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Calculate force on atoms given the energy of configuration, trajectory of simulation and position of atoms in each frame
Background
I am using a neural network to calculate the potential energy of atoms in a configuration and then adding energy of all atoms to compare it with the true energy of the configuration(label) to update the weights.
I am doing so for 10000 configurations of 100 atoms each for liquid Argon at 100K.
Note I don't have the actual energy of each atom and only the energy of the configuration.
Problem
I need to calculate the forces on each atom to see if the sum of forces for all the atoms is zero in X,Y,Z axis. In case it is not I want to find out how off is it from zero.
How do I calculate the force on atoms using their position and potential energy for a given trajectory. I know I need to do $\frac{dE}{dr}$ but what should be my E and r. Also since I don,t a formula for E(r), what should my $dE$ and $dr$ be respectively?

I'm not sure this will be possible if you only know the energy of the configuration as a whole.

I would agree with a cyclohexane. I would imagine there would be any number of possible forces on each atom that would be consistent with a particular total energy.

@fireball.1 Actually, thinking about it more, this might be possible. I'm no expert on machine learning so correct if I'm wrong, but your network should take in the x,y,z coordinates of each atom in order to compute the energy. I believe a common feature in many machine learning libraries is the ability to compute derivatives with respect to an arbitrary input parameter. So, I would think it would be possible to obtain derivatives of the total energy with respect to the x,y,z coordinates of a given atom. This should give the force on the whole assembly of atoms.

@fireball.1 Possibly useful reference: https://arxiv.org/pdf/1502.05767.pdf

As a starting point why not use the Lennard-Jones 6-12 equation for Argon, the parameters for which  are known. This gives the energy vs r . You may need to include a long range potential additionally, as this potential is quite short range.

@porphyrin dude read the question properly. I'm using my own potential function in terms of the NN. I need the net force on system to check its accuracy. Involving LJ makes 0 sense

@a-cyclohexane-molecule is it at least possible to get the force on the whole system (as in 1 configuration) ?

@fireball.1, do you expect that there is a net force on the entire system? With regards to what you've posted in the comments, if you have a potential function, then surely you can differentiate that to get the force. I don't know how you can calculate the energy of your configuration without some sort of known potential function, but I also don't know much about neural networks either. I think porphyrin's suggestion is quite valid and would be my first choice for investigating such a system.

In addition, I'm not sure checking the accuracy of the potential function by obtaining the force from the potential is valid. You could check for self-consistency that way, but accuracy suggests that you need some other method for so doing.

@fireball.1 In regard to your comment to porphyrin: be nice.

@a-cyclohexane-molecule From what I do know about machine learning, he would have a potential function technically, the issue is that it is expressed in terms of the neural network (essentially a composition of a bunch of activator functions associated with given nodes). However, as I mentioned in my comment, I think its becoming more common for machine learning libraries to include derivatives wrt to the inputs, so even if he doesn't know the form of his potential, it should still be possible to obtain the derivatives.

@porphyrin sorry for my conduct. you were just trying to help

@a-cyclohexane-molecule so the NN acts like a black box potential function. And If using that potential i get that the net external force on the system is 0, that would mean my potential values are correct in that 1 aspect.

@Tyberius actually I cant't use that derivative thing because my actual input to the neural network is 2 symmetry functions using the x,z,y coordinates of atoms in my configuration. If I use the small change in energy between 2 configurations as $dE$ then what can i possibly use as my $dr$ ?

@fireball.1 are the symmetry functions differentiable with respect to the coordinates of a single atom?

symmetry function of each atom involves the position of atoms in vicinity too

apology accepted.

tl,dr: Forces are equivalent to the negative of the gradient, use small, arbitrary $\mathrm{d}r$ for each atom
The problem, as you mention, with many neural network methods in chemistry at the moment, is that the $E = f(x,y,z)$ is a black box and thus analytical gradients (and thus forces) are often not available.
First, as you clearly know, $F = -\nabla E(x,y,z)$ so you need to solve $\frac{\mathrm{d}E}{\mathrm{d}r}$ for your neural network. (I indicate this for anyone reading who doesn't realize that the force will inherently be the negative of the gradient of the energy function.
So what you need is a finite difference method. This is, of course, a common problem (e.g., on Math Stack Exchange or Wikipedia):
$$ \frac{\partial E}{\partial x} \approx \frac{f(x+\epsilon)-f(x)}{\epsilon}$$
Picking a good step $\epsilon$ is a bit tricky, but I'd probably pick something like 0.01 Å or 0.001 Å, as a start. In your case, it sounds like you want Cartesian gradients, so you'll take your $f(x,y,z)$ and perform the partial derivatives for all three axes (i.e., the initial point $(x,y,z)$ plus 3 extra energy evaluations per atom.
That is, you'd loop through each atom, move it a bit in the x, y, and z directions and evaluate the system to get the gradient/force on that atom. You'll, therefore, have $3N$ energy evaluations to get all the forces on all the atoms.
You can perform better (and more computationally expensive) finite difference methods if you choose.

Since we are using a constant $\epsilon$ to calculate all forces, that means to calculate the total force i am just doing $f(x+n.\epsilon)-f(x)$ divided by $\epsilon$ and it's value doesn't really matter if the force on system is to be $0$.  Isn't this faulty in a way?

If you're sure that the forces are zero, then in principal $\epsilon$ doesn't matter. In practice, there are numerical issues that do matter, because forces are rarely exactly zero. Moreover, numerical accuracy issues often come into play.

Could you possibly point me to a more reliable reference for this kind of technique? It feels a little too easy you see

@fireball.1 - this is a simple implementation of numerical gradients. If you don't trust me, you can see the links to Math.SE and Wikipedia

@fireball.1 numerical differentiation is a very common technique, particularly in this type of case where it would it be difficult or impossible to compute an analytical derivative.

I'm fine with numerical differentiation. The constant $\epsilon$ is the only point I'm not so sure about.

I'm not sure your question - in any numerical derivative, you pick a small interval $\epsilon$. Picking one is dependent on your problem. Based on my experience, I made some suggestions in my answer.


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Why does Ca(CH3CH2CH2COO)2 dissolve in HCl and not in NaOH?
I do not understand why a strong base, such as NaOH would not dissolve a solid such as this. There would be a double displacement in which you form Ca(OH)2 and NaCH3CH2COO (aq).
Why would HCl dissolve this better?

Is it $\ce{CH3CH2COO-}$ like in the text, or $\ce{CH3CH2CH2COO-}$ like in the title?

At what temperature? Both are solids at room temperature.

Propionic acid is a liquid at room temperature and miscible with water.
Sodium propionate only has a solubility of 1 gram per milliliter. It is common to "salt out" carboxylic acids by adding an alkali metal. I think potassium is more commonly used than sodium.
Calcium hydroxide is relatively insoluble in water, especially in NaOH with a $K_{sp} =5.5\times10^{−6}$. So any $\ce{Ca^{2+}}$ freed from the calcium propionate would form a gelatinous mass around the rest of the calcium propionate thus sealing it off.
In HCl calcium propionate would yield calcium chloride and propionic acid, both of which are very soluble in water.

A solubility of $\pu{1 g/mL}$, or $\pu{1 kg/L}$, doesn't seem very small to me. Sodium chloride, which I consider as well soluble, only comes to $\pu{0.359 kg/L}$ according to Wikipedia.

@Martin-マーチン - It is all relative.  Propionic acid is miscible with water. So relative to  propionic acid, sodium propionate has a lower solubility.

source: Stack Exchange

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

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Electrophilic nature of bromine free radical
My chemistry teacher told me that Bromine radical is a weak electrophile, and that it is more likely to join the more electron rich part of a molecule.
However I could not verify his statement from any book or Google.
Hence I am asking this question on this website: Is bromine radical electrophilic in nature or is there some other reason as to why bromine radical joins the more electron rich part of the molecule?

yes, Br• can be considered 'electrophilic', see e.g. Carey/Sundberg 5th ed, Part A, pg 1004, or also this Q&A on radical chlorination. Similar arguments apply to bromination. https://chemistry.stackexchange.com/q/49850/16683 Note, however, that because Br• is more stable than Cl•, typically the energy of the next radical that is formed plays a greater role in determining selectivity, rather than the energy of the transition state leading to that radical

The statement "Because Br radical is more stable than Cl radical, energy of next radical that is formed plays greater role in determining selectivity" brings up another question. If 1-(1-Bromoethyl)-4-methylbenzene reacts with 1 equivalent of NBS, on which carbon will the bromination occur? The benzylic carbon with three hydrogen or the benzylic carbon that has a bromine as well as a CH3 group attached to it (the CH3 group can stabilise the radical formed by hyperconjugation)?

@orthocresol can you please answer the question I asked in my comment? :)

It's not clear cut (to me, at least). Please don't ask full questions in comments - either add it into your existing question by editing the post (since it provides context for the original question), or ask a new one.

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How to optimise battery usage on modern electronics?
What is the best way to use an external power bank with a modern smart phone or similar device if the goal is to get as much usage out of the combination?
Premises:

The devices has a battery of 2 Ah and the power bank's capacity is 10 Ah
As I understand it, a rule of thumb is that the thermal efficiency η when charging a battery is about 50 %. That is, charging the power bank from (1) requires 20 Ah.

With 1 and 2 in mind, the power bank could charge the phone from 0 to 100 % 2.5 times.
Therefore, my question is: wouldn't it be better to keep the power bank attached to the phone all the time to keep the phone at 100 % as long as possible rather than using it to charge the phone when the battery is half empty or so? My idea is that keeping the power bank attached all the time would make η ≈ 100 % since the phone's battery would just "pass by" the current from the power bank. Am I right?

"That is, charging the power bank from (1) requires 20 Ah." Is that what you mean to state? How would one charge a 10 Ah bank from a 2 Ah phone?

@DrMoishePippik No, I mean that charging the power bank with 50 % efficiency requires 20 Ah

There is always some internal resistance in a battery. All other things being kept constant, the larger the battery, the less internal resistance, so less energy is wasted inside a larger battery due to internal heating and energy loss.
This is particularly important in high-drain devices such as phones and laptops, where batteries are discharged in just a few hours. By having the power bank in parallel, the load is shared and batteries are being drained over a longer time.
For example, a once-cell penlight with an incandescent lamp might burn for an hour. Two cells, one replacing the other, would then last two hours. But two cells in parallel might last 2.2 hours, because of the lower internal resistance and decreased energy waste in the cells.


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Organic reaction with Na2S2O4
Question:

In the first step due to cooperating Inductive and resonance effect C6H5N2 + will be attached at the ortho position of methyl group  for which para position is occupied by Amine group  but I am confused with the second reaction with  Na2S2O4.

Thanku that means C6h5N-  is separated and remaing =N got 2 hydrogen as a process of reduction the pdf is not opening

Sodium dithionite Na2S2O4 (also called sodium hydrosulfite) is a flexible reducing agent that reduces and cleaves ArN=NAr bonds to give ArNH2. It is also good for ArNO2 to ArNH2 and reductions of enones to ketones. Further reading here 1

Thanku that means C6h5N-  is separated and remaing =N got 2 hydrogen as a process of reduction           am i correct in understandingthis way and           the pdf is not opening

You are correct - the other product is aniline. If you are happy with the answer please accept it. PDF should open now

You are correct. Both nitrogens from the N=N system end up with 2 hydrogens so the reaction product is D and aniline (C6H5NH2)

When we are going to add C6H5 N2+to the ring it gets add upto that postion otho to methyl grp bcoz    1 ) +M effect of NH2 is equally activating at both ortho and para postion but position ortho to nh2 would lead to decrease in activating affect of +I by CH3 grp so i thought the position ortho to Ch3 is decided in this way.... pls correct me if i am wrong somewhere.

The directing effect of the -NH2 is stronger than than the effect of the Me group, structure C is never going to be formed as it requires meta addition. Structure D is the only possible answer. Other isomers may be formed but that is beyond the scope of the question, addition to the 2 sides ortho to the NH2 is possible.


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Why doesn't diborane act as an oxidising agent
I read that diborane has 3 center's sharing 2 electrons.
Shouldn't it be easier for diborane to accept one more electron and act as an oxidising agent than a reducing agent .
Could someone tell me what's wrong my reasoning?

A sodium cation has a positive charge. Shouldn't it want to accept some electrons and act as an oxidizing agent? If you don't think that makes sense, then consider if the situation with diborane is analogous.

In this case, the electron acceptor is a  Lewis acid, not an oxidizer.


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Fluorine and fossils

They are now assisted by chemists who are able to fix dates through the analysis of pollen, radioactive carbon, and fluorine found in connection with fossils..

Here the text is talking about basic paleontology.
Could anyone teach me in a simple way how fluorine is connected with fossils findings?
(This is originally about English test (TOEFL), but the subject in question does not in my opinion belong to ELL)

Apparently, fossils absorb fluorine https://en.wikipedia.org/wiki/Fluorine_absorption_dating

I checked the literature, it has been reported that fluorine ions replace hydroxyl ions in osseous tissue (bone) so by making a measurement of the flourine content of bone it is possible to estimate the age of the bones.
One early paper on the subject is Cook, S. F.; Ezra-cohn, H. C. An Evaluation of the Fluorine Dating Method. Southwestern Journal of Anthropology 1959, 15 (3), 276–290. DOI: 10.1086/soutjanth.15.3.3628977.
One method would be to use neutron activation anaylsis on bones to measure the flourine-19, using the n,gamma reaction. This method does have a problem as sodium-23 can form flourine-20 by a n,alpha reaction. As a result a correction will have to be made for the sodium content of the bones.
This is discussed in Cheng, T. P.; Anderson, H. D.; Mills, D. S.; Spate, V. L.; Baskett, C. K.; Morris, J. S. Determination of the fluoride distribution in rabbit bone using instrumental neutron activation analysis. J. Radioanal. Nucl. Chem. 1997, 217 (2), 171–174. DOI: 10.1007/BF02034437.


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What is the more stable conjugate base of cyclohex-2-en-1-one?
We have to identify the structure of its more stable conjugate base obtained by proton transfer of cyclohex-2-en-1-one.

In the following question, there are two possibilities:
We can either remove a hydrogen nearest to carbonyl group or remove a hydrogen next to carbon carbon double bond.
I think the later should be correct as it involves higher conjugation, but the answer is the opposite, they have removed hydrogen which is nearest to carbonyl group. Why is this position preferred?

The level of conjugation is the same, so inductive effects are most important here.

Under kinetic conditions, deprotonation occurs at C6 to give an enolate that is not conjugated with the double bond. Under equilibration conditions, deprotonation occurs at C4 to give an extended, vinylogous enolate. The more stable conjugate base is the one formed under equilibrating conditions (leveling effect).

You may want to look at: https://chemistry.stackexchange.com/questions/43007/what-is-the-most-acidic-proton-of-2-methylcyclopent-2-en-1-one?rq=1


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Effect of salt bridge on Nernst potential
I learned 2 things:

A salt bridge is necessary in a galvanic cell so charge does not build up in the halfcells countering the potential difference created by the separate redox pairs. Indeed, this would stop the flow of electrons eventually.
The potential of the galvanic cell depends on the concentrations of ions of both redox pairs through the Nernst equation.

Now looking up the salt bridge on Wikipedia, I get 'It maintains electrical neutrality within the internal circuit, preventing the cell from rapidly running its reaction to equilibrium.'
So if the salt bridge influences the equilibrium, it must affect the Nernst equation, I was wondering how you could work this in the equation? Is the original Nernst valid when the salt bridge is there? And how do you write it in absence of the salt bridge?

You have the wrong concept.
(1) below is an image of a Cu/Zn galvanic cell (a battery...). The reaction is:
$$\ce{Zn + Cu^{2+} -> Zn^{2+} + Cu}$$
The problem is that if the copper ions, $\ce{Cu^{2+}}$, get into the Zn half-cell then the copper would plate directly on the zinc electrode. So the salt bridge is necessary to keep the the copper ions away from the zinc electrode. That way the electrons flow through the wire.

(2) In measuring the voltage with the Nernst equation it is dependent on the fact that "no" current is flowing. Of course a very very tiny amount must flow, say something on the order of microamps or smaller.
When there is significant current flow then all sorts of nasty things start happening inside the galvanic cell and the voltage drops. So the Nernst equation by itself isn't sufficient to calculate the cell voltage of a galvanic cell when it has a significant current flow.

Thanks for that! Although I see your point that the Nernst equation fails when significant current flows from a practical standpoint, it does predict what will happen when you connect the 2 half cells from your example-> DeltaE=1.10-(0,059/2)*Log[(Zn2+)/Cu2+)]. This means that if we start with 2 solutions of 1M the initial voltage will be 1,10V. Once the reaction starts the concentration of Zn2+ will increase and that of Cu2+ will decrease so the voltage will drop until equilibrium is reached en DeltaE=0. Now what would happen then in absence of a salt bridge? How far could the reaction go?

@Stikke - If you took away the salt bridge on the cell shown the electrons will not flow through the wire. Electricity is about charge flow. Electrons flow through the wire, but ions flow through the salt bridge. You also have to realize that you can't create charged solutions. So $\ce{Cu^{2+}}$ ions lost have to be balanced by losing some anion. Likewise "excess" $\ce{Zn^{2+}}$ ions have to go somewhere.

My idea was that charges would build up creating a field exactly counteracting the potential created by the Redox couple, so a small flow of electrons would be allowed for a short while, after which the reaction would stop.

@Stikke - Actually you're right. An itty bitty teeny weeny amount of current would flow for an even ittier bittier teenier weenier amount of time polarizing each electrode. I don't know if anyone has ever been able to measure that infinitesimal current pulse. You have to realize that solutions don't want to have a net charge. The key point however is that current flow would not form a galvanic cell. In a true galvanic cell a very very small amount of current could flow at a constant rate and at a constant voltage for a long time.

You read that wrong. The salt bridge doesn't influence the equilibrium (provided you're using normal salts and not something insane). If you start transferring electrons from anode to the cathode, you're going to quickly build up excess negative charge that opposes the potential of the cell. Having a salt bridge simply closes the circuit so that charges can continue to move.


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Odorless smoke from 1812 Overture cannon fire
I went to an outdoor performance of the 1812 Overture last night. At the finale, a sequence of 'cannon fire' occurred.
In past years the smoke has had the expected sulfur odor but this year the dense cloud of white smoke that drifted over the crowd had no (to my nose) detectable odor at all. In all other ways the detonations were normal- extremely loud, jet of flame, dense white smoke. Any ideas about what explosives were used?

Most likely they replaced the black powder with nitrocellulose also known as smokeless powder (although it does generate some smoke especially when used in artillery) which does not contain sulfur and hence the lack of smell.

Possibly... But the white smoke produced was pretty dense.

Yes, smokeless gun powder is not exactly smokeless, just produces less smoke than black powder. The guy here: https://www.youtube.com/watch?v=yceuluh_pJE is comparing smokeless gunpowder with black powder and both produce quite a bit of smoke.


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Mechanism of sodium hypochlorite and hydrochloric acid
The overall reaction for the reaction between NaClO and HCl is: $\ce{NaClO + 2HCl -> NaCl + Cl2 + H2O}$

To try and understand how this reaction occurs, I've thought of a simple mechanism for it. However, I don't even understand my own mechanism. Here's what I proposed:

$\ce{NaClO + HCl -> NaCl + HClO}$
$\ce{HClO + HCl -> Cl2 + H2O}$

The mechanism looks like it could work, but whether it would work is another question.
One problem I have with understanding it is the second step: how exactly would that work? My first thoughts were that since HCl is the stronger acid out of the two, H2ClO+ and Cl- would be produced in the second step, and a third step would have something happen between both of these products to finally produce H2O and Cl2. However, I scrapped that thought since surely an ionic compound would then be produced and nothing would happen afterwards.
Now, I'm lost on what would really happen in the second step.

Maybe I've got it all wrong, but does anyone have any other thoughts on this? Or are there any actual proposed mechanisms for this out there?

For step 2, write HClO as HOCl. HOCl is a weak acid (pKa=7.5) that can be protonated on oxygen by HCl to give H2OCl+ (+ charge on oxygen) and Cl-. Have Cl- displace H2O from H2OCl+ to afford Cl2 (think SN2 rxn.)

@user55119 That's what I was thinking. After you said SN2 reactions, my thoughts are that a lone pair on the oxygen atom accepts a proton from HCl. The Cl- ion then attacks the chlorine atom in H2ClO+, hence Cl2 and H2O are formed.


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Concentration of excess reactant
If increasing/decreasing the concentration of an excess reactant influenced the percentage yield of a product experimentally, how could this be explained?
I know that in theory, the excess shouldn't affect the yield- since it's the limiting that decides how much is produced. But is there any possible explanation for this result?

RE: I know that in theory, the excess shouldn't affect the yield... Says who?!?

@MaxW Says almost every other high school textbook that I've read. If the given statement in the question is true, I personally think it would make for a good question to clear the misconception of high schoolers.

See Le Chatelier's principle https://courses.lumenlearning.com/boundless-chemistry/chapter/factors-that-affect-chemical-equilibrium/

Yes, but Le Chatelier's principle only applies to systems at equilibrium. If I am reacting some organic goo there is no guarantee that the system reached equilibrium. Kinetics could play a big factor in the yield.

But even if we use Le Chatelier's principle, wouldnt it still result in the same amount of yield since the amount of limiting reactant is kept the same??

@MaxW  Could you please help me suggest how I could use Kinetics to explain this??? My teacher said that more concentration of acetic anhydride and the increase in yield of aspirin may be because of the solubility in room temperature (some just returned back to its solid form without reacting)? Do you think this is valid?

@RaoulKessels gave a good example in his answer.

A very typical situation in organic chemistry is the following:

$\ce{A -> undisired products}$
$\ce{A + B -> target product}$

Here it is easy to see that increasing the concentration of $\ce{B}$, we will get a higher yield of the target product.
The other reaction can be of many kinds.
$\ce{A}$ can decompose in the reaction medium, can react with itself, with the solvent, with byproducts form the desired reaction, etc.

Thank you! But does the "increasing conc of B results in higher yield of target product" take into account that there is a limiting and excess reactant?

Here the idea is that A is limiting and B in excess

Doesn't:
A⟶undesired products
A+B⟶target product
just mean that the addition/inclusion of B would produce the target product?
If there is an equal amount of B with A, this would make the same amount as when there is an excess of B.
How could this prove that increasing concentration of B would result in a higher yield?


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Nucleoside/nucleobase chemical synthesis
I'm interested in the chemical synthesis of nucleoside phosphoramidites, but all the procedures I can find have "start with a protected nucleoside" as the first step. Searches for chemical synthesis of nucleosides all seem to find process that start with existing nucleobases, or in vivo processes, or speculative origin-of-life processes. Searching for chemical synthesis of nucleobases ONLY seems to find speculative origin-of-life processes.
What I'm looking for (just out of curiosity) are processes for chemical synthesis of nucleosides from a non-biological basis (not using pre-existing nucleobases) as used in typical commercial/industrial supply of nucleoside phosphoramidites. I'm starting to wonder if they even are chemically synthesized, or if they are instead harvested in some way, but it's hard to find a smoking gun that outright says it one way or the other without access to whitepapers anymore (no more university account).
Does anybody know?

Why would any practical man make something that's lying all around?

Hypothetically, a business might do so if it was cheaper to obtain a salable product by e.g. mixing some petrochemicals in a vessel than by fermenting and sufficiently purifying a biological product (apparently not, by your implication). Even if I knew all the options I wouldn't be able to judge the economics of each at commercial scale to arrive at the answer deductively. Hence my question.

So something like this: https://en.wikipedia.org/wiki/Purine#Laboratory_synthesis ? Well, over century ago guys made nucleobases just to see they could...

Yes, like that. Thanks, that's great context. A feasible process but it sounds like it isn't used to produce commercial quantities.

Finally found what I was looking for, especially paired with Mithoron's comments.
"Currently, purine nucleotides are obtained at industrial level mainly by microbial fermentation".
https://microbialcellfactories.biomedcentral.com/articles/10.1186/s12934-015-0234-4
Ledesma-Amaro, Rodrigo & Jiménez, Alberto & Santos, María de los Ángeles & Revuelta, José. (2013). Biotechnological production of feed nucleotides by microbial strain improvement. Process Biochemistry. 48. 1263-1270. 10.1016/j.procbio.2013.06.025.


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Does a stoichiometric coefficient represent molecules or moles of a substance?
I understand that the one mole of atoms is equal to $6.02\times10^{23}$ atoms. However, when I used various resources to learn about stoichiometry, they showed me a balanced equation:
$$\ce{2H2O->2H2 + O2}$$
and said "2 moles of water" decomposed to form "2 moles of molecular hydrogen" and "1 mole of molecular oxygen". It is my understanding that there are only 2 molecules of water, not $2 \times 6.02 \times 10^{23}$ molecules of water and so on with the other parts of the equation as well. Could someone please clear up my confusion?

Balanced equations represent ratios of reacting chemical species, not absolute quantities. What we mean when we say $$\ce{2H2 +O2->2H2O}$$ is not literally that two molecules of hydrogen react with one of oxygen, but rather that two units of hydrogen react. These units could be moles or molecules or something else entirely.

I'll just add that the units can't be weights or masses. The units ultimately have to be proportional to the number of molecules.

The balanced equations are just ratios. Read your example closely, it states that:

2 moles of water decomposes to form 2 moles of hydrogen and 1 mole of oxygen.

Note that all of the units are in moles and the statement does NOT say:

2 moles of water decomposes to form 2 molecules of hydrogen and 1 molecule of oxygen.

The mole unit is just a multiplier.  The question could have alternatively said:

2000 molecules of water decomposes to form 2000 molecules of hydrogen and 1000 molecules of oxygen.

or

$12.04\times10^{23}$ molecules of water decomposes to form $12.04\times10^{23}$ molecules of hydrogen and $6.02\times10^{23}$ molecules of oxygen.

and the implication of either would be the same. In all cases the ratio of water:hydrogen:oxygen is 2:2:1 respectively.


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What does it mean for an equillibrium to move?
Le Chatelier's principle was described to me as:

If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

I do not understand the phrase, "...the position of the equilibrium moves to counteract the change."
How can the equilibrium "move?" I know that the equilibrium constant cannot change so that cannot be it.

For starters, it should use the term "in response to" instead of "to counteract".

Also, I have an answer to a related question that may help provide clarity.
https://chemistry.stackexchange.com/questions/75619/can-we-force-reactions-with-positive-delta-g/98708#98708

@A.K. I'm sorry but I don't understand how that answer helps with this question. It seems beyond anything I'm learning.

Hi Leif Metcalf, welcome to Chem.SE! Your question seems to be possibly a duplicate of What does it mean to shift equilibrium?. Have a look at that question, and if it doesn't help, feel free to [edit] your question to offer more clarification, as to how your question is different from the one I linked. Thanks!

Also related: What does “to the left” mean? and What is the difference between the equilibrium position and the equilibrium constant?

Not "conditions," just changing concentrations.

Let's keep this simple. Think about adding vinegar (acetic acid) to water.
$$\ce{CH3COOH + H2O <=> H3O^+ + CH3COO^-}$$
The equilibrium equation is
$$\ce{K_a} = \dfrac{\ce{[H^+][CH3COO^-]}}{\ce{[CH3COOH]}}$$
Le Chatelier's principle simply says that if I add more acetic acid to the water then the solution gets more acidic. However if I add more water, then the solution gets less acidic. Hence the equilibrium moves.(In this case it works to a point. Adding a drop of water to glacial acetic acid isn't an aqueous solution...)

It "moves", huh? ;-)

I think it is better depicted by $K_\mathrm{a}$ (acid dissociation constant) instead of $K_\mathrm{\alpha}$, isn't it?

@GaurangTandon - Thanks for pointing out blooper...


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Why are the concentrations in the equilibrium constant multiplied?
$$\alpha\,\mathrm{A}+\beta\,\mathrm{B}+\cdots\rightleftharpoons \rho\,\mathrm{R}+\sigma\,\mathrm{S}+\cdots$$
The equilibrium constant for this reaction can be calculated using the following formula:
$$\mathrm K_c = \frac{{{[\mathrm{R}]}}^\rho {{[\mathrm{S}]}}^\sigma ... } {{{[\mathrm{A}]}}^\alpha {{[\mathrm{B}]}}^\beta ...}$$
In this formula for the equilibrium constant of a reaction, why are the quantities R and S (and likewise A and B) multiplied?
I understand the equilibrium constant to be a ratio of the amounts of the products and the amounts of the reactants. But if this is the case, wouldn't a formula like $\mathrm K_c = \frac{[\mathrm{R}] + [\mathrm{S}] ... }{[\mathrm{A}] + [\mathrm{B}] ...}$ make more sense? I think this question is really a result of not understanding the meaning of the equilibrium constant.

It's a standard result from classical thermodynamics, any university level Physical Chemistry text book should cover the derivation - or look at http://www2.mcdaniel.edu/Chemistry/ch307.notes/Chemical%20Equilibrium.html

As discussed in comments, you can find a good derivation in a standard university level physical chemistry or thermodynamics textbook.
Let's instead consider why it should be true.
From a microscopic perspective, what we're trying to balance, in essence, is the density of states on both sides of the equation, in essence, the count of ways to rearrange the various pieces. These are heavily dependent on the relative spacing of energy levels and degrees of freedom on both sides. Fortunately, this complex information is captured in the value of $K$.
But the point is that if you're trying to find the total number of permutations of two different collections, you multiple the values. For example, if there are two dice, the total number of outcomes is 36.
And concentrations essentially measure the number of things to rearrange, so the total number of possibilities should scale with the product, not the sum.
This is all a bit hand-wavy, but again, I'm not trying to derive the answer. I'm just trying to suggest that even without a formal derivation, it seems very reasonable that the correct operation to use is multiplication.
EDIT:
After reading this again, I realized I can extend this with another hand-wavy justification.
Remember that we have:
$$\Delta G = -RT \ln K$$
If you have two reactions that you combine together, you want the energies to sum. But that translates into products inside the log! Once again, the proper operation for equilibrium constants is multiplication.


text:
What is the chemical name of this derivative of aniline?
What is the chemical name of the following compound?
$$\ce{(C6H5)-N=CH2}$$
I doubt that it might be N-phenyl methanal oxime . But  , it can't be oxime of OH ion is absent.  What would it be?

It is N-phenylmethanimine being an imine.

The preferred name is N-phenylmethanimine, but often it is named N-methyleneaniline (or N-methylenaniline, more correct one would be N-methylideneaniline though).
UPDATE a typo (excessive "e" in bogus "phenylmethaneimine") fixed. (thanks, Loong)

+1 If you want, you can include this link in your answer (https://pubchem.ncbi.nlm.nih.gov/compound/N-Methyleneaniline#section=Top). Your call though :-)

@paracetamol I don't :) already seen it, but kinda intentionally did not mention it, since the bunch of various names (sometimes bad) is not prioritised (i did little literature search what name is often used); it's a helpful comment anyway


text:
Does hydration show kharasch effect?
In the conversion of propene to 1-propanol, can we just use water in presence of peroxide instead of first converting it into 1-bromopropane and then reacting with $\ce{NaOH}$?

Please keep in mind isomers when talking about organic compounds.  your question is not immediately obvious that you are talking about 1-propanol until a reader reaches your discussion of 1-bromopropane. also knowing the specific/general peroxide you have in mind would be helpful.

No, you can't.
If you observe the mechanism of Kharash effect, an $\ce{HBr}$ molecule first undergoes homolytic bond cleavage to produce a $\ce{Br}$ free radical. To undergo that homolytic bond fission, the electronegativities of the two elements, attached by that bond, have to be very close, which is possible for $\ce{H}$ and $\ce{Br}$ ($\chi _ {\ce{H}} = 2.1; \chi_{\ce{Br}} = 2.96$). The benzoyl peroxide is just a free radical initiator.
Now if you come to water, electronegativity difference between $\ce{H}$ and $\ce{O}$ is huge $(\chi _ {\ce{H}} = 2.1; \chi_{\ce{O}} = 3.44)$. So, it is not possible at all to undergo homolytic bond cleavage of $\ce{H-O}$ bond even in presence of peroxides.
To do hydration according to anti-Markownikoff's rule (without any rearrangement), the most widely used method is Hydroboration-Oxidation. This is an alternative and better than doing first bromination in presence of peroxides and then hydrolysis, because in many other cases chances of rearrangement also remains there.


text:
Why electron configuration of tungsten doesn't obey the full & half full (stability) rule?
So I was reviewing chemistry lesson and I am curious when I saw the electron configuration in the VIB group:
For $\ce{_24Cr}$ we write the electron configuration $\ce{_18[Ar] 4s^1 5d^5}$ instead of $\ce{_18[Ar] 4s^2 5d^4}$,
For $\ce{_42Mo}$ we write $\ce{_36[Kr] 5s^1 4d^5}$ instead of $\ce{_36[Kr] 5s^2 4d^4}$,
But why for Tungsten $\ce{_74W}$ we do write $\ce{_54[Xe] 6s^2 4f^14 5d^4}$ and not $\ce{_54[Xe] 6s^1 4f^14 5d^5}$?
Is it because after the $\ce{f}$ subshell the rule is not valid anymore?

Chemistry 'rules' are more often suggestions. The more electrons, the more interactions there are. Remember that the 'shells' we use are solutions to the single-electron Schrodinger equation, not the full 74-electron problem. Perhaps what should be surprising is that they work even half as well as they do.

https://chemistry.stackexchange.com/questions/41304/why-tungsten-does-not-have-stable-configuration


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How to determine charge of compounds in solution with many amphoteric functional groups?
I am curating a metabolic model that has compounds such as NADH/NADPH/Etc. that have multiple ionize-able hydrogens. Normally with amino acids there are only a few hydrogen with varying pKas that makes it easy to find what the charge the of the molecule is and the formula at a given pH. However with NAD, I can't get a good idea of what the actual formula is at pH 7ish. I have been using Kegg, ModelSEED, pubChem, etc. There is alot of conflicting information. What is the charge/formula of NAD in particular and more important, what resources are best to determine molecular compound charges at various pHs.

NAD = nicotinamide adenine dinucleotide?

Yes that is the molecule.


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Length of NaH bond from the edge length of unit cell
If the edge length of $\ce{NaH}$ unit cell is $\pu{488 pm}$, what is the length in $\pu{pm}$ of an $\ce{NaH}$ bond? The answer is $\pu{244 pm}$, I thought that $d = 2 \sqrt{2}\times r$ , do you not need to know the radius for this?

Please elaborate what you mean by radius.  Also note the edits made to format number as equations "$ $" for equations, "$ \ce{} $" for chemical formulas and equations, and "$ \pu{} $" for units.

$\ce{NaH}$ has a rock salt type structure meaning each edge has two $\ce{Na-H}$ bonds. When working out bond lengths in a unit cell, you assume the atoms are point charges (i.e. they take up no volume) and so you do not need to know the radius, only the unit cell parameter which as you have said is $\pu{488 pm}$. Since there are two bonds along each edge, the bond length is simply half of the unit cell parameter, therefore $\pu{244 pm}$.
Finally it dawned on me, why you may have been concerned about knowing atomic radii, yes, you need to know the radii to determine whether $T_\mathrm{d}$ or $O_\mathrm{h}$ sites are preferred. However, given the context of the question, there must have been something previously alluding to the octahedral sites being occupied (and hence rock salt structure) rather than the tetrahedral sites (which would give you a ccp structure with half of $T_\mathrm{d}$ sites occupied). However, you still wouldn't need to use the ionic radii in any calculations pertaining to bond length. I have also included a drawn picture of the unit cell of Sodium Hydride to help you visualise how to calculate the bond length.


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Where to find information about galvanic deposition of various metals?
I am looking at the process to deposit Cu and Ag onto metallic surfaces using galvanic techniques. Unfortunately this works not as expected.
For example I use 2 rods of Cu in a bath of CuSO4 with some drops H2SO4. When applying a voltage, I get -- depending on the voltage -- either a block cover that can be wiped with a towel of a very rough Cu surface. I did not try with Ag/AgNO3 yet.
Anyway I have the problem, that I am looking for information or a form where I can post my problems and get useful tips, what I can do to enhance my baths.


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Hydrogen molecule potential energy graph

In the above graph, I was confused at the point where the internuclear distance increases and potential energy become zero. Though internuclear distance is very small and potential energy has increased to zero.
Another question that though the internuclear distance at a particular point is constant yet potential energy keeps on increasing.
I think the point where potential energy become zero we can say that attractive forces between electron and nucleus of bonding hydrogen atoms just equals the repulsive force between bonding hydrogen electrons. It's just what I am assuming not sure either its true or not.

Normally, when the atoms are separated by a huge amount (infinity) we take this energy to be zero. Bringing the atoms closer to one another has an attractive potential and the bond has formed on reaching the lowest point on your curve.  Moving the atoms closer together increases the potential energy because here we push electrons closer to one another and this energy increases rapidly  with each small decrease in separation due to repulsion of similar charges. The slope of the curve gives the force on the atoms, so at small separation they experience a large force and repel one another. At large separation, where the slope is positive, there is a smaller attractive force.

Why there is no activation energy for this reaction?

There does not have to be an activation barrier, it just depends on the reaction and how the energy varies with separation of the atoms. Some electron transfer reactions are activationless for example. Many other reactions are so fast because the activation barrier is tiny and the rate constant is limited by how quickly the reactants diffuse together.

@porphyrin How a potential diagram like this be also used for describing intermolecular interactions? I mean in the example of the OP it describes the formation of hydrogen molecule. But we also use Lennard-Jones potential for intermolecular forces. What I can't understand is how to interpret the minima in such cases. Is there a "bond formation" when talking about intermolecular potentials?

@Anton, there is no electron sharing in this case, polarisability and dipoles provide
the interaction and this can have a minimum vs separation of species. Ask a question if you want a complete answer :)

How to find the equation of this graph?

@Shub If you want an accurate plot then you will have to calculate exactly for say H$^+$ but if you look for a Morse potential ora Lennard -Jones one these give a good approximation.


text:
Change of first order rate constant with temperature

In the decomposition of $\ce{N2O5}$, the time taken to reduce the concentration of $\ce{N2O5}$ half of the original amount is 12 min at $\pu{50^\circ C}$, while 5 h at $\pu{25^\circ C}$ and 10 days at $\pu{0^\circ C}$.

How can we understand that the given assertion is correct because the temperature is different every time?

Because the temperature affects the decomposition kinetics (as is generally expected for any kinetic process). What is your question?

First of all, decomposition of $\ce{N2O5}$ is a first order reaction, not a zero order reaction. So, your question headline needs to be corrected.
You can justify the given assertion by simple mathematics. For a first order reaction, we know that, $$t_{1/2} = \frac{\ln \space 2}{k}$$ which is independent of initial concentration. It only depends on the rate constant which will change if you change the temperature.
From Arrhenius equation also, we know the relation,
$$\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} -\frac{1}{T_2}\right)$$
So here we have three rate constants for three different temperature. Thus we will have at least two sets of above equations. Dividing one by another we will have,
$$\frac{ln\left(\frac{k_2}{k_1}\right)}{ln\left(\frac{k_3}{k_2}\right)} = \frac{\left(\frac{1}{T_1} -\frac{1}{T_2}\right)}{\left(\frac{1}{T_2} -\frac{1}{T_3}\right)}$$
Now, from the relation between half life and rate constants we can have
$$\frac{\ln\left(\frac{\left(t_{1/2}\right)_1}{\left(t_{1/2}\right)_2}\right)}{\ln\left(\frac{\left(t_{1/2}\right)_2}{\left(t_{1/2}\right)_3}\right)} = \frac{\left(\frac{1}{T_1} -\frac{1}{T_2}\right)}{\left(\frac{1}{T_2} -\frac{1}{T_3}\right)}$$
Now from the given assertion by putting the values you have to just verify whether both the sides are giving the same value or not. In this case both sides, gives the value ≈ 1.2. So, I guess this assertion is thus correct.


text:
DNA dye: Are there any natural dye for DNA (instead of artificial ones)?
In order to make easy and low-cost experiments, we are searching for natural dyes (or easy to make) that can color DNA, instead of for instance Ethidium Bromide that is hard to find and is expensive.
We are new to this field, but we learned that DNA dyes need to intercalate in between DNA bases. To do so, the molecule must be preferably poly-cyclic, aromatic and planar.
So the question is, is there any chemical that can be readily used for this purpose or if not, any lead about a molecule that can be easy to find by anyone.

Rhodamine dyes are very cheap (used as dyes in fishing and easily purified) or
you could synthesise some porphyrin.

Thank you both for your answers. Clarification: we are trying to get dyes that can be used by children or non-scientific persons. Methylene blue might be the best option, but we are trying to get even more accessible chemicals, like lawson, cephalopod ink or beetroot (we did not try them yet).

In that case why not try some food dyes or writing inks. You can probably look up their structures on line and choose flat aromatic types that can  intercalate, although binding is also possible in the major and minor grooves.

Though not a naturally occurring dye, methylene blue has been used as a DNA stain. It is inexpensive and readily available. It is also comparatively nontoxic, and is used as a medication in pisciculture, so it can be found in pet stores.
If you choose to use methylene blue, save some for the blue bottle demonstration, a sensitive test for oxygen. Vitamin C or other reducing agents can be used instead of glucose.

Ellipiticine is a molecule that meets most of your criteria.

a "natural" molecule, produced by trees native to Hawaii and/or Australia.  It therefore meets your ask for "natural" molecules.

a DNA intercalator, binding to the major groove of the DNA helix.

a UV absorber whose spectrum changes on binding DNA.

a fluorophore whose fluorescence properties are modulated by solvent and/or binding to DNA.

However it is not easy to get.  Ethidium bromide is probably easier.


text:
What is the specific heat and latent heat of an ethanol-water mixture?
How do I find the specific heat and latent heat of ethanol-water mixture? Let’s say the mole fraction of water is 0.3 and ethanol is 0.7.
I am working on a project and needed values for these and couldn’t find it.

So did you try googling 'specific heat ethanol water mixture'? That results in several good looking hits...


text:
Can an ant etch a hole in a tent with formic acid?
Given a polymer foil tent, could ant's formic acid help to etch a hole in it? I thought of a polyester or polyamid foil. If one ant's acid is not enough, they come as team.
This is not a theoretical question beacause I have observed it on a camping in South France. I am not just sure whether they just bite through, or indeed use their acid.
https://ibb.co/m5u5B8

I would be VERY surprised if it did. The answer depends on what material the tent is made of. Most tents now days are made of polyester and according to the table below (taken from: http://rfelektronik.se/manuals/Datasheets/solventguide.pdf) polyester has good resistance in formic acid (Note, S stands for satisfactory in table).
However, even if it was nylon, unsatisfactory resistance means that the material will deteriorate in hours to weeks when used with the chemical. I dont think an ant or some ants would be able to produce enough acid to seriously damage the thick fibers of the tent. Maybe if a lot of ants were attacking the same spot one after the other you would eventually dissolve some of the nylon to make a small hole, you never know.
Also, an ant attacking a tent is a very unlikely scenario (ants use formic acid to defend themselves).

Thanks for your answer; actually, they really did it! https://ibb.co/m5u5B8 may be just biting through, if you say the acid is rather not relevant.

@J.Doe So, assuming the hole was not already there and the ants just found a way in it, i wouldnt be surprised at all if they chew through. That is their job after all! For example carpenter ants chew through wood and also see here: https://www.reddit.com/r/ants/comments/8flc4y/can_ants_eat_through_plastic/


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Does Henry's law depend on curvature of gas/liquid interface?
Could you tell me Henry's law depends on curvature of gas/liquid interface or not? It would be better with references.

This could be a good quesiton if you edit it to show an attempt to solve it.  Otherwise it will likely be closed as a homework question.  Also see the Kelvin equation

No, but not because the vapour pressure of mixtures doesn't depend on curvature (it does), but because Henry is a narrowly-defined law that doesn't contemplate surface curvature.
Henry's law is, after all, a barely better description of vapour pressure in mixtures than Raoult, and that's because it introduces an experimental value (the Henry constant) for each solute/solvent pair. The only reason why it's (rarely) used is that it's extremely easy to operate with. And pretty much the only reason why it's taught is because it's a simple, mathematically trivial treatment that moves away from ideality; it's sort of the minimum, simplest step away from ideality, which is convenient for a classroom.
But if you attempt to combine Henry with the effect of surface curvature, say by combining it with Kelvin's equation, you run into a serious problem. The equilibrium difference of pressure between a liquid, $p^L$, and its vapour, $p^V$, when the liquid has a curvature of radius $r$ is
$$ p^L = p^V + 2\sigma/r$$
where $\sigma$ is the surface tension of the liquid. And that's when you run into a problem - surface tension of a liquid mixture also depends on the composition, so the entire point of Henry's law (that concentration of a species in solution is proportional to its partial pressure with a proportionality constant $H^{cp}$) is lost. Even if you accept the simplest case (that surface tension behaves ideally with composition following Gibb's isotherm) you are stuck with a $H^{cp}$ that is itself a function of $c$, and so the mathematical model has definitely moved away from Henry.
By which point you are better off using other, more realistic models for your non-ideal mixture. Note that even then, you are stuck with iteratively computing the crossed effects of surface tension on vapour pressure and vice versa, as the relationship can't be solved analytically as far as I know.


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IUPAC Nomenclature for aldehyde

I was trying to name the following compound and I thought that if both the aldehyde group carbons should be included in the parent chain or I should go with the longest chain . I saw the solution for this and got two answers as I) 3-(Formylmethyl)hexanal and 2) 3-Propylpentanedial.
Can someone please help me in naming it correctly??

IUPAC (2013) rule

P-44.1.1 The senior parent structure has the maximum number
  of substituents corresponding to the principal characteristic
  group (suffix) or senior parent hydride …

(more e.g. here) is applied in the phase of selecting "parent structure", before than the "longest chain rule" (P-44.3.2, e.g. here) in the phase of selecting the "principal chain".
Therfore the latter name, 3-propylpentanedial, is correct.

So the first name is wrong?Also I was thinking if it should be 3-Propylpentane-1,5-dial or just 3-Propylpentanedial

@user395062 yes, in this case I'd call it wrong, because it does not indicate very well the presence two equivalent aldehydic groups. And "1,5-" is not necessary at all, as aldehyde group(s) can be only at the end(s) of the chain, like in dicarboxylic acids (e.g. butanedioic, not butane-1,4-dioic acid).

Can you please explain what do you mean by equivalent aldehydic groups?

@user395062  Note that the propyl group is attached to the central carbon (C3) atom of the pentanedial molecule. So, the molecule is symmetric. But in fact this does not have to do much with IUPAC names. I rather just meant that the …-1,5-dial name indicates the presence of two aldehyde groups, but it is not obvious from the 3-(formyl… name. But this name is not created correctly according the rules and their order, anyway.

If there had been -OH group in place of second aldehydic group then also we would have included that part of the compound in the parent chain?


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When to include or exclude lone pair counting electrons according to Hückel's rules?
When counting the number of pi electrons, in furan only one lone pair was included, but the other one was not.
In the cyclopentadienyl anion (2nd aromatic compound in the picture) the lone pair was counted.
When to include and exclude lone pairs in Hückel's rules?

As written on Chemistry Libre Texts,

Perhaps the toughest part of Hückel's Rule is figuring out which electrons in the compound are actually π electrons. Once this is figured out, the rule is quite straightforward.  π  electrons lie in p orbitals and  sp2 hybridized atoms have 1 p orbital each. So if every molecule in the cyclic compound is sp2 hybridized, this means the molecule is fully conjugated (has 1 p orbital at each atom), and the electrons in these p orbitals are the π electrons.

In furan, there are two lone pairs, one in the sp2-orbital, and the other in the p-orbital. The latter is hence the set of π electrons.
In the cyclopentadienyl anion, the lone pair is present in the p-orbital, as all the sp2-orbitals are involved in bonding.
Thus, π electrons and not just lone pairs are to be counted in Huckel's Rule.

hybridization state cannot be easily deduced from the structure formula though..

@mykhal Hybridisation can, in most cases, be deduced from the structure directly, especially for organic compounds

Argh, how I dislike the Chemistry Libre Texts. For the following reasons: (1) π electrons lie in p orbitals is not necessarily true, as long as the symmetry fits the pi orbitals, others can participate, too. (2) sp2 hybridized atoms - only orbitals can hybridise. (3) So if every molecule in the cyclic compound is sp2 hybridized, this means the molecule is fully conjugated - No. Point in case: Why is cyclooctatetraene non planar but the cyclooctatetraenide anion planar?.

@Martin Thanks... Good to know... What's a more trusty website to refer?

chemistry.stackexchange.com :P


text:
Flexible & thermally conductive materials?
Quick question: Are there any common or easily obtainable materials that are thermal conductors, but also flexible? By flexible, I mean along the lines of a plastic bag, with similar or greater durability.

Aluminium foil. Dunno where one can get one thick enough to be durable.

Depends on how good of a thermal conductor you need. For example, silver-loaded epoxies are better thermal conductors than plain epoxy, so are often used in microelectronic packaging. But, the silver-loaded epoxy will never be confused with pure silver for thermal conductance.

There is a reasonable answer that does not involve aluminium. Thin silicone sheets are extremely available, but an even better solution to your problem would be a polyurethane elastomer with a thermal conductive filler however these are less readily available. Obviously they are not going to be as thermally conductive as metals but that is the sacrifice you make (currently) for flexibility.
Hopefully in the future graphene will be cheap and readily available but with the green energy crisis, I doubt it!

Welcome to CHEM SE!If you would like to know more about how the site works then take a [tour].We also support $\LaTeX$ for formatting chem/math expression see this link for more details.

I always assumed silicone was an insulator, considering some water bottles/bladders are made of it. Figured manufacturers would want to avoid thermal transfer. It's actually the opposite of what I'm looking into; essentially an expansion chamber type of water bladder in a closed loop system that can still radiate heat. If it is the case that all silicon can transfer heat more effectively than plastic, then perhaps I could just use one of those water bladders as initial experiments.

In general it is, but some silicones are designed for the purpose of heat dissipation and are commonly used in electronics for such a purpose.


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A simple equilibrium problem
Take the following simple equilibrium:
$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}$$
Ethanoic acid will act as a typical Brønsted-Lowry acid and donate a proton to water, forming a conjugate acid-base pair. In this case, H3O+ is the conjugate acid, and CH3COO- is the conjugate base.
Logic tells us that if we have a solution of ethanoic acid and water, increasing the amount of acid will decrease the solution's pH. This makes sense, since there'd literally be a higher acid concentration.
However, the interpretation I get due to the equilibrium is the opposite. I see it as increasing the concentration of ethanoic acid will mean there's a larger concentration of conjugate base, hence the solution will become more basic.
Is this flawed logic or simply a misunderstanding of Le Chatelier's principle?

Flawed logic. Ignoring autoionization of water, $\ce{[H^+] = [CH3COO^-]}$. $\ce{H^+}$ is a strong acid, $\ce{CH3COO^-}$ is a weak conjugate base, so the solution is acidic.

@MaxW That doesn't explain much. $\ce{[H+] = [CH3COO-]}$ is meant to suggest what exactly? How does that statement suggest that the solution is acidic?

Sorry that you don't understand. Assuming that $\ce{[H+] =[CH3COO−]}$ implies that the concentration of $\ce{CH3COOH}$ should be above about $1\times 10^{-5}$ molar. In reality $\ce{[H+] =[CH3COO−] + [OH^-]}$.

@MaxW I've been reading about how conjugate acids/bases affect equilibria. I read that weak acids produce strong conjugate bases; in this case, ethanoate is a strong conjugate base. Strong conjugate bases shift equilibria to the left since the tendency of them to accept protons is much greater than the tendency of weak acids to donate them. So is this why the solution will become more acidic?

$\ce{CH3COO−}$ is a weak base compared to $\ce{OH−}$.

Do you understand now, or still not get it? In aqueous solution there is no stronger acid than $\ce{H^+}$, nor any stronger base than $\ce{OH^-}$.

There is an example calculation given here https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566#95566

You are neglecting the hydronium ions produced during dissociation of acetic acid.
We can consider the following sequence of events when we add more acetic acid to solution. In reality they occur simultaneously, but it's easier to think about it sequentially.

We have more than the equilibrium concentration of acetic acid, so some of it dissociates to produce $\ce{H+}$ and acetate ion.
Now we have more than the equilibrium concentration of acetate ion, so some of it reacts with $\ce{H+}$ to form acetic acid again.

Your argument is that step 2 makes the solution more basic, but you are neglecting the effect of step 1. Every extra acetate ion produced in step 1 also involves the production of a $\ce{H+}$. Furthermore, step 2 has to occur to a lesser degree than step 1, because step 2 consumes the acteate ions produced in step 1. The acidity of the solution must therefore go up as a whole.*

*Of course it's also possible that steps 1 and 2 both occur at nearly the same rates, and then the pH of the solution doesn't really change much as a result. This is what happens in a buffer.

Could you explain in a bit more detail as to why step 2 occurs at a lesser degree than step 1? You only said that it 'has to'. My interpretation of your answer is that since step 2 is happening less than step 1, there will be an increasingly greater $\ce{[H+]}$ in the solution. However, the ethanoate produced will form more ethanoic acid, which in turn produces even more $\ce{[H+]}$... Basically, $\ce{[H+]} > \ce{[OH-]}$ or something?

@holbo2, I clearly stated that it has to "because step 2 consumes the acetate ions produced in step 1." Your interpretation is incorrect. Here, I have consolidated the entire equilibrium process into a two-step sequence. If instead you consider a sequence of steps that goes 1 2 1 2 1 2 1 2 ..., then by a similar argument as mine you can say that each successive step partially counteracts the act of the step before. Certainly there will not be "an increasingly greater $\ce{[H+]}$ in the solution."

Fair enough, but even the fact that you 'clearly stated' it doesn't explain why it occurs lesser; to someone who clearly doesn't understand it well, that's a statement and not an explanation. I see it simply because the equilibrium will be shifted towards the right in this case, since this is the side of acid dissociation. I get that the steps counteract each other, but would one step happen at a faster rate than the other?

(> Would one step happen at a faster rate than the other?) @holbo2, yes, that's the basis of my answer. To elaborate on my explanation, step 1 must occur at a faster rate than does step 2 to produce the excess acetate ions required in step 2. Otherwise, there wouldn't be excess acetate ions for step 2 to consume.

@a-cyclohexane-molecule Makes sense. Am I correct in saying - the forward reaction happens at a faster rate than the reverse reaction, hence the equilibrium is shifted towards the right which means that the rate of $\ce{[H+]}$ production is happening faster than $\ce{[CH3COOH]}$ production? Is this why stronger acids are even more acidic in solution... Something like that, anyway? That kinda correlates with how buffer systems stay in equilibrium with small additions of acid/base, right?

@holbo2, sequentially, yes, yes, and yes, as long as you keep in mind that buffer systems represent the limit where the two rates are almost equal, and strong acids as the opposite limit where 1 dominates 2.

@a-cyclohexane-molecule I get it now; makes perfect sense. It's just rates and equilibria really. I suppose that's why aqueous $\ce{HCl}$ has a low pH - the rate of 1 'dominates' the rate of 2, hence the equilibrium shifts accordingly. Heck, my initial post sounds really stupid now... Thanks!

Ok, let me come at this from a different angle using math.
As you say, take the following simple equilibrium:
$$\ce{CH3COOH + H2O <=> H3O+ + CH3COO-}\tag{1}$$
Now the equilibrium equation is given by:
$$K_\mathrm{a} = \frac{\ce{[H^+][CH3COO^-]}}{\ce{[CH3COOH]}}\tag{2}$$
Let's assume some initial concentration of acetic acid $x$. Let $\ce{[H^+]_x}$ and $\ce{[CH3COO^-]_x}$ be the corresponding concentration of $\ce{H^+}$ and $\ce{CH3COO^-}$.
$$K_\mathrm{a} = \frac{\ce{[H^+]_x[CH3COO^-]_x}}{x}\tag{3}$$
Now let's double the initial concentration of acetic acid to $2x$, using $\ce{[H^+]_{2x}}$ and $\ce{[CH3COO^-]_{2x}}$ so:
$$K_\mathrm{a} = \frac{\ce{[H^+]_{2x}[CH3COO^-]_{2x}}}{2x}\tag{4}$$
thus
$$\frac{\ce{[H^+]_x[CH3COO^-]_x}}{x} = \frac{\ce{[H^+]_{2x}[CH3COO^-]_{2x}}}{2x}\tag{5}$$
and
$$\ce{[H^+]_{2x}[CH3COO^-]_{2x}} = 2\times (\ce{[H^+]_x[CH3COO^-]_x})\tag{6}$$
now if we assume that $\ce{[H^+] = [CH3COO^-]}$ then
$$\ce{[H^+]_{2x} = \sqrt{2}\times [H^+]_x}\tag{7}$$
Since acetic acid is a weak acid, the assumption $\ce{[H^+] = [CH3COO^-]}$ is valid roughly so long as $[\ce{CH3COOH}] > 1\times10^{-5}$. In reality the charge in the solution must be neutral so:
$$\ce{[H^+] = [CH3COO^-] + [OH^-]}\tag{8}$$
So another way to look at the assumption is that
$$\ce{[H^+]} \gg \ce{[OH^-]}\tag{9}$$

That's a nice way of looking at it - the mathematics from steps 1 to 7 make sense. I have one question though: you said that in reality, the assumption will be $\ce{[H+] = [CH3COO-] + [OH-]}$. However, how does this mean that $\ce{[H^+]} \gg \ce{[OH^-]}$? Sorry if this is a silly question.

The solution is acidic so $\ce{[H^+] \gg  [OH^-]}$ must be true. $\ce{K_w = 1\times10^{-14} = [H^+]  [OH^-]}$

While you are right that by increasing the concentration of acetic acid in a solution we also increase the concentration of acetate, this does not factor into the pH calculation. Since:$$\mathrm{pH=-log[\ce{H3O+}]}$$
It only matters that by adding more acetic acid, there is more $\ce{H3O+}$ in the solution. This is why the pH decreases.

I will do my best here, If more acid is added, the pH should remain the same due to the continuous reactions with water creating base so that equilibrium is once again reached. This should remain the scenario until such time as more acid is added than the amount of water. When this stage is reached the additional acid will no longer be able to react with water and so will remain is its acidic state thus decreasing the pH.

That makes sense, however I have one problem. You said that the pH should remain the same as the equilibrium is always restored. However, adding water to an acid will dilute it, hence its pH will increase, so this doesn't agree with your statement and instead kind of agrees with what I said about a more basic solution being formed?

Sorry your best isn't good enough. This is wrong. The more acetic acid you add, the more acidic the solution gets.


text:
IUPAC name of C₆(CH₃)₆²⁺
What is the correct IUPAC name (or least incorrect one, if the current rules can't precisely describe that) of the interesting species $\ce{[C6(CH3)6]^{2+}}$ (which is extensively discussed here), containing exotic hexacoordinate pyramidal carbocation* ?
The structure can be drawn as

or

My guess is methyl(pentamethyl-η5-cyclopentadienyl)carbon(2+).

*) Some structures in the Wikipedia article have the two positive charges placed into the cyclopentadienyl (aromatic) ring, which I think is incorrect.

Rather preferred (PIN) or one of various others (general etc.) then (in)correct.

This article on Wikipedia is based on stuff that was published 40 or more years ago :( There has to be 2+ formal charge on apical carbon, but not real charge.

@Mithoron .. and the "least incorrect" variant request was for the case if the current rules are incapable of that.. my question rewording edit covers that as well, sort of. I think organic nomenclature cannot handle this, and inorganic nomenclature does not have PINs.

From the IUPAC Gold Book:

Carbocations may be named by adding the word 'cation' to the name of the corresponding radical. Such names do not imply structure (e.g. whether three-coordinated or five-coordinated carbon atoms are present).

So since this is the dication of hexamethylbenzene, it would probably  least incorrectly be called hexamethylbenzene dication. At the very least, this seems to be how a lot of people refer to it (or more verbosely, pentagonal-pyramidal hexamethylbenzene dication).


text:
When is a reaction said to be under kinetic control, is it not easily reversible?
If we consider this reaction: Compound 1 + Compound 2 --> Compound 3
I understand that if a reaction is under kinetic control, the less stable (e.g. a mono-substituted diene) is formed and if the reaction is under thermodynamic control the more stable stable (disubstituted diene) is formed.
However, if we wanted to go from the kinetic product to the thermodynamic product, would this reaction be considered as being under kinetic control? In other words, are the reactants under kinetic/thermodynamic control, or is it the products?
I think this is more of a terminology question.

You are considering a reaction Compound A --> Compound B --> Compound C or two alternatives, A <--> B compared to A --> C. Whatever way B --> C goes, there is no point in asking wether that's thermodynamic or kinetic, because it's both. And you don't address reversibility in your question text, why is it in the heading?

A product formed under kinetic conditions may or not be reversible under the conditions of the reaction. Here is an example of one that is reversible under the reaction conditions. Kinetic control is determined by the difference in activation energy (ΔEa) of the two pathways. The thermodynamic equilibrium is determined by ΔΔGo. Acetoacetic ester (AAE 1) is a mixture of keto and enol forms. Bromination of AAE in an inert solvent rapidly gives α-bromoAAE 2 in addition to HBr. This is the product that is formed faster having a lower activation energy. This reaction is also  reversible, initiated by HBr (microscopic reversibility). Eventually, the more stable γ-bromoAAE 3 is formed from AAE 1 by enolization toward C4 (γ-position) of AAE under the influence of HBr.

I think I got i! But, if with the addition of HBr, the more stable (thermodynamic) product was not formed from the less stable product (less stable --> intermediate --> more stable), would this reaction (or lack of) be considered to be under Kinetic Control?

There is no HBr added. It is produced as a by-product of the bromination. In this regard, the rxn. is also auto-catalytic. For a non-reversible kinetic/thermodynamic process, see the deprotonation of 2-methylcyclohexanone at https://en.wikipedia.org/wiki/Thermodynamic_versus_kinetic_reaction_control

If you run the reaction for a short period of time or at low temperature and monitor the progress of the reaction to optimize bromide 2, then the rxn. is kinetically-controlled. If on the other hand, you wait longer and/or operate at room temperature, then it will be thermodynamically-controlled.

@user55119 What if the products was higher in energy than the reatants? Would it make sense then to speak about "kinetic" or "thermodynamic" control?

@ado sar: If the products are higher in energy, then the reaction is endothermic and moves to the left. The higher peak is rate limiting.


text:
Detailed description of hydroxy(phenyl)methanesulfonic acid?
Hydroxy(phenyl)methanesulfonic acid, or benzaldehyde adduct is the product of reacting benzaldehyde with sodium metabisulfite ($\ce{NaS2O5}$) solution.
It is obviously a well-known and often used reaction, and yet, I'm not able to find much info on the adduct, or the reaction itself.
One thing I managed to dig out is its corresponding PubChem entry.
I want to find out the exact mechanism of the reaction with an equation, a bunch of videos of someone performing it will be nice too...
But most importantly - the hydroxy(phenyl)methanesulfonic acid: What is its color, texture and so on? What is its crystalline structure? Is it supposed to smell of benzaldehyde, how soluble it is in different solvents, how should it be stored, is it harmful or dangerous, does it decompose in some conditions, melting point, and so on.

looking for Mandelic acid?

No. I have updated with a pic of the reaction. The adduct at step 1 seems to be the same though.

To clarify: the benzaldehyde adduct at step one of the mandelic acid preparation in that Wikipedia article is the Hydroxy(Phenyl)Methanesulfonic Acid that I'm interested in(or at least I think so)

Hydroxy(phenyl)methanesulfonate is a compound of a class commonly called "bisulfite addition products" or "bisulfite adducts", since bisulfites are commonly used for the reaction you have mentioned. But there's an equilibrium in aqueous environment between metabisulfites and bisulfites
$$\ce{S2O5^{2-} + H2O \rightleftharpoons 2HSO3^{-} }$$
in favor of the bisulfite, this is why metabisulfites can be used as well, probably.
The bisulfite anion is proposed to have two tautomers:

The form at the left has a lone electron pair at the sulfur atom, and can act as a nucleophile.
The carbonyl/aldehyde group $\ce{>C=O}$ has a partial positive charge at the carbon atom.
The reaction in now no mystery, it's a nucleophilic addition to the carbonyl group.
If I may cite a crude reaction mechanism from Clayden/Greeves/Warren's Organic Chemistry, short chapter Bisulfite addition compounds:

And from the March's Advanced Organic Chemistry:

The reaction is reversible (by treatment of the addition product
  with either acid or base) and is useful for the purification of the starting compounds, since the addition products are soluble in water
  and many of the impurities are not.

If you want more details (kinetics, etc.), there is freely available couple of scientific articles, Sulphur isotope effects in the bisulphite addition reaction of aldehydes and ketones

I. Equilibrium effect and the structure of the addition product
II. Bond-formation effect

Thank you, really detailed and straight-forward. But how about the adduct's properties, is it storable for example?

@Nigel This can be the issue. I have added short citation from March's book, which might help.


text:
Amount concentration versus number concentration
The IUPAC Gold Book defines

Amount concentration
Amount of a constituent divided by the volume of the mixture. Also called amount-of-substance concentration, substance concentration (in clinical chemistry) and in older literature molarity. For entities B it is often denoted by B. The common unit is mole per cubic decimetre or mole per litre sometimes denoted by M.

and

Number concentration
Number of entities of a constituent in a mixture divided by the volume of the mixture.

I can't see a distinction between the two terms since it seems to me that "Amount of a constituent" is the same as "Number of entities of a constituent".
Could you please explain the difference between them? Is it a conceptual difference or just a difference in the context in which they are most commonly used?

The "amount" of a substance is the proper term for the "number of moles" (Wikipedia; Gold Book). So, the two quantities are related by a factor of the Avogadro constant, $\pu{6.022\times 10^23 mol-1}$.
If you have $\pu{1 mol}$ of helium in a $\pu{1 m3}$ box, the amount concentration is $\pu{1 mol m-3}$, or $\pu{0.001 mol dm-3}$ if you prefer those units.
The corresponding number concentration is $\pu{6.022\times 10^23 m-3}$.

Isn't mol just a specific unit for amount? According to the International System of Units, amount of substance is a quantity, while mol is a unit. I think one can use other units for amount if one wishes: tens, thousands, pi, dozens, etc.


text:
What is the arrangement of water molecules in a hydration shell around an anion?
I believe I read somewhere that the hydration shell of ions always has hydrogen on the outside, but I can't relocate the exact whereabouts of this. Thus, I decided to ask here, as to what the arrangement of the water molecules around a anion is (is it both hydrogens pointed inwards to the ion or if one is facing outside, one is facing inside).

Maybe of interest: J. Phys. Chem. B, 2007, 111 (48), pp 13570–13577
DOI: 10.1021/jp075913v Also: Angew. Chem. Int. Ed. 2011, 50, 352 – 353 DOI: 10.1002/anie.201004501 http://139.30.122.11/paschek/PAPERS/DPRLAngewandteInt2011.pdf

Also: Faraday Discuss. Chem. Soc., 1988,85, 137-146; 10.1039/DC9888500137; https://pubs.rsc.org/en/content/articlelanding/1988/dc/dc9888500137; And this: http://www1.lsbu.ac.uk/water/ion_hydration.html ; And this thesis: An ab initio study of ion solvation in water by Jasper Martijn Heuft http://www.science.uva.nl/research/its/molsim/Publications/thesis_heuft.pdf (<- look at Figure 1.5 and 1.6)


text:
How to calculate a molecule's oscillator strength?
I was wondering how to calculate a molecule's (e.g. biphenyl) oscillator strength using a very simplified model.
If I want to calculate a molecule's excited states a good start is something like a particle in a box model, which won't come up with even nearly exact energies but with an idea of energy values. Is there anything similiar for the oscillator strength which would deliver simplified solutions?

The simplest would be to use Huckel theory in one of its variants. Using symmetry  one can then determine which transitions are allowed or forbidden.  Knowing electron density approximations to transition moments can be calculated. Btw, biphenyl may not be a good starting point as its geometry is variable depending the the angle between rings.


text:
How long does it take for nitric oxide to turn into dinitrogen and dioxygen?
I'm new to chemistry and am studying nitric oxide as a combustion accelerant. Nitric oxide is said to be a free radical.
People say that nitric oxide, by itself, is pretty stable. But I saw that $$\ce{2NO -> N2 + O2}$$ so in a room of nitric oxide, will it decompose to nitrogen and oxygen?

Theoretically, it will decompose, but the reaction is rather slow, on a human timescale. One experiment ran for almost forty years (J. Phys. Chem. 1958, 62 (3), 360–361), from 1918 to 1957, without noticeable decomposition.
That said, in aqueous solutions it breaks down more rapidly. In vivo, the half-life is a few seconds.
It is amazing to me that a free radical is so stable!

Then again, the decomposition pathway in vivo is not the one that leads to O2 and N2.

@IvanNeretin, true, and its physiological reaction (as opposed to decomposition) has engendered a large pharmaceutical market, which is still rising.


text:
Why is density fitting not used to reduce four-center integrals further down to two-center integrals?
In density fitting, why don't researchers turn all the 4-center integrals into 2-center ones instead of stopping at composing them from 3-center integrals?

related: 3-index integral schemes for density fitting

I try to answer your question, although I'm not completely sure if I understand what you are asking for.
You want a decomposition like for example:
$$
 (pq|rs) = \sum_P A^p_P A^q_P A^r_P A^s_P
$$
am I correct?
In the framework of density fitting this type of decomposition is not possible without further steps/tricks.
If we look at the two electron integrals
\begin{equation}
(pq|rs)=\int d\pmb{r}_1 \int d\pmb{r}_2 \phi_p(\pmb{r}_1) \phi_q(\pmb{r}_1) \frac{1}{\pmb{r}_{12}} \phi_r(\pmb{r}_2) \phi_s(\pmb{r}_2) \quad,
\end{equation}
we can consider this electron repulsion integral as repulsion between two generalized electron densities
\begin{equation}
(pq|rs)=\int d\pmb{r}_1 \int d\pmb{r}_2 \rho_{pq}(\pmb{r}_1)  \frac{1}{\pmb{r}_{12}} \rho_{rs}(\pmb{r}_2) \quad,
\end{equation}
with $\rho_{pq}(\pmb{r}) = \phi_p(\pmb{r}) \phi_q(\pmb{r})$ and $\rho_{rs}(\pmb{r}) = \phi_r(\pmb{r}) \phi_s(\pmb{r})$.
This density maybe approximated with the aid of an auxiliary basis set:
\begin{equation}
 \tilde{\rho}_{pq}(\pmb{r}) = \sum^{N_{aux}}_P d_P^{pq} \chi_P(\pmb{r})
\end{equation}
Hence the name "density fitting". The coefficients $d_P^{pq}$ are determined by minimizing the functional
\begin{equation}
  \Delta_{pq} = \int d\pmb{r}_1 \int d\pmb{r}_2 \frac{\left[ {\rho}_{pq}(\pmb{r_1}) - \tilde{\rho}_{pq}(\pmb{r_1}) \right] \left[ {\rho}_{rs}(\pmb{r_2}) - \tilde{\rho}_{rs}(\pmb{r_2}) \right]}{r_{12}}
\end{equation}
and leads according to Dunlap [1] to:
\begin{equation}
 d_Q^{pq} = \sum_{P} \left(pq|P\right) \left[\pmb{J}^{-1}\right]_{PQ} \quad,
\end{equation}
with
\begin{equation}
 \left(pq|P\right) = \int d\pmb{r}_1 \int d\pmb{r}_2 \phi_p(\pmb{r}_1) \phi_q(\pmb{r}_1) \frac{1}{\pmb{r}_{12}} \chi_P(\pmb{r}_2)
\end{equation}
\begin{equation}
 J_{PQ} = \int d\pmb{r}_1 \int d\pmb{r}_2 \chi_P(\pmb{r}_1) \frac{1}{\pmb{r}_{12}} \chi_Q(\pmb{r}_2)
\end{equation}
Then we can re-write the electron repulsion integral as
\begin{equation}
 (pq|rs)=\sum_{PQ} \left(pq|P\right) \left[\pmb{J}^{-1}\right]_{PQ} \left(Q|rs\right)
\end{equation}
The framework of density fitting does not give us the possibility to decompose $\left(pq|P\right)$ further,
but it is possible in combination with more general tensor decomposition techniques like the canonical decomposition (CP)[2],
in which we approximate a tensor in the format you were fishing for (as long as I understood you correctly)
\begin{equation}
 X_{i,j,...d} = \sum_{r}^{R} \omega_r  a_{ir}^{(1)} a_{jr}^{(2)} ... a_{dr}^{(d)} \quad,
\end{equation}
where $R$ is the rank of the tensor. Without losing generality, we will in the following absorb the weighting factor $\omega_r$
in one of the factors.
We can apply this general tensor decomposition to the density fitting representation of the two electron integral.
For the $[J^{-1}]_{PQ}$ matrix we can simply us a SVD
\begin{equation}
 [J^{-1}]_{PQ}  \approx \sum_{r_x}^{R_x} \bar{U}^{P}_{r_x} \bar{V}^{Q}_{r_x}  \quad,
\end{equation}
where the singular values $\sigma^{1/2}$ are absorbed into the vectors
For the 3-index integrals we do a CP decomposition
\begin{equation}
  \left(pq|P\right) \approx \sum_{r}^{R} A_{r}^{p} A_{r}^{q} A_{r}^{P}
\end{equation}
Thus the 4-index integrals can be written as
\begin{equation}
 \left(pq|rs\right)  = \sum_{PQ} \left( \sum_{r_1}^{R_1} A_{r_1}^{p} A_{r_1}^{q} A_{r_1}^{P} \right)
                                 \left( \sum_{r_x}^{R_x} \bar{U}^{P}_{r_x} \bar{V}^{Q}_{r_x} \right)
                                 \left( \sum_{r_2}^{R_2} A_{r_2}^{r} A_{r_2}^{s} A_{r_2}^{Q} \right) \quad.
\end{equation}
In some sense we are done since only two index quantities occur, but we can simplify the expressions a bit.
The sums can be rearranged to
\begin{equation}
  \left(pq|rs\right) = \sum_{r_1}^{R_1}\sum_{r_2}^{R_2}\sum_{r_x}^{R_x} \left( \sum_{P} A_{r_1}^{P}  \bar{U}_{r_x}^P  \right) \left( \sum_{Q} \bar{V}_{r_x}^Q   A_{r_2}^{Q} \right)
                         A_{r_1}^{p} A_{r_1}^{q} A_{r_2}^{r} A_{r_2}^{s} \quad,
\end{equation}
and a summation over $P$ and $Q$ yields
\begin{equation}
  \label{eqn:thc1}
  \left(pq|rs\right) = \sum_{r_1}^{R_1}\sum_{r_2}^{R_2}\sum_{r_x}^{R_x} P_{r_1 r_x} Q_{r_x r_2}
                         A_{r_1}^{p} A_{r_1}^{q} A_{r_2}^{r}  A_{r_2}^{s} \quad.
\end{equation}
Only $P_{r_1 r_x}$ and $Q_{r_x r_2}$ depend on $r_x$ and via summation we introduce a new coefficient matrix $B_{r_1 r_2}$, which
combines the expansions length $R_1$ and $R_2$.
\begin{equation}
  \left(pq|rs\right) = \sum_{r_1}^{R_1}\sum_{r_2}^{R_2} B_{r_1 r_2} A_{r_1}^{p} A_{r_1}^{q} A_{r_2}^{r} A_{r_2}^{s} \quad,
\end{equation}
which is quite similar to the THC format introduced by Martinez and co-workers.[3]
In one additional step we obtain a CP factorization for the full electron repulsion integral
\begin{equation}
  \left(pq|rs\right) = \sum_{P}^{R_1\cdot R_2} \bar{A}_{P}^{p}  \bar{A}_{P}^{q} \bar{A}_{P}^{r} \bar{A}_{P}^{s}
\end{equation}
So finally we arrived at a decomposition in only two index quantities and there are several groups[4,3,5], who looked
at this integral format(s). It is not of widespread use since so far the available algorithms for the required CP decomposition
are quite costly. In combination with density fitting the costs will scale with $\mathcal O(N^4)$, which is not that bad,
but the prefactor is so large that for a long range of molecular sizes a CCSD(T) calculation will finish quicker than the decomposition
of the two electron integral tensor.
The story is a bit different for the intermediate step, which can be identified as the THC format. Martinez and co-workers
are heavily working on using this format and enabled $\mathcal O(N^4)$ scaling implementations for MP2 and CCSD. However,
for MP2 it is to the best of my knowledge not really compatible with DF-MP2 implementations.
So after this long answer the short conclusion: Such integral formats exist in form of the THC format or CP decomposed integrals,
but further work is necessary to increase the efficiency and resolve some other problems.

References:

Dunlap, B. I., Connolly, J. W. D. & Sabin, J. R. On Some Approximations in Applications of X Alpha Theory. The Journal of Chemical Physics 71, 3396–3402 (1979).
Carroll, J. D. & Chang, J.-J.   Analysis of individual differences in multi-dimensional scaling via an n-way generalization of “eckartyoung” decomposition. Psychometrika 35, 283–319 (1970). URL http://dx.doi.org/10.1007/BF02310791
Hohenstein, E. G., Parrish, R. M. & Martínez, T. J. Tensorhypercontraction density fitting. I. quartic scaling second- and third-order Møller-plesset perturbation theory. The Journal of Chemical Physics 137 (2012). URL http://scitation.aip.org/content/aip/journal/jcp/137/4/10.1063/1.4732310.
Benedikt, U., Auer, A. A., Espig, M. & Hackbusch, W. Tensor decomposition in post-Hartree Fock methods. I. Two-electron integrals and MP2. The Journal of Chemical Physics 134 (2011). URL http://scitation.aip.org/content/aip/journal/jcp/134/5/10.1063/1.3514201.
Schmitz, G., Madsen, N. K. & Christiansen, O. Atomic-batched tensor decomposed two-electron repulsion integrals. The Journal of Chemical Physics 146, 134112 (2017). URL https://doi.org/10.1063/1.4979571. https://doi.org/10.1063/1.4979571


text:
Confusion in IUPAC Nomenclature
I was trying to name the following compound but I got stuck in naming the aldehydic part of the compound.

I looked for the answer and found it as 6-bromo-2-oxocyclohexanecarbeldehyde. I had come to know earlier that carbeldehyde suffix is used when there are more than two aldehydic group in the compound but why carbeldehyde is used here? I am not able to understand this. If someone can explain which rule is used here. I will be very grateful for this.

The suggested name is "6-bromo-2-oxocyclohexanecarbeldehyde" is wrong. The correct name is 2-bromo-6-oxocyclohexane-1-carbaldehyde.

Related: When is the carbaldehyde suffix correct? and Why is the suffix carbaldehyde valid here?

Okay. But why the suffix carbeldehyde is used here?

Can you please explain why hexane is used instead of hexyl?

Why …carbaldehyde?
It's similar to $\ce{C6H11-COOH}$ cyclohexanecarboxylic acid (not cyclohexanoic acid). It's the case where the functional group's carbon is not included in the parent structure. (Such case are also nitriles/carbonitriles, $\ce{R-CN}$)
Why not …hexyl… or rather …cyclohexyl… ?
This (wrong name) would be constructed similarly to a different, non-preferred type of nomenclature, a radical name, which names e.g. ethanol as ethyl alcohol.
There are more than two aldehydic group in the compound.
They are not. Alhedyde group is ${\ce{-CHO}}$. The other one, $\ce{>C=O}$ is an oxo group (ketone). The group for suffix is chosen by the pre-defined priority, ${\ce{-CHO}}$ > $\ce{>C=O}$.
Why 2-bromo-6-oxo-… and not 6-bromo-2-oxo-… ?
The groups named by prefixes (after highest priority - suffix group is assigned lowest locant possible) are not numbered by their priority, but such that the assigned locant set is lowest possible. If there are more equivalent possibilities, then they are numbered in the alphabetically sorted order (which is used in the name).

Carbaldehyde is not particulary used when there are two aldehydic groups in a molecule.
To describe an aldehyde you either use "-al" or "-carbaldehyde". The difference is that when you use "-carbaldehyde" you also count the carbon, meaning that carbaldehyde stands for "CHO" group including the carbon.
(That is why it is 2-oxo.)

I'm afraid, but that doesn't really answer the question. Please also see our guidelines on how to ask and answer nomenclature questions.


text:
How to evacuate a glass vessel containing powders?
I would like to evacuate a vessel (a glass round-bottomed flask) and which contains a few grams of platinum. However, I am worried about the platinum powder being sucked up by the vacuum.
A few people told me that this is possible if I take some care, but I am still not convinced. I use a turbo pump in combination with a diaphragm pump for the rough vacuum. So in the end a very expensive vacuum cleaner! I know that the powder will destroy the turbo pump if the powder is sucked in. Do any of you have some experience with powders in vacuum?

Two questions: In what order do you intend to connect and switch on the pumps? And what is the reason a vacuum cleaner works? Hint: There is no vacuum involved.

Why are you using a turbopump? I would imagine that the off-gassing of ground glass oil would be continuous. Is there a need for ultra high vacuum.

I often high-vac powders on Schlenk. Issues only arise when going between positive and negative pressures though to minimise any risk I often lightly plug the sidearm of the glass stopper/adapter(?) with cotton wool. Glass wool would also suffice and I am certain there are already stoppers/adapters which have a porous mesh build in to prevent powder from entering the vacuum system.

It is standard procedure for air-sensitive chemistry to apply a vacuum to a flask containing a solid. Most of the equipment used routinely for this involves a vacuum/gas line that allows the connection of the flask either to a source of an inert gas (nitrogen or argon) or a vacuum using a switchable stopcock (so you can evacuate all the gas and then fill the vessel with something inert). Nobody connects a vessel directly to a vacuum pump and the gas vacuum line is shielded from the pump with traps (usually cooled to liquid nitrogen temperatures) to collect any material entering the line (both solids and volatiles) that you don't want to get into your pump.
Using such an apparatus requires a certain amount of skill. One such skill is not applying a vacuum too quickly (which might cause some material in the flask to be blown into the line). Careful and slow application of the vacuum will avoid this and careful observation of the flask while applying the vacuum will help you learn just how fast you can afford to apply the vacuum. But it might be worth learning this with a cheap powder before you try it with platinum.


text:
Why does a reaction take time?
A reaction in nature is, as we know, generally an interaction between the electron clouds of different atoms and nuclei. Why does this takes time?

The (large) thermal excitations that allow for productive interactions of this kind are rare and infrequent.

Besides, it takes time for atoms to meet.

Firstly, some reactions have a very low probability, secondly, the probability depends on temperature (activation energy!), and thirdly, diffusion is not exacly a fast process, and takes time even in a gas phase. Once the reaction progresses, the remaining educt particles have an increasingly hard time finding each other even if you thoroughly mixed everything at the beginning.

Well diffusion in a gas is faster than a liquid which is faster than a solid. But even in gas diffusion takes time.

Not all reactions are bimolecular $A+B\to $ products but if the reaction is bimolecular  diffusion  comes into play in these reactions; i.e molecule have to meet before they can react. For a bond dissociation or cis-trans isomerisation a collision is not necessary.
The fastest a bimolecular reactions can occur at is when there is a low energy barrier between reactants and products compared to the thermal energy available at, say, room temperature and then reaction occurs at first encounter. In solution this clearly depends on the viscosity, but in water or other low viscosity solvents the rate constant is typically between $10^{10}$ and $10^{11}\mathrm{dm^3mol^{-1}s^{-1}}$. In viscous solvents such as glycerol the diffusion rate constant is about ten thousand times smaller. In some electron transfer reactions, the reactant and product (electron donor and acceptor) have been synthesised to be held adjacent to one another and the reaction started by a short pulse of light, then reaction rates can now be even faster as they are not limited by diffusion, typically reaction occurs in $\approx 10^{-13}$ s.
If there is a big energy barrier between reactants and products there is no lower limit to the rate constant at, say, room temperature; eg. diamonds in air, because here the barrier to reaction is so high that according the the Boltzmann distribution of energies virtually no molecules will gain enough energy to react.
In an isomerisation or dissociation reaction no collision is necessary since a photon can provide the energy to cause reaction. In this case the fastest the bond can break in, say, a diatomic molecule is half a vibrational period or typically $10^{-14}$ s. In a more complex molecule energy has to find its way around the molecule to the bond that has to be broken and this can take a bit longer or a lot longer, it just depends on the molecule....