra shell in rubidium?

$0.82$ is the Pauling electronegativity. Several other ways exist for calculating the electronegativity, for example according to Mulliken, to Allred-Rochow, and to Sanderson.  On those scales, Rubidium is a little bit  smaller than Potassium. See Electronegativity scales on Wikipedia.

There are a few ways that electronegativity can be measured/calculated, the original being that of Pauling, followed by others scales such as the Allred-Rochow and Mulliken electronegativity. When looking at the values obtained by Pauling, one must keep in mind that these values aren't grounded in quantum mechanics (i.e. electronegativity does not have an observable that can be measured). However, these values seen on the Pauling electronegativity charts are relative values estimated from bond energies.
Anyways, defining Pauling's electronegativity as the "ability of an atom to attract electrons" (Matthew Salem, Chem Libre Texts), one can look at two factors to estimate these values.
1. Nuclear Charge/Atomic radius - As we know from Coulombs law,
$\large F=k\frac{q_{1}q_{2}}{r^2}$
the attraction between to oppositely charged points is a function of the magnitude of each charge (q) and the distance between them ($r^2)$. So the number of protons in the nucleus along with the distance of the electron from the nucleus determine the attractive force it feels from the nucleus.
2. Electron Shielding –  The electrons that are closer to the nucleus can shield bonding electrons from the  attractive forces of the nucleus. The ability to shield outer electrons is greatest in the s orbitals, a little weaker in the p orbitals, and poor in the d/f orbitals.
With these factors in mind, one can approximate the effective nuclear charge $(Z_{eff})$ of an atom, which is an estimation of the attraction certain electrons will feel. Effective nuclear charge is estimated with the equation below,
$(Z_{eff}) = Z - S$
where Z is the atomic number and S is the shielding constant (number of non-valence electrons).
If you calculate the $(Z_{eff})$ for rubidium and potassium, one would find that they are equal, suggesting that the electronegativity of each atom is are approximately the same.
This link discusses $(Z_{eff})$ in more depth if you are interested.


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Fish Fertilizer acid / base reactions
Hi Everyone new to chemistry and learning lots but I had a question that I am having trouble working out.
I work for a vegetable grower and we make our own fertilizer.
My question is around the fish fertilizer. The brief description of the process is we use fish meal and mix it with phosphoric acid and enough water to make a paste. Let it sit a week then add, depending on volume a good amount of KOH. We also add a some skim milk and a small amount of molasses. This process among other things converts the protein in the fish to plant available nitrogen and gives us phosphorus and potassium.
I would love to understand all the chemical processes that are happening here and would like to understand how we are converting the protein in the fish meal to nitrogen. Is it phosphoric acid or KOH that is doing the work and how does the amount of these acids effect the process? Why is such a small amount of acid but a large amount of base used?
I have done my own research but coming up short in understanding what is going on and what the acid and base are doing.

Cells of fish contain proteins, and proteins are made of long chain of amino-acids, linked together by peptide bonds. In the presence of an acid (like phosphoric acid in your experiment or hydrochloric acid in our stomach), the peptide bonds are destroyed and the free aminoacids are released, so that they can be immediately assimilated by living organisms, like roots of plants if these acids are added to the ground as fertilizers.
But the phosphoric acid is not destroyed in the process. It still remains in the aminoacids mixture. As most acids, it is a corrosive substance that must be neutralized before being disposed in the ground. This neutralization is made with KOH, by one of the reactions $$\ce{H3PO4 + 2 KOH -> K2HPO4 + 2 H2O}$$ $$\ce{H3PO4 + 3 KOH -> K3PO4 + 3 H2O}$$ Whatever the reaction chosen, phosphoric acid is now transformed into a potassium phosphate, which can be assimilated. The advantage of this process is that now the fertilizer contains amino acids (bringing Nitrogen N to the soil), plus a salt bringing two important elements for fertilizing a soil : phosphorus P and potassium K. The three fundamental elements (N, P, K) for the agriculture are present in this fertilizer
To go back to the use of KOH, the equation shows that the amount of KOH to be added is twice or three times the amount of the initial phosphoric acid. That is why you have to add a huge amount of KOH to your mixture.

What role do the molasses play? Seems to serve no purpose.

Apparently the molasses are useless.

Seems like that including the skimmed milk :-)

Plants and many bacteria can metabolize simple sugars (such as in sucrose or galactose) directly. This is why plants manufacture sugars in the first place (good solar energy battery). Possibly these are like "energy drinks" for the plants, or fuel for symbiotic soil bacteria.


text:
Intramolecular Claisen-Condensation
My task was to do an intramolecular Claisen condensation with this molecule.

I already draw the product to the reaction, but I'm not sure, if I did it right. Could anyone help me out or confirm the solution? There should be only one product since there is symmetry.
Here is the way how I got the product in my question:

What exactly is your question? Do you want ChemSE to provide reagents or a mechanism? If so, let's see what you think. Then help will be offered. In any case, welcome to ChemSE! Your question will likely be considered as homework. You need to make some effort to answer the question yourself or, at least, make some effort. Take a look here about homework policy. https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange

@user55119 I want to know if the product, which I got is right.

It is necessary to show how you got the product in your question. So, we may able to see you are in right path or not.

Here is the way how I got the product in my question:

Looks all right to me... I think this is called 'Dieckmann condensation'.

Maybe a couple of hints for further reflection: 1) do you think the product you drew may react further in the reaction environment (sodium ethoxide etc)? 2) what stops the initial deprotonated intermediate from undergoing retro-Michael?

@user6376297 1) the beta carbon of the product is actually well sheilded by the carbonyl , ester and geometry of the six membered ring so it will be very sterically costly to react at this center 2) I think amine group is a very bad leaving group so retro-michael is not much preferred

The mechanism you have drawn is catalytic in base. However, the Dieckmann reaction is stoichiometric in base. Why?

Did you just made a different account for writing an answer? If so, request the mods to merge both the accounts.

@user55119 : that was the point of my question 1 :)


text:
Do hydrogen sulfide and oxygen produce pure sulfur or sulfur dioxide? If both, under which circumstances does each scenario occur?
If both of the following hydrogen sulfide and oxygen chemical equations can occur, under what circumstances is the harmless pure sulfur and the toxic sulfur dioxide produced?
$\ce{2H2S + O2 -> 2H2O + 2S}$
$\ce{2H2S + 3O2 -> 2H2O + 2SO2}$
Specifically I have a fresh water aquarium with sand, plants, and small creatures I collected from a creek. I want to prevent hydrogen sulfide building up by oxygenation with an air pump, but I want to make sure the environment does not result in the harmful sulfur dioxide in the tank.

IMHO, it would end as sulphate $\ce{SO4^2-}$

Okay, so how do I get rid of $\ce{SO4^2-}$ in the aquarium?

Sulphates are natural ions in water. Anything that comes by sulphan oxidation is harmless, compared to the equivalent amount of original sulphan.

Per my review of conceivable chemistry, harsh conditions are not required, but likely acidic conditions could increasingly be problematic! Per the context of the question, to quote:

Specifically I have a freshwater aquarium with sand, plants, and small creatures I collected from a creek. I want to prevent hydrogen sulfide building up by oxygenation with an air pump, but I want to make sure the environment does not result in the harmful sulfur dioxide in the tank.

So, consider the scenarios if one passes air/O2, in the presence of say a photocatalyst (B-12, dye,..) or perhaps sonolysis (sound vibrations), into:
$\ce{H2S (aq) = H+ (aq) + HS- (aq)}$
possible issues in time depending on pH.
For example, with sonolysis:
$\ce{H2O + Sonolysis-> .H + .OH}$
albeit, any vibrations (say, from air-pump) are not that likely effective here.
However, continuing with the illustrative chemistry (source of sulfur related radical chemistry, see Page 7, this source):
$\ce{.OH + H2S -> .HS + H2O }$
$\ce{.HS + O2 -> .SO2- + H+ }$
$\ce{.SO2- + O2 -> SO2 + .O2- }$
which is one path to SO2 not contingent on pH, but elevated pH would remove formed sulfur dioxide!
$\ce{.H + O2 -> .HO2 }$
$\ce{.HO2 <=> H+ + .O2- \text{pK}_a 4.88}$
Now, if water in the fish tanks gets too acidic (pH < 4.88), then possible further aqueous reactions include:
$\ce{ .HO2 + .HO2 -> H2O2 + O2 (Slow) }$
$\ce{ .HO2 + HS- -> HO2- + .HS }$
And importantly:
$\ce{ H+ + HO2- = H2O2 }$
which readily converts H2S into H2O + S. Expect with time some aqueous SO2 and perhaps also sulfate as well.
More likely, however, is strong sunlight exposure acting on natural coloration present in the fish tank serving as a weak photocatalyst, generating some electron holes (h+) and also solvated electrons (e-(aq)):
$\ce{Dye + hv -> [Dye]{*} -> Dye + h+ + e- }$
$\ce{H2O + h+ -> H+ + .OH }$
$\ce{.OH + H2S -> .HS + H2O }$
where the reaction could continue (per the sonolysis scheme above with one path to SO2 not pH dependent) where pH < 4.88 further promotes sulfur species, $\ce{ S, SO2 {and} SO4(2-)}$.
So, my review of possible chemistry suggests keep the pH near neutral to be more safe from SO2 (but not entirely), but interestingly if H2S concentrations become dangerous, more acidic pH could be a solution with sunlight, accomodating glass, and some dye.
[EDIT] Confirming source on the scenario of employing light (and also ferrous) in the removal of hydrogen sulfide, to quote:

Such oxidation reactions are catalyzed both by soluble metals such as iron and by light. Hydrogen sulfide also can combine with metals such as iron (Fe++) to precipitate as black iron sulfide (Figure 1 bottom; FeS and FeS2).

I am curious on why the only answer that actually answers the question and has a confirming source on the action of light on H2S (with advanced content on how this could, in effect occur) would not be viewed/accepted as a correct/acceptable answer.

Sulfur dioxide is not a problem ; in an aquarium it will make very dilute sulfurous acid which will likely neutralize with some carbonate mineral/gravel. H2S is lethal , extended exposure at 50 ppm for a day , or short exposure at 500 ppm. On the other hand , what you smell in aquarium sand is less then 1 ppm, the sense of smell is very sensitive to H2S. There are several answers on the internet but I expect most people can smell H2S at 0.1 ppm. An aerator will have no significant affect on H2S in an aquarium. When I clean my pond , which is well aerated ,the smell of H2s is strong from decomposing leaves, etc.
Regarding the first part of your question ; Hundreds of tons a day of elemental sulfur are made ,world wide, every day by burning H2S with limited air at high temperature. By recycling and control of gas compositions the only products are elemental sulfur and water,

So what can I do in my aquarium to reduce $\ce{H2S}$ without changing water? There are many micro-organisms in the aquarium which is from the creek water that I prefer not to be taken out in a water change.

I don't think you need to do anything. It is normal  for small amounts of H2S to be present in the debris on the bottom of natural  bodies of water. And even aquifers have sulfate reducing bacteria that put H2 S into well water.

Please note, this answer has not technically answered the question: "Do hydrogen sulfide and oxygen produce pure sulfur or sulfur dioxide? If both, under which circumstances does each scenario occur?" That would require an exploration of the chemistry which is not provided. But, I enjoyed reading your reply (no rating action taken).


text:
Relation of polarity of organic functional groups and whole molecules
I know this should be a simple chemistry question, but I am currently very confused on how polar/nonpolar molecules work. For example, since the ether group is insoluble in water, does this mean that all molecules containing ether would be insoluable/nonpolar? Likewise, are all molecules containing alcohol group polar then?

Polarity is an ill-defined term which does not mean much except to organic chemists. You should rather ask yourself by looking at the structure, "Does this molecular have a dipole moment?" Learn to identify this part first.

Naming the question on chemistry forum "Simple chemistry question" is like naming a book "Simple book". Also, such a name works like a click bait, what is very disliked. I have changed it.

To further complicate things, molecules can be both polar and nonpolar. Or more precisely, have polar and nonpolar parts. Detergents are good examples of this. They have a polar group such as a sulfate, and a nonpolar group, such as a long hydrocarbon tail.

@Poutnik Wow, and here I was unsatisfied with current title, but yeah, that's a big progress actually! :D Maybe I should lower expectations...

Groups are neither soluble neither insoluble. They contribute on substance solubility or insolubility.
Their polarity relates to dipole moments of respective chemical bonds which relate to the difference of atom electronegativities.
The effect of a group must be related to whole molecule, considering overall structure.
E.g. methanol or acetic acid are polar with unlimited miscibility with water. But $\ce{OH}$ or $\ce{COOH}$ group attached to a long alkyl chain do not make the whole molecule polar and notably soluble.
Fatty acids are not called fatty without reason. Because polarity of the group is overruled by nonpolarity of the long alkyl chain.
Note the diethyl ether solubility in water is about $\pu{6 g/100 mL}$, so the ether group is at least moderately polar, compart to the formal methylene group $\ce{CH2}$ of  n-pentane, that is practically insoluble.


text:
What is "activated red phosphorus"?
[WP = white phosphorus ; RP = red phosphorus ; ARP = activated red phosphorus]

A YT channel, Extraction & Ire has a video from two years ago (2018) on converting WP to RP by keeping the former in sunlight for several days. Chloroform was chosen as solvent to dissolve WP in a glass vial. The solution had gone from deep orange to red on slight when kept in sunlight for 5 days. After 5 days, the solution was tested with sodium hydroxide solution to confirm that if it is RP (RP doesn't react with $\ce{NaOH}$, WP does). On dumping it to the $\ce{NaOH}$ solution, it immediately started reacting (phosphine was evolving which had a dreadful smell). After 15 minutes on adding, the orange-red color vanished, that means all of the phosphorus has reacted. Some white crystals formed in the solution which is presumed to be sodium formate ($\ce{HCOONa}$) from the chloroform (isn't it soluble in water?).
$$\ce{P4_{(?)} + 3NaOH + 3H2O -> PH3 ^ + 3NaH2PO2}  \\
\ce{2NaH2PO2 -> PH3 ^ + Na2HPO4}   \\
\ce{CHCl3 + 4NaOH -> HCOONa (\downarrow ?) + 3NaCl + 2H2O}$$
So, the conversion failed. He apparently named the species "ARP". He made the following conclusion:
\begin{array}{c|c|c|c}
\mathbf{~} & \mathbf{WP} & \mathbf{RP} & \mathbf{ARP}  \\\hline
\text{color} & \text{white} & \text{red} & \text{red}\\
\text{soluble in chloroform} & \text{Yes} & \text{No} & \text{No}\\
\text{Reactive towards NaOH} & \text{Yes} & \text{No} & \text{Yes}\end{array}
So, apparently WP converted to an unknown species(allotrope?) which was assumed to be ARP. We actually don't know its composition nor we can find any convincing literature. Google didn't give any relevant hits except two papers:

Sukhov, Boris & Malysheva, S. & Kuimov, Vladimir & Smetannikov, Yu & Tarasova, N. & Lupanov, A. & Gusarova, N. & Trofimov, BA. (2004). Reaction of Activated Red Phosphorus with Allyl Bromide under Phase-Transfer Catalysis. Russian Journal of General Chemistry - RUSS J GEN CHEM. 74. 1128-1129. 10.1023/B:RUGC.0000045878.42411.31 (it says it is  polymerized organophosphorus chains)
Nucleophilic Activation of Red Phosphorus for Controlled Synthesis of Polyphosphides, Minyoung Jo, Alina Dragulescu-Andrasi, L. Zane Miller, Chongin Pak, and Michael Shatruk,Inorganic Chemistry 2020 59 (8), 5483-5489 DOI: 10.1021/acs.inorgchem.0c00108 (it says it is series of polyphosphides)

Questions:

What is ARP? What is the actual composition of ARP?
Is it a different allotrope of phosphorus?

The 2020 paper says "RP is readily activated by refluxing it with different potassium alkoxide to give soluble polyphosphide ions". Polyphosphides is known and has been studied a lot (first paper I could find dates way back in 1967). So, what does "readily activation of red phosphorus" mean in this context? Could it mean ARP?

Another fancier allotrope of P is black phosphorous. Since nitrogen and P are of the same group, scientists have been able to make black nitrogen as well. There is a violet allotrope of P as well.

@M.Farooq Yes indeed, all those allotropes were discovered way back (violet - 1865, black -1914). Also, this question discusses about allotropes of grp 15 elements: https://chemistry.stackexchange.com/questions/28124/allotropy-in-nitrogen-and-bismuth-of-group-15

Black nitrogen: https://www.sciencealert.com/a-new-kind-of-black-nitrogen-finally-resolves-a-periodic-table-mystery ... https://phys.org/news/2020-06-scientists-long-sought-after-nitrogen-allotrope-black.html ... https://newatlas.com/materials/black-nitrogen-allotrope-periodic-table/

Well if ARP is "a series of polyphosphide ions and polymerized organophosphorus chains" it isn't an allotrope of phosphorous

@IanBush Even if it is not an allotrope, why was this phosphorus species unknown for a long time? This species doesn't seem exotic nor this requires drastic conditions/equipment/chemicals to synthesize. Also do note that I have given a question mark beside the word allotrope.

I can't answer that. But I will comment that the complex allotropy of phosphorous always amazes me, particular in the bizarre set of experimental conditions required to make this subtly different allotrope from that one. I mean recrystallise from molten lead - why did Hittorf think of that? So maybe these guys were simply the first to "wake up this morning, got the dissolve it in chloroform and irradiate for 5 days blues" As for the composition, well from what you have presented that looks like an active research question.


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Hybridization of orbitals and forming of bonds in the nitrogen dioxide molecule

I considered the nitrogen atom hybridizing its orbitals and then ending up having 3 $\mathrm{sp^2}$ orbitals and the oxygen with 2 $\mathrm{sp}$ orbitals. The thing which bothers me is when people draw a radical (single electron) over the Nitrogen atom after there is the bond formation between nitrogen and 2 oxygen atoms.
Let us consider the Nitrogen, it has one full $\mathrm{p}$ orbital (AKA lone pair) and 3 single electrons in 3 $\mathrm{sp^2}$ orbitals, so that means that it can form 3 bonds – first it bonds with one oxygen forming a σ bond, leaving the nitrogen with 2 electrons, now the nitrogen forms one π bond with the same oxygen and the remaining single electron forms another σ bond with another oxygen atom.
Where is the single radical on the nitrogen after all the bonding has taken place?

Moreover, as in the above picture, people draw the other oxygen (the one that did not form a π bond with nitrogen) as having 3 lone pairs, whereas, after the hybridization of the oxygen atom, I can only see it having 2 lone pairs in 2 $\mathrm{p}$ orbitals.
What is actually correct? Am I going wrong somewhere?

In your simple model, you start off with:

nitrogen: $\mathrm{3(sp^2)^1 + p^2}$
left oxygen: $\mathrm{(sp)^1 + sp^2 + p^1 + p^2}$
right oxygen: $\mathrm{(sp)^1 + sp^2 + p^1 + p^2}$

The σ bonds are easy. Use $\mathrm{sp(O) + sp^2(N)}$. That leaves you with:

left oxygen: $\mathrm{sp^2 + p^2 + p^1 + (\sigma_\ce{O-N})^2}$
nitrogen: $\mathrm{p^2 + (sp^2)^1 + (\sigma_\ce{N-O(1)})^2 + (\sigma_\ce{N-O(2)})^2}$
right oxygen: $\mathrm{sp^2 + p^2 + p^1 + (\sigma_\ce{N-O})^2}$

Now the question is how to do a π system (obviously there can only be one σ bond between two atoms). The orbitals used for the π system must be orthogonal to the bond axis, since that axis must be a node. Both oxygens have two p-type orbitals which fulfill this condition. Nitrogen also must use its p-type orbital to form the π bond. The $\mathrm{sp^2}$ orbital cannot because it is in the same plane as the other two $\mathrm{sp^2}$ orbitals, angled at $120^\circ$ rather than $90^\circ$. This gives us a π system composed of two oxygen p orbitals (one on each oxygen) and one nitrogen p orbital; choosing our orbitals well will result in this π system having four electrons. Essentially, this is theme and variation of Your Average 4-electron-3-centre bonding; but you can also consider it akin to the allyl anion.
The one remaining unpaired orbital would be in $\mathrm{sp^2}$ on nitrogen, pointing away from both $\ce{N-O}$ bonds but being in the same plane.

The above are the considerations and reasonings that you should use from the starting point you had. It is not the only way. You could assume that nitrogens five electrons are actually distributed as $\mathrm{2(sp^2)^1 + (sp^2)^2 + p^1}$. In this case, the argument outlined above is essentially the same except that the π system now contains three electrons rather than four; it is an allyl radical system, if you wish. Since this ab initio construction will lead to all orbitals with s contribution ($\mathrm{sp^2}$ on nitrogen, $\mathrm{sp}$ on oxygen) being fully occupied, it is more consistent with Bent’s rule and likely to be a better proposition.
If you do it like this, you can – in the Lewis 2-electron-2-centre formalism – choose either oxygen and have the radical form an $\ce{N-O}$ π bond on that side; the other will remain a radical. By resonance, you can draw the mirrored molecule.
$$\ce{O=N-\overset{.}{O} <-> \overset{.}{O}-N=O}$$
In these structures, the radical is never centred on nitrogen although we find that dimerisation at low temperatures leads to $\ce{O2N-NO2}$.

Actual measurement shows, that the radical is centred in an orbital that transforms as $\mathrm{a_1}$. As $\mathrm{a_1}$ is symmetric with regards to all symmetry transformations, it cannot be a π type orbital which can transform either as $\mathrm{a_2}$ (node through nitrogen) or $\mathrm{b_1/b_2}$ (no node through nitrogen; depending on your choice of coordinates). You see, it gets more and more complicated the more you rely on experimental data.

Very nice answer! Interesting that CO2 has one electron less (is isoelectronic with NO2+), and nitrite ion has one electron more. Both are easier to imagine, but halfway in between is a monster!

$\ce{NO2}$ , $\ce{NO}$, $\ce{ClO2}$ and a few other are deceptively simple molecules with surprisingly sophisticated bonding. Their reasonable understanding requires use of molecular orbitals theory. Below I try to invoke resonance to provide a simpler understanding for $\ce{NO2}$
$\ce{NO2}$ is a 'hypervalent' compound, i.e. to describe it adequately, one should invoke tricks involving dative bonds. I think, it is easiest to think about it as a resonance hybrid of $\ce{O = N^{.+}-O^{-}}$ and $\ce{O^{-}- N^{.+}= O}$. Indeed, $\ce{NO2}$ fragment bound via nitrogen is fairly common in chemistry and $\ce{NO2}$ readily interacts with, for example, chlorine:
$\ce{2NO2 +Cl2 = 2 ClNO2}$
and direct ESR spectroscopy data * suggest strong spin density on nitrogen. However!
$\ce{NO2}$ radical is too stable for this simple interpretation, and same ESR data suggest that considerable spin density in NO2 is localized on oxygen atoms. This means that structures
$\ce{O^. - N = O}$ and  $\ce{O = N - O^.}$ also strongly participate.
Since no structure with unpaired electron in pi-system allows for a considerable spin density on nitrogen, it leads to conclusion that spin density is delocalized with participation of $sp^2$ hybridized lone electron pairs of oxygen, in particular ones opposed to the in-plane $sp^2$ orbital with the lone electron.  This description is apparently in agreement wit MO treatment of $\ce{NO2}$. This IS counter-intuitive, since common organic sense is to put unpaired electron into $\pi$-system

https://www.tandfonline.com/doi/abs/10.1080/00268976800100641

Thanks for your reply, but I am asking as to why the hybridization won't work here. The thing which you told about resonance, I am in the first place confused as to why does formal charges appear in the molecule, and how does the radical appear on the nitrogen, as it would not make sense wrt my diagram.

@FinalBOSS You can make it work if you use the above treatment, it implies every atom is $sp^2$ hybridized. But in general you shouldn't ask 'why hybridization doesn't work here', but 'why hybridization would work here'.

How is $\ce{NO2}$ hypervalent?

@Jan It has significant spin density on the nitrogen atom and the nitrogen forms three bonds, i.e. it has nitrogen with a 'dangling bond' and three normal bonds. But it cannot form more than three bonds when neutral.


text:
Is plane of symmetry a necessary and sufficient condition for optical inactivity in coordination compounds?
In organic chemistry, we have to check for all three types of symmetry in a compound to check its chirality, ie, plane, alternate axis and centre of symmetry. But in coordination compounds, do we need to check all the symmetries to check its chirality? The ligands are supposed to be achiral. The difference between the two being that in coordination compounds only one "chiral" center is present but in organic compounds it can be more than one.
All the achiral coordination compounds of any coordination number which I have come across have a plane of symmetry. But for those which don't have, I have to check whether it's image coincides with itself or not. So, is plane of symmetry a necessary and sufficient condition for achirality of coordination compounds?
This thing is necessary for my purpose, as I have to solve problems in limited time, so I cannot use the orignal method always.

Related: https://chemistry.stackexchange.com/questions/123121/link-between-centrosymmetry-dipole-moment-and-chirality

I remember my coordination chemistry professor at uni drawing a tetracoordinating chelate ligand with a central metal on the blackboard. The tetrahedral complex had improper rotation (which we spent two minutes proving by him drawing the mirror image). It was thus achiral. Alas, it’s not mentioned on his webpages …

Does this answer your question? What is the perfect definition for chirality?

The number $\pi$ for bowhead whales† is still 3.1415... Same thing here.
Chirality of coordination compounds is the same thing as chirality of organic compounds. If all your symmetry elements are just rotations, you have it; if some are improper rotations, then you don't. It is as simple as that, and is not going to be any simpler.
Can you just check for the planes and call it a day? Well, you kinda can, and that will do, until you come across a compound which doesn't have a plane, but still is achiral because of some other symmetry element.
Are there such compounds? Well, that's like asking whether there is a place on Earth where three bricks stand on top of each other. You may google long and hard, or you may apply for a travel grant to visit countries with rich traditions of brickwork masonry... or you may just take three bricks and put them on top of each other. That's it; problem solved.
Here are my bricks. I claim that the tetrahedral $\ce{[Zn(NH3)4}]^{2+}$ has an inversion axis $\bar4$ and hence is achiral. You may point out that it has a plane as well. Fine. Let's give it some substituents: change each ligand to $\ce{N(CH3)3}$. What? Does it still have a plane, given that everything which rotates is rotated to the appropriate position? Fine. Let's give it different substituents. Change each ligand to $\ce{CH3NHC2H5}$. Where is the plane now?
So it goes.

† There was an old anecdote about an even more old paper on biophysics which was said to include some volume calculations and some crude approximations of a whale as a cylinder, then of course the number $\pi$ showed up, followed by the hilarious phrase along the lines of "The number $\pi$ for bowhead whales is about 3.14".

But the chirality introduced due to ethyl methyl amine is because of the chirality of the ligands and not because of the arrangement of different ligands around the central atom. I have assumed the ligands as non chiral.

@siddharthkalra   If the ligands are chemically different, because you arrange around a central ion $\ce{M^{4+}}$ at once $\ce{F-}$ and $\ce{Cl-}$ and $\ce{Br-}$ and $\ce{I-}$, than the complex is chiral even if the ligands are not.  Well if the four are chemically different like methyl, ethyl, propyl, ... (which includes the «if at least one them is chiral»), the complex is chiral, too.

No. A plane of symmetry is a sufficient, but not necessary criterion to exclude optical activity of a molecular compound.  Equally sufficient, but not necessary criterion for the absence of chirality is the presence of a centre of inversion.  Note that you probe for these criteria independent from each other the whole molecule.
An easier example of organic molecules suitable for the first criterion is (2R,3S)-tartaric acid - equally known as meso-compound - where you may draw a mirror plane across the molecule even if you have two examples of atom centred chirality in the molecule present.  (It is not good to call the stereogenic atoms «chiral atoms».  Beside atom centred chirality, there equally are helical and axial chirality, too.)
By training, you are going to identify these compounds by inspection of their formulae:

(credit)
Don't hesitate to build models of the molecule in question and to redraw the structure formulae like the following to get more familiar with the topic:

(credit)

 Only the two first columns of the four depict a mesocompound.

But I was talking about coordination compounds having only one central metal atom and having non chiral ligands. I am fine with chirality in organic compounds. I just wanted to ask if there is any shorter method to check chirality in such  coordination compounds by just checking plane of symmetry, as all the coordination compounds I have come across have a plane of symmetry if they are achiral.

@siddharthkalra No.  As stated, you probe the molecule / complex for a mirror plane which either is positive (is present), or negative (is absent).  And you probe the molecule for the presence / absence of a centre of inversion.  The order of testing them may be formalized in a flowchart (like here) if aiming to determine the point group of the object (e.g., IR spectroscopy), but neither one of the two tests is a necessary and sufficient condition for the other to yield either answer.


text:
Why does hydrolysis of ADP to AMP yield the same amount of energy as ATP to ADP, if ADP is more stable?
I learned about the resonance stability of ADP and the fact that ATP is less stable due to intramolecular instability. I was surprised to see that the energy net of conversion of ADP to AMP is the same as ATP to ADP. Shouldn't the energy net of a reaction ADP to AMP+P be lower because it's not so readily hydrolysed?

https://chemistry.stackexchange.com/questions/37677/why-is-atp-but-not-adp-or-amp-a-sole-source-of-energy

I learned about the resonance stability of ADP and the fact, that ATP is less stable due to intramolecular instability.

What you are probably refering to is an explanation why ADP and phosphate are more stable than ATP (and water). The explanation goes something like this:

[Tiffany Lui, University of California, Davis on Libretexts] Resonance stabilization of ADP and of Pi is greater than that of ATP. The oxygen molecules of the ADP are sharing electrons. Those electrons are constantly being passed back and forth between the oxygens, creating an effect called resonance. This stables the ADP. Resonance does not occur in ATP; therefore, it is a more unstable molecule.

Source: Libretexts
This argument has gone awry. First, you can't compare the stability of two species unless they contain the same set of atoms and have the same charge (see e.g. https://chemistry.stackexchange.com/a/146497). Second, the resonance stabilization on the terminal phosphate is similar for ADP and ATP. The better argument is that there is more resonance in ("inorganic") phosphate than in a phosphate ester or phosphoanhydride.
The experimental data shows that the energetics of ATP hydrolysis yielding ADP are very similar to ADP hydrolysis yielding AMP. In fact, there is an enzyme that turns ATP and AMP into two ADP, and the equilibrium constant for that reaction is close to 1:
$$\ce{ATP + AMP <=> 2ADP}$$

ATP has two high energy bonds which are phosphoanhydride bonds. Both bonds release the same amount of energy. Therefore, when ATP converts to ADP, breaks one phosphoanhydride bond, and releases -30.5 kJ/mol. When the 2nd bond is broken in ADP releases -30.5 kJ/mol because it is the same phosphoanhydride bond. However, AMP has only a phosphate ester bond which releases -61 kJ/mol.
The "less stability" of ATP doesn't affect energy yield.

You are quoting Gibbs energy changes for the reaction with water at the biochemically defined standard state (neutral pH, some magnesium ions), $$\Delta_R G^{\circ \prime}$$

The term "high energy bond" is confusing to students as soon as they learn a bit of chemistry. It might be less confusing to say that ATP has a high phosphate group transfer potential, or simply that hydrolysis is an exergonic reaction. It is confusing because no bond "releases energy"; if it did, it would not form. Instead, the bonds formed after hydrolysis are stronger (electrons at lower energy) than those in place before hydrolysis.

The Gibbs energy for a phosphate ester hydrolysis (e.g. AMP to adenosine) is on the order of -15 kJ/mol, not the -61 kJ/mol claimed. That value might be for ATP hydrolized to AMP and two phosphates.


text:
Will group 2 chlorides dissolve in water in water or react to produce distinct products? How does this differ by the type of salt?
I thought it'd dissolve... but am i wrong? I added calcium chloride to water, and it was smoking, i was so confused. Can someone explain happened? I am doing an experiment where i dissolve salts in  water, but im very confused now.
PS - my other salts are barium chloride, strontium chloride and magnesium chloride. Are these safe to use?

Have you used anhydrous CaCl2,, dihydrate or hexahydrate ?

You should not experiment with substances if you have lack of knowledge of their chemical properties and behaviour. It may forgive you few times, but it may cause severe injuries if not so lucky. Note that soluble baryum salts are poisonous.

Say the experiments should be done having some experience and proper safety precautions. For basic chemistry better reading first

I must emphasize once again, do not perform experiments if you don't know what you are doing. Prior knowledge is very important. Chemicals are not candies. It can cause serious injuries if you are not careful.

anhydrous calcium chloride. pretty sure strontium chloride and barium chloride are dihydrate or hexahydrate?

Have you used anhydrous $\ce{CaCl2}$, dihydrate $\ce{CaCl2 . 2 H2O}$  or hexahydrate $\ce{CaCl2 . 6 H2O}$ ( 3 most common forms) ? It is a big difference in the resulting thermal effect.
Hexahydrate causes cooling down of the solution while being dissolved. If ice is used instead of water,  as the mixture hexahydrate : ice  2 : 1, it forms the famous freezing mixture, cooling itself down to $\pu{ -50 ^{\circ}C}$.
Anhydrous calcium chloride causes heating up while being dissolved. It is because of releasing energy of ion hydration. Rather then smoking, it was condensed water vapor.
Dihydrate is somewhere between, with roughly neutral thermal effect.
Interaction of the group 2 element chlorides with water is generally safe. There is very slight hydrolysis, growing progressively upwards along the group. The beryllium chloride hydrolysis is significant, releasing hydrogen chloride. Note that barium and especially beryllium are poisonous.

"Aside of dissolution enthalpy, group II chlorides do not react with water, so they are safe.".  Beryllium chloride would react significantly, producing hydrochloric acid/HCl gas.  But then again, our aspiring chemist is not at a good stage to work with beryllium compounds anyway.  Clarified this paragraph.

Well, I included just elements OP considered, but you are obviously right.   But beryllium is rather a honourable member of the group 13.

So, to clarify, barium chloride dihydrate/hexahydrate are not safe?

I have provided no information about existence of particular BaCl2 hydrates. Saying A is OK/exists says nothing about if B is OK/exists. Soluble barium  salts are poisonous and therefore not safe. The best way for you is to buy/borrow of a textbook of anorganic chemistry. A good one is 2 volume "Chemistry of elements" by Greenwood/Earnshaw ( I hope I remember the title/authors well). Search also for book references withing CH SE. The second-best is reading the element articles on Wikipedia and following mentioned references.

@Raze XIX :  P.S.: See resources-for-learning-chemistry


text:
Remove sticky PDMS from glasses slides
I prepared some PDMS mixing a Sylgard 186 (elastomer) and a curing agent with 10:1 ratio. I sucked all the bubbles and I sandwiched the PDMS between two glass slides to get flat surfaces. Finally I heated up at 60C for 6 hours. Now I would like to get my PDMS by removing one of the two glass slides, but it is practically impossible...
Does anyone have some suggestions to overcome this issue?


text:
What do "USP 26" and "USP 42" mean?
I am being asked by a colleague if the Polyvinylpyrrolidone (PVP) we buy in our lab is "USP 26" or "USP 42". I've never heard those terms before. I searched up online but found nothing.
I'm sure its something concerning the US Pharmacopeia, but I've found their site to be extremely walled up and unhelpful.
Can anyone tell me what do those terms mean, and how to find out if my PVP is either? The manufacturer's Safety data sheet didn't seem to mention it.
Note: I don't work at a food grade or drug grade lab. I'm also not in the USA.

Likely refers to a USP version number. For example, here is the USP monograph for povidone and there are numerous subscripted references to USPXX where XX is the revision number of the "harmonized" and not "national" text. My guess is that you're being asked which USP version of the reference standard applies to your material.

I was able to reach the manufacturer and ask her directly. USPXX is the "edition" of the US Pharmacopeia's "exams" that a product must pass in order to be considered pharmaceutical grade. As time goes on, these exams may get updated and improved. So a higher number USPXX means that a product passed a more recent (and likely, more demanding) USP set of exams.
Therefore, USP 26 is a cheaper PVP, that may not be able to pass the USP 42 standards. Or at least, wasn't tested on them.
On the other hand, USP 42 is a more expensive PVP,  that can and did pass the standard.
As for the actual differences between both standards, I don't know them. So I can't say if the difference between then it's significant or not. It seems USP guidelines are not available freely.
Edit: Todd Minehardt's comment had the right answer it seems.


text:
qv and qp relationship
Is there any relationship between q at constant volume and q at constant pressure that I could derive with just the reaction equation, and the enthalpy of formation and Cp of the products and reactants?
I am aware that q at constant pressure is enthalpy of formation of products - enthalpy of formation of reactants.

You are assuming constant temperature, right?

I had not assumed that but I believe we would have to as one of our state variables.

Let us assume that the initial and final temperatures in both the isobaric and isochoric processes are same. We will limit this discussion to ideal gases only. To calculate heat transfered we will use the expression $$q = nC\Delta T$$ Here C is molar heat capacity of the gas. For isobaric process, this will be $$q_p = nC_p\Delta T$$ For the isochoric process $$q_v=nC_v\Delta T$$
$\Delta T$ for both these processes same. Substracting these two equations we get $$q_p - q_v = n(C_p - C_v)\Delta T$$ The value of $C_p - C_v$ is $R$. Therefore $$q_p - q_v = nR\Delta T$$

No, there are some issues. Firstly, you should use $C_{p,m}$ and $C_{v,m}$ which are the molar heat capacities at constant pressure and volume. Secondly, the equation $C_{p,m} - C_{v,m} = R$ is only true for an ideal gas. If you want to limit the scope to just ideal gases, that's fine, but you should make it clear. My biggest confusion, however, if you heat an ideal gas at constant pressure, its volume is going to increase. How can you then have a constant-volume process that has the same initial and final state...?

Alright i will make the edits. About your confusion, you made a valid point. I considered here that only initial and final temperature are same for both processes. So if you heat an ideal gas isobarically and heat the same gas isochorically in two similar containers raising the temperature of both gases by a same amount, then my equation gives the relation between the heat required for both the cases.

Ah, well, OK. Then I suggest also changing "initial and final state variables" to "initial and final temperature", since "state variables" include $p$ and $V$, too.

Yes I've done that.

I don't think op was asking for derivation of mayers relation

@Buraian How is thia derivation of Mayer's relation. In fact I used Mayer's relation to derive the equation in consideration.

I misread, good answer!