text:
Methylnitrobenzene ring formation Synthetic Route

I know the malonic ester is probably used to make propanoic acid or some carboxylic acid, and then joined to the methylnitrobenzene. But I just can't figure out what glues to the methyl. I know the nitro can be reduced to amine and bonded to a carboxylic acid, but then what about the ends?

What are we missing? What other reagents are specified? How many steps?

@Waylander It's just another synthetic route question in which reagents provided on the left hand side must be used to synthesize the compound on the right.

This is a challenging question as we have so few details.
One method would be to convert the orthonitrotoluene into the benzyl bromide with NBS or another free radical bromination reagent. Then you could react the dimethyl malonate with sodium methoxide in methanol. Next react the sodium salt of dimethyl malonate with 2-nitrobenzyl bromide.
Then filter off the sodium bromide thus formed, next stir it with the product of reacting Merifield's resin with tris-2-aminoethyl amine to scavenge anyunreacted nitrobenzyl bromide. Next subject the mixture to hydrogenation with a palladium on carbon cataylst. This will convert the nitro group into an amine.
Next evapourate it down to form a oil, heat this to make the amine react with the methyl ester to form methanol and an amide. When you are done with this step. Then react it with dilute sodium hydroxide to convert the remaining methyl ester into the sodium carboxylate. Make it acidic again and it will decarboxylate to give you a cyclic amide.

For the next step I would convert the amide to a thioamide with Lawesson's reagent and then treat the thioamide with rainey nickel to convert it into am amine. Between some of the steps it would be a good idea to purify the intermidates, but this is how I would (off the top of my head) try to make the conversion. I suspect that there will be cheaper ways of making the final product.

LiAlH4 or Borane.DMS will reduce the amide. Conditions for the bromination here: https://patents.google.com/patent/US5092971

Are you sure about the LiAlH4, I am a little wary of it. Have you got a good refernece for it working. Also what do you think of the thioamide route ?

Yes, I've used LiAlH4 many times to reduce amides to amines. I prefer Borane.dms as the workup is easier. The thioamide route will work but is an extra step

Tet. Letts (1999) vol40 3673 for a reduction using 9-BBN

Well I tend to be a bit careful about LiAlH4, one of my classmates had a nasty near miss with it. He managed to start a rather nasty fire with it. The thioamide method involves some smelly and toxic things, but it is a bit harder to burn the lab down with Lawessons's reagent. I would also like to use not true raniey nickel but the nickel boron material obtained from NiCl2 and NaBH4.

I've seen a colleague have a fire with LiAlH4. However I have used it safely many times. The procedures for safe workup are widely described.

True I have used it safely a few times as a PhD student and as a associate Prof, but on a large scale I would rather use sulfur chemistry which has a lower risk of fire


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Why does adding salt to boiling water cause it to flare up suddenly?
I was expecting that the water would become less "boily" since the ions in $\ce{NaCl}$ would require energy to disassociate. Instead it turned whitish and the water level suddenly rose and then subsided. I think the white color can be attributed to the fact that table salt is white but why did the water suddenly flare up ?

related https://chemistry.stackexchange.com/questions/57706/how-does-the-mentos-coke-explosion-work/57726?s=4|28.8734#57726 https://chemistry.stackexchange.com/questions/58301/why-do-raisins-bob-up-and-down-in-carbonated-water https://chemistry.stackexchange.com/questions/89081/potassium-citrate-in-water

The salt does not instantly dissolve so the surface of the crystals suddenly provides a lot of nucleation sites for the water to form vapour - hence the surge as it boils from these surfaces. The white is due to turbulence from the boiling just like waterfalls look white.

It seems that water does not begin boiling at 100.001 degrees C. A pot of water keeps getting hotter and hotter until it produces $H_2O$ vapor at a rate sufficient to produce a visible bubble in one location, and then turbulence shakes some more vapor free. The extra nucleation sites proposed by Waylander suggests that most pots don't have many nucleation sites. Of course! Who wants a rough pot that is hard to clean? Much preferred is a smooth pot. Other nucleation sites (or turbulence) can cause the same boil-up: pasta added to a slightly boiling pot will flare up too.
And this suggests that the hot water is actually hotter than equilibrium, either because of a hotter temperature than 100 degrees, and perhaps the vapor bubbles in the water are actually at a little more than atmospheric pressure, or that the vapor space just above the hot water is not 100% water vapor (i.e., not 1 atmosphere of $H_2O$), so more water vapor produced by more nucleation sites can be instantly accommodated if a boil-up occurs - this is not an adjustment of an equilibrium, but a collapse of a non-equilibrium situation.
The same thing can be done at room temperature with Diet Coke. I opened a cold fresh bottle very carefully and heard a little hiss - I know that the pressure in a bottle of Diet Coke is about 45 psi, because I have hooked up an automobile tire gauge to a bottle cap and screwed it on, then shook the bottle to get an equilibrium. But total equilibrium was not achieved, because the Coke in the bottle did not fizz. I poured some into a glass carefully - small amount of fizz, which stopped. Then I salted the liquid - viola! FIZZ! just because I added some nucleation sites. That reminded me of the Mentos addition videos.

source: Stack Exchange

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

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What should I do to prevent cellulose from turning black?
In my research on producing ethanol from grass, I have boiled 4g of grass in a beaker on a Bunsen burner with 30ml of sulfuric acid (bottle of 1 mol/l). However, within a few minutes the solution turned black and thus was burnt with release of smoke.
In my next attempt to try out this experiment this Monday, I want to make a 2.5 % sulfuric acid solution using a bottle of sulfuric acid with a concentration of 1 mol/l and add it to 4g of grass and boiling it for one hour.
What is your advice on this? Should I add a larger volume of sulfuric acid solution, and if so, how much?

Hi Tofighi, welcome to Chem.SE! Take the [tour], enjoy your stay! However it is frowned upon to ask multiple questions in one question. If you could split this question up, that would be nice! Thanks!

I hope it is specific now.

Because of the edit, it should go back into the reopen queue. Have patience and it should be reopened.

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


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How come when it rusts iron can undergo oxidation without the prescence of a another metal to undergo reduction?
It seems to me rusting is an electrochemical process where oxygen, which has particularly high electronegativity, fills the role of cathode or is that an incorrect way to put it?

The main problem you're having is that electrochemical processes don't have to have a cathode/anode.

Rusting is a redox reaction.
$\ce{4 Fe + 3 O2 -> 2 Fe2O3}$
Redox reactions mean one chemical is being reduced and one is being oxidized. Reduction = electron gain, oxidation = electron loss. The reason it is called oxidation is because oxygen is so good at reducing other chemicals (it wants electrons). How do you tell what gets electrons in a compound? Oxidation numbers.
Oxidation numbers signify how many electrons each atom has. -n means it has n more than neutral state, +n means it has n less than neutral state. In molecules which have only one element, oxidation state is 0. So Fe/O2 both have oxidation state 0. There are some rules on oxidation states in compounds which are complicated, basically usually group 1 is +1, group 2 is +2, O is -2, and group 17 is -1. In this case each O in the end is -2, and Fe is +3. So each Fe lost 3 electrons.

Some rules: https://socratic.org/questions/how-do-you-calculate-the-oxidation-number-of-an-element-in-a-compound


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Can ethylene glycol give victor mayer test?
I think there should be no problem for it to give red coloration in victor mayer test as a standard primary alcohol but a problem I encountered showed it otherwise.
Therefore, I would like to have a clear cut view of possibilities of what could happen, what actually happens and why?

Do you have a source for link about this test? Seems it's largely forgotten stuff from XIX century.

I am sorry as a friend of mine sent me this, this is the best i can give-
https://imgur.com/a/f596kXF

@Mithoron These are the references Wikipedia gives: https://en.wikipedia.org/wiki/Viktor_Meyer#cite_note-2

@GaurangTandon I know, but these are ancient German texts. I more or less found out what's that supposed to be and it seems this: https://chemistry.stackexchange.com/questions/68546/reaction-of-ethylene-glycol-with-pi3p-i2 is related as halogenation using P is first step.


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The Salt Effect stops working in extreme cases?
I and three friends did an experiment where we investigate the salt effect, also known as the uncommon ion effect.
The specifics are that we made a $\pu{1L}$ solution saturated with $\ce{CaCO3}$ and $\ce{NaCl}$, filtrated it, and measured the $\ce{Ca^2+}$-ion concentration and compared it with a pure solution saturated with only $\ce{CaCO3}$. The $\ce{Ca^2+}$-ion concentration was measured using AAS.
We expected the solution with the $\ce{NaCl}$ to have more $\ce{Ca^2+}$-ions than the one without, but that wasn't the case. The opposite was true. The concentration dropped by about 15% when the salt effect was supposed to occur.
Is it possible that there were so many $\ce{Na+}$ and $\ce{Cl-}$ ions that there weren't enough $\ce{H2O}$ molecules anymore to surround the $\ce{Ca^2+}$ and carbonate ion, causing their concentration to drop? Or is it far more likely that we simply made a mistake somewhere.

related https://chemistry.stackexchange.com/questions/95884/is-there-a-methodology-for-calculating-the-co-solubility-of-two-salts-with-diffe


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Find amount of benzoic acid in dissolution

Solubility of benzoic acid in water at $\pu{0^\circ C}$ is $\pu{1.7 g/L}$ and given the fact that $\pu{37 mL}$ was used to dissolve $\pu{1 g}$ of benzoic acid. How much of the benzoic acid would be left over given these conditions?

I have tried using $c = m/vM$ to convert $\pu{1.7 g/L}$ to a form given the $\pu{1 g}$ and the $\pu{37 mL}$ actually used. How would I find out how much is left over?

$\pu{0.037L}$ of water dissolves $0.037\times \pu{1.7g}$ of benzoic acid at $\pu{0^\circ C}$ = $\pu{0.0629g}$.
Therefore $1 - 0.0629 = \pu{0.9371g}$ benzoic acid remains undissolved


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Trying to understand the contradiction in the data of Proust/Dalton/Berzelius vs Gay-Lussac
I am trying to understand some of the historical context behind the discovery of atoms (pre-Avogadro).  In particular, I am getting stuck in reading the following documents:
http://physics.unl.edu/~klee/phys261/readings/class09.pdf
http://physics.unl.edu/~klee/phys261/readings/class11.pdf
My understanding is that "atoms" (as they understood them at the time) were discovered to react in whole-number ratios by Proust/Dalton/Berzelius (who measured by mass) and by Gay-Lussac (who measured by volume).  While this was highly suggestive of some sort of "atom" concept, there was a discrepancy between the two sets of measurements.  But I'm having trouble following the specific discrepancy pointed out by the above PDFs.
The first PDF gives the following mass data:
                         Mass of Nitrogen    Mass of Oxygen
Nitrous Oxide (N₂O)             100 g             58 g
Nitric Oxide (NO)               100 g            127 g
Nitrogen Dioxide (NO₂)          100 g            239 g

The text doesn't explicitly mention how this data was interpreted by Proust/Dalton/Berzelius.  Looking at the data, I don't see that much can really be said:
? N + 1 O = ? Nitrous Oxide (N₂O)
? N + 2 O = ? Nitric Oxide (NO)
? N + 4 O = ? Nitrogen Dioxide (NO₂)

The question marks are there since they didn't know what was "in" 100 g of nitrogen, which would mean that they didn't really know what was in the product, either, right?
So that's my first question:  Am I correct in thinking that the above is about as far as Proust/Dalton/Berzelius could go with that data?  Or were they pretty sure they knew how to fill in the question marks?
Now, continuing on to the second PDF, when Gay-Lussac measured by volume, he had:
                        Volume Nitrogen  Volume Oxygen  Volume Product
Nitrous Oxide (N₂O)             100           49.5             100
Nitric Oxide (NO)               100          108.9             200
Nitrogen Dioxide (NO₂)          100          204.7             200

Now, the document goes on to say that the above data might suggest the following formulas:
2 N + 1 O = Nitrous Oxide (N₂O)
1 N + 1 O = Nitric Oxide (NO)
1 N + 2 O = Nitrogen Dioxide (NO₂)

The text then says that the equation for nitrous oxide represents a contradiction.
But I don't see how that contradiction follows from the data.  As the text suggests, if we say that a volume of "50" represents one "atom" (so that volume of "100" nitrogen represents two atoms), wouldn't we say:
2 N + 1 O = 2 Nitrous Oxide (N₂O)
2 N + 2 O = 4 Nitric Oxide (NO)
2 N + 4 O = 4 Nitrogen Dioxide (NO₂)

Wouldn't that mean really it's the other two equations that don't make sense based on the number of atoms not working out?  I feel like I'm missing a basic idea in the text.
(FYI, I am not in school and so I'm not a student, nor is this homework.)

maybe the contradiction would be that nitrogen and oxygen should have theoretically equal volumes

Today the PDFs have disappeared from what I assume to be a supplemental reading material for a class.

In regards to your first question, I would think that Dalton would have a sense of the molecules' formulas based on the information at hand. Not only did Dalton have the mass ratios for these compounds, but he had also developed a measure of the relative mass of the atoms, as shown in your first document. Given this information, it would not have been difficult to compute that there were $\sim2$ atoms of nitrogen for every one of oxygen in nitrous oxide, nitric oxide was one to one, etc.
In regards to your second question, I think you are misinterpreting the contradiction. The authors proposal for the reactions was that if the compound formed was $\ce{N2O}$, it should only form one volume, since it should only form as much volume as there was volume of $\ce{O}$ originally. This is because there can only be as many molecules formed as there were atoms of the limiting reactant. The two volumes of $\ce{N2O}$ that are actually produced runs counter to this, as it would seem to require having $2$ volumes of $\ce{O}$ to start with.
For all the reactions, the products and reactants were related by whole number ratios, but not the same ratios suggested by their predicted molecular formulas.

Thanks for the clarification on the known ratios. So going back to the first part to fill in those question marks, I would think that since Proust etc didn't know about diatomic molecules, they would say something like 2N + 1O = 1N₂O, 2N + 2O = 2NO, and 2N + 4O = 2NO₂. As for the second part, however, I'm not totally sure I follow. If Guy-Lussac thought that we should "only form as much volume as there was volume of O originally," then aren't both the N₂O and NO forumulas inconsistent? Because in both of those cases, the volume of the product does not match the volume of oxygen?

@anon I phrased that confusingly, focusing too much on the one example. I will fix it, but to clarify I believe they had a concept of a limiting reactant at the time. So Gay-Lussac would predict that all of those would form the same volume as the limiting reactant, so in fact all of the reaction schemes he proposed are inconsistent, they give twice the volume we would expect based on their proposed ratios. I think the author might have only highlighted N2O as an example.

@anon it's also worth noting that the author doesn't say that Gay-Lussac proposed these reactions, but just that someone might, so it isn't clear if he interpreted the data this way.


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Can nitrogen dioxide react with plastics of any kind?
I have searched a lot on the internet but couldn't find any helpful thing. My question is that can nitrogen dioxide react with plastics of any kind?

Nitrogen dioxide is a free radical but its energy is rather low so it will be unlikely to be able to abstract a hydrogen from most polymers. If we consider polypropylene as the "standard plastic" then I would not expect nitrogen dioxide to be able to corrode it by abstracting hydrogens from the plastic.
But if we choose something else such as an epoxy as the standard plastic then nitrogen dioxide when combined with water could start to chemically alter the plastic. It can react with amines to form things such as N-nitroso compounds.
When nitrogen dioxide dissolves in water it forms nitric and nitrous acids. It can be regarded as a mixed acid anhydride for these two nitrogen oxy acids. The nitric acid could also start to degrade and alter plastics such as epoxy plastics by nitration of electron rich rings.
The acid formed when nitrogen dioxide dissolves in water could degrade plastic by increasing the rate at which esters such as the phthalate diesters react with water. It is important to keep in mind that many plastics are a combination of a polymer (or several polymers) and some additives such as plasticisers. If you degrade the plasticisers some plastics will stop being "plastic" and become brittle.


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Can a galvanic electrochemical cell produce current only one time without a salt bridge?
I think that galvanic cell produce current only one time without salt bridge. Potentional difference of both the electrodes are different and the electrons flow till their difference becomes same.Salt bridge is used to continue the reaction.
The cell can generate can generate current without salt bridge.Please help me to solve the problem.....

Yes it will .. But it will stop in short period due to inbalance of charges in both beakers


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Why is the IUPAC name of 1H-benzo[d]naphthalene not 1H-benzo[i]naphthalene?
In the nomenclature of fused arenes you need to add a 'fusion locant'. The first face should correspond to positions 1 and 2 and is lettered ‘a’, continuing alphabetically clockwise around the exterior of the parent structure.
In this case, the parent is naphthalene and using numbering of naphthalene the benzo is fused on position d so the name would be 1H-benzo[d]naphthalene

But IUPAC also has some rules about the orientation of such structures:
(a) maximum number of rings in a horizontal row (b) the maximum number of rings in the upper right quadrant. (See the drawing on the right.) Subsequent the numbering of peripheral atoms in the preferred orientation starts from the uppermost ring. (If there is more than one uppermost ring, the ring furthest to the right is chosen.)
 Using that numbering scheme the face of fusion should be i.

Apparently, you are mixing two different numbering schemes.
For the construction of a fusion name, the two components are numbered separately (if necessary), using letters for the parent component and numbers for the attached component. The corresponding rule P-25.3.1.3 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

(…)
  Isomers are distinguished by lettering, continuously, each peripheral side of the parent component (including sides whose locants are distinguished by letters, for example, 2a,3a) using the italic letters a, b, c, etc., beginning with a for the side numbered ‘1 ,2’, b for ‘2,3’ etc. To the letter as early in the alphabet as possible that denotes the side where the fusion occurs are prefixed, if necessary, the numbers of the positions of attachment of the other component. These numbers are chosen to be as low as is consistent with the numbering of the compound and their order conforms to the direction of lettering of the parent component. (…)

The parent component for the example that is given in the question is naphthalene (not the complete benzonaphthalene). The letter as early in the alphabet as possible that denotes the side where the fusion occurs is d. Therefore, the correct name of the complete compound is
1H-benzo[d]naphthalene.

The other numbering scheme described in P-25.3.3.1.1 applies to the complete fused ring system.

P-25.3.3.1.1 The numbering of peripheral atoms in the preferred orientation starts from the uppermost ring. If there is more than one uppermost ring, the ring furthest to the right is chosen. Numbering starts from the nonfused atom most counterclockwise in the ring selected and proceeds in a clockwise direction around the system, including fusion heteroatoms but not fusion carbon atoms. (…)

Thus, for the example that is given in the question, this numbering applies to the complete
1H-benzo[d]naphthalene. In particular, this numbering is used for the locant of the indicated hydrogen. It would also be used for any substituents of the compound.


text:
I have an idea through utilizing soda lime or other basic substances to help solve climate change
I have an idea through utilizing soda lime or other basic substances to help solve climate change. I want to see if there are some possible errors or other negative implication that I am disregarding.
I came across soda lime during one of my biology textbooks and noticed that it has the ability to absorb $\ce{CO2}$. The first thing that popped into my head was utilising soda lime ($\ce{CaHNaO2}$) to help solve global warming.
My idea is having large or wide buildings that are made with soda lime on the edge that will "suck up" the carbon dioxide. This carbon dioxide can then be used for various different ways in which the community and the society can benefit from.

convert it to electricity or fuel (it has been said that there is a way of doing this as in Cornell University Research - YouTube video). Essentially, the electricity made would be redistributed to the city  to use.
export the gas to bread industries (the yeast produces carbon dioxide that may be used for brewing and for bakery)
if possible, change the carbon dioxide to oxygen and release oxygen.

I know I might not be aware of some factors and it may seem idealistic, but I want to hear from the community. Is what I am proposing realistic enough? Or are there some practical problems with this approach?

Limestone already contains CO2. It won't take up more. Especially removing it from limestone consumes a lot of energy, and further converting it to fuel would need even more energy. Yeast produces CO2, and the amount is totally negligible in comparison to our greenhouse problem. Lastly, converting CO2 to carbon and O2 also takes up a huge amount of energy.

Soda-lime deteriorates unless kept in an airtight container leading to waste of resources

My worry is that the energy requirements for making soda lime, caustic soda or some other carbon dioxide adsorber will be so high that overall it will result in the emission of more green house gases.
Years ago I was touring a waste disposal site where they react waste sulfuric acid with limestone. The boss of the site told me that some years ago someone had asked if he could scrub out the carbon dioxide from the reaction tank with sodium hydroxide or a similar reagent. He gave the same answer as I am giving to us when he recalled the tale.


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Does CN4 with the following structure exist?
I am a mathematician, not a chemist, so my questions are rather combinatorial. I am intrigued by the possibility of the following binary compound:
$$\ce{\overset{-}{N}=\overset{+}{N}=C=\overset{+}{N}=\overset{-}{N}}$$
which ought to be isoelectronic with carbon suboxide, and thus possibly stable. Does this compound exist, even in interstellar space?

Now that's an interesting question. After all, diazomethane is a thing.

There were 0 hits on ChemSpider Structure Search.

Looks horrendously unstable

Cyanogen azide, $\ce{(N$!\equiv!$C-N=N^+=N^-)}$, is pretty similar but not quite.

Does not sound like something anyone would want to handle http://blogs.sciencemag.org/pipeline/archives/2008/07/14/things_i_wont_work_with_cyanogen_azide

Diazidomethane is a thing too - though not a thing you want to make. Org. Process Res. Dev.  (2008) 12 1285

I suspect it is going to be hard to make this substance, no mention in Reaxys. I think it might be an impossible compound to make. If you do make it take care work using a long pair of tongs to handle it, a kevlar jacket, heavy gloves, good eye protection and ear defenders.

@Oscar Lanzi maybe on top of the ear defenders, face shield and arc flash glasses (UV blocking) one should wear the Ned Kelly helmet (Steel bucket with a small eye slot), maybe one should work from behind a sheet of thick steel using a master slave manipulator or maybe choose a different synthetic target to work on.

I was hoping that you chaps, knowing more about this than I, could supply illuminating observations about the stability of the molecule, and the parallels with carbon suboxide.  I can get banter about protective lab gear without going to the lengths of signing up to chemistry stack-exchange.  Come on - what are you made of?

@ThomasForster as far as I can tell, this compound has never been synthesized, so it is difficult to say for certain whether it would be stable. However, based on the the existence of similar compounds (diazomethane, cyanogenazide) that are at least thermodynamically stable, the simplicity of compound, and the caveat of accepting interstellar space, where the compound is likely to be kinetically stable due to the low temperature, I would say it is likely it does exist. For further supporting evidence, QM calculations could be done to confirm its thermodynamic stability.

@ThomasForster Confirmation that this compound isn't known experimentally: https://books.google.com/books?id=FA7VAwAAQBAJ&pg=PT178&lpg=PT178&dq=carbon+suboxide+isoelectronic&source=bl&ots=XBXyf1RRqV&sig=2z1T5kuGniPUtUZTfKlYBZGP7bU&hl=en&sa=X&ved=0ahUKEwjZmKTItJfbAhUJSa0KHTkXBasQ6AEIcTAJ#v=onepage&q=carbon%20suboxide%20isoelectronic&f=false From what I have found about similar compounds, it is likely that your compound is bent into a V shape, similar to pentazenium

Carbon suboxide has a very shallow bonding potential. It is commonly written as $\ce{O=C=C=C=O}$. There are a couple of resonance forms which need to be accounted for. Another interpretation of the bonding patter is as a carbon-zero compound: $\ce{^{\oplus}O#C^\ominus\bond{->}C^0\bond{<-}C^\ominus=O^\oplus}$. This description makes somewhat sense because of 2-electron-donor nature of carbon monoxide. I would not expect this to be similar with the nitrogen derivative. It would, however, be an interesting point of study.

Could someone confident with nomenclature name this compound?  I am tempted to call it "diazocarbene", but PubChem uses that for something else.  The azo bridge is known to tautomerize to hydrazone if there is a hydrogen at either end, but that supposes it is attached to a carbon with a hydrogen to tautomerize.  My feeling is that naming the functional groups might find some data about this molecule, or give an understanding of its problems.

@MikeSerfas bis(diazo)methane

This compound CN$_4$ is a subject of theoretical study, with a theoretically stable structure. Literature on the compound is scant, but this has been considered before and calculated with post Hartree-Fock ab initio molecular calculations.
Stable is a relative term, and I use it here to just mean "has an energy minimum." I have calculated the geometry and charge balance at the CCSD/aug-cc-pVTZ level of theory for this molecule as a proof of concept. In terms of viability in interstellar space, this calculation should represent that closely. It is possible that the transition-state is low enough in energy that even low-temperature vibrations of the molecule could disassociate it quickly. Regardless, to the mathematician this can reasonably be considered stable.
The N=N bond length I calculated was 1.135Å, and the C=N bond length was 1.231Å in my calculations. The partial charges are depicted here:

Although $\ce{CN4}$, called bis(diazo)methane, has not been experimentally observed, it has been investigated theoretically and found to be vibrationally stable.

Structure
Energy
Source
Remarks

$\pu{499.79 kJ mol^{-1}}$
MolCalc
Non-linear geometry doesn't seem to work

$\approx -\pu{672 kJ mol^{-1}}$
Reference 1
Bond orders may be incorrect

$\pu{647.29 kJ mol^{-1}}$
MolCalc
-

Protonated bis(diazo)methane $\ce{HCN4+}$
As investigated by Papakondylis A. and Mavridis A. (2005), the protonated form is also stable.

Structure
Energy
Source
Remarks

$\approx\pu{-675 kJ mol^{-1}}$
Reference 2
Also isoelectronic to bis(diazo)methane

Isoelectronic Structures
$\ce{CN4}$ belongs to a family of isoelectronic compounds with 34 electrons which have been called pseudohalide ions.$^\text{1}$. I don't fully understand why these were called pseudohalide ions; because halide ions have a complete shell (10, 18, 36, etc.). A few others to mention are:

Name
Compound
Remarks

Carbon Suboxide
$\ce{O=C=C=C=O}$
Known, stable, linear

Dicynamide
$\ce{[N=C=N=C=N]-}$
Known, stable, bent

Pentazenium
$\ce{[N#N-N=N=N]+}$
Known, stable, bent

-
$\ce{[NNBNN]-}$
Unknown, stable (predicted), bent

-
$\ce{[NNONN]+}$
Unknown, stable (predicted), bent

-
$\ce{[NNONN]+}$
Unknown, stable (predicted), bent

-
$\ce{[N#C-N=N=N]}$
Unknown, stable (predicted), bent

Boron trioxide
$\ce{O=B-O-B=O}$
Known, stable, bent

Thus, $\ce{CN4}$ isoelectronic analogs maybe linear, bent, or zig-zag (when the ends are also bent). MolCalc (MP2) shows a linear geometry with a formation energy of $\pu{499.79 kJ mol^{-1}}$.
Further Reading
Perhaps the most extensive study of these structures was carried out by Pyykkö, P and Runeberg, N. (1991)$^\text{1}$ who carried out HF and MP2 calculations with a 6-31G* basis set and found the stable geometries, energies, and vibrational levels of such compounds.
References

Pyykkö, P and Runeberg, N. (1991). Ab initio studies of bonding trends: Part 9. The dicyanamide-carbon suboxide-dicyanoether-cyanogen azide isoelectronic series A=B=C=D=E. Journal of Molecular Structure: THEOCHEM, 234, 279-290. 10.1016/0166-1280(91)89018-V.
Papakondylis A. and Mavridis A. (2005). J. Phys. Chem. A, 109(29), 6549-6554.  10.1021/jp0580975.
Dissertation by Susanne Klein. Quantenchemische Untersuchungen an zweiwertigen E(0)-Verbindungen der 14. Gruppe und deren Analoga der Gruppe 15. Available as PDF.

Thank you for including the references here

I think the pseudohalide point seems a bit irrelavent, as plenty of things that are not isoelectronic to this structure or are similar to it at all count as pseudohalogen ions. It's a bit misleading, I almost thought pseudohalides somehow referenced this specific class of compound until I double checked

@thelocalsage you're most welcome. Yes! The pseudohalide part is a little confusing. However, I was hoping someone would clarify why these compounds have been called pseudohalides.


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Why does a six-membered ring contract in this SN1 reaction?
I was watching this video on Khan Academy (YouTube link) regarding SN1 mechanisms and carbocation rearrangements and came across this mechanism.
At around timestamp 4:20 in the video, why does step 2 take place where the ring contracts to 5 carbons? Surely it would be far easier to simply have a methyl shift from the carbon in blue which would also result in a more stable tertiary carbocation.

A cursory look through several organic chemistry texts (Clayden, Carey, March, Vollhardt) as well as SciFinder doesn't reveal any instances of the same reaction used in the Khan Academy video - this would imply that either the reaction is made up to illustrate a point (albeit not very well) or comes from somewhere more obscure (historical literature in a language other than English).
SN1 vs SN2
The first question that it would be useful to answer is whether the reaction proceeds via an SN1 or SN2 pathway, or, to phrase this a more accurate way, to what extent is the halide dissociated at the time when the rearrangement occurs.
In general, secondary halides can undergo both SN1 or SN2 reactions depending on the conditions used (substitutions at secondary centres are always hard to predict, and often go with neither complete inversion nor complete racemisation). In March's Advanced Organic Chemistry,  the chapter on cationic  rearrangements summarises:

The SN1-type process occurs mostly when B is a tertiary atom or has one aryl group and at least one other alkyl or aryl group. In other cases, the SN2-type process is more likely.

Stereoelectronics
If we accept that the reaction goes via an SN2-like pathway, we must then acknowledge the stereoelectronic requirement that the migrating group must be able to interact with the C-Br σ* orbital.
This requirement immediately rules out the migration of the exocyclic methyl group, leaving only the internal C-C bond to migrate (assuming that bromine adopts the equatorial position and the ring sits on a chair-like conformation - both of which are reasonable assumptions).

Migrating group ability
Even ignoring the steroeelectronic issues, migratory aptitude would favour rearrangement of the internal C-C bond as it is marginally more able to stabilise the developing positive charge during the rearrangement.
tl;dr
In the absence of the original literature, it's hard to say with certainty which group migrates (though it would be an easy experiment to carry out....). That said, following the general trends for rearrangements/migrations found in undergraduate organic chemistry texts, it would seem most likely that the rearrangement takes place as shown rather than the competing 1,2-methyl shift (we're also ignoring the possibility of a proton elimination to give an alkene).


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Which has stronger hydrogen bonds: water or ice?
As ice is the solid form of water and it has more hydrogen bonds than water, because it's oxygen atoms are precisely tetrahedrally positioned and each oxygen is hydrogen bonded by four neighbouring oxygen atoms.
This changes the density of ice by expanding it, but does it change the strength of hydrogen bonds of ice? Or more precisely, which one has strong hydrogen bonds, water or ice?

The molecules are still the same; why would the bonds be any different?

@IvanNeretin because bond strength changes with distance?

This is pretty much based on false premise there's one single bond strength for ice or for water.

@Mithoron would you care to add an answer of yourself elaborating on this?

I recommend editing the question to include the temperature and pressure of the ice and liquid water. Agreed with @Mithoron that there is probably a distribution of bond energies here; so an answer ought to discuss the mean energies, plus possibly relative contributions. I'm sure there are complications here as well, especially differentiating H-bonds from other intermolecular interactions.

@PaŭloEbermann To do it properly I'd need to write a lot on how this isn't such a good question and I'm afraid it's too broad now. Even adding particular temp./pressure wouldn't make it really good. In such case it would be rather a thing for experimental or computational analysis.

While I cannot answer the question (right now, neither literature nor lab access to a suitable spectrometer), bond strength is experimentally accessible via vibrational spectroscopy (IR, Raman): that looks at the strength of individual bond as opposed to the sum of all hydrogen bonds in @Amish Dua's reference (if my glance read that correctly). I'd expect that the ice spectrum shows a well defined vibrational band, i.e. a narrow distribution of bond strengths, whereas for liquid water I'd expect a very broad band, i.e. a very wide distribution of bond strength, and I'd expect that this ...

... distribution ranges from much weaker to stronger compared to hydrogen bonds in ice.

From x-ray scattering experiments the average  O - O bond length in water is a little larger than that in ice ($I_h$)  $0.281 $ vs $0. 276$ nm, but there is a  range of bonds lengths which has a distribution the peak of which is the same for both ice and water.

The quick answer is that ice has stronger hydrogen bonds than liquid water on average.$\require{mediawiki-texvc}%for angstrom unit$
As has already been brought up, this question is not easy to answer comprehensively as it may very well depend on temperature and pressure in ways that are not immediately obvious. That being said, I will answer this question for the most obvious case which is looking at hydrogen bond strength when water is just barely a liquid (just above $\pu{0~^\circ C}$) and when water is just barely frozen (just below $\pu{0~^\circ C}$). Also, as you may or may not know, ice freezes into many different forms which all have different lattice arrangements. I cannot speak to unusual forms of ice such as the one with a cubic lattice, but what I say is immediately applicable to forms like proton-ordered hexagonal ice (I think this is ice IX). Regular ice is called ice Ih.
It is worth looking at the very comprehensive reference [1], which discusses details of water from the dimer all the way to the many forms of ice which can form. The important points for us are only the following structural values, however:
In ice Ih, the average $\ce{O-H}$ bond length is $\pu{1.01 \AA}$ and the $\ce{HOH}$ bond angle is $109.5^\circ$. The trend which has been observed experimentally is that in going from the gas-phase water monomer, through the clusters, to the liquid, and finally to ice, the bond angle slowly increases to this tetrahedral value. Additionally, the average bond length in the liquid is shorter than in the solid, but you will have a hard time finding reliable reported values of this number from experiment because it is very difficult to determine the position of hydrogen atoms from scattering experiments in non-solid phases (this is mostly because of an ambiguity as to which oxygen atoms a hydrogen should be associated with in the scattering data). Nonetheless, there are reliable average $\ce{O-O}$ distances from scattering experiments.
In the liquid, the average $\ce{O-O}$ distance is $\pu{2.85 \AA}$ while in ice Ih it shrinks to $\pu{2.74 \AA}$. Intuitively, it is very hard to understand how this would happen if it is not simply due to increased hydrogen bond strength.
We can look at an even simpler case though and see that increased hydrogen bond strength causes the $\ce{O-O}$ distance to decrease in ice. This example is a comparison of the water dimer and water trimer. In the water dimer, the $\ce{O-O}$ distance is $\pu{2.976 \AA}$. In the water trimer, the $\ce{O-O}$ distance is about $\pu{2.82 \AA}$. This decrease in distance and the corresponding red-shift in vibrational frequency are given physical meaning in the Ref. [2].
Ref. [2] gets at a very important point for us, one which Ref. [1] also uses to explain the shrinking in $\ce{O-O}$ distances in ice. This is that hydrogen-bonding in water highly cooperative. What do I mean by this? This is exactly the topic of Ref. [2]. The key idea is this: when you allow three water molecules to mutually hydrogen bond, as in the ring shape of the water trimer, the increase in attractive interaction strength is more than simply three times the interaction strength of a water dimer. By attractive interaction strength, I mean that if you subtract out the initial decrease in binding energy which comes from distorting the water monomers. Again, all of this is laid out in detail in Ref. [2], but we learn the general rule that many-body interactions in water are very important.
So, finally, I feel we have said enough to explain why the knowledge that a shrinking $\ce{O-O}$ distance and widening $\ce{HOH}$ angle tells us that ice has stronger hydrogen bonds than liquid water. In the first case, the decreasing $\ce{O-O}$ distance means that every pair of monomers experiences greater nuclear repulsion. This must be at least offset by a greater hydrogen bond strength in order for this geometry to be favorable. Part of the way this repulsion is offset is that as the system cools and forms a solid, the hydrogen-bonds are allowed to become more ideal. This is why the bond angle widens. The tetrahedral arrangement of atoms means that the $\ce{OH--H}$ hydrogen-bond angle is much closer to 180 degrees on average than in the liquid where it frequently deviates fairly significantly.
There is one last point to make. It can be estimated both from accurate simulations and from experiment what is the average dipole moment of a water monomer in bulk liquid and in ice. It is found that the dipole increases going from the liquid to ice. This additionally tells us that the average charge on each hydrogen atom must have increased from in the liquid. We are guaranteed this because we know the bond angle increased, which should generally decrease the dipole moment (a linear configuration has no dipole). Instead, the bond angle increases, so we get more ideal hydrogen bonds, and the charge increases, so the hydrogen bonds would be stronger even if they were no more ideal than in the liquid!
Hopefully this rather long-winded answer helps you understand how we can take the evidence at hand to arrive at the conclusion that ice has stronger hydrogen bonds than liquid water.

Now to the question of different forms of ice and different temperatures and pressures. I would think that pretty much across the board, every form of ice will follow the same trends we have seen here and hence have stronger hydrogen bonds. One strange case is cubic ice, which has $\ce{OHO}$ angles somewhere close to 90 degrees. This is very strange, but should lead to a large increase in the dipole of each molecule and probably increase in positive charge on each hydrogen atom, so even then the hydrogen bonds should be stronger than in the liquid.
There may be temperatures and pressures where some of these arguments break down, but the $\ce{O-O}$ distance should smoothly decrease to that in the liquid and the angle should close to that in the liquid as the pressure increases, so one would expect the hydrogen bond strength to decrease to that in the liquid and never overshoot it.

References:

Ludwig, R. Water: From Clusters to the Bulk. Angew. Chem. Int. Ed. 2001, 40 (10), 1808-1827. DOI: 10.1002/1521-3773(20010518)40:10<1808::AID-ANIE1808>3.0.CO;2-1.
Xantheas, S. S. Cooperativity and hydrogen bonding network in water clusters. Chemical Physics 2000, 258 (2-3), 225-231. DOI: 10.1016/S0301-0104(00)00189-0.

TL:DR: it has been demonstrated both thermodynamically and computationally that the hydrogen bonding in hexagonal ("ordinary") ice is greater than in liquid water.
Ford and Falk [1] used spectroscopic data (IR absorption frequencies) associated with hydrogen bond strength (intermolecular potential energy).
Namely, the relationship between the measured frequency shift $\Delta\nu$ of OH stretching vibration and correlation curves for the variety of frequencies allows to plot the distribution of hydrogen bond energies in ice and water.
Liquid water possesses a much broader variety of energies (only about $15\,\%$ molecules lie within $\pu{± 1 kcal mol^-1}$ range at maxima) in comparison to ice $(90\,\%).$
The authors partially attribute greater hydrogen bonding energy for ice to the heat of fusion $(\pu{1.435 kcal mol^-1}).$

Huš and Urbic [2] mention in their review (reference numbers have been adapted to current references list) that

the HB strength in ice is greater than in liquid water [3–5]

and by performing own DFT calculations (6-31++G(df) and aug-cc-pVTZ basis sets) for the water cluster with the ﬁrst and second solvation shells also back this observation up.
It was found that the addition of water molecule results in additive change in bond strength, about $\pu{± 0.7 kcal mol^−1}$ per added molecule being a basis-agnostic value.
As the solvation shell used in calculation expands, the average hydrogen energy increases, too (alongside with the number of highly unlikely coordination environments with extremely high/low values):

When all possible hydrogen bonds are formed, the structure of the cluster resembles a highly ordered structure of ice.
In that instance HB strength was calculated to be $\pu{7.84 kcal mol−1}$ (basis set 6-31++G(df)) or $\pu{6.66 kcal mol−1}$ (basis set aug-cc-pVTZ).

Authors also reference experimentally measured hydrogen bond strength of $\pu{3.6 ± 0.5 kcal mol^−1}$ using thermal conductivity of the vapor [6].
References

Ford, T. A.; Falk, M. Hydrogen Bonding in Water and Ice. Can. J. Chem. 1968, 46 (22), 3579–3586. DOI: 10.1139/v68-591.
Huš, M.; Urbic, T. Strength of Hydrogen Bonds of Water Depends on Local Environment. Chem. Phys. 2012, 136 (14), 144305. DOI: 10.1063/1.3701616.
White, J. C.; Davidson, E. R. An Analysis of the Hydrogen Bond in Ice. Chem. Phys. 1990, 93 (11), 8029–8035. DOI: 10.1063/1.459332.
Hamann, D. R. $\ce{H2O}$ Hydrogen Bonding in Density-Functional Theory. Phys. Rev. B 1997, 55 (16), R10157–R10160. DOI: 10.1103/PhysRevB.55.R10157.
Smith, J. D.; Cappa, C. D.; Wilson, K. R.; Messer, B. M.; Cohen, R. C.; Saykally, R. J. Energetics of Hydrogen Bond Network Rearrangements in Liquid Water. Science 2004, 306 (5697), 851–853. DOI: 10.1126/science.1102560.
Feyereisen, M. W.; Feller, D.; Dixon, D. A. Hydrogen Bond Energy of the Water Dimer. J. Phys. Chem. 1996, 100 (8), 2993–2997.
DOI: 10.1021/jp952860l.

In liquid state of water the atoms will be having so much kinetic energy that they cannot form permanent bonds.
although bonds will be formed but they will break simultaneously until the temperature hence the kinetic energy is reduced and bonds become stable
Although the strength will be almost the same but due to constant breaking and joining the net interaction will be weaker than in ice
The hydrogen bonding in ice(hex) is about 3 kJ ˣ mol-1 stronger than liquid water.
According to this paper
http://www1.lsbu.ac.uk/water/water_hydrogen_bonding.html

this is a nice video i found on youtube :)
https://www.youtube.com/watch?v=UukRgqzk-KE

This is completely incorrect and I'm not gonna waste time on youtube movie which is no proof whatsoever. Constant breaking and reforming of bonds says exactly nothing about their strength.

"In liquid state of water the atoms will be having so much kinetic energy that they cannot form strong bonds.":  This is incorrect because, as the image shows, water molecules are constantly forming and breaking bonds; sure they're temporary, but this doesn't imply weakness.  As far as the conservation of energy is concerned, sure breaking a strong bond takes a lot of energy, but the creation of a new strong bond satisfies the balance, such that merely having bonds breaking in favor of new, similar bonds doesn't imply energetic weakness.

Water has stronger hydrogen bonds than ice does.
Liquid water is denser than ice. Since water and ice are both made of H2O molecules, the fact that water is denser means the H2O molecules are closer together in water than they are in ice. This means the intermolecular forces attracting one H2O molecule to another must be stronger in water than they are in ice. Since hydrogen bonds are the primary intermolecular forces in H2O, the hydrogen bonds in liquid water are stronger than those in ice.
Remember that hydrogen bonds are an electrostatic attraction (an attraction between a positive charge and a negative charge). The strength of an electrostatic attraction depends on the magnitude of the charges and on the distance between the charges. Since the distance between molecules is greater in ice but the magnitude of the charge is the same, the strength of the attraction is lower.

This is maybe even more incorrect.  Density says nothing about bond length, in this case as demonstrated in picture added to another incorrect, but slightly more useful post.

So now,What will be the correct answer to this?Water or ice?I can't get a clear answer from above discussion.

To Mithoron: if you're talking about two different substances that have two different masses, then you're right that density doesn't say anything about how close the particles are to one another. But in this case we're talking about the same substance: the H2O molecules in ice and the H2O molecules in water have the exact same mass, so the only way water can be denser than ice is if the H2O molecules in water are closer together than the ones in ice.


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Why does butter get soft with heat, but not ice?
Sometimes a recipe asks me to soften the butter before I use it. This can be accomplished with a few seconds in the microwave, or just by leaving it out at room temperature for a while.
Ice, on the other hand, does not get soft with heat. It just immediately starts melting.
What chemical differences between the two would account for this?

The main difference between warming 'solid' butter and warming solid ice is that one is a mixture and one is a pure substance.
Ice has a crystal structure set in place by the shape and polarity of the water molecule. It requires quite a bit of energy to be removed from 0 degree water to cause the formation of the solid phase. The structure has two options - remain water, or, once cooled enough, become solid.
Butter is a mixture of fat and water and other components. As this mixture warms, the fat component changes viscosity and becomes more liquid and less sludgy. The water does not change much, but is trapped in the layers of the fats. At 40 F, just above refrigerator temperature, butter is 'hard' because the fats are not flowing. As the temperature warms, the fats start to move, thus softening the butter.
The butter softening behavior can be seen in some metals as they get heated close to the melting point. Iron becomes 'soft' and malleable, and can be forged. Ice has a particular, rigid structure, so never has a 'soft solid' stage.

Would it be accurate to say that butter has a variety of states it can be in and still be "solid", while ice has only one state, and this accounts for the difference?

@EliRose Yes, that is a good description. Solids can have different crystal structures, etc.. The word 'state' though, in strictest terms, is a particular term with particular meaning. At some point butter is a very high viscosity liquid that seems like a soft solid.


text:
How does light emission depend on temperature by fluorescence?
I have a glass, which emits green light after charging it with sunlight. I though this is fluorescence, and only photons with shorter wave length (than the green light has) can excite the electrons of this material and can be reemitted as green light. At least I remember on something like that from my studies. Now the interesting part that when I wash it with warm water, it starts to emit light too. How do you explain that?

It is phosphorescence. The excited state exists for some time after the initial excitation. Heating the plate causes the excited state to decay more quickly.

@MaxW I think I can verify this. If it is true, then it should be depleted relative fast after applying warm water. Just a min. :-)

@MaxW Hmm, it will take more than that. It can glow for several hours, and I guess this would just cut that time to half or third. I'll do an experiment with it tomorrow.

This phenomenon is thermoluminescence. The incoming photons excite electrons of the material to a higher level, a metastable state, which fall back to a lower state and emit photons. The metastable state may last for only nano- or picoseconds, in which case it is termed fluorescence. If the electrons need a bit of a "nudge" and hang around for a few seconds or hours, it is termed phosphorescence.
In the extreme case of some materials, the metastable state can last for millenia, which can be useful for dating the age of the item. For example, when pottery is fired, atoms in a metastable state relax. From then on, energy accumulates from cosmic rays and radioactive decay of nearby minerals such as thorium or potassium. By heating the ceramic and measuring the amount of light emitted, one can approximate the age of the pottery.
Ceramics need to be heated to a few hundred Celsius, but your glass works in just warm water. There are similar infrared detection cards with phosphors that are charged using visible or UV light, and that emit visible light when triggered by low-energy IR.


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Mechanism for reaction of aryl diazonium with potassium iodide
I’m working on a synthesis problem that requires me to draw the arrow mechanism for the second half of the Sandmeyer reaction. Basically it’s the halogenation of an aryl diazonium compound. Typically it’s done using a copper catalyst, but it can also be done using just potassium or sodium iodide. Problem is, I can’t find any reference showing how this works in an arrow pushing mechanism. Does anyone know where I could find an example to show me how it works?

Well, it's probably SN1 so not much of writing ;)

I was taught that it's a radical mechanism, and that makes sense because if it was a pure SN1 then this would work with chloride ion too, but you need CuCl (not just NaCl) for that

It follows a radical mechanism (SET  = Single Electron Transfer)

Your source for this?

I found the image online, but I studied this mechanism on the course of Mechanisms for Organic Reactions.

Where does the .I2- come from in 1)?

@Andrea can you tell where the image was found so we can trace it?

@Waylander once the mechanism is going then the di-iodine radical anion is generated by Step 3.  But yes, we need a startup path.  I would guess the cation reacts with iodide ion to generate di-azo and mono-iodine radicals, then the latter adds another iodide ion while the former jumps into Step 2.


text:
If the human body consists of 60% water why can't we put out fire with our body?
I have often heard of people being burned at the stake, but if the body is 60% water shouldn't the fire just be put out?

An aluminium kettle on the fire might be 95% water, it's still dry and melts after an hour.

You know what happens to the water if you put out fire with it?

You can put out small fires with your body.  You know the "stop, drop and roll" advice for when your clothes are on fire?  Doing that cuts off the air to the flames but it also dumps all the heat into your body.  Since it's not a lot of heat when your clothes have just caught fire, that's okay.

And in the example @zwol mentioned the water we have in our bodies helps us with its high heat-capacity

@zwol Being the question mainly about water, I admit that when you said "You can put out small fires with your body" I initially thought of a... less dignified example than "dropping and rolling".

@xDaizu There is actually an English idiom about that ;-)

The body tissues first get cooked because of presence of the water. If that's not bad enough, once the water evaporates, the tissue gets burnt. If it's hot enough, the water gets superheated before evaporating from the tissue which means more damage even if it stops short of charring. Sounds familiar?

I'm voting to close this question as off-topic because it's not really about chemistry and rather unclear.

Lick your fingers (water from in your body) and pinch a candle -- it goes out.  If you try this at home, don't hang about with your fingers in the flame.  Or don't try it at home.

Main reason is human bodies do not fit through fire hoses all that well, which is why we use pythons and boa constrictors instead. ;-)

Is this supposed to be some kind of a 4chan-cess joke?

We need to consider something. Are the people being truly burnt at the stake or are they being killed by the effects of fire? If the fire is very prolonged then their bodies will dry out and then finally be calcined to ashes.
It is an interesting (but a rather morbid) fact that cremation furnaces typically run at about 800 degrees Celsius for many hours, yet they tend to be unable to burn bodies to dust. Typically after cremation, some large fragments remain. These have to be crushed.
There are tales of spontaneous human combustion, these have been circulating for many years. I recall that the Victorians thought it was associated with extreme alcohol abuse. They thought a build-up of alcohol in the body would lead to the body being able to catch fire. A modern-day explanation is that humans wear clothing, the clothing can act as a wick.
A person dies as a result of something, then their clothing comes into contact with an ignition source. The clothing starts to burn, this heats up fat in their body. The fat melts and runs into the clothing. The clothing then acts like a candle wick and then the burning fat makes more heat which melts more fat thus turning the person into a candle.
I think that this type of human candle effect is very rare, it is likely to require a combination of several things at once. Many people who are killed in serious fires (who suffer burns as well as being exposed to hot and toxic smoke) leave behind bodies which have not been burnt to ashes.

Water doesn't inherently beat fire. Throw a bit of water into a big fire and it will evaporate very quickly.
To find an answer, we have to take a look at what is required for a fire, and how water generally is used to fight fires. You might have heard of the fire triangle. Fire needs:

Fuel
Oxygen
Heat

Take away one of these, and the fire will die. Throwing water on a fire does two things:

It covers the fuel (at least during the initial splash, whether it will afterwards depends on the relative size of the fire), lowering availability of oxygen
It lowers the temperature, since the water is (I hope) colder than the fire itself. Whether this is sufficient to douse the flames depends again on the relative size of the fire.

Here we can see the things throwing water on a fire does, that putting a human on a stake does not. A human on a stake does not (make any attempt to) cover the fire. Since the human is on there since the start of the fire, warming up along with it, it doesn't accomplish a whole lot in the removing-heat-department either. If anything, as Nuclear chemist noted, it'll only add more fuel.

Evaporation also uses a significant amount of energy. I believe you have to heat water over 500 K to use the same amount of energy that evaporation of the same mass of water requires.

@Matt hmm, this is where my knowledge of thermodynamics falls short. I can imagine that there is a difference between a sudden introduction of a colder body versus having the water slowly heat up along with the fire. I'm not sure though.

In a chemist's perspective, the fire is a high-temperature mixture of extremely reactive radical species, most importantly reactive oxygen species (ROS). Every method to reduce the movement of radical transfer to surrounding will stop the fire. Blocking the oxygen is also a way since throughout combustion reactions ROS are continuously depleted.
Simply put, the human body's water does not have any role in doing this. Our body has not only fatty parts of cells but also dry materials like hair. And whenever water of our body is in contact with fire around, it will either evaporate and escape or become another ROS source.
I can refer you to a book like Introduction to Fire Dynamics to really grasp what is fire, it is a very crucial topic in terms of our livelihood as well, yet in general, is covered in neither undergraduate nor graduate chemistry.

On an average, a 70-kg man is made up of about 42 L of total water.

28 litres is intracellular water
14 L is found in the extracellular fluid of which
3 L is blood plasma
1 L is the transcellular fluid (cerebrospinal fluid, ocular, pleural, peritoneal and synovial fluids)
10 L is the interstitial fluid (including lymph), which is an aqueous medium surrounding cells.

So the majority (66%) of the body's water is inside the cells. When you're burned, you don't really have the ability to extract the water from the cells.

Cells will lyse (burst) quite quickly at the temperatures reached in a fire. Therefore the water from cells will be "extracted".

Could you provide a reference for these numbers?

@JackAidley Yes, but that water will be "extracted" to the already liquid interstitial fluid.

Probably the simplest, most intuitive answer possible is to directly compare this to a very familiar instance of dousing fire with water.  Did you ever see a team of firefighters spraying massive amounts of water with their fire hose?  The simple answer to why a human being made mostly of water can't put out a good-sized fire with their body is that it takes way more than a single human being worth of water to douse a good-sized fire.

Water puts out fire by coating the fuel and preventing it from getting oxygen. The fact that our bodies have water is irrelevant because we are not the source of fuel. We are the innocent victims of dissipating heat and noxious gasses. That's the simple explanation. In more detail...
How fire works - an overview
Fire is an energetic reaction where electrons transfer from a reducer (electron donor) to an oxidizer (an electron acceptor) in what is called a redox (reduction-oxidation) reaction. The acceptor is typically molecular oxygen, but oxygen is already moderately stable and needs a little "kick" in order to break its bonds and become willing to accept electrons. This kick is provided by heat*, and sufficient heat provides the activation energy necessary for the reaction to take place. As soon as the electrons are transferred, more heat is released, far more than what it took to initiate the reaction (making fire exothermic). Fire invests in heat, with heat.
Fire starts as soon as the reactants become so hot that their reaction rates become high enough to sustain the reaction without the input of any external heat. If the reaction rate is too low, more heat is lost than is produced, and all you get is smoldering. If the rate of heat production is significantly higher than the rate of heat loss, the excess heat manifests itself as a fire. The fire will die if you take away its heat, fuel, or oxygen. Exotic fires can be created with different reactants, giving them reaction rates that vary from barely fast enough to sustain itself, to so fast that the entirety of fuel is used up in a fraction of a second and a high-velocity explosion results. In this answer, I will be assuming a traditional fire. That is, a fire with a hydrocarbon fuel and molecular oxygen.
Humans are not very flammable
With very few exceptions, people just don't burn. We call heat-related injuries burns, but our bodies are not particularly good at sustaining a fire. If you were to hold your hand over a candle, your hand would not burst into flame, but you would be injured by the heat. Fire is only harmful to us if we are near another source of flammable fuel, for example the wood used to burn someone at the stake, or our cotton clothes.
Despite this, there are three reasons fire is harmful to humans:

Fire generates a lot of heat, and most of that heat is dissipated into the environment over quite a distance. Even if we are a small distance from the reaction itself, the heat is intense enough to damage us. The heat hurts, but it is not the most harmful thing fire has to offer.

Fire is not a pure, clean reaction. A perfect traditional fire would take a hydrocarbon, oxygen, and heat, and convert it into pure carbon dioxide, water, and more heat, but fire is not perfect. It burns uncleanly, releasing toxic gasses like carbon monoxide and cyanide.

Finally, fire kicks up particles of burnt material in smoke which, in an ironic twist, coats our lungs with burning, mucosa-irritating ash and prevents us from absorbing oxygen. We may be able to put fire out with water, but fire can put us out with smoke.

When someone is killed by fire, they are typically killed by the smoke inhalation, or by the toxic gases that are released when certain materials burn. Only rarely is it the heat alone which is fatal, in which cases it kills in a few days by damaging our protective epidermis and causing us to dehydrate rapidly and lose almost all our resistance to infection. Unlike plant material, we do not die because we ignite. We die because we need our skin intact and our lungs clean.
Water creates a barrier between fuel and oxygen
There is nothing magic about water that makes it able to put out fires, and its mere presence is of no concern to fire. In fact, traditional fires create water, but in a reaction so energetic that the water is in gas form. There are three reasons water is useful for extinguishing flame:

Water is a liquid, so it will evenly coat the fuel†, restricting access to oxygen.

Water is not flammable, so it will not serve as additional fuel for traditional fires.

Water has a high heat capacity, so it will be harder for the fire to evaporate it.

All of these require the water coat the source of fuel, acting as a barrier between the fuel and the oxygen. The water will absorb heat from the fire, all the while preventing it from gathering more oxygen, until the water can't absorb any more and evaporates. With enough water, the burning material will not have enough heat to get rid of the water and take another "breath". As the heat gets lower, it quickly becomes unable to sustain a reaction, and it rapidly cools.
Even if the fire is so big that the water evaporates before it hits the ground, it will still soak up heat that would otherwise go to increasing the ambient temperature and bringing nearby flammable materials closer to their ignition point. The water doesn't have to be cold, it just has to start out significantly cooler than its boiling point. This will decrease the overall heat, letting any more water put on the fire have a better chance of lasting long enough to coat the fuel.
Our water is nowhere near the fire
Humans do have water, but it is tucked deep inside us. We cannot put out fires near us (or right under us, or on our clothing) because the water doesn't leave our bodies in significant quantities. If we were somehow able to release a deluge of sweat and direct it anywhere we please, we might be able to put out small fires on our own (at the expense of severe dehydration). As it is however, the vast majority of our internal water is not in any position to blocking a fire's access to fuel.
Even if we are injured by the fire, our water is not kept in some big sack where rupture will result in it all being released at once. The water is inside and in between our cells, and if fire damages our cells, the water will simply be released into the already-aqueous interstitial medium. When our skin is sufficiently damaged (to a fatal extent), we will very slowly lose water through our skin. The rate will be high enough that we can die of dehydration in a matter of days, but far, far too low to put out a fire of any size. We will die of the effects of fire before we ever dry up.
* Life survives on the same reaction found in fire, using a hydrocarbon fuel, oxygen, and heat. Enzymes reduce the required energy of this "kick", making it possible to sustain a controlled exothermic reaction at body temperature.
† If the fire in question is an oil fire, water can make things much worse by causing the oil to float.

Human bodies, most living creatures in fact, don’t burn like wood immediately or at all.
Human bodies technically boil.
Sometimes rapidly, or even explosively; other times human bodies “burn” or boil more slowly, progressing from 1-3rd degree burns, which spread with ongoing intensity and exposure.
When I was on active duty, I had to sit through many, many gross pictures and videos of war victims, via bombs, vehicle fires, fuel fires, electrical fires, etc... in order to learn how to provide combat/trauma medical aid to people who have basically been boiled. Some people I knew saw it in person, and it was basically the same as the pictures and videos. Boiled, but burnt crispy on some parts, usually extremities and membranes.
Like steak, or pork.
When grilling, if partially direct exposure to flames which cooks meat, but it’s mostly the fluids inside being heated to a point when the flesh is cooked.

"Human bodies technically boil." Boil in what sense? In the culinary sense of "be cooked by being surrounded by boiling water?" In the physical/chemical sense of "be transformed from a liquid to a gas"? And I'm not sure what you mean  by claiming that meat is cooked by the fluids inside being heated. The whole piece of meat, including both the liquids and the cellular structures that contain them, are heated simultaneously. It's not like, say, boiling penne pasta where the hot water moves freely through the solid pasta and heats it. The water in meat is mostly not mobile.


text:
Does tetraoxidovanadium (VO4+) ion exist?
In my book (NCERT India), it is mentioned that: (page 224)

...$\ce{Mn2O7}$ gives $\ce{HMnO4}$ and $\ce{CrO3}$ gives $\ce{H2CrO4}$ and $\ce{H2Cr2O7}$. $\ce{V2O5}$ is, however, amphoteric though mainly acidic and it gives $\ce{VO4^3–}$ as well as $\ce{VO2+}$ salts. In vanadium there is gradual change from the basic $\ce{V2O3}$ to less basic $\ce{V2O4}$ and to amphoteric $\ce{V2O5}$. $\ce{V2O4}$ dissolves in acids to give $\ce{VO^2+}$ salts. Similarly, $\ce{V2O5}$ reacts with alkalies as well as acids
  to give $\ce{VO4^3-}$ and $\ce{VO4+}$ respectively. The well characterised $\ce{CrO}$ is basic but $\ce{Cr2O3}$ is amphoteric.

The sentence under scrutiny has been boldened. I wonder how the $\ce{VO4+}$ ion can exist.
Going by the usual oxidation numbers, four oxide anions should contribute a total of -8 charge. To balance it, vanadium needs to be in +9 oxidation state. But, vanadium has only five valence electrons $(\ce{[Ar] 3d^3 4s^2})$, so how can it even attain the +9 oxidation state? Though, it may be possible that this ion has a different structure (possibly a peroxide linkage?) that allows it to exist with a +5 oxidation state on vanadium, but I am not sure.
I have read relevant documents on the internet, including the Wikipedia page for the vanadate ion and a Scholar search, but neither yield any relevant results. I can't search it on ChemSpider Structure Search either because I don't know its structure, and a textual search yields only unrelated partial matches.
So, is it true that the $\ce{VO4+}$ ion dost not exist, or have I missed something? Or is it that there is a misprint in the book, and they intended to write another ion in place of $\ce{VO4+}$?

Since the comment by Ivan pointing to a possible misprint got four votes and no answer yet, I'll post it as an answer. In short:  this is a misprint in the book, and the book probably meant $\ce{VO2+}$.
Here's a citation to back it up: Bertrand, G. L.; Stapleton, G. W.; Wulff, C. A.; Hepler, L. G. Thermochemistry of Aqueous Pervanadyl and Vanadyl Ions. Inorg. Chem. 1966, 5 (7), 1283-1284. DOI: 10.1021/ic50041a048.

...it does appear that below $\mathrm{pH} \approx2$ the principal species are the pervanadyl and vanadyl ions represented by $\ce{VO2+(aq)}$ and $\ce{VO^2+(aq)}$ (respectively).

Also worth noting that the formation of the pervanadyl ion is more favourable with a free energy of formation of $\pu{-140.3kcal/mol}$, as opposed to $\pu{-106.7kcal/mole}$ for the vanadyl ion (as mentioned in the source above)


text:
Is this approach to finding density in crystal lattice acceptable?
In answering a question in an elementary chemistry text that asks for the density of gold ($\ce{Au}$) based on crystal dimensions and weight per atom, I found the approach I used was a little different from that of the book and wanted to see if this is reasonable.
Gold crystal cells are face-centered cubes of a given length edge and a given weight per atom. The book reasons that since each of the 8 corner atoms is shared by 8 cells and each of six face atoms is shared by 2 cells, each cell adds $8/8+6/2 = 4$ atoms to a crystal structure - which gives the correct density.
My approach was to count all the atoms in a cube of cells and disregard sharing, since as long as the count was accurate the average number of atoms per cell would be correct in the limit.
For example: in a cube of 8 cells, there are $3^3$ corner atoms and $3\cdot 2^2(2+1)$ face atoms. So the general formula is $(n+1)^3+3n^2(n+1)$ atoms in an $n \times n$ cube. But in $n^3$ cells the ratio $r$ of atoms per cell to cells is
$$r=\lim_{n\to\infty} \frac{(n+1)^3+3n^2(n+1)}{n^3}=4$$

So this also gives the correct density, and my question is whether there is anything wrong with this approach?

(In an earlier version of this question I failed to count all the face atoms; hopefully this is easier to follow.)
[Edited to correct formula.]

Yes of course, if you do the math right, that limes gives the exact same number.
Don't forget to look out for atoms that are completely within the unit cell.


text:
Capillary forces in horizontal plane
In a physical chemistry course, we went to the lab and made a microstamp out of an elastomer. This stamp has gaps of 25 µm in width. When the elastomer had hardened, we placed it on a microscope slide and dropped a single drop of epoxy glue to one side of the stamp. Due to capillary forces, the glue migrated slowly to the other side of the stamp.
In my script, there is an equation to calculate the rising height of a liquid due to capillary forces. Our goal is to calculate the surface tension of the glue, knowing the migration speed, the density and the contact angle.
As our stamp is in the horizontal plane, there is no rising height, so how do I manipulate the formula to calculate the surface tension of the epoxy glue in the horizontal plane?
Here is the formula of my script:
$$
h = \frac{2 \sigma \cos \theta}{\rho g r}
$$
Where $\sigma$ is the surface tension, $\theta$ the contact angle, $\rho$ the density of the liquid, $g$ the gravitational acceleration and $r$ the radius of the gap.
And a little picture of our experiment. The glue migrates from the right to the left.


text:
What are the products formed when phenol is reacted with benzyl bromide?
What are the products formed when phenol is reacted with $\ce{PhCH_2Br}$ (benzyl bromide)?
I can think of two ways in which the reaction would proceed. One would be the formation of an ether through SN2 mechanism. The other is formation of 4-benzylphenol (Friedel–Crafts alkylation)
Which one of the two products is likely to be formed?
If both are formed, is there any preference for one being the major product?

Please specify the solvent, amount, catalyst absence or presence, temperature and then your reasoning considering all of them. If there is still a point for a question, then helping you is possible.

I believe the SN2 mechanism will predominate. The benzyl bromide is not sufficiently electrophilic to perform a Friedel-Crafts Alkylation, as that often requires the use of a metal catalyst.

Hey Alex, the OP didn't say we aren't permitted to use a metal catalyst. In fact, I think saying  that we are performing FC alkylation assumes that we have to use the metal catalyst. So, is there any other reason why you would want to reject the EAS substitution?

Normally to do a Friedel Crafts alkylation a lewis acid or some other means of generating a carbocation is needed. I am also sure that it will be mainly SN2 as long as the system is alkaline or basic. If we were to add aluminium bromide or aluminium chloride things could get interesting as the phenol could coordinate to the aluminium. This will withdraw some electron density from the benzene ring and thus make the phenol less reactive.


text:
What is first protonated aldehyde or lactone?
In a six membered lactone with carbaldehyde in beta position, which oxygen gets protonated first? the lactone or the aldehyde?
(Using concentrated HCl)

You should assume both will be protonated and deprotonated reasonably quickly. If you're trying to get good selectivity between the two, you're out of luck.


text:
Basicity comparison of ammonia derivatives (and guanidine)

Arrange the following in decreasing order of basicity:

$\ce{N(CH3)3}$
Guanidine
$\ce{N(C6H5)3}$
$\ce{NH(CH3)2}$

My answer - 2 > 1 > 4 > 3
Option '2' due to equivalent resonance structures on protonation. Then option
 '1' due to thrice +I effect. Followed by option '4' having +I effect twice which causes the lone pair electrons to be repelled from the nitrogen atom. Finally option '3', because lone pair electrons are delocalised on three rings.
Is my logic for the question correct or not? I am not sure if my answer is correct.

So sorry for the option 4 .I have now edited that.

Then your logic is just fine.

In triphenyl amine (3), the lone pair of nitrogen is in conjugation with phenyl groups and thus would be least available for protonation.
Due to resonance structures of guanidine (2), it should be the strongest base.
Although number 4 doesn't exist as it is written. Let's suppose it is $\ce{N(CH3)2}$ with a negative charge (the only possible way to have two methyl groups on a nitrogen).
If so, it is more basic than guanidine, because charged compounds are generally more reactive than neutral ones and will attract $\ce{H+}$ more quickly.
So if number 4 is charged, then the basicity will decrease in order
4 > 2 > 1 > 3

Thanks for the answer .The option 4 was $\ce{NH(CH3)2}$ but anyways thanks for the knowledge that if nitrogen was negative then it would be more basic than Guanidine

The comparison of the basicity of amines can be quantitatively done by comparing the $\mathrm{p}K_\mathrm{a}$ values of their conjugate acids because $\mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b} = \mathrm{p}K_\mathrm{w} = 14$.  Accoedingly, the higher the $\mathrm{p}K_\mathrm{a}$ of amine's conjugate acid, the stronger the basicity of amine:

$\ce{[NH(CH3)3]+}$: $\mathrm{p}K_\mathrm{a}=9.8$ - any $3^\circ$ amine is weaker than its corresponding  $2^\circ$ amine (4) due to the steric hindrance.
$\ce{[H2N=C(NH2)2]+}$: $\mathrm{p}K_\mathrm{a}=13.6$ - strongest base
$\ce{[NH(C6H5)3]+}$: $\mathrm{p}K_\mathrm{a}\lt 0.78$ - weakest base ($\mathrm{p}K_\mathrm{a}$ of $\ce{[NH2(C6H5)2]+}$ is $0.78$)
$\ce{[NH2(CH3)2]+}$: $\mathrm{p}K_\mathrm{a}=10.7$

Based on the relevant $\mathrm{p}K_\mathrm{a}$ values, the order of the basicity of given amines should be 2 > 4 > 1 > 3.

How to solve this when you don't know the values of pKa?

How to compare between 4 and 1

It is a rule of thumb that the strength of the basicity between $2^\circ$ and $3^\circ$  amines with same substitution group is given as   $2^\circ \gt 3^\circ$. For instance, $\ce {NH(Et)2} \gt \ce {N(Et)3}$.


text:
Does expansion of an ideal gas not always represent an increase in entropy of the system?

Expansion of an ideal gas not always represent an increase in entropy of the system. Statement is True or False?

What I thought is that whenever an ideal gas expands the randomness of gas molecules increases. So, entropy must increase.

What if, during the expansion, the temperature decreases?

Entropy does not depend on just the volume the gas has available.  It also depends on the energy the gas has to exploit that available volume.
Suppose you expand the gas by pushing against a piston.  The gas is doing work to "earn" its extra volume, so even as volume increases the gas is expending energy.  Thus unless heat flow into the gas is allowed the gas will cool.  This loss in energy tends to decrease random motion, and thus tends to decrease entropy, even as having more volume would have the opposite effect.  We have a "tug of war" here.
In an idealized version of such an expansion,  where we aallow no heat flow into or out of the gas (adiabatic) and we maintain equilibrium with no "losses" due to friction ("reversible") then the loss of energy to doing work must effectively balance the gain in volume, leaving zero entropy change.  With no heat flows through the insulation and no heat generation due to friction in this idealized scenario, there must be zero entropy change in both the gas and the surroundings.
The lesson:  Any process (such as expansion  of an ideal gas) that is  reversible (passing through equilibrium states) and adiabatic (no heat flow) has zero entropy change.


text:
Assay of stainless steel
I frequently need to discriminate between 304 and 316 stainless steel which cannot be done by eye.
304 SS has 18% chromium and 8% nickel
316 SS has 16% chromium, 10% nickel and 2% molybdenum
How can I do a chemical assay to discriminate between them and ideally identify their precise composition to 1% or better accuracy? I assume the basic procedure is to create a solution, then do a titration of some kind.
If the first step in your answer is... "dissolve the sample in acid," be aware that stainless steel is pretty resistant to most acids and I need a procedure that will work within 24 hours.
Also, please no answers like "use an xrf machine" or some surface detection technology. I need a chemical process that assays the whole sample, not just its surface.

Why is your supplier not marking them prominently and correctly? Note that 316 is not magnetic, although 304 is not guaranteed to be magnetic. As for assaying the 'whole sample' - how do you plan to do that?

On what order of magnitude size is that "whole sample" piece? 1g? 50g?

316 is much more likely to show ferromagnetism than 304. Cold worked 316 will be about half as magnetic as carbon steel .

Why do you think a blank surface of your material would have a different composition than the interior? Saw off a piece and analyse the cut if you think that would be necessary.

Digest a sample of the metal in aqua regia, I have found that the gauze in a kitchen sieve can be digested in a very short time with a mixture of 16 M nitric acid and concentrated hydrochloric acid.
Next you should dilute it with water and filter it to remove any particles.
I would use either flame AAS or ICPOES to measure the metals after dilution to a suitable concentration in plastic ware. I would want to measure the Fe, Ni, Cr and Mo in the sample. The Fe to Ni and Fe to Cr ratio may help you distingish between a lot of grades of stainless steel.
One alternative for iron is to use 2,2'-bipy and hydroxylamine to form the deep red iron(II) tris bipy complex.
This can be measured with a UV / vis machine in a 1 cm cuvettes, you will have to make a series of solutions in volumetric flasks. You make 0, 1, 2, 5, 10 and 20 ppm iron by adding a 1000 ppm standard to the flasks, then add a solution of 2,2'-bipy in water, the hydroxylamine and then dilute to the line. Mix them well and then measure the red colour with the UV / vis machine.
I am sure that there are colourmetric methods for Mo and Cr but sadly I do not know them.

I am looking for a chemical solution. I do not want to buy a big complicated machine.

I did some more searching around and found that there is a standard spot test for molybdenum in ASTM publication 550 that can be used on stainless steels. It is not ideal in that it does not determine the proportions of chromium and nickel, but just tells whether molybdenum is present:
a) Place equal drops of concentrated hydrochloric acid and concentrated nitric acid on the metal surface. Allow 3 to 4 minutes for reaction and transfer 2 drops to a spot plate.
b) Add 2 drops of 10% potassium thiocyanate.
c) Add 25% sodium thiosulfate drop-wise stirring until the red color begins to disappear. A violet color will appear if molybdenum is present.

Asking this question, I assume you are not a chemist, and neither have a lab or a chemist available in the place you want to do this, right?
Buy yourself a handheld XRF analyser. They look like a toy pistol, come nicely coloured and with touchscreen displays, and there are a dozen or so companies building them. Google gives you many hits, although no prizes. I'll guess you can get one in the 2-3000 € region? Definitely cheaper than hiring a chemist, or starting to fiddle around with acids and other stuff yourself, if you need to do it on a daily basis, and professionally.
(The machines on wikipedia https://en.wikipedia.org/wiki/X-ray_fluorescence look very intimidating, google has much nicer pics https://www.google.com/search?q=handheld+xrf . ;-)

Sorry, didn't see the last line in your question that you don't want XRF. ;-) Still, instead of dissolving a sizeable sample pice, can't you just file it to a fine powder and do XRF on it then? Or just file off anything that might be inhomogeneous surface before holding the XRF on it?

I want to analyze the sample, not scrapings off the surface of the sample. I thought I made that clear in my question. I want a chemical assay solution, not xrf/spectroscopic solutions.

@ShakaBoom I noticed. Hence my above comment. :-/


text:
Polluting agents generated by road traffic in a city
I want to measure the air quality according to road traffic in a city.
What are the polluting agents (generated by fuel combustion) strictly related and generated by road vehicles that is worth measuring?
Among $\ce{CO}$, $\ce{CO2}$, $\ce{NO_$x$}$, PM2.5, PM10, $\ce{NH3}$, $\ce{O3}$, $\ce{SO2}$ which are the most significative? Should I add or remove any agent?

You might also want consider the metals which appear in road dust, I recall having seen some work on neutron activation anaylsis of dusts collected near roads. The rubber tyres are likely to contain some metals such as zinc.

That's a homework assignment, isn't it?

No, I am asking for a real project. There are different electronic devices capable of measuring different agents and I want understand what I should look for.

Why has it been tagged as opinion based?

@FrancescoBoi Because it's very arbitrary what pollutant is significant. For whom? People who are astmathic? Pregnant? Pedestrians or people in nearby flats? Etc.

Even for pollution coming from road vehicles like cars and motorikes?

There are cities which periodically monitor the air pollution such as Seoul to inform their inhabitants: what agents do they measure?

Ask Seoul. Or the internet, more conveniently.

Thank you very much. so useful, never thought about that.

You might also want consider the metals which appear in road dust, I recall having seen some work on neutron activation anaylsis of dusts collected near roads. The rubber tyres are likely to contain some metals. Cars will be likely to release metals in other ways. I would consider the dust from the operation of disk brakes, the wear on tyres, fuel additives and also corrosion of bodywork and engine parts. I know that zinc and cadmium plated parts can release both ionic and particulate forms of these metals.
One paper on the subject was by Siddique et. al. im Radiochimica Acta, 2013, volume 101, pages 273-277.
It is interesting that lead levels in the air have fallen since leaded petrol was banned in the UK, but some workers have reported finding platinum group metals in the air near to roads.
Using ICPMS Simitchiev et. al, in TALANTA, 2008, volume 77, pages 889-896 found Pt, Rh and Pd in road dust.


text:
Would 99,9% Isopropyl alcohol damage plastic of Bratz doll boxes?
I am a Bratz doll collector for quite some time now. I collect them boxed, most of the boxes and packagings are mint enough for me but I also have a few older ones which are quite dirty.
I’ve read some things online that cleaning with Isopropyl alcohol will take off most of the dirt and at the same time it will desinfect it as well. It seems perfect!
But since I invested a lot of money in my collection over the years, I would like to know for sure if Isopropyl alcohol 99,9% could harm my plastic Bratz doll boxes in any way?
I enclosed a picture of a Bratz doll box, if needed.
Thanks in advance for answering my question! =)

I suspect you are going to have to try it. Without knowing eactly what the packaging is made of it, is difficult to answer.

Though it probably wont harm it, cleaning collectibles or antiques usually reduces resale value. Only do this is it's just a keepsake.

Depends on the materials your collectibles are made of. Use a 33% v/v solution for a relatively harmless wash. Two parts of water to one part of Isopropyl alcohol that you have


text:
Why can't beryllium react with hydrogen gas directly to give hydride?
I read in a textbook that all alkali metal and alkaline earth metal can react with hydrogen directly to give hydride except for beryllium. I want to know the reason why because I know that $\ce{BeH2}$ exists in a polymeric state.

See this paper: https://pubs.acs.org/doi/pdf/10.1021/ja00078a057

@NilayGhosh this q never got an answer that could stand up to criticism. Maybe summarize key findings from the reference in an answer that will work?

@OscarLanzi Please go ahead and make into an answer. I can see you had deleted your answer 4 years ago due to lack of reference(?)

That paper shows evidence for BeH, BeH2, and several gaseous oligomers under heroic conditions.  BeH is endothermic deltaG = 67 kcal. that combined with the strong bond energies of H2 and crystal lattice of Ge direct reaction would be kinetically and thermally difficult.  The answer seems to be they do react if all the activation energies are removed at a low temperature to prevent decomposition.

As described in Ref. 1, the proposed reaction is endothermic and thus unlikely to be favored. In contrast the other alkaline earth metals give exothermic reactions that ought to proceed at least at low enough temperature. So then the question becomes why does beryllium give an endothermic reaction?
One clue lies in the unusually high melting and boiling points of this metal versus other alkaline-earth metals, which do react directly with hydrogen. From the respective Wikipedia articles we render the following melting and boiling points for the first three alkaline earth metals:
Beryllium: 1287°C, 2469°C
Magnesium: 650°C, 1091°C
Calcium: 842°C, 1484°C
This implies that metallic Beryllium has much stronger bonds than other metallic alkaline earth elements, and a reaction of beryllium metal with hydrogen is thereby hindered by those stronger elemental bonding.
In this respect beryllium resembles not the other alkaline earth metals or even post-transition metals in Group 12, but higher-melting transition metals such as titanium or iron. The lack of a reaction between beryllium and hydrogen thereby fits a more general pattern: high-melting metals, which require breaking strong bonds to form salt-like compounds, do not do so with hydrogen. They either fail to react directly with hydrogen at all (unless they are somehow atomized first, as in Ref. 1), like beryllium or iron; or else they instead form interstitial hydrides that allow most of the metal-metal bonding to be preserved like titanium.
Beryllium is also not alone in being atypically hard to melt or boil. The same thing happens with boron versus the rest of Group 13 and carbon versus the rest of Group 14. In all these cases the small size of the electronic core, which is only a $1s$ subshell, allows strong and deep overlap in the delocalized bonds between the atoms, whereas the larger cores of the later-period congeners limits overlap and makes the heavier elements potentially more reactive.
References

Thomas J. Tague Jr. and Lester Andrews (1993). "Reactions of beryllium atoms with hydrogen: Matrix infrared spectra of novel product molecules". J. Am. Chem. Soc. 115, 25, 12111–12116.
https://doi.org/10.1021/ja00078a057

It really is fairly simple the product is not particularly stable and reacting from the elements has a large activation energy to supply Ge and H atoms. Ge and H atoms react nicely at temperatures near 0 K, the activation energy is supplied elsewhere.

More is involved than activation energy. Breaking the strong bonds implied by the high melting point is also a thermodynamic barrier.

That is what the activation energy is composed of. Add in a strong H-H bond and finally an endothermic compound why does anyone suspect this reaction to happen at ambient temperatures


text:
Can a copper(II) chloride be added to liquified butane?
What would happen to either chemical if the two were mixed together, specifically what would happen if the dust was mixed in when the Butane was in an aqueous state.
I have a limited knowledge of chemistry so I apologize if my terms are incorrect.

Copper chloride will just drop to the bottom and remain there.

"Butane in an aqueous state".  This has got to be good.

If we had butane under pressure in a vessel as liquid and this was to be in contact with a solution of copper chloride in water then I think that nothing will happen. The butane and the water will form two seporate layers.
However if we were to add a copper extraction agent to the butane such as LIX-860 (a phenolic oxime reagent) then it would be possible to convert some of the copper from aqueous copper chloride into an organic soluble copper complex. Then the copper would be extracted into the butane layer.
Aother option would be to add copper chloride powder to the butane solution of LIX-860, I think that this could react but the reaction would be slower. Also I think that the hydrogen chloride formed as side product from the reaction would have a greater inhibiting effect on the reaction.
Another option would be to use a lipophilic chloride ionic liquid such as aliquat 336 to dissolve the copper chloride. This could react (slowly) with powdered copper(II) chloride to form 2(R4N+).[CuCl4]2- which would have some solubility in liquid butane. From my own experience of doing solvent extraction I know that aliquat 336 in alkanes is not a great system. When the metal loading becomes moderate the organic phase can separate into the organic solvent (diluent) and a liquid metal bearing layer.
In short I hold the view that pure butane will not be able to dissolve copper chloride, but with a suitable additive it will be possible to get a solution of copper in butane. The copper will be there as a copper complex.
I think that if we were to place water in a sealed container and put pure butane gas in the head space that the concentration of the butane in the aqueous layer will be very low. The copper chloride is likely to stay in the aqueous layer as hydrated copper ions and chloride anions if the concentration is low. If the concentration is high then it will be able to form chlorocomplexes of copper.

phenanthroline is a good reagent for complexing Copper salts and getting them into the organic phase.


text:
White compound, insoluble in water, turns pink and then white again on adding water

Which white, ionic compound(s) that is insoluble in water turns pink when you add only a little of water but white again when you add more water?

I know that the cation is one of the following:
$$\ce{Mg^2+, Ca^2+, Sr^2+, Ba^2+, Fe^3+, Ni^2+, Co^2+, Zn^2+, Al^3+, or NH4^+}$$
I was physically examining such a compound and determined it does not change the color of litmus paper (red or blue). I thought that it could be zinc carbonate because it also precipitates hydroxide (when then compound is added to dilute nitric acid).
EDIT: I found out that the compound is lithium carbonate, and it just took a long time to dissolve. How would lithium carbonate change color like this?

Reversible color change reminds me immediately complexation, and in this case maybe a complex only stable when there is very less amount of water around. However, if it is really insoluble as stated I can't know.
And the question is a riddle like one, do you have any effort on solving this question, if so edit and share it after the question, please.

I landed here because i am looking for some ingredient of ash (Wood fire, possibly some metal) that turns rose/pink if a fresh twig smoulders on top (my guess was that this is due to water vapor), but stays ash-colored in the rain. So i hope somebody will answer this.

I found out it was lithium carbonate. Why did this happen?

@Bob Where'd you get the answer from? The sources may help us.

I was told. The compound was known by someone else.

https://chemistry.stackexchange.com/questions/96539/white-compound-insoluble-in-water-turns-pink-and-then-white-again-on-adding-wa?rq=1


text:
Why don't I get the same value of percentage ionic character of a particular molecule from different equations?
About the ionic character of a polar covalent compound Pauling gave two equations as

[1-$e^{.25(x_a - x_b)}] $%

[18$(x_a-x_b)^{1.4}$]%
Hanary and Smith gave the equation

[$16(x_a-x_b)+3.5(x_a-x_b)^2$]%

Where $x_a$ and $x_b$ stands for the electronegativity of $a$ and $b$ atom in bond $a-b$.
Now if I put $x_a-x_b = 2,$ The 1st, 2nd & 3rd equations respectively give the value as
     -.648%, 47.5% & 46%. What is the reason behind these  different values of percentage ionic character of the same molecule? At where       am I wrong?

You are wrong in thinking that percentage of ionic character  is a real observable thing, like mass or energy. It isn't. (Neither is electronegativity, BTW.)

@IvanNeretin then how Pauling gave the equations and how do we have an electronegativity table?

When Pauling gave the equations, he knew their limitations. As for the electronegativity table, we surely have not one, but many such tables, all subtly different.


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Can you ignite the thermite reaction in a solution?
As far as I know thermite is a typical redox reaction where $\ce{Al + Fe^{3+} -> Fe + Al^{3+}}$ happens. I guess the activation energy is very high, because you need to split the $\ce{Fe_2O_3}$ into $\ce{Fe^{3+}}$ and $\ce{O^{2-}}$. Probably you need molten aluminium too at $\pu{660°C}$ to ignite it.
I was wondering whether it is possible to ignite thermite in a water solution. Since $\ce{O^{2-} + H_2O -> 2OH^-}$, I guess a very acidic solution could help to start the reaction, since $\ce{OH^- + H^+ <=> H_2O}$. On the other hand it could mean a $\ce{2Al + 6H^{+} \rightarrow 2Al^{3+} + 3H_2\uparrow}$ reaction too.
Any idea whether thermite can be ignited this way, for example with a drop of concentrated acid? Or will it just result hydrogen gas, because that reaction has a lot lower activation energy?

Thermite I usually activated by extreme heat not chemically. And it is used in underwater welding, so yes it can be done is a liquid.

Why do you think igniting water solution might work?