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How to make a mixture of a certain volume with x/y ratio mass?
For example for 1mL to get 3:7 mass ratio sand/water.
I have multiple samples to make with different ratios but the same volume and I can't seem to find an ecuation for this.
Thank you!

What keeps you from the deriving the formula yourself? For additive volumes scenario like (quartz) sand + water, it is trivial. Total volume is sum of sand and water volume. Volume of each component is then ratio of its mass and density. Keeps masses in desired ratio and involve proportionality constant to reach target volume.

Because it's more complicated than that. My main issue is I have a mixture of 1mL with 30wt% SiO2 and the rest water; and I can't figure out the weight of SiO2 in this mixture and/or the volume it occupies.

Where is the complication?

Well, I don't know the weight of the mixture and of the compounds (SiO2 and water). All I know is that out of 1mL of mixture, there is 30wt% of SiO2. And also the density of the compounds.

Let's admit that the mixture contains $\pu{x}$ grams sand, and $\pu{y}$ grams water (and also $\pu{y}$ mL water). x and y are two unknowns. Let's take the density of sand $\pu{2.66 g/mL}$. The total volume of the mixture is : $\pu{V = \frac{x}{2.66 g/ml}  + y = 1.00 mL}$. This is the first equation. The second equation is the mass ratio : $\pu{\frac{x}{x+y} = 0.3}$.
The solution of this system is :  $\pu{x = 0.3681 g}$ sand, and $\pu{y = 0.8615 g}$ water. As the total mass is $\pu{1.2296 g}$, the fraction of sand is $\pu{\frac{0.3681 g}{ 1.2296~g} = 0.300 = 30}$%
If this is of interest, the volume of the sand can be calculated : $\pu{V_{sand} = 0.3681 g/(2.66 g/mL) = 0.1385 mL }$. With $\pu{0.8615 mL}$ water, it makes a total volume $\pu{1.000 mL}$

$$V = m ( \frac {w}{\rho_\mathrm{sand}}+ \frac {1-w}{\rho_\mathrm{water}})$$
$$m = \frac {V}{ \frac {w}{\rho_\mathrm{sand}}+ \frac {1-w}{\rho_\mathrm{water}}}$$
$$m_\mathrm{sand} = \frac {Vw}{ \frac {w}{\rho_\mathrm{sand}}+ \frac {1-w}{\rho_\mathrm{water}}}=\frac{\pu{1 mL} \cdot 0.3}{\frac{0.3}{\pu{2.65 g mL-1}}+\frac{0.7}{\pu{1 g mL-1}}}\approx \pu{0.369 g}$$
$$m_\mathrm{water} = \frac {V(1-w)}{ \frac {w}{\rho_\mathrm{sand}}+ \frac {1-w}{\rho_\mathrm{water}}}=\frac{\pu{1 mL} \cdot 0.7}{\frac{0.3}{\pu{2.65 g mL-1}}+\frac{0.7}{\pu{1 g mL-1}}} \approx \pu{0.861 g} \equiv \pu{0.861 mL}$$
There would be obvious issue with maintaining the desired local mass ratio  within the mixture because of sedimentation, but I suppose the mixture is prepared in-situ from components.

Quantity
Legend

V
total volume

m
total mass

w
mass fraction <0..1> of sand

$\rho$
density. Density of water is trivial (1 g/mL) unless high accuracy is needed. Knowledge of sand density is mandatory. For pure enough quartz sand it is 2.65 g/mL.

mw
sand mass

m(1-w)
water mass


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Relation between Solubility Product and Equilibrium Constant
I had read about Solubility, and I found that Solubility is not Molarity (because previously I believe that both are same thing). But, they have a big difference (source : https://en.wikipedia.org/wiki/Solubility ).  Thus, now I'm in confusion that if solubility and Molarity is different terms, then the Solubility Product and Equilibrium Constant will also be different, for the same reaction, because while writing the Equilibrium constant we use Molarity (source: https://www.calstatela.edu/sites/default/files/dept/chem/11winter/201/jan20.pdf ) So, now I had tried to find out the relation between two of them. But, I am unable to find any such relations. So, anyone can help me to find out the relation between the Solubility product and Equilibrium Constant for a reaction.

New Question @Poutnik sir, please see this question (pasted below)

So, in this question they had asked about the mass of Silver Iodate (AgIO3) in 100 mL saturated solution. But, when I saw the answer of this question (pasted below), then they are defining Solubility as the Moles of Salt present in 1000 mL of solution. But, you had defined solubility as the maximum mass of substance in 100 mL solvent.

Solution of the question.

So, I'm confused in the definition of Solubility, and which I can use always to get correct answers. Please help.

Solubility product is an equilibrium constant in a specific form for precipitation reaction of respective ions. Both can be expressed in terms of molarity, but more exactly in unitless activities.
Solubility is maximum mass of substance able to be dissolved in the given volume off solvent, usually 100mL.
Molarity is molar concentration as (molar) amount of substance in 1 L of the solution.

Sir, please check my question, I had added some new data that will help you to understand my query more clearly now.

@UJJWALMISHRA
The analytical composition of a saturated solution, expressed in terms of the proportion of a designated solute in a designated solvent, is the solubility of that solute. The solubility may be expressed as a concentration, molality, mole fraction, mole ratio, etc.  https://goldbook.iupac.org/terms/view/S05740

See also https://en.wikipedia.org/wiki/Solubility_table // There is multiple equivalent ways how to express solubility. The way of g/100 mL of solvent is one of the most common.

Solubility is a constant at given temperature. Molarity can be any numerical value between zero and the solubility.

source: Stack Exchange

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

sort: votes


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Would magnesium react with acyl halides to form a variant of grignard reagent just like alkyl and aryl halides?
We were having a chemistry class in our school and we were discussing the answer to some question related to acyl halides when this popped up. Our teacher mused that it may be possible but he hadn't seen any source mention it ever. We all tried looking it up online but did not get anything. Does anyone have any information or arguments in support or opposition of this?

Welcome to ChemSE! Your question will likely be considered as homework and be voted down. You need to make some effort to answer the question yourself. Take a look here about homework policy. https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange. The lack of information may be telling you something. What would be the fate of such a "Grignard reagent" in solution during its formation? You might figure that Victor tried it. ;)

@OscarLanzi   Noted, comment deleted

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


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How can I heat up a separatory flask?
I currently am working on a vacuum brewing system for coffee (which is not super useful but pretty cool I have to say) such as this one (image).

However I wondered if there was a way of pouring the coffee from the Florence flask directly to a mug? I'm thinking of using a separatory funnel instead of the Florence flask but can't seem to find any way of heating it up, is that a good option? By the way I'm still a high school student who likes chemistry.
Thanks a lot, Hugo
EDIT - more precise :

This is one of the way I was thinking of doing it. On the picture :

1 is a Florence flask where I would heat up the water that would go directly into 2
2 is a beaker where the water would brew with coffee

When I turn the heat down, the brewed coffee should come back into the Florence flask 1.
However, to pour the coffee into the mug, I would have to manually pour the coffee from the Florence Flask. I therefore am looking for an option to move the coffee directly to the mug.
At that point, I thought of a separatory funnel - which probably isn't a chemically sound way of doing it - (terribly) drawn on the picture. But obviously I need a way of heating the separatory funnel.
My question is : is using a separatory funnel a good option to move around the coffee, if not is there an easier/better way? If it is, how can I effectively heat a separatory funnel?
Thanks for the help!

There are heating mantles, such as these t Fisher: https://www.seco.us/2582254/Product/Glas-Col-O-Series-Fabric-60deg-Funnel-Heating-Mantle-115-V-40-w . However, perhaps Cory Doctorow can convince you of the virtues of cold brewed coffee: https://boingboing.net/2013/07/20/cheap-easy-no-mess-cold-brew.html

if want it to look science-y and stuff and you were down to create some mechanical apparatus to move things around, perhaps you could get creative and use a retort flask (you can get them used sometimes for an OK price, I think) but I'm a little confused about what direction everything in your ideal apparatus goes so maybe try clarifying your proposed system for better insight.

@thelocalsage hey, thanks for your help. I edited the post to be more precise, see if it helps. Also don't get how the retort flask can be used here, could you explain more about it?

Your setup is not clear. Where is the coffee powder ? In 1 or in 2 ? Why is the bridge full of a red liquid ? Will you heat up or cool down flask 1 ? Why do you call mug a beaker ? Why do you heat up both containers 1 and 2 ? Why do you want to put a separator flask ? And where ?

Cool project. Check out this video, his setup is very close to yours and he uses tongs for grabbing the neck of the Florence for transfer to the cup.

@Maurice the coffee powder is in 2: the beaker. When heated up in the Florence Flask 1, the water will go to the beaker 2 where it will brew with the coffee. When I stop the heating, the brewed coffee located in 2 will go to 1 because of the vacuum. This is the red/brownish liquid in the bridge. That's the basic system. My point is I want to "upgrade it" so that once the coffee is in the flask, I can move it to my mug. This is why I thought of a separatory funnel, so that I could just turn the tap and pour the coffee in my mug. Hope you get it, sorry if my explanations are confusing

@ToddMinehardt Thanks for the video, the tong does work just fine but I wonder if there is a way of not having to manipulate that much for the transfer. Anyways thanks!

@Hugosh Understood. Maybe a side arm round bottom? Although I have to admit it's not clear to me how you'd get the coffee to flow out of the side arm without some suction. Anyway, good luck!

@Hugosh i don't believe there are great ways of heating separatory flasks simply because they don't tend to be used in situations that require heating. and they tend to have plastic stopcocks that could melt at higher temperatures...i suggested the retort flask as a possibility because if it was mounted on a rotational mechanism you could plug the top once done and use the arm as a spout but that may mess with the vacuum suction and it may be a bit too complex... but it would be science-y lol.

@hugosh i have an idea: what if the tube instead of being a single tube had a closable branch in the middle (like a stopcock) that fed out into your cup. you would prepare the coffee as normal, and once it's brewed and you begin the cooling process at 1 you wait for the coffee to travel back up the tube. Once the coffee reaches and fills the stopcocked prong and a little of it has begun to dribble back to 1, you open the stopcock and begin to let coffee out. the coffee forms a seal at the stopcock so the pressure differential isn't compromised, and the extra prong acts as a siphon. thoughts?

@localsage that does sound very cool and (kind of) easy to make. However, I gotta say I don't completely understand the apparatus. So if I recap, there would be a stopcock in the middle of the tube. Once the coffee is brewed and starts coming back to 1, I open the stopcock which leads the coffee to the cup? I don't get what's the prong part. Sorry for the loss of time, but I'm wondering if you could do a 2-second drawing on Paint or whatever because I really don't get it all here. Thanks for your help

OK I love this coffee setup and I want one. My suggestion might be too expensive, but if you could get your hands on a schlenck flask, you could heat and dispense from that. It still requires tipping it, but you wouldn't have to take the top off if the tube connecting 1 and 2 were flexible. Or you could dispense through a cannula. That's assuming that I'm understanding the pressure changes correctly when the coffee comes back over.


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What makes a molecule lipopholic?
I know that non polar (Hydrophobic) molecules are usually lipophilic.
Is there a general explanation for this phenomenon, or does the explanation differ depending on the molecule itself?

Please use appropriate tags and correct spelling (in the title for instance). Also, textbooks and many online resources would provide a sufficient answer, for instance the Wikipedia: https://en.wikipedia.org/wiki/Hydrophobe

To be dissolved into water, a substance has to break hydrogen bonds existing between water molecules. This requires energy. So the dissolution into water is only possible if enough new hydrogen bonds are created between the solute and water. If no or not enough new bonds can be created in solution, Mother Nature is reluctant to accept these newcomers, and the arriving substance is not soluble in water. This is the case for non polar substances, which are also called hydrophobic. Nevertheless, these substances can be dissolved in non polar solvents, which have no hydrogen bonds, because the van der Wals forces existing between non polar molecules are much weaker than hydrogen bonds in polar solvents. Van der Wals forces requires few energy to be broken. In principle, any new non polar molecule can be dissolved in any non polar solvent, provided their dimensions are not too big.


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Is it possible for a long carbon chain to have both acidic and basic groups?
I was reading about a zwitter ion and I had a thought that is the same possible for an acidic and basic groups. Like a very long carbon chain with a basic group and an acidic group on either sides, I am guessing it’d be called amphoteric but amphoteric compounds themselves do not contain acidic and basic group, so is it possible to synthesis one? And if it is done already what is it called?

There is no reason why basic and acidic groups could not coexist being attached to different carbons of the carbon chain, when they can coexist on the same carbon of aminoacids $\ce{H2N-CHR-COOH}$, occuring free or condensed in proteins in living cells.
These aminoacids exchange protons via $\ce{H+(aq)}$, forming zwitterions $\ce{H3N^{+}-CHR-COO-}$, neutralizing themselves.
For longer chains, consider e.g. $\ce{H2N-(CH2)_n-COOH}$ with an acidic and a basic group at the ends (simple example is GABA, $\gamma$-aminobutyric acid, a natural neurotransmitter where $n=3$). similarly as above, in water solution they would exist as zwittterions. In strongly acidic/basic solutions, only amino/carboxy group would have charge, respectively.
For $n=5$, it is epsilon-aminokaproic acid with an acidic and a basic group. Its cyclical lactam with amidic bond $\ce{-CO-NH-}$ is used as the monomer for the original polyamide Silon $\ce{[-NH-(CH2)_5-CO-]_n}$, with more simple production, compared to the Nylon-66. It was produced in the post-WW-II Czechoslovakia. For the curiosity, it was invented by famous Czech organic chemist Otto Wichterle, who invented also soft contact lenses.
Analogically, the above occurs for other acidic and basic groups, e.g.: $\ce{R1-NH-R2-SO3H ->[H2O]R1-NH2^{+}-R2-SO3-}$.

Can you cite a reference for Wichterle as well as inventor? If Wikipedia is to be believed both IG Farben and Toray first developed nylon 6 in 1938. Wichterle is not mentioned

@Andrew https://en.wikipedia.org/wiki/Otto_Wichterle // https://en.wikipedia.org/wiki/Nylon_6 // https://en.wikipedia.org/wiki/Silon // (CZ2EN)  https://cs-m-wikipedia-org.translate.goog/wiki/Silon?_x_tr_sl=cs&_x_tr_tl=en&_x_tr_hl=cs&_x_tr_pto=wapp

those pages give a date of 1940 for Wichterle’s nylon work. Farben’s patent was filed in 1938: https://patents.google.com/patent/US2241321A/en

It could be parallel work. He worked on it secretly to hide it from nacists. Perhaps he improved the synthesis. I assume he was not in position and circumstances to get it patented.


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What is the major product in the following reaction?
Reaction:

My thought-process:
According to me, this reaction is like the bromination of alkene but here instead of $\ce{CCl4}$ solvent, we have $\ce{H2O}$ as a solvent. For the first step a $3$-Membered Chlorinium Ion Intermediate is formed:

Now how do I proceed further? The solvent is $\ce{H2O}$ so it cannot act as a good nuclophile, whereas we still have the Cl- ion as a nucleophile available, so why doesn't the reaction proceed just like it would in $\ce{CCl4}$ medium, giving us a vicinal dichloride product.
According to the solution I have, a carbocation (the more stable out of the two possible carbocations) is formed and then $\ce{H2O}$ acts as a nucleophile  and according to that the product is formed. But can a carbocation form when the intermediate is a 3 membered halonium ion? I don't know about this. I'm confused, isn't this a SN$1$ type reaction, so how can a carbocation form?
The answer given is:

P.S: This question was asked in JEE Mains $12$ April $2019$ Shift $1$

You underestimate how good a nucleophile water is.

@Waylander Negatively charged species are better nucleophile than neutral species. In this case, we have bromide ion, so comparitively, isn't water a not so good nucleophile here?

You are incorrect about how good a nucleophile Chloride ion is, and there is far more water present than chloride.

Once you have the chloronium ion, you can think of it as having either of two carbocation structures, which are formed by breaking one or the other carbon-chlorine bond. Whichever carbocation is more stable forms with lower activation energy, so the nucleophile (water) will preferentially attack that carbocation center.
Here, if you break the carbon-chlorine bond on the right a primary carbocation is formed, whereas breaking the left bond gives a secondary cation. The secondary cation is more stable, so the water is directed accordingly.

Part of my question is why does bromide ion not attack? Is it becuase water is present in large quantity as compared to bromide ion?

If you mean chlorine (rather than bromine), yes - assuming you are using chlorine water for the reaction there is much more water than chlorine.


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Can AMBER as low-layer be used to model reaction mechanisms using ONIOM? If yes, do you need to parameterize the intermediates and Transition states?
I'm debating moving from (M06-2X:PM7:UFF) to (M06-2X:PM7:AMBER) for modelling a reaction mechanism in an enzyme. I've already examined the errors introduced by my partitioning and whether they cancel sufficiently.
However, as dispersion/H-bonding is quite vital for describing this mechanism, I'm tempted to switch from UFF to AMBER. However, due to the way ONIOM works (it does an MM calc for the entire system, a SE calc for the mid + high layers and a DFT for the high layer), I'll need to parameterize my ligands due to them not being available in Gaussian16's Amber parameter bank.
For the cofactor, there already exist online parameters for the product and reactant structures alike and the hypothesized mechanisms do not have intermediates for the cofactor, only the substrate.
For the substrate, no cofactors exist. In literature, people parameterized this substrate using RESP charges obtained through DFT calculations - something I could do myself. There would be the reactant and product structures that need parameterization, and I am not sure if I also need to parameterize the intermediate carbocation (reaction path 1) or the intermediate oxyanion (reaction path 2).
My supervisor is not too familiar with AMBER, and I don't want to bother him with questions that probably lead nowhere. His past publications with ONIOM used UFF as the low layer. Myself I'm mostly familiar with the theory of HF, MPn, B3LYP, Minnesota functionals and coupled cluster methods.
My hunch is that I can't use AMBER as low-layer in ONIOM to model transition states or intermediates. However, I'd rather ask those who may have done so before giving up on this thought.


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Formal way to say bubble / bubbling
I know this question might seem too simplistic but I've been looking everywhere for a better than bubble but I have not found it. The most similar word would be effervescence but that's slightly different. I am "bubbling oxygen" in water and testing for oxygen dissolved but it's a formal paper and "bubbling" sounds very informal

I think "bubbling" is already formal enough. Other terms are more specific, i.e. you could use "oxygenation" or "aeration".

If OTOH the bubbling purpose is to eliminate or transfer volatile components, "stripping" or "purging" could be used.

The term is used in scientific papers, so it's probably good enough:

Reactive Deposition of Cobalt Electrodes: II . Role of Bubbling Oxygen
Ultrasonic cavitation erosion of Ti in 0.35% NaCl solution with bubbling oxygen and nitrogen

"Sparging" is a common name given to bubbling a gas stream under the surface of a liquid, often with the purpose of saturating the liquid with the gas from the stream and/or displacing almost all other gasses or other volatiles from the liquid (e.g., solvents are often sparged with nitrogen or argon gas to reduce their oxygen content and allow oxidation-sensitive chemistry).


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How to use DSC to differentiate first and second order phase transition?
The DSC trace below was recorded for a thermotropic calamitic liquid crystal. The heating rate was 5 $^\text{o}$C min$^{-1}$; a heating scan is followed by a cooling scan.

Based on the previously reported result, there are two phase transitions to be expected in the temperature range of interest: from crystal to a nematic phase, then to an isotropic phase.
From the DSC trace, how can we say something about the order of the phase transition?
1. The most popular answer from this site says:

If you want to see whether your transitions are first or second order,
you must run modulated DSC, where the non-reversing and reversing part
of the heat flow is related to the first and second-order transitions,
respectively.

What is the reason for first-order phase transition being non-reversing while second-order being reversing?
2. Second-order phase transition $\implies \Delta S = 0 \implies \Delta H = 0$ at transition temperature. Based on this, shouldn't any second-order phase transition not shown in the DSC trace (because no heat exchanged at constant pressure during that transition)?


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Discrepancies in using pOH vs pH to solve H+/OH- concentration change problem
I found an extremely confusing problem in one of the problems we did in class for general chemistry. This was the question:

How many moles of $\ce{H3O+}$ or $\ce{OH-}$ must you add per liter of $\ce{HA}$ solution to adjust its $\mathrm{pH}$ from 3.15 to 3.65? Assume a negligible volume change.

The answer given was as follows:

Going from $\mathrm{pH}$ 3.15 to $\mathrm{pH}$ 3.65, you are decreasing in the amount of acid in solution.
$\mathrm{pH}$ 3.15 is for a solution with concentration of $\ce{[H3O+]} = 7.08 \times 10^{-4}$.
$\mathrm{pH}$ 3.65 is for a solution with concentration of $\ce{[H3O+]} = 2.24\times 10^{-4}$.
One must add the difference of these in base $\ce{(OH-)}$ in order to
neutralize the hydronium ions.
Add $(7.08\times 10^{-4} - 2.24\times 10^{-4}) = 4.84\times 10^{-4}$ of $\ce{OH-}$

Now this seems simple enough. But I tried to do this another way, that is, by converting these $\mathrm{pH}$ values to $\mathrm{pOH}$: 10.85 to 10.35. And then I followed the same procedure outlined above, to get a completely different answer:

$\mathrm{pOH}$ 10.85 gives a concentration of $\ce{[OH-]} = 1.41\times 10^{-11}$
$\mathrm{pOH}$ 10.35 gives a concentration of $\ce{[OH-]} = 4.47\times 10^{-11}$
The difference of which gives a $\ce{[OH-]} = 3.05\times 10^{-11}$

I'm copy-pasting the answer I was given for my query as to how this can be:

Logarithmic scales are not linear.  Amounts of base need to affect
change at $\mathrm{pH}$ 3.15 are not the same that are needed to cause change at $\mathrm{pOH}$ 10.85.
$\mathrm{pH}$ 3.15 means you have a lot of acid and not much base.  When you are considering changing the $\mathrm{pOH}$ you are thinking (incorrectly) that you just have a solution of hydroxide in water. That is not the case!  At $\mathrm{pOH}$ 10.85 there is a lot of acid present!
The reason is that at $\mathrm{pH}$ 3.15 (or 3.65) the concentration of $\ce{H3O+}$ is a million fold greater than $\ce{OH-}$, and the amount of hydroxide (at $\mathrm{pH}$ 3.15 or $\mathrm{pOH}$ 10.85) is 10,000 times less than what is present in pure water.  Thus by adding just the $3.07 \times 10^{-11}$ moles of $\ce{NaOH}$ that would appear (on paper) to raise the $\mathrm{pOH}$ from 10.85 to 10.35, you are forgetting that the solution is not just water and hydroxide, but there is a lot of hydronium ion that is present.  Actually that small amount of $\ce{OH-}$ would immediately be neutralized by the huge amount of acid and you would still have a low / acidic $\mathrm{pH}$. See this worked out below:
$\ce{[H3O+]} \text{ (at $\mathrm{pH}$ 3.15) } = 7.08 \times 10^{-4}$.  By adding $3.07 \times 10^{-11}$ moles of $\ce{OH-}$, an equal amount of $3.07 \times 10^{-11}$ moles of $\ce{H3O+}$ is neutralized. Now the $\ce{[H3O+]} = (7.08 \times 10^{-4} – 3.07 \times 10^{-11}) = 7.08 \times 10^{-4}$.  You see, nothing very much has changed when you are dealing with such a small amount of base and a large amount of acid.
This is the reality because at low $\mathrm{pH}$ you don’t have to add very much acid to change $\mathrm{pH}$ significantly. At high $\mathrm{pH}$ you don’t have to add much base to change the $\mathrm{pH}$ significantly.  But, on the other hand, at low $\mathrm{pH}$ you have to add a lot of base to neutralize the large amount of acid present, and similarly at high $\mathrm{pH}$ you have to add a lot of acid to neutralize the large amount of base present.
Bottom line, just because the $\mathrm{pH}$ and $\mathrm{pOH}$ signify the same solution (in regard to relative amounts of hydronium or hydroxide present), how much base you have to add to change the $\mathrm{pH}$ versus the $\mathrm{pOH}$ is very different.

I understood that we need a lot of base to neutral the acid at a very low $\mathrm{pH}$. And I did not pursue this further in class because the mathematical calculations for this discrepancy were beyond the scope of what needed to be covered in our class; but I couldn't let this go until I understood how the math works here, since by my above logic, I should have gotten the same answer using $\mathrm{pOH}$ as when I did using $\mathrm{pH}$, if not for the nonlinear logarithmic discrepancy. I don't know how to work this out.

The question is poorly formulated, as the task has the solution dependent on the HA. If assuming it is fully dissociated strong acid, than the task  is easy ( unless we introduce the hell by involving activities )

The explanation is also not so clear. The key is that as you add base, the $\ce{H+ + HO- <=> H2O}$ equilibrium must be maintained. With a large excess of H+ relative to HO-, that means that when you add HO-, essentially all of the HO- reacts with H+, so the HO- concentration changes only a tiny bit while the H+ concentration changes a lot. You can work this out with an ICE table. You'll see that the approximation that the H+ concentration changes by the amount of HO- added is nearly correct, and that the HO- concentration changes only a tiny bit

When the pOH is greater than 7, the concentration of $\ce{OH-}$ ions has nothing to do with the amount of added $\ce{NaOH}$ solution. These ions have only one origin : the autoprotolysis of water.

The most precise answer to a problem like this requires consideration of the equilbrium of water dissociation: $\ce{H2O <=> H+ + HO-}$, where $K_w = [\ce{H+}][\ce{HO-}]=10^{-14}$.
Although it is not specified in the problem, we also have to assume that HA is a strong acid that is fully dissociated at the initial pH.
As you correctly calculated, the initial pH of 3.15 indicates that $[\ce{H+}]=7.08\times10^{-4}$ M and $[\ce{HO-}] = 1.41\times 10^{-11}$ M. The final pH of 3.65 indicates that $[\ce{H+}]=2.24\times10^{-4}$ M and $[\ce{HO-}] = 4.47\times 10^{-11}$ M.
Now let's consider what happens when you add base to the initial solution. There are two possibilities for every $\ce{HO-}$ that goes in. It can either react with $\ce{H+}$ to make water (eliminating both ions from solution and decreasing the $\ce{H+}$ concentration) or it can remain unreacted and result in an increase in the $\ce{HO-}$ concentration. We'll thus define two variables: let $x$ be the total amount of $\ce{HO-}$ added (that is, the answer to the problem) and let $a$ be the amount that reacts with $\ce{H+}$.
It is then easy to see that change in $[\ce{H+}]$ between the initial and final conditions is simply $-a$, and the change in $[\ce{HO-}]$ must be $x-a$. Since we already have numerical values for the initial and final concentrations, solving for $a$ and $x$ is trivial algebra and gives us a final result of $$a= 4.84\times10^{-4}\text{ M and. . . }$$ $$x=4.84\times10^{-4}\text{ M}$$
The fact that both values are the same to three digits tells us that the difference $x-a$ is so small relative to $x$ or $a$ that it is negligible and thus $x\approx a$. That is why you can calculate the amount of base added ($x$) simply using the change in the pH (after converting to $-a$, the change in $[\ce{H+}]$). You're implicitly assuming that $x\approx a$.
When you try the same thing with pOH, it doesn't work because you are implicitly assuming that $x=x-a$, which is obviously false.

This makes a lot of sense now. Thank you so much @Andrew!!

@jj_a: If you like the answer you should accept it. Just use the green check mark to do so.

The explanation is simple, and has nothing to do with logarithms. [Here I assume we are working with a strong acid and a strong base.]:
You are thinking the change in the amount of $\ce{OH^-}$ should be equal to the amount of $\ce{OH^-}$ you added.  But that can't be the case, since most of the $\ce{OH^-}$ is used up in neutralizing the $\ce{H^+}$!  I.e., most of the $\ce{OH^-}$ you add combines with $\ce{H^+}$ to give you $\ce{H2O}$. The increased amount of $\ce{OH^-}$ $(\pu{3.05*10^{-11} mol})$ is merely what's left over following the neutralization, and is determined by the final $\ce{[H^+]}$ and the value of $K_w$.
So you can see your misunderstanding has nothing to do with logarithms.  The logarithms are only used to convert concentrations to pH, i.e., to change the mode of presentation of the concentrations.  You would have had the same misunderstanding if you had calculated the change in $\ce{[OH^-]}$ without using pH or pOH. You still would have been confused why the change in the amount of $\ce{OH^-}$ present was much less than the amount of $\ce{OH^-}$ you added.

If you don’t want to consider the autodissociation of water explicitly, you have to figure out what the major species is. In the absence of weak acids and bases, the major species (other than water) is hydronium ions at acidic $\mathrm{pH}$ and hydroxide ions at basic $\mathrm{pH}$.
It gets a bit more complicated if your question asks you to go from acidic to basic, or basic to acidic, e.g. from $\mathrm{pH \ 3}$ to $\mathrm{pH \ 12}$. In that case, you would add $\pu{0.001 mol L-1}$ strong base to get to neutral, and another $\pu{0.01 mol L-1}$ base to get to $\mathrm{pH \ 12}$.
In all of these considerations, you are ignoring the minor species, hoping that you are making a negligible mistake in doing so.


text:
Why is instant coffee powder not as dark as the coffee resulting from it?
Instant coffee granules generally are of a light brown colour, while coffee made from it is a deep black. Pigments are far more concentrated in the dry form, to add water is to dilute them. Still, this results in a much deeper colour. Why is it?

Many solids are darker when wet than when dry. Might not be a matter of concentration, but some other physical effect regarding light scattering or something. I am sure some physicist can explain this. Also, consider than many things like foams (even coffee foam, actually) are lighter than the liquid they are made from, because they incorporate air.

How do you determine "darkness", and, more importantly, compare optical properties (Which ones? Are the lighting conditions the same?) of the liquid and solid (consider the amount of light reflected back at observer)? I suppose you would agree that there is a certain concentration threshold when you even subjectively would agree that the powder is darker than the diluted coffee.

@andselisk  I think the observation is somewhat trivial. I will see to add two pictures anyway.

Also, please refrain from introducing new tags like coffee or melanoides, especially when it has a typo and refers to a snail instead of an actual chemical compound melanoidin.

@andselisk  I thought the funktion to add new tags was there by intention. Was that as far off as the compound name?

As @user6376297 states, the difference is likely that between a surface phenomenon and that of the whole amount, dissolved in water. Salt is opaque\ white as a powder, and clear in large crystals and when dissolved.

It's not actually a chemical reaction (so this question is more appropriate for Physics.SE, not Chemistry).
You can see exactly the same phenomenon when you spill water on your coloured shirt; the wet spot appears darker.

…
The light blue cloth reflects a lot of the light that hits it.
That's caused by the high index of refraction of the cloth.
The reflected light of course can't get absorbed by the dye in the cloth.
Water has an intermediate index of refraction, between air and many solid materials.
That reduces the reflection, allowing more light to get to the dye where it's absorbed.
The material looks darker.
This seems particularly effective on porous materials, perhaps because much of the light is reflected at various angles several times before bouncing out.
Water reduces the reflection at all the surfaces.
On my glossy white paint, the water doesn't matter since the paint surface reflects the light anyway.
I think the black cloth has so many dye molecules within a wavelength or so of the surface that it absorbs almost all the light that hits it even without a water layer to help.
(The boundary between the outside and inside of the material is in effect a bit fuzzy, just as the images of things seen in a light microscope are fuzzy by about a wavelength of light.)
— Q & A: why are wet things darker? | Department of Physics | University of Illinois at Urbana-Champaign


text:
Combining two reactions in equilibrium with same reactants but different products: What would be the new K?
If I have the reactions :

A(g) ⇌ B(g)     K1
A(g) ⇌ C(g)     K2

Am I able to add the equations like done in Hess's law despite nothing canceling out? And if I can do so, is the new equilibrium constant equal to K1$\cdot$K2?
In other words, is the following equation true?

2A(g) ⇌ B(g) + C(g)     K = K1$\cdot$K2

Using the law of mass action, this seems valid as K would equal $\mathrm{[B][C]/[A]^2}$ which can be rewritten as $\mathrm{[B]/[A] \cdot [C]/[A]}$ which equals K1$\cdot$K2; however, this implies that the concentration of B is always equal to that of C (assuming the initial concentration of both were equal) which seems unreasonable.

There is no new K - that's two different reactions.

You can calculate a K to get from B to C.

Or build an enzyme that couples the two reactions in a 1:1 stoichiometry.

@Mithoron But can´t two different reactions be added to produce a new K? Like as  Karsten Theis said the two reactions can be subtracted to get B<-->C with K = K2/K1, so how come the two reactions in the question can not be added?

If 1 A forms  B or C it does not mean 2 A react together forming both. If it would, then initially pure A would lead to B:C ratio 1:1 , what is not the case unless accidentally both original constants are the same

Do you have real example one reactant producing two different products under the same conditions as showed in your first two equations? I don't think it is real. $K$ is only depending on temperature so two reactions have to be in two different temperatures, at least.

You are correct in saying that the law of mass action applies in this case. The two equations (1 and 2 respectively)
$$\ce{A(g) <=>[K_\mathrm{1}] B(g)}$$
$$\ce{A(g) <=>[K_\mathrm{2}] C(g)}$$
can be combined to form an equation with $K_\mathrm{eq}$ = $K_\mathrm{1}K_\mathrm{2}$:
$$\ce{2A(g) <=>[K_\mathrm{1}K_\mathrm{2}] B(g) +C(g)}$$
However, do be wary that just because the the right side of the equation has both products B and C with the same stochiometric coefficients, this does not mean that at equilibrium, the two will have the same concentration. Why? The "reactants" - the two molecules of A - don't actually react with each other. Instead, each molecule of A can independently be converted into either B or C - if $K_\mathrm{1}$ is larger than $K_\mathrm{2}$, the chances that B will be the produced molecule is higher.
To mathematically illustrate this, write out the equilibrium expressions for reactions 1 and 2, and solve for the concentration of A:
$$ \ce{K_\mathrm{1} = \frac{[B]}{[A]} -> [A] = \frac{[B]}{K_\mathrm{1}}} $$
$$ \ce{K_\mathrm{2} = \frac{[C]}{[A]} -> [A] = \frac{[C]}{K_\mathrm{2}}} $$
Set the two equations for [A] equal to each other to obtain:
$$ \ce{[A] = \frac{[B]}{K_\mathrm{1}} = \frac{[C]}{K_\mathrm{2}} ->$$
$$\ce {\frac{[B]}{[C]} = \frac{K_\mathrm{1}}{K_\mathrm{2}}}} $$
Thus, we can see that the ratio of the concentrations of B and C is the ratio of their respective equilibrium constants of formation from A.
I believe the problem you have here is that you are confusing elementary steps with the overall equation. Reactions 1 and 2 are elementary steps, but the overall reaction is not. Thus, you have to be much more careful and ask yourself the following question: are the reactants actually reacting together, or are the two reactions completely independent of each other? A good way to intuitively see this would be to consider the following equilibria:
$$\ce{A(g) <=>[K_\mathrm{3}] B(g)}$$
$$\ce{C(g) <=>[K_\mathrm{4}] D(g)}$$
Combining these, we obtain:
$$\ce{A(g) + C(g) <=>[K_\mathrm{3}K_\mathrm{4}] B(g) +D(g)}$$
I hope you can see that although B and D have the same stochiometric coefficients, the amount of each at equilibrium will be different as reactions 3 and 4 are completely independent of each other. In fact, the above equation isn't even really a reaction, as the two reactants do not react with each other. Rather, it serves more of a general observational purpose, such as, I put "A" and "C" into a container and they produce "B" and D". They could be reacting with each other (if they were, it would be an elementary step, and [B] = [D]), or they could just be in independent equilibriums. The above equation cannot be used to find the relative amounts of B or D; you must break down the reaction into the elementary steps to discover that, just as with the equation you derived.
For those that would like a concrete example of a reaction where "A" can turn into several different products without reacting with other molecules of "A", one that comes to mind from experience is the isomerization of β-pinene at temperatures above 773 K:
$$\ce{β-pinene <=>[K_\mathrm{x}] p-menthadiene}$$
$$\ce{β-pinene <=>[K_\mathrm{y}] limonene}$$
$$\ce{β-pinene <=>[K_\mathrm{z}] myrcene}$$
I've left out the organic structures/mechanisms here for sake of simplicity. There are three products formed instead of two, but the point remains the same. Reactions where one molecule isomerizes to several different products at high temperatures with competing mechanisms are quite common in industrial chemistry. Similar to reactions 1 and 2, the ratios of products in the isomerization of β-pinene will be the ratios of the equilibrium constants. Something interesting to note is that since these equilibrium constants change with temperature, higher or lower temperatures will favor certain isomerization products (see van't Hoff equation), although having just one produced is notoriously difficult.
Another more common example of a reaction where $\ce{2A(g) -> B(g) +C(g)}$ would be that of a nuclide that emits radiation through both alpha and beta decay, such as 212Bi. In many of these cases, the rates of alpha and beta radiation of nuclide A are different, indicating that decay products B and C will be different as well. However, because radioactive decay is not an equilibrium, its applicability is limited in this case, and rate constants of decay are used instead of equilibrium constants.


text:
Has a fully methylated pentagonal dodecahedrane been synthesized or at least attempted? If not, is there some reason it could not exist?
I first came across dodecahedrane in comments below What dodecahedral molecule is Linus Pauling likely holding in this photograph? Does it have 40 carbon atoms?
Wikipedia's Dodecahedrane; Derivatives; Hydrogen substitution says:

Substitution of all 20 hydrogens by fluorine atoms yields the relatively unstable perfluorododecahedrane $\ce{C20F20}$, which was obtained in milligram quantities. Trace amounts of the analogous perchlorododecahedrane $\ce{C20Cl20}$ were obtained, among other partially chlorinated derivatives, by reacting $\ce{C20H20}$ dissolved in liquid chlorine under pressure at about 140°C and under intense light for five days. Complete replacement by heavier halogens seems increasingly difficult due to their larger size. Half or more of the hydrogen atoms can be substituted by hydroxyl groups to yield polyols, but the extreme compound $\ce{C20(OH)20}$ remained elusive as of 200613 Amino-dodecahedranes comparable to amantadine have been prepared, but were more toxic and with weaker antiviral effects.
Annelated dodecahedrane structures have been proposed.
13Fabian Wahl et al. (2006)Towards Perfunctionalized Dodecahedranes—En Route to C20 Fullerene (also readable here)

The abstract of Wahl et al. (2006):

“One-pot” substitution of the twenty hydrogen atoms in pentagonal dodecahedrane ($\ce{C20H20)}$ by OH, F, Cl, and Br atoms is explored. Electrophilic insertion of oxygen atoms with DMDO and TFMDO as oxidizing reagents ended, far off the desired $\ce{C20(OH)20}$, in complex polyol mixtures (up to $\ce{C20H10(OH)10}$ decols, a trace of $\ce{C20H(OH)19}$?). Perfluorination was successful in a NaF matrix but (nearly pure) $\ce{C20F20}$ could be secured only in very low yield. “Brute-force” photochlorination (heat, light, pressure, time) provided a mixture of hydrogen-free, barely soluble $\ce{C20Cl16}$ dienes in high yield and $\ce{C20Cl20}$ as a trace component. Upon electron-impact ionization of the $\ce{C20Cl16}$ material sequential loss of the chlorine atoms was the major fragmentation pathway furnishing, however, only minor amounts of chlorine-free $\ce{C20+}$ ions. “Brute-force” photobrominations delivered an extremely complex mixture of polybromides with $\ce{C20HBr13}$ trienes as the highest masses. The MS spectra exhibited exclusive loss of the Br substituents ending in rather intense singly, doubly, and triply charged $\ce{C20H_{4–0}^{+(2+)(3+)}}$ ions. The insoluble ∼$\ce{C20HBr13}$ fraction ($\ce{C20Br14}$ trienes as highest masses) obtained along a modified bromination protocol, ultimately allowed the neat mass selection of $\ce{C20−}$ ions. The $\ce{C20Cl16}$ dienes and $\ce{C20H_{0–3}Br_{14–12} tri-/tetraenes}$, in spite of their very high olefinic pyramidalization, proved resistant to oxygen and dimerization (polymerization) but added $\ce{CH2N2}$ smoothly. Dehalogenation of the respective cycloaddition products through electron-impact ionization resulted in $\ce{C22–24H_{4–8}^{+(2+)}}$ ions possibly constituting bis-/tris-/tetrakis-methano-$\ce{C20}$ fullerenes or partly hydrogenated $\ce{C22}$, $\ce{C23}$, and $\ce{C24}$ cages.

Neither Wikipedia nor Wahl et al. (2006) mention substitution of a  $\ce{CH3}$ group for each hydrogen, so I'd like to ask:
Question: Has a fully methylated dodecahedrane been synthesized or at least attempted? If not, is there some reason it could not exist?

companion question: Has substitution of all 20 hydrogen atoms of pentagonal dodecahedrane ever succeeded?  If not, is there any reason that it can not be done?

Option three - it could exist but would be pre-emptively difficult to make.


text:
Can all 20 H atoms of pentagonal dodecahedrane be substituted with high yield and a wide range of substituents? What limits full substitution?
I first came across dodecahedrane in comments below What dodecahedral molecule is Linus Pauling likely holding in this photograph? Does it have 40 carbon atoms?
Wikipedia's Dodecahedrane; Derivatives; Hydrogen substitution says:

Substitution of all 20 hydrogens by fluorine atoms yields the relatively unstable perfluorododecahedrane $\ce{C20F20}$, which was obtained in milligram quantities. Trace amounts of the analogous perchlorododecahedrane $\ce{C20Cl20}$ were obtained, among other partially chlorinated derivatives, by reacting $\ce{C20H20}$ dissolved in liquid chlorine under pressure at about 140°C and under intense light for five days. Complete replacement by heavier halogens seems increasingly difficult due to their larger size. Half or more of the hydrogen atoms can be substituted by hydroxyl groups to yield polyols, but the extreme compound $\ce{C20(OH)20}$ remained elusive as of 200613 Amino-dodecahedranes comparable to amantadine have been prepared, but were more toxic and with weaker antiviral effects.
Annelated dodecahedrane structures have been proposed.
13Fabian Wahl et al. (2006)Towards Perfunctionalized Dodecahedranes—En Route to C20 Fullerene (also [readable here)

The abstract of Wahl et al. (2006):

“One-pot” substitution of the twenty hydrogen atoms in pentagonal dodecahedrane ($\ce{C20H20)}$ by OH, F, Cl, and Br atoms is explored. Electrophilic insertion of oxygen atoms with DMDO and TFMDO as oxidizing reagents ended, far off the desired $\ce{C20(OH)20}$, in complex polyol mixtures (up to $\ce{C20H10(OH)10}$ decols, a trace of $\ce{C20H(OH)19}$?). Perfluorination was successful in a NaF matrix but (nearly pure) $\ce{C20F20}$ could be secured only in very low yield. “Brute-force” photochlorination (heat, light, pressure, time) provided a mixture of hydrogen-free, barely soluble $\ce{C20Cl16}$ dienes in high yield and $\ce{C20Cl20}$ as a trace component. Upon electron-impact ionization of the $\ce{C20Cl16}$ material sequential loss of the chlorine atoms was the major fragmentation pathway furnishing, however, only minor amounts of chlorine-free $\ce{C20+}$ ions. “Brute-force” photobrominations delivered an extremely complex mixture of polybromides with $\ce{C20HBr13}$ trienes as the highest masses. The MS spectra exhibited exclusive loss of the Br substituents ending in rather intense singly, doubly, and triply charged $\ce{C20H_{4–0}^{+(2+)(3+)}}$ ions. The insoluble ∼$\ce{C20HBr13}$ fraction ($\ce{C20Br14}$ trienes as highest masses) obtained along a modified bromination protocol, ultimately allowed the neat mass selection of $\ce{C20−}$ ions. The $\ce{C20Cl16}$ dienes and $\ce{C20H_{0–3}Br_{14–12} tri-/tetraenes}$, in spite of their very high olefinic pyramidalization, proved resistant to oxygen and dimerization (polymerization) but added $\ce{CH2N2}$ smoothly. Dehalogenation of the respective cycloaddition products through electron-impact ionization resulted in $\ce{C22–24H_{4–8}^{+(2+)}}$ ions possibly constituting bis-/tris-/tetrakis-methano-$\ce{C20}$ fullerenes or partly hydrogenated $\ce{C22}$, $\ce{C23}$, and $\ce{C24}$ cages.

Since Wikipedia lists nothing after Wahl et al. (2006) I'd like to ask:
Question: Can all 20 H atoms of pentagonal dodecahedrane be substituted with high yield and a wide range of substituents? What limits full substitution?

companion question: Has a fully methylated pentagonal dodecahedrane been synthesized or at least attempted? If not, is there some reason it could not exist?

I think the question title needed to be more precise. Given that fully substituted molecules are known but only in small/trace amounts and with just a few substituents, I modify the title as the reader can now see.

@OscarLanzi I've gone ahead and removed the "heading" bit as the way it was displaying seemed distracting. Please feel free to edit further.

Full substitution has been obtained, but only for very limited cases. The WP article cites successful full substitution only with fluorine and chlorine atoms. Neither heavier halogens nor small groups such as hydroxyl or amino were substituted fully in the syntheses that were carried out. This is summarized in the question and will not be repeated here.
A significant stumbling block to full substitution is steric hindrance. While the pentagonal angles are an almost perfect fit for "natural" Bond angles in tetrahedrally bonded carbon, the forced planarity of the rings leads to ligands being eclipsed relative to each other and thus creates steric repulsion between these. Isolated five-membered rings in organic compounds are generally puckered unless they are fully conjugated and aromatically coupled, as in pyrrole. The steric repulsions are small for hydrogen atoms but become worse with larger atoms or groups, explaining the decreasing success seen with heavier halogens and the lack of full substitution even with simple multi atomic groups. It would seem, therefore, that the fully methylated compound seemingly displayed by Linus Pauling would be very difficult to establish in real life.
Dodecahedrane has been dehydrogenated to make $\ce{C_{20}}$, which is perforce the smallest fullerene made up of faces with at least five sides.

The synthesis of the C20 fullerene C20 in 2000, from brominated dodecahedrane,[1] may have demoted C20H20 to second place.

Cited references

Prinzbach, H.; Weiler, A.; Landenberger, P.; Wahl, F.; Wörth, J.; Scott, L. T.; Gelmont, M.; Olevano, D.; von Issendorff, B. Gas-phase production and photoelectron spectroscopy of the smallest fullerene, C20. Nature 2000, 407 (6800), 60–63. DOI: 10.1038/35024037.

Thank you for your answer! Somewhat related:Help understanding how "steric effects" are distinct from "electronic effects"?

fyi I've just added a bounty to the companion question Has a fully methylated pentagonal dodecahedrane been synthesized or at least attempted? If not, is there some reason it could not exist?


text:
Under what conditions does manganese dioxide lose its catalytic property?
We know that under standard conditions (room temperature and 760 mmHg air pressure) we have this reaction:
$$\ce{2H2O2(l) ->[MnO2] 2H2O(l) + O2(g)}$$
Here, manganese dioxide is behaving as a catalyst, accelerating the decomposition of hydrogen peroxide into water and oxygen.
If we want manganese dioxide to instead react with hydrogen peroxide what should the conditions be? Also, what would the products be if this reaction occurs?

By catalysing hydrogen peroxide decomposition, it does react with it, but this reaction cyclically regenerate it to the initial state, with manganese reversibly switching oxidation state.
One of possible models is:
$$\begin{align}
\ce{H2O2(aq) + 2 MnO2(s) &-> H2O(l) + Mn2O3(s) + O2(g)}\\
\ce{H2O2(aq) + Mn2O3(s) &-> H2O(l) + 2 MnO2(s)}
\end{align}$$
The point is, hydrogen peroxide has both oxidative and reductive properties, with the latter being oxidized to oxygen.
Variants ( with the $\ce{MnO2}$ reduction part ongoing in the Leclanché cell) :
$$\begin{align}
\ce{H2O2(aq) + 2 MnO2(s) + H2O(l) &-> H2O(l) + 2 MnO(OH)(s) + O2(g)}\\
\ce{H2O2(aq) + MnO2(s) &-> Mn(OH)2(s) + O2(g)}\\
\ce{H2O2(aq) + 2 MnO(OH)(s) &-> 2 H2O(l) + 2 MnO2(s)}\\
\ce{H2O2(aq) + Mn(OH)2(s) &-> 3 H2O(l) +  MnO2(s)}
\end{align}$$
Other model may involve also $\ce{Mn^2+(aq)}$ at low enough $\mathrm{pH}$.
The way of suppress the catalysis may be catching eventual intermediates e.g. by a reagent forming stable complexes with $\ce{Mn^{II}}$ or $\ce{Mn^{III}}$. It would cause majority of $\ce{Mn}$ ending in these complexes after oxidation enough $\ce{H2O2}$ to $\ce{O2}$.
Altenative would be using acidic environment where $\ce{MnO2}$ would end as $\ce{Mn^2+(aq)}$.
Reaction of $\ce{MnO2}$ with citric acid combines both, with acid undergoing oxidation and complexation.

I suppose, it could stop being a catalyst if there were a set of conditions which inhibited one of the two reactions, while allowing the other one to proceed. I don't know if such conditions exist...!

@orthocresol I suppose blocking it, e.g. by forming some stable manganese complex,would block also eventual reactions forming other manganese products.

@Poutnik, Is there a reference or experimental proof that supports those equations?

@M.Farooq Not sure. Possibly. They have been meant as model reactions, following general concept of redox reactions, frequently homogenous, being catalyzed by transition metals with clipping oxidation numbers, like iron, copper or silver. // It could be written more schematically like Mn(III) <-> Mn(IV)

@Poutnik, I highly doubt that there are redox reactions on MnO2. The above reactions imply that the decomposition is not quantitative (i.e., one molecule of H2O2 is wasted in the second equation) but as far I remember it is quantitative.

@M.Farooq I do not see your implication. Check it more carefully. The net reaction is the same as eventual direct decomposition: $\ce{2 H2O2 -> 2 H2O + O2}$ // But I do agree that in context of heterogen catalysis, it would be more complicated than the model.

@Poutnik, You are right. The net result is the quantitative decomposition equation.

Mn2O3 is frequently mentioned for Leclanche galvanic cell. On the other hand, Mn^2+ is not, AFAIK, even if there is mildly acidic pH due NH4Cl.// Unfortunately, I do not have access.

Catalyst is a substance which alters the rate of reaction without being consumed in it. It doesn't mean that catalyst doesn't participate in the reaction. It does react with reactants as Poutnik already stated in his answer. I came up with an another mechanism from [reference][1] and [reference][2]:
$$ \ce{ MnO2  + H2O2 + 2H+ -> Mn^{2+} + 2H2O + O2}$$
$$ \ce{ Mn^{2+} + 2H2O <--> Mn(OH)_2 + 2H+ }$$
$$ \ce{ Mn(OH)_2 + H2O2 -> MnO2 + 2H2O }$$
giving an overall reaction:
$$ \ce { 2H2O2 ->[MnO_2] 2H2O + O2}$$

References:
[1]: D. B. Broughton, and R. L. Wentworth, "Mechanism of Decomposition of Hydrogen Peroxide Solutions with Manganese Dioxide. I," J. Am. Chem. Soc. 1947, 69(4), 741–744 (DOI: https://doi.org/10.1021/ja01196a003).
[2]: D. B. Broughton, R. L. Wentworth, and M. E. Laing,"Mechanism of Decomposition of Hydrogen Peroxide Solutions with Manganese Dioxide. II," J. Am. Chem. Soc. 1947, 69(4), 744–747 (DOI: https://pubs.acs.org/doi/10.1021/ja01196a004).

I think the relavent equations are given in reference 1. Reference 2 is their further studies. If you don't agree feel free to delete my addition.

@M.Farooq: Do you think these references are good enough to backup Poutnik's answer?

@MathewMahindaratne, I am not an expert, but the references are very outdated, not because they are not reliable, but the formation of Mn(OH)2 sounds a little dubious. In those times, chemical kinetics was not very advanced. What proof do the authors provide about the formation of Mn(OH)2 as an intermediate?

@M.Farooq: I don't have access to the papers at home but will look at them on Monday. Yet, it was stated that catalysis does not occur until the solution is saturated with manganese hydroxide. Their reserch is based on radioactive mangenese compound as well. Not sure how they come to the conclusion.

@MathewMahindaratne, I had a quick look. They have several hypotheses, but Poutnik's  is not supported anywhere in those papers. They insist on the role Mn (OH)2...but keep in mind the pH effects. In short a more recent and detailed study is needed. I did not downvote this answer.

The 2nd equation is unbalanced and it is not really Redox equation..There should be rather H2O instead of H2O2. I would say formation of Mn^2+ is questionable.

The main question is not catalytic mechanism, but catalysis suppression.

Manganese dioxide loses its catalytic activity if there is a strong acid in the reaction your mentioned in your post. Consequently, manganese dioxide simply reduces itself to colorless/light $\ce{Mn^{2+}}$ and hydrogen peroxide converts itself to water. I do not exactly recall evolution of oxygen. For the sake electron balance in a redox process, $\ce{O2}$ formation is necessary. However, the dissolution is very fast.
In fact this reaction was widely used indirectly (via $\ce{KMnO4}$ route)  for cleaning glassware long time ago in classical analytical work.

If MnO2 is reduced to Mn(II), and H2O2 reduced to water, what gets oxidised?

@orthocresol, MnO2 almost instantly dissolves in acidic solution in the presence of H2O2. I cannot find an equation at this moment. If you ever had a KMnO4 stain on the skin, a small amount of vinegar and medicinal H2O2 (< 1.5-3%) instantly removes it.


text:
Confused about single displacement Reaction
This question is about predicting reactions
Al(s) + FeO(s) ---> no reaction
Al is higher on the activity series so why does it result in no reaction instead of replacing Fe?

Once initiated there will be a thermite reaction.

Note that, generally, metal reactivity serie relates to electrochemical potentials in water environment and is just indirectly related to outcome of water free reactions between solids. Saying that, I agree with the prior comment.

I don't understand the premise of the second statement. If a metal is higher on the activity series, it should be more likely to lose electrons...

Although the electrode potential of the reaction:
$$\ce{2Al(s) + 3 Fe^2+(aq) -> 2Al^3+(aq) + 3 Fe (s) }$$
is favorable (E0 = +1.21 V), which is what I am assuming you are referring to, it does not apply in this instance as it takes place in aqueous solution. Although the reaction of solid aluminum with solid ferrous oxide is thermodynamically favorable, it is not kinetically favorable. The Fe2+ ions are locked in a lattice with O2- ions, and solid aluminum normally has a protective oxide coating on it that prevents it from oxidation. However, because the reaction is thermodynamically favorable, it will actually proceed at a very slow rate (but only if the aluminum oxide layer is removed). This rate is almost impossible to detect, and there is no reaction for all intents and purposes.
When in aqueous solution, Fe2+ ions could technically be formed and the above equation could apply in this case; however, this will not occur in this situation as FeO is insoluble in water. If the ferrous salt used instead was soluble, such as FeCl2, then the reaction could occur in aqueous solution, and iron will precipitate out of solution.
The only way for the reaction to take place with the given states of matter is for a thermite reaction to take place:
$$\ce{2Al(s) + 3 FeO(s) -> Al2O3(s) + 3 Fe (s) }$$
However, the only way for this to occur is for the reaction system to be ignited, which is not a given condition in your equation.


text:
Why aren't ionic crystals stronger than diamond?

If ionic bonds are stronger than covalent bonds, then why aren't ionic crystals stronger than diamond, which is bonded by covalent bonds?

Diamond has tetrahedral structure with carbons forming covalent bonds, whereas Sphalerite, which has ionic bonds between $\ce{(Zn, Fe)}$ and $\ce{S}$ is far softer than diamond. If ionic bonds are stronger than covalent bonds, then why does this occur?

If men are stronger than women, how is possible a woman can beat a man? Some women are stronger than some men. Some covalent bonds lead in the big picture to stronger ( btw define stronger ) large scale structures than some ionic ones.

The description of ionic bonds as “stronger” generally refers to homolytic cleavage (one electron goes to each atom). Heterolytic cleavage (separation into ions) is typically much more easily accomplished

The starting generalisation here is just wrong. Ionic bonds can be strong or weak; so can covalent bonds. There are so many exceptions, the generalisation is useless.

https://chemistry.stackexchange.com/questions/8687/is-pyrite-fes%e2%82%82-an-ionic-or-a-covalent-compound

https://chemistry.stackexchange.com/questions/11048/are-metallic-ionic-bonds-weaker-than-covalent-bonds

TL:DR I wouldn't call sphalerite ionic. In fact it's bonding is quite similar to the one in diamond, but bonds are more polarised, longer and much weaker and labile.

There are no good universal generalisations relating bond-type and bond strength in crystals
The problem with answering this question is that the assumption you start with is just wrong. There are no good universal rules relating bond-type and strength. I'm not sure why this generalisation is so common (there are a number of similar questions scattered across this site and very few good answers) but I blame naive teachers not wanting to face the issue of how complicated chemistry is.
Let me illustrate the problem with the generalisation with some examples. First, if ionic bonds are always stronger, how do you explain the existence of ionic liquids (used in some industrial processes to avoid organic solvents)?  1-butyl-3-methylimidazolium hexafluorophosphate, for example, consists of two bulky ions but the ionic attraction isn't even strong enough to hold together a crystal at room temperature. Graphite and diamond are covalent compounds that have high melting points. But diamond forms strong crystals and graphite is soft enough to be a lubricant. Sodium chloride is an ionic compound and (table salt) forms moderately strong crystals but has a high melting point. Iodine is a covalent compound but its crystals are so weakly held together than it has a notable vapour pressure at room temperature and will rapidly evaporate on modest heating.
What explains these disparate observations is not the type of bonding but the structure of the compounds. The archetypal "weak" covalent compounds (possibly the source of one bad generalisation in teaching) do not consist of crystals held together by covalent bonds at all. Iodine crystals consist of I2 molecules (with covalent bonds between iodine atoms) but the crystals are held together by (much weaker) van der Waals interactions, hence why they evaporate so easily. NaCl consists of a crystal lattice where all the bonds are ionic. It is relatively strong and the bonding explains its properties. Ionic liquids also have ionic bonds but the ions involved are large and "floppy" which means the strength of the ionic attraction is far lower than in NaCl which explains why they are liquids.
Diamond and graphite form a particularly interesting pair (both consisting only of carbon). Diamond is a massive network where every carbon is connected to every other carbon in a near-infinite tetrahedral array. It is extremely hard and doesn't melt easily. The structure of graphite has two types of bond holding its structure together. It consists of flat planes of hexagons of carbon (like a lot of benzenes fused together) held together by covalent bonds (some delocalised across the whole plane). These bonds are strong but the planes are held together by van Der Waals forces between the planes which are unusually strong because they cumulate over the whole large plane. But they are much weaker and less rigidly directional than the covalent bonds in the plane. As a result the planes can slip against each other making graphite "soft" enough to be a useful lubricant (though it is also worth noting that this softness is non-isotropic and the individual planes are strong in two dimensions which explains the extraordinary tensile strength of graphene which is, effectively, an isolate plane of graphite).
The point of these examples is to illustrate than any generalisation about the type of bond and the strength of a crystal is meaningless in general and only makes sense of you understand the structure involved.
The false generalization that ionic bonds are stronger probably arises because many simple compounds taught in schools seem to follow it. But the examples (iodine versus NaCl, for example) conflate the bonding in the bulk crystal with the bonding in the molecules in the crystal. Iodine might better be described as a van der Waals crystal: the molecules have covalent bonding but the crystal does not, being held together by much weaker interactions.
Ionic liquids have ionic bonding, but this is weak because the ions are large, vastly lowering the strength of the ionic interactions.
Some covalent solids do have covalent bonds across the whole crystal (or large parts of it). They do tend to be strong, sometimes. But, as the examples of diamond and graphite illustrate, you still need to understand the structure to explain all the properties. Individual planes of graphite are "strong" but the planes only stick together with weak forces.
All this makes the generalisation that "ionic bonds are stronger than covalent bonds" pretty useless. But, if you understand the structure of the specific crystals you can usually explain the strength.

If I understand the question correctly, it is asking why sphalerite is not as hard as diamond based on comparision of the bond strengths within the compounds.
@Andrew's comment is a good way to address this problem. When you take a hammer and break apart a crystalline salt, it cleaves along planes defined by the crystal lattice. The bonds break heterolytically; each atom either gives up or keeps the electrons that used to be between itself and it's partner(s). Breaking diamond - also crystalline - also along planes - breakes homolytically and neither carbon across the bond is going to be happy about either being an ion or a radical. Also, the diamond allotrope is made up of connected tetrahedral centers, so these strong covalent bonds are arranged symmetrically around each individual carbon. This is going to make it a lot harder to break. (Compared to the graphite allotrope, for example, which has layers of carbon that can slip past one another, which is why it's soft enough to be used in pencils.)


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%s character distribution of phosphorus hybrid orbitals
The following question was asked in the Indian Olympiad Qualifier Chemistry Part I (IOQC) today:

For the given compound, %s character of phosphorus hybrid orbitals which contribute to various bonds are given in the table below.

The difference in %'s' character of various phosphorus bonds could be due to:
(A) The large size of bromine atom
(B) The large electronegativity difference between $\ce{P}$ and $\ce{O}$
(C) Increased overlap of $\sigma$-orbitals of terminal $\ce{P-O}$ bond
(D) Stronger covalent character of $\ce{P-O}$ in cyclic oxygen atoms

This is a multiple correct question, so any number of the options can be correct.
Due to the large size of bromine, the $\ce{P-Br}$ bond length will be longer than the $\ce{P=O}$ and $\ce{P-O}$ bonds; a longer bond length means lower s character, hence (A) should be correct according to me.
I am not sure how to approach the other options, any help is appreciated.

The most possible approach to answer the question is by Bent rule.
Bromine being large, poor overlapping results in weak sigma bonding, and lower s character in the hybrid orbital of P. The oxygen of Exo or terminal P=O uses hybrid orbitals of more s-character as it overlaps to a greater extent. The oxygen of the cyclic ring therefore can not have a strong covalent sigma bond as Exo P=O.
As there are two types of P-O bonds, B could not be the reason
The answer is A and C

https://doi.org/10.1351/goldbook.BT07000 Bent's Rule does state the opposite of what you are saying.

Bent rule applied for differentiation Exp P-O bond and P-O bond of the ring. Bent’s rule appears to be failed at Br atom due to its large size

Either Bent's rule applies, or it doesn't. There are definitely other factors at play than just Bent's rule. It doesn't apply to the terminal oxygen bond; Bent's Rule would require less s-character not more. This approach to the question cannot be correct. Unfortunately.


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How do you find oxidation state of the metals of a complex with two coordination spheres?
For example suppose we have $\ce{[Pt(NH3)4][PtCl4]}.$ For the naming of this complex I need to know the oxidation number of the metals; let the oxidation number of first and second $\ce{Pt}$  be $x, y$ respectively, then one can obtain the equation $x+y=4$. How are we supposed to proceed further? Thanks.

Platinum can exist under two common  oxidation numbers : $+2$ and $+4$. The most reasonable oxidation numbers are $x = y = 2$, because if the first $\ce{Pt}$ atoms is at $+4$, the other one is at zero, which is not reasonable. And if the first $\ce{Pt}$ atom is at zero, $\ce{[Pt(NH3)4]}$ would not be charged, which is contrary to the original formula.

Do you have a list of common oxidation numbers for transition metals? How do you remember them; do you have some general trends which can ease the memorization process?

Reference : https://www.rsc.org/periodic-table/element/78/platinum

@osmium  There are 3 major ways how to reach chemical knowledge 1/Memorize data what is memory challenging. 2/Recognize or learn behaviour patterns, what is challenge in understanding 3/ Derive behaviour from principles of quantum chemistry, what is computational demanding and extremely challenging mathematically and conceptually. Reasonable approach is combination of 1/ and 2/.

@Osmium Chemistry is too wide and deep domain for a single author or book being the best in all of it. It cannot be at the same time wide enough and deep enough and correct enough. // See also  Chemistry SE: resources-for-learning-chemistry


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How can you have standard enthalpy of formation of water at 298 K?
Firstly, I want to clarify in advance that im not good at this subject.
Secondly, when i go through my chem book to search for thermodynamic data of water at 298 kelvin. I am able to find the standard enthalpy of formation of water in its gaseous state as well as solid state. How can you have water in gaseous/solid form at 298 kelvin? Water only exists in its liquid form at 298 kelvin
Thanks in advance.

Water can be gaseous at 298 kelvin, if the pressure is low enough (< 3167 Pa).

@Maurice You mean vapor partial pressure, as water evaporates even at 100 kPa.

But as far as I know, these reactions are preceded under standard state conditions, thus room temperature and 1 atm. So pressure cannot be high enough.

These reactions are wrapped by formal operations leading from and to standard conditions. This can be done for enthalpy cases by formal heating or cooling of reagents and products. By the Hess law, the enthalpy change does not depend on the path.// E.g. reaction enthalpy at 25°C = enthalpy of reagent heating 25to90°C, + reaction enth. at 90°C + enthalpy of product cooling 90to25 °C.

Water vapor at 25 C and 1 atm is a hypothetical ideal gas state.  You can get the enthalpy of this state by starting with liquid water at 25 C and 1 atm, then dropping the pressure of the liquid to the equilibrium vapor pressure at 25 C, then adding the heat of vaporization at 25 C; for an ideal gas state, going from water vapor at the equilibrium vapor pressure at 25 C to the hypothetical ideal gas sate of 25 C and 1 atm requires no enthalpy change.


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How does a freeze-thaw cycle separate a mixture?
Imagine that I defrost a tube of serum (only the yellow stuff, with blood cells removed) standing still and upright in ambient air. I have seen that as it defrosts the sample separates into different bands of compounds. Is the separation that I witness based on a particular compound trait? For example molecule weight? What is this phenomena called? And has it been studied by chemists? I will try to post a photo later if I can get a sample to show you.

Some of it happens during freezing. Just because the solution you froze was homogeneous (as all solutions should be), it does not mean the solid is homogeneous. As a kid, I discovered that if you suck on a apple juice popsicle, you get the sugar first and the water remains frozen. So the non-water part of the juice was present in cavities of a scaffold of frozen (nearly pure) water.

This might be an example of zone separation through the Czochralski technique, if the components were truly dissolved, and the tube was warmed slowly along its access.
However, blood serum, contains proteins, electrolytes, antibodies, antigens, and  hormones. Some components, such as lipoproteins, are in a colloidal suspension, rather than dissolved, and can be separated out by ultracentrifugation or, as you have done, by freezing. After freezing has destroyed the colloid, allowing components to clump, they'd segregate according to their densities and other properties, e.g., relative hydrophilic or hydrophobic nature.
You'd need to analyze those bands to see whether you've demonstrated zone refining or breaking a colloid with freezing and then performing gravitational separation.

How could I learn about other methods to separate the serum (not necessarily by density?)


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Are there any examples of chemiluminescent reaction catalyzed by Group 1/2 metals?
While researching chemiluminescent reaction that normally require enzymes but are replaced with metallic catalysts, as mentioned in this research paper: Direct and Indirect Chemiluminescence: Reactions . I've bee trying to find similar papers on this, but with non-transition metals as catalysts; specifically group 1 and 2 metals. I appreciate the help

Though I wouldn't term the use of $\ce{NaOH}$ or $\ce{NaHCO3}$ in the chemiluminescence of luminol a catalyst, per se, the reaction must proceed in an alkaline environment to produce light. However, "Transition metal complexes, such  as  the  hexacyanoferrate  ion are  used  to  catalyze  the decomposition of $\ce{H2O2}$."
Consider that transition metals have multiple oxidation states, so the metal ion can serve to transfer charges, and then return to the initial oxidation state, i.e., acting as a catalyst without being consumed. Group I and II generally have only a single oxidation state in solution, and would not do so.


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Analytical Prediction of Molecular Interactions
Is it possible to predict the molecular interaction between any two molecules analytically or through some simulation mechanism based upon analytical theory?
I see a rather big gap between, say Quantum Mechanics and chemistry and I'm wondering if you can derive chemical reactions from base principles.

There is dynamically evolving field of quantum chemistry ( search e.g. for the Praque school of Q.CH., founded by prof Rudolf Zahradník), used e.g. as preselection step of searching for potential drug candidate, before any synthesis and testing is ever done. Another application is quantum and statistical simulation of behaviour of limited atom counts, like process leading to Coulombic explosion of alkali metals with water..

So there is no way to derive the interactions possible? Or at least write out their governing equations and simulate? That seems very surprising and kind of terrible lol.


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How molarity and mass fraction are related?

Concentrated sulfuric acid contains very little water, only $5\,\%$ by mass. It has a density of $\pu{1.84 g mL-1}$. What is the molarity of this acid?

I know to use the molarity $c$ is in $\pu{mol L^-1}$ and that the solution is $95\,\%$ acid and $5\,\%$ water by mass. I am just stuck trying to solve the problem.

https://en.wikipedia.org/wiki/Molar_concentration#Related_quantities

If the density of $95\%$ $\ce{H2SO4}$ is given as $\pu{1.84 g mL-1}$ and its molar mass is $\pu{98.08 g mol-1}$, you may do following conversion since units of molarity is $\pu{mol L-1}$ (suppose it is $M$):
$$ M = \frac{\pu{1.84 g}\text{ of solution }}{\pu{ mL}\text{ of solution }} \times \frac{\pu{95 g} \text{ of } \ce{H2SO4}}{\pu{100 g}\text{ of solution }} \times \frac{\pu{1000 mL}}{\pu{L}}\times \frac{\pu{1 mol}}{\pu{98.08 g}} = \pu{17.8 mol L-1}$$

@Karsten Theis: No, I just use the data given by OP.

1840 g/L x  0.95 / 98.08 g/mol = 17.83 mol/L

@Karsten Theis: I made an error. Forgot it is $95%$. Corrected it. Thanks for the careful reading.

$M$ is the standardized symbol for molar mass, not molarity. Also, what's with all the text in… a math formula? The unit symbol turns out to be $$\pu{mol \ce{H2SO4} L^-1 solution^-1}$$ which is nonsense.

@andselisk ♦: There were some confution of readers about unit conversion so I include the text for clarity.

@MathewMahindaratne That's why it's advised to first solve the problem using symbolic algebra (following standard notations and defining lesser common variables), and only then plug in the numbers. That's how clarity is achieved. Mixing text and math is simply wrong.


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why is benzene ring ubiquitous in almost all the organic compounds
Could you please give me a reasonable cause why benzene ring is found in every organic compound in food, pharmaceuticals, dyes . In addition, benzene is a known canceriogen, why shouldn't it be toxic when found in food items for example. I hope that I've formulated my question very clearly and if someone could reformulate my question in a better fashion, it would be highly appreciated. Thanks in advance for all your answers

Benzene ring is not benzene. You would be dead without benzene ring.

Nature does not tell us why the benzene cycle is so frequent in natural products. Maybe it is due to the remarkable stability of this ring. But in fact we must admit that we do not know everything in chemistry. Nature is stronger than chemists.

Benzene is a compound, which is a liquid and toxic. Phenyl gruop is the cyclic group (what you called "benzene ring") resulting from substituting one or more hydrogen in benzene with any other group. Most of those compounds with phenyl group are not terribly toxic (e,g., tylenol and aspirin).

Benzene rings are common but not ubiquitous in organic compounds. For example, of the amino acids that make up all proteins in living things (so most of you) only 3 out of 20 have benzene rings in them.

Related: Why six C atoms are usually seen in cyclic compounds?

P.S.: Or rather not ever born....

Does this answer your question? Why six C atoms are usually seen in cyclic compounds?

Nilay Ghosh
: no it doesn't answer my question

By "benzene ring", I assume that you mean the aryl functional group:
$\hspace{8 cm}$
Molecules with the above motif - called arenes - are common throughout nature and are present in a multitude of chemical reactions. The specific group above is called a "phenyl" group since there is only one substitution on the ring, but more substitutions are possible.
This type of structure is common in molecules for a multitude of reasons. The primary one is that it is very stable; certain cyclic conjugated groups, like the one above, have a property known as aromaticity (see Huckel's rules for more). Aromatic compounds are stable because the electron density inside the ring is completely delocalized. In fact, these types of rings usually only undergo substitution mechanisms, and rarely addition mechanisms unless forcing conditions (high temperature, high pressure, etc.) are utilized. You can read more about electrophilic aromatic reactions here, but the important thing to note is that when these groups react, they rarely do so by completely breaking this delocalization (as would otherwise occur in an addition reaction).
Additionally, although benzene is carcinogenic, the molecules that are derived from it can be either carcinogenic or not carcinogenic. Most simple derivatives of benzene (with only one substitution on the ring) fall roughly equally into either category: toluene and aniline are likely not carcinogenic, while phenol is a known carcinogen.
Going further to more complicated rings, we can see that what determines a molecule's carcinogenicity is less concrete than you might think. Back in the 1960s and 1970s, there was a huge environmental controversy when a class of compounds known as polychlorinated biphenyls (PCBs), which have two phenyl groups (see example below), were found to be carcinogenic to humans and nature:
$\hspace{5 cm}$
These kinds of molecules are part of a larger class of compounds called polycyclic aromatic hydrocarbons (PAHs) - almost all of which have some adverse human effect, usually carcinogenic.
However, not all aryl-containing substances are carcinogenic. You mentioned food in your question - most aryl rings in foods are found in aromatic amino acids, such as the tyrosine molecule below:
$\hspace{5 cm}$
Heterocylic aromatic groups can also be found in nucleic acids, such as cytosine (don't be fooled by the nitrogens replacing the carbons in the ring - it's still an aromatic compound!):
$\hspace{7 cm}$
In plants, you can find aromatic compounds in most plant hormones, such as auxins (the molecule below is indole-3-acetic acid, the most common auxin):
$\hspace{6 cm}$
Aromatic rings are also a common motif in many pharmaceutical compounds of interest.
Ultimately, if you view this from a chemical evolution perspective, delocalized aromatic rings are more "favored" in nature because of their stability and nonreactivity. Whether a molecule is carcinogenic or not can be influenced by the presence of an aromatic ring, but there are a plethora of other factors at play.

To restate the question: what is being referred to as "benzene ring" is a six-membered aromatic carbon ring. The question asks why this structure is ubiquitous, despite the simplest form of this structure being a carcinogen.
The simplest answer is that properties (chemical, physical, toxilogical) are features of the entire molecule, not of a fragment of the molecule. There's a popular meme that makes this point using NaCl: sodium metal reacts spectacularly with water; chlorine is a toxic gas. Salt is neither toxic nor water-reactive. Likewise, structures with aromatic six-membered rings are not necessarily carcinogenic.
In addition to being a very stable structure, as pointed out by the previous answer, the six membered ring has geometric advantages - it is easier to form. The aromatic six-membered ring is flat and has electronic properties that can be used in the assembly of host-guest molecular interactions. Shape plays a huge part in biologically relevant chemical systems.
Benzene is carcinogenic, but the very similar molecule toluene is not.

It turns out that even this seemingly minor modification of the structure changes the shape/electronic properties enough to alter the key interactions that are necessary to change DNA. https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1469723/


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Bromine isotope peaks were observed in EI-MS, but not in ESI-MS
I have been conducting a synthesis involving 2,3,5-tribromothiophene. Because I thought that desired product doesn't have so small molecular weight, I decided to use ESI instead of EI, and so I also requested ESI-MS experiment of 2,3,5-tribromothiohpene as a standard. In that time I got somewhat weird spectrum (Why is the base peak of my MS spectra of 2,3,5-tribromothiophene at m/z=382?). Below are the ESI-MS spectra of 2,3,5-tribromothiophene, and methanol which was used as solvent to insert it to the MS instrument, respectively.

Recently, I made a hypothesis that since 2,3,5-tribromothiohpene was purchased at around 2019, it may decomposed in long-term storage and produced strange spectrum. To verify this, I requested a EI-MS experiment and in this time I got a reasonable spectrum that has clear isotope peaks.

What is the problem with ESI-MS? May the operator miss some settings of the instrument? In fact, I have never observed any isotope peaks whenever I requested ESI-MS experiments of my reaction mixture. As a client, are there any instrument settings that I can request to the analyzer? For what it's worth, the model of the instrument is Xevo TQ-S micro.

They probably have used acetic acid erozol for electrospray so malecular peak would have been (.M + AcOH).

@MathewMahindaratne But still there are no isotope peaks for 3 Br's.

Can you post your ESI-MS as well so we can look at it?

@MathewMahindaratne Added it to the post.

Cannot see peaks around 80 m/z to find out bromine is eliminated. There is anothe peak next to 382 m/z peak. My geuss is two bormine atoms must have eliminated.

Unfortunately putting a bounty on question also attracts spam and trolling behaviour. Please raise a flag for that so that we can counteract this. Apart from this ignore it for the time being, as it will be deleted in due time.

You might already know this, but her we go anyway. There is currently a proposal for a more specialised site for mass spectrometry. I think this question might be a good fit, and you might want to support this proposal.

@Martin-マーチン Thank you. I didn't know the site for mass spectrometry exist. I will delete this post about 2 hours later.

@Martin-マーチンOh... Excuse me, but how can I retrieve the bounty? Just have to wait? I can't delete the post.

Currently the site for mass spectrometry is only a proposal. Your question is fine here, as is the bounty, but for the future you might be interested in the new proposal. Please don't delete it, just wait and hope someone is able to answer it.

@Martin-マーチン Ah, I see. I misunderstood the thing you said.

https://chemistry.stackexchange.com/questions/163871/bromine-isotope-peaks-were-observed-in-ei-ms-but-not-in-esi-ms


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Radius ratio of capped octahedral coordination (CN=7)
I was asked to work out the Radius ratio of capped octahedral coordination when CN = 7. How to define a capped octahedral? Is there any restrictions on the lengths of the edges or the distances of the vertex to the center?

Actually, that link isn't helpful for working out the ratio at all. See this publication by Roald Hoffman instead.

tks,man. I will use Mathematica to have a calculation.


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How do water vapor molecules push other air molecules away?
I've read that when water vapor enters the atmosphere, the vapor molecules push the oxygen and nitrogen molecules out of the way and occupy their space.
But if water vapor molecules are lighter than oxygen and nitrogen molecules, then how do their molecules push them away? On what factors depends a gas pushing another gas away?

Welcome to the chemistry site of the StackExchange (SE) network.
Be aware that laconic, not elaborated questions are usually closed on the SE.
The site expects that you include the compact summary of your related current knowledge, involving your conclusion of searching for existing related  info or answers. It would prevent others to tell you what you already know or what you could easily find yourself. Effort not shown may be considered as effort not done, possibly leading to the question closure. How do I ask a good question?.

What is the setting we are talking about? A pressure cooker? An autoclave? An ordinary living room?

There is a lot of free space in air, not need of pushing anything away. Just to get into the picture, at normal conditions, molecules of air typically freely fly along a path several hundred times their size, until they hit other molecules.(higher tens of nm)
Imagine a large gym where is flying and bouncing 200 volleyballs, and you just add 6 basketballs. Do they need to push volleyballs away ? No, the do not.
What happens, when the molecule counts per volume increases, is the increase of pressure and the gas expansion. If you had 2 reservoirs of dry air, each 20 °C with the pressure 100 kPa, and if you let evaporate some water in one of them (while keeping the temperature), its pressure would increase. If they were interconnected, its content would partially expand to the other until pressures got equal.

Ať the given T and p, gas contains (idealized,with small error ) the same number of molecules, regardless of gas.(Avogadro law). If a molecule has a lower mass, the gas has lower density:
$$\rho=\frac{pM}{RT}$$

Air with vapor expands to have the same pressure as the surrounding air, so numbers decrease to value according to p and T.
Total number of molecules per unit volume is $\frac NV =\frac {N_\mathrm{A}.pV}{RT}\frac 1V=\frac {N_\mathrm{A}.p}{RT}$

Perhaps all started with misunderstood replacement. If you add 10 mmol of air molecules to 1 mol of air molecules do they replace them? The same for 10 mmol of water molecules.
This schematic illustration may help (not in the scale):

The first line below illustrates 2 regions of dry air.
The second line gets in the left extra water molecules, gaining higher density of molecules and therefore pressure.
The third line is the result after higher presure of the left region pushing surrounding air away and expands until equalization of molecular densities and pressures.molecules, gaining higher density of molecules and therefore pressure.

At the same temperature and pressure, assuming ideal gas behavior, there is the same average space per a single molecule of a gas, regardless of the type of this molecule ($\ce{N2}$, $\ce{O2}$, $\ce{CO2}$, $\ce{H2O}$, $\ce{CH4}$,...)

..A.....A.....A.....A..    | ..A.....A.....A.....A..
..A..W..A.....A..W..A.. -->| ..A.....A.....A.....A..
..A.....W.....A.....A.....W.....A..| ..A.....A.....A.....A..

A = nitrogen or oxygen molecules of air
W = water molecules
. = free space

But if the vapor molecules just occupy empty space in the atmosphere, then how does that decrease the overall density? Bcz even if water vapor has a lower density, it is still more dense than empty space. So, if the vapor molecules(which have low density) now occupy empty space(which have no density) then wouldn't that increase the mass, and therefore increase the density(even if by a small amount)

Let's say there's some water in a container and there's a column of air above it. Now let's say you heat the water and so it turns into water vapor, which flies into the atmosphere. The molecules of this water vapor weren't part of the atmosphere when it was water. So, how does the number of molecules (and thus mass,M) not increase when they enter the atmosphere?

@ACRafi The answer has been updated.

Think of this as a meteorological question, not as chemistry.
Suppose some area of the Earth, perhaps 30 km square, has say 10 billion tonnes of air above it.
If it is over a wet (ocean) or moist (jungle) area and some water evaporates into it, say a few hundred million tonnes of water vapour. The air will now have a mass that is several percent higher than it was before; its original mass plus the mass of the water.
The weight of that increased mass of air over any small area will be increased by several percent, increasing the air pressure (which is normally, and somewhat amazingly, over 10 tonnes per square metre).
In terms of weather, that is called a high pressure area.
Very large high pressure areas are marked with an "H" on weather maps.

The surrounding dry air will still have its original lower pressure, and so air will flow from the higher pressure area into the lower pressure area, creating wind.
In the above map, the huge high over Idaho can be seen pushing the blue-coloured front to the south and east.
Similarly, the low over Wisconsin is allowing the red front to move northward.
(The colours indicate that the moving air is relatively colder or warmer than the air in the area they are moving into.
The black lines are isobars, indicating lines with the same air pressure along them, as indicate by their numbers.)
The original 10 billion tonnes of air that was over the original area will spread out, so that over any one spot there will be a little less of the original air than there was before.
This results in the phenomenon that the question refers to as "pushing the oxygen and nitrogen molecules out of the way".

It really is Chemistry, a combination of the ideal gas law and Daltons law of partial pressures with the Laws of thermodynamics and the kinetic theory of gases thrown in for good measure.

Nice to see the meteorological fellow. Not being a pro, but after my master degree I served as an enlisted air force meteorologist in 1989-90.

Water molecules leaving a surface have a momentum directed away from the surface. This momentum of the added molecules results in an increase in pressure of the air mass. The air expands against the surrounding atmosphere increasing the volume. The collisions with other molecules push them away until pressures [almost] equalize. Since the molecular weight of water is lower than air the overall density of the air mass decreases. Surrounding more dense air is pulled down by gravity forcing the less dense air upwards where it adiabatically cools by expanding against the lower pressure air etc. Contrary to popular usage convection is caused by gravity affecting air density variations not by pressure differences, no gravity there is no convection. Convection is directional with gravity; pressure is exerted in every direction causing air masses to expand or contract until pressures equalize, resulting in changes in density.
To answer the original question the air molecules are pushed outwards by the greater number of collisions with the greater number of molecules in the air with the water molecules [higher pressure]. Temperature is also involved since higher water temperature both increases evaporation and raises the temperature of the air both of which raise the pressure increasing the rate of expansion and the decrease in density that increases convection and results in hurricanes over the tropical oceans.


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Understanding the concept of voltage in electrode reactions
Background: I'm currently reading "Electrochemical methods: fundamentals and applications," by Allen J. Bard. In chapter 1, he has the following setup:

He then states that we incorporate the power supply such that the potential of the Pt electrode is more negative than that of the Ag/AgBr electrode. This leads to reduction of H+ at the Pt electrode. Here's my question:
Let's assume that potential is kQ/r, where 'k' is the coulomb constant, 'Q' is the "constant charge" that affects 'q' (a variable charge), and 'r' is the distance between 'q' and 'Q.' I'm having issues applying that concept/formulation of potential to the setup described here.

What would 'r' refer to exactly in this context?

In the situation where H+ is reduced, the implication is that there's a net movement of electrons towards the platinum electrode. This means that's there's a net force on electrons whose direction is towards the platinum electrode. I don't know how to relate this finding to the statement that the potential at the Pt electrode is more negative than the potential at the other electrode.

Any help is appreciated.

Note that usual arrangement is 3-electrode setup, 2 working electrodes designed for passing the current and the 3rd reference electrode with minimal current. Potential of 1 or  both working electrodes  is then measured art the reference one.

It seems to me you try to  apply to the potential meaning the approach of electrostatics, which is not easily applicable in electrochemistry. Consider  it rather in sense of nonvolume work dW=U.dq.

See also Wikipedia topics Standard hydrogen electrode(SHE) and Absolute electrode potential

To restate the question as I understand it, there are two parts: 1) if you look at the platinum electrode as a capacitor, what is the distance between the "plates", and 2) what is the direction of electron flow in the reduction of H+ to H2?
I'll start with the second question: because there is reduction occuring at the platinum electrode, the electrons are flowing from the platinum surface to the hydrogen ions. In order for that to be accomplished, the electrode has a build-up of electrons, or negative charges. Because opposite charges attract, positive ions from solution will start to accumulate near the platinum electrode. This charge separation acts as a capacitor. Terms you may encounter that refer to this effect are "the electrical double layer", or "double layer region", or "double layer capacitance". On a molecular level, this effect is used to describe the regions as the inner and outer Helmholtz planes. Therefore, r is the distance between the "plates" or planes in the double layer. This is governed by the radii of the ions.
The image below, from Wikipedia https://en.wikipedia.org/wiki/Double-layer_capacitance illustrates this concept.


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Help in analysing dilatometer results
I need some help in analysing results for a dilotometer experiment. I did this experiment for the first time and don't have any prior experience in this field. The main aim of this experiment was calculating the CTE( coefficent of thermal expansion). The type of dilatometer was a push rod dilatometer.
The sample was basic gamma alumina pellet of 19 mm by 3 mm. The pellet was a green body i.e it wasn't sintered. I have attached the resultant graph.
My main doubt is as follows -

The operating temperature was at 900 deg cel. For calculating the CTE or linear CTE in this case, do I consider only the intial length at 25 deg cel and final length at 900 deg cel, or do I take into account the up and down trends occuring in between.

I am fairly certain that at 900 deg cel the gamma alumina gets converted into another phase, how do I account for the phase changes?

If anyone could guide me in how to analyse the graph, it would help me a lot. Sorry if this sounds like a basic question, I know homework questions aren't entertained here. This experiment is part of my QA/QC establishing process of my catalyst samples.

Since you used green material, it could have shrunk on sintering. Given this I suspect you can't get reliable expansion data at all.

But even if it's shrinks, I'll get a negative coefficent of thermal expansion right? I have seen people in literature reporting negative CTE values

Why even if? Why not because of ?After 110 min, it starts to turn toward negative, possibly due sintering ( and/or phase change), crossing zero near 155 min. // IMHO, dilation measurement makes sense either for bulk material, either for sintered one within T range of no observer sintering/phase changing.

@Poutnik Thank you for replying. After around 130 mins i.e when it reached 900 deg cel,  cooling was started.  Yes, I have planned do a similar experiment for a sintered body.

Hmm, then I have misinterpreted the chart, thinking 900 °C was reached at the end at about 175 min.