text:
Is it possible to make synthetic oil without fossil fuel?
If human ran out of crude oil and all other fossil fuel, are we still able to create synthetic oil? I've searched online about the raw material of synthetic oil, but I'm still not sure if none of the raw materials is from crude oil.
Also is there any synthetic oil with viscosity less than 10cSt?

Fischer–Tropsch process

Without fossils not. The processes as in FT  still requires  carbon monoxide / hydrogen from fossils. Its value its contingent to situations (embargoes, poor quality sources, ,,,,,). In other words of technological relevance but still requires energy and material from fossils.

@Alchimista Wood, water and solar power and here it comes :D All "fossil fuels" were made by natural processes from actual fossils.

@Mithoron. What in my comment makes to think I am not aware of it?

@Alchimista I mean you can make fuel not from "fresh"organic matter - it's not "fossil".

@Mithoron. Sure. One can make charcoal then FT :) But I took the Q on practical feasibility rather than chemical one.  Therefore I was a bit critic on the given answer . Without fossil fuels a lot of chemistry stops. We are just thinking at fuels because the rest is a corollary :)

Yes, it is being theorized and it's being done:
Example A:
https://www.researchgate.net/publication/264810861_Engineering_cyanobacteria_for_production_of_a_cyclic_hydrocarbon_from_CO2_and_H2O
Example B:
https://www.sciencedirect.com/science/article/pii/S2211926416304982
Example B is more theoretical with economic considerations, and example A is more methodical. In these papers, we see cyanobacteria being used to make the aromatic hydrocarbon limonene with the limonene synthase gene derived from a plant.
As for can we make synthetic oil that is chemically proportional to fossil fuel, Probably not ... but by adjusting the ratios of aromatic hydrocarbons, we can measure some useful properties like the heat value of the fuel for which limonene has a heat value very close diesel fuel and the density is close as well as you can see if you read the aforementioned Life Cycle Analysis paper. If you want to explore the chemical composition of fossil fuel then you need to do some spectroscopy work.
But fossil fuel isn't perfect. There is residual incombustible material, and perhaps by adjusting the proportion of combustible aromatic hydrocarbons in fossil fuel maybe we can get more energy from fossil fuel. This is also a function of the funding that these research projects get and also industrial support which is also a function of money ... which is why you need high yield in these reactions, hence genetically engineered cyanobacteria.

Ok is limone counts for fuel at least we will have polluted towns but smelling fresh. The problem in your answer is that every combustible can be called fuel, then.  Tree cells are of course doing so. And almost every organic chemist working at the bench. Tough in the latter case he/she uses energy likely from fossils as well as solvents and reactants.


text:
When two negatively charged molecules are hydrogen bonding with each other, why do the charges increase?
Context
[EDIT: As some commentators have suggested, I  should conclude that the energy minimize feature is not intended for the purpose of Hydrogen-bonding, and primarily for refining the modelling of molecules.]
Using the Chem3D automatic Huckel charge calculator, I was able to calculate the charges of the atoms in the citrate and borate ion.
Individually, the borate ion's oxygen atoms have a small negative charge, i.e. -0.07. Individually, the citrate ion's oxygen atoms have a large negative charge, i.e. -0.56.
I use the energy minimize function to make the two hydrogen bond. I then do the charge calculation, and I found that some of the citrate's oxygens' charges increased to ~-0.67, while some of them decreased, and the borate ion's increased to ~-0.5. The hydrogen atoms do increase positive charge.

I actually noticed another particular feature, the oxygen atom of the hydroxyl involved in both hbond donating and accepting has a positive charge now. The other charges remain as I've explained above. The highlighted yellow atom was just an oxygen atom that I clicked.
Question
For my own explanation, I would suggest that the hydrogen bonding is the cause of the charges being induced and increased. Yet, due to the size of the molecules, only two or four atoms seem to be involved in the hydrogen bonding. Then, why should the OTHER atoms still increase their charge? The atoms not involved in hydrogen bonding between the two molecules?
Would it be a feature of the calculator, or would this be the scenario in real life?
I am interested in how different ions are compatible with borate in terms of hydrogen bonding and screening, based on their charges. This will aid my experiments.

How are you making them hydrogen bond? Which hydrogen on which molecule are you making bond to which atom on which other molecule. You should edit the question to specify because it makes a difference in how things are set up.

@R.M. Updated, I am making them hydrogen bond using the Energy Minimize Function.  It automatically aligns the molecules to the greatest stability possible.

As a rule of thumb, I wouldn't trust anything from the 'energy minimise function' on Chem3D - it uses a very low level of calculation, and is prone to converging local but not global minima. You can interface GAMESS (free) to run with Chem3D, which offers more... computationally valuable calculations

@NotEvans. Thank you, yet I am doing this at a high-school level. Could you please explain this further? What is the difference between local and global minima? Do you mean that it is inefficient to use the function for the simulation of more than a single molecule?

Dear Guna, description of "non-covalent" interactions (hydrogen bonds, van der Waals and London forces) will be rather poor unless you are using a computationally expensive method (which you aren't) or a force field, that has been specifically designed to consider them (which I doubt).
This means, the energy, and the geometry will be most likely wrong - in some cases the geometry wil be significantly wrong (>0.1 A).
As suggested above, you can use GAMESS, or psi4 (www.psicode.org), to get a more accurate electron density and structure. I'd suggest the BLYP-D3M(BJ) functional as a start.

@tetrahydrofuran I was able to test charges with GAMESS.  I don't think my laptop is able to handle simulating two molecules together, with GAMESS.

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text:
Why does ammonium chloride form white crystals?
Why does ammonium chloride ($\ce{NH4Cl}$) form white crystals on top of a test tube when heated?

I suppose you are not asking why crystalline $\ce{NH4Cl}$ is white.
Ammonium chloride decomposes upon heat
$$\ce{NH4Cl(s) ->[\Delta] NH3(g) + HCl(g)}$$
The top of the test tube is cold, and they recombine to generate the original salt.
$$\ce{NH3(g) + HCl(g) -> NH4Cl(s)}$$
This is the white crystal you see.

RE: "I suppose you are not asking why crystalline $\ce{NH4Cl}$ is white." // That is the part that you didn't explain... So why white?

@MaxW Because there are no chromophores (which account for the color in organic dyes) or $d-d$ $f-f$ transitions (which account for the color in metal aqua ions for transitional metals) or charge-transfer transitions (which account for the color in e.g. $\ce{MnO4^-}$). These things simply can't occur in such simple structure, and that's why simple ions of the main group elements and a molecule simple as $\ce{NH4^+}$ does not have color, even when hydrolyzed.

Almost there... but it does have a "color". It is white. Why is it not clear like a pane of glass?

@MaxW It can be if it is pure and compose of a single crystal. However when it is composed of huge amount of very small crystals they will show white color. It is the same for $\ce{NaCl}$ crystal carefully grown in its saturated solution vs. $\ce{NaCl}$ powder.

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text:
Reversed trends in stability of transition metal ions with respect to oxidation state
Why is titanium more stable with a lower oxidation state, but vanadium more stable with a larger oxidation state?
More specifically, why is $\ce{Ti^{+}}$ more stable than $\ce{Ti^{3+}}$, but $\ce{V^{2+}}$ is less stable than $\ce{V^{4+}}$ in aqueous solution?
Both $\ce{Ti^{+}}$ ($\mathrm{3d^{2}4s^{1}}$) and $\ce{V^{2+}}$ ($\mathrm{3d^{3}4s^{0}}$) have a half filled $t^3_{\mathrm{2g}}$ orbital configuration which imparts stability during the formation of aqua complexes, assuming the $\mathrm{4s}$ electron of $\ce{Ti^{+}}$ can go to the $\mathrm{3d}$ sub-shell.
Is this discrepancy related to the charge/size ratio, which is larger in $\ce{Ti^{3+}}$ than in $\ce{Ti^+}$?

Ti+ is not a thing at all.


text:
Stoichiometry of incomplete combustion

Methane is burned with air in a continuous steady-state combustion reactor to yield a mixture of carbon monoxide, carbon dioxide and water.
The feed to the reactor contains: $\pu{7.8 mole \% }\ce{CH4}$, $\pu{19.4 mole \% } \ce{O2}$, $\pu{72.8 mole \% } \ce{N2}$. The conversion of methane is $90\%$ and the gas leaving the reactor contains a ratio of $\pu{8 mol}\,\ce{CO2}/\pu{1 mol}\,\ce{CO}$.

My attempt:

$\ce{2 CH4 + 3 O2 -> 2 CO + 4 H2O}$

$\ce{CH4 + 2 O2 —> CO2 + 2 H2O}$

Both of these reactions have $90\%$ conversion.
From the ratio, can I say that approximately $88.88\%$ of the methane in the feed produces $\ce{CO2}$ while $11.11\%$ produces $\ce{CO}$?

Complicated little rascal!
What we know:
1) The N2 is inert.
2) The oxygen is in excess.
3) 90% of the 7.8 mole % CH4 reacts; this is 7.02 mole % CH4.
4) This gives 8 moles CO2 per mole of CO, or 88.89% CO2 plus 11.11% CO.
The resultant output contains CO2: 7.02 mole % of the original CH4 x 88.89% = 6.24 mole % CO2.
The resultant output contains CO: 7.02 mole % of the original CH4 x 11.11% = 0.78 mole % CO.
The resultant output also contains 10% unreacted CH4. The 90% conversion refers to the starting material, not to the products. (Yield % would refer to one product.)
So the stoichiometry is 80% of the methane goes to CO2, 10% goes to CO, and 10% is unreacted.


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For two compounds that are symmetrical, why does the more compact one have a higher melting point?
In this post, I refer to two molecules B (Di-tert-butyl ether) and C (Dibutyl ether).

Wikipedia: Di-tert-butyl ether (B)
b.p.  107.2 C
m.p. - 61   C

ChemSpider: Dibutyl ether (C)
b.p.  142 C
m.p. - 96 C

I understand that Compound C has a higher boiling point than compound B, because of its greater surface area and therefore stronger intermolecular VDWs.
I don't understand, however, why Compound B has a higher melting point than compound C. A lecturer told me that this is due to compound B being more compact and symmetrical. I think that I understand the concept that some structures pack into a crystalline lattice better than others, such that the structure of the packing influences the strength of intermolecular forces within the solid, therefore affecting the melting point.
However, it seems to me that both B and C above are equally symmetrical in at least one conformation. Beyond that, C has conformations of greater surface area than B, which should further strengthen its intermolecular bonds in a solid state. I would expect the melting point of C to be higher than B; yet this is not the case.
I feel like the answer may lie in the fact that C has many more conformers than B. Even so, I can't seem to draw a link anywhere. Shouldn't the solid formed take the most stable crystalline structure, much like how water tends to just one of many crystal structures?

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I comment because I can't give you solid answer. The teacher refer to a kind of compactness and if you let me say so "sphericity". This reduces surface forces but has a positive effect on paking. Most important is the conformational spread of the linear chain. They bring an entropic term that favours melting and solubility. In other words in your solid there is potential for the molecule to be not only in the long confirmation sketched. Even if they adopted it, to get there they have already got energy at the expense of the wdw forces. Ergo the latter are less effective in binding the solid

Melting is a pretty tricky thing, unlike boiling. You can't really compare it like that. True, one would expect higher melting point in a molecule having central symmetry, but your molecules don't have it, and other kinds of symmetry are less relevant. As for the number of conformers, it hardly matters much, since the crystal is typically built of only one conformer.

You basically see it in DG = DH - TDS, if DS is big enough.

@Alchimista Thank you, your comment has been very helpful! I can at least kind of grasp the intuition behind it now.

@IvanNeretin Could you qualify why melting is trickier than boiling? I can immediately imagine that that is true, but some elaboration will help me learn. Why only central symmetry? Why not bilateral? It seems like Alchimista has an answer based rotational isomerism, but in contrast you say that that hardly matters. Could you elaborate on that?

The crystal is built by one conformer .But at expense of energy to overcome the entropy term . It packs but less strongly than reasonably comparable molecule.  Like in this case.  Obviously adding a substituent other than alkyl will make a different scenario and we won't argue why mp bp are so.

The terms used by your teacher are quite in line with at least a qualitative explanation.
I really suspect that there are various ways to express compactness and sort of "sphericity" of molecules. I am not a specialist so forgive me for being imprecise on this.
However, if we forget a rigorous classification in symmetry group, a potato is more spherical symmetrical than a banana.
In such a sense the ramified isomer is definitely more  symmetric and "round" than its linear counterpart.
This is even more evident if all conformers are taken into account: in the linear molecule a chain end can even turn toward the center while the opposite being extended.
Now this does not necessarily explain the observed behaviour as for the more spherical molecule can pack easily but at expense of VdW forces that depend on molecular surface.
However, it relates to the fact that the linear isomer has a number of conformers much higher than the ramified one (rotation about 4 bonds versus 1 for each side of the ether)*** and results in the first having a bigger conformational entropic term.
Although the crystals will generally contain one isomer, the crystal formation take place at expense of the intramolecular forces. As such, when keeping entalpy terms comparable, bigger is the entropy term then less stronger is the binding.
$\Delta G = \Delta H - T\Delta S$,
For comparable enthalpies and very different conformational entropy terms, than the melting occurs at lower T for the entropy rich molecule, i.e. that one for wich more conformations become accessible.
This obviously is not the case for the liquid to gas transition, as the conformational S term is similar in both phases and negligible faced to the gas entropic content.
***Conformers due to the rotation around the bonds in the terbutyl groups do not really alter the shape of the molecules (connection with the compactness mentioned above). More rigorously at each side we should have (calling x the number of conformers around a bond) that the number of conformations is circa $x^4$ and about $3x^2$ for the linear and the branched molecules, respectively.
Additional note: conformational entropic terms are crucial to many phenomena in polymers and proteins (solubility, crystallization, protein folding/unfolding, etc).
A treatment of this requires statistics that I have forgotten. So take my numerical estimated about conformers with a grain of salt.  I am satisfied if I did convey the message.
Add. Note 2. Another useful example is, for instance, a stiff and insoluble chain such as polythiophene.
It is insoluble as for electron delocalisation makes it flat and rigid. Once side substituted with alkyl chains, the entropic term brought in by the flexible chain is the main responsible for the acquired solubility.


text:
Lab Methods for Determination of Phosphoric Acid Concentration
Dear stack exchange,
For my next lab session we have been tasked to come up with two different methods for determining the concentration of an unknown sample of phosphoric acid.

Design at least two different experimental methods to determine the
  concentration of the phosphoric acid solution using the materials and
  equipment provided. (NB: two different volumetric titrations are not two
  different methods.)

For reference, the "materials and equipment provided" are (along with the standard burettes, pipettes, etc.):

A sample of the phosphoric acid solution
  A selection of pH indicators:
  - Bromocresol green (range 3.8-5.4)
  - Phenolphthalein (range 8.0-10.0)
  - Thymolphthalein (range 9.3-10.5)
  A standardised NaOH (approximately 0.1 M) solution
  A pH meter

One obvious method is an acid-base titration, which is fine, but I have no idea what to do as a second method since we cannot do two different volumetric titrations - perhaps there is some finnicky definition of "volumetric" which means there is another type of titration we can do? It seems as though it has to be something along those lines based on the provided equipment.
Many thanks!

Specific gravity would be a useful technique; perhaps not as accurate as titration. Weigh 50 mL of the unknown and check vs tables in the CRC Handbook.
Do a reaction: add weighed amounts of NaHCO3 (slowly, carefully) until the foaming stops. I think this will just neutralize 2 of the 3 acidic protons on H3PO4. Knowing how much NaHCO3 reacts, you can determine how much H3PO4 there was.
Evaporation to "dryness": in a porcelain dish, weigh a decent amount of the unknown and heat at about 110C for an hour or so. The liquid at the end will be highly concentrated, but perhaps not 100% H3PO4. Check with a known concentration of H3PO4.
If you can get the apparatus, you could check conductivity. You could even do it yourself (roughly) with a multimeter and stainless steel wire (of course, platinum wire is preferable).
You could try to determine viscosity; check vs known standards. Very rough for analysis.
You could offer the lab director $10 for the correct answer.

I think the $10 might be the way forward ;) thank you very much for the extent of the reply! Should be able to find the equipment for at least one of those :D

You can add a single dose of NaOH to a sample of the acid and measure the pH with the pH meter. You can easily calculate the charge on a mole of phosphate at a given pH and as the moles of NaOH you added is equal to the moles of phosphate multiplied by the charge change per mole you can get the number of moles of phosphate.

Ooh thank you! Hadn't even thought of that, would be an interesting write-up


text:
Can carbon monoxide from a floor-heater in a poorly ventilated room eventually be ignited by the fire and explode?
I have often wondered about carbon monoxide accidents caused by free-standing heaters burning in a poorly ventilated room causing fatalities. I am aware that CO is explosive between 12 and 75% volume air and wonder why it doesn't eventually ignite from the flames or glow present when the CO level reaches 12%? (This never seems to happen). Or does the decreasing oxygen content in the air due to the combustion cause the fire to die out before the critical level of 12% is reached?

Your last sentence seems reasonable.

I would be much more concerned about dying in CO, it takes less that required for an explosion. If you have any" CO" in your room, you will get a headache.

In a completely unventilated room, with only CO being produced, to get 12% CO implies that you have used up 6% of the O2 (out of ~21% in the air). The remaining gas mix besides CO is no longer 'air' as understood in explosive/flammability ranges.

Maybe it happens but nobody ever notices as anyone in the room died a long time earlier from CO poisoning.

Another possibility:  the "carbon monoxide" formed by combustion is not pure enough CO to get to flammability (12% total CO) in air.  The gas could be mostly carbon dioxide but still enough CO to poison you.

I know from experience that carbon monoxide/air has a very low fame velocity. I think that air/carbon monoxide mixtures would make a very poor fuel air explosive.
I think however that by the time you get to the point of having sufficient carbon monoxide in a room to allow a flame front to start to travel from one side of the room to the other that unless you are wearing self containing breathing apparatus you will be dead of carbon monoxide poisoning.
Jon Custer is very right, if we consider methane as the fuel then we can write the equation.
$\ce{2CH4 + 3O2 -> 2CO + 4H2O}$
Thus to reach 11% carbon monoxide in a sealed room with a methane flame burning under very special and rich conditions to only form water and carbon monoxide. I think that all the oxygen in the room will be used up long before we reach 11% $\ce{CO}$ in the air. I think that the highest $\ce{CO}$ concentration we could get in the room would be 6.67% by volume. I think that as the oxygen level in the room will be so low that even with additional $\ce{CO}$ added to the room air then the room gas/air mixture will be unable to explode.
The only time I imagine you could get a carbon monoxide/air mixture to "explode" would be if a carbon monoxide cylinder was to leak and then an ignition source was to be added.

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text:
Can I use a primary amine to substitute (deprotect) a tosylate at high yields?
I am trying to use tosyl chloride to allow for a easy substitution of the primary alcohol on glucose then use a BOC-diamine to substitute the tosylate to form a primary amine with a short alkyl chain. But I am having trouble finding any methods in the literature that use primary or secondary amines as the nucleophile for the substitution reaction. Most use relatively stronger nuc like sodium azide or sodium ethoxide.
Is a primary amine not a strong enough nuc? And if so what methods/reagents should I use to obtain a relatively high yield?

The problem with amine nucleophiles is usually overreaction.

Quite so! Your best chance here is to use a large excess of the amine to reduce the amount of disubstituted amine.

Should I use a non nuc-base like dipea or even K2CO3?

Potassium carbonate in acetonitrile would be a good choice.


text:
How does the hexaaquacopper (II) form?
So the hexaaquacopper (II) ion has an octahedral structure with 6 aqua ligands datively bonded to the central Cu2+ cation:

My question is: how is this possible if the electronic configuration of the Cu2+ cation is [Ar]4s-0 3d-9. This only leaves one empty 4s orbital and 3 4p orbitals available for dative bonding which would suggest that a tetraaqua complex should form but this is not the case?

4d is participating.

How is this possible? Wouldn't the 4d be at too high of a level to participate?

@WeijunZhou It's not. OP's right.

@Mithoron This is from a highly phenomenological description that describes the central atom as undergoing $sp^3d^2$ hybridization (as in this figure which contains Chinese: $sp^3d^2$ for Fe and that is what I have been taught years ago. I know that CFT and LFT are far better alternatives and I should be more careful next time in making comments.

@LemuelCrentsil The link above already pretty much answers your question, but you may wish to also learn about the alternative theory as listed here, which is still quite commonly used in some elementary courses. It's simple to understand, but not correct, and resulted in my comment above. You should always use CFT or LFT whenever possible.

@Weijin Zhou Thank you for your answer I will also have a look at CFT and LFT.


text:
Explaining stereochemistry in a sigmatropic 1,3-alkyl migration
I have quite a bit of difficulty in explaining sigmatropic rearrangement reactions. Consider the following reaction:

How do I explain the formation of the shown product? When simply moving the electron pair in my head, I come to the other enantiomer, where the $\ce{-CH3}$ group shows in our direction.
I've repeated all my notes on sigmatropic rearrangements but unfortunately, I don't get it. I suppose that I should treat this like an alkyl migration. The latter should happen over a suprafacial intermediate state as we have an uneven number of electron pairs (1) and we have thermal conditions. This results in a suprafacial reaction. Knowing this, I went on and added the HOMO and LUMO orbitals and postulated intermediate steps. Note: Yes, I am aware, that pericyclic reactions happen in a concerted fashion, but somehow I need to see, why the groups on the bridging carbon have this conformation.

So, let's go on to my question...

Is my proposed explanation okay? If not, please provide some information. I really have trouble with this kind of reactions.

Both enantiomers are produced in equal amounts.  If you break the bond on the "other" side of the methylated carbon you produce the enantiomer.

Problem #1 is that if you interchange the methyl and hydrogen groups, you get diastereomers, not enantiomers. But this is a relatively minor point. I assume your question is about the diastereoselectivity, not enantioselectivity (as ron rightly pointed out there is no enantioselectivity in this reaction).

The analysis you have drawn is somewhat similar to a Fukui frontier molecular orbital analysis. In that picture, for a reaction to be thermally allowed, you need to check for constructive bonding overlap between the HOMO and the LUMO, within geometrical reason. This is achieved by connecting lobes of "like" shading: i.e. shaded lobe with shaded lobe, and unshaded with unshaded.
You don't necessarily have to draw it stepwise (although I gather it's your own way of doing it). You can simply draw both interactions in the starting material, like this. The bottom interaction is not an issue. The top interaction, however, is an issue.

Problem #2 is that you tried to connect the large unshaded lobe of the σ* orbital with the white lobe of the π orbital. That is not geometrically reasonable: if you connect these two lobes, then you need to somehow end up with the new σ bond being formed underneath the ring. In other words, you would have to have an antarafacial migration of the alkyl group.
Problem #3 is that you have concluded that the reaction must be suprafacial based on electron counting. Note that in this context, the reaction being suprafacial means that the migrating alkyl group stays on the same face of the ring. In fact, the reaction must occur in a suprafacial manner simply because of geometrical constraints: there's simply no way that the alkyl group is going to migrate to the opposite side. Technically it could, but you would have to break so many bonds in the transition state that it would not be possible.
[Note that this has a different meaning from the suprafacial or antarafacial components in a Woodward–Hoffmann analysis.]
So what is the solution? Well, because the reaction is suprafacial, you need to use the shaded lobe of the π HOMO, which is on the top face of the ring – the same face as the migrating alkyl group. So then for constructive bonding you need to also use the shaded lobe of the σ* LUMO.

Now one might argue that just like how the unshaded lobe of the π HOMO was geometrically inaccessible, so should the shaded lobe of the σ* LUMO. But we already know that the C–C σ bond is going to break in the course of the reaction, so you're going to go through a pathway whereby the bond behind it (with an arrow pointing towards it) can rotate such that the shaded lobe of the σ* can overlap with the shaded lobe of the π* orbital. In fact, that's not too different from what you drew. The transition state is pretty similar to the intermediate in your stepwise reaction – it just has a few more dotted lines.

If you now "join" the shaded lobes then you can see how the correct stereochemistry arises.

In general my preferred way of analysing pericyclic reactions is to use Woodward and Hoffmann's original rules. Here, there is no shading of lobes: you connect the lobes within geometrical reason, then count the number of $4q+2$ suprafacial components, and the number of $4r$ antarafacial components. If the total is odd, the process is thermally allowed; if it is even, the process is thermally forbidden.
I am not going to explain this approach in detail - it is too long for here and it is adequately covered in textbooks which you should have access to - but I will upload the two diagrams I drew:

Thank you for this very helpful answer! Your sketches did really help to understand this. One little question: How do you count the supra- and antarafacial components? I don't get it. My textbook says, that a [1,j] supra-supra sigmatropic rearrangement is thermally allowed if j + 1 = 4n + 2. In this case j should be 3, so we have 3 + 1 = 4 = 4n + 2. So n would be 1/2...

You mean for the WH analysis at the bottom?

Yes, exactly. I can't combine this with the definition I've read.

I'll say that the way you and I have been taught pericyclics seems to be different. Anyway, I'll try to make some sense of it. The first WH analysis has two suprafacial components, which is probably what your book means by suprafacial-suprafacial. In this case, suprafacial component means that the dotted lines I draw go to the same "side". For the π bond the meaning of this is immediately obvious, for the σ less so: here since both lines to the σ bond go to the larger "inside" lobe, the σ bond is a suprafacial component. Now you see that for $j = 3$ this suprafacial+suprafacial

component is thermally forbidden by the WH rules, just as your textbook predicted: since 4 cannot be expressed by $4n+2$ with $n$ an integer, it is not thermally allowed. Instead the thermally allowed version has two components, one suprafacial and one antarafacial.

Oh, I see. I don't know how to thank you in an appropriate manner. You REALLY helped me a lot in understanding all this! Thank you very much!

No problem! Asking an interesting question is enough reward. And accepting the answer if it helped, of course, but you already did that.

Unfortunately, I have to come back to the WH rule. I now checked in a different textbook. They say that a reaction is thermally allowed if the total of 4q+2 suprafacial and 4r antarafacial components is an uneven number. 4q+2 is the number of electrons in the suprafacial process and 4r the number of electrons in the antarafacial process. q and r is the number of components resulting. In our example (the allowed one) we have 2 π electrons that are in a suprafacial process and 2 σ electrons of an antarafacial process. So 4q+2 = 2 and 4r = 2.

With this, I can't reproduce your result of 1 component for the suprafacial and 0 components for the antarafacial processes which would result in a total of 1 + 0 = 1 and thus be thermally allowed.

"q and r is the number of components resulting" - that is not true. $q$ and $r$ themselves have no meaning, except that they must be integers. We agree that there is one 2-electron suprafacial component, and because there is an integer $q$ for which $4q+2 = 2$, there is therefore one $(4q+2)$ suprafacial component. There is also one 2-electron antarafacial component, but because there is no integer $r$ for which $4r = 2$, there is no $(4r)$ antarafacial component.

Okay, well then their explanation is a piece of... But if q = 1, your equation is 41 + 2. That's 6, isn't it? But didn't we just say, that it should be 2? So in fact q has to be 0: 40 + 2 = 2.

You are right $q = 0$, but $q$ is not the number of components. There is one suprafacial $(4q+2)$ component which has $q = 0$ (i.e. it is a 2-electron component).

Oh, okay, now I really got it. Again, thanks for your patience! You deserve a medal!


text:
How can I interpolate between two densities at different temperatures?
I'm trying to find the temperature of methanol ($\ce{CH3OH}$) when its density is equal to $\pu{780 kg m-3}$.
I know that when $T = \pu{30^\circ C}$ the density is equal to $\pu{783 kg m-3}$, and when $T = \pu{40^\circ C}$ the density is equal to $\pu{774  kg m-3}$.
Is there any formula or a trick to find the temperature corresponding to an intermediate density? Likewise is it possible to find the density at a temperature intermediate between the two?

There are no simple relation, but if you assume the expansion coefficient does not change with temperature you can use a linear relation for the volume given the same mass.

You could use tabulated literature data and then interpolate between the data points. For example, the following plot was created using data from NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9. It shows the liquid saturation line as well as the vapour saturation line.

In this case, interpolation for a liquid density of $780.00\ \mathrm{kg/m^3}$ yields a temperature of $31.643\ \mathrm{^\circ C}$.
By way of comparison, the calculated liquid density is $781.55\ \mathrm{kg/m^3}$ for a temperature of $30.000\ \mathrm{^\circ C}$, and $772.10\ \mathrm{kg/m^3}$ for a temperature of $40.000\ \mathrm{^\circ C}$.

Hello and thank you for your answer , would you mind providing a link for this database? Because I would like to be informed about the liquid saturation line as well as the vapour saturation line not just for the methanol but for most popular liquide


text:
What acid can dissolve thorium dioxide?
In one of my projects (which is about alpha-voltaic cells, don't worry, I know that thorium isn't a strong alpha source, it's only for measurements) I would need to get water soluble thorium compound from a small sample of thorium dioxide  for purification with selective reactions and further measurements. Would any mineral acid work? I found some sources and publications on the internet by google searching but the biggest amount of information I could get is that

"...Several hot acids can solve it..."

Which is not nearly as accurate as I would need it because I don't want to spend the budget of the project on expensive acids which doesn't even work instead of better equipments. (Their price is from their necessary purity)

Thorium dioxide can be dissolved in the mixture of hydrofluoric acid and hydrochloric acid, as described by Kluger and Lieser in 1981.1
Proceed with caution since hot condensed acid are known to be highly corrosive and hydrofluoric acid is also highly toxic and penetrating.

Kluge, E.; Lieser, K. H. Dissolution of ThO2 in HCl/HF-mixtures. Radiochim. Acta 1981, 29 (2–3), 167–168. DOI: 10.1524/ract.1981.29.23.167.

Hot hydrofluoric acid. That gives me the willies just thinking about it.

@zeta-band Actually it's drops of $\ce{NH4F}$ solution added to hot $\ce{HCl}$ ... but I agree with you.

$\color{red}{WARNING:DO \ NOT \ ATTEMPT \ TO \ REPEAT \ THE  \\ FOLLOWING \ EXPERIMENTS \ WITHOUT \ EXPERIENCE \\
 IN \ WORKING \ WITH \ HF \ OR \ WITHOUT \ THE \ PROPER \\
WORK \ ENVIRONMENT \ AND \ PERMISSIONS\ !}$   In the procedure of refining used nuclear fuel pellets which are typically consisting of $(Th,U)O_{2}$ with 1-4% $UO_{2}$ content The most common acid mixtures which are used are $HCl/HF,HNO_{3}/HF,H_{2}SO_{4}/HF.$ If the Thorium oxide was previously heated above $1000°C$ It makes the process much more slower. The least dangerous way to obtain a water soluble thorium compound is to dissolve the Thorium oxide in a solution of $8-13 \ mol\cdot l^{-1} \ \ HNO_{3}$ and  $0.02-0.05 \ mol\cdot l^{-1} \ \ F^{-}$ at temperatures of $50-70°C$ (The fluoride ion concentration can be achieved by adding any fluoride salts which will produce it or by adding hydrofluoric acid to it). This method does not needs the acids to be at the temperature of boiling.The produced Thorium nitrate is much more soluble in water than it's halides so this is a good procedure from this side too.

Reference: DISSOLUTION OF  $(Th,U)O_{2}$  IN NITRIC  ACID - HYDROFLUORIC   ACID   SOLUTIONS


text:
Rule of thumb of how much vacuum does lab glass stand?
Sometimes I have to apply vacuum to all kind of different borosilicate glass equipment like:

round bottom flasks
liebig condensers
fractional distillation apparatus
spiral condensers
chromatography column
and so on...

While I know that a thick walled round bottom flask will pretty much stand any vacuum my vacuum pump can produce, I am not so sure for something as fragile as a spiral condenser.
Is there a kind of rule of thumb how much mBar thin walled glass equipment like spiral condensers or chromatography columns can stand without risk of implosion?

The risk of implosion is always here, even properly certified glassware may get a tiny crack and implode. Always assume that the risk exists. Be especially suspicious with flasks above 100 ml except for very thick-walled ones for vacuum filtering. That said, there is no much difference from 0.01 kPa and 10 kPa  as the outside pressure is 99+ and 90+ kPa respectively (10% difference), so if a piece of glassware can be evacuated by water ejector to meaningful degree, it usually can withstand any vacuum within.

That said, you should be suspicious of large flat and almost flat surfaces and sharp angles.Small tubes are fine. Glass condensers are usually meant to be used with vacuum.

How can you use a chromatography column "in vacuum"? Are you meaning to clean/dry it?

Scratches and cracks have a much larger effect on strength for internal pressures than for external pressures, due to the nature of compression strain verses tensile strain.

From a manufacturer https://www.aceglass.com/dpro/kb_article.php?ref=9849-EPAB-3259

The basic rule is that the stress on an evacuated glass vessel is pretty much the same at 0.01 mbar (oil pump), 1 mbar (membrane) or 25 mbar (water jet) absolute internal pressure. 0.99999, 0.999 or 0.975 bar $\Delta p$ to the outside make very little difference if the thing goes pop.
And basically everything made out of borosilicate and with no flat surface is safe. Condensors definitely, they're not delicate at all. It's basic procedure to make them oxygen-free by evacuating.
Be careful with chromatography columns and flasks in the > 1 liter range. They're principally safe to be evacuated, too, but then are also a bomb, armed and ready to drop. If it goes off, you better not stand next to it. The danger from evacuated vessels is more or less proportional to their volume.

The primary danger of imploding flasks is the hot/cold/toxic/corrosive/reactive substances inside. The glass itself will make a startling noise and small chips could get in the eyes if you are very close with no goggles but the nature of implosion is not nearly as dangerous as explosion.

@Karl: Is regular glass much less resistant to vacuum vs borosilicate (I think the lower elasticity plays a big role for internal pressure, but I'm wondering if it makes much of a difference when the pressure is external). I have a 50,000ml flask with medium thick walls I'll like to take down to -25Hg.

@Hans 50 liters? Wow, no idea. -25 Hg means ~35 mbar below ambient? That could occor by itself within 12h if you put a cork on during a thunderstorm. Don´t think anything could happen there.

@Karl -25mmHg (~ 125 Torr I think). Yes, 50L

If you go using ancient units, you´d better know what they mean. ;) Torr and mmHg are the same thing.

@Karl I think -25mmHg (under atm) is ~125 Torr (abs).

@Hans Impossible. 1000 hPa are approx. 760 Torr. -25mmHg relative would be around 735 Torr.


text:
Colorimetry at home?
Is there a reasonably priced option for colorimetry at home? Is this even a good option to accomplish my goal?
I'd like to roughly determine the concentration of a chemical in an ethanol extract from a natural product. It has a definite color. I would be able to buy small quantities of the purified product from a chemical supplier. Based on my recollections of college chemistry, I should be able to make a solution of known concentration and compare them with a colorimeter to determine how much I've extracted. Would this be reasonable, or should I be looking at another way of doing this?

If you have a colorimeter yes. If you are a hobbyist with electronics, yes. In the worst case you can try to get where your solution stand in between two known concentration. For all this you need an analytical scale and some volumetric flasks.

I appreciate the help.

I appreciate the help. I have looked into different options to try to get an idea of costs and availability. For anyone else looking at doing home colorimetry an analytical scale is less than 20 USD on Amazon, as are small volumetric flasks. A used spectronic 20 is less than 200 USD on ebay (sometimes substantially less), and it appears that there are several website guides to 'build your own spectrophotometer'.

To make the solutions, the volumetric flasks are ok, but to compare them you will need Nessler cylinders.


text:
Acetic acid odor while testing CO2 sensor; can it be easily removed, is there a better way?
A friend asked me to help test a hobby-quality $\ce{CO2}$ sensor MG-811 (updated link). It uses a solid electrolyte and should provide a sensitivity of something like -40 mV per decade. I though I'd just use a small amount of acetic acid + baking soda to make roughly 4,000 ppm in a large container, and compare to "fresh air" which should be roughly 400 ppm, or at least within the 350 to 600 ppm range, as a very rough initial test.
But then I noticed/remembered (duh!) that the acid smells quite strong. Are these simply molecules of $\ce{CH3COOH}$ that are evaporating, or are they bringing along molecules of water in some kind of cluster?
I realized that they could potentially contaminate the sensor, so I'd like to know what I'm up against in terms of removing them from the gas before immersing the sensor.
I've tried to look up the vapor pressure of acetic acid, and it seems to be quite large, about 10 Torr at 20C. See for example here and here. So I'm thinking of using some kind of ad-hoc cold-trap to try to remove it. Might there be another way to do this without using special laboratory equipment? Would passing the gas over dry $\ce{NaHCO3}$ or a solution of it help to "getter" the remaining acetic acid vapor?
Or, is there a better way to produce a controlled amount of $\ce{CO2}$ to begin with?

A cold trap would be better than trying to pass the gas stream through more baking soda and potentially altering the concentration of the stream to be measured by an unknown amount. Some more details on the proposed test and setup would be helpful.

@J.Ari Thanks, I wasn't aware that gaseous  CO2 would react with baking soda in any significant way.

But the remaining acetic acid would unless you make sure that the acetic acid is the limiting reagent to make your CO2. The odor will dissipate in time if the acetic acid is reacted away.

@J.Ari oh for sure. I think you are right, the acid should be the limiting reagent. (I recently learned about limiting reagents a few weeks ago).

You could use a less volatile acid, like oxalic. But I think the reaction with NaHCO3 probably gives off a mist as the reaction bubbles, and that might be why the odor is so strong.
How about using seltzer water (or Perrier? or Diet Coke!) as the source of CO2? I estimate that the pressure of CO2 is about 45 psi in the unopened bottle, and if you attach a balloon to a freshly opened bottle and let it outgas slowly, the balloon will fill up and be essentially pure CO2. If a latex balloon is unsuitable or too hard to handle, a mylar balloon ($1 at the Dollar Store) will work just as well. The gas will be saturated with water vapor.

This is probably a much better way to go. The humidity won't be a problem since I'll have to dilute the gas by a factor of 250 anyway to get ~4,000 ppm.  I like this method very much, and feel silly not having thought of it. That's probably because I haven't actually had a carbonated beverage out of a bottle or can in years. Sad. :-)


text:
Best Settings for Electrolysis of water
I want to electrolyze water as a personal project, but I have found conflicting power recommendations on google. Some sources recommended a higher voltage, claiming that higher amps will just destroy the electrodes, while others insist a higher amperage is required, and extra voltage will be wasted heating up the water. Unable to find one absolute result, I'm turning here- assuming not a crazy amount of amps or volts, what is a good ratio to get the best results? (an explanation of the answer would be appreciated.)

Ideally, it takes 1.23 volts to split water. Less voltage than this will give essentially no current, no electrolysis. If you want more H2 + O2, you raise the potential. (Of course, you use an electrolyte like H2SO4 or NaOH with electrodes that will not corrode. Platinum is OK, also gold. Oh, yeah, also stainless steel for ordinary people.) If the surface becomes covered with bubbles of gas, the effective area of the electrode is reduced, so the current will drop. So you raise the voltage, and bubbles come pouring off. Then the electrolysis uses more energy (volts x amps). 1 amp for 96,500 seconds (27 hours) will give you 1 gram of hydrogen and 8 grams of oxygen. The product is dependent on the current, and the current is dependent on how high you raise the voltage.
It's not as slow as watching grass grow, but even at furious bubbling, it's surprising how slowly the volumes build up.

Thanks for your answer! I could use some clarification, though. What actually causes the bubbles, amperage or voltage?

When an electron flows to/from an ion, it is a current. The voltage is how hard you have to push it. So the quantity of product (H2, O2) is proportional to the electron flow, which is current in amperes. 1 ampere for 96,500 seconds is one Faraday = 1 mole of electrons.


text:
How to find the molar mass of an iron compound based on its mass and mole fraction in the compound?
I'm trying to solve this question:

We have a molecule composed of $3$ iron atoms, and $4$ atoms of another element. We are given the following information: it has $\pu{2.36 g}$ of iron for $\pu{3.26 g}$ of molecule.

I want to find the molar mass of the compound, I have tried so far:
$$ m = \pu{3.26 g} = \pu{0.00326 kg}$$
Since it has $3$ atoms of $\ce{Fe}$ and $4$ atoms of an unknown substance, therefore:
$$3 + 4 = 7~\text{atoms},\\
  \pu{1 mol} = \pu{6.022* 10^23 atoms}\\
\frac{7}{\pu{6.022* 10^23}} = \pu{1.16 * 10^-23}$$
As we know: $M = m / n,$ I tried to divide $0.00326$  by $\pu{1.16 * 10^-23}$ and I obtained $\pu{2.79429 * 10^19}$, but the correct answer is $\pu{231.43 g/mol}$.
What I have done wrong?

(3.26-2.36)/(2.36/55.845/3*4)4+55.8453

@WeijunZhou thank you but what 55.845 represent?

@napi15 The molar mass of iron.

Let $\ce Z$ denote the unknown element.
The total amount of iron atom in the compound is,
$$n(\ce{Fe})=\frac{\pu{2.36g}}{\pu{55.845 g\cdot mol^{-1}}}=\pu{0.0423 mol},$$
The molecule comprises of 3 iron atoms and 4 other atoms, so the amount of the unknown atom is,
$$n(\ce Z)=\pu{0.0423 mol}\times \frac43=\pu{0.0564 mol},$$
The molar mass of $\ce Z$ is,
$$M(\ce Z)=\frac{\pu{3.26g}-\pu{2.36g}}{\pu{0.0564 mol}}=\pu{15.97 g\cdot mol^{-1}}$$
So $\ce Z$ is $\ce{O}$. The molar mass is
$$M(\ce{Fe3O4})=\pu{55.845 g\cdot mol^{-1}}\times 3 + \pu{15.97 g\cdot mol^{-1}}\times 4=\pu{231.43 g}.$$

Before I start explaining it, I would say that the data provided in the question and the correct answer corresponds to the compound $\ce{Fe3O4}$, whose molar mass is $\pu{231.43 g}$. But, I am not proceeding with this known fact rather I'm considering only the given data and the molar mass of iron, $\pu{55.845 g}$.
In $\pu{1 mol}$ of the compound, $\ce{Fe3X4}$, there are $\pu{3 mol}$ of $\ce{Fe}$ and $\pu{4 mol}$ of the unknown element $\ce{X}$.
Let molar mass of the compound be $M$ grams.
There is $3\times\pu{55.845 g}$ of $\ce{Fe}$ in $M$ grams of the compound.
Given, in $\pu{3.26 g}$ of the compound, $\pu{2.36 g}$ of $\ce{Fe}$ is present.
Therefore, $\pu{1 g}$ of $\ce{Fe}$ is present in $\pu{(3.26 / 2.36) g}$ of the compound.
Therefore, $3\times\pu{55.845 g}$ of $\ce{Fe}$ is present in
$(3.26 / 2.36) \times 3 \times \pu{55.845 g}$ of the compound.
Therefore, $M = (3.26 / 2.36) \times 3 \times 55.845 = 231.43$, the molar mass is $\pu{231.43g/mol}$.

The correct unit for (relative) molar mass is  $\pu{g/mol}$, you are a bit inconsistent with that. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

Yes I knew since begining it was the oxygene , but I must follow the flow of the question


text:
Making para-dibromobenzene
What steps do I need to take to make the compound in the picture from just a benzene ring? Thank you.

According to US Patent 3062899A p-dibromobenzene is the major product of the bromination of benzene with 2 equivalents of bromine at ambient temperature. With catalytic aluminium trihalide this remains the outcome below 40C.

According to this forum:

To benzene, (add) $\ce{Br2}$ and $\ce{HOAc}$ to get bromobenzene.
Then (add) $\ce{ZnBr2}$ to get p-bromobenzene as the major isomer. Or
from aniline, NBS and $\ce{HOAc}$ to get p-bromoaniline.  Then
diazotonization of aniline with $\ce{NaNO2}$ and subtitution with
$\ce{NaBr}$ to get p-bromobenzene.

As the source is a forum, I found it highly unreliable but then I found a page which shows the process of synthesis of bromobenzene. It says that the para isomer is formed as a minor by-product and there is also a note on the isolation and purification of the para isomer:

.....Now distill the crude bromobenzene slowly, rejecting the fraction
boiling up to 150°C, and collecting that of bp 150-160°C. Yield,
28-29g (about 19ml). A small quantity of crude p-dibromobenzene
remains in the flask.
[....]
The p-dibromobenzene formed as a by-product in the above reaction
usually solidifies when the undistilled residue obtained in the first
distillation is chilled. It may be isolated by adding about 10ml of
methylated spirit and some activated charcoal to the flask, boiling
for a few minutes, and filtering hot. On cooling the filtrate in
ice-water, crystals of p-dibromobenzene, mp 89°C separate:
recrystallise a second time if necessary to obtain colourless
crystals.

There is also a patent(@Waylander) which says that p-bromobenzene can be prepared by using bromine and aluminum trihalide as catalyst.

In general, the bromination of benzene with two moles of bromine per
mole of benzene at ambient temperature produces predominantly
p-dibromobenzene. If the bromination is conducted in the presence of
an aluminun trihalide catalyst at low temperature, e.g., below about
40 C., the para-dibromo isomer is still the principal product.


text:
Calculating Compressibility factor from the Van der Waals' Gas equation
So this problem has been bugging me for a long time. According to Wikipedia the compressibility factor $Z$ is defined as the ratio of the volume occupied by a real gas divided by the volume occupied by an ideal gas at the same temperature and pressure.
$$Z = \frac{V_\text{real}}{V_\text{ideal}}$$
Now the ideal gas equation is
$$P_\text{ideal} V_\text{ideal} = nRT$$
Now from here we can substitute the value of $V_\text{ideal}$ into the $Z$ expression to get $$Z = \frac{P V_\pu{m}}{RT}$$
Where $V_\pu{m}$  is the molar volume of the gas and $P$ is the value of $P_\pu{ideal}$ that was in the ideal gas equation.
OK so far so good.
Deriving the compressibility factor value from the van der Waals' gas equation:
We know for real gases the pressure exerted by them is always less than the pressure exerted by an ideal gas so we correct the pressure by the term
$an^2/V^2$
$$P_{\text{ideal}}= P_{\text{real}} + \frac{an^2}{V^2}$$
And the volume occupied by the gas molecules can be written as
$$ V_{\text{ideal}} = V_{\text{real}} - nb$$
On substituting the above values of $P_\text{ideal}$ and $V_\text{ideal}$ in the ideal gas equation, we get our van der Waals' Gas equation
$$\left(P+\frac{an^2}{V^2}\right)(V-nb) = nRT$$
Please note that in the above equation $P$ is $P_\text{real}$ and not $P_\text{ideal}$. Now I'm going a step further to calculate the value of $Z$ from this equation by shifting the terms and I got this $$Z = \frac{P V_\pu{m}}{RT} = \frac{V_\pu{m}}{V_\pu{m} - b} - \frac{a}{RTV_\pu{m}}$$
Now here's the problem that I'm facing: In the above expression the value of $P$ is not the $P_\text{ideal}$ but it is the value of $P_\pu{real}$ that we had in the van der Waals' gas equation. Doesn't that contradict the whole definition of Compressibility factor that we had learned in the first place? How this expression for $Z$ can be correct?
Thanks for reading.

A better definition of the compressibility factor is  $$Z=\frac{PV}{nRT}$$This removes any ambiguity.

I actually have a query here. "same temperature and pressure" So a is defined only when pressures are same so cant we just put preal=pideal?

$$Z = \frac{V^{\mathrm{r}}_{\mathrm{m}}}{V^{\circ}_{\mathrm{m}}}\tag{1}\label{1}$$
$V^{\mathrm{r}}_{\mathrm{m}} = \text{Volume of 1 mol real gas.}$
$V^{\circ}_{\mathrm{m}} = \text{Volume of 1 mol perfect gas.}$
The van der Waals equation is $$ \left( P^{\text{r}} + \frac{an^2}{(V^{\text{r}})^2}\right) (V^{\text{r}} - nb ) = nRT \tag2\label{2}$$
Now, consider you have a container containing a $1~\mathrm{mol}$ real gas. You know its pressure $P^{\text{r}}$, volume $V^{\text{r}}$ and temperature $T$ and you wish to find the compressibility factor of this gas.
So, for calculating $Z$, we know the real volume of the gas, i.e $V^{\mathrm{r}}$ and now we need to calculate  $V^{\circ}$ which is the volume it should have occupied if it behaved like a perfect gas, i.e. it obeyed the perfect gas law.
Thus, $$V^{\circ} = \frac{RT}{P^{\text{r}}}$$
Therefore, substituting this value of $V^{\circ}$ in $\eqref{1}$ $$Z = \frac{V^{\mathrm{r}} P^{\text{r}}}{RT}\tag3\label{3}$$
From $\eqref{2}$, by rearranging the terms, we get $$P^{\text{r}} = \frac{RT}{(V^{\text{r}} - b)} - \frac{a}{(V^{\text{r}})^2} \tag4\label{4}$$
Substituting this value of $P^{\text{r}}$ in $\eqref{3}$, we get $$Z = \frac{V^{\text{r}}}{(V^{\text{r}} - b)} - \frac{a}{V^{\text{r}}(RT)} \tag5\label{5}$$

OK! Please reply me if I'm right- We knew Vr already and for calculating the value of V(ideal) we simply used the ideal gas equation. Right?

@Serotonin Exactly.

Thanks dude! Yours was a perfect explanation. I wonder why this is not elaborated in any of the books.

Two good physical chemistry books are Elements of Physical Chemistry and Chemical Principles. Both are authored by Peter Atkins.

Actually I checked that book before asking this question.

@Serotonin Yeah. This thing is not explained clearly but the books are otherwise good.

From van der Waals' equation: $$(p + \frac{an^2}{V^2})(V - nb) = nRT\\
\color{brown}{p} = \frac{nRT}{V - nb} - \frac{an^2}{V^  2}$$
Compressibility factor is the ratio of volume of real gas to ideal gas:
$$Z= \frac{V}{V_{\text{ideal}}} = \frac{\color{brown}{p}V}{nRT} = \frac{V}{V - nb} - \frac{an}{VRT}$$


text:
What exactly is an "electron-sponge"?
What exactly an "electron-sponge" [behavior/action/property/system] nickname is, and what makes a material an "electron-sponge" (preferably, quantitatively)?
From what I found, it's typically a molecular cluster or a metal complex which is capable of undergoing multiple oxidation and reduction waves, or is able of accepting several electrons without losing their chemical identity. However, this definition seems to be rather qualitative and quite vague. Also, it would be interesting to find out who and when coined this term.


text:
What is the Lewis structure of Ozonide ion?
What will be the Lewis structure of the ozonide ion, including formal charges?
Do all oxygen atoms have filled octate?

Look at all the resonance structures of the ozone molecule to start. If you find the resonance structures of less importance, you will be able to find the structure of the ozonide ion.

Related: https://chemistry.stackexchange.com/questions/20086/why-dont-unstable-odd-electron-species-dimerize-so-that-they-become-stable

From a cursory e-book search:

Ozonide ion is paramagnetic and is shown to have magnetic
  susceptibility. Bond length is 135 Å and bond angle is 108°. The
  structure is:


text:
Thermodynamic stability of meta-xylene over ortho- and para-isomers
When talking about the example of the alkylation of toluene by chloromethane in the presence of $\ce{AlCl3}$, Hepworth, Waring and Waring (2002) mentioned that:

At room temperature, a mixture of 1,2-dimethylbenzene and 1,4-dimethylbenzene results, but at 80 °C the product is mainly 1,3-dimethylbenzene. In fact, heating either of the 1,2- or 1,4-isomers in the presence of $\ce {AlCl_3}$/$\ce {HCl}$ results in rearrangement to the more stable 1,3-dimethylbenzene.

This leads me to question: Why would the 1,3-isomer be more thermodynamically stable than the 1,2- and 1,4-isomers? Clearly, there must be an activation energy barrier to overcome as heat is required for the conversion. The methyl group being an ortho/para-director would lead to the formation of 1,2- and 1,4- isomers. Thus, wouldn't these isomers also be the more thermodynamically stable isomers?
Reference
Hepworth, J. D., Waring, D. R., & Waring, M. J. (2002). Aromatic Chemistry. United Kingdom: The Royal Society of Chemistry.

Methyl group being o/p directing is a kinetic property based on the energy of a transition state (or the energy of an intermediate by the Hammond postulate), whereas thermodynamic stability has to be based on the energy of the product.

All of these Friedel-Crafts reactions are reversible.  Hence, if the reaction is run for a long enough time, the thermodynamic product (m-xylene) will predominate over the kinetic products (o- and p-xylene). It is curious though, that m-xylene appears to be slightly more stable that p-xylene.  Why that is, is a good question.

The thermodynamic stability of m-xylene over o-xylene or p-xylene can be deduced by considering the hyperconjugative effect. More specifically, resonance forms with a positive charge on methyl-substituted carbons are more important than resonance forms with a negative charge on them, as the methyl group stabilises the positively charged carbon by hyperconjugation.
In the o- or p- isomers, the hyperconjugation of one methyl group produces negative charge at ortho and para position where another methyl group resides. This is highly destabilising. But, in the m-isomer, hyperconjugation does not destabilise the structure and both methyl groups' hyperconjugation effects act in a cooperative way with each other.

@SoumikDas This may seem very basic, but I find the representation of hyperconjugation in your diagram a little weird. Traditionally, when we make hyperconjugation structures in alkenes, don't we break the C-H bond and shift the π bond one place to get a sort of allylic carbanion with the release of a proton?

@YusufHasan This is just one step away from breaking the C–H bond; all you need to do would be to detach that H+ from the CH3 group and make a double bond between the CH2 and the adjacent positive charge: $$\ce{H-CH2-C+ <-> H+$\cdots$CH2=C}$$ It doesn't affect the subsequent analysis that the negative charge is delocalised onto C-2 and C-4 of the benzene ring.

Can anyone please explain how does methyl groups' hyperconjugation works?


text:
How to choose alcohol solvent for organic extraction?
Assuming I have an organic substance that is known to be "soluble in alcohol" (e.g. capsaicin) I want to extract, which alcohol should/can I use? I know that both ethanol and methanol are commonly used, but what about:

isopropanol
glycerol
n-Butanol
1,4-Butanediol
tert-Butyl alcohol
1-Pentanol
tert-Amyl alcohol

In general, which alcohols are suitable/unsuitable for organic compound extraction (e.g. which should be the best solvent in terms of amount of solute dissolved in given volume of solvent), and why? Is there any a priori way to determine e.g. if isopropanol extraction will be OK for a given "soluble in alcohol" organic substance?

Note: I am aware of this discussion, but it doesn't cover the specific nature of alcohol extraction.

Need to consider how you're going to isolate the extract once you've done the extraction. Some of your possible solvents have high boiling points and will not be easy to remove.

@Waylander going up to 300C is OK with me; most of them are boiling @ <150C, one way or the other.

'Soluble in alcohol' probably just means soluble in ethanol. Of course it may well be soluble in other alcohols as well.

What are the foreseen uses of the extraxted? How many solvents are available to you? How do you suppose to recover the extracted?  Once one know details as these,  then one might can easily answer. Cost and toxicity can matrer. And so on.  You mention that you can heat up to 300 °C , but capsaicin does not care of that as it melts at 62°. And so on.

The optimal solvent will extract the greatest amount of desired product and the least amount of difficult-to-remove substances.
The extraction is only the first step; separation from solvent and other extractives follows. Try one, then another solvent to see if there is a difference. Maybe number of carbons or hydroxyls is important; you can also mix alcohols, and/or add water.
Capsaicin has a phenolic group; perhaps it would be extracted from its source with alkali (NaHCO3 or Na2CO3).


text:
Setting to basis set for electron transfer in type II CdS/TiO2 heterostructre for PYXAID package
I am studying electron injection from $\ce{CdS}$ into $\ce{TiO2}$ using NAMD approach. Now I'm using pyxaid package.
Would be possible  someone kindly help me with construction of basis set for this process?
30 bands selected for Hamiltonian construction and from this selected band set number and active space of [6-13] has been made. 7 is the valence band and 8 to 13 are conduction states.
For initial electronic state I selected -11,-12,-13 (one of which could be $\ce{CdS}$'s lumo) as I found from some preliminary decoherence rate calculations (12 decoherence more rapidly than 9-12).
params["active_space"] = [6,7,8,9,10,11,12,13]

# Generate basis states
params["states"] = []
params["states"].append(["ES1",[6,-6,7,-8]])
params["states"].append(["ES2",[6,-6,7,-9]])
params["states"].append(["ES2",[6,-6,7,-10]])
params["states"].append(["ES2",[6,-6,7,-11]])
params["states"].append(["ES2",[6,-6,7,-12]])
params["states"].append(["ES2",[6,-6,7,-13]])

# Initial conditions
nmicrost = len(params["states"])
ic = []
i = 0
while i<100:
    j = 3      # started from -11,-12,-13 one of which I believe belongs to CdS (donor semiconductor)
    while j<nmicrost:
        ic.append([2*i,j])
        j = j + 1
    i = i + 1

params["iconds"] = ic

So you are using Quantum Espresso for your electronic structure? And your question is essentially how to construct the input file? What's wrong with the default settings?


text:
Is a beaker/reaction vessel like the one pictured available commercially?
We're doing photocatalysis with the catalyst fixed to a 2" Si wafer, washing an aqueous solution (methylene blue, the concentration varies but in the picture I believe it's 0.1 mMol/l) over the top in beakers that look like this

(the lid is the heatsink for the 365 nm UV LED that drives the reaction; that part is pretty much optimised).
The few we have are custom made, and we need some more.  I can't find a source of anything comparable, but I'm only a physicist who doesn't necessarily know what's available or what terms to search for.
It needs two hose barbs as near the bottom as possible, and a flat base wide enough to lift a 2" wafer out (i.e. about a 60 mm minimum internal diameter).
They're also visible in this timelapse video of the reaction.
If such a thing exists, what's it called? Or is there something we could repurpose? Both the cost and the lead time of custom glassware are a bit high.

Does the vessel have to be glass? You could modify a plastic beaker of the right size with bulkhead unions and some O-rings/grommets.

@J.Ari I wondered about that. But I might have to go through the base because of the curvature, which is greater than the thickness of the o rings in my bulkhead fittings over the width of the fitting

A beaker / reactor of such type would be rather expensive as well. If design and elegance do not matter you can easily adapt every suitably sized beaker by holding two pipes at the levels at which the hose barbs are supposed to be. Perhaps you choose a beaker a bit wider so there will be no turbulence or differences of sort as compared to what you are using.
Any flask or other container, combined with two small tubes and a couple of stands will effectively let you save money and time. Perhaps you must get space for the inlet and outlet through the led holder lid.

I wrote a rather detailed comment but it hasn't appeared, neither has my upvote. In summary: I assume you mean run the hoses from the top.  A wider beaker would be needed to avoid shadowing the wafer, which would increase the system volume or decrease depth, but might be workable.  Turbulence is good for mixing.  Priming the pump (not shown) would be tricky.  Modifying the metal lid is much easier the modifying the glass.  I'm not sure about this approach but it's worth a shot.

I think it must work with limited adjustment. Really. I mentioned a wider container so you can have space to go in with the in/out pipes even without touching the lid (avoid drill holing it ). I assume you can work in system open to air. Glad it may help

A non-vented system would be another experiment, and we'd machine a reaction vessel out of metal for that. Drilling holes in the lid changes less than a bigger beaker, plus that way I can mock something up tomorrow using one of these beakers with the hoses clamped off - I can make some assumptions about the light pattern to account for shadowing.

I don't think these are UV LED. Most likely vanilla blue LED. You can buy strips of blue LED light and wrap it around a 500 mL beaker. Run your reaction in a second vessel inside that larger beaker. or else, you can use a aluminium sheet, cut vertical perforations and wrap the LED strips around it. Place the photocage around your reaction vessel.

I bought the LEDs. They're 365 nm.  But the beaker is full of about 1 mMol methylene blue solution.


text:
Mechanism for reaction of an allene with the Grignard Reagent(RMgX)
How does the Grignard Reagent($\ce{RMgX}$) react with an allene $\ce{CH_2=C=CH_2}$ and what are the products formed?
This is exactly what I tried to do.

Any help will be appreciated.

Consider the stability of potential carbanions.

@Mithoron I very well know how Grignard reagent reacts with a ketone, but that certainly doesn't me help here.

Well, I think it should at least give you good idea. If it doesn't react in a very similar way then it would be tricky question, but  Weijun Zhou seems to think ^^^ the same.

I think it's fine. You didn't say that it will later be treated with $\ce{H3O+}$ though so that is probably another step, not part of the mechanism of the reaction with $\ce{RMgX}$.

@WeijunZhou in almost all the reactions of Grignard reagent, the compound is finally treated with dilute $H_3O^+$

So that's it...what I did is the only way for Grignard reagent to react with an allene?

Why do you think these react at all?

@Zhe I have no idea. Our teacher once told us to predict the reaction mechanism. Will they not react?

I wouldn't think so, but I haven't worked with allenes much. There are really only two pathways to look at. Deprotonation, which you've looked at, and migratory insertion.

@Zhe can you show the second one(migratory insertion) please?

Look up migratory insertion on wikipedia and then scroll down to where C-C double bonds are involved. I don't think this is very likely with just magnesium though...

@zhe I found a new way, I'll have to answer my own question...tell me if you think it's correct..


text:
Sodium Carbonate saturation curve
Where can I find the saturation curve for Na2CO3 (Sodium Carbonate) versus temperature?
Thanks,
ECD40

Google, I guess..

Wikipedia has a table in the article on sodium carbonate. You could plot the numbers or Google it. I got a neat curve (attached below)but you have to do a little figuring to get the temperatures. Or you could Google images for sodium carbonate solubility to get other curves.
Washing soda is the decahydrate; soda ash is anhydrous.
Decahydrate:
7 g/100 mL (0 °C)
16.4 g/100 mL (15 °C)
34.07 g/100 mL (27.8 °C)
Heptahydrate:
48.69 g/100 mL (34.8 °C)
Monohydrate:
50.31 g/100 mL (29.9 °C)
48.1 g/100 mL (41.9 °C)
45.62 g/100 mL (60 °C)
43.6 g/100 mL (100 °C)


text:
How do we indicate a volume fraction like 0.05 in procedure descriptions?
I'm translating a Russian biotech text, and there's this sentence with a number without measurement units:

The contents of the wells were discarded; the wells were washed three times with 200 µL of Buffer containing 0.05 of Tween 20.

I was told that this figure indicates a volume fraction, and is dimensionless, hence no units after it. Will this be understood by the native speaker of English as-is, without any units, as in my quote?
Or is it better to write it down as:

The contents of the wells were discarded; the wells were washed three times with 200 µL of Buffer containing 5% (v/v) of Tween 20.

Or maybe like this:

The contents of the wells were discarded; the wells were washed three times with 200 µL of Buffer containing 5 vol% of Tween 20.

Is there a way to retain 0.05 and somehow indicate that this is a volume fraction?

Your ways are the correct ones.  "0.05 v/v" won't pose any doubts, too. Or "0.05 volume fraction" as you did here. But I vote for your two options.

Your ways are the correct ones.  "0.05 v/v" won't pose any doubts, too. Or "0.05 volume fraction" as you did here. But I vote for your two options.

I think it would be wise to include units, because how would you distinguish between mass and volume fractions? Options 2 and 3 that you present are both seen in the scientific literature and text books. My opinion is that option 3 is the least ambiguous way to present both the numeric value and the context (i.e. the units).


text:
What is 150 degree test kerosene?
I saw an old ad for kerosene in which it was advertised at "150 degree test".
What does that mean?

This seems to be what you are looking for, page 282 if it doesn't go automatically. https://books.google.com/books?id=3wQSAQAAMAAJ&pg=PA282&lpg=PA282&dq=150+degree+fire+test+kerosene&source=bl&ots=jkpt6yW-J1&sig=229KLWed3sgTOS6dJSskNWzb0FE&hl=en&sa=X&ved=0ahUKEwjryKPPjIDZAhWl5IMKHXmQA3QQ6AEITDAK#v=onepage&q=150%20degree%20fire%20test%20kerosene&f=false


text:
Diels-Alder reaction question
In class, we learned that Diels-Alder reactions only occurs with s-cis conformations and that conformations like those in cyclopentadiene are permanently locked in a cis-transformation and that is why they are so reaction. We also learned that the reaction proceeds at a higher rate with an electro-withdrawing substituent (eg. carbonyl) . Why then is a molecule like the one in the figure not reactive while a similar structure, a cyclopentadienone, is?

Whoa, carbonyl isn't a substituent here, but part of ring, which isn't a flat one... Also why do you even think it is or isn't reactive?

Compared to cyclopentadienone? Well, since both molecules have s-cis bonds, thats probably not it. Also, according to my textbook, carbonyl groups increase rxn rate because they are electron withdrawing due to resonance and both structures have resonance structures. Perhaps because cyclopentadienone has more ring strain? Im not really sure though cause I can't picture the structure of the molecule in the figure.

Actually, why would have a electron withdrawing carbonyl group and the fact that they produce resonance structures increase the rate of reaction?

Cycloundecapentaenon is so different and incomparable that I have no idea why you picked this compound.

You need to draw this structure in a 3D visualizer program, eg, Avogadro. While this structure looks similar to cyclopentadiene, in 3D it looks vastly different.

Is that the all-cis isomer? One of the double bonds at the back looks like it’s trans to me.

You are correct. The all cis conformation does have the similar problem as well.

You should probably edit the image. (Btw, it’s configuration not conformation.)


text:
Comparing the Hückel and extended Hückel methods
I'm very confused about the differences between these methods. From my textbook, it states that the Hückel method only takes into account the π bonding interactions, while the extended Hückel method takes into account all the valence electrons. But what exactly does this do?
To my knowledge, the s, pz, px, and py orbitals are all orthogonal to one another and wouldn't mix anyways. Are shielding effects the only thing accounted for by extended Hückel theory?

Ever heard about sigma bonds? That's the difference.

s, pz, px, and py on the same site are orthogonal. However, for example, an s on one site is not orthogonal to a px on a different site.

In the simple Hückel method (SHM) the basis set is limited to p orbitals. This set is limited in a great extent to pz orbitals which constraints the molecular plane to be the xy plane. Basically, you are limited to planar molecules.
  The inclusion of all valence s and p orbitals in the extended Hückel method (EHM) naturally lifts the spacial constraints and you can work with non-planar molecules.
These are the two first differences between the two methods. The other points are:
SHM:

Orbital energies are limited to same-atom interactions, adjacent-atom interactions while all other interactions are 0.
Fock matrix elements are not actually calculated.
Overlap integrals are limited to 1 or 0.

EHM:

Orbital energies are calculated and vary smoothly with geometry.
Fock matrix elements are actually calculated.
Overlap integrals are actually calculated.

You can look up the derivation and steps for the implementation of these two methods in this book that I used as a reference:
Errol G. Lewars; Computational Chemistry, Introduction to the Theory and Applications of Molecular and Quantum Mechanics, Second Edition;
Springer: 2011. DOI: 10.1007/978-90-481-3862-3

Would you be willing to comment on the Fock matrix elements? More specifically, I assume it's still $h + 2J - K$, but are there simple approximations for J and K?

@pentavalentcarbon that's where the empirical part of the method comes in. We obtain the whole matrix elements basically directly from experimental ionization potentials for their corresponding orbitals.

Pentavalentcarbon brought up an interesting point that I thought I could elaborate on: How are the matrix elements of this simplified Hamiltonian obtained? I'll basically just summarize the relevant parts of this Chem-Libre section on the Extended Hückel method
This is where the empirical part of this method comes into play. If you have the overlap matrix elements $S_{ij}$, then the elements of the Hamiltonian $H_{ij}$ can obtained using experimentally obtained values of a given orbital's ionization potential. The diagonal elements are simply taken to the negative of the ionization potential of the corresponding atomic orbital, $$H_{ii}=-\mathrm{IP}$$ The off-diagonal elements take only slightly more effort with $$H_{ij}=\frac{K}{2}(H_{ii}+H_{jj})S_{ij} \text{ , } i\not=j$$ $K$ is just a proportionality constant and a commonly used value is $1.75$, based on a study by Hoffman on the orbital energies of ethane.

Interesting. That expression is identical to what Q-Chem would call the Generalized Wolfsberg-Helmholtz guess, though its matrix elements are in the AO basis, and the diagonal elements are simply the diagonal from $T + V$. For other programs that can use EHM (Dalton comes to mind), I wonder what they are actually doing.


text:
Regioselectivity in Claisen condensation and aldol reaction
I'm having a hard time determining regio-selectivity in those two reactions.

At the image above I've drawn one example for each on of them. My question is how do we choose where the deprotonation will occur. My approach was that the base will deprotonate the substrate in order to produce the most stable enolate. In the Claisen condensation, A should be the main product since the -CH2- group is present thus corresponding the mechanism. For B there will be no acidic hydrogens so it is not favoured, or at least that's what I thought.
Moving to the aldol condensation I can't figure out which product will be the main. Theoretically in C, the enolate is more stable than that of D, but the final unsaturated ketone isn't more stable in D?
Either way, I can't comprehend how can we determine what enolate will be produced each time.

Claisen and aldol condensations are thermodynamically controlled,* so it is not a question of which enolate forms, it is a question of which product forms.
In your first example, product A can be deprotonated by ethoxide as there is still an acidic hydrogen between the two carbonyls. Product B, on the other hand, has a quaternary carbon between the two carbonyls and cannot be deprotonated.
Because the pKa of a 1,3-dicarbonyl (~10) is significantly smaller than the pKa of ethanol (~15),** this deprotonation of A is almost complete and it therefore pulls the equilibrium position over to one which favours product A, simply by le Chatelier's principle. In fact if you somehow make product B via a different route and then treat it with ethoxide, you should exclusively get product A out of your flask, via retro-Claisen and Claisen condensation.
For the second example I would expect product D (tetrasubstituted alkene) to be preferentially formed over C (trisubstituted).

* Unless you use kinetic deprotonation conditions such as LDA. In this case you will then have to look at the stability of the transition states leading to enolate formation, but that's a story for another day.
** Very rough values.

Thank you for your answer. Another way I think to justify the Claisen product A is the stability of the anion. The one produced in A has one alkyl group while the B has two. Given that alkyl are electron donors, A should be favoured.
Regarding aldol condensation, some of my colleagues support that while D forms a more stable C=C bond it is also the most sterically hindered so C is favoured, basically two things competing each other.

No, enolate stability really has nothing to do with it (and even if it did, the more substituted enolate is more stable: note that the predominant resonance form of an enolate is $\ce{C=C-O-}$, not $\ce{^-C-C=O}$).

Why do you say that these reactions are thermodynamically controlled? Do you mind elaborating on that?

@TanYongBoon all steps are generally considered to be reversible under the typical conditions - NaOEt and heat for the Claisen, for example. Except for the deprotonation of the a,b-unsaturated ketone product - that is irreversible. Likewise for the aldol the simplest procedures just add Ba(OH)2 or NaOH to the substrate and heat. Everything there is reversible.

Just a quick question. I understand that the aldol reaction usually involves a carbonyl compound reacting with itself under basic conditions. However, I have also seen instances where another carbonyl compound is added and the enolate of one attacks the newly-added carbonyl compound. However, under such conditions, we would also be getting the self-reacted products right?

@TanYongBoon, that's what I was very briefly alluding to with LDA. Generally, you add carbonyl A to LDA to generate enolate A, and then you add carbonyl B to enolate A. Funnily enough, the rate of the aldol addition is faster than the rate of proton exchange between A and B, which allows you to get good yields of the desired aldol product; C–H deprotonations are generally slower than one might think. I've never actually had to do an aldol myself, so I don't know how much enolisation of B you see (if any), but there's an example procedure here: http://www.orgsyn.org/demo.aspx?prep=CV7P0185

authors: Αντώνιος Κελεσίδης


text:
Using drip extraction, how much caffeine is extracted from coffee as a function of the amount of water used?
When using a drip coffee maker, there are basically two choices we make: the amount of water and the amount of coffee grounds.
If we use the same amount of coffee grounds but double the water, how much extra caffeine is in the resulting brew? Does all the available caffeine in the grounds get quickly liberated, so that there's not much of a difference between brewing with more water and diluting the brew with water afterwards? Or is it closer to the other extreme, where twice the water means twice the caffeine?
I saw on wikipedia that only about 20% of the caffeine is extracted for an 'optimal brew,' but I wasn't sure if that means we can extrapolate and say that the remaining 80% can be extracted just as easily.
I have the same question about the amount of coffee grounds. I only care to consider the 'normal brewing regime', that is with water at ~80-90C, 3-6 cups of water, and 1/4-1/2 a cup of grounds.
Thanks!

Caffeine has a solubility in water of 20% at 80 $^\circ$C, while the coffee has between 1 and 4% caffeine.
That means that you have more than enough water to extract all the caffeine in the coffee.
With the quantities you have given, I estimate that you are using from 12 to 24 mL of water per gram of coffee (bulk density 0.25 g/mL).
So, it is not very likely that a change in volume will affect the total quantity of extracted caffeine much. Only the concentration per unit volume will be lower if more water is used.
But then, solid-liquid extraction is not that perfect and possibly you will extract some more caffeine with more water. The main property that influences the extraction is the size of the milled coffee. The smaller the better is the extraction.
Many other components of coffee are extracted fairly fast.
Therefore, here we do a percolation (expresso) which extracts everything with very less water. The concentration of caffeine per unit volume seems to be higher, but the dose of coffee is much smaller.


text:
What is the relation between Rate of reaction and rate of consumption of reactants?

The rate of the following reaction is 0.300M/s. What is the relative rate of change of each species in the reaction?$$\ce{A + 3B -> 2C}$$

We are supposed to find A and B and I understand that we have to use $\frac{ [\delta]A}{[\delta]t}$ and then use a negative for the coefficient. But what I don't understand is, why for B we had to multiply -3 but not -1/3? And is it the same for A?
Also, I am confused on how I can differentiate between using these two equations of rates. The Delta of reactants/delta of time, versus the rate $v(nu)=k[A]m[B]n$
equation.

For your first part you're supposed to multiply it by -3 instead if $\frac{-1}{3}$  as the rate of reaction is given for one mole. That is if $R$ is the rate of reaction it is defined as .
$$\frac{-d[A]}{dt} = \frac{-d[B]}{3dt} = \frac{d[C]}{2dt} = R$$
So the rate of consumption of B is $-3R$.

As for your second part it's a tad unclear for me if you could further edit your question.I could take a shot at answering your second part as well

Thank you very much! I have another question. I hope you can answer it please. How did you find that the rate of the reaction is given for one mole? For the second question I meant that are we treating [A] the same way we are treating the coefficient of [B]?

@WEA I didn't find out that the rate of reaction is for one mole. It is defined like that only. Also we are not treating the rate of consumption  of [A] the same as that of [B]. There rate of consumption of B is thrice of that of A.

Your help was very helpful and clear. I have one more question if you don't mind. So if the coefficient of A is always one does it mean that we have to multiply by the original coefficient with a negative sign?

If the reaction is balanced, then yes. In general Reactants are multiplied with a negative sign to show that they are being consumed in the reaction i.e their concentration is decreasing with the passage of time.


text:
Where am I wrong in my calculations?
I am revisiting the concept of mole and I decided to figure out random values (for instance: mass of 1 molecule of water) to see how much I understand the concept of moles and how it relates to amu. So here is how I started:
There is 1 mole of C atoms in 12g of C and 1 C atom is 12 amu. Therefore, there is 1 mole of 1 amu in 1g. That means there are 6.02e23 nucleons in 1 gram. So, 1 nucleon is 1g/6.02e23= 1.66e-24g. A molecule of water has 14 nucleons (6 neutrons and 6 protons from oxygen and 2 protons from both H); thus, 1.66e23 x 14 should be the g of a water molecule, which turns out to be 2.33e-23. However, this is not the correct mass of water molecule (2.992 x 10¯23 grams).

Who said oxygen contains 6 neutrons and 6 protons?

8............ wow. I had the valence electrons number in mind.Thanks

Note that the atomic mass unit (amu) is deprecated, and instead the newer definition should be used, which is the unified atomic mass unit (u). Also note, that the molar masses of atoms and compound reflect their relative abundance, too. There are different isotopes and it is generally not true to say one carbon atom has the mass 12 u (and especially not 12 amu).

As Ivan pointed out in the comments, I had the molar mass of water incorrect. By fixing the molar mass to 18g/mol, I get the correct answer.


text:
Will the carboxylate groups of the citrate anion undergo hydrogen bonding?
The citrate anion of trisodium citrate has three carboxylate groups. I am concerned with whether or not there will be hydrogen bonding in water, with hydroxyl groups of other ions or alcohols.

It has been suggested to me that the negative carboxylate oxygens won't hydrogen bond, due to resonance delocalisation of the charge. If it mattered whether or not the oxygen was partially charged, I would hypothesize that the oxygen is charged, because carbon has a lower electronegativity.
The oxygen should have two lone pairs, yet it has either been implied that they are not there, or for some reason that they will not interact in hydrogen bonding.

My question is: will the citrate anion hydrogen bond with other molecules or polymers with hydroxyl groups, in aqueous solution?
It has been made aware to me that there is an article that has done research to show that carboxylates are strong hydrogen bond acceptors in anhydrous solvents.

Yes it will form hydrogen bonds all right. Indeed, oxygen has some lone pairs, they are very much there and won't go anywhere, even if it were not charged (though in fact it is charged, which makes hydrogen bonding all the more effective).

@IvanNeretin I completely agree. It will hydrogen bond for sure. I think that it will form hydrogen bonds even if it is chelating some ion (calcium or something like that)

As the comments indicate, you don't need a formal charge on oxygen at all.  "Neutral" oxygen atoms are usually negatively charged anyway just from bond polarity, as in water.  That and the lone pairs on oxygen are quite enough to enable hydrogen bonds to protic hydrogen.


text:
Step up reactions in elementary organic chemistry
What are some elementary (high school level) step up reactions? (in which the length of the carbon chain of an organic molecule is increased by one)
Background: I was looking for a way to methyl chloride to acetone (spoilers: I still haven't found one :( I could use Wurtz reaction to form ethane from methyl chloride, but I don't know of a way to add one more carbon to this alkane. I found this question, but it only focuses on the ketene reagent's action on alkanes. I suppose there might be many more such "step up" reagents and their many more compounds which can be "stepped up". Hence, my question.

PS: A term related to "Step up" is "Step down", which observed in Hunsdiecker reaction and Hoffmann Bromamide reaction.

Cyanide SN2 on halide is a basic one.

@orthocresol Oh, that's good! You also reminded me of addition of grignard reagent to carbonyl compound.

In addition to cyanide,dithiane anion or nitromethane anion on halide are possible

Arndt-Eistert? It's basically high school level reactions in certain order, for stepping up of carboxylic acid


text:
Paint removal from metal
What chemical product works well to strip paint off to reveal bare metal in this instance a small tin (guessing to be aluminium or steel) with as good a finish as possible.

N-Methyl-2-pyrrolidone - typical paint stripper.


text:
Is Dacron and Glyptal the same polymer?
I know Dacron or Terylene is a polymer of ethylene glycol and terephthalic acid . But is Glyptal also a name for Dacron? Glyptal has almost similar monomers.

According to the Polymer Properties Database, Glyptal is a manufacturer of alkyd resins formed from polyhydric alcohols with polybasic acids. Certainly, they might make polyethylene terephthalate as well.


text:
What's this mystery acid?
Today, my dad told me about a cleaning product for membranes he bought a while ago. He told me he put his finger in it and it didn't hurt him at all, nor did it seem to damage any tissues.
But later that day, he brought the substance and checked it's pH with a digital pH tester. It showed that it was pH 0. after diluting it to a concentration of around 20%, he said the pH meter reads 0.68.
I googled "strong acid non corrosive" and it came up with carborane superacid ($\ce{H(CHB11Cl11)}$) as an answer.
Is there a specific answer or is there a number of possible answers?

And what tests can we carry out to identify this acid?
Edit: My dad kept the acid for 15 seconds before rinsing

Acids (with few nasty exceptions) are not immediately corrosive to human skin.

my dad kept the acid on his skin 15 seconds before rinsing it with water

So what? I did the same on more than one occasion (not that I recommend it, but there were no consequences). Strong acids are very much not like what you think they are.

Suggested: https://www.youtube.com/watch?v=XeVZQoJ5FdE

Superacids like the one mentioned in the question are not for sale in stores etc.

This acid is actually not that concentrated.
Suppose it were $\ce{HCl}$, then:
$\frac{10^{-0.68}\cdot 36.45}{0.20}\frac{100}{1000}=4\%\ \ce{HCl}$
I am not going to try on my skin but it is not like the 37% which we have in the bottle.
I guess that since it is for cleaning membranes, it is some chelant for the calcium with a bit of acid for the carbonate.
Edit
I could not resist...
I had some 5% $\ce{H2SO4}$ prepared for adjusting pH so I have tried it after all.
So, no more speculation.

20 seconds with the tip of the finger in it, does not itch;
20 seconds on the hand, does not itch either;
on the tongue, it has definitively an acid taste, but it does not burn. It is, however, too acidic to take it.

Warning
Do never ever touch chemicals, nor taste them.
Always work with proper attire, lab coat, gloves, goggles, etc.

I up voted but probably you should change into that concentrated.... I should change my answer as well as for I am not so sure that my finger started hitching in seconds with just 4% HCl.

@Alchimista I have edited ;-)

Funny ediing :)) But I referred to your incipit . As for you use a strong acid for your demonstration ;)

Don't try this at home kids! :D

@Alchimista Ohh, right. Now I get. I will edit again. Thank you.

I cannot tell you what acid is in the product.
However burning and corrosion of skin and living tissues is not necessarily and directly due to fast acid/base reactions.
It involves reaction such as oxidation, nitration, dehydration,  etc. or denaturing proteins.
For instance the damaging power of concentrated sulphuric acid resides mostly in its dehydrating and oxidizing power.
As such, the skin can get in touch with very acidic media without suffering immediate and permanent damage - and particularly so if the area is immediately and throughly rinsed.
This stated, I am not sure if such a low pH should cause an immediate hitching sensation. In my direct experience less than 15 seconds were enough to always start feeling hitches, independent of the nature and concentration of the acid handled.
On the other hand, we can exclude acid such as HCl, sulphuric acid, nitric acid, phosphoric acid. Plugging a finger on their solutions for 15 seconds will definitively be felt by the operator, unless very diluted,  and burn its skin if conversely they are concentrated enough.
Final note. It is not a good idea to stick a finger in unknown chemicals as for some of them are very nasty, being them acid or alkali or none of the two.

"exclude acid such as HCl, sulphuric acid, nitric acid, phosphoric acid" - not really. pH = 0 is mentioned and it's quite diluted - around 1M.

@Mithiron. I edited it upon reading the answer of R. Kessels. It is because I think to have felt hitching with all of them, including the about HCl 4 % we have in the bench. Then is matter of seconds 5 , 10, 15 etc. More diluted means  no idea as for I am a chemist not an homeopath :))

I think the main danger from sulfuric acid isn't its oxidizing power, it is its dehydrating power.  The reaction of water and concentrated sulfuric acid is extremely exothermic.  Thus, sugar mixed with concentrated sulfuric acid results in decomposition of the sugar into water and carbon, with the water dissolved in the sulfuric acid.  The reaction is pretty spectacular.

@AlaskaRon. Yes, a spectacular example of oxidation as you can see from gas and water evolution plus the formation of tar. However I will edit for clarity, even more as for conc. Sulph. Ac. can suck out water even without reacting.

@AlaskaRon you had me searching for p-nitroaniline videos. Quantity or quality?