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What is going on in and below a yellow or blue gas flame?
Unsure if this is more chemistry or physics, but here goes...
Consider a Bunsen burner and the flame it produces when the air inlets are closed vs open.
According to the description on the linked Wikipedia page, combustion is incomplete when the air inlets are closed, but essentially complete when the air inlets are suitably adjusted.
This raises a number of questions about what is going on in each scenario.
Why is combustion incomplete when the inlets are closed? Wouldn't gases mix as they rise up allowing most or all of the fuel to burn? Where exactly is combustion occurring relative to the incandescence? Obviously no lower than the tip of the burner where the fuel gas meets air, but how far up the visible flame? According to the answer to this question, combustion is only occurring at the outer shell, but how does fuel remain unexposed to air until it is too cool to react?
It would seem that even when the inlets are open, combustion only starts at the tip of the burner, not inside the barrel... why? What prevents the flame from migrating down inside the barrel where fuel and air are already mixing?
Motivating my question is a desire to understand how my gas fireplace can make a pretty yellow flame while the flames of a gas cooktop or gas barbecue are blue.

This is a remarkably complicated set of inter-related questions! There is a large literature on flames, their chemistries and properties. Michael Faraday even wrote the classic book on candle flames. I am upvoting, but you really should go for one tight question at a time! By the way, having the flame propagate down the barrel can lead to a memorably loud event called a flashback. The burner heads in flame atomic absorption spectrometers are tied down to prevent burner head launching if a flashback happens in the pre-mixed acetylene and air combustion mixture.

Does this answer your question? Why are hydrocarbon flames yellow or blue?

Considering that there are already very high quality answers here specific to the historically important  invention of the Bunsen burner, it's principle of operation and observed behaviors, I think serves the site and future readers better to keep this question open rather than close as duplicate and direct views away from it. Therefore I'm voting to leave open Related in meta and so far unanswered: Have there been analyses to see if views of well-received answers are reduced by closure as duplicate which try to control for other factors?

Not sure if the fireplace is supposed to look like that, but the yellow / blue colour can indicate an incomplete combustion (eventually too much CO, not only soot). If the fireplace is something related to camping, then it can be also because you burn butane instead of methane.

@Alchimista I'm talking about a gas fireplace designed to mimic the look and ambience of a wood-burning fireplace but can be turned on and off at the flick of a switch - ceramic imitation logs, burning natural gas, does provide useful heating but displays a "natural"-looking yellow flame by design.

@AnthonyX I was thinking that the yellow colour is a wanted feature. The comment was mostly due to the fact that CO is too dangerous because sneaky ;)

@Mithoron Why flame is yellow or blue is interesting, but not quite the focus of my question, I'm more interested in what I suppose is the physics/engineering concerning the promotion or inhibition of mixing and/or combustion in various zones of the device - why the combustion in a yellow flame is incomplete - what stops all the fuel from being consumed vs what causes the fuel to be so fully consumed in the blue flame scenario.

@Alchimista These devices are designed with either strong draft or sealed glass barrier with outdoor inlet/exhaust to keep combustion products from infiltrating a living space

@AnthonyX I just saw this comment and I think it's an excellent suggestion. Do you think you might? If not, I will, but I think you're on a roll here so maybe you can consider it? :-)

I say leave open, vote other question as duplicate.

The yellow emission has a continuous spectrum. So it is not due to an atom or of a molecule. It is due to a solid black body, hot enough to emit light (about $1000$°C). The solid stuff is solid carbon $\ce{C_n}$ with an unknown but rather high $n$ value. The presence of solid carbon may be demonstrated by inserting a cold glass tube for a while in the yellow flame. It soon becomes covered by black soot.
The blue flame has been analyzed by Herzberg in $1939$. Its spectrum is made of thin lines belonging to the unstable molecule $\ce{C2}$. The distance between the lines is typical of the vibration spectrum of the diatomic molecule $\ce{C2}$.
So when burning, the scenario is the following : The methane (or any other alcane) molecule begins by loosing its $\ce{H}$ atoms, which burn to make $\ce{H2O}$. Then the remaining $\ce{C}$ atoms start polymerization to produce first $\ce{C2}$ which emits a blue light. Then if enough oxygen is available, the $\ce{C2}$ molecules are oxidized into $\ce{CO2}$. But if Oxygen is lacking, the polymerization goes on to produce a black body (soot) which is warm enough to emit a continuous light. At the end, when the hot gas gets out of the flame, this hot carbon polymer is oxidized by outer air into $\ce{CO2}$ without producing light.
Of course, if oxygen is seriously lacking, the flame may produce black deposit of soot, but this never happens in a Bunsen burner.

This answer is an excellent compliment to M. Farooq's answer, providing the chemical reaction details that suppliment M. Farooq's fluid dynamics analysis.

@electronpusher Entirely agree with you and I upvoted both answers and the question. Now maybe someone will ask about the reverse flames mentioned by M.Farooq or ask about the flame experiments that were conducted on the ISS!

This is a good answer about what happens in flames, but it misses the actual questions posed.

@matt black. The question asked was : What is going on in and below the yellow and the blue gas flame ? I have given the answer : The blue flame is made of $\ce{C2}$ molecules. The yellow flame is made of solid carbon polymer $\ce{Cn}$, with $n$ is a huge number.

@Maurice But the context given in the full question also pointed out "a desire to understand how my gas fireplace can make a pretty yellow flame" and asked "Why is combustion incomplete when the inlets are closed?"

@matt black. When the inlets are closed, the gas getting out of the burner is made of methane, without oxygen. The only oxygen available is coming from the outer atmosphere. The reaction occurs in the intermediate zone where they mix, where Oxygen is lacking to react with both $\ce{H}$ and $\ce{C}$ atoms. That is why the reaction produces much polymer $\ce{C_n}$ , much soot, hot enough to emit plenty of yellow light from the black body radiation.

As Ed V has told you, there are plenty of mixed questions, but very interesting ones. I will try to address your key question.

Consider a Bunsen burner and the flame it produces when the air inlets
are closed vs open. According to the description on the linked
Wikipedia page, combustion is incomplete when the air inlets are
closed, but essentially complete when the air inlets are suitably
adjusted.

The key point is the Bernoulli's principle in action in a Bunsen burner. If look at the base of the burner's, fuel enters the vertical tube via a very narrow orfice (see part "a" in the picture). Therefore, the velocity of the fuel is pretty high at the base of the burner. This also creates vacuum at the base of the burner. Depending on the size of the air opening at the base (adjustable in modern burners), air is drawn in.
Now the diameter of the Burner tube is not very narrow so there is no turbulence (my own speculation) rather the gases (fuel and oxidant) have a laminar flow. In perfect laminar flow, fluids should not mix at all. The blue cone in the flame reflects the shape of laminar flow of the gases. Fastest in the center of the tube and slower near the tube periphery. The amount of oxygen entering the orifice is still not sufficient to burn all the fuel.
You have to rely on the diffusion of oxygen in the air to burn the remaining fuel.
Important point: In a pure blue flame, there is no incandescence. This blue color is molecular emission.
When the inlet is closed, air supply is definitely limited. The flame is relatively yellowish. The cone vanishes and now you have incandescence from soot particles forming from incomplete combustion.
What is rather interesting, and I saw a video long time ago on YouTube was the question, what if you supply oxygen and burn the flame in the atmosphere of the fuel gas. All these flame structures vanish in a so-called reverse flame.
Picture taken from J. Chem. Educ. 2000, 77, 5, 558

EDIT: A reader asks the evidence of laminar flow in the comment. Think of the diameter of the Bunsen burner tube, and think of the gas velocities needed to create an Reynold's number greater than 2000. With turbulene, you will not get a beautiful blue cone (indicative of laminar flow). See the example here Turbulence in flames

I think your speculation on turbulence is exactly wrong. The setup in the bunsen encourages effective fuel/air mixing and that, most likely, involves turbulence (unless you have some evidence otherwise).

Yes turbulence in flames exist but not in normal conditions. You need very high gas velocities. Added pictures. If you have flame spectroscopy with nitrous oxide, you will see the flame is crazily turbulent with an eerie sunset like glow!

The issue isn't about what the flow in the flame looks like: it is about mixing gases in the bunsen tube far below the flame. The flow at the exit if far slower and less turbulent than the flow around the gas nozzle at the bottom.

Did you read the linked paper in the edited section? There is no turbulence in ordinary flame or in the Bunsen burner tube! In a Bunsen burner tube which is about 1 cm wide, can you calculate the gas velocities required to generate a Reynold's number of 2000?

The abstract talks about flames, I can't tell if it talks about what happens at the bottom of the tube. The fuel in a bunsen in emitted from a very narrow orifice and is, therefore, moving a great deal faster than the (mixed) gas at the top of the tube. (ratios are maybe 100 sq mm at the top to perhaps 1/100 sq mm at the orifice).

BTW if you consider that the orifice from which methane flows is about 10000 or 1000 times smaller than the area of the burner head, it isn't hard to get a big reynolds number at the orifice even if the flow at the top is not turbulent.

No, the gas orifice dimension is not like a hypodermic needle orifice. It is about ~ 1 mm wide or slightly less in a typical Bunsen burner. If there were turbulence near the air holes, you would not see Bernoulli effect or Venturi's effect. Once a fluid enters a wider bore, it forgets its turbulent behavior. Anyway, feel free to come up with solid numbers by measuring the dimensions of a real Bunsen burner and calculate the Reynold's number and gas velocities.

Bunsen Burners are specifically designed for the flame not to backpropagate into the tube
The other answers give good descriptions of what is happening in the flame under different conditions. But what is missing is that Bunsen Burners are designed to achieve the properties you observe.
For a "blue" flame where the air flows into the burner at the bottom and is well mixed with the fuel (which is what the gas nozzle and the tube are designed to achieve), the rate of gas flow is part of the design. And that rate is intended to produce a hot, well mixed flame where the gas flows fast enough to prevent the flame propagating down the tube.
When the air inlet is closed, the gas in the tube is mostly fuel and mixing only occurs above the top of the tube and is much less efficient, hence the yellow flame resulting from only partial combustion. The flame can't propagate down the tube as there is no air to enable the fuel to burn. The mixing is less efficient, by the way, because it is limited by the slower speed of the gas at the top of the tube compared to the fast speed of the fuel emerging from the small aperture in the fuel inlet which causes turbulent mixing inside the tube with air when the bottom inlet is open.
Most burners in cookers and fires are designed to work for a specific fuel and create the ideal well-mixed fuel/air flow at the outlet of their burners (and the mix depends on the fuel which is why different burners are needed for different fuels–as when UK gas cookers had to be retrofitted with new burners when goal gas (a mix of hydrogen and CO) was replaced by natural gas (mostly methane) in the 1960s).
So, if you want to adjust a flame in your fire to be yellow for a pleasant effect, you will have to fiddle with the burner heads to adjust the incoming flow of air in those burner heads. This might be very hard if the burners are not designed to be adjustable and will have the (possibly undesired) effect of greatly reducing the amount of heat given out.

It still leaves a bit of mystery as to why well-mixed fuel and air starts burning exactly at the very tip of the Bunsen tube and not over a region inside or above. Same goes for any other clean-burning gas appliance. I guess the question might be how does a flame holder so neatly hold the flame of already-well-mixed fuel and air that one might think could already start burning somewhere else?

@AnthonyX It isn't that much of a mystery if you recognise that these devices are designed. A generic burner can have all sorts of pathologies including the flame back propagating down the tube. But a Bunsen burner or cooker burner is designed to get the flow rates right so the gas flow balances the speed of flame propagation so the edge of the flame stays close to the tip of the burner.


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How to experimentally determine the dissociation energy (electronegativity) between identical atoms?
I was reading on Wikipedia about how Pauling electronegativities are determined through
$$ \vert \chi_\ce{A} - \chi_\ce{B} \vert = \sqrt{E_\mathrm{d}(\ce{AB}) - \frac{E_\mathrm{d}(\ce{AA}) + E_\mathrm{d}(\ce{BB})}{2}}$$
At first, I thought I understood this since for an element like hydrogen the meaning of $E_\mathrm{d}(\ce{HH})$ is clear and easy enough to measure experimentally.
But I don't see how I would do the measurement in general. Do I have to make (unstable) diatoms and then try to measure the dissociation energy very quickly? Even the electronegativity for atoms with $Z > 100$ are known which baffles me. How do they measure these dissociation energies?

Off the top of my head, I believe the answer is that the element simply needs to bond covalently or metallically with itself when it is in its standard state.  Its standard state doesn't need to be a diatomic.  The metals form metallic bonds.  The metalloids and non-metals (except the noble gases) form covalent bonds.   For instance, carbon's standard state is graphite.   The BDE for the C-C bond would then be the average energy it takes to break apart the C-C bonds in graphite.  Other than the noble gases, I don't know of any elements that don't chemically bond with themselves.

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What is the meaning of a “halide component” for Friedel–Crafts reaction?

Among the given halides, which one can be used as halide component for Friedel–Crafts reaction?
(a) Isopropyl chloride
(b) Bromobenzene
(c) Chlorobenzene
(d) Chloroethene

I don't understand what a "halide component" is. Is it the reactant that we use with existing molecule (a simple benzene ring, for example) on which we want to perform FC reaction on, or the reagent itself (i.e. anhyd. $\ce{AlCl3}$)?

Obviously, none of these is AlCl3.

Can you please elaborate?

Friedel–Crafts reactions proceed through the alkylation or acylation of an aromatic ring using an alkyl halide. Considering the alkyl halide compound of the form
$$\ce{R-X}$$
where $\ce{R}$ is the alkyl group and $\ce{X}$ is the halide (whatever it may be), I assume the question is asking what compound, of the previously noted form, could be added to an aromatic ring using electrophilic aromatic substitution.
Given this, I would suggest imagining which of these four compounds are actually alkyl halides at all — only one of them will work for Friedel–Crafts alkylation. It's also important to understand one of the intermediate steps in the alkylation mechanism. Formation of the carbocation (if this is preferred) or formation of the complex with the Lewis acid catalyst is preferable only for alkyl halides.
This is due to the fact that carbocations are increasingly unstable as hybridization shifts (when there's double bond character).

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Why isn't the aniline hydrolysed in Sanger's protein sequencing method?
Recently I was reading about identifying N-terminal amino acid residues using Sanger's reagent (1-fluoro-2,4-dinitrobenzene). The following image showing the reaction is taken fro Wikimedia Commons:

The free N-terminus amino group performs a nucleophilic aromatic substitution to get the first product; after that we hydrolyse the adduct. Why doesn't the Ar–N bond in the final product also undergo hydrolysis to give 2,4-dinitroaniline, though? There are many examples (e.g. isocyanides, cyanides) where heating with strong acid causes cleavage of the C–N bond. Is it because resonance with the aromatic ring causes the nitrogen lone pair to be not basic?

Please note my edit to your question: (1) The chemical names are wrong; substituents (fluoro, dinitro) are joined to the parent chain (benzene) by hyphens, not separated by spaces. (2) The $\mathrm{S_NAr}$ mechanism is not an "ary $\mathrm{S_N2}$". (3) No need to put three or four consecutive question marks; one is enough.

@orthocresol brother can you please check my understanding in the comments that i made on your answer?

You raise the examples of isonitriles and nitriles undergoing C–N bond cleavage via hydrolysis. That is true, but it is worth thinking about how this happens. Both isonitriles and nitriles are converted to amides first when reacted with aqueous acid. Now, amides will certainly be hydrolysed at the last stage of the peptide sequencing: you can see that in the image itself, as the amide bonds holding the polypeptide together are broken (so that the labelled residue can be detected).
However, amides behave quite differently from amines. The C–N bond you're asking about ($\ce{R-NH-Ar}$, Ar = 2,4-dinitrophenyl) is an amine, not an amide. The comparison to isonitriles / nitriles is therefore not relevant. Amines simply don't undergo typical $\mathrm{S_N2}$ reactions (quaternary ammonium salts do, but that's a different matter entirely), so the C–N bond doesn't get broken via hydrolysis. It is true that there is resonance with the electron-withdrawing dinitrophenyl group, which reduces the availability of the nitrogen lone pair, but this is actually irrelevant: even if that aryl group were just a methyl group, the conclusion would not be affected.
Note that this has nothing to do with the C–N bond strength, either. The C–N bond in an amide has partial double bond character, and is stronger than the C–N bond in an amine (which is a plain single bond). The question is not which is stronger, it is which has an available pathway for cleavage that doesn't involve too high an activation energy. Amides can be cleaved via a nucleophilic acyl substitution mechanism: although this is not easy to accomplish (you need strong acid and lots of heat), it is still possible. Amines, on the other hand, could conceivably be cleaved via a $\mathrm{S_N2}$ or $\mathrm{S_N1}$ mechanism. Neither is very favourable in this case (amines are very poor leaving groups, and the carbocation above is not stable at all, with that electron-withdrawing carbonyl group next to it).

I think what I got confused on was thinking that R-NH-R' If the Nitrogen in this formed a coordinate bond with a proton then the quaternary amine so formed will be a good leaving group but it turns out even then it doesn't. amines form salts with acids and this doesn't change the fact that they are bad leaving groups. Here the only possibility was nucleophilic acyl substitution. That is infact not related to Rnh2 being a good leaving group or not. It is just a property of acid derivatives to show this reaction whenever surrounding medium is deficient of the group that is attached to carbonyl

Even if RNHR' form salts with some acid(ie donates a lp to $H^+$) even then we wont expect it to break the original C-N bonds that are present in the salt. kindly confirm these thoughts of mine.

from the last line of my first comment sir I wanted to say that that the equilibrium of acyl sn2 will be drawn to the products only if there is a deficiency of the group being displaced in the surrounding med( to prevent reverse reaction)(le charterlier principal)


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How baking soda help in indigestion of stomach?
We have learnt that baking soda can be used for neutralizing excessive acid in stomach formed during indigestion.
And after some time we get a burp(CO2) after drinking baking soda solution.
So why we got that burp?
I got 2 answers in whom I am totally confused which one is correct :-
Answer 1. When we dissolve baking soda in water we know it as a basic salt(strong base weak acid salt) that's why it form H2CO3 by taking a proton (Hydrogen ion) from water , thereby releasing OH- ion in solution that help in neutralizing excess H+(hydrogen ion)  in stomach from HCl. And simultaneously keep forming H2CO3 upto the time we are drinking baking soda solution.
NaHCO3 + H2O---> Na+(spectator ion) + H2CO3
This H2CO3 dissociate to form H2O and CO2 , causing us burp.
H2CO3---> H2O + CO2(cause burp)
Answer 2. Sodium carbonate react with Hydrochloric acid.
NaHCO3(aq) + HCl (aq)----> NaCl (aq)+ H2CO3(aq)
Now this carbonic acid dissociate to form
Carbon dioxide cause burp.
H2CO3 ---> H2O + CO2 (cause burp)
Now my doubt is " which is correct reason for release of carbon dioxide as burp after drinking baking soda?"
1 or 2
NaHCO3 is stomach is reacting with Water with which it is drinked with (dissolve in ) or HCl present in stomach to give CO2?As NaHCO3 can react with water and HCl both , so in stomach which reaction is going on ? Why?
HCl referring here Hydrochloric acid present in stomach for digestion and activation of pepsin

Essentially the same thing as happens with baking soda and vinegar: you get carbon dioxide and water and sodium ions as spectators and the conjugate base of the weak acid.

@Maurice, why not 1? Bcuz we when  dissolve NaHCO3 in water  it will start forming H2CO3 which go along with water in stomach and dissociating simultaneously cause burp??? Also read the edit I made recently

@EdV sorry it's typing error , thanks and also can you elaborate your comment

@Ed V i think I have done all HCL into HCl

Dissolve NaHCO3 in water and observe it. Does it release a gas ? No, it does not. // You may find useful these links for text formatting ( not to be applied to titles ):
notation ,  text + formula formatting  and
upright vs italic

@ poutnik, it should create a little fizz according to equation , shouldn't it? Look
https://bakinghow.com/baking-soda-and-water/#:~:text=Essentially%2C%20the%20baking%20soda%20reacts,acid%2C%20ultimately%20creating%20carbon%20dioxide.      Read it you will find that there written it produce CO2. Than how it is possible No fizz. Please explain sir.

BTW thanks for link. I will use it next time

Please note that hydrogen chloride ($\ce{HCl}$) and hydrochloric acid are not the same. See here for example: https://en.m.wikipedia.org/wiki/Hydrochloric_acid. You will see that both your options are incorrect as soon as you understand the difference.

@snijderfrey, I never mention anything about hydrogen chloride in question or have I??....And I know the difference sir. ..

Well, hydrogen chloride is written in your first equation below answer 2.

Oo.....look at edit in question

Hydrogen chloride is HCl(g,l,s). Hydrochloric acid is HCl(aq), fully dissociated to H+(aq) and Cl-(aq).

@poutnik , that what I have written in 1st eq. Below second answer

The essential reactions of baking soda neutralizing an excessive stomach acid, irritating the upper digestive system, are:
$$\ce{HCO3-(aq) + H+(aq) <=>> H2CO3(aq) <=>> H2O + CO2(aq) <=> H2O + CO2(g)}$$
$\ce{Na+(aq)}$ ( from $\ce{NaHCO3(s)}$ ) and $\ce{Cl-(aq)}$ ( from $\ce{HCl(aq)}$ ) ions are "spectator" ions, forming solution of ionized $\ce{NaCl(aq)}$.

(+1) Finally! I thought this tempest in a teapot would go on and on. Thanks for putting it out of our misery!

@poutnik, nice answer but what about water in which I mix baking soda to drink can also react with NaHCO3 (1st ans) in stomach along with HCl  . Shouldn't it?

@SureshChandraPal Time for own thinking. Does water contain H+(aq) ? How much ?

@poutnik, yes it does . In a very small amount(if asking about self ionization of water)  But I have learnt that basic salt  can snatch off H+ from water leaving behind OH-   ,   so as much water is there  that much H+  NaHCO3can take from it......isn't it?    Look : Answer to Is NaHCO3 acidic or basic? by Mike Jones https://www.quora.com/Is-NaHCO3-acidic-or-basic/answer/Mike-Jones-2528?ch=15&oid=174685333&share=328c4e7f&srid=hrHyD7&target_type=answer .   Here you will find NaHCO3 can take H+ from H2O ,

no where mention it take H+ from self ionization..and a small amount of water contain a lot H2O thus lot of H+ it can snatch. Right??.

@SureshChandraPal Time to learn about quantitative acidobasic equilibrii.

@SureshChandraPal, Please find on the internet just how basic the bicarbonate anion is, and how much more acidic H+(aq) is compared to water. You really could accept this answer as the correct one.

Your first equation is based on a non-existant reaction. I repeat it here : $\ce{NaHCO3 + H2O -> Na+ + H2CO3 + OH-}$.
No ! $\ce{NaHCO3}$ will very rarely produce $\ce{OH-}$ ions in water : concentration lower than $10^{-5}$ M, as pH is around $8.3 - 8.4$. On the contrary, it may react with $\ce{OH-}$ ions, but not produce them.
So your second answer is correct : $\ce{NaHCO3}$ reacts with $\ce{HCl}$ to produce $\ce{H2CO3}$, then $\ce{CO2}$

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How to dissolve and homogenize DNA
I have about 20-30 grams of purified mammalian DNA preserved in EDTA. It is clumped up and solid. I need to dissolve and homogenize it, so it can be aliquoted, without damaging or lysing or fragmenting the DNA. What is the best way to do this?

I think you can do this with TE buffer or water, the latter of which needs to be 18 M$\Omega$ RO.

How large are the DNA pieces (eg number of base pairs)? That’s a big factor in how susceptible they are to mechanical damage. If it’s a small plasmid, you can pipette vigorously without much worry, but if it’s genomic DNA you need to be much gentler

@Andrew I updated the questions. It's mammalian genomic DNA.

I suspect you'll get a much better answer if you migrate this question to biology SE, but here's what I've done:

add your chosen volume of nuclease-free TE buffer or water (assuming the DNA is precipitated with salts).

Leave it in the fridge (~4 C) overnight to allow for slow hydration.

Gently resuspend the DNA by very slowly pipetting the solution up and down using a 1 mL pipette with about 1 cm of the tip cut off so that the opening is substantially larger than normal.

Make aliquots for future use so as to minimize handling.


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How Does Peroxide + UV = Oxidation?
I restore shoes, & we have this method for oxidizing the translucent soles to get them to look new again.
The process is:
clean surface
Apply 40 Volume peroxide cream
Place under UV Growlight.
I follow the process. I get the results. But I don’t actually understand what happening. What does the peroxide & passing uv do to remove the yellowing? Can I concentrate the peroxide for more thorough & faster results? How big of a factor is the UV light? If I hade to make a compromise in either direction, would I be better off with more concentrated peroxide cream or more powerful UV source? I want to really understand what’s going on so that I can improve the service that I provide. Thank you for any time you put into this question.

You are not "de-oxidising" the shoes, you are oxidising them. The yellow colour comes from conjugated double bonds and oxidation destroys the double bonds and hence removes the colour. The oxidation is done by free oxygen radicals produced by photolysis of the peroxide molecules. This method is widely used by the retro computer enthusiast community, and is know as retr0briting.

@JohnRennie Maybe convert your comment into an answer?

@EdV There's way too much questions here.

@Mithoron You are probably right, if I understand your point, but some open questions attract very low quality answers, which would not be the case if the first commenter posts an answer. Just my opinion, anyway.

I appreciate that. I understand that there’s a lot of questions, but if you could help me with at least one, I’d be very grateful.

I do not consider myself qualified to answer the questions you have and @Mithoron put his finger on it: lots of questions. But there are many following questions: what is the polymeric translucent sole material?, what is 40 volume peroxide cream?, what is the wavelength range and intensity of the UV light?, how long does the treatment last?, and on and on. So you have a treatment method you feel works well, but it is empirical, as I see it, and I have no idea how you would improve it, given what I said, except by further trial and error experimentation. Sorry this is not so helpful.


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How to adjust selectivity of gas-permeable PDMS membranes for environmental applications?
I have an idea about making an environmental sensor that uses a gas-permeable membrane (GMP) and how its probe could be submerged in a water source with an off the shelf gas sensor.
So far, my research has pointed me to using polydimethylsiloxanes (PDMS), but I am not able to get a straight answer on several aspects of GPMs from the distributers and manufacturers on one specific question I have. From what I understand, these membranes can be tuned by changing the thickness to increase or optimize permeability for certain gasses.
Is it possible to tune them to a specific set of gases, or to use a single membrane that blocks moisture but allows at least ammonia, carbon dioxide, oxygen and hydrogen sulfide gases through?
Would I be better served by membranes tuned to these specific gases, or is there a way to use a single membrane, tunable to the smallest of these $(\ce{O2}?),$ or maybe the largest $(\ce{NH3}?)$ that also has enough permeability for the other gasses, at least in a way that can be compensated for with time for their diffusion into the sensor chamber?
Any clarification or details about how this might work would be appreciated.


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Hydrogens of cyclohexanol are enantiotopic or diastereotopic, which is true?
I am studying spin-spin coupling theory in NMR. Below is an exercise from Mcmurry's organic chemistry.

The textbook says that * stands for chiral center. But is it true? When I searched for it some said that there is no chiral center in cyclohexanol. Then, Aren't this hydrogens enantiotopic?

The substitution of one of the two hydrogens leads to the cis or the trans isomer of the disubstituted cyclohexane. Thus, they are certainly diastereotopic.
Perhaps, what confused you, as well me in a previous answer focused on 1,3-cyclohexandiol, is that substitution with an OH group leads to two diastereoisomers one of which consists of a couple of enantiomers (the other is the meso form).
A prochiral H on a stereogenic center does not imply its enantiotopic character. The obtained diastereoisomers can be optically active in fact, but not enantiomers.

Cyclohexanol does not have a chiral carbon but hydrogen atoms are diasteotopic because the substitution for one of the two hydrogen atoms (in alternate fashion) would give diastereomers, as the $\ce{C-1}$ carbon of cyclohexanol will now become either chiral (for C-2 and C-3 H-substitution) or will have specific cis or trans geometry(for C-4 H-substitution).
The enantiotopic hydrogen word is used only when one of the hydrogen replacement generates an enantiomer with only one chiral carbon (at the site of hydrogen replacement).


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Applied potential to Saturated silver electrode
I recently did my first cyclic voltammetry. I used a saturated silver reference electrode and two platinum working and counter electrode. This was done with an IKA electrasyn 2.0 device, my parameters I can choose are my maximum applied voltage and sweep rate. When I get my data back I can only find the current vs the applied voltage (so not vs the Ag/AgCl reference electrode even though I used it in the experiment). How do i convert this data, maybe I was going to add ferrocene so I have a reference but maybe that won't be necessary.

I am afraid you cannot. Are you sure you have configured the experiment properly ? As both potentials of Pt electrodes toward Ag/AgCl electrode are "floating".

Maybe have a quick look here: https://chemistry.stackexchange.com/a/128616/79678. Your reference electrode is already taken into account on the abscissa of the plot. That said, it is always good to test known electroactive standards, e.g., ferrocene.


text:
Can osmium react with oxygen at room temperature?
I’m an avid collector of elements (I love shiny rocks - don’t judge).
I’m looking to buy some osmium, as it’s one of the heaviest and hardest metals out there.
However, people online told me to avoid buying it, because it could react with oxygen naturally and become its acute toxic oxide.
However, from doing my own research, it seems it only oxidizes at 200 °C, which I doubt my metal vials will ever reach.
Can this be confirmed? Would it be safe to just store 99.99% pure osmium out in the open?

See also http://episodictable.com/osmium/    //  ... it’s [OsO4] also extremely poisonous. The molecule easily binds with oils in our body, irritating the skin, nose, and throat. It’ll also burn the eyes, but it won’t stop there: it also permanently stains the corneas, rendering a person blind. Even at concentrations too low to smell — roughly two parts per billion — it can cause a fatal buildup of fluid in the lungs or damage the kidneys.... you’ll need to store it carefully — inside a sealed glass container, preferably with secondary containment on top of that ... in an inert environment..

"you’ll need to store it carefully — inside a sealed glass container, preferably with secondary containment on top of that ... in an inert environment..". Or put it in an alloy, which is usually done in industrial applications. The base metal of the alloy effectively barricades the oxidation by forming nonvolatile oxides instead.

The name of the element comes from the Greek word osme, which means smell. The smell is osmium tetroxide. That might be a clue as to its reactivity. BTW solid osmium cubes are available (https://luciteria.com/metal-cubes/osmium-cube) but are very far from being cheap.

Osmium powder is clearly dangerous but it is less clear whether bulk osmium is a major hazard (but probably best to only store in a closed container.)

@matt_black That was where I was going to buy them, but based on the comments and answers so far, itd be a toxic metal to have, so I will not.

@MisterSirCode I think the cheap 1g piece comes in a sealed box (I'm tempted too).

@matt_black I have family members and many small animals. Id rather not endanger them. Ill just stick to chromium, tungsten and gold.
Honestly it was very tempting, extremely heavy blue-hued metal. It just looks like something alien. However, I already have heavy metals and blue metal (silicon) so Im good.

If you change your mind about osmium: https://www.elementsales.com/pl_element.htm#os

OsO4: https://chemistry.stackexchange.com/a/94692/17368

From Encylopedia Britannica:

Of the platinum metals, osmium is the most rapidly attacked by air. The powdered metal, even at room temperature, exudes the characteristic odour of the poisonous, volatile tetroxide, OsO4.

Osmium tetroxide is remarkably toxic and is not good stuff to have out in the open. Do not do this!

thanks for the information. Thats too bad. I guess tungsten and gold will have to do then, since I cant have osmium

Why OsO4 is toxic? https://chemistry.stackexchange.com/questions/151649/why-is-osmiumviii-oxide-so-toxic/152022#152022 (@MisterSirCode)

Sometimes I wonder how many crisis-aversions can be directly attributed to Stack Exchange users.

What Britannica misses but Wikipedia includes is the point that the reaction rate of oxidation matters so what is true of a powder with high surface area is not that relevant to the bulk compound. To quote from the Wikipedia entry "the rate [of oxidation] depends on temperature and the surface area of the metal.  As a result, bulk material is not considered hazardous."

As the others already stated, handling pure Osmium is too dangerous at home, but there are shops that offer small samples of elements sealed in acrylic glass, which is supposed to be safe. Maybe this would satisfy your collectory needs ;-)
They may look like these cubes:

Taken from:
https://www.smart-elements.com/shop/acrylic-element-cube-osmium-os-50mm-3/
I'm not promoting this specific shop, this was just the first google hit.

Real life periodic table has the elements store in glass cubes: https://www.youtube.com/watch?v=G9rH20lOE1c

According to Wikipedia:

Finely divided metallic osmium is pyrophoric[1] and reacts with oxygen at room temperature, forming volatile osmium tetroxide. Some osmium compounds are also converted to the tetroxide if oxygen is present.[1]

Osmium does not really differ from most metals by reacting with air at room temperature. But in most cases such as iron the oxide remains solid on the metal surface and -- if we avoid conditions that may break down the oxide chemically such as moisture, or trigger electrochemical reactions such as residual salts -- the solid oxide film will provide protection against extensive reaction. In contrast, the volatile oxide formed by osmium when it reacts with air vaporizes off and provides no such protection, and thus formation of the toxic oxide fumes continues. Most osmium is used in the form of alloys whose base metals form solid oxides.
You may just want to extract the writing point from a fountain pen and call it a day:

Due to volatility and toxicity of its oxide, osmium is rarely used in its pure form for high-wear applications. Its alloys with platinum and iridium are used in fountain pen tips, electrical contacts and instrument pivots.

Cited reference

Smith, Ivan C.; Carson, Bonnie L.; Ferguson, Thomas L. (1974). "Osmium: An Appraisal of Environmental Exposure". Environmental Health Perspectives 8: 201–213. https://doi.org/10.2307/3428200, JSTOR 3428200. PMC 1474945. PMID 4470919.

thanks, I was sincerely hoping itd be a little safer to have, but it appears not. I can use tungsten and gold instead

But the wikipedia entry also says "the rate [of oxidation] depends on temperature and the surface area of the metal. As a result, bulk material is not considered hazardous"

It can! Please think twice before ordering osmium. The oxide is volatile and extremely toxic: one good huff is enough to give you severe pulmonary edema, and make you drown in your own body fluids several hours after the exposure. Contact with eyes will make you permanently blind by turning your corneas black and opaque.
However, the most danger is in handling osmium powder. The real danger is in handling osmium powder. Meanwhile, the question explicitly suggests that the bulk form of osmium is being considered: OP says "I love shiny rocks", not "I love dull gray powders". Also, all the sellers of elements for collection that I know of tend to sell osmium in the bulk form: cubes, pellets and nuggets, not in powdered form. I wish more people were actually reading the questions, instead of just quickly glancing over the title and copy-pasting two sentences from the first Google hit.
Formation of $\ce{OsO4}$ from osmium metal in room temperature is thermodynamically unfavorable. Handling bulk osmium is not nearly as dangerous. Be careful, though! While the majority of metals is ductile and malleable, some exceptions -- including osmium -- are brittle. Instead of deforming under mechanical stress, they shatter into small pieces, greatly increasing the surface area that could react with oxygen in the air!
Here is a video of a man handling bulk osmium nugget. Notice that he is wearing gloves. He has since been recently active, and is alive and well.
Wikipedia says:

$\ce{OsO4}$ is formed slowly when osmium powder reacts with $\ce{O2}$ at ambient temperature. Reaction of bulk solid requires heating to 400 °C.

It is a bit of a shame that the accepted answer does not really make any effort to actually answer your question and does not even differentiate between powdered and bulk osmium. Instead, it just copies two short sentences from an external source.

Hmmm - while I agree that the accepted answer could be better for the reasons you give, the reference used, Encyclopedia Britannica,  is not one I would characterise as "of questionable quality and reputation"

In the English speaking world it has arguably been the premier publication of its type for over two centuries https://en.wikipedia.org/wiki/Encyclop%C3%A6dia_Britannica#Critical_and_popular_assessments

I don't think that starting an answer with "It can!" is a good way to introduce the horrors that come after. || "The claim of osmium being able to react with air at room temperature is about the osmium powder." That is incorrect. You might want to read the quote again. || I also wouldn't want to bank on the comparison with aluminium or iron. These metals form (in absence of moisture) quite strong protective surface oxides. It is one key difference to osmium. || So while this answer states facts, I find it particularly less helpful than the answers already given.

From the link of the 1st comment under Q:  In the 1980s, NASA wanted to conduct spectroscopy in the extreme ultraviolet range using instruments on board the space shuttle. Element 76 is one of the few materials that’s capable of reflecting light in this highly energetic range — more than twice as good as gold or platinum. So the folks in charge of the mission sent up mirrors with a coating of pure osmium… and when they came back down, the osmium was gone.

@IanBush - that wiki link is impressive; I didn't realize Britannica's been around 250+ years... although I'd argue that Wikipedia has itself far surpassed all other encyclopedias.  Britannica covers 288,274 topics, while Wikipedia has 54,739,246 pages (6,417,116 in English)...and that's in only 20 years. (Can you imagine what Wikipedia will be like, a couple centuries from now!)

it wouldn't be stored anywhere necessarily safe, so I still think it wouldnt be worth it to test my luck and possibly poison myself and the people around me. At most I might get an acrylic sealed one, or resin sealed one, as those are a lot safer.

@ashleedawg those are not good metrics here. "surpasses" in size does not imply better accuracy. Britanica is incredibly well curated and stable, whereas Wikipedia is alive, constantly changing, and is and will always be chock full of errors and omissions no matter how big it gets. They are two totally different kinds of beasts.


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Bond-order parameter $g_6(r)$
I would like to know the explicit steps to compute the bond correlation function
$$g_6(r) = \langle\Psi_6^*(0) \cdot \Psi_6(r)\rangle$$
for colloids in my experiment (2D). I have the position of all my particles, I know how to compute the individual hexagonal order parameter for particle $k$
$$ \Psi_{6,k} = \sum_{j=1}^N\frac{\exp[6i\theta_{kj}]}{n_j}$$
where $n_j$ is the number of neighbors of the particle $k$ (usually computed via Voronoi Tessellation that define the neighbor without cutoff) and $\theta_{kj}$ is the angle between $\vec{r}=\vec{r}_j-\vec{r}_k$ (the bond) and an arbitrary axis (x-axis for example)
But I cannot see how $\Psi_{6,k}$ depends on the position to apply the formula for $g_6(r)$. How can I compute $\Psi_{6k}(r)$? Is it some kind of average or density function?

Please define $\theta_{kj}$ and $n_j$, and what is the subscript $k$ as well?

https://physics.stackexchange.com/questions/461153/bond-order-correlation-function may be of use

Rushing out but looking at the link of the physics forum it seems (in the Questions language) k is the index of a particle, and thus you get position dependence via that index

@IanBush I have updated the question. If $\psi_{6k}(r)$  is the order parameter for the particle $k$, I know the value of $\psi_{6k}(r_k)$ but not the value for all $r$. Then, how can I compute $\psi_{6k}(r)$?

See this which might help (or references therein). Looks to me like they compute the value by summing over angles ($\theta$).

If I have understand properly from all the references that I have read, let said that I choose particle $ i $ as reference, then $\vec{r}i = \vec{0}$ and $| \vec{r}j | = r$. I have to compute the probability distribution function for $ \Psi_i(r) $. Then in 2D the value $ \Psi_j $ at distance $ r $ on a shell of radius $ r $ and $ r + dr $ should be $ d \Psi (r) = 2 \pi , r , \rho\psi (r) , dr $, and the correlation have to be compute as $g{6} (r) = <\rho_\psi(0) \cdot \rho_\psi(r)>$, where <> means ensemble average ... Isn 't it? It's a bit confused to me ..

@user239504 Yes, that looks correct.

@IanBush I have try to compute it in this way, but I cannot find the same results that another old program that I want to replace with this new program that I'm writing (that's why I want to understand every step). Could you please explain more explicitly the process that I have to follow? I have try to do something similar to this link for $g(r)$ but with $g_6(r)$, but I'm not able to do it properly [http://www.physics.emory.edu/faculty/weeks//idl/gofr2.html]

OK, that's not totally informative is it. Few things to do today but will try to find time to write an answer later.

I would say it would be more fitting to have your code as another answer rather than as an edit to the question.

I have removed the code from the question. Please add it as an answer instead.

Assume we are given a configuration of atoms $\{{\bf r_i}\}$ and any other data required to characterise the atomic layout, e.g. any lattice vectors if we have periodic boundary conditions in any of the 3 possible directions. With each of the atoms let us associate a property that can be characterised by a scalar quantity  $\{p_i\}$. For example $p_i=1 \space \forall i$ is one possibility, this will lead to something closely related to the radial distribution function. Another possibility is $p_i=q_i$ where $q_i$ is the charge on atom $i$. Another possibility is $\Psi_{6,k}$ as defined in the question. It doesn't really matter what, as long as you can calculate a scalar quantity associated with each of the atoms of interest the recipe below will work [1]
So the first step in the process is for the given configuration calculate $\{p_i\}$ for the property of interest. Now we want to calculate the average of $p_i^\dagger . p_j$ where atom $i$ and atom $j$ are separated by a given distance $d$. In practice on a computer we divide the distance range into a number of different bins, each of which cover the range $d$ to $d + \delta d$. So the method will be something like the following - but note I have had to make assumptions about how you wish to normalise the accumulated values; you may have to modify that according to what the relevant literature says. Anyway the essence of the method

Set to zero two sets of bins, one to accumulate the correlation terms to, one to count how many atom pairs are in each distance range
Loop over all the first particles $i$
Loop over all the second particles $j$[2]
Calculate $v=p_i^\dagger . p_j$
Calculate the distance between atom $i$ and atom $j$
Work out which bin that distance corresponds to
Add $v$ to the appropriate correlation accumulator bin
Increment the appropriate atom pair counter bin
Finish the loop over $j$
Finish the loop over $i$
The average correlation at each distance is the value of the correlation accumulator bin divided by the atom pair counter bin corresponding to the distance of interest

This generates the contribution for a single configuration - the final step is if appropriate to average over a large number of configurations.

[1] For the more mathematically minded I would normally convolute $\{p_i\}$ with delta functions at the atoms positions to represent the distribution of the property of interest. This complication is not needed here, but makes the argument more easily generalised to non-point particles
[2] Looping over all atoms in the $j$ loop is not the most efficient way to do this, and will be very slow for large systems. If you care about time to solution you should set up a neighbour list for each atom before the loop, and then in the $j$ loop only go over those atoms within the maximum distance of interest from the reference atom $i$. this reduces the scaling of the algorithm from $O(N^2)$ to $O(N)$ where $N$ is the number of atoms in the system

I see! Thank you very much for the explanation, now it is more clear for me. Respect to the normalization I think that I just need to divide for the total number of particles to find the correct function. I'm gonna add at the end of my question the code (write in python) that I'm doing for people that maybe need it in the future!

I almost wrote some code, but as a Fortran or C guy I didn't want to frighten people ...

It looks like $p_i$ and $p_j$ become vectors, rather than scalars in step 4.

A dot can be used to mean multiply, it's not just for dot products. I'll remove if it causes confusion

It's fine for me, a presume that $v=...$ is just the expression for the value in brackets, it could be for scalars $v=\Psi(\vec{r}_i) , \Psi(\vec{r}_j)$, vectors  $v=\vec{a}(t_1) \cdot \vec{a}(t_2)$, etc

@IanBush - Fortran-77 and C guy here, don't hold back!

Fortran 2003 nowadays - nothing earlier is worth using. Too late in the UK now, maybe tomorrow


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How would I find what specific wavelength I need to measure for a specific chemical in spectrophotometry (calcium perchlorate)?
I am currently doing a research project, and for it, I would like to use a photospectrometer to find the concentration of Calcium Perchlorate in my sample solutions. My current method of data collection relies on using an indicator (methalyne blue) to form a purple precipitate to observe under a microscope as a qualitative means to collect data, but if I can use a spectrophotometer (96 well) to directly measure absorbance and in turn concentration, that wold be heavily preferred.
How would I be able to find out what specific wavelength I need to measure for my Calcium Perchlorate? I can't find any preexisting values online for Calcium Perchlorate, so any help would be greatly appreciated. Thank you!

Apparently, calcium perchlorate is a yellow-white crystalline solid, so just prepare a solution at some reasonable concentration and run a UV-Visible absorbance scan from 200 nm to 900 nm. If you get absorbances that are very small, use a higher concentration.

@EdV, seems like a good answer, why not make it one?

@Snijderfrey Nothing particularly insightful, frankly: if the OP has the instrument available, it is an easy thing to try. I did a little searching and evidently the IR might be better for this one. The yellow tint, though, might indicate a usable absorbance either side of yellow. Apparently, this chemical might exist on Mars, so that would be cool. If you want to write up an answer, go for it!

@edV funny you say that! i'm actually doing an experiment on phytoremediation of calcium perchlorate concentrations seen on potential human Mars landing zones :).

@EdV sorry I didn't realize that's return comments on mobile, but i'm definitely gonna give it a shot in the upcoming days, thank you so much for the insight, it really helps!


text:
What are the safest non-toxic elements I can own?
As in a recent question, I mentioned I collect elements. My collection already has quite a few elements and a rare alloy
(I have silicon, molybdenum, tungsten, gold, platinum, silver, copper, aluminum, chromium, cobalt, and magnesium, and a rare gold alloy (1 part gold, 2 parts aluminum) that has a bright purple hue)
Im trying to find more elements to add to my collection but I wanted to know which would be the safest to own?
Probably my only slightly dangerous element is magnesium, as it can burn, and quite hot at that.
Im talking in terms of elements that can exist at room temperature, or near it, and not oxidize a lot, or combust / release toxic oxides / radioactive particles / act as a poison (such as mercury or lead)
As for ones I already know / think are safe enough, and that I want to buy, Im currently looking at titanium, iridium, niobium, and scandium at the moment

As a chemist I am used to all kind of dangerous stuff that is kept in sealed glass vials, and have no problem with this. Nothing happens if you can make sure the glass remains in one piece. A block of lead is not particularly dangerous. Magnesium burns once ignited,  but so does wood... Would you like to touch the stuff? What do you want to do with it?

I used to have an element collection, but after retirement I donated some elements to my former department and the environmental health and safety folks there and I also sent back, gratis, some elements I purchased from Metallium. You might look at the linked website to see how the more dangerous non-radioactive elements, e.g., bromine, are made safely available to collectors. And, of course, what @Snijderfrey said is good. Iridium, by the way, is insanely expensive now.

Have you thought of pure carbon $\ce{C}$(graphite), iron $\ce{Fe}$, zinc $\ce{Zn}$, tin $\ce{Sn}$, sulfur $\ce{S}$  ? They are all easy to obtain, and not specially toxic. The gases like oxygen $\ce{O2}$, nitrogen $\ce{N2}$, hydrogen $\ce{H2}$ or noble gases, can be stored in closed vessels

Nickel and cobalt should be easy. Nirod- filler metal for cast iron welding , is 95 % Ni. Unalloyed cobalt not so easy . Vitallium used for dentures is Co + Cr , tell people it is Co, they won't know the difference.

Rare earth magnets have some odd elements , but you would need to find the compositions., and unlikely any are unalloyed.

Take a tour of a steel mill; Yon should be able "find" ferro vanadium, manganese, nickel, chromium Nb ( or Cb) and of course many you already have.

Iodine and bromine are likely available ( in glass) from supply houses but today they may want permits or something ( unless you can find an old lab that has supplies) .

This is a good scavenger hunt. How about beryllium window from an old X-ray tube?  Lithium batteries but not pure.

@blacksmith37 Don’t forget David Hahn, the radioactive boy scout!

Im not wanting bulk of these elements really, just like having shiny things. I already have gold and silver bars (for investments) but I do like collecting smaller amounts of other elements.

Red Phosphorus and Gold not mentioned yet,

@AJKOER Red phosphorus looks neat, might get some. I already have gold though

What is safe very much depends on your experience with handling of dangerous substances, your equipment, storage conditions etc. It is hard to give a general answer, but from your question it appears that you have little experience with chemicals and assessing how hazardous they are in certain conditions. So my advice would be: Stay on the safe side and do not take any risks you cannot assess. However, I think it would be better if you made the decision yourself, rather than based on a list or recommendation by anonymous people on some random internet forum.
One approach could be to visit a website of one of the major chemical suppliers. They usually sell quite a few elements, but also are obliged to supply the material safety data sheet (MSDS), at least in Europe. Have a look at them, they are informative. For example, an MSDS for silver I just had a look at states on the first page:

Not a hazardous substance or mixture according to Regulation (EC) No 1272/2008.

You can sefely conclude that silver will not be dangerous. For lead, the MSDS says:

Reproductive toxicity (Category 1A), H360FD
Effects on or via lactation, H362
Specific target organ toxicity - repeated exposure, Oral (Category 1), Central nervous
system, Blood, Immune system, Kidney, H372
For the full text of the H-Statements mentioned in this Section, see Section 16.

So yes, lead is dangerous in some ways, especially if it enters your body, but for one big solid lump that does not happen accidentally so easily, in contrast to powder for example which you could breathe in. You can look at the H and P (hazard and precautionary) statements in the MSDS if you want to get a better impresssion. But for a proper assessment, you would have to find some competent person you trust. I would say that a piece of lead that you only look at and do not manipulate in any other way is harmless, but would suggest that you make up your own mind.
One note for MSDSs: They read rather drastic usually, and I assume this is on purpose. So if you want to be on the safe side, they are a good indicator.

Just to let you know: Cobalt, one of the elements you already have, reads like this:

Flammable solids (Category 1), H228
Eye irritation (Category 2), H319
Respiratory sensitization (Category 1), H334
Skin sensitization (Category 1), H317
Germ cell mutagenicity (Category 2), H341
Carcinogenicity (Category 1B), H350
Reproductive toxicity (Category 1B), H360F
Long-term (chronic) aquatic hazard (Category 4), H413
For the full text of the H-Statements mentioned in this Section, see Section 16.

(+1) Good answer, though I hear silver bullets are bad for vampires! ;-)

@EdV, I would think vampires are left out of Regulation (EC) No 1272/2008 (but I would have to check, to be honest).

In my best old office, I had lots of stuff on the oak shelves, including 5 bottles containing C, Si, Ge, Sn and Pb. A student saw them and asked if I was allowed to have lead! Good I did not show him the pound of As or half pound of Tl.

I have used lead strips to weight aquarium plants for 70 years. I will let you know if there is ever a problem . Because plumbium was the material of choice for drinking water pipes for about 2000 years , I am not very worried.

Thanks, I appreciate all the info, and yeah, my storage conditions are "casual" at best. Im just a programmer / modeler who happens to enjoy collecting things, especially shiny metals and rare / expensive things.
Ill just stay on the safe side and not buy anything that could harm me or anyone else in any way.


text:
Reactivity of Alkenes with HBr
I am new in the Organic Chemistry (World of Reactions). I had got one question in my test. But, when I had seen it, I got no proper logic to tick the correct answer, and then when I saw solution, then there is only written, "More Substituted Alkene, is more Reactive." But, how they are more reactive ? I mean, that if they are substituted with ALKYL GROUPS, then they will get more stable (due to + Inductive Effect) and will not React. I think, I am wrong somewhere, and when I searched on the internet, there is No Proper Logic, for this question.
So, I just want to know the Correct Logic and Mechanism of this question. Please help.

What you should be looking at is the stability of the carbocation intermediate formed when the strong acid $\ce{HBr}$ protonates the alkene. This stability is more sensitive to alkyl substitution than the stability of the uncharged alkene, so the net effect is that alkyl substitution lowers the activation energy versus having no substitution. Ergo a faster reaction, e.g.
2,3-dimethyl-but-2-ene (two methyl groups at each double-bonded carbon, tertiary cation when protonated)
$>$ but-2-ene (one methyl group, secondary cation)
$>$ ethene (no substitution, primary cation).


text:
Removing bugs and tar from my car with Bounce fabric softener sheets
I just used Bounce fabric softener moistened with water to remove stuck tar and dried bugs from my car.  It worked like a charm, wiping it clean in seconds.  Why did this work?  The web says it contains dipalmethyl hydroxyethylammoinum methosulfate, among other chemicals.

Good experiment! You should write a letter to Bounce. They might include this information on their label and send you a lifetime supply.

Thanks for the answer.  Makes sense.  I hardly had to wipe at all.  I'd tried it before with vigorous scrubbing and just water to no avail.  Bounce took it all off in no time!  Who knew???

Water is the universal solvent. But it doesn't do a great job dissolving road tar (i.e., asphalt), or bugs. Wiping is an important part of removing dirt from a hard surface - in addition to obtaining a lower surface tension so the surfactant solution can actually wet the dirt, and the surface.
In many washing applications (like dishes or laundry), nonionic or anionic surfactants are used to emulsify oils and suspend them in the wash water or foam. The low surface tension of the solution is the major actor.
Somewhat similar cationics (quaternary ammonium salts) are commercially available in squirt bottles for cleaning hard surfaces. And cationics can be very effective because they often adsorb or adhere to surfaces better than nonionics or anionics. This could be a key to your success.
We don't know what kind you have on your car, but water-based acrylic polyurethane enamel paint is currently the most widely used auto paint (https://en.wikipedia.org/wiki/Automotive_paint).
These common paints would have polar amine and carbonyl groups; the cationic surfactants would better attach to the polar paint surface - not necessarily emulsifying the road tar and bugs - but preventing redeposition of tar and bug parts. So wiping would knock the dirt off, and the cationic surfactant would help keep it off.


text:
Mass spec mass calculation difference
This is not a homework question, but for research.
I am performing mass spec on serum metabolites using Thermo Q Exactive PLUS with a HESI source which was set to a spray voltage of -2.7kV under negative mode. During the metabolite identification step: I am curious why the mass was used (supplied by my collaborator) differs from the mass I calculated.
Sorry, I am new to chemistry or mass spec: but in the software, we can enter values up to 7 decimal places for picking the peak.
I would use glucose as an example, which has the formula C6H12O6.
What I used is the isotopic mass found on the internet:
C12: 12
H: 1.007825
O:15.99491
proton: 1.007276
I ran a negative mode (I believe the most abundance adduct is -H), I calculated the adduct should be 179.0561121, but the reference list supplied by my collaborator is 179.05603.
It would be nice if you have any idea of what mass can I use in order to get the same calculation. I have tried to search online for are there is any mass list that should be used for mass spec, but seems does not help (as I think the mass I use can calculate the mass before ionization, which should be accurate and correct). Thus, I would like to ask for your opinion here before challenging the collaborator.
Thanks!
below are an additional example (I am indeed doing an isotopic tracing experiment):
here is a pyruvate (C3H4O3) with 2 carbon labeled with 13C. My calculation is:
12 * 1 + 2 * 13.00335 + 1.007825 * 4 + 3 * 15.99491 - 1.007276 = 89.015478
However, the reference list supplied by my collaborator said it is 89.015396.

What type of mass spec is that? Do you have five correct decimal places?

This must be a million dollar mass spec if it is accurate to five decimal places?

Well, a hydrogen is a proton and an electron; a proton is just a proton, so the masses should differ by that of the electon. Using a value of $1.6735575\times10^{-27}\pu{kg}$ for hydrogen and $1.67262192369\times 10^{−27}\pu{kg}$ for proton, when I subtract the two I get a difference of $9.35576\times 10^{-31}\pu{kg}$, which is quite close to the (resting) mass of an electron, which is reported as $9.10938356 × 10^{-31}\pu{kg}$.

I think only Fourier transform ion cyclotron resonance mass spectrometers can reach this type of high resolution. I don't think someone would analyze glucose there. May be the problem is just a lab exercise. I hope the OP adds more details.

Thanks @M.Farooq. I have tried to add more details, I hope I can get the question reopened soon.
In short, we are using Thermo Q Exactive PLUS with a HESI source. I believe the adduct is [M-H]-. I have also deleted the hydrogen vs proton question (which is just a side question, and sorry for my ignorance). Ultimately, I just want to know what is the correct mass to use for calculating mass spec weight (or if there is some online calculator). ( indeed, we performed MS in two independent collaborators, and they seem used the same mass for identifying the molecule)

@WilliamWong, Have you checked this website? https://fiehnlab.ucdavis.edu/staff/kind/metabolomics/ms-adduct-calculator/

@M.Farooq yes, the mass for proton is derived from the excel given on that website. otherwise, the differences seem even larger (e.g. glucose, I got 179.0555631)

For [M-H] in negative ion mode, it is M - 1.007276. For monoisotopic glucose in hypothetical negative ion mode, the neutral isotopic mass is 180.0634, so in negative mode as per the link, 180.0634 - 1.007276 =179.0561(2). It matches very closely with your collaborator's number.


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Has anyone added dyes to a mixture about to separated by chromatography in order to visualize where the desired product lies?
Usually in preparative chromatography one needs to know when the desired substance to be separated is coming down the pike. Often this required some sophisticated instrumentation. But, it occurred to me that another way would be to two find dyes that (1) won't combine with anything in the mixture to be separated, (2) have Rf values that bracket the Rf value of the desired compound w/o including any others. Then one can simply visually inspect the chromo column (assuming it is glass)to tell where the product is and when it is about to come out. So, if this has been done then really my question is: is a table of dyes, and what kinds of functional groups they are safe with, and their Rf values.
In answer to comments below:
I know a TLC is relatively fast. I was thinking of preparative chromatography.
Even approximate Rf values of dyes, run with maybe 2-3 different, pure solvents, in dry silica, would be useful. One could then guess which dyes would be useful, run test TCLs to find dyes that are either just above or below the desired substance X on the TLC plate.
Ideally the dye spots would not overlap the X spot at all so no separation would be required after.
Then in the preparative step I don't need to auto-sample or use a UV detector: just glance over at the column (thinking of  low pressure flash chromo in a glass column) to see if the first dye is approaching the bottom - while doing other stuff. When the first dye just runs out, then I change the collection flask until the post-X dye starts showing up. Sounds easy! But I know there would be complications.

I think you could spend a lot of time trying to find suitable dyes. Running a tlc of your fractions really does not take very long.

Rf values are not tabulated. They vary strongly with the solvent (e.g. you Cyclo:EE mixture ratio, it's water content), and the specific column material, including the age/storage conditions of the material. Even your packing.

And you'd need a big cupboard of dyes, not all of which are very stable.

If you have easy separation problems, this might work. Often enough, separation problems are difficult, and you would collect many fractions and analyze them afterwards anyway, e.g. by thin layer chromatography. So you lose your desired advantage.

Assuming everything going smoothly and no interference occurs you can perhaps do that. But you need a dye which can be removed easily, say by decanting active coal or filtering through it. Alternatively you might find the perfect scenario in which the dye just signal the incoming target. It depends on the complexity of the crude and the easy of separation. Principles apart. I did upvote the answer by Farook.

Thanks for the comments. I have added to the question to address these.

the idea is good for academic purposes and good for a lecture hall demo. One may also try a UV lamp, if your compound fluoresces in the column. Caution UV is very harmful for the eyes (protective goggles)!
As far as I know, no this is not the practice because preparative chromatography in pharmaceutical industry is automated beyond imagination. The sample collection in vials etc. are actuated and triggered by UV-Vis or mass spectrometry detectors.
The basic purpose of chromatography is to keep the mixture as simple as possible rather than adding more compounds.

I the UV lamp idea as it essentially broadens the dyes I could use to fluorescent ones. Pyrex is opaque to shortwave UV, I think. But long wave would work.


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Why are anti bonding orbitals distorted towards the less electronegative element?
Bonding molecular orbitals are distorted towards the more electronegative atom and so the atomic orbitals from this atom contribute more to those bonding orbitals. This implies that anti bonding molecular orbitals are distorted in the same way towards the more electropositive atom. In the case of bonding molecular orbitals being distorted, the reason for the distortion is intuitively that electrons are attracted to the more electronegative atom and so are closer in energy, but it is not intuitive to me for the anti bonding orbitals why the more electropositive atom is the one which attracts the anti bonding electrons to a greater extent - they are still electrons after all so why should the more electropositive atom distort their electron density towards it?

It could be that electrons in antibonding orbitals are repelled from the higher electron density in the bonding orbitals that is higher in the vicinity of the more electronegative atom.  In effect antibonding electrons force the atoms apart weakening bonds so are not as strongly attracted to both atoms.

If that is not intuitive, why just above  you said "it implies"? Because of the reason explained by Hans Wurst? If so, we are at the starting point....

The relevant atomic orbitals of an electronegative element are typically lower in energy than the relevant atomic orbitals on the element with lower electronegativity.
This means that the bonding molecular orbital, which is the molecular orbital with low energy, should include a higher coefficient of the atomic orbital that is lower in energy, i.e. the atomic orbital of the electronegative element. The coefficient of an atomic orbital over all molecular orbitals has to sum to 1. Which means that you can't mix as much into the antibonding molecular orbital and you have to use more of the atomic orbital of the more electro-positive element atomic orbital in the antibonding molecular orbital construction. This means that the bonding molecular orbital is "dominated" by the atomic orbital of the electronegative element, while the antibonding molecular orbital is "dominated" by the atomic orbital of the less electronegative element. Leading to the observation that the bonding molecular orbitals is distorted towards the more electronegative element, while the antibonding molecular orbital is distorted towards the other element.


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Do ferrofluid nanoparticles come out of solution when exposed to a solenoid DC field current?
As the electrically-charged ionic tails of surfactants used to attach to the magnetic nanoparticles require the particle to be magnetized to stick to them; if a ferrofluid is passed has a solenoid DC current field, which causes demagnetization, will this cause the surfactants to leave the particles and in effect, make the particles insoluble again?

I struggle to understand what you are asking here. Surfactants are not attracted to magnetic particles by magnetic effects. Static magnetic fields don't cause demagnetisation. And simple observation will demonstrate that ferrofluids don't fall apart under static magnetic fields (the spectacular 3D patterns the fluids produce in strong fields would not be possible if the solutions fell apart).

Ferrofluids are not true solutions, but are stable colloidal suspensions. Brownian motion is sufficient to keep the tiny particles, on the order of 10 nm, from settling out. Other examples of colloids are milk (fatty globules in liquid) and colloidal gold in water.
Though an alternating magnetic field in an AC-powered solenoid would only stir up the particles and keep them  suspended, a strong-enough steady magnetic field would separate the ferrofluid particles from the liquid, as would a centrifuge, just as whole milk is separated into cream and skimmed milk.
It is also possible to disrupt colloids by changing pH+. For example, cheeses are made by adding citric, acetic or lactic acids (or letting friendly microorganisms do that for you).
Bon appétit, but don't eat the ferrofluid.

Ferrofluids would be pretty useless if static fields caused the fluids to break down and separate. None of their practical applications would even be possible if this were true.

@matt_black, read: a strong-enough field. Given sufficient strength, the iron particles adhere strongly enough to squeeze out the suspending fluid. It might take something from the Francis Bitter Magnet Lab (q.v., http://web.mit.edu/fbml/) or a magnetar with field 10e11 T, but at some point the ferrofluid would precipitate.

I'm happy to clarify my claim to "under static magnetic field strengths routinely achievable on earth" ferrofluids won't separate. The record is currently ~50T which is 10 times the typical big fields in NMR or MRI machines. I don't think we know much about what happens to any material when fields go much above 10T.


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Relationship between the principal axes of inertia of an enantiomeric pair
If an object is chiral, its mirror image and itself are non-superimposable and represent two distinct versions of the same object. A hand, for instance, can be either the right hand or the left hand. Or a helix, which can be winding clockwise or counter-clockwise.
With molecules is the same. A molecule for which its mirror image is non-superimposable with itself is chiral, whether or not it contains any carbon. Whenever that is the case, the molecule and its mirror image form an enantiomeric pair and they are each other's enantiomer.
I have been playing with the principal axes of inertia as tools in attempts at superimposing two different molecules when I asked myself what would be the relationship between the axes of inertia of each molecule in an enantiomeric pair. I was thinking that the length of each vector might be the same since the relative distances of atoms and their angles are the same, but maybe the axes themselves are distinct since the spatial disposition changed, but nevertheless they have a certain relationship between one another?
So my question is: Is there a known relationship between the principal axes of inertia of molecules in an enantiomeric pair?

Chiral pairs are mirror symmetric, so are all their (vectorial) properties. Now what is your question?

@Karl What do you mean they are mirror symmetric? I thought the entire point of being chiral was that the object does not have any $S_n$ axis of symmetry (which would include inversion centers as they are equivalent to $S_1$ and regular reflections, which are the same as $S_2$).

Mirror symmetric to each other, of course.

Your reasoning concerning the same bond lengths and angles giving the same moments of inertia and therefore the same principal axes is correct. Chiral enantiomers have indistinguishable rotational constants and moments of inertia. However, the projection of the dipole moment on these axes is different. This feature was exploited in this paper to distinguish between the different enantiomers using three-wave mixing with microwaves.

Wow, I had heard of this paper as an incredibly complex form of microwave spectroscopy to distinguish enantiomers. An easier and recent way is chiral tagging in microwave spectroscopy. A small chiral alcohol is injected along with the analyte. It forms a diastereoisomer and diasterisomers have a different center of mass hence distinguishable spectra.

It's probably one of the coolest applications of MW spectroscopy, but indeed not for the faint of heart  The method you refer to is indeed more practical to distinguish enantiomers (although I believe that the author's are/were working on a commercialization of the technique), I added the link mostly as an illustration of the effect asked for by the OP.

I agree this paper is not for the faint of heart, which includes myself.

The paper is brilliant. However, is there a simple computation of a system of vectors (akin to the moments of inertia) that I could perhaps make that would behave like the dipole moments?

The principal axes don't provide information about the positive or negative direction, but I suppose you could define a random vector in your molecule, mirror it and look at it's projections on the principal axes in both enantiomers (similar to Fig. 1 of the paper). This quantity does not have any physical meaning, but could perhaps help identify the different enantiomers.


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Retrosynthesis of {6-[2-(benzylamino)ethyl]-1,4-dihydropyrimidin-4-yl}methanol
I have {6-[2-(benzylamino)ethyl]-1,4-dihydropyrimidin-4-yl}methanol (at the top) and have to make the retrosynthesis steps for my class.
I tried different things but I'm not sure about it. In 1 you see I broke off a C-C bond (I know that is difficult to break, do you think it is possible?) Otherwise in 2 is the most obvious way to break but then I would have to make double bonds in the aromatic ring which is also quite difficult.
So you see my problem, do you have any ideas?

I mean, those double bonds are still gonna be there even if you manage to find a C–C bond formation reaction in your first choice.

@orthocresol yeah but then breaking a C-C bond is not favorable in physiological conditions right?

Why do physiological conditions matter? Is the synthesis supposed to be done in a human body? I would have assumed you were doing this in the lab.

Also, breaking a C–C bond isn't an issue here at all; "breaking" a bond in the retrosynthesis corresponds to the formation of a C–C bond in the forward synthesis. Making C–C bonds is not easy, but not impossible either. My original point, though, was that even if you make that disconnection, you can't avoid the issue of how to make that ring. You might want to think more about that part of it, because it seems like the key to your entire problem.

Reductive amination is the best way to install the benzyl group. Consider using an alkyl nitrile as the source of the amine sidechain.

Seems to me that you will struggle to keep the dihydropyrimidine at that oxidation level. It wants to be aromatic


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Why is it difficult to calculate ΔS of the surroundings (said a teacher)?
My chemistry teacher told me that the Gibbs Free Energy (G) function was created because determination of ΔSsurroundings, and thus ΔStotal was difficult, and we needed something better to explain spontaneity of a reaction.
However, I didn't exactly understand why it is so.
ΔSsurroundings = $\frac{ΔH(surr)}{T}$
And we were told ΔHsurroundings = -ΔHsystem
And ΔHsystem can be obtained experimentally.
So can't I simply say ΔSsurroundings = $\frac{-ΔH(sys)}{T}$, maybe with this T being a different, yet deducible temperature ?
What exactly am I missing? And what is the true need for the G function anyway?

Good thinking. Now, using your reasoning, take $\Delta G = \Delta H - T\times\Delta S$, divide it by T, and what do you get?

Your teacher is right.  The surroundings are assumed to be vast—they need to be to act as a quasi-infinite heat bath, such that any heat flow into or out of them can be treated as reversible, because it changes the surroundings only infinitesimally.  Thus how would you directly measure $\Delta S_{surr}$?  Generally, you can't.  So you instead determine $\Delta S_{surr}$ indirectly from $\Delta H_{sys}$, which acts as a surrogate for $–T\Delta S_{surr}$.  And $\Delta G_{sys}$ is useful because it equals  $–T\Delta S_{surr}–T\Delta S_{sys} = –T\Delta S_{univ}$ => the condition for spontaneity.

Basically you've answered yourself

@Snijderfrey This may be a silly question, but what if the temperatures of the system and surroundings aren't equal? Then how can I get ΔS(surr) by dividing ΔH by T of the system?

This is a comment on your last sentence,  '... and what is the true need for the G function anyway ?'
It arose out  of experimental necessity. In the dim, distant past, it was naturally accepted that it was possible to convert all the heat of a reaction into work, so that $=-\Delta H$ was all that was needed to determine this. Many reactions were studied and the $-\Delta H$ for unmeasured reactions calculated, however, when these reactions were eventually measured serious discrepancies became clear, i.e. much larger than experimental error.
According to the first law the external work done must be equal to the heat loss of the system, unless some heat is given to or taken from the surroundings.  This was the point first clearly seen by Gibbs.
When an isothermal reaction occurs reversibly $T\Delta S$ is the heat absorbed from the surroundings, and if positive more work will be done than that given by the heat of reaction, and vice versa.
So the Gibbs function was not created as your teacher suggests but as a result of experimental data forcing us to consider how entropy affects reactions.


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What do we define as "surroundings" in thermodynamics?
I was thinking about the 2nd law of thermodynamics, and how ΔStotal=ΔSsystem+ΔSsurrounding.
You could potentially have a reaction with negative ΔStotal occurring- a non-spontaneous reaction.
But by the 2nd law, ΔSuniverse , which a teacher defined as ΔStotal or ΔSsystem+ΔSsurrounding , should always increase.
An example of a non-spontaneous reaction would be electrolysis of water- I read that it is so as water would remain water forever and not decompose on its own. Would this decomposition not be a non-spontaneous process, thus violating the 2nd law?

Well, you are right: non-spontaneous reactions don't occur, as they violate the Second Law. As for the apparent contradiction, this may lend some insight: what kind of non-spontaneous reactions are you thinking of that occur in the real world?

It is possible to carry out a reaction of ideal gases reversibly, at constant temperature and pressure, and, in such a case, the sum of the entropy changes of the system and surroundings is zero.  Google Van't Hoff equilibrium box.

The second sentence makes an incorrect assumption, therefore the premise of your line of argumentation is flawed.

@orthocresol Ah, I read somewhere that certain reactions were coupled with others as they are non-spontaneous. Would a reaction like this count as a non-spontaneous reaction?
Sorry for this maybe basic question, I've just started learning thermodynamics.

I think the duplicate question will help you.


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Is this plastic clip made of polypropylene or polyethylene?
I am trying to re-attach an automotive clip (painted plastic with a yellow (epoxy?) hard adhesive backing) which secures a trim piece back onto the vehicle's roof sheet metal (unpainted (primed?) area).

I am thinking of using Gorilla Glue, a polyurethane adhesive, which says it bonds many dissimilar surfaces, but specifically says to avoid PP and PE:

Gorilla Glue will work well on many types of plastic; however, we do not recommend for use on polypropylene (PP) or polyethylene (PE) plastics or any type of rubber with high oil or plasticizer content.

Any clues as to whether the clip and backing are PP or PE?
General thoughts on whether Gorilla Glue is suitable, or if another type of adhesive would be better?

I realize this isn't the typical question on the Chemistry Stack Exchange Community, but thought the technical chemistry aspect is more suited here than the Motor Vehicle Maintenance & Repair Community.

One often can identify PE or PP by the "waxy" feel and appearance. See https://resin-expert.com/en/guide/best-glue-for-polypropylene for gluing PP. However, you might choose an adhesive that handles a variety of materials. Be aware a car roof can get quite hot (so it's doubtful lower-melting PE is used), and the cement would have to handle a wide temperature range.

@DrMoishePippik The clip itself feels very glassy or metallic, maybe partly due to the metallic paint. The yellow backing does have some "waxy" feel, and fingernails can mark it fairly easily. Thanks for the link and PE/PP identification tips. I've applied Gorilla Glue and will test it's strength (doesn't have to be super strong for this application) after letting it cure 1-2 days. If it doesn't hold, I will try JB Weld next since the article suggests epoxy is best.

As far as question is concerned, edit should be much more thorough for reopening - with details besides photo and improved clarity. Still, I think the answer probably is no, and this glue would probably work. Why? If it's painted then glue should stick too, but for solid answer, solid data are needed.

@adatum For the record, none of the closers are moderators. It appears the question will be reopened. Personally, I have used Gorilla Epoxy for everything from repairing a cracked rear quarter panel to filling in cracks on bicycle shifter enclosures. A light sanding with high grit sandpaper before application wouldn't hurt. If this doesn't work, JB Weld will - that's the go-to for many bike techs (I was one).

@ToddMinehardt Thanks for the feedback, epoxy seems most recommended. Still not sure if the clip's backing is PE or PP, but the polyurethane Gorilla Glue is holding solidly.  I did sand the backing with 400 grit sandpaper, but not the vehicle roof. We'll see about the long term and temperature variations. One concern with rigid adhesives and dissimilar surfaces is how they will handle contraction/expansion in winter and summer. Note that Gorilla Glue has better moisture resistance and wider service temperature than Gorilla Epoxy according to the manufacturer's specs.


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What is meant by "charge separation" for resonance structures?
I'm confused about the definition of "charge separation" in the context of resonance structures. One of the rules of stability of resonance structure is "the structure with least separation of formal charge is more stable". Does "separation" in this context mean the distance between two charges or does it refer to the difference in the magnitude of two charges? Or it does it mean something entirely different?
And I have searched this site and others for an explanation but there were no clear explanations on what it was or of what it implies in a resonance structure.

or this one https://chemistry.stackexchange.com/questions/87491/charge-separation-and-stability-in-resonating-structures/140022

I had already seen the first link before I posted my question which did not really help me understand but after visiting the second link i feel like now I have understood to some extent the meaning of charge separation. Thanks a lot!

The good old googling site:chemistry.stackexchange.com charge separation resonance

Simple classic electrostatic should help. I would say that if there must be formal charge separation, then the less unstable limiting form is that whit minimal distance between them, ie minimal "formal electrical dipole". So you have to look for what minimise separation, both for a formal physical distance as well as the overall amount of separated formal charges.


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Reversible and Irreversible processes
In an ideal world, is moving a piston quickly, a reversible process as there is no heat loss due to friction? (as opposed to a quasi static process)

Quickly excludes reversible, friction or not.

Why would moving a piston quickly reduce frictional heating?

Moving a piston rapidly results in an adiabatic process, not an irreversible process.  In a rapid compression or expansion, viscous dissipation of mechanical energy to internal energy is irreversible.

Hm, IMHO, fast piston movement may or may not lead to reversible process even without such dissipation. If I compress an ideal gas system adiabatically and expand isothermally  to the initial state, surrounding will not return to the initial state, as I have converted some work to heat.

Additionally, are not reversible processes always in equilibrium state, therefore idealistic regardless of dissipation?

@Poutnik There is negligible viscous dissipation in a reversible process.

Still, lack of dissipation is essential but not sufficient condition for process reversibility. Dissipationless cycle of adiabatic compression, cooling and isothermal expansion converts work to heat. But you cannot reverse it to convert heat to work. But I will be happy if I realize I am wrong.

In my understanding, reversible process, if a system is brought to the initial state, surrounding is brought to initial state as well ( in absence of own external processes ). Or vice versa, bringing surrounding to initial state brings the system to the initial state too.

In thermodynamics, a reversible process is a process whose direction can be reversed to return the system to its original state by inducing infinitesimal changes to some property of the system's surroundings.  Throughout the entire reversible process, the system is in thermodynamic equilibrium with its surroundings. Wikipedia reversible process


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Microscopically, how does temperature increases in simple exothermic reactions?
Let’s say we have a simple exothermic reaction $\ce{A_{(g)} + B_{(g)} -> C_{(g)}}$
Since it’s exothermic, that means that kinetic energy of C is greater than the kinetic energy of the reagents. However, momentum still has to be conserved : $p_C = p_A + p_B$. In the center of mass reference frame, we have $p_A = -p_B$ (before the reaction) and $p_C = 0$ (after the reaction). But we just said that $\frac{p_C^2}{2m_C} > \frac{p_A^2}{2m_A} + \frac{p_B^2}{m_B} > 0$ !
What is the way out of this apparent contradiction ? What actually happens in an exothermic reaction ?

No contradiction: think of two heavy metal blocks on the ground with some TNT between them. Detonate the TNT. BOOM! The two blocks fly apart in opposite directions: lots of kinetic energy and momenta in opposite directions.

And for the opposite, synthesis case,  the energy is transferred to other colliding molecules/atoms, or a photon is emitted.  If 2 H atoms meet to form H2 and do not encounter the 3rd party to pass energy of bonding they separate again.

Similarly, nuclear fusions usually end up emitting small particle ( alpha, p, n, e, neutrino, photon ), carrying extra energy and momentum.

I did flagged for closing. I want to add two simple remarks useful for future confused users. 1) No, the KE of individual molecules is not T. Concept from statistics do not apply to two colliding particles. 2) Chemical reactions aren't elastic collisions at all.


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Wave Dispersive X-ray Fluoresence and Bragg's Law
Thermo Fisher website explains that "WDXRF uses crystals to disperse the fluorescence spectrum into individual wavelengths of each element, providing high resolution and low background spectra for accurate determination of elemental concentrations." Additionally, it goes on to say

WDXRF systems are based on Bragg’s law, which states that crystals will reflect x-rays of specific wavelengths and incident angles when the wavelengths of the scattered x-rays interfere constructively. While the sample position is fixed, the angles of the crystal and detector can be changed in compliance with Bragg’s law so that a particular wavelength can be measured. Only x-rays that satisfy Bragg’s law are reflected.

However, I am still confused on how the crystal, in correspondence with Bragg's law, disperse the spectrum to show each residual wavelength. Because there is a large variation of wavelengths coming from the sample, wouldn't this mean that passing it through the crystal would cause some of them to destructively interfere? How does this help with analyzing the spectra?
Hopefully, someone can provide a more in depth explanation on how this technique is used to analyze fluorescence spectra.

Different 'colors' (energies) of x-rays will diffract at different angles from a crystal. That is all. Either have a position sensitive detector, or rotate the crystal to sample the different energies at one fixed detector.

Do you recall optical (visible) diffraction grating? Let us say, wavelegnths from 400-700 nm pass through a transmission grating. What do you see? A rainbow. This means that all wavelengths passed through the diffraction grating, but at different diffraction angles. A grating does not annhilate energy during destructive interference.

"X-ray monochromators are analogous to grating monochromators and spectrometers in the visible portion of the spectrum. If the lattice spacing for a crystal is accurately known, the observed angles of diffraction can be used to measure and identify unknown X-ray wavelengths.

Because of the sensitive wavelength dependence of Bragg reflection exhibited by materials such as silicon, a small portion of a continuous spectrum of radiation can be isolated. Bent single crystals used in X-ray spectroscopy are analogous to the curved line gratings used in optical spectroscopy. The bandwidth of the radiation after it has passed through a high-resolution monochromator can be as narrow as Δλ/λ = 10−4, and, by tilting a pair of crystals with respect to the incident radiation,

the wavelength of the diffracted radiation can be continuously tuned without changing the direction of the selected light. (From Encyclopedia Britannica).


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Is shampoo or soap a better emulsifier?
Made homemade shampoo using:

Sodium lauryl sulfate (SLS)

Glycerin

95% ethanol

Sodium stearate

Made homemade soap using:

20% sodium hydroxide

Vegetable oil

95% ethanol

20% sodium chloride

After doing an oil-in-water emulsion test, it showed that my shampoo was the better emulsifier, but I was expecting it to be the soap since a lot of resources have said it should be.  There's potential I could have mixed them up on accident, but I think it was okay... do my results make sense or do they seem incorrect?  Please explain.
Any help on this would be greatly appreciated.  I'm new to this.  Thank you!

I am afraid that that measuring emulsification is not that easy. The statement that soap or shampoo is a better emulsifier cannot be generalized. My suggestion is not to use homemade shampoos etc. The problem is that you do not know the purity of chemicals. Was ethanol pure? Did it contain harmful substances? You would never know.

Your raw materials are irrelevant, if all you care about is which is the better emulsifier. What counts is how you measured that: how much shampoo vs. how much soap (do you measure in moles, by volume, by weight?), what substance you're emulsifying (n-dodecane? peanut butter?), what procedure (shaken? stirred?)...

Your question is not clear. Which proportion of these four components do you mix ?


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Can you coat/ "plate" things with non metals or compounds?
I was recently watching a video on electro plating I understand that it only works with metals so I wondered if there's a method to do the same thing but with non-metals.

Which non-metal do you have in mind?

@M.Farooq this is mere  curiosity I don't have a project in mind so this is pretty much random but  idk Sulfur, carbon and I also said compounds so sugar? I haven't done chemistry since highschool sorry.

I don't think you cannot electroplate sulfur, carbon or sugar on metal electrodes.

On the other hand, you can take a brush and coat anything in paint.

If you read "electroplating is covering... by metals" that is a definition based on its common use. Everything that it is electroactive and sticks to an electrode upon reduction or oxidation can be suitable. Electrochemical deposition of coniugated polymers or donor acceptor salts both involves organic compounds, and is in fact an electroplating process.

Yes. An example of a nonmetal "plating" out is the formation of ammonium amalgam. It is also shows the surprising stability of the ammonium radical, which behaves like a metal.
ScienceMadness.org states, "Ammonium amalgam has remained a curiosity since first made in 1808 [by both Davy and Berzelius, ed.], it has the intriguing property of being seemingly comprised of free radical Ammonium $\ce{NH3•H}$."
Of course, in forming an amalgam, the $\ce{NH3•H}$ plates out onto the mercury, but is quickly absorbed inside.
Do not use ammonium amalgam to fill cavities in teeth! ;-)

Wikipedia describes it as "Ammonium amalgam is a grey, soft, spongy mass discovered in 1808 by Humphry Davy and Jöns Jakob Berzelius. It decomposes readily at room temperature or in contact with water or alcohol". Given that it is not stable and only exists with Hg, not a prime example of a "coating", per se. Link: https://en.wikipedia.org/wiki/Amalgam_(chemistry) .

Yes, you can. Here is a reference concerning an electrolysis experiment based on aqueous $\ce{H2S}$, to quote:

An analysis of the chemical and electrochemical reactions in aqueous sulfide solutions indicated that the electrolysis of hydrogen sulfide in an alkaline solution will yield elemental sulfur and hydrogen without the passivation of the electrode if the ratio of NaOH to NaHS is properly controlled.

So, barring 'proper controls', your electrode (here the target of plating) would be covered with a deposit of sulfur.
So, employing a large flat surface area as the targeted electrode appears workable.
When it is fully coated, the process conveniently stops!